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IMO-1959-1
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_1
|
Prove that the fraction \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).
|
[
"Denoting the greatest common divisor of \\(a, b\\) as \\((a,b)\\), we use the Euclidean algorithm:\n\n\\[\n(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1\n\\]\n\nTheir greatest common divisor is 1, so \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"Proof by contradiction:\n\nAssume that \\(\\dfrac{14n+3}{21n+4}\\) is a reducible fraction where \\(p\\) is the greatest common factor of \\(14n+3\\) and \\(21n + 4\\). Thus,\n\n\\[\n21n+4\\equiv 0\\pmod{p} \\implies 42n+8\\equiv 0\\pmod{p}\n\\]\n\nSubtracting the second equation from the first equation we get \\(1\\equiv 0\\pmod{p}\\) which is clearly absurd.\n\nHence \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"Proof by contradiction:\n\nAssume that \\(\\dfrac{14n+3}{21n+4}\\) is a reducible fraction.\n\nIf a certain fraction \\(\\dfrac{a}{b}\\) is reducible, then the fraction \\(\\dfrac{2a}{3b}\\) is reducible, too. In this case, \\(\\dfrac{2a}{3b} = \\dfrac{42n+8}{42n+9}\\).\n\nThis fraction consists of two consecutives numbers, which never share any factor. So in this case, \\(\\dfrac{2a}{3b}\\) is irreducible, which is absurd.\n\nHence \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"We notice that:\n\n\\[\n\\frac{21n+4}{14n+3} = \\frac{(14n+3)+(7n+1)}{14n+3} = 1+\\frac{7n+1}{14n+3}\n\\]\n\nSo it follows that \\(7n+1\\) and \\(14n+3\\) must be coprime for every natural number \\(n\\) for the fraction to be irreducible. Now the problem simplifies to proving \\(\\frac{7n+1}{14n+3}\\) irreducible. We re-write this fraction as:\n\n\\[\n\\frac{7n+1}{(7n+1)+(7n+1) + 1} = \\frac{7n+1}{2(7n+1)+1}\n\\]\n\nSince the denominator \\(2(7n+1) + 1\\) differs from a multiple of the numerator \\(7n+1\\) by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that \\(\\frac{21n+4}{14n+3}\\) is irreducible.\n\nQ.E.D",
"By Bezout's Lemma, \\(3 \\cdot (14n+3) - 2 \\cdot (21n + 4) = 1\\), so the GCD of the numerator and denominator is \\(1\\) and the fraction is irreducible.",
"To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with \\(\\frac{21n}{14n}\\) + \\(\\frac{4}{3}\\). Simplifying the fraction, we end up with. \\(\\frac{3}{2}\\). Now combining these fractions with addition as shown before in the problem, we end up with \\(\\frac{17}{6}\\). It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us \\(\\frac{25}{17}\\). Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). Trying 2, we end up with \\(\\frac{46}{31}\\). Again, both results are 1 off from being multiples of 15. Trying 3, we end up with \\(\\frac{67}{45}\\). Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:\n\n1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but\n\n2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us \\(\\frac{319}{199}\\), we keep ending up with results that at the end will only have a common divisor of 1.",
"Let \\(\\gcd (21n+4,14n+3)=a.\\) So for some co-prime positive integers \\(x,y\\) we have\n\n\\[\n21n+4 = ax \\qquad (1)\n\\]\n\n\\[\n14n+3 = ay \\qquad (2)\n\\]\n\nMultiplying \\((1)\\) by \\(2\\) and \\((2)\\) by \\(3\\) and then subtracting \\((1)\\) from \\((2)\\) we get\n\n\\[\n42n+9-(42n+8)=3ay-2ax\n\\]\n\n\\[\n\\implies a(3y-2x)=1.\n\\]\n\nWe must have \\(a=1\\), since \\(a\\) is an positive integer. Thus, \\(\\gcd(21n+4,14n+3)=1\\) which means the fraction is irreducible, as needed."
] |
IMO-1959-2
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2
|
For what real values of \(x\) is
\[
\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,
\]
given (a) \(A=\sqrt{2}\), (b) \(A=1\), (c) \(A=2\), where only non-negative real numbers are admitted for square roots?
|
[
"The square roots imply that \\(x\\ge \\frac{1}{2}\\).\n\nSquare both sides of the given equation:\n\n\\[\nA^2 = \\Big( x + \\sqrt{2x - 1}\\Big) + 2 \\sqrt{x + \\sqrt{2x - 1}} \\sqrt{x - \\sqrt{2x - 1}} + \\Big( x - \\sqrt{2x - 1}\\Big)\n\\]\n\nAdd the first and the last terms to get:\n\n\\[\nA^2 = 2x + 2 \\sqrt{x + \\sqrt{2x - 1}} \\sqrt{x - \\sqrt{2x - 1}}\n\\]\n\nMultiply the middle terms, and use \\((a + b)(a - b) = a^2 - b^2\\) to get:\n\n\\[\nA^2 = 2x + 2 \\sqrt{x^2 - 2x + 1}\n\\]\n\nSince the term inside the square root is a perfect square, and by factoring 2 out, we get\n\n\\[\nA^2 = 2(x + \\sqrt{(x-1)^2})\n\\]\n\nUse the property that \\(\\sqrt{x^2}=|x|\\) to get\n\n\\[\nA^2 = 2(x+|x-1|)\n\\]\n\nCase I: If \\(x \\le 1\\), then \\(|x-1| = 1 - x\\), and the equation reduces to \\(A^2 = 2\\). This is precisely part (a) of the question, for which the valid interval is now \\(x \\in \\left[ \\frac{1}{2}, 1 \\right]\\)\n\nCase II: If \\(x > 1\\), then \\(|x-1| = x - 1\\) and we have\n\n\\[\nx = \\frac{A^2 + 2}{4} > 1\n\\]\n\nwhich simplifies to\n\n\\[\nA^2 > 2\n\\]\n\nThis tells there that there is no solution for (b), since we must have \\(A^2 \\ge 2\\)\n\nFor (c), we have \\(A = 2\\), which means that \\(A^2 = 4\\), so the only solution is \\(x=\\frac{3}{2}\\).\n\n~flamewavelight (Expanded)\n\n~phoenixfire (edited)",
"Note that the equation can be rewritten to\n\n\\[\n\\sqrt{(\\sqrt{2x-1}+1)^2} + \\sqrt{(\\sqrt{2x-1}-1)^2}=A\\sqrt{2}\n\\]\n\ni.e., \\(\\sqrt{2x-1}+1 + |\\sqrt{2x-1}-1|=A\\sqrt{2}\\).\n\nCase I: when \\(2x-1\\ge 1\\) (i.e., \\(x\\ge 1\\)), the equation becomes \\(2\\sqrt{2x-1}=\\sqrt{2}A\\). For (a), we have \\(x=1\\); for (b) we have \\(x=\\frac{3}{4}\\); for (c) we have \\(x=\\frac{3}{2}\\). Since \\(x\\ge 1\\), (b) \\(x=\\frac{3}{4}\\) is not what we want.\n\nCase II: when \\(0\\le 2x-1 <1\\) (i.e., \\(1/2\\le x <1\\)), the equation becomes \\(2=\\sqrt{2}A\\), which only works for (a) \\(A=\\sqrt{2}\\).\n\nIn summary, any \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\) is a solution for (a); there is no solution for (b); there is one solution for (c), which is \\(x=\\frac{3}{2}\\).",
"For real \\(x\\), \\(x \\ge \\frac{1}{2}\\). Also,\n\n\\[\n\\sqrt{x+\\sqrt{2x-1}} = \\frac{1}{\\sqrt{2}} |\\sqrt{2x-1}+1|\n\\]\n\n\\[\n= \\frac{1}{\\sqrt{2}} (\\sqrt{2x-1}+1),\n\\]\n\n\\[\n\\sqrt{x-\\sqrt{2x-1}} = \\frac{1}{\\sqrt{2}} |\\sqrt{2x-1}-1|\n\\]\n\nIt is easy to know:\n\n(1) When \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\),\n\n\\[\n|\\sqrt{2x-1}-1| = 1-\\sqrt{2x-1}.\n\\]\n\n(2) When \\(x \\ge 1,\\)\n\n\\[\n|\\sqrt{2x-1}-1| = \\sqrt{2x-1}-1.\n\\]\n\nLet's define\n\n\\[\nA = \\sqrt{x+\\sqrt{2x-1}} + \\sqrt{x-\\sqrt{2x-1}} =\n\\]\n\n\\[\n= \\frac{1}{\\sqrt{2}} (\\sqrt{2x-1}+1+|\\sqrt{2x-1}-1|).\n\\]\n\nHence, for \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\), \\(A = \\sqrt{2};\\) \\(x \\ge 1,\\) \\(A \\ge \\sqrt{2}.\\)\n\nTo sum up, the solution for (a) \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\); (b) No solution; (c) \\(x=\\frac{3}{2}\\).\n\n~EUCLID_2003"
] |
IMO-1959-3
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_3
|
Let \(a,b,c\) be real numbers. Consider the quadratic equation in \(\cos{x}\) :
\[
a\cos ^{2}x + b\cos{x} + c = 0.
\]
Using the numbers \(a,b,c\), form a quadratic equation in \(\cos{2x}\), whose roots are the same as those of the original equation. Compare the equations in \(\cos{x}\) and \(\cos{2x}\) for \(a=4, b=2, c=-1\).
|
[
"Let the original equation be satisfied only for \\(\\cos{x}=m, \\cos{x}=n\\). Then we wish to construct a quadratic with roots \\(2m^2 -1, 2n^2 -1\\).\n\nClearly, the sum of the roots of this quadratic must be\n\n\\[\n2 (m^2 + n^2) - 2 = 2 \\left(\\frac{b^2-2ac}{a^2}\\right) - 2 = \\frac{2b^2 - 4ac - 2a^2}{a^2},\n\\]\n\nand the product of its roots must be\n\n\\[\n4m^2 n^2 - 2(m^2 + n^2) + 1 = \\frac{4c^2}{a^2}+\\frac{4ac - 2b^2}{a^2} + \\frac{a^2}{a^2} = \\frac{(a+2c)^2 - 2b^2}{a^2}\n\\]\n\nThus the following quadratic fulfils the conditions:\n\n\\[\na^2 \\cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\\cos{2x} + (a+2c)^2 - 2b^2 = 0\n\\]\n\nNow, when we let \\(a=4, b=2, c= -1\\), our equations are\n\n\\[\n4 \\cos^2 {x} + 2 \\cos {x} - 1 = 0\n\\]\n\nand\n\n\\[\n16 \\cos^2 {2x} + 8 \\cos {2x} - 4 = 0,\n\\]\n\nThis simplifies to previous equation. The first root of the first equation \\(\\cos {x} = \\frac{-1 + \\sqrt{5}}{4}\\) corresponds to \\(\\cos {2x} = \\frac{-1 - \\sqrt{5}}{4}\\) and the second root of the first equation \\(\\cos {x} = \\frac{-1 - \\sqrt{5}}{4}\\) corresponds to \\(\\cos {2x} = \\frac{-1 + \\sqrt{5}}{4}\\). These are relatively well-known trigonometric values, yielding roots \\(x = \\boxed{72^\\circ, 144^\\circ}\\)."
] |
IMO-1959-4
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_4
|
Construct a right triangle with a given hypotenuse \(c\) such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.
|
[
"We denote the catheti of the triangle as \\(a\\) and \\(b\\). We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)",
"The conditions of the problem require that\n\n\\[\nab = \\frac{c^2}{4}.\n\\]\n\nHowever, we notice that twice the area of the triangle \\(abc\\) is \\(ab\\), since \\(a\\) and \\(b\\) form a right angle. However, twice the area of the triangle is also the product of \\(c\\) and the altitude to \\(c\\). Hence the altitude to \\(c\\) must have length \\(\\frac{c}{4}\\). Therefore if we construct a circle with diameter \\(c\\) and a line parallel to \\(c\\) and of distance \\(\\frac{c}{4}\\) from \\(c\\), either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.",
"We denote the angle between \\(b\\) and \\(c\\) as \\(\\alpha\\). The problem requires that\n\n\\[\nab = \\frac{c^2}{4},\n\\]\n\nor, equivalently, that\n\n\\[\n2 \\frac{ab}{c^2} = \\frac{1}{2}.\n\\]\n\nHowever, since \\(\\frac{a}{c} = \\sin{\\alpha};\\; \\frac{b}{c} = \\cos{\\alpha}\\), we can rewrite the condition as\n\n\\[\n2\\sin{\\alpha}\\cos{\\alpha} = \\frac{1}{2},\n\\]\n\nor, equivalently, as\n\n\\[\n\\sin{2\\alpha} = \\frac{1}{2}.\n\\]\n\nFrom this it becomes apparent that \\(2\\alpha = \\frac{\\pi}{6}\\) or \\(\\frac{5\\pi}{6}\\); hence the other two angles in the triangle must be \\(\\frac{ \\pi }{12}\\) and \\(\\frac{ 5 \\pi }{12}\\), which are not difficult to construct. Q.E.D.\n\nNote. It is not difficult to reconcile these two constructions. Indeed, we notice that the altitude of the triangle is of length \\(c \\sin{\\alpha}\\cos{\\alpha}\\), which both of the solutions set equal to \\(\\frac{c}{4}\\) .",
"If we let the legs be \\(a\\) and \\(b\\) with \\(a < b\\), then \\(c^2 = a^2 + b^2\\). Because \\(c^2 = 4 a b\\) as well, we immediately deduce via some short computations through quadratic formula that \\(a = (2 - \\sqrt{3})b\\) and \\(a = (2 + \\sqrt{3})b\\). Thus, \\(\\frac{a}{b} = \\tan 15^\\circ\\) and \\(\\frac{a}{b} = \\tan 75^\\circ\\), and so one of the angles of the triangle must be \\(15^\\circ\\) and the other must be \\(75^\\circ\\). A \\(15^\\circ\\) angle is easily constructed by bisecting a \\(30^\\circ\\) angle (which is formed by constructing the altitude of an equilateral triangle), and \\(75^\\circ\\) angle is constructed by constructing a 15 degree angle on top of the 60 degree point.",
"We draw perpendicular \\(h\\) on \\(c\\) and denote the angle between the median and \\(c\\) as \\(\\alpha\\). The area of the triangle is \\(\\frac{ab}{2}\\) or \\(\\frac{ch}{2}\\)\n\nWe can write \\(c\\) as \\(2\\sqrt{ab}\\) So we have,\n\n\\[\n\\frac{ab}{2} = \\frac{2\\sqrt{ab}h}{2}\n\\]\n\nWhich simplifies to be\n\n\\[\n\\frac{\\sqrt{ab}}{h} = \\frac{1}{2}\n\\]\n\nOr\n\n\\[\n\\sin\\alpha = \\sin\\frac{\\pi}{6}\n\\]\n\nSo, we draw \\(c\\) and from it's midpoint, we draw a segment \\(\\frac{c}{2}\\) forming an angle \\(\\frac{\\pi}{6}\\) Join the vertices to get the required triangle.",
"This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c\n\nNow go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c."
] |
IMO-1959-5
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_5
|
An arbitrary point \(M\) is selected in the interior of the segment \(AB\). The squares \(AMCD\) and \(MBEF\) are constructed on the same side of \(AB\), with the segments \(AM\) and \(MB\) as their respective bases. The circles about these squares, with respective centers \(P\) and \(Q\), intersect at \(M\) and also at another point \(N\). Let \(N'\) denote the point of intersection of the straight lines \(AF\) and \(BC\).
(a) Prove that the points \(N\) and \(N'\) coincide.
(b) Prove that the straight lines \(MN\) pass through a fixed point \(S\) independent of the choice of \(M\).
(c) Find the locus of the midpoints of the segments \(PQ\) as \(M\) varies between \(A\) and \(B\).
|
[
"## Part A\n\nSince the triangles \\(AFM, CBM\\) are congruent, the angles \\(AFM, CBM\\) are congruent; hence \\(AN'B\\) is a right angle. Therefore \\(N'\\) must lie on the circumcircles of both quadrilaterals; hence it is the same point as \\(N\\).\n\n\\[\n1IMO5A.JPG\n\\]\n\n## Part B\n\nWe observe that \\(\\frac{AM}{MB} = \\frac{CM}{MB} = \\frac{AN}{NB}\\) since the triangles \\(ABN, BCM\\) are similar. Then \\(NM\\) bisects \\(ANB\\).\n\nWe now consider the circle with diameter \\(AB\\). Since \\(ANB\\) is a right angle, \\(N\\) lies on the circle, and since \\(MN\\) bisects \\(ANB\\), the arcs it intercepts are congruent, i.e., it passes through the bisector of arc \\(AB\\) (going counterclockwise), which is a constant point.\n\n## Part C\n\nDenote the midpoint of \\(PQ\\) as \\(R\\). It is clear that \\(R\\)'s distance from \\(AB\\) is the average of the distances of \\(P\\) and \\(Q\\) from \\(AB\\), i.e., half the length of \\(AB\\), which is a constant. Therefore the locus in question is a line segment.",
"\\[\nIMO19595A.png\n\\]\n\n## Part a)\n\nNotice that \\(\\angle BAF = \\arctan (\\frac{MF}{AM}) = \\frac{MB}{AM}\\) and \\(\\angle ABC = \\arctan (\\frac{MC}{MB}) = \\frac{AM}{MB}\\).\n\n\\(\\implies\\) \\(\\angle BAF + \\angle ABC = 90^{\\circ}\\).\n\nNow notice that \\(\\angle FNB = 90^{\\circ}\\). Considering \\(\\angle BAF\\) as \\(\\theta\\), this gives \\(\\angle MBN = 90^{\\circ} - \\theta\\) and thus \\(\\angle MFN = 90^{\\circ} + \\theta\\). But notice that \\(\\angle MFA = 90^{\\circ} - \\theta\\), which means that \\(\\angle AFN = 180^{\\circ}\\). Therefore points \\(A, F, N\\) are collinear. Now \\(\\angle BNF = 90^{\\circ}\\) and \\(\\angle ANC = 90^{\\circ}\\). Therefore, \\(\\angle BNC = 180^{\\circ}\\) and thus points \\(B, N, C\\) are collinear. Therefore, AF and BC intersect at N.\n\n## Part b)\n\nConstruct the bisector of arc AB above AB. Call it X. \\(\\angle ANM = \\angle ACM = 45^{\\circ}\\). Now \\(\\angle ANB = 90^{\\circ}\\) which means N lies on the circle with AB as diameter.\n\n\\(\\implies\\) \\(\\angle ANX = \\angle ABX = 45^{\\circ} = \\angle ANM\\). Therefore since M and X are on the same side of \\(N\\), \\(NM\\) passes through \\(X\\) wherever we choose \\(M\\) on \\(AB\\).\n\n## Part c)\n\nLet the midpoint of \\(PQ\\) be \\(R\\). Let \\(G\\) be the midpoint of \\(AM\\). Let \\(F\\) be the midpoint of \\(MB\\). Let \\(I\\) be the foot of the perpendicular from \\(R\\) onto \\(AB\\). Therefore by Midpoint Theorem, \\(RI = \\frac{PG + HQ}{2} = \\frac{AG + HB}{2} = \\frac{AB}{4}\\). Therefore the distance \\(RI\\) is a constant and thus the locus is a straight line parallel to \\(AB\\) at a distance (to \\(AB\\), of course) of \\(\\frac{AB}{4}\\)."
] |
IMO-1959-6
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_6
|
Two planes, \(P\) and \(Q\), intersect along the line \(p\). The point \(A\) is in the plane \(P\), and the point \({C}\) is in the plane \(Q\); neither of these points lies on the straight line \(p\). Construct an isosceles trapezoid \(ABCD\) (with \(AB\) parallel to \(DC\)) in which a circle can be inscribed, and with vertices \(B\) and \(D\) lying in the planes \(P\) and \(Q\), respectively.
|
[
"We first observe that we must have both lines \\(AB\\) (which we shall denote \\(a\\)) and \\(DC\\) (which we shall denote \\(c\\)) parallel to \\(p\\), since if one of them is not, then neither can be and they must both intersect \\(p\\) (since they are both coplanar with \\(p\\)), making them skew.\n\nNow we note since a circle can be inscribed in the trapezoid, we must have \\(AB + DC = AD + BC\\), and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the average of the lengths of the bases.\n\nWe can find this average by dropping perpendicular \\(AA'\\) to \\(c\\) such that \\(A'\\) is on \\(c\\). The average will be \\(A'C\\), which is one of the sides of the rectangle with sides on \\(a\\) and \\(c\\) with vertices at \\(A\\) and \\({C}\\).\n\nWe now draw a circle with center \\({C}\\) that contains \\(A'\\). The intersections of this circle with \\(a\\) are the two possible values of \\(B\\), from either of which it is trivial to determine the corresponding location for \\(D\\). It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all. Q.E.D."
] |
IMO-1960-1
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_1
|
Determine all three-digit numbers \(N\) having the property that \(N\) is divisible by 11, and \(\dfrac{N}{11}\) is equal to the sum of the squares of the digits of \(N\).
|
[
"Let \\(N = 100a + 10b+c\\) for some digits \\(a,b,\\) and \\(c\\). Then\n\n\\[\n100a + 10b+c = 11m\n\\]\n\nfor some \\(m\\). We also have \\(m=a^2+b^2+c^2\\). Substituting this into the first equation and simplification, we get\n\n\\[\n100a+10b+c = 11a^2 +11b^2 +11c^2\n\\]\n\nFor an integer divisible by \\(11\\), the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by \\(11\\). Thus we get: \\(b = a + c\\) or \\(b = a + c - 11\\).\n\nCase \\(1\\): Let \\(b=a+c\\). We get\n\n\\[\n100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2\n\\]\n\n\\[\n10a+c = 2a^2+2ac+2c^2\n\\]\n\nSince the right side is even, the left side must also be even. Let \\(c=2q\\) for some \\(q = 0,1,2,3,4\\). Then\n\n\\[\n10a+2q=2a^2+4aq+8q^2\n\\]\n\n\\[\n5a+q=a^2+2aq+4q^2\n\\]\n\nSubstitute \\(q=0,1,2,3,4\\) into the last equation and then solve for \\(a\\).\n\nWhen \\(q=0\\), we get \\(a=5\\). Thus \\(c=0\\) and \\(b=5\\). We get that \\(N=550\\) which works.\n\nWhen \\(q=1\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=2\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=3\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=4\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nCase \\(2\\): Let \\(b = a + c - 11\\). We get\n\n\\[\n100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)\n\\]\n\n\\[\n10a+c=2a^2+2c^2+2ac-22a-22c+131\n\\]\n\n\\[\n2(a-8)^2+2(c-\\frac{23}{4})^2+2ac-\\frac{505}{8}=0\n\\]\n\nNow we test all \\(c=0\\rightarrow10\\). When \\(c=0,1,2,4,5,6,7,8,9\\), we get no integer solution to \\(a\\). Thus, for these values of \\(c\\), there is no valid \\(N\\). However, when \\(c=3\\), we get\n\n\\[\n2(a-8)^2+2(3-\\frac{23}{4})^2+6a-\\frac{505}{8}=0\n\\]\n\n\\[\n2(a-8)^2+6a-48 = 0\n\\]\n\nWe get that \\(a=8\\) is a valid solution. For this case, we get \\(a=8,b=0,c=3\\), so \\(N=803\\), and this is a valid value. Thus, the answers are \\(\\boxed{N=550,803}\\).",
"Define a ten to be all ten positive integers which begin with a fixed tens digit.\n\nWe can make a systematic approach to this:\n\nBy inspection, \\(\\dfrac{N}{11}\\) must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.\n\nFor a given ten, the sum of the squares of the digits of \\(N\\) increases faster than \\(\\dfrac{N}{11}\\), so we can have at most one number in every ten that works.\n\nWe check the first ten:\n\n\\[\n11*11=121\n\\]\n\n\\[\n1^2+2^2+1^2=4\n\\]\n\n\\[\n12*11=132\n\\]\n\n\\[\n1^2+3^2+2^2=14\n\\]\n\n11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.\n\nWe try the second ten:\n\n\\[\n21*11=231\n\\]\n\n\\[\n2^2+3^2+1^2=14\n\\]\n\n\\[\n22*11=242\n\\]\n\n\\[\n2^2+4^2+2^2=24\n\\]\n\nTherefore, no numbers in the second ten work.\n\nWe continue, to find out that 50 and 73 are the only ones that works.\n\n\\(N=50*11=550\\), \\(N=73*11=803\\) so there are two \\(N\\) that works.",
"Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting \\(\\boxed{N=550,803}\\)."
] |
IMO-1960-2
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_2
|
For what values of the variable \(x\) does the following inequality hold:
\[
\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?
\]
|
[
"Set \\(x = -\\frac{1}{2} + \\frac{a^2}{2}\\), where \\(a\\ge0\\). \\(\\frac{4\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)^2}{\\left(1-\\sqrt{1+2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)}\\right)^2}<2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)+9\\)\n\nAfter simplifying, we get \\((a+1)^2<a^2+8\\)\n\nSo \\(a^2+2a+1<a^2+8\\)\n\nWhich gives \\(a<\\frac{7}{2}\\) and hence \\(-\\frac{1}{2} \\le x<\\frac{45}{8}\\).\n\nBut \\(x=0\\) makes the LHS indeterminate.\n\nSo, answer: \\(-\\frac{1}{2} \\le x<\\frac{45}{8}\\), except \\(x=0\\).",
"If \\(x \\neq 0\\), then the LHS is defined and rewrites as follows:\n\n\\begin{align*} \\frac{4x^2}{(1 - \\sqrt{2x + 1})^2} &= \\biggl(\\frac{2x}{1 - \\sqrt{2x + 1}}\\biggl)^2 \\\\ &= \\biggl( \\frac{2x}{1 - \\sqrt{2x + 1}} \\cdot \\frac{1 + \\sqrt{2x + 1}}{1 + \\sqrt{2x + 1}} \\biggl)^2 \\\\ &= (1 + \\sqrt{2x + 1})^2 \\\\ &= 2x + 2\\sqrt{2x + 1} + 2. \\end{align*}\n\nThe inequality therefore holds if and only if\n\n\\[\n2x + 2\\sqrt{2x + 1} + 2 < 2x + 9.\n\\]\n\nor\n\n\\[\n\\sqrt{2x + 1} < \\frac{7}{2}.\n\\]\n\nSo \\(2x + 1 < 49/4\\) and therefore \\(x < 45/8\\). But if \\(x < -1/2\\) then the inequality makes no sense, since \\(\\sqrt{2x + 1}\\) is imaginary. So the original inequality holds iff \\(x\\) is in \\([-1/2, 0) \\cup (0, 45/8).\\)"
] |
IMO-1960-3
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_3
|
In a given right triangle \(ABC\), the hypotenuse \(BC\), of length \(a\), is divided into \(n\) equal parts (\(n\) an odd integer). Let \(\alpha\) be the acute angle subtending, from \(A\), that segment which contains the midpoint of the hypotenuse. Let \(h\) be the length of the altitude to the hypotenuse of the triangle. Prove that:
\[
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
\]
|
[
"Using coordinates, let \\(A=(0,0)\\), \\(B=(b,0)\\), and \\(C=(0,c)\\). Also, let \\(PQ\\) be the segment that contains the midpoint of the hypotenuse with \\(P\\) closer to \\(B\\).\n\n\\[\n[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NW); label(\"P\",P,ENE); label(\"Q\",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); [/asy]\n\\]\n\nThen, \\(P = \\frac{n+1}{2}B+\\frac{n-1}{2}C = \\left(\\frac{n+1}{2}b,\\frac{n-1}{2}c\\right)\\), and \\(Q = \\frac{n-1}{2}B+\\frac{n+1}{2}C = \\left(\\frac{n-1}{2}b,\\frac{n+1}{2}c\\right)\\).\n\nSo, \\(\\text{slope}\\)\\((PA)=\\tan{\\angle PAB}=\\frac{c}{b}\\cdot\\frac{n-1}{n+1}\\), and \\(\\text{slope}\\)\\((QA)=\\tan{\\angle QAB}=\\frac{c}{b}\\cdot\\frac{n+1}{n-1}\\).\n\nThus, \\(\\tan{\\alpha} = \\tan{(\\angle QAB - \\angle PAB)} = \\frac{(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})-(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}{1+(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})\\cdot(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}\\) \\(= \\frac{\\frac{c}{b}\\cdot\\frac{4n}{n^2-1}}{1+\\frac{c^2}{b^2}} = \\frac{4nbc}{(n^2-1)(b^2+c^2)}=\\frac{4nbc}{(n^2-1)a^2}\\).\n\nSince \\([ABC]=\\frac{1}{2}bc=\\frac{1}{2}ah\\), \\(bc=ah\\) and \\(\\tan{\\alpha}=\\frac{4nh}{(n^2-1)a}\\) as desired.",
"Let \\(P, Q, R\\) be points on side \\(BC\\) such that segment \\(PR\\) contains midpoint \\(Q\\), with \\(P\\) closer to \\(C\\) and (without loss of generality) \\(AC \\le AB\\). Then if \\(AD\\) is an altitude, then \\(D\\) is between \\(P\\) and \\(C\\). Combined with the obvious fact that \\(Q\\) is the midpoint of \\(PR\\) (for \\(n\\) is odd), we have\n\n\\[\n\\tan {\\angle PAR} = \\tan (\\angle RAD - \\angle PAD) = \\frac{\\frac{PR}{h}}{1 + \\frac{DP \\cdot DR}{h^2}} = \\frac{PR \\cdot h}{h^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{AQ^2 - DQ^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{\\frac{a^2}{4} - PQ^2} = \\frac{\\frac{a}{n} \\cdot h}{\\frac{a^2}{4} - \\frac{a^2}{4n^2}} = \\frac{4nh}{(n^2-1)a}.\n\\]",
"Let \\(\\angle ACB = x\\), and \\(\\angle ABC = 90^\\circ - x\\). Let \\(M\\) be the midpoint on the hypotenuse \\(BC\\), and \\(Q\\) and \\(P\\) be points such that \\(PQ\\) contains \\(BC\\), with \\(Q\\) closer to \\(C\\) and \\(P\\) closer to \\(B\\). The midpoint will always be in the middle of line \\(QP\\), unless \\(n\\) is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as:\n\n\\[\nh = a \\cos(x) \\sin(x)\n\\]\n\nNext, we shall denote line \\(AM\\) as \\(f\\), where \\(AM\\) is the median to the hypotenuse. This means that line \\(AM = BM = CM\\), and as \\(BM = \\frac{a}{2}\\), we have:\n\n\\[\nf = \\frac{a}{2}\n\\]\n\nWe know that \\(\\angle MAB = 90^\\circ - x\\), and \\(\\angle MAC = x\\). This means that \\(\\angle AMB = 2x\\) and \\(\\angle AMC = 180^\\circ - 2x\\). The length of \\(QP\\) is \\(\\frac{a}{n}\\). Let \\(\\angle QAM = k\\) and \\(\\angle PAM = z\\), such that \\(\\angle QAP\\) (or \\(\\alpha\\)) equals \\(k + z\\). This means that \\(\\angle AQM = 2x - k\\), and \\(\\angle APM = 180^\\circ - 2x - z\\).\n\nAs \\(M\\) is in the middle of \\(QP\\), we have \\(QM = PM = \\frac{a}{2n}\\). Applying the sine law on triangle \\(AQM\\), we get:\n\n\\[\n\\frac{\\sin(k)}{\\frac{a}{2n}} = \\frac{\\sin(2x - k)}{\\frac{a}{2}}\n\\]\n\nSimplifying:\n\n\\[\n\\frac{2n \\sin(k)}{a} = \\frac{2 \\sin(2x - k)}{a}\n\\]\n\n\\[\nn \\sin(k) = \\sin(2x - k)\n\\]\n\nUsing the identity \\(\\sin(2x - k) = \\sin(2x) \\cos(k) - \\cos(2x) \\sin(k)\\), and since \\(\\sin(2x) = 2 \\sin(x) \\cos(x)\\), we substitute:\n\n\\[\n\\sin(2x) = \\frac{2h}{a}\n\\]\n\nThus:\n\n\\[\nn \\sin(k) = \\frac{2h}{a} \\cos(k) - \\cos(2x) \\sin(k)\n\\]\n\nNow, we know that:\n\n\\[\n\\cos(2x) = \\frac{\\sqrt{a^2 - 4h^2}}{a}\n\\]\n\nSubstituting this into the equation:\n\n\\[\nn \\sin(k) + \\sin(k) \\frac{\\sqrt{a^2 - 4h^2}}{a} = \\cos(k) \\frac{2h}{a}\n\\]\n\nFactoring out \\(\\sin(k)\\):\n\n\\[\n\\sin(k) \\left( n + \\frac{\\sqrt{a^2 - 4h^2}}{a} \\right) = \\cos(k) \\frac{2h}{a}\n\\]\n\nThus:\n\n\\[\n\\tan(k) = \\frac{2h}{an + \\sqrt{a^2 - 4h^2}}\n\\]\n\nBy performing similar steps with \\(\\tan(z)\\), we can use the addition formula for \\(\\tan(z+k)\\) to find \\(\\tan(\\alpha)\\), where \\(\\alpha = z + k\\).\n\nCourtesy of EobardThawne",
"Let \\(D\\) and \\(E\\) be points on \\(BC\\) such that \\(\\alpha=\\angle DAE\\), \\(D\\) is closer to \\(B\\) than to \\(C\\), and \\(E\\) is closer to \\(C\\) than \\(B\\). Furthermore, let \\(\\alpha_1=\\angle BAD\\) and \\(\\alpha_2=\\angle CAE\\), and let \\(\\beta=\\angle B\\) and \\(\\gamma=\\angle C\\). Applying Law of Sines on \\(\\triangle BAD\\) yields \\begin{align*} \\frac{\\sin\\alpha_1}{\\frac{a(n-1)}{2n}}&=\\frac{\\sin(180-(\\alpha_1+\\beta))}{c}\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin(\\alpha_1+\\beta)\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin\\alpha_1\\cos\\beta+\\cos\\alpha_1\\sin\\beta\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin\\alpha_1\\cdot\\frac{c}{a}+\\cos\\alpha_1\\cdot\\frac{b}{a}\\\\ \\frac{2cn}{n-1}\\sin\\alpha_1&=c\\sin\\alpha_1+b\\cos\\alpha_1\\\\ \\frac{cn+c}{n-1}\\sin\\alpha_1&=b\\cos\\alpha_1\\\\ \\tan\\alpha_1&=\\frac{b(n-1)}{c(n+1)} \\end{align*} Similarly, \\(\\tan\\alpha_2=\\frac{c(n-1)}{b(n+1)}\\). Next, we use the sum of tangents formula: \\begin{align*} \\tan(\\alpha_1+\\alpha_2)&=\\frac{\\frac{b(n-1)}{c(n+1)}+\\frac{c(n-1)}{b(n+1)}}{1-\\frac{b(n-1)}{c(n+1)}\\cdot\\frac{c(n-1)}{b(n+1)}}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)\\cdot\\frac{n-1}{n+1}}{1-\\frac{(n-1)^2}{(n+1)^2}}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)(n^2-1)}{(n+1)^2-(n-1)^2}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)(n^2-1)}{4n}\\\\ &=\\frac{(b^2+c^2)(n^2-1)}{4nbc}\\\\ \\end{align*} Since \\(\\alpha+\\alpha_1+\\alpha_2=90^\\circ\\), we need to find \\(\\tan\\alpha=\\tan\\left(\\frac{\\pi}{2}-(\\alpha_1+\\alpha_2)\\right)\\). This is equal to \\(\\cot(\\alpha_1+\\alpha_2)\\), which is the reciprocal of \\(\\tan(\\alpha_1+\\alpha_2)\\), or \\(\\frac{4nbc}{(b^2+c^2)(n^2-1)}\\). Since \\(bc=ah\\) due to triangle areas and \\(a^2=b^2+c^2\\) due to the Pythagorean Theorem, we conclude that the expression is equal to \\(\\frac{4nah}{a^2(n^2-1)}\\), which simplifies to \\(\\frac{4nh}{(n^2-1)a}\\).\n\n~eevee9406"
] |
IMO-1960-4
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_4
|
Construct triangle \(ABC\), given \(h_a\), \(h_b\) (the altitudes from \(A\) and \(B\)), and \(m_a\), the median from vertex \(A\).
|
[
"Let \\(M_a\\), \\(M_b\\), and \\(M_c\\) be the midpoints of sides \\(\\overline{BC}\\), \\(\\overline{CA}\\), and \\(\\overline{AB}\\), respectively. Let \\(H_a\\), \\(H_b\\), and \\(H_c\\) be the feet of the altitudes from \\(A\\), \\(B\\), and \\(C\\) to their opposite sides, respectively. Since \\(\\triangle ABC\\sim\\triangle M_bM_aC\\), with \\(M_bM_a=\\frac12 AB\\), the distance from \\(M_a\\) to side \\(\\overline{AC}\\) is \\(\\frac{h_b}{2}\\).\n\nConstruct \\(AM_a\\) with length \\(m_a\\). Draw a circle centered at \\(A\\) with radius \\(h_a\\). Construct the tangent \\(l_1\\) to this circle through \\(M_a\\). \\(\\overline{BC}\\) lies on \\(l_1\\).\n\nDraw a circle centered at \\(M_a\\) with radius \\(\\frac{h_b}{2}\\). Construct the tangent \\(l_2\\) to this circle through \\(A\\). \\(\\overline{AC}\\) lies on \\(l_2\\). Then \\(C=l_1\\cap l_2\\).\n\nConstruct the line \\(l_3\\) parallel to \\(l_2\\) so that the distance between \\(l_2\\) and \\(l_3\\) is \\(h_b\\) and \\(M_a\\) lies between these lines. \\(B\\) lies on \\(l_3\\). Then \\(B=l_1\\cap l_3\\)."
] |
IMO-1960-5
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_5
|
Consider the cube \(ABCDA'B'C'D'\) (with face \(ABCD\) directly above face \(A'B'C'D'\)).
a) Find the locus of the midpoints of the segments \(XY\), where \(X\) is any point of \(AC\) and \(Y\) is any point of \(B'D'\);
b) Find the locus of points \(Z\) which lie on the segment \(XY\) of part a) with \(ZY = 2XZ\).
|
[
"Let \\(A=(0,0,2)\\), \\(B=(2,0,2)\\), \\(C=(2,0,0)\\), \\(D=(0,0,0)\\), \\(A'=(0,2,2)\\), \\(B'=(2,2,2)\\), \\(C'=(2,2,0)\\), and \\(D'=(0,2,0)\\). Then there exist real \\(x\\) and \\(y\\) in the closed interval \\([0,2]\\) such that \\(X=(x,0,2-x)\\) and \\(Y=(y,2,y)\\).\n\nThe midpoint of \\(XY\\) has coordinates \\(((x+y)/2, 1, (2-x+y)/2)\\). Let \\(a\\) and \\(b\\) be the \\(x\\)- and \\(z\\)-coordinates of the midpoint of \\(XY\\), respectively. We then have that \\(a+b=y+1\\) and \\(a-b=x-1\\), so \\(a+b\\in [1,3]\\) and \\(a-b\\in [-1,1]\\). The region of points that satisfy these inequalities is the closed square with vertices at \\((1,1,2)\\), \\((2,1,1)\\), \\((1,1,0)\\), and \\((0,1,1)\\). For every point \\(P\\) in this region, there exist unique points \\(X\\) and \\(Y\\) such that \\(P\\) is the midpoint of \\(XY\\).\n\nIf \\(Z\\in XY\\) and \\(ZY=2XZ\\), then \\(Z\\) has coordinates \\(((2x+y)/3, 2/3, (4-2x+y)/3)\\). Let \\(a\\) and \\(b\\) be the \\(x\\)- and \\(z\\)- coordinates of \\(Z\\). We then have that \\(a+b=(4/3)+(2/3)y\\) and \\(a-b=(4x-4)/3\\), and \\(a\\in (4/3,8/3)\\) and \\(b\\in (-4/3,4/3)\\). The region of points that satisfy these inequalities is the closed rectangle with vertices at \\((0,2/3,4/3)\\), \\((2/3,2/3,1)\\), \\((1,2/3,2/3)\\), and \\((4/3,2/3,0)\\). For every point \\(Z\\) in this region, there exist unique points \\(X\\) and \\(Y\\) such that \\(Z\\in XY\\) and \\(ZY=2XZ\\)."
] |
IMO-1960-6
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_6
|
Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let \(V_1\) be the volume of the cone and \(V_2\) be the volume of the cylinder.
a) Prove that \(V_1 \neq V_2\);
b) Find the smallest number \(k\) for which \(V_1 = kV_2\); for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
|
[
"Part (a):\n\nLet \\(R\\) denote the radius of the cone, and let \\(r\\) denote the radius of the cylinder and sphere. Let \\(l\\) denote the slant height of the cone, and let \\(h\\) denote the height of the cone.\n\nConsider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle \\(\\omega\\) inscribed in an isosceles triangle \\(T\\).\n\nThe area of \\(T\\) may be computed in two different ways:\n\n\\[\n[T] = \\frac{1}{2} \\times r \\times (2l + 2R) = r(l+R)\n\\]\n\n\\[\n[T] = \\frac{1}{2} \\times 2R \\times h = Rh\n\\]\n\nFrom this, we deduce that \\(r = \\frac{Rh}{l+R}\\).\n\nNow, we calculate our volumes:\n\n\\[\nV_1 = \\frac{1}{3}\\pi R^2 h\n\\]\n\n\\[\nV_2 = \\pi r^2 \\times 2r = 2\\pi r^3 = \\frac{2R^3 h^3}{(l+R)^3}\n\\]\n\nNow, we will compute the quantity \\(\\frac{3(l+R)^3}{\\pi R^5 h} (V_1 - V_2)\\) and prove that it is always greater than \\(0\\). Let \\(x = \\frac{h}{R}\\). Clearly, \\(x\\) can be any positive real number. Define \\(W_1 = \\frac{3(l+R)^3}{\\pi R^5 h} V_1\\) and \\(W_2 = \\frac{3(l+R)^3}{\\pi R^5 h} V_2\\). We will calculate \\(W_1\\) and \\(W_2\\) in terms of \\(x\\) and then compute the desired quantity \\(W_1 - W_2\\).\n\nIt is easy to see that:\n\n\\[\nW_1 = (\\sqrt{x^2+1} + 1)^3\n\\]\n\n\\[\nW_2 = 6x^2\n\\]\n\nNow, let \\(u = \\sqrt{x^2+1}\\). Since \\(x > 0\\), it follows that \\(u > 1\\). We now have:\n\n\\[\nW_1 = (u + 1)^3\n\\]\n\n\\[\nW_2 = 6(u^2 - 1)\n\\]\n\nDefine \\(f(u) = W_1 - W_2\\). It follows that:\n\n\\[\nf(u) = (u+1)^3 - 6(u^2 - 1)\n\\]\n\n\\[\nf(u) = u^3 - 3u^2 + 3u + 7\n\\]\n\n\\[\nf(u) = (u-1)^3 + 8 > 8 > 0\n\\]\n\nWe see that \\(f(u) > 8\\) for all allowed values of \\(u\\). Thus, \\(V_1 - V_2 > \\frac{8\\pi R^5 h}{3(l+R)^3}\\), meaning that \\(V_1 > V_2\\). We have thus proved that \\(V_1 \\ne V_2\\), as desired.\n\nPart (b):\n\nFrom our earlier work in calculating the volumes \\(V_1\\) and \\(V_2\\), we easily see that:\n\n\\[\n\\frac{V_1}{V_2} = \\frac{(u+1)^3}{6(u^2 - 1)}\n\\]\n\nRe-expressing and simplifying, we have:\n\n\\[\n\\frac{6V_1}{V_2} = \\frac{(u+1)^2}{u-1} = (u-1) + 4 + \\frac{4}{u-1}\n\\]\n\nBy the AM-GM Inequality, \\((u-1) + \\frac{4}{u-1} \\ge 4\\), meaning that \\(\\frac{V_1}{V_2} \\ge \\frac{4}{3}\\). Equality holds if and only if \\(u-1 = \\frac{4}{u-1}\\), meaning that \\(u=3\\) and \\(x = 2\\sqrt{2}\\).\n\nIf we check the case \\(x = \\frac{h}{R} = 2\\sqrt{2}\\), we may calculate \\(V_1\\) and \\(V_2\\):\n\n\\[\nV_1 = \\frac{1}{3} \\pi R^2 h = \\frac{2\\sqrt{2}}{3}\\pi R^3\n\\]\n\n\\[\nV_2 = 2\\pi r^3 = 2\\pi \\left(\\frac{R\\times 2\\sqrt{2} R}{4R}\\right)^3 = \\frac{\\pi}{\\sqrt{2}} R^3\n\\]\n\nIndeed, we have \\(\\frac{V_1}{V_2} = \\frac{4}{3}\\), meaning that our minimum of \\(k = \\frac{4}{3}\\) can be achieved.\n\nThus, we have proved that the minimum value of \\(k\\) such that \\(V_1 = kV_2\\) is \\(\\frac{4}{3}\\).\n\nNow, let \\(\\theta\\) be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:\n\n\\[\n\\tan\\left(\\frac{\\theta}{2}\\right) = \\frac{1}{x} = \\frac{\\sqrt{2}}{4}\n\\]\n\nFrom the double-angle formula for tangent,\n\n\\[\n\\theta = \\arctan\\left(\\frac{4\\sqrt{2}}{7}\\right)\n\\]\n\nThis angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths \\(4\\sqrt{2}\\) and \\(7\\). This is straightforward, and the angle opposite the leg of length \\(4\\sqrt{2}\\) will be the desired angle \\(\\theta\\).\n\nIt follows that we have successfully constructed the desired angle \\(\\theta\\).",
"Part a):\n\nLet \\(r\\) and \\(a\\) denote the radius of the sphere and the radius of the base of the cone, respectively.\n\nConsider a plane that contains a cross section of the cone. We get a circle (the sphere) inscribed in both an isosceles triangle (the cone) and a rectangle (the cylinder). Let \\(A\\) and \\(A_1\\) be the vertices at the base of the triangle and let \\(H_1 H_2 = h\\) be the altitude through the third vertex \\(H_1\\). Let \\(\\theta\\) be the angle with arms \\(AO\\) and \\(AA_1\\), where \\(O\\) is the center of the circle; \\(0 < \\theta < \\frac{\\pi}{4}\\) .\n\nWe have \\(\\tan \\theta = \\frac{r}{a}\\) ; \\(0 < \\tan \\theta < 1\\). For the two volumes we have \\(V_2 = \\pi r^2 \\times 2r = 2\\pi r^3\\) and \\(V_1 = \\frac{\\pi a^2 \\times H_1 H_2}{3} = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1) }{3}\\) .\n\nKnowing \\(\\angle A_1 A H_1 = 2 \\times \\angle A_1 A O = 2 \\theta\\), and using the formula for a double-angle tangent, we can simplify the result:\n\n\\[\nV_1 = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1)}{3}\n\\]\n\n\\[\nV_1 = \\frac{2 \\pi r^3 \\tan \\theta}{3 \\tan^3 \\theta (1 - \\tan^2 \\theta)}\n\\]\n\n\\(V_1 = k V_2\\), where \\(k = \\frac{1}{3\\tan^2 \\theta (1 - \\tan^2 \\theta)}\\)\n\nWe see that the biquadratic equation \\(3 \\tan^4 \\theta - 3 \\tan^2 \\theta + 1 = 0\\) has no real roots for \\(\\tan \\theta\\), which means that \\(k \\neq 1\\). Therefore \\(V_1 \\neq V_2\\) .\n\nPart b):\n\nLet \\(\\tan^2 \\theta = x; 0 < x < 1\\). Consider the function \\(f(x) = 3x - 3x^2\\). The coefficient in front of \\(x^2\\) is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of \\(f(x)\\)). Using the vertex formula for a quadratic function, we get \\(x = -\\frac{b}{2a} = \\frac{1}{2}\\) , which gives us \\(f(x) = y_{max} = \\frac{3}{4}\\) . From there \\(k_{min} = \\frac{4}{3}\\) .\n\nFor \\(\\tan^2 \\theta\\) we have \\(\\tan^2 \\theta = \\frac{1}{2}; \\tan \\theta = +/- \\frac{\\sqrt{2}}{4}\\) . We defined \\(\\tan \\theta\\) to be in the interval (0;1), so \\(\\tan \\theta = \\frac{\\sqrt{2}}{4}\\) . That gives us \\(\\tan \\angle A_1 A H_1 = \\frac{4 \\sqrt{2}}{7}\\) . To find \\(2\\theta\\) we construct a right triangle with legs equal to \\(7\\) and \\(4\\sqrt{2}\\) using a unit length reference."
] |
IMO-1960-7
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_7
|
An isosceles trapezoid with bases \(a\) and \(c\) and altitude \(h\) is given.
a) On the axis of symmetry of this trapezoid, find all points \(P\) such that both legs of the trapezoid subtend right angles at \(P\);
b) Calculate the distance of \(P\) from either base;
c) Determine under what conditions such points \(P\) actually exist. Discuss various cases that might arise.
|
[
"(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.\n\n(b) Let \\(x\\) be the distance from \\(P\\) to one of the bases; then \\(h - x\\) must be the distance from \\(P\\) to the other base. Similar triangles give \\(\\frac{x}{\\frac{a}{2}} = \\frac{\\frac{c}{2}}{h - x}\\), so \\(x^2 - hx + \\frac{ac}{4} = 0\\) and so \\(x = \\frac{h \\pm \\sqrt{h^2 - ac}}{2}.\\)\n\n(c) When \\(h^2 \\ge ac\\).\n\n[1]\n\nIn our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).\n\nLet our point P be on the axis of symmetry at z distance from the origin O.\n\nThe coordinates of the points A,B,C,D,E,F and P are given in the figure.\n\nNow,\n\nSlope of the line \\(PC= (z-0)/(0-c/2) = -2z/c\\) Slope of the line \\(PB= (z-h)/(0-a/2) = -2(z-h)/a\\)\n\nSince the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.\n\ni.e\n\n\\[\n4z(z-h)=-ac\n\\]\n\nor \\(z^2 - zh + ac/4= 0\\)\n\nNow, solving for z, we get, \\(z= [(h + ( h^2 - ac ) ^1/2 ]/2\\) and \\([(h - ( h^2 - ac ) ^1/2 ]/2\\)\n\nSo, z is the distance of the points from the base CD..\n\nAlso the points are possible only when , \\(h^2 - ac >= 0\\).. and doesn't exist for \\(h^2 -ac <0\\)"
] |
IMO-1961-1
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_1
|
Solve the system of equations:
\[
\begin{matrix} \quad x + y + z \!\!\! &= a \; \, \\ x^2 +y^2+z^2 \!\!\! &=b^2 \\ \qquad \qquad xy \!\!\! &= z^2 \end{matrix}
\]
where \(a\) and \(b\) are constants. Give the conditions that \(a\) and \(b\) must satisfy so that \(x, y, z\) (the solutions of the system) are distinct positive numbers.
|
[
"Note that \\(x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2\\), so the first two equations become\n\n\\[\n\\begin{matrix} \\quad (x + y) + z \\!\\!\\! &= a \\; \\; (*) \\\\ (x+y)^2 - z^2 \\!\\!\\! &=b^2 (**) \\end{matrix}\n\\]\n\n.\n\nWe note that \\((x+y)^2 - z^2 = \\Big[ (x+y)+z \\Big]\\Big[ (x+y)-z\\Big]\\), so if \\(a\\) equals 0, then \\(b\\) must also equal 0. We then have \\(x+y = -z\\); \\(xy = (x+y)^2\\). This gives us \\(x^2 + xy + y^2 = 0\\). Mutiplying both sides by \\((x-y)\\), we have \\(x^3 - y^3 = 0\\). Since we want \\(x,y\\) to be real, this implies \\(x = y\\). But \\(x^2 + x^2 + x^2\\) can only equal 0 when \\(x=0\\) (which, in this case, implies \\(y,z = 0\\)). Hence there are no positive solutions when \\(a = 0\\).\n\nWhen \\(a \\neq 0\\), we divide \\((**)\\) by \\((*)\\) to obtain the system of equations\n\n\\[\n\\begin{matrix} (x+y)+z &= a \\; \\quad \\\\ (x+y)-z &= b^2/a \\end{matrix}\n\\]\n\n,\n\nwhich clearly has solution \\(x+y = \\frac{a^2 + b^2}{2a}\\), \\(z = \\frac{a^2 - b^2}{2a}\\). In order for these both to be positive, we must have positive \\(a\\) and \\(a^2 > b^2\\). Now, we have \\(x+y = \\frac{a^2 + b^2}{2a}\\); \\(xy = \\left(\\frac{a^2 - b^2}{2a}\\right)^2\\), so \\(x,y\\) are the roots of the quadratic \\(m^2 - \\frac{a^2 + b^2}{2a}m + \\left(\\frac{a^2 - b^2}{2a}\\right)^2\\). The discriminant for this equation is\n\n\\[\n\\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2 = \\frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}\n\\]\n\n.\n\nIf the expressions \\((3a^2 - b^2), (3b^2 - a^2)\\) were simultaneously negative, then their sum, \\(2(a^2 + b^2)\\), would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when \\(3a^2 > b^2\\) and \\(3b^2 > a^2\\). But we have already replaced the first inequality with the sharper bound \\(a^2 > b^2\\). It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from \\(\\left(\\frac{a^2 + b^2}{2a}\\right)^2 > \\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2\\) or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if \\(a\\) is positive and \\(3b^2 > a^2 > b^2\\). Q.E.D.",
"Obviously, \\(a=x+y+z>0\\). The third equation implies that \\(x,z,y\\) is a geometric sequence. Then let \\(x=\\frac{z}{r}\\) and \\(y=rz\\), with \\(r,z>0\\) and \\(r\\neq1\\). Then the first two equations become:\n\n\\[\n\\left(r+1+\\frac{1}{r}\\right)z=a~~~(1)\n\\]\n\nand\n\n\\[\n\\left(r^2+1+\\frac{1}{r^2}\\right)z^2=b^2~~~(2)\n\\]\n\nTaking \\(\\frac{(2)}{(1)}\\) (since \\(z>0\\)), we get:\n\n\\[\n\\frac{(r^2+1+\\frac{1}{r^2})z^2}{(r+1+\\frac{1}{r})z}=\\left(r-1+\\frac{1}{r}\\right)z=\\frac{b^2}{a}~~~(3)\n\\]\n\nWe can then take \\((1)^2-(2)\\) and \\((2)-(3)^2\\) to get:\n\n\\[\n\\left(r^2+2r+3+\\frac{2}{r}+\\frac{1}{r^2}\\right)z^2-\\left(r^2+1+\\frac{1}{r^2}\\right)z^2=2z^2\\left(r+1+\\frac{1}{r}\\right)=a^2-b^2~~~(4)\n\\]\n\nand\n\n\\[\n\\left(r^2+1+\\frac{1}{r^2}\\right)z^2-\\left(r^2-2r+3-\\frac{2}{r}+\\frac{1}{r^2}\\right)z^2=2z^2\\left(r-1+\\frac{1}{r}\\right)=b^2-\\frac{b^4}{a^2}=\\frac{b^2}{a^2}(a^2-b^2)~~~(5)\n\\]\n\nLet \\(k=r+\\frac{1}{r}\\). By AM-GM, \\(k\\ge2\\) with equality at \\(r=\\frac{1}{r}\\implies r=1\\), which is impossible. Hence, \\(k>2\\). Then, \\(\\frac{(5)}{(4)}\\) becomes:\n\n\\[\n\\frac{k-1}{k+1}=\\frac{b^2}{a^2}~(6)\\implies a^2+b^2=(a^2-b^2)k>2(a^2-b^2)\\implies3b^2>a^2\n\\]\n\nFrom the above restrictions on \\(a\\) and \\(b\\), we see that there must exist some \\(k>2\\) satisfying \\((6)\\), and hence, some \\(r>0\\neq1\\) satisfying \\((6)\\). From \\((4)\\), if \\(a^2-b^2>0\\), then there must exist some positive \\(z\\) satisfying \\((4)\\), and consequently since \\((4)\\) and \\((6)\\) are equivalent to the remaining equations, they satisfy \\((1)\\) and \\((2)\\). Hence, \\(x,y,z\\) satisfy the original system, and from the restrictions on \\(r\\) and \\(z\\), they are distinct positive reals. Hence, \\(\\boxed{a>0\\text{ and }3b^2>a^2>b^2}\\). \\(\\blacksquare\\)\n\n~rhydon516"
] |
IMO-1961-2
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_2
|
Let \(a\), \(b\), and \(c\) be the lengths of a triangle whose area is S. Prove that
\[
a^2 + b^2 + c^2 \ge 4S\sqrt{3}
\]
In what case does equality hold?
|
[
"By Heron's formula, we have\n\n\\[\nS = \\sqrt{s(s-a)(s-b)(s-c)}.\n\\]\n\nThis can be simplified to\n\n\\[\nS = \\sqrt{\\left(\\frac{a+b+c}{2}\\right)\\left(\\frac{-a+b+c}{2}\\right)\\left(\\frac{a-b+c}{2}\\right)\\left(\\frac{a+b-c}{2}\\right)}.\n\\]\n\nNext, we can factor out all of the \\(2\\)s and use a clever difference of squares:\n\n\\[\nS = \\frac{1}{4}\\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\n\\]\n\n\\[\nS = \\frac{1}{4}\\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}\n\\]\n\n\\[\nS = \\frac{1}{4}\\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}.\n\\]\n\nWe can now use difference of squares again:\n\n\\[\nS = \\frac{1}{4}\\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}\n\\]\n\n\\[\n4S\\sqrt{3} = \\sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}\n\\]\n\nWe must prove that the RHS of this equation is less than or equal to \\(a^2 + b^2 + c^2\\).\n\nLet \\(a^2 = A\\), \\(b^2 = B\\), \\(c^2 = C\\). Then, our inequality can be reduced to\n\n\\[\nA + B + C \\geq \\sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.\n\\]\n\nWe now have to prove\n\n\\[\n(A + B + C)^2 \\geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.\n\\]\n\nWe can simplify:\n\n\\[\nA^2 + B^2 + C^2 + 2AB + 2BC + 2CA \\geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2\n\\]\n\n\\[\n4A^2 + 4B^2 + 4C^2 \\geq 4AB + 4BC + 4CA\n\\]\n\n\\[\nA^2 + B^2 + C^2 \\geq AB + BC + CA.\n\\]\n\nFinally, we can apply AM-GM:\n\n\\[\n\\frac{A^2 + B^2}{2} \\geq AB\n\\]\n\n\\[\n\\frac{B^2 + C^2}{2} \\geq BC\n\\]\n\n\\[\n\\frac{C^2 + A^2}{2} \\geq CA\n\\]\n\nAdding these all up, we have the desired inequality\n\n\\[\nA^2 + B^2 + C^2 \\geq AB + BC + CA,\n\\]\n\nand so the proof is complete.\\(\\square\\)\n\nTo have \\(A + B + C = 4S\\sqrt{3}\\), we must satisfy\n\n\\[\n\\frac{A^2 + B^2}{2} = AB,\n\\]\n\n\\[\n\\frac{B^2 + C^2}{2} = BC,\n\\]\n\n\\[\n\\frac{C^2 + A^2}{2} = CA.\n\\]\n\nThis is only true when \\(A = B = C\\), and thus \\(a = b = c\\). Therefore, equality happens when the triangle is equilateral.\n\n~mathboy100",
"We firstly use the duality principle. \\(a=x+y~~b=x+z~~c=y+z\\) The LHS becomes \\((x+y)^2+(x+z)^2+(y+z)^2\\) and the RHS becomes \\(4\\sqrt{3}\\sqrt{(x+y+z)xyz}\\) If we use Heron's formula. By AM-GM \\(\\frac{(x+y+z)^3}{27} \\ge xyz\\) Making this substitution \\([ABC]\\) becomes \\(\\sqrt{(x+y+z)^4\\frac{1}{27}}\\) and once we take the square root of the area then our RHS becomes \\(\\frac{4\\sqrt{3}}{3\\sqrt{3}}(x+y+z)^2=\\frac{4}{3}(x+y+z)^2\\) Multiplying the RHS and the LHS by 3 we get the LHS to be \\(3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).\\) Our RHS becomes \\(4(x^2+y^2+z^2)+8(xy+yz+xz).\\) Subtracting \\(4(x^2+y^2+z^2)+6(xy+yz+xz)\\) we have the LHS equal to \\((2(x^2+y^2+z^2))\\) and the RHS being \\(2(xy+xz+yz)\\) If LHS \\(\\ge\\) RHS then LHS-RHS\\(\\ge 0\\) LHS-RHS=\\(2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.\\) \\((x-y)^2+(x-z)^2+(y-z)^2 \\ge 0\\) by the trivial inequality so therefore, \\(a^2 + b^2 + c^2 \\ge 4S\\sqrt{3}\\) and we're done.\n\n~PEKKA",
"Let \\(\\theta\\) be the angle between edges \\(a\\) and \\(b\\) subtending \\(c\\).\n\nWe notice that \\(0\\le(a-b\\cos x)^2=a^2-2ab\\cos x+b^2\\cos^2x\\le a^2-2ab\\cos x+b^2\\) holds for all \\(a,b,x\\in\\mathbb R\\).\n\nIf we let \\(x=\\theta+30^\\circ\\),\n\n\\[\n\\begin{align*}a^2-2ab\\sin(\\theta+30^\\circ)+b^2&\\geq0\\\\ a^2-2ab(\\sin\\theta\\cos30^\\circ+\\cos\\theta\\sin30^\\circ)+b^2&\\geq0\\\\a^2-ab\\sin\\theta\\sqrt3-ab\\cos\\theta+b^2&\\geq0\\\\a^2-ab\\cos\\theta+b^2&\\geq ab\\sin\\theta\\sqrt3\\\\a^2-ab\\cos\\theta+b^2&\\geq2\\left(\\frac12ab\\sin\\theta\\right)\\sqrt3\\\\a^2-ab\\cos\\theta+b^2&\\geq2T\\sqrt3\\\\2a^2-2ab\\cos\\theta+2b^2&\\geq4T\\sqrt3\\\\a^2+b^2+(a^2-2ab\\cos\\theta+b^2)&\\geq4\\sqrt3\\ T\\\\\\boxed{a^2+b^2+c^2\\geq4\\sqrt3\\ T}\\ \\blacksquare\\end{align*}\n\\]\n\nFor equality, it is necessary for \\(a^2-2ab\\sin x+b^2=0\\implies a=b\\) and \\(\\sin x=\\sin(\\theta+30^\\circ)=1\\implies\\theta=60^\\circ\\), which implies that \\(\\triangle ABC\\) is equilateral.\n\n~ztilB",
"Again, we use Heron's formula but we also set \\(a=y+z\\), \\(b=z+x\\), \\(c=x+y\\). Then \\(S=\\sqrt{xyz(x+y+z)}\\) and\n\n\\[\n\\begin{align*} 4\\sqrt{3xyz(x+y+z)} &= \\frac{4}{3}\\sqrt{27xyz(x+y+z)} \\\\ &\\le \\frac{4}{3}(x^2+y^2+z^2+2xy+2yz+2zx) \\end{align*}\n\\]\n\nand if we set \\(A=xy+yz+zx\\) and \\(B=x^2+y^2+z^2\\) then it suffices to prove that \\(A+B \\le \\frac{2}{3}(A+2B)\\) which is equivalent to \\(A\\le B\\) and is obviously true. ~alexanderchew2021"
] |
IMO-1961-3
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_3
|
Solve the equation
\[
\cos^n{x} - \sin^n{x} = 1
\]
where \(n\) is a given positive integer.
|
[
"Since \\(cos^2x + sin^2x = 1\\), we cannot have solutions with \\(n\\ne2\\) and \\(0<|cos(x)|,|sin(x)|<1\\). Nor can we have solutions with \\(n=2\\), because the sign is wrong. So the only solutions have \\(sin (x) = 0\\) or \\(cos (x) = 0\\), and these are: \\(x =\\) multiple of \\(\\pi\\), and \\(n\\) even; \\(x\\) even multiple of \\(\\pi\\) and \\(n\\) odd; \\(x\\) = even multiple of \\(\\pi + \\frac{3\\pi}{2}\\) and \\(n\\) odd.",
"First consider \\(n=1\\). Squaring both sides yields \\(\\cos^2{x}-2\\cos{x}\\sin{x}+\\sin^2{x}=1\\), and utilizing identities yields \\(\\sin{2x}=0\\). This results in solutions \\(x=\\frac{\\pi}{2}k\\) for integers \\(k\\). After manually checking the solutions in the original equation, we conclude that only \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) are solutions to the equation.\n\nNext, we consider \\(n=2\\). This is equivalent to \\(\\cos{2x}=1\\), so the solutions are \\(x=k\\pi\\) for integers \\(k\\). Once again, substituting into the equation results in a valid identity, so these are the only solutions.\n\nLet \\(f(x)=\\cos^n{x}-\\sin^n{x}\\). Taking the derivative yields \\(f'(x)=-n\\sin{x}\\cos{x}(\\sin^{n-2}{x}+\\cos^{n-2}{x})\\) for \\(n>2\\). We wish to consider extrema, so we find all \\(x\\)-values such that \\(f'(x)=0\\) by checking the zeros of each factor. In \\(x\\in[0,2\\pi]\\), \\(f'(x)=0\\) for even \\(n\\) when \\(x=0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2},2\\pi\\), and for odd \\(n\\) when \\(x=0,\\frac{\\pi}{2},\\frac{3\\pi}{4},\\pi,\\frac{3\\pi}{2},\\frac{7\\pi}{4},2\\pi\\). The differences between the two come from the \\(\\sin^{n-2}{x}+\\cos^{n-2}{x}\\) factor, reaching zeros when \\(\\tan^{n-2}{x}=-1\\). This is impossible when \\(n\\) is even, and if \\(n\\) is odd, it implies that \\(\\tan{x}=-1\\), hence the solutions.\n\nWe then calculate the value of \\(f(x)\\) at every such \\(x\\). For even \\(n\\), the \\(x\\)-values \\(0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2},2\\pi\\) result in \\(f(x)\\)-values of \\(1,-1,1,-1,1\\). Since these are the only possible extrema, we conclude that \\(1\\) is always a maximum and \\(-1\\) is always a minimum; thus the only solutions are the maxima, which occur at \\(x=0,\\pi,2\\pi\\) when \\(x\\in[0,2\\pi]\\); more generally, we have \\(x=k\\pi\\) for integers \\(k\\).\n\nNext, for odd \\(n\\), the \\(x\\)-values \\(x=0,\\frac{\\pi}{2},\\frac{3\\pi}{4},\\pi,\\frac{3\\pi}{2},\\frac{7\\pi}{4},2\\pi\\) yield \\(f(x)\\)-values of \\(1,-1,-2\\left(\\frac{1}{\\sqrt{2}}\\right)^n,-1,1,2\\left(\\frac{1}{\\sqrt{2}}\\right)^n,1\\). Obviously \\(x=0,\\frac{3\\pi}{2},2\\pi\\) are all solutions; we show that on our closed domain these are the only solutions. On \\(x=0\\) to \\(x=\\frac{\\pi}{2}\\), the function is always decreasing; if it were increasing at any point, an extremum would be required to change the sign of the derivative; however, there does not exist an extremum between these two points (since we have already found all the extrema!); thus only \\(x=0\\) satisfies the equation. We can repeat this argument for all values between two extrema; eventually, we can conclude that the only times that \\(f(x)=1\\) are the above (since no segment contains \\(1\\) at any point other than an endpoint; as a result, the function cannot equal \\(1\\) at any point other than an endpoint of some segment, which are the extrema). As a result, extending this to all \\(x\\), the solutions are \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) for integers \\(k\\).\n\nSince these results match our \\(n=1\\) and \\(n=2\\) cases, we conclude that when \\(n\\) is odd, the solutions are \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) for integers \\(k\\), and when \\(n\\) is even, the solutions are \\(x=k\\pi\\) for integers \\(k\\).\n\n~ eevee9406"
] |
IMO-1961-4
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_4
|
In the interior of triangle \(P_1P_2P_3\) a point \(P\) is given. Let \(Q_1,Q_2,Q_3\) be the intersections of \(PP_1, PP_2,PP_3\) with the opposing edges of triangle \(P_1P_2P_3\). Prove that among the ratios \(\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}\) there exists one not larger than \(2\) and one not smaller than \(2\).
|
[
"Let \\([ABC]\\) denote the area of triangle \\(ABC\\).\n\nSince triangles \\(P_1P_2P_3\\) and \\(PP_2P_3\\) share the base \\(P_2P_3\\), we have \\(\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_1}{P_1Q_1}\\).\n\nSimilarly, \\(\\frac{[PP_1P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}, \\frac{[PP_1P_2]}{[P_1P_2P_3]}=\\frac{PQ_3}{P_3Q_3}\\).\n\nAdding all of these gives \\(\\frac{[PP_1P_3]}{[P_1P_2P_3]}+\\frac{[PP_1P_2]}{[P_1P_2P_3]}+\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}+\\frac{PQ_3}{P_3Q_3}+\\frac{PQ_1}{P_1Q_1}=1\\).\n\nWe see that we must have at least one of the three fractions not greater than \\(\\frac{1}{3}\\), and at least one not less than \\(\\frac{1}{3}\\). These correspond to ratios \\(\\frac{PP_i}{PQ_i}\\) being less than or equal to \\(2\\), and greater than or equal to \\(2\\), respectively, so we are done.",
"Let \\(K_1=[P_2PP_3], K_2=[P_3PP_1],\\) and \\(K_3=[P_1PP_2].\\) Note that by same base in triangles \\(P_2PP_3\\) and \\(P_2P_1P_3,\\)\n\n\\[\n\\frac{P_1P}{PQ_1}+1=\\frac{P_1Q_1}{PQ_1}=\\frac{[P_2P_1P_3]}{[P_2PP_3]}=\\frac{K_1+K_2+K_3}{K_1}.\n\\]\n\nThus,\n\n\\[\n\\frac{P_1P}{PQ_1}=\\frac{K_2+K_3}{K_1}\n\\]\n\n\\[\n\\frac{P_2P}{PQ_2}=\\frac{K_1+K_3}{K_2}\n\\]\n\n\\[\n\\frac{P_3P}{PQ_3}=\\frac{K_1+K_2}{K_3}.\n\\]\n\nWithout loss of generality, assume \\(K_1\\leq K_2\\leq K_3.\\) Hence,\n\n\\[\n\\frac{P_1P}{PQ_1}=\\frac{K_2+K_3}{K_1}\\geq \\frac{K_1+K_1}{K_1}=2\n\\]\n\nand\n\n\\[\n\\frac{P_3P}{PQ_3}=\\frac{K_1+K_2}{K+3}\\leq \\frac{K_3+K_3}{K_3}=2,\n\\]\n\nas desired. \\(\\blacksquare\\)"
] |
IMO-1961-5
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_5
|
Construct a triangle ABC if the following elements are given: \(AC = b, AB = c\), and \(\angle AMB = \omega \left(\omega < 90^{\circ}\right)\) where M is the midpoint of BC. Prove that the construction has a solution if and only if
\[
b \tan{\frac{\omega}{2}} \le c < b
\]
In what case does equality hold?
|
[
"Prolong BA to a point D such that \\(BD = 2AB\\). Take circle through B and D such that the minor arc BD is equal to \\(2*\\omega\\) so that for points P on the major arc BD we have \\(\\angle BPD = \\omega\\). Draw a circle with center A and radius AC, and the point of intersection of this circle and the major arc BD will be C. In general there are two possibilities for C.\n\nLet X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require \\(AX \\geqslant AC > AB\\). But \\(\\frac{AB}{AX} = \\tan{\\frac{\\omega}{2}}\\), so we get the condition in the question"
] |
IMO-1961-6
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_6
|
Consider a plane \(\epsilon\) and three non-collinear points \(A,B,C\) on the same side of \(\epsilon\); suppose the plane determined by these three points is not parallel to \(\epsilon\). In plane \(\epsilon\) take three arbitrary points \(A',B',C'\). Let \(L,M,N\) be the midpoints of segments \(AA', BB', CC'\); Let \(G\) be the centroid of the triangle \(LMN\). (We will not consider positions of the points \(A', B', C'\) such that the points \(L,M,N\) do not form a triangle.) What is the locus of point \(G\) as \(A', B', C'\) range independently over the plane \(\epsilon\)?
|
[
"We will consider the various points in terms of their coordinates in space. We have \\(L=\\frac{A+A^\\prime}{2},M=\\frac{B+B^\\prime}{2},N=\\frac{C+C^\\prime}{2}\\). Since the centroid of a triangle is the average of the triangle's vertices, we have \\(G=\\frac{1}{3}\\left(L+M+N\\right)=\\frac{1}{6}\\left(A+B+C+A^\\prime+B^\\prime+C^\\prime\\right)\\). It is clear now that \\(G\\) is midpoint of the line segment connecting the centroid of \\(ABC\\) and the centroid of \\(A^\\prime B^\\prime C^\\prime\\). It is obvious that the centroid of \\(A^\\prime B^\\prime C^\\prime\\) can be any point on plane \\(\\epsilon\\). Thus, the locus of \\(G\\) is the plane parallel to \\(\\epsilon\\) and halfway between the centroid of \\(ABC\\) and \\(\\epsilon\\)."
] |
IMO-1962-1
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_1
|
Find the smallest natural number \(n\) which has the following properties:
(a) Its decimal representation has 6 as the last digit.
(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number \(n\).
|
[
"As the new number starts with a \\(6\\) and the old number is \\(1/4\\) of the new number, the old number must start with a \\(1\\).\n\nAs the new number now starts with \\(61\\), the old number must start with \\(\\lfloor 61/4\\rfloor = 15\\).\n\nWe continue in this way until the process terminates with the new number \\(615\\,384\\) and the old number \\(n=\\boxed{153\\,846}\\).",
"We know from the two properties that for some string \\(x\\), \\(n=10x+6\\). Let the number of digits in \\(x\\) be \\(a\\); then moving the \\(6\\) to the front would give it place value \\(10^a\\); as a result, \\(4n=6\\cdot10^a+x\\). Multiplying this by \\(10\\) gives \\(40n=6\\cdot10^{a+1}+10x\\), and subtracting the former yields \\(39n=6(10^{a+1}-1)\\), or \\(13n=2(10^{a+1}-1)\\). As a result, \\(13|(10^{a+1}-1)\\). By Fermat's Little Theorem, we know that \\(10^{12}-1\\) divides \\(13\\), so it isn't difficult to try values of \\(a+1\\) less than \\(13\\) to find the smallest such \\(a\\).\n\nEventually, we notice that \\(10^6-1\\) divides \\(13\\), so \\(a=5\\). Then \\(\\boxed{n=153846}\\), and since the number ends in \\(6\\), we know that \\(x\\) is also an integer, so this is the solution.\n\n~eevee9406"
] |
IMO-1962-2
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_2
|
Determine all real numbers \(x\) which satisfy the inequality:
\[
\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}
\]
(Note that the problem as written in the official PDF https://www.imo-official.org/year_info.aspx?year=1962 does not have the outer \(\sqrt{\ }\))
|
[
"Obviously we need \\(\\sqrt{3-x} \\geq \\sqrt{x+1}\\) for the outer square root to be defined, \\(x\\leq 3\\) for the first inner square root to be defined, and \\(x\\geq -1\\) for the second inner square root to be defined. Solving these we get that the left hand side is defined for \\(x\\in \\left[ -1,1 \\right]\\).\n\nNow obviously the function \\(f(x)=\\sqrt{\\sqrt{3-x}-\\sqrt{x+1}}\\) is continuous on \\(\\left[ -1,1 \\right]\\), with \\(f(-1)=\\sqrt 2\\) and \\(f(1)=0\\). Moreover, as \\(3-x\\) is a decreasing and \\(x+1\\) an increasing function, both \\(\\sqrt{3-x}\\) and \\(-\\sqrt{x+1}\\) are decreasing functions, and hence \\(f(x)\\) is a decreasing function. Therefore there is exactly one solution to \\(f(x)=\\dfrac{1}{2}\\).\n\nWe can now find this solution:\n\n\\[\n\\begin{align*} \\sqrt{\\sqrt{3-x}-\\sqrt{x+1}} &= \\dfrac{1}{2} \\\\ \\sqrt{3-x}-\\sqrt{x+1} &= \\dfrac{1}{4} \\\\ \\sqrt{3-x} &= \\dfrac{1}{4} + \\sqrt{x+1} \\\\ 3-x &= \\dfrac 1{16} + x+1 + \\dfrac{\\sqrt{x+1}}2 \\\\ 2 - 2x - \\dfrac 1{16} &= \\dfrac{\\sqrt{x+1}}2 \\\\ 31 - 32x &= 8\\sqrt{x+1} \\\\ 1024 x^2 - 1984x + 961 &= 64(x+1) \\\\ 1024 x^2 - 2048x + 897 &= 0 \\end{align*}\n\\]\n\n(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)\n\nSolving the quadratic equation for \\(x\\), we get\n\n\\[\nx_{1,2}=\\dfrac{ 2^{11} \\pm \\sqrt{ 2^{22} - 2^{12}\\cdot 897} }{2^{11}} = 1 \\pm \\dfrac{\\sqrt{127}}{32}\n\\]\n\nThe reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than \\(1\\) is the additional solution, and \\(x=1-\\dfrac{\\sqrt{127}}{32}\\) is the root we need.\n\nHence the solutions to the given inequality are precisely the reals in the interval \\(\\boxed{ \\left[ ~ -1,\\quad 1-\\dfrac{\\sqrt{127}}{32} ~ \\right) }\\).\n\n\\[\n[asy] import graph; size(300,300,IgnoreAspect); real f(real x) {return sqrt( sqrt(3-x) - sqrt(x+1) );}; real g(real x) {return 1/2;}; draw(graph(f,-1,1),blue,\"$f(x)=\\sqrt{\\sqrt{3-x}-\\sqrt{x+1}}$\"); draw(graph(g,-1,1),red,\"$g(x)=1/2$\"); xaxis(\"$x$\",BottomTop,Ticks); yaxis(\"$y$\",LeftRight,Ticks); attach(legend(),point(E),20E,UnFill); [/asy]\n\\]"
] |
IMO-1962-3
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_3
|
Consider the cube \(ABCDA'B'C'D'\)(\(ABCD\) and \(A'B'C'D'\) are the upper and lower bases, respectively, and edges \(AA'\), \(BB'\), \(CC'\), \(DD'\) are parallel). The point \(X\) moves at constant speed along the perimeter of the square \(ABCD\) in the direction \(ABCDA\), and the point \(Y\) moves at the same rate along the perimeter of the square \(B'C'CB\) in the direction \(B'C'CBB'\). Points \(X\) and \(Y\) begin their motion at the same instant from the starting positions \(A\) and \(B'\), respectively. Determine and draw the locus of the midpoints of the segments \(XY\).
|
[
"First we prove a small lemma: If the particles \\(X\\) and \\(Y\\) move along straight lines at constant velocities, then the locus of the midpoint of \\(XY\\) is also a line. This is rather trivial, since all lines in 3D space may take the parametric form \\((at+n, bt+m, ct+p)\\), with \\(t\\) being time, and the average of two such lines must also have a linear parametric form.\n\nThe locus clearly starts at the midpoint of \\(AB'\\), or the center of face \\(ABB'A'\\). As \\(X\\) moves from \\(A\\) to \\(B\\), and \\(Y\\) moves from \\(C'\\) to \\(C\\), both \\(X\\) and \\(Y\\) move along straight lines, so the midpoint of \\(XY\\) traces out a line segment, starting at the midpoint of \\(AB'\\) and ending at the midpoint of \\(BC'\\), or the center of face \\(BCC'B'\\). This concludes the first phase of motion. A quick check reveals that the locus goes nowhere during the second and fourth phases of motion, and only moves backward on the third. Thus the locus is just the segment connecting the centers of sides \\(ABB'A'\\) and \\(BCC'B'\\).\n\nsolution is actually wrong, it's not segment it's parallelogram- author mixed the movement of points (even if he clearly understands what he is doing- just a little mistake) Here is correct soluton:\n\nAnswer: the rhombus CUVW, where U is the center of ABCD, V is the center of ABB'A, and W is the center of BCC'B'.\n\nTake rectangular coordinates with A as (0, 0, 0) and C' as (1, 1, 1). Let M be the midpoint of XY. Whilst X is on AB and Y on B'C', X is (x, 0, 0) and Y is (1, x, 1), so M is (x/2 + 1/2, x/2, 1/2) = x (1, 1/2, 1/2) + (1-x) (1/2, 0, 1/2) = x W + (1-x) V, so M traces out the line VW.\n\nWhilst X is on BC and Y is on C'C, X is (1, x, 0) and Y is ( 1, 1, 1-x), so M is (1, x/2+1/2, 1/2 - x/2) = x (1, 1, 0) + (1-x) (1, 1/2, 1/2) = x C + (1-x) W, so M traces out the line WC.\n\nWhilst X is on CD and Y is on CB, X is (1-x, 1, 0) and Y is (1, 1-x, 0), so M is (1-x/2, 1-x/2, 0) = x (1, 1, 0) + (1-x) (1/2, 1/2, 0) = x C + (1-x) U, so M traces out the line CU.\n\nWhilst X is on DA and Y is on BB', X is (0, 1-x, 0) and Y is (1, 0, x), so M is (1/2, 1/2 - x/2, x/2) = x (1/2, 0, 1/2) + (1-x) (1/2, 1/2, 0) = x V + (1-x) U, so M traces out the line UV."
] |
IMO-1962-4
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_4
|
Solve the equation \(\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1\).
|
[
"First, note that we can write the left hand side as a cubic function of \\(\\cos^2 x\\). So there are at most \\(3\\) distinct values of \\(\\cos^2 x\\) that satisfy this equation. Therefore, if we find three values of \\(x\\) that satisfy the equation and produce three different \\(\\cos^2 x\\), then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that \\(\\frac{\\pi}2\\), \\(\\frac{\\pi}4\\), and \\(\\frac{\\pi}6\\) all satisfy the equation, and produce three different values of \\(\\cos^2 x\\), namely \\(0\\), \\(\\frac12\\), and \\(\\frac34\\). So we solve \\(\\cos^2 x = \\text{each of these}\\). Therefore, our solutions are:\n\n\\[\nx = \\frac{(2k+1)\\pi}2,\\, \\frac{(2k+1)\\pi}4,\\, \\frac{(6k+1)\\pi}6,\\, \\frac{(6k+5)\\pi}6 \\quad \\forall k\\in Z\n\\]"
] |
IMO-1962-5
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_5
|
On the circle \(K\) there are given three distinct points \(A,B,C\). Construct (using only straightedge and compass) a fourth point \(D\) on \(K\) such that a circle can be inscribed in the quadrilateral thus obtained.
|
[
"Aviso: La siguiente solución está en español\n\ngeneralidad ya que es simétrico para todas las combinaciones posibles)\n\nCircunferencia K en 2 puntos, uno de ello se nombrará con E (va a estar del mismo lado que B con respecto a AC) y el otro se nombrará F\n\niguales, abrir con el compás la distancia BE y se traza la Circunferencia de Centro en F y Radio BE, así esta nueva circunferencia corta a la circunferencia K en 2 puntos, llámese D al punto que está del mismo lado de A con respecto a la mediatriz de AC, ese es el punto pedido que cumple la"
] |
IMO-1962-6
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_6
|
Consider an isosceles triangle. Let \(r\) be the radius of its circumscribed circle and \(\rho\) the radius of its inscribed circle. Prove that the distance \(d\) between the centers of these two circles is
\[
d=\sqrt{r(r-2\rho)}
\]
.
|
[
"<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> Instead of an isosceles triangle, let us consider an arbitrary triangle \\(ABC\\). Let \\(ABC\\) have circumcenter \\(O\\) and incenter \\(I\\). Extend \\(AI\\) to meet the circumcircle again at \\(L\\). Then extend \\(LO\\) so it meets the circumcircle again at \\(M\\). Consider the point where the incircle meets \\(AB\\), and let this be point \\(D\\). We have \\(\\angle ADI = \\angle MBL = 90^{\\circ}, \\angle IAD = \\angle LMB\\); thus, \\(\\triangle ADI \\sim \\triangle MBL\\), or \\(\\frac {ID}{BL} = \\frac {AI} {ML} \\iff ID \\cdot ML = 2r\\rho = AI \\cdot BL\\). Now, drawing line \\(BI\\), we see that \\(\\angle BIL = \\frac {1}{2}\\angle A + \\frac {1}{2}\\angle ABC, \\angle IBL = \\frac {1}{2}\\angle ABC + \\angle CBL = \\frac {1}{2}\\angle ABC + \\frac {1}{2}\\angle A\\). Therefore, \\(BIL\\) is isosceles, and \\(IL = BL\\). Substituting this back in, we have \\(2r\\rho = AI\\cdot IL\\). Extending \\(OI\\) to meet the circumcircle at \\(P,Q\\), we see that \\(AI\\cdot IL = PI\\cdot QI\\) by Power of a Point. Therefore, \\(2r\\rho = PI \\cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)\\), and we have \\(2r\\rho = r^2 - d^2 \\iff d = \\sqrt {r(r - 2\\rho)}\\), and we are done."
] |
IMO-1962-7
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_7
|
The tetrahedron \(SABC\) has the following property: there exist five spheres, each tangent to the edges \(SA, SB, SC, BC, CA, AB\), or to their extensions.
(a) Prove that the tetrahedron \(SABC\) is regular.
(b) Prove conversely that for every regular tetrahedron five such spheres exist.
|
[
"\\[\nIMO 1962 P7 01.png\n\\]\n\nPart (a)\n\nLet points \\(P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}\\) be the points where the smallest sphere is tangent to the edges \\(SA, SB, SC, BC, CA, AB\\) respectively.\n\nFor each face of the tetrahedron there is a circular cross section of the smallest sphere. Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.\n\nFrom the properties of an incircle we know that:\n\n\\(|AP_{AB}|=|AP_{AC}|\\), \\(|BP_{BC}|=|BP_{BA}|\\), and \\(|CP_{CA}|=|CP_{CB}|\\) in \\(\\Delta ABC\\)\n\nThere is a larger circle that is tangent to \\(AB\\) and the extensions of \\(CA\\) and \\(CB\\).\n\nThis larger circle is a cross section of the sphere that's also tangent to the extension of \\(CS\\) away from \\(S\\) and the edges of \\(\\Delta ABC\\)\n\nTherefore, this larger circle and the incircle of \\(\\Delta ABC\\) are part of that same sphere.\n\nIn order for these two circles to be part of the same sphere and also tangent to line \\(AB\\), then the point of the tangent of this larger circle needs to be the same as point \\(P_{AB}\\)\n\nThe only way these to circles can share the same tangent point on edge \\(AB\\) is if \\(|AP_{AB}|=|BP_{AB}|\\)\n\nUsing the same argument with the larder circle of edge \\(BC\\) then \\(|BP_{BC}|=|CP_{BC}|\\)\n\nand with the larger circle of edge \\(AC\\) then \\(|CP_{AC}|=|AP_{AC}|\\)\n\nThis results in:\n\n\\(|AP_{AB}|=|AP_{AC}|=|BP_{BC}|=|BP_{BA}|=|CP_{CA}|=|CP_{CB}|\\) in \\(\\Delta ABC\\)\n\nwhich means that\n\n\\(|AB|=|BC|=|AC|\\) in \\(\\Delta ABC\\), thus \\(\\Delta ABC\\) is an equilteral triangle.\n\nLikewise,\n\n\\(|AB|=|BS|=|AS|\\) in \\(\\Delta ABS\\), thus \\(\\Delta ABS\\) is an equilteral triangle.\n\n\\(|AS|=|SC|=|AC|\\) in \\(\\Delta ASC\\), thus \\(\\Delta ASC\\) is an equilteral triangle.\n\n\\(|SB|=|BC|=|SC|\\) in \\(\\Delta SBC\\), thus \\(\\Delta SBC\\) is an equilteral triangle.\n\nSince all four faces are equilateral triangles, then tetrahedron \\(ABCS\\) is a regular tetrahedron.\n\nPart (b)\n\nLet's consider a regular tetrahedron in the cartesian space with center at \\((0,0,0)\\) with \\(r_c\\) as the circumradius.\n\nWe can write the coordinates for the four vertices as:\n\n\\[\nV_1=\\left( \\frac{2\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nV_2=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)\n\\]\n\n\\[\nV_3=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;-\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)\n\\]\n\n\\[\nV_4=\\left(0 ,\\;0,\\;r_c\\right)\n\\]\n\nAnd the midpoints of all edges as:\n\n\\[\nM_{12}=\\frac{V_1-V_2}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{13}=\\frac{V_1-V_3}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{23}=\\frac{V_2-V_3}{2}=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{14}=\\frac{V_1-V_4}{2}=\\left( \\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{24}=\\frac{V_2-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{34}=\\frac{V_3-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)\n\\]\n\nNow we calculate the following six dot products:\n\n\\[\n(V_1-V_2) \\bullet M_{12}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_1-V_3) \\bullet M_{13}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_2-V_3) \\bullet M_{23}=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0\n\\]\n\n\\[\n(V_1-V_4) \\bullet M_{13}=\\left( 2\\frac{\\sqrt{2}}{3}r_c \\right)\\left( \\frac{\\sqrt{2}}{3}r_c \\right)+0+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{4}{9}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\n\\[\n(V_2-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( \\frac{\\sqrt{6}}{3}r_c \\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\n\\[\n(V_3-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( -\\frac{\\sqrt{6}}{3}r_c \\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\nSince all these dot producs equal to \\(0\\) that means that the lines from the center to each of the midpoints are all perpendicular.\n\nNow we calculate the distances from \\((0,0,0)\\) to all the midpoints:\n\n\\[\n|M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{14}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{24}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{34}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\nSince all distances are all the same and all dot products are \\(0\\), then we have our first sphere at \\((0,0,0)\\) with a radius of \\(\\frac{\\sqrt{3}}{3}r_c\\)\n\nNow we will look at the other 4 spheres.\n\nLet \\(E_{ij}\\) be the point of tangent of the larger sphere on the extension of line \\(V{i}V{j}\\) in the direction of \\(V{i}\\) to \\(V{j}\\) and beyond \\(V{j}\\) for \\(i=1,2,3,4\\); \\(j=1,2,3,4\\); and \\(i \\ne j\\)\n\nSince \\(|V_j E_{ij}|=|V_j M_{ij}|=\\frac{1}{2} |V_j V_i|\\), then \\(\\frac{1}{2}(V_j-V_i)=E_{ij}-V_j\\), thus \\(E_{ij}=\\frac{3V_j-V_i}{2}\\)\n\nUsing this formula we calculate the following:\n\n\\[\nE_{41}=\\left( \\sqrt{2} r_c ,\\;0,\\;-r_c \\right)\n\\]\n\n\\[\nE_{42}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)\n\\]\n\n\\[\nE_{43}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;-\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)\n\\]\n\nWe will start with the sphere below the base of the tetrahedron opposite of vertex \\(V_{4}\\) below \\(\\Delta V{1}V{2}V{3}\\)\n\nThe center of this larger sphere is at \\(C_1=(0,0,-2r_c)\\) and it is tangent at points \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\)\n\nWe calculate the following dot products:\n\n\\[\n(V_1-V_2) \\bullet (M_{12}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_1-V_3) \\bullet (M_{13}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_2-V_3) \\bullet (M_{23}-C_1)=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0\n\\]\n\n\\[\n(E_{41}-V_4) \\bullet (E_{41}-C_1) = \\left( \\sqrt{2}r_c \\right)^2+0+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\n\\[\n(E_{42}-V_4) \\bullet (E_{42}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( \\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\n\\[\n(E_{43}-V_4) \\bullet (E_{43}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( -\\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\nSince all these dot producs equal to \\(0\\) that means that the lines from the center \\(C_1\\) to each of the point \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\) are all perpendicular.\n\nNow we calculate the distances from \\(C_1=(0,0,-2r_c)\\) to points \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\)\n\n\\[\n|C_1M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{41}|=\\sqrt{\\left( \\sqrt{2}r_c\\right)^2+\\left(0\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{42}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{43}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\nSince all distances are all the same and all dot products are \\(0\\), then we have one of the larger spheres at \\((0,0,-2r_c)\\) with a radius of \\(\\sqrt{3}r_c\\)\n\nThen, the other three larger spheres which are the same size as the sphere with center at \\(C_1\\) are congruent and tangent to their respective sides near the other faces of the tetrahedron.\n\nand this proves that a tetrahedron with any circumradius \\(r_c\\) will have these 5 spheres, one with radius \\(\\frac{\\sqrt{3}}{3}r_c\\), at the center of the tetrahedron and the other 4 with radius \\(\\sqrt{3}r_c\\) at centers that are at a distance of \\(3r_c\\) away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.\n\nThus these five spheres exist for any regular tetrahedron.\n\n~Tomas Diaz. [email protected]"
] |
IMO-1963-1
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_1
|
Find all real roots of the equation
\[
\sqrt{x^2-p}+2\sqrt{x^2-1}=x
\]
,
where \(p\) is a real parameter.
|
[
"Assuming \\(x \\geq 0\\), square the equation, obtaining\n\n\\[\n4\\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2\n\\]\n\n. If we have \\(p + 4 \\geq 4x^2\\), we can square again, obtaining\n\n\\[\nx^2 = \\frac {(p - 4)^2}{4(4 - 2p)} \\implies x = \\pm\\frac {p - 4}{2\\sqrt {4 - 2p}}\n\\]\n\nWe must have \\(4 - 2p > 0 \\iff p < 2\\), so we have\n\n\\[\nx = \\frac {4 - p}{2\\sqrt {4 - 2p}}\n\\]\n\nHowever, this is only a solution when\n\n\\[\np + 4 \\geq 4x^2 = \\frac {(p - 4)^2}{4 - 2p} \\iff (p + 4)(4 - 2p)\\geq(p - 4)^2 \\iff 0\\geq p(3p - 4)\n\\]\n\nso we have \\(p\\geq 0\\) and \\(p \\leq \\frac {4}{3}\\)\n\nand \\(x = \\frac {4 - p}{2\\sqrt {4 - 2p}}\\)"
] |
IMO-1963-2
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_2
|
Point \(A\) and segment \(BC\) are given. Determine the locus of points in space which are the vertices of right angles with one side passing through \(A\), and the other side intersecting the segment \(BC\).
|
[
"Let \\(\\omega_1\\) be the circle with diameter \\(AB\\), and let \\(\\omega_2\\) be the circle with diameter \\(AC\\). Then the locus is simply the set of points inside either \\(\\omega_1\\) or \\(\\omega_2\\), but not both.\n\nTo see this, suppose the right angle's ray that does not pass through \\(A\\) intersects segment \\(BC\\) at \\(X\\). Then the right angle's vertex must lie on the circle with diameter \\(AX\\). So, for a particular \\(X\\), the desired locus is a circle with diameter \\(AX\\). Accounting for all possible \\(X\\), the total locus is the union of the circumferences of all circles that have a diameter \\(AX\\), where \\(X\\) is some point on \\(BC\\).\n\nAs \\(X\\) moves from \\(B\\) to \\(C\\), the motion of the circle with diameter \\(AX\\) is continuous and fluid. Any point \\(P\\) lying within \\(\\omega_1\\) but outside \\(\\omega_2\\) will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside \\(\\omega_2\\) but outside \\(\\omega_1\\). Also, the points inside both \\(\\omega_1\\) and \\(\\omega_2\\) are never intersected by this moving circle, as it always stays inside.\n\n(This proof sucks and needs some formalism)"
] |
IMO-1963-3
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_3
|
In an \(n\)-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
\[
a_1\ge a_2\ge \cdots \ge a_n.
\]
Prove that \(a_1=a_2=\cdots = a_n\).
|
[
"Let \\(a_1 = p_1p_2\\), \\(a_2 = p_2p_3\\), etc.\n\nPlot the \\(n\\)-gon on the cartesian plane such that \\(p_1p_2\\) is on the \\(x\\)-axis and the entire shape is above the \\(x\\)-axis. There are two cases: the number of sides is even, and the number of sides is odd:\n\n\\[\n\\textbf{Case 1: Even}\n\\]\n\nIn this case, the side with the topmost points will be \\(p_{\\frac{n}{2}+1}p_{\\frac{n}{2}+2}\\). To obtain the \\(y\\)-coordinate of this top side, we can multiply the lengths of the sides \\(a_1\\), \\(a_2\\), ... \\(a_{\\frac{n}{2}}\\) by the sine of the angle they make with the \\(x\\)-axis:\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = 1}^{\\frac{n}{2}}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (1)}\n\\]\n\nWe can obtain the \\(y\\)-coordinate of the top side in a different way by multiplying the lengths of the sides \\(a_{\\frac{n}{2}+1}\\), \\(a_{\\frac{n}{2}+2}\\), ... \\(a_n\\) by the sine of the angle they make with the \\(x\\)-axis to get the \\(\\emph{negated}\\) \\(y\\)-coordinate of the top side:\n\n\\[\n-y\\textrm{-coordinate} = \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot -\\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-\\frac{n}{2}-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = 1}^{\\frac{n}{2}}a_{k+\\frac{n}{2}} \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (2)}\n\\]\n\nIt must be true that \\(\\textbf{(1)} = \\textbf{(2)}\\). This implies that \\(a_k = a_{k+\\frac{n}{2}}\\) for all \\(1 \\leq k \\leq \\frac{n}{2}\\), and therefore \\(a_1=a_2=\\cdots = a_n\\).\n\n\\[\n\\textbf{Case 2: Odd}\n\\]\n\nThis case is very similar to before. We will compute the \\(y\\)-coordinate of the top point \\(p_{frac{n+3}{2}}\\) two ways:\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = 2}^{\\frac{n+1}{2}}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (3)}\n\\]\n\n\\[\n-y\\textrm{-coordinate} = \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot -\\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot \\sin \\frac{2\\pi(n - k + 1)}{n}\n\\]\n\n\\[\n= \\sum_{k = 2}^{\\frac{n+1}{2}}a_{n-k+2} \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (4)}\n\\]\n\nIt must be true that \\(\\textbf{(3)} = \\textbf{(4)}\\). Then, we get \\(a_k = a_{n-k+2}\\) for all \\(2 \\leq k \\leq \\frac{n+1}{2}\\). Therefore, \\(a_2=a_3=\\cdots = a_n\\). It is trivial that \\(a_1\\) is then equal to the other values, so \\(a_1=a_2=\\cdots = a_n\\). This completes the proof. \\(\\square\\)\n\n~mathboy100",
"Define the vector \\(\\vec{v_i}\\) to equal \\(\\cos{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{i}+\\sin{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{j}\\). Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length \\(a_i\\) is parallel to \\(\\vec{v_i}\\). We then have that\n\n\\[\n\\sum_{i=1}^{n} a_i\\vec{v_i}=\\vec{0}\\Rightarrow \\sum_{i=1}^{n} a_i\\cos{\\left(\\frac{2\\pi}{n}i\\right)} = \\sum_{i=1}^{n} a_i\\sin{\\left(\\frac{2\\pi}{n}i\\right)} =0\n\\]\n\nBut \\(a_i\\geq a_{n-i}\\) for all \\(i\\leq \\lfloor \\frac{n}{2}\\rfloor\\), so\n\n\\[\na_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)} = -a_i\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)} \\geq -a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)}\n\\]\n\nfor all \\(i\\leq \\lfloor \\frac{n}{2}\\rfloor\\). This shows that \\(a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}+a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)}\\geq 0\\), with equality when \\(a_i=a_{n-i}\\). Therefore\n\n\\[\n\\sum_{i=1}^{n} a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}=\\sum_{i=1}^{\\lfloor \\frac{n}{2}\\rfloor} a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}+a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)} \\geq 0\n\\]\n\nThere is equality only when \\(a_i=a_{n-i}\\) for all \\(i\\). This implies that \\(a_1=a_{n-1}\\) and \\(a_2=a_n\\), so we have that \\(a_1=a_2=\\cdots =a_n\\). \\(\\blacksquare\\)"
] |
IMO-1963-4
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_4
|
Find all solutions \(x_1,x_2,x_3,x_4,x_5\) of the system
\[
\begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*}
\]
where \(y\) is a parameter.
|
[
"Notice: The following words are Chinese.\n\n首先,我们可以将以上5个方程相加,得到:\n\n\\[\n2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)\n\\]\n\n当\\(x_1+x_2+x_3+x_4+x_5=0\\)时,因为\\(x_1,x_2,x_3,x_4,x_5\\)关于原方程组轮换对称,所以\n\n\\[\nx_1=x_2=x_3=x_4=x_5=0\n\\]\n\n若反之,则方程两边同除以\\((x_1+x_2+x_3+x_4+x_5)\\),得到\\(y=2\\),显然解为\n\n\\[\nx_1=x_2=x_3=x_4=x_5\n\\]\n\n综上所述,若\\(y=2\\),最终答案为\\(x_1=x_2=x_3=x_4=x_5\\),否则答案为\\(x_1=x_2=x_3=x_4=x_5=0\\)\n\nThe solution in English (translated by Google Translate):\n\nFirst of all, we can add the five equations to get:\n\n\\[\n2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)\n\\]\n\nWhen \\(x_1+x_2+x_3+x_4+x_5=0\\), Because \\(x_1,x_2,x_3,x_4,x_5\\) is symmetric in the original equations,\n\n\\[\nx_1=x_2=x_3=x_4=x_5=0\n\\]\n\nOtherwise, dividing both sides by \\((x_1+x_2+x_3+x_4+x_5\\), we get \\(y=2\\), and clearly\n\n\\[\nx_1=x_2=x_3=x_4=x_5\n\\]\n\nSummarizing, if \\(y=2\\), then the answer is of the form \\(x_1=x_2=x_3=x_4=x_5\\). Otherwise, \\(x_1=x_2=x_3=x_4=x_5=0\\).\n\n## Mistake\n\nWhile doing this question, I found out that the answer is actually wrong, \\(y\\) can equal \\(\\frac{-1-\\sqrt{5}}{2}\\) and \\(\\frac{-1+\\sqrt{5}}{2}\\) and still produce an infinite number of solutions in the form \\((n,n,-\\frac{ny}{y+1},-2ny,-\\frac{ny}{y+1})\\) where \\(n\\) is a real number and the set is cyclic (Ex: The set can correspond to \\((x_{1},x_{2},x_{3},x_{4},x_{5})\\) or \\((x_{2},x_{3},x_{4},x_{5},x_{1})\\), either works. Order matters, but not starting position.). For example, if \\(n=1\\) and \\(y=\\frac{-1+\\sqrt{5}}{2}\\) the set will be \\((1,1,\\frac{\\sqrt{5}-3}{2},1-\\sqrt{5},\\frac{\\sqrt{5}-3}{2})\\), which you can test and find out that it still works even though the set isn't symmetric.\n\nCan someone change this answer so it's correct?\n\nEdit: 亲爱的中国盆友，我找到错误了。 If \\(x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0\\), y can be anything(不一定要轮换对称)."
] |
IMO-1963-5
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_5
|
Prove that \(\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}\).
|
[
"Because the sum of the \\(x\\)-coordinates of the seventh roots of unity is \\(0\\), we have\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\frac{8\\pi}{7}} + \\cos{\\frac{10\\pi}{7}} + \\cos{\\frac{12\\pi}{7}} + \\cos{\\frac{14\\pi}{7}} = 0\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\frac{8\\pi}{7}} + \\cos{\\frac{10\\pi}{7}} + \\cos{\\frac{12\\pi}{7}} = -1\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\left(-\\frac{6\\pi}{7}\\right)} + \\cos{\\left(-\\frac{4\\pi}{7}\\right)} + \\cos{\\left(-\\frac{2\\pi}{7}\\right)} = -1.\n\\]\n\nNow, we can apply \\(\\cos{x} = \\cos{\\left(-x\\right)}\\) to obtain\n\n\\[\n2\\left(\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}}\\right) = -1\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} = -\\frac{1}{2}.\n\\]\n\nFinally, since \\(\\cos{x} = -\\cos{(\\pi-x)}\\),\n\n\\[\n\\cos{\\frac{2\\pi}{7}} - \\cos{\\frac{3\\pi}{7}} - \\cos{\\frac{\\pi}{7}} = -\\frac{1}{2}\n\\]\n\n\\[\n\\cos{\\frac{\\pi}{7}} - \\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{3\\pi}{7}} = \\frac{1}{2}\\textrm{. }\\square\n\\]\n\n~mathboy100",
"Let \\(\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=S\\). We have\n\n\\[\nS=\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=\\cos{\\frac{\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}+\\cos{\\frac{5\\pi}{7}}\n\\]\n\nThen, by product-sum formulae, we have\n\n\\[\nS \\cdot 2 \\sin{\\frac{\\pi}{7}} = \\sin{\\frac{2\\pi}{7}}+\\sin{\\frac{4\\pi}{7}}-\\sin{\\frac{2\\pi}{7}}+\\sin{\\frac{6\\pi}{7}}-\\sin{\\frac{4\\pi}{7}}=\\sin{\\frac{6\\pi}{7}}=\\sin{\\frac{\\pi}{7}}\n\\]\n\nThus \\(S = 1/2\\). \\(\\blacksquare\\)",
"Let \\(a=\\sin{\\frac{\\pi}{7}}\\) and \\(b=\\cos{\\frac{\\pi}{7}}\\). From the addition formulae, we have\n\n\\[\nS=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)\n\\]\n\nFrom the Trigonometric Identity, \\(a^2=1-b^2\\), so\n\n\\[\nS=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1\n\\]\n\nWe must prove that \\(S=1/2\\). It suffices to show that \\(8b^3-4b^2-4b+1=0\\).\n\nNow note that \\(\\cos{\\frac{4\\pi}{7}}=-\\cos{\\frac{3\\pi}{7}}\\). We can find these in terms of \\(a\\) and \\(b\\):\n\n\\[\n\\cos{\\frac{4\\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1\n\\]\n\n\\[\n\\cos{\\frac{3\\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3\n\\]\n\nTherefore \\(8b^4-8b^2+1=-(3b-4b^3)\\Rightarrow 8b^4+4b^3-8b^2-3b+1=0\\). Note that this can be factored:\n\n\\[\n8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0\n\\]\n\nClearly \\(b\\neq -1\\), so \\(8b^3-4b^2-4b+1=0\\). This proves the result. \\(\\blacksquare\\)",
"Let \\(\\omega=\\mathrm{cis}\\left(\\frac{\\pi}{14}\\right)\\). Thus it suffices to show that \\(\\omega+\\omega^{-1}-\\omega^2-\\omega^{-2}+\\omega^3+\\omega^{-3}=1\\). Now using the fact that \\(\\omega^k=\\omega^{14+k}\\) and \\(-\\omega^2=\\omega^9\\), this is equivalent to\n\n\\[\n\\omega+\\omega^3+\\omega^5+\\omega^7+\\omega^9+\\omega^{11}+\\omega^{13}-\\omega^7\n\\]\n\n\\[\n\\omega\\left(\\frac{\\omega^{14}-1}{\\omega^2-1}\\right)-\\omega^7\n\\]\n\nBut since \\(\\omega\\) is a \\(14\\)th root of unity, \\(\\omega^{14}=1\\). The answer is then \\(-\\omega^{7}=1\\), as desired.\n\n~yofro",
"We let \\(\\omega = \\mathrm{cis} \\ \\left(\\dfrac{2\\pi}{7} \\right)\\). We therefore have \\(w^i\\), where \\(0 \\leq i \\leq 6\\), are the \\(7^{\\text{th}}\\) roots of unity. Since \\(\\sum_{i = 0}^6 \\omega^i = 0\\), then \\(\\sum_{i = 1}^6 \\omega^i = -1\\), so \\(\\sum_{i = 1}^3 \\mathrm{Re}(\\omega^i) = -\\dfrac{1}{2}\\). Therefore, because \\(\\mathrm{cis} \\ \\alpha = \\cos \\alpha + i \\sin \\alpha, \\mathrm{Re}(\\mathrm{cis} \\ \\alpha) = \\cos \\alpha\\), so\n\n\\[\n\\cos \\left(\\dfrac{2\\pi}{7} \\right) + \\cos \\left( \\dfrac{4\\pi}{7} \\right) + \\cos \\left(\\dfrac{6\\pi}{7} \\right) = -\\dfrac{1}{2}\n\\]\n\n\\[\n\\implies -\\cos \\left(\\dfrac{2\\pi}{7} \\right) - \\cos \\left(\\dfrac{4\\pi}{7} \\right) - \\cos \\left(\\dfrac{6\\pi}{7} \\right) = \\dfrac{1}{2}\n\\]\n\nSince \\(\\cos \\alpha = - \\cos(\\pi - \\alpha)\\), we have \\(\\cos \\left(\\dfrac{\\pi}{7} \\right) - \\cos \\left(\\dfrac{2\\pi}{7} \\right) + \\cos \\left(\\dfrac{3\\pi}{7} \\right) = \\dfrac{1}{2}\\) and we are done \\(\\blacksquare\\)\n\n~Yiyj1"
] |
IMO-1963-6
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_6
|
Five students, \(A,B,C,D,E\), took part in a contest. One prediction was that the contestants would finish in the order \(ABCDE\). This prediction was very poor. In fact no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order \(DAECB\). This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.
|
[
"We are given that no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so in order \\(ABCDE\\). None of them finished in that order. Also only two of them had their actual positions in \\(DAECB\\). After imposing these two conditions the list of possible outcomes is: (1)\\(CAEBD\\), (2)\\(DCAEB\\), (3)\\(DCEBA\\), (4)\\(EDACB\\). One more condition is that two disjoint pairs of students predicted to finish consecutively actually did so. Out of the above four in the list, (1) and (2) have \\(AE\\) as the correctly predicted consecutive finishers(but only 1 pair), (3) has no correctly predicted consecutive finishers. But (4) has 2 disjoint correctly predicted consecutive finishers who are \\(DA\\) and \\(CB\\). Hence, order is \\(EDACB\\)."
] |
IMO-1964-1
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_1
|
(a) Find all positive integers \(n\) for which \(2^n-1\) is divisible by \(7\).
(b) Prove that there is no positive integer \(n\) for which \(2^n+1\) is divisible by \(7\).
|
[
"We claim \\(2^n\\) is equivalent to \\(2, 4,\\) and \\(1\\) \\(\\pmod{7}\\) for \\(n\\) congruent to \\(1\\), \\(2\\), and \\(0\\) \\(\\pmod{3}\\), respectively.\n\n(a) From the statement above, only \\(n\\) divisible by \\(3\\) will work.\n\n(b) Again from the statement above, \\(2^n\\) can never be congruent to \\(-1\\) \\(\\pmod{7}\\), so there are no solutions for \\(n\\).",
"This solution is clearer and easier to understand.\n\n(1) Since we know that \\(2^n-1\\) is congruent to 0 (mod 7), we know that \\(2^n\\) is congruent to 8 mod 7, which means \\(2^n\\) is congruent to 1 mod 7.\n\nExperimenting with the residue of \\(2^n\\) mod 7:\n\n\\(n\\)=1: 2\n\n\\(n\\)=2: 4\n\n\\(n\\)=3: 1 (this is because when \\(2^n\\) is doubled to \\(2*2^n\\), the residue doubles too, but \\(4*2=8\\) is congruent to 1 (mod 7).\n\n\\(n\\)=4: 2\n\n\\(n\\)=5: 4\n\n\\(n\\)=6: 1\n\nThrough induction, we easily show that this is true since the residue doubles every time you double \\(2^n\\).\n\nSo, the residue of \\(2^n\\) mod 7 cycles in 2, 4, 1. Therefore, \\(n\\) must be a multiple of 3. Proved.\n\n(2) According to part (1), the residue of \\(2^n\\) cycles in 2, 4, 1.\n\nIf \\(2^n+1\\) is congruent to 0 mod 7, then \\(2^n\\) must be congruent to 6 mod 7, but this is not possible due to how \\(2^n\\) mod 7 cycles. Therefore, there is no solution. Proved.\n\n~hastapasta"
] |
IMO-1964-2
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_2
|
Suppose \(a, b, c\) are the sides of a triangle. Prove that
\[
a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.
\]
|
[
"Let \\(b+c-a = x\\), \\(c+a-b = y\\), and \\(a+b-c = z\\). Then, \\(a = \\frac{y+z}{2}\\), \\(b = \\frac{x+z}{2}\\), and \\(c = \\frac{x+y}{2}\\). By AM-GM,\n\n\\[\n\\frac{x+y}{2} \\geq \\sqrt{xy},\n\\]\n\n\\[\n\\frac{y+z}{2} \\geq \\sqrt{yz},\n\\]\n\n\\[\n\\textrm{and }\\frac{x+z}{2} \\geq \\sqrt{xz}.\n\\]\n\nMultiplying these equations, we have\n\n\\[\n\\frac{x+y}{2} \\cdot \\frac{y+z}{2} \\cdot \\frac{x+z}{2} \\geq xyz\n\\]\n\n\\[\n\\therefore abc \\geq (a+b-c)(b+c-a)(c+a-b).\n\\]\n\nWe can now simplify:\n\n\\[\n(a+b-c)(b+c-a)(c+a-b) \\leq abc\n\\]\n\n\\[\n(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \\leq abc\n\\]\n\n\\[\na(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \\leq abc\n\\]\n\n\\[\n-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \\leq abc\n\\]\n\n\\[\na^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \\leq abc\n\\]\n\n\\[\na^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\\le{3abc}\\textrm{. }\\square\n\\]\n\n~mathboy100",
"We can use the substitution \\(a=x+y\\), \\(b=x+z\\), and \\(c=y+z\\) to get\n\n\\[\n2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\\leq 3(x+y)(x+z)(y+z)\n\\]\n\n\\[\n2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz\n\\]\n\n\\[\nx^2y+x^2z+y^2x+y^2z+z^2x+z^2y\\geq 6xyz\n\\]\n\n\\[\n\\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\\geq xyz=\\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}\n\\]\n\nThis is true by AM-GM. We can work backwards to get that the original inequality is true.",
"Rearrange to get\n\n\\[\na(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \\ge 0,\n\\]\n\nwhich is true by Schur's inequality."
] |
IMO-1964-3
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_3
|
A circle is inscribed in a triangle \(ABC\) with sides \(a,b,c\). Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from \(\triangle ABC\). In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of \(a,b,c\)).
|
[
"Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!\n\n\\[\n[BHM]/[BCA] = \\text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2\n\\]\n\nDenote \\([ABC]\\) the area of \\(\\triangle ABC\\) and \\((ABC)\\) the perimeter of \\(\\triangle ABC\\).\n\nThen \\(\\frac{[BHM]}{[ABC]} = \\left(\\frac{(BHM)}{(ABC)}\\right)^{2} =\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}\\).\n\nSo \\([BHM]=\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}\\cdot [ABC]\\).\n\nWe know, \\(r_{1}\\) is the radius of the incircle of \\(\\triangle BHM\\): \\(r_{1}= 2 \\cdot \\frac{[BHM]}{(BHM)}\\).\n\nArea of the incircle of \\(\\triangle BHM\\)\n\n\\[\n= \\pi \\cdot 4 \\cdot (\\frac{[BHM]}{(BHM)})^{2}= 4\\pi \\frac{(c+a-b)^{2}}{(c+a+b)^{4}} \\cdot ([ABC])^{2}\n\\]\n\nArea of the incircle of \\(\\triangle ABC\\):\\(4\\pi (\\frac{[ABC]}{a+b+c})^{2}\\).\n\nSum of the area of the 4 incircles:\n\n\\[\n4 \\pi ([ABC])^{2}\\left[\\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}}{(a+b+c)^{4}}\\right]+4\\pi (\\frac{[ABC]}{a+b+c})^{2}\n\\]\n\n\\[\n=4 \\pi \\frac{([ABC])^{2}}{(a+b+c)^{2}}\\left[\\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}+(a+b+c)^{2}}{(a+b+c)^{2}}\\right]\n\\]\n\n\\[\n=16 \\pi \\frac{([ABC])^{2}(a^{2}+b^{2}+c^{2})}{(a+b+c)^{4}}\n\\]"
] |
IMO-1964-4
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_4
|
Seventeen people correspond by mail with one another - each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three people who write to each other about the same topic.
|
[
"Lemma: Consider a complete graph with 6 vertices colored with 2 colors. There exists a monochromatic triangle.\n\nProof: Consider one vertex and all connections leading out from it. Call it \\(V_1\\). It has 5 edges coming out from it. By the Pigeonhole Principle, there are at least 3 of the same color. Call this color red. Call those vertices \\(V_2\\), \\(V_3\\) and \\(V_4\\). If any of the segments \\(V_2V_3\\), \\(V_2V_4\\), or \\(V_3V_4\\) are red, then we have a monochromatic triangle with vertices \\(V_1\\) and the other two that are also red. If they are all the other color, then we have a monochromatic triangle with vertices \\(V_2\\),\\(V_3\\), and \\(V_4\\). \\(\\blacksquare\\)\n\nMain Problem: Represent these people as vertices on a connected graph with 17 vertices and colored with 3 colors, one corresponding with each topic. So this problem is reduced to showing that on a connected graph with 17 vertices and colored with three colors, there exists some monochromatic triangle. Look at an arbitrary vertex. Call it \\(V_1\\). Look at the 16 other vertices that it is connected to. By the Pigeonhole Principle, there are at least 6 vertices connected to \\(V_1\\) that are all one color. Call this color 1. If any of the connections inbetween these six vertices are in color 1, then we are done. If none of them are color 1, we know that that there are only 2 colors in those 6 vertices. By Lemma 1, we know that there is a monochromatic triangle in those 6 vertices. So we are done."
] |
IMO-1964-5
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_5
|
Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have.
|
[
"Suppose, those five points are \\((A, B, C, D, E)\\). Now, we want to create some special structure. Let, we take the line \\(BC\\) and draw a perpendicular from \\(A\\) on \\(BC\\), andd call it \\(P_1\\). We can do this set up in \\(\\binom{5}{1}\\binom{4}{2}=30\\) ways. There will \\(30\\) such \\(P_i\\) s.\n\nNow, we will find how many other perpendiculars intersect the line. We can do this in total \\(20\\) ways. Why? See, can draw perpendiculars from \\(B\\) and \\(C\\) to other lines( we haven't counted the perpendicular from \\(B\\) to \\(AC\\) and perpendicular from \\(C\\) on \\(AB\\) , as they intersect \\(P_1\\) at the same point) in \\(5\\) ways for each. So, total \\(10\\) ways.\n\nNow, \\(5\\) perpendiculars from each \\(D\\) and \\(E\\) on the other lines except on \\(BC\\)( because in this case teh perpendiculars from \\(D\\) and \\(E\\) will be parallel to \\(P_1\\) , and so shall not intersect). So,total \\(10\\) cases. From, these two cases we get \\(P_1\\) will be intersected at at most \\(5*6*(5+5+5+5)=600\\) points.\n\nBut, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total \\(\\frac{600}{2}=300\\) ways.\n\nNow, as we had excluded the orthocentres, we have to add now. There are total \\(\\binom{5}{3}=10\\) orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are \\(5\\) such.\n\nSo, total ways \\(300+10+5=315\\)."
] |
IMO-1964-6
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_6
|
In tetrahedron \(ABCD\), vertex \(D\) is connected with \(D_0\), the centroid of \(\triangle ABC\). Lines parallel to \(DD_0\) are drawn through \(A,B\) and \(C\). These lines intersect the planes \(BCD, CAD\) and \(ABD\) in points \(A_1, B_1,\) and \(C_1\), respectively. Prove that the volume of \(ABCD\) is one third the volume of \(A_1B_1C_1D_0\). Is the result true if point \(D_o\) is selected anywhere within \(\triangle ABC\)?
|
[
"\\[\nIMO 1964 P6 01.png\n\\]\n\nLet \\(A_{2}\\) be the point where line \\(AD_{0}\\) intersects line \\(BC\\)\n\nLet \\(B_{2}\\) be the point where line \\(BD_{0}\\) intersects line \\(AC\\)\n\nLet \\(C_{2}\\) be the point where line \\(CD_{0}\\) intersects line \\(AB\\)\n\nFrom centroid properties we have:\n\n\\[\n|AA_{2}|=3|D_{0}A_{2}|\n\\]\n\n\\[\n|BB_{2}|=3|D_{0}B_{2}|\n\\]\n\n\\[\n|CC_{2}|=3|D_{0}C_{2}|\n\\]\n\nTherefore,\n\n\\[\n\\frac{|AA_{2}|}{|D_{0}A_{2}|}=3\n\\]\n\n\\[\n\\frac{|BB_{2}|}{|D_{0}B_{2}|}=3\n\\]\n\n\\[\n\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3\n\\]\n\nSince \\(\\Delta D_{0}A_{2}A_{1}\\sim \\Delta AA_{2}A_{1}\\), then \\(|AA_{1}|=|DD_{0}| \\frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|\\)\n\nSince \\(\\Delta D_{0}B_{2}B_{1}\\sim \\Delta BB_{2}B_{1}\\), then \\(|BB_{1}|=|DD_{0}| \\frac{|BB_{2}|}{|D_{0}B_{2}|}=3|DD_{0}|\\)\n\nSince \\(\\Delta D_{0}C_{2}C_{1}\\sim \\Delta CC_{2}C_{1}\\), then \\(|CC_{1}|=|DD_{0}| \\frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|\\)\n\nSince \\(|AA_{2}|=|BB_{2}|=|CC_{2}|\\) and \\(AA_{1} \\parallel BB_{1} \\parallel CC_{1} \\parallel DD_{0}\\),\n\nthen \\(\\Delta A_{1}B_{1}C_{1}\\parallel \\Delta ABC\\), and \\(Area_{\\Delta A_{1}B_{1}C_{1}}=Area_{\\Delta ABC}\\)\n\nLet \\(h_{D}\\) be the perpendicular distance from \\(D\\) to \\(\\Delta ABC\\)\n\nLet \\(h_{\\Delta A_{1}B_{1}C_{1}}\\) be the perpendicular distance from \\(\\Delta A_{1}B_{1}C_{1}\\) to \\(\\Delta ABC\\)\n\n\\[\n\\frac{h_{\\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\\frac{|AA_{2}|}{|D_{0}A_{2}|}=\\frac{|BB_{2}|}{|D_{0}B_{2}|}=\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3\n\\]\n\n\\[\nh_{\\Delta A_{1}B_{1}C_{1}}=3h_{D}\n\\]\n\n\\[\n\\frac{1}{3}h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}=3\\frac{1}{3}h_{D}Area_{\\Delta ABC}\n\\]\n\n\\[\n\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}=3\\frac{h_{D}Area_{\\Delta ABC}}{3}\n\\]\n\nSince \\(Volume_{ABCD}=\\frac{h_{D}Area_{\\Delta ABC}}{3}\\) and \\(Volume_{A_{1}B_{1}C_{1}D_{0}}=\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}\\)\n\nthen, \\(Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}\\)\n\nthus, \\(Volume_{ABCD}=\\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}\\)\n\nthis proves that the volume of \\(ABCD\\) is one third the volume of \\(A_1B_1C_1D_0\\)\n\nThe result is NOT true if point \\(D_o\\) is selected anywhere within \\(\\triangle ABC\\) as ratios of \\(\\frac{|AA_{2}|}{|D_{0}A_{2}|}\\), \\(\\frac{|BB_{2}|}{|D_{0}B_{2}|}\\), and \\(\\frac{|CC_{2}|}{|D_{0}C_{2}|}\\) will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios \\(\\frac{|AA_{2}|}{|D_{0}A_{2}|} \\ne \\frac{|BB_{2}|}{|D_{0}B_{2}|}\\ne \\frac{|CC_{2}|}{|D_{0}C_{2}|}\\) which means that \\(\\Delta A_{1}B_{1}C_{1} \\nparallel \\Delta ABC\\) and the volume relationship will no longer hold true.\n\n~Tomas Diaz. [email protected]"
] |
IMO-1965-1
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_1
|
Determine all values \(x\) in the interval \(0\leq x\leq 2\pi\) which satisfy the inequality
\[
2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.
\]
|
[
"We shall deal with the left side of the inequality first (\\(2\\cos x \\leq \\left| \\sqrt{1+\\sin 2x} - \\sqrt{1-\\sin 2x } \\right|\\)) and the right side after that.\n\nIt is clear that the left inequality is true when \\(\\cos x\\) is non-positive, and that is when \\(x\\) is in the interval \\([\\pi/2, 3\\pi/2]\\). We shall now consider when \\(\\cos x\\) is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. \\(4\\cos^2{x}\\leq 1+\\sin 2x+1-\\sin 2x-2\\sqrt{1-\\sin^2 2x}=2-2\\sqrt{\\cos^2{2x}}\\). This inequality is equivalent to \\(2\\cos^2 x\\leq 1-\\left| \\cos 2x\\right|\\). I shall now divide this problem into cases.\n\nCase 1: \\(\\cos 2x\\) is non-negative. This means that \\(x\\) is in one of the intervals \\([0,\\pi/4]\\) or \\([7\\pi/4, 2\\pi]\\). We must find all \\(x\\) in these two intervals such that \\(2\\cos^2 x\\leq 1-\\cos 2x\\). This inequality is equivalent to \\(2\\cos^2 x\\leq 2\\sin^2 x\\), which is only true when \\(x=\\pi/4\\) or \\(7\\pi/4\\).\n\nCase 2: \\(\\cos 2x\\) is negative. This means that \\(x\\) is in one of the interavals \\((\\pi/4, \\pi/2)\\) or \\((3\\pi/2, 7\\pi/4)\\). We must find all \\(x\\) in these two intervals such that \\(2\\cos^2 x\\leq 1+\\cos 2x\\), which is equivalent to \\(2\\cos^2 x\\leq 2\\cos^2 x\\), which is true for all \\(x\\) in these intervals.\n\nTherefore the left inequality is true when \\(x\\) is in the union of the intervals \\([\\pi/4, \\pi/2)\\), \\((3\\pi/2, 7\\pi/4]\\), and \\([\\pi/2, 3\\pi/2]\\), which is the interval \\([\\pi/4, 7\\pi/4]\\). We shall now deal with the right inequality.\n\nAs above, we can square it and have it be true whenever the original right inequality is true, so we do that. \\(2-2\\sqrt{\\cos^2{2x}}\\leq 2\\), which is always true. Therefore the original right inequality is always satisfied, and all \\(x\\) such that the original inequality is satisfied are in the interval \\([\\pi/4, 7\\pi/4]\\).",
"Manipulate the inequality so that it becomes:\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\sqrt{\\frac{1+\\cos(\\frac{\\pi}{2}-2x)}{2}} - \\sqrt{\\frac{1-\\cos(\\frac{\\pi}{2}- 2x)}{2}} \\right| \\leq 1\n\\]\n\nInside the absolute value, we identify the half-angle formulas. However, since we do not know the sign of the resultant expression, we have to use absolute value signs since the principal square root is always positive; then the inequality becomes\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\left|\\cos\\left(\\frac{\\pi}{4}-x\\right)\\right| - \\left|\\sin\\left(\\frac{\\pi}{4}-x\\right)\\right|\\right| \\leq 1\n\\]\n\nwhich, considering absolute value, is the same as\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\right| - \\left|\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\right| \\leq 1\n\\]\n\nSince both inner absolute values range between \\(0\\) and \\(1\\), their positive difference also ranges from \\(0\\) to \\(1\\), so the right inequality is always true. Thus we take a look at the left inequality. For \\(x\\in\\left[\\frac{\\pi}{2},\\frac{3\\pi}{2}\\right]\\), the left-hand side is never positive; however, the right-hand side is always nonnegative due to absolute value, so the inequality holds for these \\(x\\)-values. As a result, we consider the remaining two sections of the interval.\n\nFor \\(x\\in\\left[0,\\frac{\\pi}{4}\\right]\\) and \\(x\\in\\left[\\frac{7\\pi}{4},2\\pi\\right]\\), when considering the \\(-\\frac{\\pi}{4}\\), inside the absolute value, cosine is positive but sine is negative; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2} +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2}\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\sin\\left(\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\sin\\left(\\left(x-\\frac{\\pi}{4}\\right)+\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le |\\sin x|\\\\ \\end{align*} If \\(x\\in\\left[0,\\frac{\\pi}{4}\\right]\\), then due to the interval, \\(\\sin x\\) and \\(\\cos x\\) are both never negative; thus, \\begin{align*} \\cos x &\\le \\sin x\\\\ 1&\\le\\tan x\\\\ \\end{align*} But this is never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.\n\nIf \\(x\\in\\left[\\frac{7\\pi}{4},2\\pi\\right]\\), then due to the interval, \\(\\sin x\\) is negative and \\(\\cos x\\) is positive; thus, \\begin{align*} \\cos x &\\le -\\sin x\\\\ 1&\\le-\\tan x\\\\ -1&\\ge\\tan x\\\\ \\end{align*} But this is also never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.\n\nFor \\(x\\in\\left[\\frac{\\pi}{4},\\frac{\\pi}{2}\\right]\\), inside the absolute value, when considering the \\(-\\frac{\\pi}{4}\\), both cosine and sine are positive; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2} -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2}\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cos\\left(\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\sin\\left(\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(\\left(x-\\frac{\\pi}{4}\\right)+\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le |\\cos x|\\\\ \\end{align*} This is always true, so the inequality holds.\n\nFor \\(x\\in\\left[\\frac{3\\pi}{2},\\frac{7\\pi}{4}\\right]\\), inside the absolute value, when considering the \\(-\\frac{\\pi}{4}\\), both cosine and sine are negative; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|-\\cos\\left(x-\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\sqrt{2}\\cos x &\\le \\left|-\\left(\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right)\\right|\\\\ \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\end{align*} This is the same as the case for \\(x\\in\\left[\\frac{\\pi}{4},\\frac{\\pi}{2}\\right]\\), so the inequality holds.\n\nCombining our findings, we find that the solutions are \\(\\boxed{x\\in\\left[\\frac{\\pi}{4},\\frac{7\\pi}{4}\\right]}\\).\n\n~eevee9406"
] |
IMO-1965-2
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_2
|
Consider the system of equations
\[
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0
\]
\[
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0
\]
\[
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0
\]
with unknowns \(x_1\), \(x_2\), \(x_3\). The coefficients satisfy the conditions:
(a) \(a_{11}\), \(a_{22}\), \(a_{33}\) are positive numbers;
(b) the remaining coefficients are negative numbers;
(c) in each equation, the sum of the coefficients is positive.
Prove that the given system has only the solution \(x_1 = x_2 = x_3 = 0\).
|
[
"Clearly if the \\(x_i\\) are all equal, then they are equal to 0. Now let's assume WLOG that \\(x_1=0\\). If \\(x_2\\) or \\(x_3\\) is 0, then the other is clearly zero, so let's consider the case where neither are 0. \\(a_{12}\\) and \\(a_{21}\\) are negative, so exactly one of \\(x_2\\) or \\(x_3\\) is positive. Unfortunately this means that one of \\(a_{22}x_2 + a_{23}x_3\\) or \\(a_{32}x_2 + a_{33}x_3 = 0\\) is positive and the other is negative, so the equation couldn't possibly be satisfied if \\(x_2\\) or \\(x_3\\) isn't 0. We have covered the case where one of the \\(x_i\\) is 0, now let's assume that none of them are 0.\n\nIf two are positive and one is negative, then when the negative \\(x_i\\) is paired with one of the positive \\(a_i\\), the corresponding equation is negative. This is bad. If two are negative and one is positive, then when the positive \\(x_i\\) is paired with one of the positive \\(a_i\\), the corresponding equation is positive. This is also bad. Therefore the \\(x_i\\) all have the same sign.\n\nCase 1: The \\(x_i\\) are all positive. WLOG \\(x_1\\leq x_2\\leq x_3\\). Now consider the third equation, \\(a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0\\). Therefore \\(x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0\\), but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.\n\nCase 2: The \\(x_i\\) are all negative. WLOG \\(x_1\\geq x_2\\geq x_3\\). Consider the third equation, \\(a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0\\). Therefore \\(x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0\\), but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.\n\nTherefore at least one of the \\(x_i\\) is 0, which implies all of them are 0.",
"We will prove that the matrix\n\n\\[\na_{11}\\ \\ \\ \\ a_{12}\\ \\ \\ \\ a_{13}\n\\]\n\n\\[\na_{21}\\ \\ \\ \\ a_{22}\\ \\ \\ \\ a_{23}\n\\]\n\n\\[\na_{31}\\ \\ \\ \\ a_{32}\\ \\ \\ \\ a_{33}\n\\]\n\nhas its determinant \\(> 0.\\)\n\nThe determinant is\n\n\\[\na_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{32}a_{21} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{33}a_{12}a_{21}.\n\\]\n\nAfter a little algebraic manipulation we can rewrite this as\n\n\\[\n[a_{33}(a_{11} + a_{13}) - a_{13}(a_{31} + a_{33})](a_{21} + a_{22} + a_{23}) - (a_{21}a_{33} - a_{23}a_{31})(a_{11} + a_{12} + a_{13}) - (a_{11}a_{23} - a_{13}a_{21})(a_{31} + a_{32} + a_{33})\n\\]\n\n(or we can just verify that this is true).\n\nNote that \\(a_{11} + a_{12} + a_{13} > 0\\) implies \\(a_{11} + a_{12} > 0\\) and \\(a_{11} + a_{13} > 0.\\) The expression above is clearly \\(> 0\\). To show this in a simple way, I will just write out the sign of each factor:\n\n\\[\n[(+)(+) - (-)(+)]\\ (+) - ((-)(+) - (-)(-))\\ (+) - ((+)(-) - (-)(-))\\ (+)\n\\]\n\nso now we can see that the end expression is \\((+).\\)\n\n[Solution by pf02, November 2024]"
] |
IMO-1965-3
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_3
|
Given the tetrahedron \(ABCD\) whose edges \(AB\) and \(CD\) have lengths \(a\) and \(b\) respectively. The distance between the skew lines \(AB\) and \(CD\) is \(d\), and the angle between them is \(\omega\). Tetrahedron \(ABCD\) is divided into two solids by plane \(\varepsilon\), parallel to lines \(AB\) and \(CD\). The ratio of the distances of \(\varepsilon\) from \(AB\) and \(CD\) is equal to \(k\). Compute the ratio of the volumes of the two solids obtained.
|
[
"Let the plane meet \\(AD\\) at \\(X\\), \\(BD\\) at \\(Y\\), \\(BC\\) at \\(Z\\) and \\(AC\\) at \\(W\\). Take a plane parallel to \\(BCD\\) through \\(WX\\) and let it meet \\(AB\\) in \\(P\\).\n\n\\[\nProb 1965 3.png\n\\]\n\nSince the distance of \\(AB\\) from \\(WXYZ\\) is \\(k\\) times the distance of \\(CD\\), we have that \\(AX = k \\cdot XD\\) and hence \\(AX/AD = k/(k+1).\\) Similarly \\(AP/AB = AW/AC = AX/AD.\\) \\(XY\\) is parallel to \\(AB\\), so also \\(AX/AD = BY/BD = BZ/BC.\\)\n\nvol \\(ABWXYZ =\\) vol \\(APWX +\\) vol \\(WXPBYZ.\\) \\(APWX\\) is similar to the tetrahedron \\(ABCD.\\) The sides are \\(k/(k + 1)\\) times smaller, so vol \\(APWX = k^3/(k + 1)^3\\) vol \\(ABCD.\\)\n\nThe base of the prism \\(WXPBYZ\\) is \\(BYZ\\) which is similar to \\(BCD\\) with sides \\(k/(k + 1)\\) times smaller and hence area \\(k^2/(k + 1)^2\\) times smaller. Its height is \\(1/(k + 1)\\) times the height of \\(A\\) above \\(ABCD,\\) so vol prism \\(= 3 k^2/(k + 1)^3\\) vol \\(ABCD.\\)\n\nThus vol \\(ABWXYZ = (k^3 + 3k^2)/(k + 1)^3\\) vol \\(ABCD.\\)\n\nWe get the volume of the other piece as vol \\(ABCD\\ -\\) vol \\(ABWXYZ,\\) and hence the ratio is (after a little manipulation) \\(k^2(k + 3)/(3k + 1).\\)\n\n## Remark (added by pf02, November 2024)\n\nNote that the problem is untypically sloppy or misleading. It mentions the sizes \\(a, b, d, \\omega\\) as if they are needed. In fact, as the solution above shows, they are not needed either in expressing the result, or in solving the problem. A thorough problem solver might worry about not having used all the data in the problem.\n\nHowever, one can imagine other solutions, where these quantities would be used in the process of solving the problem. For example one could break up the doubly truncated triangular prism \\(ABWXYZ\\) into pyramids \\(APWX, PXYZW, BPZY\\). Computing the volume of each of these pyramids would require all the data in the problem. The end result should of course be the same, but the thorough problem solver would not have the uneasy feeling of not having used all the data in the problem."
] |
IMO-1965-4
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_4
|
Find all sets of four real numbers \(x_1\), \(x_2\), \(x_3\), \(x_4\) such that the sum of any one and the product of the other three is equal to \(2\).
|
[
"Let \\(P = x_1x_2x_3x_4\\) be the product of the four real numbers.\n\nThen, for \\(i = 1,2,3,4\\) we have: \\(x_i + \\prod_{j \\neq i}x_j = 2\\).\n\nMultiplying by \\(x_i\\) yields:\n\n\\(x^2_i + P = 2x_i \\Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \\Longleftrightarrow x_i = 1 \\pm t\\) where \\(t = \\pm \\sqrt{1-P} \\in \\mathbb{R}\\).\n\nIf \\(t=0\\), then we have \\((x_1,x_2,x_3,x_4)=(1,1,1,1)\\) which is a solution.\n\nSo assume that \\(t \\neq 0\\). WLOG, let at least two of \\(x_i\\) equal \\(1+t\\), and \\(x_1 \\ge x_2 \\ge x_3 \\ge x_4\\) OR \\(x_1 \\le x_2 \\le x_3 \\le x_4\\).\n\nCase I: \\(x_1 = x_2 = x_3 = x_4 = 1+t\\)\n\nThen we have:\n\n\\[\n(1+t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+4t = 0 \\Longleftrightarrow t(t^2+3t+4) = 0\n\\]\n\nWhich has no non-zero solutions for \\(t\\).\n\nCase II: \\(x_1 = x_2 = x_3 = 1+t\\) AND \\(x_4 = 1-t\\)\n\nThen we have:\n\n\\((1-t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+2t = 0\\) \\(\\Longleftrightarrow t(t+1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,-1,-2\\}\\)\n\nAND\n\n\\((1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0\\) \\(\\Longleftrightarrow -t(t-1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,1,-2\\}\\)\n\nSo, we have \\(t = -2\\) as the only non-zero solution, and thus, \\((x_1,x_2,x_3,x_4) = (-1,-1,-1,3)\\) and all permutations are solutions.\n\nCase III: \\(x_1 = x_2 = 1+t\\) AND \\(x_3 = x_4 = 1-t\\)\n\nThen we have:\n\n\\((1-t)+(1-t)(1+t)^2 = 2 \\Longleftrightarrow -t^3-t^2 = 0\\) \\(\\Longleftrightarrow -t^2(t+1) = 0 \\Longleftrightarrow t \\in \\{0,-1\\}\\)\n\nAND\n\n\\((1+t)+(1+t)(1-t)^2 = 2 \\Longleftrightarrow t^3-t^2 = 0\\) \\(\\Longleftrightarrow t^2(t-1) = 0 \\Longleftrightarrow t \\in \\{0,1\\}\\)\n\nThus, there are no non-zero solutions for \\(t\\) in this case.\n\nTherefore, the solutions are: \\((1,1,1,1)\\); \\((3,-1,-1,-1)\\); \\((-1,3,-1,-1)\\); \\((-1,-1,3,-1)\\); \\((-1,-1,-1,3)\\).",
"We have to solve the system of equations\n\n\\[\nx_1 + x_2x_3x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1) \\\\ x_2 + x_1x_3x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (2) \\\\ x_3 + x_1x_2x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (3) \\\\ x_4 + x_1x_2x_3 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (4)\n\\]\n\nSubtract (2) from (1) and factor. We get\n\n\\((x_1 - x_2)(1 - x_3x_4) = 0\\),\n\nwhich implies \\(x_1 - x_2 = 0\\) or \\(1 - x_3x_4 = 0\\).\n\nSimilarly, subtracting (3) and then (4) from (1) and factoring, we get\n\n\\[\n(x_1 - x_3)(1 - x_2x_4) = 0 \\\\ (x_1 - x_4)(1 - x_2x_3) = 0\n\\]\n\nThey imply \\(x_1 - x_3 = 0\\) or \\(1 - x_2x_4 = 0\\), and \\(x_1 - x_4 = 0\\) or \\(1 - x_2x_3 = 0\\).\n\nWe will consider four possibilities:\n\n1. \\(x_2 = x_1, x_3 = x_1, x_4 = x_1\\)\n\n2. \\(x_2 = x_1, x_3 = x_1\\) and \\(x_2x_3 = 1\\)\n\n3. \\(x_2 = x_1\\) and \\(x_2x_3 = 1, x_2x_4 = 1\\)\n\n4. \\(x_2x_3 = 1, x_2x_4 = 1, x_3x_4 = 1\\)\n\nNote that in fact, there are four more possibilities, but they just correspond to permutations in \\(x_1, x_2, x_3, x_4\\) of cases 2. and 3., so there is no harm in not dealing with them explicitly.\n\nCase 1. Plug \\(x_2, x_3, x_4\\) in equation (1). We get \\(x_1 + x_1^3 = 2\\). This is an equation of degree \\(3\\) whose only real root is \\(x_1 = 1\\). We get the solution \\(x_1 = x_2 = x_3 = x_4 = 1\\).\n\nCase 2. Plug \\(x_2, x_3\\) into \\(x_2x_3 = 1\\), and get \\(x_1^2 = 1\\). We get \\(x_1 = 1\\) or \\(x_1 = -1\\). The first solution, \\(x_1 = 1\\), yields \\(x_2 = x_3 = 1\\), and using (4), \\(x_4 = 1\\). The second solution, \\(x_1 = -1\\), yields \\(x_2 = x_3 = -1\\), and using (4), \\(x_4 = 3\\). We get the solution \\(x_1 = x_2 = x_3 = -1, x_4 = 3.\\) Because of the permutations of \\(x_1, x_2, x_3, x_4\\), we also get the solutions \\((-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)\\).\n\nCase 3. Plug in \\(x_2x_3, x_2x_4\\) in (4) and (3), and get \\(x_4 + x_1 = 2\\) and \\(x_3 + x_1 = 2\\). Now plug \\(x_2, x_3, x_4\\) in (1), and get \\(x_1 + x_1(2 - x_1)^2 = 2\\). This equation becomes \\((x_1 - 1)^2(x_1 - 2) = 0\\). \\(x_1 = 1\\) yields \\((1, 1, 1, 1)\\) which we already know, and \\(x_1 = 2\\) yields \\((2, 2, 0, 0)\\) (or some other values, depending on where we plug \\(x_1 = 2\\)), which do not give a possible solution.\n\nCase 4. plugging \\(x_2x_3, x_2x_4, x_3x_4\\) into (4), (3), (2), we get \\(x_4 + x_1 = 2, x_3 + x_1 = 2, x_2 + x_1 = 2\\). Plugging \\(x_2, x_3, x_4\\) into (1), we get \\(x_1 + (2 - x_1)^3 = 2\\). The solutions of this equation are \\(1, 2, 3\\). Note that \\(x_1 = 1\\) and \\(x_1 = 3\\) yield solutions we already know, and \\(x_1 = 2\\) is impossible.\n\nThus, the five solutions to the problem are \\((1, 1, 1, 1), (3, -1, -1, -1), (-1, 3, -1, -1), (-1, -1, 3, -1), (-1, -1, -1, 3)\\).\n\n[Solution by pf02, November 2024]"
] |
IMO-1965-5
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_5
|
Consider \(\triangle OAB\) with acute angle \(AOB\). Through a point \(M \neq O\) perpendiculars are drawn to \(OA\) and \(OB\), the feet of which are \(P\) and \(Q\) respectively. The point of intersection of the altitudes of \(\triangle OPQ\) is \(H\). What is the locus of \(H\) if \(M\) is permitted to range over (a) the side \(AB\), (b) the interior of \(\triangle OAB\)?
|
[
"Let \\(O(0,0),A(a,0),B(b,c)\\). Equation of the line \\(AB: y=\\frac{c}{b-a}(x-a)\\). Point \\(M \\in AB : M(\\lambda,\\frac{c}{b-a}(\\lambda-a))\\). Easy, point \\(P(\\lambda,0)\\). Point \\(Q = OB \\cap MQ\\), \\(MQ \\bot OB\\). Equation of \\(OB : y=\\frac{c}{b}x\\), equation of \\(MQ : y=-\\frac{b}{c}(x-\\lambda)+\\frac{c}{b-a}(\\lambda-a)\\). Solving: \\(x_{Q}=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right]\\). Equation of the first altitude: \\(x=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right] \\quad (1)\\). Equation of the second altitude: \\(y=-\\frac{b}{c}(x-\\lambda)\\quad\\quad (2)\\). Eliminating \\(\\lambda\\) from (1) and (2):\n\n\\[\nac \\cdot x + (b^{2}+c^{2}-ab)y=abc\n\\]\n\na line segment \\(MN , M \\in OA , N \\in OB\\). Second question: the locus consists in the \\(\\triangle OMN\\).",
"This solution is a simplified version of the previous solution, it fills in some gaps. and it provides more information.\n\nJust like in the previous solution, we use analytic (coordinate) geometry, but we don't care how the axes are chosen.\n\nLet \\(A, B, O\\) have coordinates \\((a_1, a_2), (b_1, b_2), (c_1, c_2)\\), and let \\(M = (\\lambda a_1 + (1 - \\lambda b_1, \\lambda a_2 + (1 - \\lambda b_2)\\) with \\(\\lambda \\in [0, 1]\\). The idea is to just follow the degrees of the expressions and equations in \\(\\lambda, x, y\\) involved as we make the computations for obtaining the coordinates of \\(H\\), and the equation of the curve \\(H\\) is on. We will see that the equation for \\(H\\) is an equation of degree \\(1\\), so we will know that it is a line. We don't need to write out the equation explicitly.\n\nThe coordinates of \\(M\\) are expressions of degree \\(1\\) in \\(\\lambda\\).\n\nThe equation for \\(MP\\) (the perpendicular from \\(M\\) to \\(OA\\)) is an equation of degree \\(1\\) in \\(x, y\\) with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\).\n\nThe coordinates of \\(P\\) (the foot of the perpendicular from \\(P\\) to \\(OA\\)) are expressions of degree \\(1\\) in \\(\\lambda\\).\n\nThe equation of the perpendicular from \\(P\\) to \\(OB\\) is of degree \\(1\\) in \\(x, y\\), with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\). This corresponds to equation (2) in the above solution.\n\nSimilarly, the equation of the perpendicular from \\(Q\\) to \\(OA\\) is of degree \\(1\\) in \\(x, y\\), with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\). This corresponds to equation (1) in the above solution.\n\nNow, in principle, we would have to solve the system of two equations (1) and (2) to obtain the coordinates of \\(H\\) as expressions of \\(\\lambda\\), and then eliminate \\(\\lambda\\) to obtain the equation in \\(x, y\\) for \\(H\\). As a shortcut, we can eliminate \\(\\lambda\\) directly from the two equations (1) and (2). Either way, the result is an equation of degree \\(1\\) in \\(x, y\\).\n\nThis tells us that the locus is on a line. We just need to specify which set of points on this line is the locus. And, we want to make the line explicit.\n\nThe previous solution, with a good amount of hand waving, tells us that the solution is \"a line segment \\(B_1A_1, B_1 \\in OA, A_1 \\in OB\\)\". (On top of the hand waving the solution uses the unhappy notation \\(M\\) for \\(B_1\\) and \\(N\\) for \\(A_1\\), which is bad because \\(M\\) has already been used!) We will do better than that.\n\nLet \\(A_1\\) be the foot of the perpendicular from \\(A\\) to \\(OB\\), and \\(B_1\\) be the foot of the perpendicular from \\(B\\) to \\(OA\\). (For this paragraph see the picture shown in Solution 3.) Consider the limit situation when \\(M = A\\). Then \\(Q = A_1\\), and \\(P = A\\). It follows that the intersection \\(H\\) of the perpendiculars from \\(P\\) to \\(OB\\) and \\(Q\\) to \\(OA\\) is \\(A_1\\). Similarly, the limit situation when \\(M = B\\) yields \\(H = B_1\\). Now it is reasonable to say that when \\(M\\) moves from \\(A\\) to \\(B\\), \\(H\\) moves from \\(A_1\\) to \\(B_1\\). So, the locus is the line segment joining the feet \\(A_1, B_1\\) of the perpendiculars in \\(\\triangle OAB\\) from \\(A, B\\). This answers question (a).\n\nFor part (b) of the problem, with a good amount of hand waving, the previous solution says \"the locus consists in the \\(\\triangle OB_1A_1\\)\". We justify this by pointing out that if \\(M\\) is inside \\(\\triangle OAB\\), then we can take the triangle \\(\\triangle OA'B'\\), such that \\(A' \\in OA\\), \\(B' \\in OB\\), \\(A'B'\\) going through \\(M\\) and parallel to \\(AB\\). Then \\(H\\) will be on the corresponding segment \\(A_1'B_1'\\) determined by the feet of the perpendiculars in \\(\\triangle OA'B'\\). Conversely, it is easy to see that any point \\(H \\in \\triangle OA_1B_1\\) is on a segment \\(A_1'B_1'\\) obtained from a triangle \\(\\triangle OA'B'\\), and \\(H\\) is obtained from a point \\(M \\in A'B'\\). This answers question (b).\n\n[Solution by pf02, October 2024]",
"This solution is elementary, it does not use analytic geometry.\n\nLet \\(A_1\\) be the foot of the perpendicular from \\(A\\) to \\(OB\\), and \\(B_1\\) be the foot of the perpendicular from \\(B\\) to \\(OA\\). Construct \\(H\\) as described in the statement of the problem.\n\n\\[\nProb 1965 5.png\n\\]\n\nWe will prove that \\(H \\in A_1B_1\\).\n\nFirst, as shown in the second picture, take \\(MQ \\perp OB\\), and let \\(H_1\\) be the intersection of \\(A_1B_1\\) with the perpendicular from \\(Q\\) to \\(OA\\). From the triangle \\(\\triangle BAA_1\\) we have \\(\\frac{AM}{MB} = \\frac{A_1Q}{QB}\\) because \\(AA_1 \\parallel MQ\\). From the triangle \\(\\triangle A_1BB_1\\) we have \\(\\frac{A_1Q}{QB} = \\frac{A_1H_1}{H_1B_1}\\) because \\(QH_1 \\parallel BB_1\\). So, \\(\\frac{AM}{MB} = \\frac{A_1H_1}{H_1B_1}\\).\n\nSecond, as shown in the third picture, take \\(MP \\perp OA\\), and let \\(H_2\\) be the intersection of \\(A_1B_1\\) with the perpendicular from \\(P\\) to \\(OB\\). By a similar argument, we have \\(\\frac{AM}{MB} = \\frac{A_1H_2}{H_2B_1}\\).\n\nIt follows that \\(H_1 = H_2\\) since the two points divide \\(A_1B_1\\) in the same ratio. This is the point \\(H\\) from the statement of the problem.\n\nThe solution to the problem is now completed by repeating the last two paragraphs from Solution 2.\n\n[Solution by pf02, October 2024]"
] |
IMO-1965-6
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_6
|
In a plane a set of \(n\) points (\(n\geq 3\)) is given. Each pair of points is connected by a segment. Let \(d\) be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length \(d\). Prove that the number of diameters of the given set is at most \(n\).
|
[
"Image of problem Solution. Credits to user awe-sum.\n\n## Remarks (added by pf02, October 2024)\n\n1. As a public service, I will upload the \"Image of problem Solution\" mentioned above to this web page. That way, a reader can see the \"Solution\" immediately, without having to go to another web site, and we are not subjected to the possibility of the imgur.com website being taken down, or Imgur's parent company deciding to delete this particular image. Credits for the image are due to user awe-sum, as pointed out above.\n\n2. This \"Solution\" is presented very badly, and edited very badly. Indeed, some terms are undefined, left to the reader to make sense of (e.g. \"incident\", \"degree\"). But let us be forgiving, and let us do our best to make sense of the \"Solution\".\n\n3. The \"Solution\" is incomplete, to the point of not being a solution at all. There are some serious gaps, which raise questions which are not addressed, and a reader can not be expected to fill in the details. These are:\n\na. The author says \"... at least one point is incident to at least three diagonals\". But, it could also happen that two points are \"incident\" to two \"diagonals\" (i.e. diameters) each. The author does not address this possibility at all.\n\nb. The author says \"consider all points at distance exactly d from point Q (the green circle). We see that all of them are outside the locus, the exception being P.\" This is far from obvious. It assumes that all the other k-4 points (those points of the k given points which are not highlighted in the picture) are inside the \"locus (the blue shaded region)\". In fact, it seems to this reader that this is not necessarily true.\n\nc. Another issue with this \"Solution\" is that it assumes \\(n \\ge 4\\), while the statement of the problem has \\(n \\ge 3\\). This shortcoming is easy to fix, unlike the previous two I mentioned.\n\n4. I will give two solutions below, in the section \"Solution 2\" and \"Section 3\".",
"\\[\nProblem 1965 6 sol by awe-sum.png\n\\]",
"\\(\\mathbf{Definition:}\\) For the purpose of this proof, let us define an \\(equilateral\\ arc\\ triangle\\) as the shape we obtain when we take a triangle \\(\\triangle ABC\\) in which we replace the side \\(AB\\) by the \\(\\pi/6\\) arc of the circle centered at \\(C\\) with radius \\(CA = CB\\), going from \\(A\\) to \\(B\\), and similarly for the other two sides \\(BC\\) and \\(AC\\).\n\nSee the picture below.\n\n\\[\nProb 1965 6 fig1.png\n\\]\n\nNote that if the sides of the original equilateral triangle were of length \\(d\\), then the distance from a vertex to any point on the opposite arc is \\(d\\). Also, the distance between any two points on the arcs, such that none of them are vertices is \\(< d\\). And finally, the distance from a vertex to any point on the arcs adjacent to it (which is not another vertex) is \\(< d\\).\n\n\\(\\mathbf{Lemma:}\\) A configuration of \\(n\\) points has exactly \\(n\\) diameters if an only if \\(3\\) of the points are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary of the triangle.\n\n\\[\nProb 1965 6 fig2.png\n\\]\n\nIn the configuration shown above, we have \\(9\\) points with \\(9\\) diameters.\n\n\\(\\mathbf{Proof\\ of\\ the\\ Lemma:}\\) It is clear that if we have a configuration as described in the lemma then there are \\(n\\) diameters. We will prove the converse by induction. We want to prove that if we have a configuration of \\(n\\) points with \\(n\\) diameters, then the points are in a configuration as described above.\n\nIf \\(n = 3\\) it is clear that the \\(3\\) points have to be the vertices of an equilateral triangle since the sides are equal (to \\(d\\)). Assume we know the statement to be true for \\(n\\), and prove it for \\(n + 1\\). Assume that we have \\(n + 1\\) points with \\(n + 1\\) diameters \\(d\\). Consider \\(n\\) of the points, and let \\(Q\\) be the remaining point. It follows from the induction hypothesis that \\(3\\) of the \\(n\\) are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary.\n\nNow let us examine where \\(Q\\) may be. It can not be outside of the equilateral arc triangle because then the distance to at least one of the vertices would be \\(d' > d\\), and the diameter of the set would be \\(d'\\) rather than \\(d\\). It can not be inside the triangle because then all the distances from \\(Q\\) to the \\(n\\) points considered are \\(< d\\), and we would not have \\(n + 1\\) diameters, as assumed. So \\(Q\\) must be on the boundary of the equilateral arc triangle, which proves the induction step, and the lemma.\n\nNow the solution to the problem is easy to finish. Assume we have \\(n\\) points given, and the configuration has \\(m\\) diameters. I claim that we can not have \\(m > n\\). We prove this by contradiction. If we did, let us consider a set \\(S\\) consisting of only \\(n\\) of the diameters. Because of the lemma, this implies that the \\(n\\) points form a configuration as described in the lemma, i.e. \\(3\\) of them are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary. Now consider a diameter \\(D\\) which is not in \\(S\\). It is the distance of a segment between two of the points \\(P_1, P_2\\) in the configuration. If one of \\(P_1, P_2\\) is a vertex, then this diameter \\(D\\) is in \\(S\\), which is a contradiction. If none of \\(P_1, P_2\\) are vertices, then the length of \\(P_1P_2 < d\\), so \\(D = P_1P_2\\) is not a diameter, which is again a contradiction.\n\nIt follows that \\(m \\le n\\) which proves the problem.\n\n[Solution by pf02, October 2024]",
"See Proposition 2.3 on page 19 in \"Lecture Notes: Combinatorics in the Plane\" by Torsten Ueckerdt, from March 12, 2015 at https://www.math.kit.edu/iag6/lehre/combplane2013s/media/lecture_notes.pdf ."
] |
IMO-1966-1
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_1
|
In a mathematical contest, three problems, \(A\), \(B\), and \(C\) were posed. Among the participants there were \(25\) students who solved at least one problem each. Of all the contestants who did not solve problem \(A\), the number who solved \(B\) was twice the number who solved \(C\). The number of students who solved only problem \(A\) was one more than the number of students who solved \(A\) and at least one other problem. Of all students who solved just one problem, half did not solve problem \(A\). How many students solved only problem \(B\)?
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[
"Let us draw a Venn Diagram.\n\n\\[\nIMO1966.png\n\\]\n\nLet \\(a\\) be the number of students solving both B and C. Then for some positive integer \\(x\\), \\(2x - a\\) students solved B only, and \\(x - a\\) students solved C only. Let \\(2y - 1\\) be the number of students solving A; then \\(y\\) is the number of students solving A only. We have by given\n\n\\[\n2y - 1 + 3x - a = 25\n\\]\n\nand\n\n\\[\ny = 3x - 2a.\n\\]\n\nSubstituting for y into the first equation gives\n\n\\[\n9x - 5a = 26.\n\\]\n\nThus, because \\(x\\) and \\(a\\) are positive integers with \\(x-a \\ge 0\\), we have \\(x = 4\\) and \\(a = 2\\). (Note that \\(x = 9\\) and \\(a = 11\\) does not work.) Hence, the number of students solving B only is \\(2x - a = 8 - 2 = \\boxed{6}.\\)"
] |
IMO-1966-2
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_2
|
Let \(a\), \(b\), and \(c\) be the lengths of the sides of a triangle, and \(\alpha,\beta,\gamma\) respectively, the angles opposite these sides. Prove that if
\[
a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}),
\]
the triangle is isosceles.
|
[
"We'll prove that the triangle is isosceles with \\(a=b\\). We'll prove that \\(a=b\\). Assume by way of contradiction WLOG that \\(a>b\\). First notice that as \\(\\gamma = \\pi -\\alpha-\\beta\\) then and the identity \\(\\tan\\left(\\frac \\pi 2 - x \\right)=\\cot x\\) our equation becomes:\n\n\\[\na+b=\\cot \\frac{\\alpha +\\beta}{2}\\left(a\\tan \\alpha + b\\tan \\beta \\right)\n\\]\n\n\\[\n\\iff a\\tan\\frac{\\alpha +\\beta}{2}+b\\tan \\frac{\\alpha +\\beta}{2}=a\\tan \\alpha + b\\tan \\beta\n\\]\n\n\\[\n\\iff a\\left(\\tan \\alpha -\\tan \\frac{\\alpha +\\beta}{2}\\right)+b\\left(\\tan \\beta -\\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\nUsing the identity \\(\\tan (A-B)=\\frac {\\tan A-\\tan B}{1+\\tan A\\tan B}\\) \\(\\iff \\tan A-\\tan B=\\tan(A-B)(1+\\tan A\\tan B)\\) and inserting this into the above equation we get:\n\n\\[\n\\iff a\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\alpha \\tan \\frac{\\alpha +\\beta}{2}\\right)+b\\tan \\frac{\\beta -\\alpha}{2}\\left(1+\\tan \\beta \\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\n\\[\n\\underbrace{\\iff}_{\\tan -A=-\\tan A}a\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\alpha \\tan \\frac{\\alpha +\\beta}{2}\\right)-b\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\beta \\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\n\\[\n\\iff \\tan \\frac{\\alpha -\\beta}{2}\\left(a-b+\\tan \\frac{\\alpha +\\beta}{2}(a\\tan\\alpha -b\\tan \\beta) \\right)=0\n\\]\n\nNow, since \\(a>b\\) and the definitions of \\(a,b,\\alpha,\\beta\\) being part of the definition of a triangle, \\(\\alpha >\\beta\\). Now, \\(\\pi >\\alpha -\\beta >0\\) (as \\(\\alpha+\\beta +\\gamma = \\pi\\) and the angles are positive), \\(\\tan \\frac{\\alpha -\\beta}{2}\\neq 0\\), and furthermore, \\(\\tan \\frac{\\alpha+\\beta}{2}>0\\). By all the above,\n\n\\[\n\\left(a-b+\\tan \\frac{\\alpha +\\beta}{2}(a\\tan\\alpha -b\\tan \\beta) \\right)>0\n\\]\n\nWhich contradicts our assumption, thus \\(a\\leq b\\). By the symmetry of the condition, using the same arguments, \\(a\\geq b\\). Hence \\(a=b\\).",
"First, we'll prove that both \\(\\alpha\\) and \\(\\beta\\) are acute. At least one of them has to be acute because these are angles of a triangle. We can assume that \\(\\alpha\\) is acute. We want to show that \\(\\beta\\) is acute as well. For a proof by contradiction, assume \\(\\beta \\ge \\frac{\\pi}{2}\\).\n\nFrom the hypothesis, it follows that \\((a + b) \\tan \\frac{\\alpha + \\beta}{2} = a \\tan \\alpha + b \\tan \\beta\\).\n\nFrom \\(\\alpha < \\frac{\\pi}{2} \\le \\beta\\) it follows that \\(a < b\\). So,\n\n\\[\nb \\tan \\beta = (a + b) \\tan \\frac{\\alpha + \\beta}{2} - a \\tan \\alpha > 2a \\tan \\frac{\\alpha + \\beta}{2} - a \\tan \\alpha \\ge a (2 \\tan \\left( \\frac{\\alpha}{2} + \\frac{\\pi}{4} \\right) - \\tan \\alpha) =\n\\]\n\n\\[\n2a \\left( \\frac{\\tan \\frac{\\alpha}{2} + 1}{1 - \\tan \\frac{\\alpha}{2}} - \\frac{\\tan \\frac{\\alpha}{2}}{1 - \\tan^2 \\frac{\\alpha}{2}} \\right) = 2a \\cdot \\frac{\\tan^2 \\frac{\\alpha}{2} + \\tan \\frac{\\alpha}{2} + 1} {1 - \\tan^2 \\frac{\\alpha}{2}} > 0\n\\]\n\nbecause the numerator is \\(> 0\\) (because \\(Y^2 + Y + 1 > 0\\) for any real \\(Y\\)), and the denominator is also \\(> 0\\) (because \\(\\alpha < \\frac{\\pi}{2}\\), so \\(\\tan \\frac{\\alpha}{2} < \\tan \\frac{\\pi}{4} = 1\\)).\n\nIt follows that \\(\\tan \\beta > 0\\), so it can not be that \\(\\beta \\ge \\frac{\\pi}{2}\\).\n\nNow, we will prove that \\((a + b) \\tan \\frac{\\alpha + \\beta}{2} = a \\tan \\alpha + b \\tan \\beta\\) implies \\(\\alpha = \\beta\\).\n\nReplace \\(a = \\sin \\alpha \\cdot 2R\\) and \\(b = \\sin \\beta \\cdot 2R\\) (in fact, we don't care that \\(R\\) is the radius of the circumscribed circle), and simplify by \\(2R\\). We get\n\n\\((\\sin \\alpha + \\sin \\beta) \\tan \\frac{\\alpha + \\beta}{2} = \\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta\\).\n\nThis becomes\n\n\\[\n\\left( \\sin \\frac{\\alpha + \\beta}{2} \\tan \\frac{\\alpha + \\beta}{2} \\right) \\cdot \\cos \\frac{\\alpha - \\beta}{2} = \\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta)\n\\]\n\nWe will show that the function \\(f(x) = \\tan x \\sin x\\) is convex on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\). Indeed, the first derivative is \\(f'(x) = \\frac{\\sin x}{\\cos^2 x} + \\sin x\\), and the second derivative is \\(f''(x) = \\frac{\\cos^4 x - \\cos ^2 x + 2}{\\cos^3 x}\\).\n\nWe have \\(f''(x) > 0\\) on \\(\\left( 0, \\frac{\\pi}{2} \\right)\\) since the numerator is \\(> 0\\) (because \\(Y^2 - Y + 1 > 0\\) for any real \\(Y\\)), and the denominator is \\(> 0\\) on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\). It follows that \\(f(x) = \\tan x \\sin x\\) is convex on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\).\n\nUsing the convexity we have \\(f \\left( \\frac{x + y}{2} \\right) \\le \\frac{1}{2} (f(x) + f(y))\\). In our case, we have\n\n\\(\\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta) = \\left( \\sin \\frac{\\alpha + \\beta}{2} \\tan \\frac{\\alpha + \\beta}{2} \\right) \\cdot \\cos \\frac{\\alpha - \\beta}{2} \\le \\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta) \\cdot \\cos \\frac{\\alpha - \\beta}{2}\\).\n\nWe can simplify by \\(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta\\) because it is positive (because both \\(\\alpha, \\beta\\) are acute!), and we get\n\n\\(1 \\le \\cos \\frac{\\alpha - \\beta}{2}\\). This is possible only when \\(\\cos \\frac{\\alpha - \\beta}{2} = 1\\), i.e. \\(\\alpha = \\beta\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-3
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_3
|
Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.
|
[
"We will need the following lemma to solve this problem:\n\n\\(\\emph{Lemma:}\\) Let \\(MNOP\\) be a regular tetrahedron, and \\(T\\) a point inside it. Let \\(x_1, x_2, x_3, x_4\\) be the distances from \\(T\\) to the faces \\(MNO, MNP, MOP\\), and \\(NOP\\). Then, \\(x_1 + x_2 + x_3 + x_4\\) is constant, independent of \\(T\\).\n\n\\[\n\\emph{Proof:}\n\\]\n\nWe will compute the volume of \\(MNOP\\) in terms of the areas of the faces and the distances from the point \\(T\\) to the faces:\n\n\\[\n\\textrm{Volume}(MNOP) = [MNO] \\cdot x_1 \\cdot \\frac{1}{3} + [MNP] \\cdot x_2 \\cdot \\frac{1}{3} + [MOP] \\cdot x_3 \\cdot \\frac{1}{3} + [NOP] \\cdot x_4 \\cdot \\frac{1}{3}\n\\]\n\n\\[\n= [MNO] \\cdot \\frac{(x_1 + x_2 + x_3 + x_4)}{3}\n\\]\n\nbecause the areas of the four triangles are equal. (\\([ABC]\\) stands for the area of \\(\\triangle ABC\\).) Then\n\n\\[\n\\frac{3\\cdot\\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.\n\\]\n\nThis value is constant, so the proof of the lemma is complete.\n\n\\[\n\\emph{Proof of problem statement:}\n\\]\n\nLet our tetrahedron be \\(ABCD\\), and the center of its circumscribed sphere be \\(O\\). Construct a new regular tetrahedron, \\(WXYZ\\), such that the centers of the faces of this tetrahedron are at \\(A\\), \\(B\\), \\(C\\), and \\(D\\).\n\nFor any point \\(P\\) in \\(ABCD\\),\n\n\\[\nOA + OB + OC + OD = \\sum \\textrm{Distances from }O\\textrm{ to faces of }WXYZ\n\\]\n\n\\[\n= \\sum \\textrm{Distances from }P\\textrm{ to faces of }WXYZ \\leq PA + PB + PC + PD,\n\\]\n\nwith equality only occurring when \\(AP\\), \\(BP\\), \\(CP\\), and \\(DP\\) are perpendicular to the faces of \\(WXYZ\\), meaning that \\(P = O\\). This completes the proof. \\(\\square\\)\n\n~mathboy100\n\n## Remarks (added by pf02, September 2024)\n\n1. The text of the Lemma needed a little improvement, which I did.\n\n2. The Solution above is not complete. It considered only points \\(P\\) inside the tetrahedron, but the problem specifically said \"any other point in space\".\n\n3. I will give another solution below, in which I will also fill in the gap of the solution above, mentioned in the preceding paragraph.",
"We will first prove the problem in the 2-dimensional case. We do this to convey the idea of the proof, and because we will use this in one spot in proving the 3-dimensional case. So let us prove that:\n\nThe sum of the distances of the vertices of an equilateral triangle \\(\\triangle ABC\\) from the center \\(O\\) of its circumscribed circle is less than the sum of the distances of these vertices from any other point \\(P\\) in the plane.\n\nWe will do the proof in three steps:\n\n\\(\\mathbf{1.}\\) We will show that if \\(P\\) is in one of the exterior regions, then there is a point \\(P_1\\) on the boundary of the triangle (a vertex, or on a side), such that \\(PA + PB + PC > P_1A + P_1B + P_1C\\).\n\n\\(\\mathbf{2.}\\) Then we will show that if \\(P\\) is on the boundary, then \\(PA + PB + PC > OA + OB + OC\\).\n\n\\(\\mathbf{3.}\\) For the final step, we will show that if \\(P\\) is a point of minimum for \\(PA + PB + PC\\) inside the triangle, then the extensions of \\(PA, PB, PC\\) are perpendicular to the opposite sides \\(BC, AC, AB\\). This implies that \\(P = O\\).\n\n\\(\\mathbf{Proof\\ of\\ 1:}\\) If the point \\(P\\) is outside the triangle, it can be in one of six regions as seen in the pictures below.\n\n\\[\nProb 1966 3 fig1.png\n\\]\n\nIf \\(P\\) is in a region delimited by extensions of two sides of the triangle, as in the picture on the left, we notice that by taking \\(P_1 = A\\), \\(PA + PB + PC > P_1A + P_1B + P_1C\\) (because \\(P_1A = 0\\) and \\(P_1B < PB\\) as sides in an obtuse triangles, and similarly \\(P_1C < PC\\)).\n\nIf \\(P\\) is in a region delimited by a segment which is a side of the triangle and by the extensions of two sides, as in the picture on the right, take \\(P_1 =\\) the foot of the perpendicular from \\(P\\) to \\(AB\\). Then \\(PA + PB + PC > P_1A + P_1B + P_1C\\) (because the triangle \\(\\triangle PP_1C\\) is obtuse, and because the triangles \\(\\triangle PP_1A, \\triangle PP_1B\\) are right triangles).\n\n\\(\\mathbf{Proof\\ of\\ 2:}\\) Now assume that \\(P_1 = A\\). A direct, simple computation shows that \\(P_1A + P_1B + P_1C > OA + OB + OC\\) (indeed, if we take the side of the triangle to be \\(1\\), then \\(P_1A + P_1B + P_1C = 2\\), and \\(OA + OB + OC = 3 \\cdot \\frac{\\sqrt{3}}{3} = \\sqrt{3}\\)).\n\nNow assume that \\(P_1\\) is on \\(AB\\). If \\(P_1\\) is not the midpoint of \\(AB\\), let \\(P_2\\) be the midpoint. Then \\(P_1A + P_1B + P_1C > P_2A + P_2B + P_2C\\) (because \\(P_1A + P_1B = P_2A + P_2B = AB\\) and \\(P_1C > P_2C\\)). A direct, simple computation shows that \\(P_2A + P_2B + P_2C > OA + OB + OC\\) (indeed, if we take the side of the triangle to be \\(1\\), \\(P_2A + P_2B + P_2C = 1 + \\frac{\\sqrt{3}}{2}\\) and \\(OA + OB + OC = \\sqrt{3}\\)).\n\n\\(\\mathbf{Proof\\ of\\ 3:}\\) Assume that \\(P\\) is inside the triangle \\(\\triangle ABC\\). In this case, we make a proof by contradiction. We will show that if \\(P\\) is a point where \\(PA + PB + PC\\) is minimum, then the extensions of \\(PA, PB, PC\\) are perpendicular to the opposite sides \\(BC, AC, AB\\). (This statement implies that \\(P = O\\).) If this were not true, at least one of \\(PA \\perp BC, PB \\perp AC, PC \\perp AB\\) would be false. We can assume that \\(PC\\) is not perpendicular to \\(AB\\). Then draw the ellipse with focal points \\(A, B\\) which goes through \\(P\\).\n\n\\[\nProb 1966 3 fig2.png\n\\]\n\nNow consider the point \\(P_1\\) on the ellipse such that \\(CP_1 \\perp AB\\). Because of the properties of the ellipse, \\(CP_1 < CP\\), and because of the definition of the ellipse \\(PA + PB = P_1A + P_1B\\). We conclude that \\(PA + PB + PC > P_1A + P_1B + P_1C\\), which contradicts the assumption that \\(P\\) was such that \\(PA + PB + PC\\) was minimum.\n\nThis proves the 2-dimensional case.\n\nNOTE: a very picky reader might object that the proof used that a minimum of \\(PA + PB + PC\\) exists, and is achieved at a point \\(P\\) inside the triangle. This can be justified simply by noting that \\(PA + PB + PC > 0\\) and quoting the theorem from calculus (or is it topology?) which says that a continuous function on a closed, bounded set has a minimum, and there is a point where the minimum is achieved. Because of the arguments in the proof, this point can not be on the boundary of the triangle, so it is inside.\n\nNow we will give the proof in the 3-dimensional case. We will do the proof in three steps. It is extremely similar to the proof in the 2-dimensional case, we just need to go from 2D to 3D, so I will skip some details.\n\n\\(\\mathbf{1.}\\) We will show that if \\(P\\) is in one of the exterior regions, then there is a point \\(P_!\\) on the boundary of the tetrahedron (a vertex, or on a edge, or on a side, such that \\(PA + PB + PC + PD > P_1A + P_1B + P_1C + P_1D\\).\n\n\\(\\mathbf{2.}\\) Then we will show that if \\(P\\) is on the boundary, then \\(PA + PB + PC + PD > OA + OB + OC + OD\\).\n\n\\(\\mathbf{3.}\\) For the final step, consider the plane going through the edge \\(CD\\) perpendicular to the edge \\(AB\\), the plane going through \\(AB\\) perpendicular to \\(CD\\), the plane going through \\(CA\\) perpendicular to \\(BD\\), etc. There are six such planes, and they all contain \\(O\\), the center of the circumscribed sphere. We will show that if \\(P\\) is a point of minimum for \\(PA + PB + PC + PD\\) inside the tetrahedron, then \\(P\\) is in each of the six planes described above. This implies that \\(P = O\\).\n\n\\(\\mathbf{Proof\\ of\\ 1:}\\) Let \\(P\\) be in one of the exterior regions. Assume \\(P\\) is in a prism shaped region delimited by extensions of three sides meeting in a vertex (there are 4 of them). Assume it is at vertex \\(A\\), the sides being the extensions of planes \\(ABC, ABD, ACD\\). Then take \\(P_1 = A\\). We have \\(PA + PB + PC + PD > P_1A + P_1B + P_1C + P_1D\\) because of obtuse triangles formed with \\(PP_1\\).\n\nNow assume \\(P\\) is in one of the wedge shaped regions, formed by an edge and the extensions of two sides going through them. (there are six such regions.) Assume this is the line \\(AB\\) and the extensions of \\(ABC, ABD\\). Then take \\(P_1\\) to be the foot of the perpendicular from \\(P\\) to \\(AB\\). Again, we have the desired inequality because \\(PP_1\\) formed some right and obtuse triangles.\n\nNow assume \\(P\\) is in the truncated prism region delimited by a side and the extensions of the faces going through the edges of this side. (There are four such regions.) Assume this is the side \\(ABC\\), and extensions of the sides \\(DAB, DBC, DCA\\). Then take \\(P_1\\) to be the foot of the perpendicular from \\(P\\) to the plane \\(ABC\\). Again, we have the desired inequality because of right and obtuse triangles formed by \\(PP_1\\).\n\n\\(\\mathbf{Proof\\ of\\ 2:}\\) Assume \\(P_1 = A\\). If we take the edge of the tetrahedron to be \\(1\\), a direct computation gives us that \\(P_1A + P_1B + P_1C + P_1D = 3\\), and \\(OA + OB + OC + OD = 4 \\cdot \\frac{\\sqrt{6}}{4} = \\sqrt{6}\\).\n\nAssume \\(P_1\\) is on \\(AB\\). If \\(P_1\\) is not the midpoint of \\(AB\\), take \\(P_2\\) to be the midpoint of \\(AB\\). Then \\(P_1A + P_1B + P_1C + P_1D > P_2A + P_2B + P_2C + P_2D\\) because of right triangles formed by \\(P_2C, P_2D\\). And, if we take the edge of the tetrahedron to be \\(1\\), a direct computation yields that \\(P_2A + P_2B + P_2C + P_2D = 1 + 2 \\cdot \\frac{\\sqrt{3}}{2} = 1 + \\sqrt{3}\\), which is bigger than \\(OA + OB + OC + OD = \\sqrt{6}\\).\n\nAssume \\(P_1\\) is on \\(ABC\\). If \\(P_1\\) is not the circumcenter of \\(\\triangle ABC\\) then take \\(P_2\\) to be the circumcenter. We have \\(P_1D > P_2D\\) because \\(P_2D \\perp ABC\\). We also have \\(P_1A + P_1B + P_1C > P_2A + P_2B + P_2C\\) because we proved the 2-dimensional analogue of the problem. And, if we take the edge of the tetrahedron to be \\(1\\), we have \\(P_2A + P_2B + P_2C + P_2D = \\sqrt{3} + \\frac{\\sqrt{6}}{3}\\), which is bigger than \\(OA + OB + OC + OD = \\sqrt{6}\\).\n\nNOTE: In the above paragraph, we used that the similar result is true in the 2-dimensional case, with an equilateral triangle instead of a regular tetrahedron.\n\nNOTE: This part of the proof concludes filling in the gap in the first \"Solution\", written above. (A reader may complain that the proof in Solution 2 is very long (compared to the first \"Solution\"), but the first \"Solution\" should have done this too, one way or another.)\n\n\\(\\mathbf{Proof\\ of\\ 3:}\\) Now consider the six planes going through one edge, perpendicular to the opposite edge. They intersect at the circumcenter of the tetrahedron. Assume \\(P\\) is a point in the interior of the tetrahedron where \\(PA + PB + PC + PD\\) achieves its minimum value. Then \\(P\\) is in each of the six plane.\n\nProve this statement by contradiction. Assume that there is a plane among the six, so that \\(P\\) is not on it. Assume the plane is the one going through \\(CD\\), perpendicular to \\(AB\\). To make it more explicit, this is the plane going through \\(C, D, E\\), where \\(E\\) is the midpoint of \\(AB\\).\n\nConsider the ellipsoid with focal points \\(A, B\\) going through \\(P\\). This can be obtained as the set of points \\(Q\\) in space so that \\(QA + QB = PA + PB\\). It can also be obtained as the surface obtained when we form the ellipse with focal points \\(A, B\\) in the plane \\(ABC\\) (as the set of points \\(Q\\) so that \\(QA + QB = PA + PB\\)), and we rotate this ellipse from the plane \\(ABC\\) around its axis \\(AB\\). Let \\(P_1\\) be the foot of the perpendicular from \\(P\\) to the plane \\(CDE\\). We have \\(PC > P_1C, PD > P_1D\\) because \\(PP_1 \\perp CDE\\). We also have \\(PA + PB > P_1A + P_1B\\) because \\(P_1\\) is in the interior of the ellipsoid. (Indeed, the intersection of the plane \\(CDE\\) and the ellipsoid is the circle generated by rotating the ends of the small axis of the ellipse in the plane \\(ABC\\). Since the point \\(P\\) is not on the plane CDE, it must be on a smaller circle, so its projection to the plane \\(CDE\\) will be inside.)\n\nThis concludes the proof of the problem.\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-4
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_4
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Prove that for every natural number \(n\), and for every real number \(x \neq \frac{k\pi}{2^t}\) (\(t=0,1, \dots, n\); \(k\) any integer)
\[
\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}
\]
|
[
"First, we prove \\(\\cot \\theta - \\cot 2\\theta = \\frac {1}{\\sin 2\\theta}\\).\n\nLHS\\(\\ =\\ \\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}\\)\n\n\\[\n= \\frac{2\\cos^2 \\theta}{2\\cos \\theta \\sin \\theta}-\\frac{2\\cos^2 \\theta -1}{\\sin 2\\theta}\n\\]\n\n\\[\n=\\frac{2\\cos^2 \\theta}{\\sin 2\\theta}-\\frac{2\\cos^2 \\theta -1}{\\sin 2\\theta}\n\\]\n\n\\[\n=\\frac {1}{\\sin 2\\theta}\n\\]\n\nUsing the above formula, we can rewrite the original series as\n\n\\(\\cot x - \\cot 2x + \\cot 2x - \\cot 4x + \\cot 4x + \\dots + \\cot 2^{n-1} x - \\cot 2^n x\\).\n\nWhich gives us the desired answer of \\(\\cot x - \\cot 2^n x\\)."
] |
IMO-1966-5
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_5
|
Solve the system of equations
\[
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1
\]
where \(a_1, a_2, a_3, a_4\) are four different real numbers.
|
[
"Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:\n\n\\[\n- x1 + x2 + x3 + x4 = 0.\n\\]\n\nSubtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:\n\n\\[\n- x1 - x2 - x3 + x4 = 0.\n\\]\n\nHence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:\n\n\\[\n- x1 - x2 + x3 + x4 = 0.\n\\]\n\nHence \\(x2 = x3 = 0\\), and \\(x1 = x4 = 1/(a1 - a4)\\).\n\n## Remarks (added by pf02, September 2024)\n\nThe solution above is in the realm of flawed and incorrect. It is flawed because you can not claim to have solved a system of equations by having solved a particular case. It is correct in solving the particular case, but it is incorrect in stating that the result obtained is a solution to the system in general.\n\nBelow I will give a complete solution to the problem. The first few lines will be a repetition of the \"solution\" above, and I will repeat them for the sake of completeness and of a more tidy writing.",
"There are 24 possibilities when we count the ordering of \\(a_1, a_2, a_3, a_4\\), and each ordering gives a different system of equations. Let us consider one of them, like in the \"solution\" above.\n\nAssume \\(a_1 > a_2 > a_3 > a_4\\). In this case, the system is\n\n\\[\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_1 - a_2) x_2 + (a_1 - a_3) x_3 + (a_1 - a_4) x_4 = 1 \\\\ (a_1 - a_2) x_1 + \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_2 - a_3) x_3 + (a_2 - a_4) x_4 = 1 \\\\ (a_1 - a_3) x_1 + (a_2 - a_3) x_2 + \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_3 - a_4) x_4 = 1 \\\\ (a_1 - a_4) x_1 + (a_2 - a_4) x_2 + (a_3 - a_4) x_3 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ = 1\n\\]\n\nSubtract the second equation from the first, and divide by \\((a_1 - a_2)\\). Also, subtract the fourth equation from the third, and divide by \\((a_3 - a_4)\\). We obtain\n\n\\[\n-x_1 + x_2 + x_3 + x_4 = 0 \\\\ -x_1 - x_2 - x_3 + x_4 = 0\n\\]\n\nIt follows that \\(x_1 - x_4 = 0\\) and \\(x_2 + x_3 = 0\\).\n\nSubtract the third equation from the second, and divide by \\((a_2 - a_3)\\). We obtain\n\n\\[\n-x_1 - x_2 + x_3 + x_4 = 0\n\\]\n\nSince \\(x_1 - x_4 = 0\\), it follows that \\(x_2 - x_3 = 0\\). Combining with \\(x_2 + x_3 = 0\\), we get \\(x_2 = x_3 = 0\\). Replacing these in the first equation of the system, we get \\(x_4 = \\frac{1}{a_1 - a_4}\\), so we also have \\(x_1 = \\frac{1}{a_1 - a_4}\\).\n\nNow we have two ways of proceeding. We could consider each of the other 23 cases, and solve it by a similar method. The task is made easy if we notice that each case is obtained from the first case by a permutation of indices, so it can be viewed as a change of notation. With some care, we can just write the solution in each case. For example, in the case \\(a_2 > a_1 > a_3 > a_4\\), we will obtain \\(x_1 = x_3 = 0\\) and \\(x_2 = x_4 = \\frac{1}{a_2 - a_4}\\).\n\nWe will proceed differently, but we will use the same idea. Let \\(m, n, p, q\\) be the indices such that \\(a_m > a_n > a_p > a_q\\). Written in a compact way, our system becomes\n\n\\(\\sum_{\\substack{i = 1 \\\\ i \\ne j}}^4 |a_j - a_i| x_i = 1, \\ \\ \\ \\ j = 1, 2, 3, 4\\).\n\nMake the following change of notation: \\(a_m \\rightarrow b_1 \\\\ a_n \\rightarrow b_2 \\\\ a_p \\rightarrow b_3 \\\\ a_q \\rightarrow b_4\\)\n\nand\n\n\\[\nx_m \\rightarrow y_1 \\\\ x_n \\rightarrow y_2 \\\\ x_p \\rightarrow y_3 \\\\ x_q \\rightarrow y_4\n\\]\n\nIn the new notation we have \\(b_1 > b_2 > b_3 > b_4\\) and the system becomes\n\n\\(\\sum_{\\substack{k = 1 \\\\ k \\ne l}}^4 |b_l - b_k| y_k = 1, \\ \\ \\ \\ l = 1, 2, 3, 4\\).\n\nThis is exactly the system we solved above, just with a new notation (\\(b, y\\) instead of \\(a, x\\)). So the solutions are \\(y_2 = y_3 = 0, y_1 = y_4 = \\frac{1}{b_1 - b_4}\\).\n\nReturning to our original notation, we have \\(x_n = x_p = 0, x_m, x_q = \\frac{1}{a_m - a_q}\\).\n\nIn conclusion, here is a compact way of giving the solution to the system: let \\(m\\) be the index of the largest of the \\(a_i\\)'s, and q = the index of the smallest of the \\(a_i\\)'s, and let \\(n, p\\) be the other two indices. Then \\(x_n = x_p = 0, x_m, x_q = \\frac{1}{a_m - a_q}\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-6
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_6
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In the interior of sides \(BC, CA, AB\) of triangle \(ABC\), any points \(K, L,M\), respectively, are selected. Prove that the area of at least one of the triangles \(AML, BKM, CLK\) is less than or equal to one quarter of the area of triangle \(ABC\).
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[
"Let the lengths of sides \\(BC\\), \\(CA\\), and \\(AB\\) be \\(a\\), \\(b\\), and \\(c\\), respectively. Let \\(BK=d\\), \\(CL=e\\), and \\(AM=f\\).\n\nNow assume for the sake of contradiction that the areas of \\(\\Delta AML\\), \\(\\Delta BKM\\), and \\(\\Delta CLK\\) are all at greater than one fourth of that of \\(\\Delta ABC\\). Therefore\n\n\\[\n\\frac{AM\\cdot AL\\sin{\\angle BAC}}{2}>\\frac{AB\\cdot AC\\sin{\\angle BAC}}{8}\n\\]\n\nIn other words, \\(AM\\cdot AL>\\frac{1}{4}AB\\cdot AC\\), or \\(f(b-e)>\\frac{bc}{4}\\). Similarly, \\(d(c-f)>\\frac{ac}{4}\\) and \\(e(a-d)>\\frac{ab}{4}\\). Multiplying these three inequalities together yields\n\n\\[\ndef(a-d)(b-e)(c-f)>\\frac{a^2b^2c^2}{64}\n\\]\n\nWe also have that \\(d(a-d)\\leq \\frac{a^2}{4}\\), \\(e(b-e)\\leq \\frac{b^2}{4}\\), and \\(f(c-f)\\leq \\frac{c^2}{4}\\) from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields\n\n\\[\ndef(a-d)(b-e)(c-f)\\leq\\frac{a^2b^2c^2}{64}\n\\]\n\nThis is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.",
"Let \\(AR : AB = x, BP : BC = y, CQ : CA = z\\). Then it is clear that the ratio of areas of \\(AQR, BPR, CPQ\\) to that of \\(ABC\\) equals \\(x(1-y), y(1-z), z(1-x)\\), respectively. Suppose all three quantities exceed \\(\\frac{1}{4}\\). Then their product also exceeds \\(\\frac{1}{64}\\). However, it is clear by AM-GM that \\(x(1-x) \\le \\frac{1}{4}\\), and so the product of all three quantities cannot exceed \\(\\frac{1}{64}\\) (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to \\(\\frac{1}{4} [ABC]\\).\n\n## Remarks (added by pf02, September 2024)\n\nSolution 2 is written in a very sloppy way. However, an interested reader can make sense of it. More importantly, the two solutions are identical. If it wasn't for the sloppy writing, Solution 2 could be obtained from the first Solution after applying a word by word translation which replaces line segments by ratios.\n\nBelow I will give another solution. It is formally different from the previous solutions, even if not at a deep level.",
"Let \\(\\triangle ABC\\) and \\(K, L, M\\) be as in the problem. Denote \\(x = \\frac{AM}{AB}, y = \\frac{BK}{BC}, z = \\frac{CL}{CA}\\) as in Solution 2. Note that \\(x, y, z, \\in (0, 1)\\) because \\(K, L, M\\) are in the interior of the respective sides.\n\n\\[\nProb 1966 6.png\n\\]\n\nUsing the fact that the area of a triangle is half of the product of two sides and \\(\\sin\\) of the angle between them (like in the first Solution), we have that \\(\\mathbf{area} AML = x(1 - z) \\mathbf{area} ABC, \\mathbf{area} BKM = y(1 - x) \\mathbf{area} ABC, \\mathbf{area} CLK = z(1 - y) \\mathbf{area} ABC\\).\n\nNow the problem has nothing to do with geometry anymore: we just have to show that given three numbers \\(x, y, z, \\in (0, 1)\\), at least one of \\(x(1 - z), y(1 - x), z(1 - y)\\) is \\(\\le \\frac{1}{4}\\).\n\nIf \\(y(1 - x) \\le \\frac{1}{4}\\), we are done. Otherwise, we have \\(y(1 - x) > \\frac{1}{4}\\). It follows that \\(y > \\frac{1}{4(1 - x)}\\) (recall that \\(0 < x, y, z < 1\\)). In particular, it follows that \\(\\frac{1}{4(1 - x)} < 1\\), which implies \\(3 - 4x > 0\\).\n\nIf \\(z(1 - y) \\le \\frac{1}{4}\\), we are done. Otherwise, we have \\(z(1 - y) > \\frac{1}{4}\\). Using the inequality on \\(y\\) from the previous paragraph, we have \\(z \\left( 1 - \\frac{1}{4(1 - x)} \\right) > \\frac{1}{4}\\), or after a few computations, \\(z \\cdot \\frac{3 - 4x}{1 - x} > 1\\). Using the observation about \\(3 - 4x\\) from the preceding paragraph, we get \\(z > \\frac{1 - x}{3 - 4x}\\).\n\nNow consider \\(x(1 - z)\\). Using the inequality on \\(z\\) from the previous paragraph, we have that \\(x(1 - z) < x \\left( 1 - \\frac{1 - x}{3 - 4x} \\right)\\). To finish the solution to the problem, it is enough to show that \\(x \\cdot \\left( 1 - \\frac{1 - x}{3 - 4x} \\right) \\le \\frac{1}{4}\\).\n\nAfter some easy computations (and using again that \\(3 - 4x > 0\\)), this becomes \\(3(4x^2 - 4x + 1) \\ge 0\\), which is true because. \\(4x^2 - 4x + 1 = (2x - 1)^2\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1967-1
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https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_1
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Let \(ABCD\) be a parallelogram with side lengths \(AB = a\), \(AD = 1\), and with \(\angle BAD = \alpha\). If \(\Delta ABD\) is acute, prove that the four circles of radius \(1\) with centers \(A\), \(B\), \(C\), \(D\) cover the parallelogram if and only if
\(a\leq \cos \alpha+\sqrt{3}\sin \alpha\) \(\ \ \ \ \ \ \ \ \ \ (1)\)
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[
"To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of \\(a\\) must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.\n\nTo prove our conjecture we draw a parallelogram with \\(a=2\\) and draw a segment \\(DB\\) so that \\(\\angle ADB=90^{\\circ}\\)\n\nThis is the parallelogram which we claim has the maximum length on \\(a\\) and the highest value on any one angle.\n\nWe now have two triangles inside a parallelogram with lengths \\(1, 2\\) and \\(x\\), \\(x\\) being segment \\(DB\\). Using the Pythagorean theorem we conclude:\n\n\\[\n1^2+x^2=2^2\\\\x=\\sqrt{3}\n\\]\n\nUsing trigonometric functions we can compute:\n\n\\[\ncos\\alpha=\\frac{1}{2}\\\\sin\\alpha=\\frac{\\sqrt{3}}{2}\n\\]\n\nNotice that by applying the \\(arcsine\\) and \\(arccos\\) functions, we can conclude that our angle \\(\\alpha=60^{\\circ}\\)\n\nTo conclude our proof we make sure that our values match the required values for maximum length of \\(a\\)\n\n\\[\na\\leq\\cos\\alpha+\\sqrt{3}\\sin\\alpha\\\\\\\\a\\leq\\frac{1}{2}+\\sqrt{3}\\cdot \\frac{\\sqrt{3}}{2}\\\\\\\\a\\leq 2\n\\]\n\nNotice that as \\(\\angle\\alpha\\) decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as \\(\\angle\\alpha\\) increases, the value of (1) decreases below 2, confirming that (1) is only implied when \\(\\Delta ABD\\) is acute.\n\n--Bjarnidk 02:16, 17 May 2013 (EDT)\n\n## Remarks (added by pf02, September 2024)\n\n\\(\\mathbf{Remark\\ 1}\\). I am sorry to be so harshly critical, but the solution above is deeply flawed. Not only it has errors, but the logic is flawed.\n\nIt shows that when \\(a = 2, \\alpha = \\frac{\\pi}{3}\\) the parallelogram is covered by the circles of radius \\(1\\) centered at \\(A, B, C, D\\), and the inequality in the problem is true. (Even this is incomplete, while giving too many, unnecessary details.) (Note that this is not a case which satisfies the conditions of the problem since \\(\\triangle ABD\\) is right, not acute.)\n\nIn the last two lines it gives some reasoning about other values of \\(\\alpha\\) which is incomprehensible to this reader.\n\nIn one short sentence: this is not a solution.\n\n\\(\\mathbf{Remark\\ 2}\\). The problem itself is mildly flawed. To see this, denote \\(S1, S2\\) the following two statements:\n\nS1: The parallelogram \\(ABCD\\) is covered by the four circles of radius \\(1\\) centered at \\(A, B, C, D\\).\n\nS2: We have \\(a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\nThe problem says that if \\(\\triangle ABD\\) is acute, \\(S1\\) and \\(S2\\) are equivalent, i.e. they imply each other.\n\nNotice that \\(S2\\) can be rewritten as \\(a \\le 2 \\cos \\left( \\alpha - \\frac{\\sqrt{\\pi}}{3} \\right)\\).\n\nNow notice that if \\(a \\le 1\\) then S1 is obviously true. See the picture below:\n\n\\[\nProb 1967 1 fig1.png\n\\]\n\nAlso, notice that if \\(a \\le 1\\) and \\(\\alpha \\in \\left( 0, \\frac{\\sqrt{\\pi}}{2} \\right)\\) then \\(S2\\) is true as well. Indeed \\(\\left( \\alpha - \\frac{\\sqrt{\\pi}}{3} \\right) \\in \\left( -\\frac{\\sqrt{\\pi}}{3}, \\frac{\\sqrt{\\pi}}{6} \\right)\\), so \\(\\cos\\) is \\(> \\frac{1}{2}\\) on this interval, so the right hand side of \\(S2\\) is \\(> 1 \\ge a\\).\n\nWe see that if \\(a \\le 1\\) and \\(\\triangle ABD\\) is acute, both \\(S1\\) and \\(S2\\) are true. We can not say that one implies the other in the usual meaning of the word \"imply\": the two statements just happen to be both true.\n\nIf we take \\(a > 1\\) then the problem is a genuine problem, and there is something to prove.\n\n\\(\\mathbf{Remark\\ 3}\\). In the proofs I give below, we will see where we need that \\(\\triangle ABD\\) is acute. We will see that \\(\\alpha < \\frac{\\pi}{2}\\) is needed for the technicalities of the proof. The fact that \\(\\angle ADB\\) is acute will be needed at one crucial point in the proof.\n\nIn fact, it is possible to modify \\(S2\\) to a statement \\(S3\\) similar to \\(S2\\) so that \\(S1\\) and \\(S3\\) are equivalent without any assumption on \\(\\alpha\\). I will not go into this, I will just give a hint: Denote \\(\\beta = \\angle ABC\\). If \\(\\alpha\\) is acute, \\(\\beta\\) is obtuse, and we can easily reformulate \\(S2\\) in terms of \\(\\beta\\).\n\n\\(\\mathbf{Remark\\ 4}\\). Below, I will give two solutions. Solution 2 is one I carried out myself and relies on a straightforward computation. Solution 3 (it happens to be similar to Solution 2) is inspired by an idea by feliz shown on the web page https://artofproblemsolving.com/community/c6h21154p137323 The author calls their text a solution, but it is quite confused, so I would not call it a good solution. The idea though is good and nice, and it yields a nice solution.",
"We can assume \\(a > 1\\). Indeed, refer to Remark 2 above to see that if \\(a \\le 1\\) there is nothing to prove.\n\nNote that instead of the statement \\(S1\\) we can consider the following statement \\(S1'\\):\n\n\\(S1'\\): the circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\).\n\nThis is equivalent to \\(S1\\) because of the symmetry between \\(\\triangle ABD\\) and \\(\\triangle BCD\\).\n\nLet \\(F\\) be the intersection above \\(AB\\) of the circles of radius \\(1\\) centered at \\(A, B\\). The three circles cover \\(\\triangle ABD\\) if an only if \\(F\\) is inside the circle of radius 1 centered at \\(D\\).\n\n\\[\nProb 1967 1 fig2.png\n\\]\n\nThis needs an explanation: Let \\(H\\) be the midpoint of \\(BD\\), and consider \\(\\triangle FHD\\). All the vertices of this triangle are in the circle centered at \\(D\\), so the whole triangle is inside this circle. It is obvious that \\(\\triangle FHB\\) is inside the circle centered at \\(B\\), and that \\(\\triangle FAD, \\triangle FAB\\) are inside the circles centered at \\(A, B\\).\n\nWe will now show that \\(F\\) is inside the circle of radius 1 centered at \\(D\\) if an only if \\(DF \\le 1\\).\n\nThe plan is to calculate \\(DF\\) in terms of \\(a, \\alpha\\) and impose this condition. Let \\(FG \\perp AB\\), \\(DE \\perp AB\\) and \\(FF' \\parallel GE\\). From the right triangle \\(\\triangle AFG\\) we have \\(FG = \\sqrt{1 - \\left( \\frac{a}{2} \\right)^2} = \\frac{\\sqrt{4 - a^2}}{2}\\). From the right triangle \\(\\triangle DFF'\\) we have\n\n\\[\nDF = \\sqrt{(DF')^2 + (FF')^2} = \\sqrt{(DE - FG)^2 + (AG - AE)^2} = \\sqrt{\\left( \\sin \\alpha - \\frac{\\sqrt{4 - a^2}}{2} \\right)^2 + \\left( \\frac{a}{2} - \\cos \\alpha \\right)^2}\n\\]\n\n(Note that here we used the fact that \\(\\alpha\\) is acute. These equalities would look slightly differently otherwise.)\n\nNow look at the condition \\(DF \\le 1\\), or equivalently \\(DF^2 \\le 1\\). Making all the computations and simplifications, we have \\(\\sqrt{4 - a^2} \\sin \\alpha \\ge 1 - a \\cos \\alpha\\).\n\nNow I would like to square both sides. In order to get an equivalent inequality, we need to know that \\(1 - a \\cos \\alpha \\ge 0\\). This follows from the fact that \\(\\angle ADB\\) is acute. Indeed, denote \\(\\angle ADB = \\beta\\). From the law of sines in \\(\\triangle ADB\\) we have \\(\\frac{1}{\\sin \\angle ABD} = \\frac{a}{\\sin \\beta}\\). Successively this becomes \\(\\frac{1}{\\sin (\\pi - \\alpha - \\beta)} = \\frac{a}{\\sin \\beta}\\) or \\(\\frac{1}{\\sin (\\alpha + \\beta)} = \\frac{a}{\\sin \\beta}\\) or \\((1 - a \\cos \\alpha) \\sin \\beta = a \\sin \\alpha \\cos \\beta\\). From here we see that \\(\\beta < \\frac{\\pi}{2}\\) implies the right hand side is positive, so \\(1 - a \\cos \\alpha > 0\\).\n\nGoing back to our inequality, we can square both sides, and after rearranging terms we get that \\(DF \\le 1\\) if and only if\n\n\\(a^2 - 2a \\cos \\alpha + (1 - 4 \\sin^2 \\alpha) \\le 0\\).\n\nView this as an equation of degree \\(2\\) in \\(a\\). The value of the polynomial in \\(a\\) is \\(\\le 0\\) when \\(a\\) is between its solutions, that is\n\n\\(\\cos \\alpha - \\sqrt{3} \\sin \\alpha \\le a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\nNote that \\(\\cos \\alpha - \\sqrt{3} \\sin \\alpha = 2 \\cos \\left( \\alpha + \\frac{\\pi}{3} \\right)\\). If \\(\\alpha \\in \\left( 0, \\frac{\\pi}{2} \\right)\\) then \\(\\left( \\alpha + \\frac{\\pi}{3} \\right) \\in \\left( \\frac{\\pi}{3}, \\frac{5\\pi}{6} \\right)\\), and it follows that \\(2 \\cos \\left( \\alpha + \\frac{\\pi}{3} \\right) \\le 1\\).\n\nOn the other hand, remember that we are in the case \\(a > 1\\), so the left inequality is always true. It follows that\n\n\\(DF \\le 1\\) (i.e. the three circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\)) if an only if \\(a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\n[Solution by pf02, September 2024]",
"This solution is very similar to Solution 2, except that we choose another point instead of \\(F\\). This will in fact simplify the proof. Start like in Solution 2.\n\nWe can assume \\(a > 1\\). Indeed, refer to Remark 2 above to see that if \\(a \\le 1\\) there is nothing to prove.\n\nNote that instead of the statement \\(S1\\) we can consider the following statement \\(S1'\\):\n\n\\(S1'\\): the circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\).\n\nThis is equivalent to \\(S1\\) because of the symmetry between \\(\\triangle ABD\\) and \\(\\triangle BCD\\).\n\nLet \\(O\\) be the center of the circle circumscribed to \\(\\triangle ABD\\). Let \\(M, N, P\\), be the midpoints of \\(AB, AD, BD\\). The three circles cover \\(\\triangle ABD\\) if an only if \\(O\\) is inside the circle of radius 1 centered at \\(D\\).\n\n\\[\nProb 1967 1 fig3.png\n\\]\n\nThis needs an explanation. In fact, since \\(OA = OB = OD\\), the point \\(O\\) is inside or on the circle centered at \\(D\\) if and only if \\(OD \\le 1\\), if and only if \\(O\\) is in or inside the circles centered at \\(A, B, D\\). Since \\(OP \\perp BD\\), we have \\(DP < OD, PB < OB\\), so the triangles \\(\\triangle OPD, \\triangle OPB\\) are inside the circles centered at \\(D, B\\) respectively. By drawing \\(OA, OM\\) we can easily verify that the whole triangle is inside the circles centered at \\(A, D, B\\).\n\nNote that in the above argument we used that \\(O\\) is inside \\(\\triangle ABD\\), which is true because the triangle is acute.\n\nDenote \\(R\\) the radius of the circle circumscribed to \\(\\triangle ABD\\). From the law of sines, we have \\(\\frac{BD}{\\sin \\alpha} = 2R\\), and from the law of cosines we have \\(BD^2 = 1 + a^2 -2a \\cos \\alpha\\). So \\(R = OD \\le 1\\) translates to \\(\\frac{\\sqrt{1 + a^2 -2a \\cos \\alpha}}{2 \\sin \\alpha} \\le 1\\)\n\nSince \\(\\sin \\alpha > 0\\), we can simplify this inequality, and get\n\n\\(a^2 - 2a \\cos \\alpha + (1 - 4 \\sin^2 \\alpha) \\le 0\\).\n\nBut this is exactly the inequality we encountered in Solution 2, so proving that this is equivalent to\n\n\\[\na \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\n\\]\n\nis identical to what has been done above.\n\n[Solution based on an idea by feliz; see link in Remark 4, or below.]\n\nA solution can also be found here [1]"
] |
IMO-1967-2
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https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_2
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Prove that if one and only one edge of a tetrahedron is greater than \(1\), then its volume is \(\le \frac{1}{8}\).
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[
"Assume \\(CD>1\\) and let \\(AB=x\\). Let \\(P,Q,R\\) be the feet of perpendicular from \\(C\\) to \\(AB\\) and \\(\\triangle ABD\\) and from \\(D\\) to \\(AB\\), respectively.\n\nSuppose \\(BP>PA\\). We have that \\(CP=\\sqrt{CB^2-BT^2}\\le\\sqrt{1-\\frac{x^2}4}\\), \\(CQ\\le CP\\le\\sqrt{1-\\frac{x^2}4}\\). We also have \\(DQ^2\\le\\sqrt{1-\\frac{x^2}4}\\). So the volume of the tetrahedron is \\(\\frac13\\left(\\frac12\\cdot AB\\cdot DR\\right)CQ\\le\\frac{x}6\\left(1-\\frac{x^2}4\\right)\\).\n\nWe want to prove that this value is at most \\(\\frac18\\), which is equivalent to \\((1-x)(3-x-x^2)\\ge0\\). This is true because \\(0<x\\le 1\\).\n\nThe above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]\n\n## Remarks (added by pf02, September 2024)\n\nThe solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.\n\nThen, I will give a second solution to the problem.\n\nA few notes which may be of interest.\n\nThe condition that one side is greater than \\(1\\) is not really necessary. The statement is true even if all sides are \\(\\le 1\\). What we need is that no more than one side is \\(> 1\\).\n\nThe upper limit of \\(1/8\\) for the volume of the tetrahedron is actually reached. This will become clear from both solutions.",
"Assume that five of the edges are \\(\\le 1\\). Take them to be the edges other than \\(CD\\). Denote \\(AB = x\\). Let \\(P, Q, R\\) be the feet of perpendiculars from \\(C\\) to \\(AB\\), from \\(C\\) to the plane \\(ABD\\), and from \\(D\\) to \\(AB\\), respectively.\n\n\\[\nProb 1967 2 fig1.png\n\\]\n\nAt least one of the segments \\(AP, PB\\) has to be \\(\\ge \\frac{x}{2}\\). Suppose \\(PB \\ge \\frac{x}{2}\\). (If \\(AP\\) were bigger that \\(\\frac{x}{2}\\) the argument would be the same.) We have that \\(CP = \\sqrt{BC^2 - PB^2} \\le \\sqrt{1 - \\frac{x^2}{4}}\\). By the same argument in \\(\\triangle ABD\\) we have \\(DR \\le \\sqrt{1 - \\frac{x^2}{4}}\\). Since \\(CQ \\perp\\) plane \\(ABD\\), we have \\(CQ \\le CP\\), so \\(CQ \\le \\sqrt{1 - \\frac{x^2}{4}}\\).\n\nThe volume \\(V\\) of the tetrahedron is\n\n\\(V = \\frac{1}{3} \\cdot (\\)area of \\(\\triangle ABD) \\cdot\\)(height from \\(C) = \\frac{1}{3} \\cdot \\left( \\frac{1}{2} \\cdot AB \\cdot DR \\right) \\cdot CQ \\le \\left( \\frac{1}{6} \\cdot x \\cdot \\sqrt{1 - \\frac{x^2}{4}} \\cdot \\sqrt{1 - \\frac{x^2}{4}} \\right) = \\frac{x}{6} \\left( 1 - \\frac{x^2}{4} \\right)\\).\n\nWe need to prove that \\(\\frac{x}{6} \\left( 1 - \\frac{x^2}{4} \\right) \\le \\frac{1}{8}\\). Some simple computations show that this is the same as \\((1 - x)(3 - x - x^2) \\ge 0\\). This is true because \\(0 < x \\le 1\\), and \\(-x^2 - x + 3 > 0\\) on the interval \\((0, 1]\\).\n\n## Note\n\n\\(V = \\frac{1}{8}\\) is achieved when \\(x = 1\\) and all inequalities are equalities. This is the case when all sides except \\(CD\\) are \\(= 1\\), \\(P = R\\) are the midpoint of \\(AB\\), and \\(Q = P\\) (in which case the planes \\(ABC, ABD\\) are perpendicular). In this case, \\(CD = \\frac{\\sqrt{6}}{2}\\), and \\(V = \\frac{1}{8}\\) as can be seen from an easy computation.\n\n[This is an edited version of the solution by jgnr.]",
"Let \\(\\mathcal{T}\\) be the set of tetrahedrons with five edges \\(\\le 1\\). This proof will show that there is a \\(T \\in \\mathcal{T}\\) with one edge \\(> 1\\) and such that \\(\\mathbf{volume} (T) = \\frac{1}{8}\\), and that for any \\(U \\in \\mathcal{T}\\) either \\(U = T\\) or there is a finite sequence of tetrahedrons \\(T_1, \\dots, T_n\\) such that\n\n\\(\\mathbf{volume} (U) = \\mathbf{volume} (T_1) < \\dots < \\mathbf{volume} (T_n) = \\mathbf{volume} (T)\\).\n\nThe statement of the problem is a consequence of these facts.\n\nWe begin with two simple propositions.\n\n## Proposition\n\nLet \\(ABCD\\) be a tetrahedron, and consider the transformations which rotate \\(\\triangle ABC\\) around \\(AB\\) while keeping \\(\\triangle ABD\\) fixed. We get a set of tetrahedrons, two of which, \\(ABC_1D\\) and \\(ABC_2D\\) are shown in the picture below. The lengths of all sides except \\(CD\\) are constant through this transformation.\n\n\\[\nProb 1967 2 fig2.png\n\\]\n\n1. Assume that the angles between the planes \\(ABD\\) and \\(ABC\\), and \\(ABD\\) and \\(ABC_1\\) are both acute. If the perpendicular from \\(C_1\\) to the plane \\(ABD\\) is larger that the perpendicular from \\(C\\) to the plane \\(ABD\\) then the volume of \\(ABC_1D\\) is larger than the volume of \\(ABCD\\).\n\n2. Furthermore, the tetrahedron \\(ABC_2D\\) obtained when the position of \\(C_2\\) is such that the planes \\(ABD\\) and \\(ABC_2\\) are perpendicular has the maximum volume of all tetrahedrons obtained from rotating \\(\\triangle ABC\\) around \\(AB\\).\n\nThese statements are intuitively clear, since the volume \\(V\\) of the tetrahedron \\(ABCD\\) is given by\n\n\\(V = \\frac{1}{3} \\cdot (\\)area of \\(\\triangle ABD) \\cdot (\\)height from \\(C)\\).\n\nA formal proof is very easy, and I will skip it.\n\n## Corollary\n\nGiven a tetrahedron \\(T\\), and an edge \\(e_1\\) of it, we can find another tetrahedron \\(U\\) such that \\(\\mathbf{volume}(U) > \\mathbf{volume}(T)\\), with an edge \\(f_1 > e_1\\), and such that all the other edges of \\(U\\) are equal to the corresponding edges of \\(T\\), \\(\\mathbf{unless}\\) the edge \\(e_1\\) stretches between sides of \\(T\\) which are perpendicular. When we chose a bigger \\(f_1\\), if \\(e_1 < 1\\) we can choose \\(f_1 = 1\\). Or, we can choose \\(f_1\\) such that it stretches between sides which are perpendicular.\n\n(By \"stretches between two sides\" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, \\(DC_2\\) stretches between the sides \\(ABD, AC_2B\\) of \\(ABC_2D\\).)\n\n## Lemma\n\nAssume we have a tetrahedron \\(T\\) with edges \\(e_1, \\dots, e_6\\), such that \\(e_2, \\dots, e_6 \\le 1\\). If there is an edge \\(e_m < 1\\) among \\(e_2, \\dots, e_6\\) then there is a tetrahedron \\(U\\) with volume bigger than the volume of \\(T\\), whose edges are equal to those of \\(T\\), except for \\(e_m\\), which is replaced by an edge of size \\(1\\).\n\n## Proof\n\nCase 1: If \\(T\\) does not have any sides which are perpendicular, then the existence of \\(U\\) follows from the corollary.\n\nCase 2: Assume \\(T\\) has exactly two sides which are perpendicular (like \\(ABC_2D\\) in the picture above). If \\(C_2D < 1\\) and it were the only edge \\(< 1\\), then all the other edges are \\(= 1\\) (because they were assumed to be \\(\\le 1\\)). In this case \\(\\triangle ABD, \\triangle AC_2B\\) are equilateral with sides \\(= 1\\), and the planes can not be perpendicular since \\(C_2D < 1\\). (Indeed, an easy computation shows that if we take two equilateral triangles \\(\\triangle ABD, \\triangle AC_2B\\) and place them perpendicular to each other, then \\(CD = \\frac{\\sqrt{6}}{2} > 1\\).) So \\(e_m\\) must be one of the other sides. Then again, the existence of \\(U\\) follows from the corollary.\n\nCase 3: Assume that three sides are perpendicular.\n\n\\[\nProb 1967 2 fig3.png\n\\]\n\nAssume the perpendicular sides are the ones meeting at \\(A\\), i.e. each pair of the planes meeting at \\(A\\) are perpendicular. Since at least two of the edges \\(BD, BC, CD \\le 1\\) it follows that \\(AB, AC, AD < 1\\) (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume \\(e_m = AB < 1\\). We can apply the corollary, and find a tetrahedron \\(U\\) with volume bigger than the volume of \\(T\\), with edges equal to those of \\(T\\), except that \\(AB\\) is replaced by an edge \\(= 1\\).\n\nNow the problem is very easy to prove. Let \\(T\\) be a tetrahedron with edges \\(e_1, \\dots, e_6\\), such that \\(e_2, \\dots, e_6 \\le 1\\). Apply the lemma as many times as necessary (up to five times), successively replacing each edge \\(< 1\\) by an edge \\(= 1\\). We obtain a tetrahedron \\(U\\) with five edges \\(f_2, \\dots, f_6 = 1\\), and one edge \\(f_1 = e_1\\). If \\(f_1\\) stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which \\(f_1\\) is replaced by a larger edge which stretches between two perpendicular sides.\n\nWe obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides \\(= 1\\), having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is \\(> 1\\) is in fact of length \\(\\frac{\\sqrt{6}}{2}\\), and the volume of this tetrahedron is \\(\\frac{1}{8}\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1967-3
|
https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_3
|
Let \(k, m, n\) be natural numbers such that \(m+k+1\) is a prime greater than \(n+1.\) Let \(c_s=s(s+1).\) Prove that the product
\[
(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)
\]
is divisible by the product \(c_1c_2\cdots c_n\).
|
[
"For any \\(m, n\\), \\(c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)\\).\n\nWe can therefore write the product in the problem as follows:\n\n\\[\n(c_{m+1}-c_k)(c_{m+2}-c_k)\\cdots (c_{m+n}-c_k)\n\\]\n\n\\[\n= \\left(\\prod_{a = 1}^n m+a+k+1\\right)\\left(\\prod{a = 1}^n m+a-k\\right).\n\\]\n\nBut, the product of any \\(n\\) consecutive integers is divisible by \\(n!\\). We can prove this as follows:\n\n\\[\n(t-n+1)(t-n+2) \\cdots (t) = \\frac{(t+n)!}{t!} = n! \\frac{(t+n)!}{t!n!} = n! {t+n\\choose t}.\n\\]\n\nTherefore, \\(\\prod_{a = 1}^n m+a-k\\) is divisible by \\(n!\\), and \\((m+k+1)\\left(\\prod_{a = 1}^n m+a+k+1\\right)\\) is divisible by \\((n+1)!\\). However, we are told that \\(m+k+1\\) is prime and therefore it is not divisible by any of the numbers \\(1\\) through \\(n+1\\). Therefore, \\(\\prod_{a = 1}^n m+a+k+1\\) is divisible by \\((n+1)!\\).\n\nFinally, it is clear that \\((c_{m+1}-c_k)(c_{m+2}-c_k)\\cdots (c_{m+n}-c_k)\\) is divisible by \\(n!(n+1)! = (1 \\cdot 2)(2 \\cdot 3)(3 \\cdot 4) \\cdots (n \\cdot (n+1)) =\\)\\(c_1c_2\\cdots c_n\\). \\(\\square\\)\n\n~mathboy100",
"We have that \\(c_1c_2c_3...c_n=n!(n+1)!\\)\n\nand we have that \\(c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)\\)\n\nSo we have that \\((c_{m+1}-c_k)(c_{m+2}-c_k)\\ldots(c_{m+n}-c_k)=\\frac{(m+n-k)!}{(m-n)!}\\frac{(m+n+k+1)!}{(m+k+1)!}\\) We have to show that:\n\n\\(\\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\\frac{(m+n-k)!}{(m-n)!n!}\\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \\frac 1{m+k+1}\\) is an integer\n\nBut \\(\\frac{(m+n-k)!}{(m-n)!n!}=\\binom {m+n-k}n\\) is an integer and \\({(m+n+k+1)!}{(m+k)!(n+1)!} \\frac 1{m+k+1}=\\binom {m+n+k+1}{n+1}\\frac 1{m+k+1}\\) is an integer because \\(m+k+1|m+n+k+1!\\) but does not divide neither \\(n+1!\\) nor \\(m+n!\\) because \\(m+k+1\\) is prime and it is greater than \\(n+1\\) (given in the hypotesis) and \\(m+n\\).\n\nThe above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [1]\n\n## Remark and correction (added by pf02, September 2024)\n\n1. The two solutions are essentially the same. They are just different expressions of the same idea. Unfortunately, both are written very sloppily, but the interested reader can easily correct the mistakes, improve on the arguments, and make sense of the proofs.\n\n2. Both solutions are incomplete. They both assume (without saying it) that \\(m + 1 > k\\). Maybe the problem meant to say this, but it didn't, so we have to see what happens if \\(m + 1 \\le k\\). There are two cases:\n\nFirst, assume \\(m + 1 \\le k \\le m + n\\). In this case, \\((c_{m + 1} - c_k)(c_{m + 2} - c_k) \\cdots (c_{m + n} - c_k) = 0\\), and we can say that \\(0\\) is divisible by anything.\n\nNext, assume \\(m + n < k\\). In this case, each factor of the first product is negative, and we will consider the product\n\n\\[\nP = (c_k - c_{m + 1})(c_k - c_{m + 2}) \\cdots (c_k - c_{m + n})\n\\]\n\nfor the sake of working with positive factors. \\(P\\) differs from the product in the problem by at most a sign, so it is enough to show that \\(P\\) is divisible by \\(c_1c_2 \\cdots c_n\\).\n\nThe proof of this fact goes along the same lines as the proof when \\(m + 1 > k\\). We will give the proof for the sake of having a complete, properly written proof.\n\nUsing \\(p(p + 1) - q(q + 1) = (p + q + 1)(p - q)\\) we get that \\(c_1c_2 \\cdots c_n = (n + 1)! \\cdot n!\\). We also get that \\(P = \\prod_{p = 1}^n (k + m + p + 1) \\cdot \\prod_{p = 1}^n (k - m - p)\\).\n\nWe have that \\(n!\\ |\\ \\prod_{p = 1}^n (k - m - p)\\) (i.e. the product is divisible by the factorial) because\n\n\\[\n\\frac{\\prod_{p = 1}^n (k - m - p)}{n!} = \\frac{(k - m - 1)!}{(k - m - 1 - n)! \\cdot n!} = {k - m - 1 \\choose n}\n\\]\n\nwhich is an integer. (Note that this all makes sense because we are in the case when \\(m + n < k\\).)\n\nLet us now show that \\((n + 1)!\\ |\\ \\prod_{p = 1}^n (k + m + p + 1)\\). Let \\(P_1 = \\prod_{p = 1}^n (k + m + p + 1)\\). We have that\n\n\\[\n\\frac{(k + m + 1) \\cdot P_1}{(n + 1)!} = \\frac{(k + m + 1) \\cdot \\prod_{p = 1}^n (k + m + p + 1)}{(n + 1)!} = \\frac{\\prod_{p = 0}^n (k + m + p + 1)}{(n + 1)!} = \\frac{(k + m + n + 1)!}{(k + m)! \\cdot (n + 1)!} = {k + m + n + 1 \\choose n + 1}\n\\]\n\nwhich is an integer. So \\((n + 1)!\\ |\\ (k + m + 1) \\cdot P_1\\). But none of the factors of \\((n + 1)!\\) is a divisor of \\((k + m + 1)\\) because \\((k + m + 1)\\) is prime, and \\(n + 1 < k + m + 1\\). So \\(P_1 = \\prod_{p = 1}^n (k + m + p + 1)\\) must be divisible by \\((n + 1)!\\)."
] |
IMO-1967-4
|
https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_4
|
Let \(A_0B_0C_0\) and \(A_1B_1C_1\) be any two acute-angled triangles. Consider all triangles \(ABC\) that are similar to \(\triangle A_1B_1C_1\) (so that vertices \(A_1\), \(B_1\), \(C_1\) correspond to vertices \(A\), \(B\), \(C\), respectively) and circumscribed about triangle \(A_0B_0C_0\) (where \(A_0\) lies on \(BC\), \(B_0\) on \(CA\), and \(AC_0\) on \(AB\)). Of all such possible triangles, determine the one with maximum area, and construct it.
|
[
"We construct a point \\(P\\) inside \\(A_0B_0C_0\\) s.t. \\(\\angle X_0PY_0=\\pi-\\angle X_1Z_1Y_1\\), where \\(X,Y,Z\\) are a permutation of \\(A,B,C\\). Now construct the three circles \\(\\mathcal C_A=(B_0PC_0),\\mathcal C_B=(C_0PA_0),\\mathcal C_C=(A_0PB_0)\\). We obtain any of the triangles \\(ABC\\) circumscribed to \\(A_0B_0C_0\\) and similar to \\(A_1B_1C_1\\) by selecting \\(A\\) on \\(\\mathcal C_A\\), then taking \\(B= AB_0\\cap \\mathcal C_C\\), and then \\(B=CA_0\\cap\\mathcal C_B\\) (a quick angle chase shows that \\(B,C_0,A\\) are also colinear).\n\nWe now want to maximize \\(BC\\). Clearly, \\(PBC\\) always has the same shape (i.e. all triangles \\(PBC\\) are similar), so we actually want to maximize \\(PB\\). This happens when \\(PB\\) is the diameter of \\(\\mathcal C_B\\). Then \\(PA_0\\perp BC\\), so \\(PC\\) will also be the diameter of \\(\\mathcal C_C\\). In the same way we show that \\(PA\\) is the diameter of \\(\\mathcal C_A\\), so everything is maximized, as we wanted.\n\nThis solution was posted and copyrighted by grobber. The thread can be found here: [1]",
"Since all the triangles \\(\\triangle ABC\\) circumscribed to \\(\\triangle A_0B_0C_0\\) we will construct are are similar (by the requirement of the problem), the one with maximum area will be the one with maximum sides, or equivalently, the one with maximum side \\(BC\\). So we will try to maximize \\(BC\\) (while keeping the angles of \\(\\triangle ABC\\) equal to the angles of \\(\\triangle A_1B_1C_1\\)).\n\nThe plan is to find the value of \\(\\alpha = \\angle (BA_0C_0)\\) which maximizes \\(BC\\).\n\n\\[\nProb 1967 4 fig1.png\n\\]\n\nNote that for any \\(\\alpha\\) we can construct the line through \\(A_0\\) which forms the angle \\(\\alpha\\) with \\(A_0C_0\\). We can construct points \\(B, C\\) on this line, and lines through these points which form the given angles \\(\\angle B = \\angle B_1, \\angle C = \\angle C_1\\) with the line, and which pass through \\(C_0, B_0\\) respectively. Since \\(\\triangle A_0B_0C_0, \\triangle A_1B_1C_1\\) are acute, \\(A_0\\) is between \\(B, C\\) and these lines will meet at a point \\(A\\) such that \\(B_0\\) is between \\(A, C\\) and \\(C_0\\) is between \\(A, B\\).\n\n(More about this later.)\n\nThe quantities \\(a_0, b_0, c_0, \\angle B, \\angle C\\) are given. From this data, \\(\\angle A_0, \\angle B_0, \\angle C_0, \\angle A\\) are known and constructible. We will compute \\(BC\\) in terms of \\(\\alpha\\) and these quantities. This will be a function in the variable \\(\\alpha\\), and we will find the value of \\(\\alpha\\) for which this function attains its maximum.\n\nWe will start by computing \\(A_0B\\). We will use the law of sines in \\(\\triangle A_0C_0B\\). We get \\(\\frac{A_0B}{\\sin (\\pi - B - \\alpha)} = \\frac{b_0}{\\sin B}\\), and a similar equality from \\(\\triangle A_0B_0C\\) (for \\(A_0C\\)). We obtain\n\n\\[\nBC = A_0B + A_0C = \\frac{b_0}{\\sin B} \\sin (B + \\alpha) + \\frac{c_0}{\\sin C} \\sin (A_0 - C + \\alpha) = f(\\alpha)\n\\]\n\nWe can now proceed in two ways. We could use the formula for linear combination of sine functions with same period but different phase shifts (see https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations or https://mathworld.wolfram.com/HarmonicAdditionTheorem.html , (13)-(23)) or use calculus to find \\(\\alpha\\) for which \\(f(\\alpha)\\) has its maximum value.\n\nWith the first method, we would obtain that \\(f(\\alpha) = D \\sin (\\alpha + \\theta)\\) for certain \\(D\\) and \\(\\theta\\), and we would choose \\(\\alpha\\) such that \\(\\alpha + \\theta = \\pi/2\\). But we will use calculus, as a more mainstream approach. Compute the derivative \\(f'(\\alpha)\\) and consider the equation \\(f'(\\alpha) = 0\\). Use the formula for \\(\\cos\\) of sum of angles, and rearrange terms.\n\nWe have\n\n\\[\n\\cos \\alpha \\cdot \\left [ \\frac{b_0}{\\sin B} \\cdot \\cos B + \\frac{c_0}{\\sin C} \\cdot \\cos (A_0 - C) \\right ]= \\sin \\alpha \\cdot \\left [ \\frac{b_0}{\\sin B} \\cdot \\sin B + \\frac{c_0}{\\sin C} \\cdot \\sin (A_0 - C) \\right ]\n\\]\n\nFinally,\n\n\\[\n\\alpha = \\arctan \\frac{b_0 \\cos B \\sin C + c_0 \\sin B \\cos (A_0 - C)} {b_0 \\sin B \\sin C + c_0 \\sin B \\sin (A_0 - C)}\n\\]\n\nIt is easy to verify that this value is valid (i.e. it can be an angle in a triangle), and it is indeed a point of maximum for \\(f(\\alpha)\\) (i.e. \\(f''(\\alpha) < 0\\); a geometric argument would work as well).\n\nNow to answer the \"and construct it\" at the end of the statement of the problem, we will show that everything we did is constructible, rather than describe a lengthy, boring step by step construction. Recall that we already discussed that we can construct \\(\\triangle ABC\\) if we know \\(\\alpha\\). We just need to show that we can construct \\(\\alpha\\). We can construct differences of angles, and given an angle we can construct two segments whose ratio is the \\(\\sin\\) (or the \\(\\cos\\), or the \\(\\tan\\)) of the given angle, and vice-versa. Given three segments \\(x, y, z\\) we can construct the segment \\(x \\cdot \\frac{y}{z}\\). Thus, the expression giving \\(\\alpha\\) is constructible.\n\n[Solution by pf02, September 2024]\n\n## Remarks (added by pf02, September 2024)\n\n1. In solution 2, I show where I use the condition that the triangles are assumed to be acute. The first solution does not make this clear. It seems intuitively true that the condition is not necessary. In other words, a circumscribed triangle \\(\\triangle ABC\\) exists, and it can be constructed even when one or both of the given triangles \\(\\triangle A_0B_0C_0, \\triangle A_1B_1C_1\\) are right or obtuse. The condition seems to be necessary only for simplifying the proof. In the general case, we may need to rearrange the labeling of the vertices.\n\n2. Solution 1 is elegant, even though its presentation would have benefited a lot from some editing. It gives a nice geometric insight into the problem. Solution 2 does not give much geometric insight, but it is computationally very explicit. The two solutions are so different that it is worth taking a little time to show that they are equivalent. I will outline the steps of the computation.\n\nTo show that they are equivalent, put the pictures together:\n\n\\[\nProb 1967 4 fig2.png\n\\]\n\nThe idea from Solution 1 is to construct the circles circumscribed to \\(\\triangle A_0C_0B\\) and \\(\\triangle A_0B_0C\\). They intersect at \\(P\\). Consider the angles \\(\\beta, \\gamma\\) shown on the picture. We have\n\n\\[\n\\frac{A_0P}{\\sin \\beta} = \\frac{A_0B}{sin (\\angle A_0PB)} = \\frac{A_0B}{\\sin (\\angle A_0C_0B)} = \\frac{b_0}{\\sin B}\n\\]\n\nThe first equality follows from the law of sines in \\(\\triangle A_0PB\\), the second follows from the equality of angles in a circle spanning the same arc, the third follows from the law of sines in \\(\\triangle A_0C_0B\\). We gat \\(A_0P = \\frac{b_0 \\sin \\beta}{\\sin B}\\). Similarly, we get \\(A_0P = \\frac{c_0 \\sin \\gamma}{\\sin C}\\) from the other side of \\(A_0P\\).\n\nOn the other hand, we have\n\n\\[\n\\beta + \\gamma = \\pi - \\angle (CPB) = \\pi - [2\\pi - \\angle (CPB_0) - \\angle (B_0PC_0) - \\angle (C_0PB)] =\n\\]\n\n\\[\n-\\pi + \\angle (CA_0B_0) + (\\pi - \\angle A) + \\angle (C_0A_0B) = -\\pi + (\\pi - \\angle A) + (\\pi - \\angle A_0) = \\pi - \\angle A - \\angle A_0\n\\]\n\nSolve for \\(\\gamma\\) and substitute in \\(\\frac{b_0 \\sin \\beta}{\\sin B} = \\frac{c_0 \\sin \\gamma}{\\sin C}\\) to obtain one equation in \\(\\beta\\).\n\nNow use the fact that \\(PB\\) is a diameter to write an equation between \\(\\alpha\\) and \\(\\beta\\):\n\n\\[\n\\alpha = \\pi - \\angle B - \\angle (BC_0A_0) = \\pi - \\angle B - \\angle (BPA_0) = \\pi - \\angle B - (\\frac{\\pi}{2} - \\beta) = \\frac{\\pi}{2} - \\angle B + \\beta\n\\]\n\nSolve for \\(\\beta\\) in terms of \\(\\alpha\\) and substitute in the equation in \\(\\beta\\) obtained before. After some straightforward computations we get exactly the equation in \\(\\alpha\\) we had in Solution 2. The interested reader can easily work out the details, they are just straightforward algebraic computations."
] |
IMO-1967-5
|
https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_5
|
Let \(a_1,\ldots,a_8\) be reals, not all equal to zero. Let
\[
c_n = \sum^8_{k=1} a^n_k
\]
for \(n=1,2,3,\ldots\). Given that among the numbers of the sequence \((c_n)\), there are infinitely many equal to zero, determine all the values of \(n\) for which \(c_n = 0.\)
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[
"\\(c_n\\) must be zero for all odd \\(n\\).\n\nProof: WLOG suppose that \\(a_1 \\geq a_2 \\geq ... \\geq a_8\\). If \\(a_1+a_8 > 0\\) then for sufficiently high odd \\(n\\), \\(c_n\\) will be dominated by \\(a_1\\) alone i.e. it will always be positive. Similarly if \\(a_1+a_8 < 0\\); hence \\(a_1=-a_8\\). Now for odd \\(n\\) these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd \\(n\\). Since some \\(a_i\\) is nonzero \\(c_n > 0\\) for even \\(n\\).\n\nThe above solution was written by Fiachra and can be found here: [1]\n\n\\(\\textbf{Note:}\\hspace{4000pt}\\) Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently."
] |
IMO-1967-6
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https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_6
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In a sports contest, there were \(m\) medals awarded on \(n\) successive days \((n > 1)\). On the first day, one medal and \(\frac{1}{7}\) of the remaining \(m - 1\) medals were awarded. On the second day, two medals and \(\frac{1}{7}\) of the now remaining medals were awarded; and so on. On the n-th and last day, the remaining \(n\) medals were awarded. How many days did the contest last, and how many medals were awarded altogether?
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[
"This is not a particularly elegant solution, but if you start from 1 and go all the way in a clever method, by only guessing those that are 1 more than a multiple of 7, you arrive at the answer of 36.\n\n## Comment (added by pf02, August 2024)\n\nIndeed, as the author says, the above is not an elegant solution. Also, it does not give any insight into the uniqueness of the answer to the problem. I would also comment that choosing to verify the statement only for multiples of \\(7\\) plus one is not a \"clever method\", it is quite an obvious thing to do. And, note that when the author says \"arrive at the answer of \\(36\\)\", they mean \"the contest lasted for \\(6\\) days, and \\(36\\) medals were awarded\".\n\nBelow, I will give another solution, which is more in the spirit and style of contemporary problem solving.",
"Denote \\(m_0 = m\\). Let \\(m_k\\) be the number of medals left on day \\(k\\) after the medals for the day have been awarded. The problem says that \\(m_k = m_{k-1} - k - \\frac{m_{k-1} - k}{7}\\) for \\(k = 1, 2, \\dots, (n - 1)\\), and \\(m_{n - 1} = n\\), and \\(n > 1\\). Simplify the recursive relation and get \\(m_k = \\frac{6}{7}m_{k - 1} - \\frac{6}{7}k\\).\n\nWe will now get an explicit formula for \\(m_k\\).\n\n\\[\nm_1 = \\frac{6}{7}m - \\frac{6}{7} \\cdot 1\n\\]\n\n\\[\nm_2 = \\frac{6}{7}m_1 - \\frac{6}{7} \\cdot 2 = \\left(\\frac{6}{7}\\right)^2m - \\left(\\frac{6}{7}\\right)^2 \\cdot 1 - \\frac{6}{7} \\cdot 2\n\\]\n\n\\[\nm_3 = \\frac{6}{7}m_2 - \\frac{6}{7} \\cdot 3 = \\left(\\frac{6}{7}\\right)^3m - \\left(\\frac{6}{7}\\right)^3 \\cdot 1 - \\left(\\frac{6}{7}\\right)^2 \\cdot 2 + \\frac{6}{7} \\cdot 3\n\\]\n\n. . . . . . . .\n\n\\[\nm_{n - 1} = \\frac{6}{7}m_{n - 2} - \\frac{6}{7} \\cdot (n - 1) = \\left(\\frac{6}{7}\\right)^{n - 1}m - \\left(\\frac{6}{7}\\right)^{n - 1} \\cdot 1 - \\left(\\frac{6}{7}\\right)^{n - 2} \\cdot 2 - \\dots - \\left(\\frac{6}{7}\\right)^2 \\cdot (n - 2) - \\frac{6}{7} \\cdot (n - 1)\n\\]\n\nTo put it in a shorter way,\n\n\\[\nm_{n - 1} = \\left(\\frac{6}{7}\\right)^{n - 1}m - \\sum_{k = 1}^{n - 1} \\left(\\frac{6}{7}\\right)^{n - k} \\cdot k\n\\]\n\n(The reader who is not happy with having obtained this result by having observed the pattern for \\(m_k\\) can easily verify it by induction.)\n\nNote that the sum in the equality above is the sum of the arithmetic-geometric sequence formed by the geometric progression \\(\\{\\ \\left(\\frac{6}{7}\\right)^{n - 1}, \\left(\\frac{6}{7}\\right)^{n - 2}, \\dots, \\left(\\frac{6}{7}\\right)^2, \\frac{6}{7}\\ \\}\\) (with common ratio \\(\\frac{7}{6}\\) (note that the progression starts from a number smaller than \\(1\\) and is increasing)), and the arithmetic progression \\(\\{\\ 1, 2, \\dots, (n - 2), (n - 1)\\ \\}\\) (with common difference \\(1\\)). The formula for the sum of the terms of such sequences is reasonably well known (and not difficult to prove) (see for example Arithmetico-geometric_series or https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence).\n\nApplying the formula for the sum of the arithmetic-geometric sequence, and using the fact that \\(m_{n - 1} = n\\), we get\n\n\\[\nn = \\left(\\frac{6}{7}\\right)^{n - 1}m - \\frac{\\left(\\frac{6}{7}\\right)^{n - 1} \\cdot 1 - 1 \\cdot n}{1 - \\frac{7}{6}} -\\frac{\\frac{7}{6}}{\\left(1 - \\frac{7}{6}\\right)^2} \\left(\\left(\\frac{6}{7}\\right)^{n - 1} - 1 \\right)\n\\]\n\nNow do all the simplifications, and rearrange terms. We have\n\n\\[\nm - 36 = 7(n - 6) \\left(\\frac{7}{6}\\right)^{n - 1}\n\\]\n\nIn order for \\(m\\) to be an integer, we have to have that \\(\\frac{n - 6}{6^{n - 1}}\\) is an integer. This is clearly not the case when \\(n = 2, 3, 4, 5\\). It is an integer when \\(n = 6\\), in which case \\(m = 36\\). The fraction is not an integer for \\(n = 7\\), and even less so for \\(n > 7\\) since the exponential at the denominator increases much faster than the linear function at the numerator.\n\nThus, \\(m = 36, n = 6\\) is the only solution to the problem.\n\n(Solution by pf02, August 2024)"
] |
IMO-1968-1
|
https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_1
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Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.
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[
"In triangle \\(ABC\\), let \\(BC=a\\), \\(AC=b\\), \\(AB=c\\), \\(\\angle ABC=\\alpha\\), and \\(\\angle BAC=2\\alpha\\). Using the Law of Sines gives that\n\n\\[\n\\frac{b}{\\sin{\\alpha}}=\\frac{a}{\\sin{2\\alpha}}\\Rightarrow \\frac{\\sin{2\\alpha}}{\\sin{\\alpha}}=2\\cos{\\alpha}=\\frac{a}{b}\n\\]\n\nTherefore \\(\\cos{\\alpha}=\\frac{a}{2b}\\). Using the Law of Cosines gives that\n\n\\[\n\\cos{\\alpha}=\\frac{a^2+c^2-b^2}{2ac}=\\frac{a}{2b}\n\\]\n\nThis can be simplified to \\(a^2c=b(a^2+c^2-b^2)\\). Since \\(a\\), \\(b\\), and \\(c\\) are positive integers, \\(b|a^2c\\). Note that if \\(b\\) is between \\(a\\) and \\(c\\), then \\(b\\) is relatively prime to \\(a\\) and \\(c\\), and \\(b\\) cannot possibly divide \\(a^2c\\). Therefore \\(b\\) is either the least of the three consecutive integers or the greatest.\n\nAssume that \\(b\\) is the least of the three consecutive integers. Then either \\(b|b+2\\) or \\(b|(b+2)^2\\), depending on if \\(a=b+2\\) or \\(c=b+2\\). If \\(b|b+2\\), then \\(b\\) is 1 or 2. \\(b\\) couldn't be 1, for if it was then the triangle would be degenerate. If \\(b\\) is 2, then \\(b(a^2+c^2-b^2)=42=a^2c\\), but \\(a\\) and \\(c\\) must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore \\(b\\) cannot divide \\(b+2\\), and so \\(b\\) must divide \\((b+2)^2\\). If \\(b|(b+2)^2\\) then \\(b|(b+2)^2-b^2-4b=4\\), so \\(b\\) is 1, 2, or 4. Clearly \\(b\\) cannot be 1 or 2, so \\(b\\) must be 4. Therefore \\(b(a^2+c^2-b^2)=180=a^2c\\). This shows that \\(a=6\\) and \\(c=5\\), and the triangle has sides that measure 4, 5, and 6.\n\nNow assume that \\(b\\) is the greatest of the three consecutive integers. Then either \\(b|b-2\\) or \\(b|(b-2)^2\\), depending on if \\(a=b-2\\) or \\(c=b-2\\). \\(b|b-2\\) is absurd, so \\(b|(b-2)^2\\), and \\(b|(b-2)^2-b^2+4b=4\\). Therefore \\(b\\) is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so \\(b\\) cannot be the greatest of the three consecutive integers. This shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6. \\(\\blacksquare\\)",
"(Note: this proof is an expansion by pf02 of an outline of a solution posted here before.)\n\nIn a given triangle \\(ABC\\), let \\(A=2B\\), \\(\\implies C=180-3B\\), and \\(\\sin C=\\sin 3B\\). Then\n\n\\[\n\\sin ^2 A = \\sin ^2 2B = 2 \\sin B \\cos B \\sin 2B = \\sin B(\\sin B + \\sin 3B) = \\sin B(\\sin B + \\sin C)\n\\]\n\nHence,\n\n\\[\na^2 = b(b + c)\\ (*)\n\\]\n\nIndeed, we know from the Law of Sines that\n\n\\(\\frac{\\sin A}{a} = \\frac{\\sin B}{b} = \\frac{\\sin C}{c}\\).\n\nDenote this ratio by \\(r\\); we have \\(\\sin A = ra, \\sin B = rb, \\sin C = rc\\). Substitute in \\(\\sin ^2 A = \\sin B(\\sin B + \\sin C)\\) and simplify by \\(r^2\\). We get \\((*)\\).\n\nAt this point, notice that \\((*)\\) is equivalent to the equality \\(a^2c = b(a^2 + c^2 - b^2)\\) from Solution 1. Indeed, the latter can be rewritten as \\(a^2(c - b) = b(c + b)(c - b)\\), and we know that \\(c \\ne b\\). So we could simply quote the fact (proven in Solution 1) that if \\(a, b, c\\) are consecutive integers and \\(a^2 = b(c + b)\\), then \\(b = 4, c = 5, a = 6\\) is the only solution which could be the sides of a triangle.\n\nFor the sake of completeness, and for fun, I give a slightly different proof here.\n\nWe have six possibilities, depending on how the three consecutive numbers are ordered. The six possibilities are:\n\n1: \\(\\ \\ a = b - 2, c = b - 1, b\\)\n\n2: \\(\\ \\ c = b - 2, a = b - 1, b\\)\n\n3: \\(\\ \\ a = b - 1, b, c = b + 1\\)\n\n4: \\(\\ \\ c = b - 1, b, a = b + 1\\)\n\n5: \\(\\ \\ b, a = b + 1, c = b + 2\\)\n\n6: \\(\\ \\ b, c = b + 1, a = b + 2\\)\n\nFor each case, we could substitute \\(a, c\\) in \\((*)\\), get an equation in \\(b\\), solve it, and get all the possible solutions. As a shortcut, notice that \\((*)\\) implies that \\(b|a^2\\). If \\(a, b\\) are consecutive integers, then they are relatively prime, so \\(b|a^2\\) can not be true unless \\(b = 1\\). In this case the triangle would have sides \\(1, 2, 3\\), which is impossible. This eliminates cases 2, 3, 4 and 5.\n\nIn case 1, \\((*)\\) becomes\n\n\\((b - 2)^2 = b(b + b - 1)\\), or \\(b^2 + 3b - 4 = 0\\).\n\nThis has solutions \\(1, -4\\). The value \\(b = -4\\) is impossible. The value \\(b = 1\\) yields \\(a = -1, c = 0\\), which is impossible.\n\nIn case 6, \\((*)\\) becomes\n\n\\((b + 2)^2 = b(b + b + 1)\\), or \\(b^2 - 3b - 4 = 0\\).\n\nThe solutions are \\(-1, 4\\). The value \\(b = -1\\) is impossible. Thus, we get the unique triangle \\(a = 6, b = 4, c = 5\\).",
"NO TRIGONOMETRY!!!\n\nLet \\(a, b, c\\) be the side lengths of a triangle in which \\(\\angle C = 2\\angle B.\\)\n\nExtend \\(AC\\) to \\(D\\) such that \\(CD = BC = a.\\) Then \\(\\angle CDB = \\frac{\\angle ACB}{2} = \\angle ABC\\), so \\(ABC\\) and \\(ADB\\) are similar by AA Similarity. Hence, \\(c^2 = b(a+b)\\). Then proceed as in Solution 2, as only algebraic manipulations are left.",
"Note: Adding this 4th solution is justified by the fact that it is extremely straightforward, and by the fact that it shows that there are exactly two triangles for which the sides differ by \\(1\\) (i.e. they are \\(x, x + 1, x + 2\\) for some \\(x\\)), and the condition on the angles is satisfied (that one is twice the other). But only one of the solutions has \\(x\\) integer.\n\nSo, let us start by assuming that two angles are \\(\\alpha, 2\\alpha\\) and the sides are \\(x, x + 1, x + 2\\). We will want to apply the Law of Sines:\n\n\\[\n\\frac{x}{\\sin A} = \\frac{x + 1}{\\sin B} = \\frac{x + 2}{\\sin C}\n\\]\n\nThe angles \\(A, B, C\\) should be so that \\(\\sin A \\le \\sin B \\le \\sin C\\), but we don't know how to map \\(\\{A, B, C\\}\\) to \\(\\{\\alpha, 2\\alpha, \\pi - 3\\alpha\\}\\). One thing we know, is that \\(\\sin \\alpha < \\sin 2\\alpha\\). Indeed, if \\(\\alpha \\le \\pi/4\\) the inequality is true because \\(\\sin\\) is increasing on \\([0, \\pi/2]\\). Now note that \\(\\alpha < \\pi/3\\) since otherwise \\(\\pi - 3\\alpha\\) could not be the angle of a triangle. So, if \\(\\pi/4 < \\alpha < \\pi/3\\) then \\(\\pi/2 < 2\\alpha < 2\\pi/3\\) and \\(\\sin \\alpha < \\sqrt{3}/2 < \\sin 2\\alpha\\).\n\nThat means we will have to consider three possibilities:\n\n1: \\(\\ \\ \\frac{x}{\\sin \\alpha} = \\frac{x + 1}{\\sin 2\\alpha} = \\frac{x + 2}{\\sin (\\pi - 3\\alpha)}\\)\n\n2: \\(\\ \\ \\frac{x}{\\sin \\alpha} = \\frac{x + 1}{\\sin (\\pi - 3\\alpha)} = \\frac{x + 2}{\\sin 2\\alpha}\\)\n\n3: \\(\\ \\ \\frac{x}{\\sin (\\pi - 3\\alpha)} = \\frac{x + 1}{\\sin \\alpha} = \\frac{x + 2}{\\sin 2\\alpha}\\)\n\nUsing the identities \\(\\sin (\\pi - \\theta) = \\sin \\theta, \\sin 2\\theta = 2\\sin \\theta \\cos \\theta\\) and \\(\\sin 3\\theta = 3\\sin \\theta - 4\\sin^3 \\theta = \\sin \\theta\\ (4\\cos^2 \\theta - 1)\\) and simplifying by \\(\\sin \\alpha\\) the three cases become\n\n1: \\(\\ \\ x = \\frac{x + 1}{2\\cos \\alpha} = \\frac{x + 2}{4\\cos^2 \\alpha - 1}\\)\n\n2: \\(\\ \\ x = \\frac{x + 1}{4\\cos^2 \\alpha - 1} = \\frac{x + 2}{2\\cos \\alpha}\\)\n\n3: \\(\\ \\ \\frac{x}{4\\cos^2 \\alpha - 1} = x + 1 = \\frac{x + 2}{2\\cos \\alpha}\\)\n\nEach case is a system of two equations in two unknowns, \\(x, \\cos \\alpha\\). We will solve each system, obtain all possible solutions, and chose those values for which \\(x, x + 1, x + 2\\) and \\(\\alpha, 2\\alpha, \\pi - 3\\alpha\\) can be the sides and angles of a triangle.\n\nCase 1: Compute \\(x\\) from \\(x = \\frac{x + 1}{2\\cos \\alpha}\\). We get \\(x = \\frac{1}{2\\cos \\alpha - 1}\\). Substitute \\(x\\) in \\(x = \\frac{x + 2}{4\\cos^2 \\alpha - 1}\\). After doing all the computations we get \\(4\\cos^2 \\alpha - 4\\cos \\alpha = 0\\). The roots are \\(\\cos \\alpha = 0\\) and \\(\\cos \\alpha = 1\\). None are acceptable if \\(\\alpha\\) is an angle of a triangle. So case 1 yields no solutions.\n\nCase 2: Compute \\(x\\) from \\(x = \\frac{x + 2}{2\\cos \\alpha}\\). We get \\(x = \\frac{2}{2\\cos \\alpha - 1}\\). Substitute \\(x\\) in \\(x = \\frac{x + 1}{4\\cos^2 \\alpha - 1}\\). After doing all the computations we get \\(8\\cos^2 \\alpha - 2\\cos \\alpha - 3 = 0\\). The solutions are \\(\\cos \\alpha = \\frac{3}{4}\\) and \\(\\cos \\alpha = -\\frac{1}{2}\\). Only \\(\\cos \\alpha = \\frac{3}{4}\\) is acceptable, and it yields \\(x = \\frac{2}{\\frac{6}{4} - 1} = 4\\). Thus \\(4, 5, 6\\) is a possible solution to the problem.\n\nCase 3: Compute \\(x\\) from \\(x + 1 = \\frac{x + 2}{2\\cos \\alpha}\\). We get \\(x = \\frac{2 - 2\\cos \\alpha}{2\\cos \\alpha - 1}\\). Substitute \\(x\\) in \\(x + 1 = \\frac{x}{4\\cos^2 \\alpha - 1}\\). After doing all the computations we get \\(4\\cos^2 \\alpha + 2\\cos \\alpha - 3 = 0\\). The roots are \\(\\cos \\alpha = \\frac{-1 \\pm \\sqrt{13}}{4} \\approx 0.65, -1.15\\). The positive value is for an \\(\\alpha\\) acceptable as an angle in a triangle (it is a little over \\(\\pi/4\\)), and yields \\(x = \\frac{\\sqrt{13} + 1}{2}\\). We can easily verify that \\(x, x + 1, x + 2\\) can be the sides of a triangle. (Indeed, they are \\(\\frac{\\sqrt{13} + 1}{2}, \\frac{\\sqrt{13} + 3}{2}, \\frac{\\sqrt{13} + 5}{2}\\) and \\(\\frac{\\sqrt{13} + 1}{2} + \\frac{\\sqrt{13} + 3}{2} > \\frac{\\sqrt{13} + 5}{2}\\)). However, they are not integer, so are not solutions to the problem.\n\nThe only solution to the problem is the triangle with sides \\(4, 5, 6\\) from case 2.\n\n[Solution by pf02, August 2024]"
] |
IMO-1968-2
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https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_2
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Find all natural numbers \(x\) such that the product of their digits (in decimal notation) is equal to \(x^2 - 10x - 22\).
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[
"Let the decimal expansion of \\(x\\) be \\(\\overline{d_1d_2d_3\\dots d_n}\\), where \\(d_i\\) are base-10 digits. We then have that \\(x\\geq d_1\\cdot 10^{n-1}\\). However, the product of the digits of \\(x\\) is \\(d_1d_2d_3\\dots d_n\\leq d_1\\cdot 10\\cdot 10\\dots 10=d_1\\cdot 10^{n-1}\\), with equality only when \\(x\\) is a one-digit integer. Therefore the product of the digits of \\(x\\) is always at most \\(x\\), with equality only when \\(x\\) is a base-10 digit. This implies that \\(x^2-10x-22\\leq x\\), so \\(x^2-11x-22\\leq 0\\). Every natural number from 1 to 12 satisfies this inequality, so we only need to check these possibilities. It is easy to rule out 1 through 11, since \\(x^2-10x-22<0\\) for those values. However, \\(12^2-10\\cdot 12-22=2\\), which is the product of the digits of 12. Therefore \\(\\boxed{12}\\) is the only natural number with the desired properties. \\(\\blacksquare\\)",
"It is pretty obvious that \\(x\\) cannot be three digits or more, because then \\(x^2 - 10x - 22\\) is way too big.\n\nWrite \\(x = 10a + b\\) where \\(a\\) and \\(b\\) are digits satisfying \\(0 \\leq a, b < 10\\). Then, we can use SFFT:\n\n\\[\n(10a + b)^2 - 10(10a + b) - 22 = ab\n\\]\n\n\\[\n(10a + b)^2 - 10(10a + b) - 24 = ab - 2\n\\]\n\n\\[\n(10a + b + 2)(10a + b - 12) = ab - 2.\n\\]\n\nWe have\n\n\\[\n(10a + b + 2)(10a + b - 12) \\geq (10a + 2)(10a - 12) = 100a^2 - 100a + 24 = 100(a^2 - a) + 24.\n\\]\n\nIt is therefore clear that \\(a\\) must be either \\(0\\) or \\(1\\). We can then split into two cases:\n\n\\[\n\\mathbf{a = 0:}\n\\]\n\nWe have \\((b + 2)(b - 12) = -2\\) or \\(b^2 - 10b - 22 = 0\\), which is only satisfied when \\(b = -2\\) or \\(12\\).\n\n\\[\n\\mathbf{a = 1:}\n\\]\n\nWe have \\((b + 12)(b - 2) = b - 2\\). This is only satisfied when \\(b = 2\\), or \\(b + 12 = 0\\). Therefore, \\(b = 2\\), and so \\(x = \\boxed{12}.\\square\\)\n\n~mathboy100",
"Let, \\(x^2-10x-22=y\\)\n\n\\[\n\\implies x^2-10+25-47=y\n\\]\n\n\\[\n\\implies (x-5)^2=47+y\n\\]\n\nNow note that, if \\(p\\) is a prime such that \\(p|y\\) then \\(7\\geq p\\).\n\nThat means, \\(y=2^a*3^b*5^c*7^d\\)\n\nBut, \\(a^2 \\not\\equiv 2 (mod3), a^2 \\not\\equiv 2 (mod5), a^2 \\not\\equiv 5 (mod7)\\) which means \\(3,5,7\\) don't divivde \\((x-5)^2-47=y.\\)\n\nSo, \\(y=2^a\\) and \\(y+17=2^a+47=(x-5)^2\\)\n\nIt is easy to see that \\(a\\) has one solution and that is \\(2.\\)( Prove it by contradiction)\n\nSo, \\((x-5)^2=47+2=49\\)\n\n\\[\n\\implies x=12\n\\]\n\n\\[\n\\blacksquare\n\\]\n\n## Remarks (added by pf02, August 2024)\n\nSolutions 2 and 3 are not satisfactory. In fact, they can not be called solutions, since they make statements which are not proven. Specifically:\n\nIn Solution 2, the author writes \"It is pretty obvious that \\(x\\) cannot be three digits or more, because then \\(x^2 - 10x - 22\\) is way too big.\" This is intuitively true, but not obvious at all. As a crucial step in the solution, it should be proven. Later, the author states\n\n\"\\((10a + b + 2)(10a + b - 12) \\ge \\cdots = 100(a^2 - a) + 24\\). It is therefore clear that \\(a\\) must be either \\(0\\) or \\(1\\)\".\n\nFirst, the last term should be \\(-24\\) instead of \\(24\\). Either way, the conclusion about \\(a\\) is not clear at all. As a second crucial step in the solution, it should be proven.\n\nIn Solution 3, the notation and writing are very confusing. However, a diligent reader can make sense of them. But in this solution as well, there are statements which beg for a proof. The first such statement is\n\n\"\\(a^2 \\not\\equiv 2\\ (mod\\ 3), a^2 \\not\\equiv 2\\ (mod\\ 5), a^2 \\not\\equiv 5\\ (mod\\ 7)\\) which means \\(3, 5, 7\\) don't divide \\((x - 5)^2 - 47 = y\\)\".\n\n(When writing \\(a^2\\) the author means the square of an arbitrary natural number, not the square of the number\\(a\\) used in the line above this statement.) Neither the modulo statements, nor the conclusion are obvious; proofs should be given. The second unproven statement is\n\n\"\\(2^a + 47 = (x - 5)^2\\). It is easy to see that \\(a\\) has one solution and that is \\(2\\). (Prove it by contradiction.)\"\n\n(The author means \"the equation has a unique solution for \\(a\\)\".) The conclusion about the uniqueness of \\(a\\) is not easy to see, and as a crucial step in the solution, it should be proven.\n\nBelow, I will give corrected, complete, and somewhat simplified versions of these two solutions.",
"Let the decimal expansion of \\(x\\) be \\(\\overline{d_1d_2d_3\\dots d_n}\\), where \\(d_i\\) are base-10 digits. Let us prove first that \\(n \\le 2\\).\n\nUsing \\(x^2 - 10x -22 = (x - 5)^2 - 47\\) and the fact that this expression equals the product \\(d_1d_2d_3 \\dots d_n\\) we have that \\((x - 5)^2 - 47 \\le 9^n\\). Since \\(x\\) has \\(n\\) digits, we also have \\((x - 5)^2 - 47 \\ge (10^{n - 1} - 5)^2 - 47\\). Thus, we have \\(9^n \\ge (10^{n - 1} - 5)^2 - 47\\). We will show that this can not be true for \\(n \\ge 3\\).\n\nDivide by \\(9^{n - 1}\\), rearrange factors and terms, and get \\(\\left[\\left(\\frac{10}{3}\\right)^{n - 1} - \\frac{5}{3^{n - 1}}\\right]^2 \\le 9 + \\frac{47}{9^{n - 1}} < 10\\). Again using \\(n \\ge 3\\), we get \\(\\left(\\frac{10}{3}\\right)^{n - 1} < \\sqrt{10} + \\frac{5}{3^{n - 1}} < 5\\). This is false for \\(n \\ge 3\\), so \\(n =\\) the number of digits of \\(x\\) can not be more than \\(2\\).\n\nLet then \\(x = 10a + b\\), with \\(0 \\le a, b \\le 9\\) the digits of \\(x\\). We have \\(x^2 - 10x - 22 = ab \\le 81\\). By solving the quadratic equation \\(x^2 - 10x - 103 = 0\\) we see that the inequality is true when \\(0 \\le x \\le 16\\).\n\nAt this point, we could simply check which of the numbers \\(x = 0, 1, \\dots, 16\\) has the product of its digits equal to \\(x^2 - 10x -22\\). It would turn out that \\(x = 12\\) is the only one. But we can take a short cut by writing \\((10a + b)^2 - 10(10a + b) - 22 = ab\\). and using that \\(a = 0\\) or \\(a = 1\\). In the case \\(a = 0\\) we get the equation \\(b^2 - 10b - 22 = 0\\) in \\(b\\) which has no integer solutions. In the case \\(a = 1\\) we get the equation \\(b^2 + 9b - 22 = 0\\) in \\(b\\), which has solutions \\(-11\\) and \\(2\\). Since \\(b\\) is a digit, \\(b = 2\\) is the only acceptable solution, so \\(x = 12\\).",
"Let \\(y = x^2 - 10x - 22\\). Since \\(y\\) is a product of digits, the only prime factors of \\(y\\) can be \\(2, 3, 5, 7\\). Write \\(y = (x - 5)^2 - 47\\).\n\nNote that if \\(A\\) is a natural number then \\(A^2 \\equiv 0\\) or \\(1\\ (mod\\ 3)\\). Indeed, \\(A = 3k\\) or \\(A = 3k + 1\\) or \\(A = 3k + 2\\) for some \\(k\\). By writing out \\(A^2\\) the statement follows immediately. Then \\((x - 5)^2 - 47 \\equiv 1\\) or \\(2\\ (mod\\ 3)\\), so \\(3\\) can not be a factor of \\(y\\).\n\nSimilarly, if \\(A\\) is a natural number then \\(A^2 \\equiv 0, 1\\) or \\(4\\ (mod\\ 5)\\). Then \\((x - 5)^2 - 47 \\equiv 3, 4\\) or \\(2\\ (mod\\ 5)\\), so \\(5\\) can not be a factor of \\(y\\).\n\nAnd finally if \\(A\\) is a natural number then \\(A^2 \\equiv 0, 1, 2\\) or \\(4\\ (mod\\ 7)\\). Then \\((x - 5)^2 - 47 \\equiv 2, 3, 4\\) or \\(6\\ (mod\\ 7)\\), so \\(7\\) can not be a factor of \\(y\\).\n\nThe only possible prime factor of \\(y\\) is \\(2\\). So \\(2^z = (x - 5)^2 - 47\\) for some \\(z \\ge 0\\). Write this as \\(2^z -2 = (x - 5)^2 - 49\\), or \\(2(2^{z - 1} - 1) = (x - 12)(x + 2)\\).\n\nClearly \\(z = 0\\) is not acceptable because the equation in \\(x\\) has no integer solutions.\n\nIf \\(z > 1\\) the left hand side is divisible by \\(2\\), but not by \\(4\\). The right hand side is not divisible by \\(2\\) if \\(x\\) is odd, and is divisible by \\(4\\) if \\(x\\) is even. So \\(z > 1\\) yields no solutions.\n\nIf \\(z = 1\\), the equation becomes \\(x^2 - 10x - 24 = 0\\) which has solutions \\(-2, 12\\). The only acceptable solution is \\(x = 12\\)."
] |
IMO-1968-3
|
https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_3
|
Consider the system of equations
\[
ax_1^2 + bx_1 + c = x_2
\]
\[
ax_2^2 + bx_2 + c = x_3
\]
\[
\cdots
\]
\[
ax_{n-1}^2 + bx_{n-1} + c = x_n
\]
\[
ax_n^2 + bx_n + c = x_1
\]
with unknowns \(x_1, x_2, \cdots, x_n\) where \(a, b, c\) are real and \(a \neq 0\). Let \(\Delta = (b - 1)^2 - 4ac\). Prove that for this system
(a) if \(\Delta < 0\), there is no solution,
(b) if \(\Delta = 0\), there is exactly one solution,
(c) if \(\Delta > 0\), there is more than one solution.
|
[
"Adding the \\(n\\) equations together yields\n\n\\[\n\\sum_{i=1}^{n} (ax_i^2+(b-1)x_i+c)=0\n\\]\n\nLet \\(s_i=ax_i^2+(b-1)x_i+c\\).\n\n(a) If \\(\\Delta<0\\), then there is no solution to the quadratic equation \\(ax^2+(b-1)x+c\\), as the determinant is negative. This implies that either \\(s_i>0\\) for all \\(i\\), or \\(s_i<0\\) for all \\(i\\). In either case the above summation cannot be 0, which implies that there are no solutions to the given system of equations. \\(\\blacksquare\\)\n\n(b) If \\(\\Delta=0\\), then there is exactly one solution to the quadratic equation \\(ax^2+(b-1)x+c\\) (let it be \\(r\\)), and either \\(s_i\\geq 0\\) for all \\(i\\), or \\(s_i\\leq 0\\) for all \\(i\\). The only way that the above summation is 0 is if \\(s_i=0\\) for all \\(i\\). As there is exactly one \\(x_i\\) that makes \\(s_i=0\\) (namely \\(x_i=r\\)), then the only possible solution to the system of equations is \\((x_1,x_2,\\dots , x_n)=(r, r, \\dots , r)\\). It's not hard to show that this works, so when \\(\\Delta=0\\) the system of equations has exactly one solution. \\(\\blacksquare\\)\n\n(c) If \\(\\Delta>1\\), then there are exactly two solutions to the quadratic equation \\(ax^2+(b-1)x+c\\). Let the roots be \\(r_1\\) and \\(r_2\\). If \\(x_i=r_1\\) for all \\(i\\), then \\(ax_i^2+bx_i+c=ar_1^2+br_1+c=r_1=x_{i+1}\\). This shows that \\((x_1,x_2,\\dots ,x_n)=(r_1,r_1,\\dots, r_1)\\) is a solution. We can show that \\((x_1,x_2,\\dots ,x_n)=(r_2,r_2,\\dots, r_2)\\) is another solution using the same reasoning, which shows that the equation has more than one solution. \\(\\blacksquare\\)"
] |
IMO-1968-4
|
https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_4
|
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle.
|
[
"Let the edges of one of the faces of the tetrahedron have lengths \\(a\\), \\(b\\), and \\(c\\). Let \\(d\\), \\(e\\), and \\(f\\) be the lengths of the sides that are not adjacent to the sides with lengths \\(a\\), \\(b\\), and \\(c\\), respectively.\n\nWithout loss of generality, assume that \\(\\max(a,b,c,d,e,f)=a\\). I shall now prove that either \\(b+f>a\\) or \\(c+e>a\\), by proving that if \\(b+f\\leq a\\), then \\(c+e>a\\).\n\nAssume that \\(b+f\\leq a\\). The triangle inequality gives us that \\(e+f>a\\), so \\(e\\) must be greater than \\(b\\). We also have from the triangle inequality that \\(b+c>a\\). Therefore \\(e+c>b+c>a\\). Therefore either \\(b+f>a\\) or \\(c+e>a\\).\n\nIf \\(b+f>a\\), then the vertex where the sides of length \\(a\\), \\(b\\), and \\(f\\) meet satisfies the given condition. If \\(c+e>a\\), then the vertex where the sides of length \\(a\\), \\(c\\), and \\(e\\) meet satisfies the given condition. This proves the statement. \\(\\blacksquare\\)"
] |
IMO-1968-5
|
https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_5
|
Let \(f\) be a real-valued function defined for all real numbers \(x\) such that, for some positive constant \(a\), the equation
\[
f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}
\]
holds for all \(x\).
(a) Prove that the function \(f\) is periodic (i.e., there exists a positive number \(b\) such that \(f(x + b) = f(x)\) for all \(x\)).
(b) For \(a = 1\), give an example of a non-constant function with the required properties.
|
[
"(a) Since\n\n\\[\nf(x+a) \\ge \\frac{1}{2}\n\\]\n\nis true for any \\(x\\), and\n\n\\[\nf(x+a)(1-f(x+a)) = \\frac{1}{4} - (f(x)-(f(x))^2) = (\\frac{1}{2}-f(x))^2\n\\]\n\nWe have:\n\n\\[\nf(x+2a) = \\frac{1}{2} + \\sqrt{(\\frac{1}{2}-f(x))^2} = \\frac{1}{2} + (f(x) - \\frac{1}{2}) = f(x)\n\\]\n\nTherefore \\(f\\) is periodic, with \\(2a>0\\) as a period.\n\n(b) \\(f(x) = 1\\) when \\(2n\\le x < 2n+1\\) for some integer \\(n\\), and \\(f(x)=\\frac{1}{2}\\) when \\(2n+1\\le x < 2n+2\\) for some integer \\(n\\)."
] |
IMO-1968-6
|
https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_6
|
For every natural number \(n\), evaluate the sum
\[
\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \Big[\frac{n + 1}{2}\Big] + \Big[\frac{n + 2}{4}\Big] + \cdots + \bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] + \cdots
\]
(The symbol \([x]\) denotes the greatest integer not exceeding \(x\).)
|
[
"I shall prove that the summation is equal to \\(n\\).\n\nLet the binary representation of \\(n\\) be \\(\\overline{d_xd_{x-1}\\dots d_1d_0}_2\\), where \\(d_i\\in\\{0,1\\}\\) for all \\(i\\), and \\(x=\\lfloor \\log_2 n\\rfloor\\). Note that if \\(d_i=0\\), then \\(\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=\\Big[\\frac{n}{2^{i+1}}\\Big]\\); and if \\(d_i=1\\), then \\(\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=\\Big[\\frac{n}{2^{i+1}}\\Big]+1\\). Also note that \\(\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=0\\) for all \\(i\\geq x+1\\). Therefore the given sum is equal to\n\n\\[\n\\sum_{k = 0}^\\infty\\bigg[\\frac{n + 2^k}{2^{k + 1}}\\bigg] =S+\\sum_{k=0}^{x} \\bigg[\\frac{n}{2^{k + 1}}\\bigg]\n\\]\n\nwhere \\(S\\) is the number of 1's in the binary representation of \\(n\\). Legendre's Formula states that \\(n-S=\\sum_{k=0}^{x} \\bigg[\\frac{n}{2^{k + 1}}\\bigg]\\), which proves the assertion. \\(\\blacksquare\\)",
"We observe\n\n\\[\nn \\equiv 2^k \\text{mod }2^{k+1} \\longrightarrow \\left[\\frac{n+1+2^k}{2^{k+1}}\\right] - \\left[\\frac{n+2^k}{2^{k+1}}\\right] = 1 \\text{ and } \\left[\\frac{n+1+2^k}{2^{k+1}}\\right] - \\left[\\frac{n+2^k}{2^{k+1}}\\right] = 0 \\text{ otherwise.}\n\\]\n\nBut\n\n\\[\nD = [d \\text{ } | \\text{ } \\exists \\text{ } s \\in \\mathbb{N} \\text{ such that } d \\equiv 2^{s-1} \\text{ mod } 2^s] = \\mathbb{N} \\longrightarrow \\sum_{k = 0}^\\infty\\bigg[\\frac{n+1 + 2^k}{2^{k + 1}}\\bigg] - \\sum_{k = 0}^\\infty\\bigg[\\frac{n + 2^k}{2^{k + 1}}\\bigg] = 1,\n\\]\n\nso the result is just \\(n\\). \\(\\square\\)\n\n~ilovepi3.14",
"By Hermite's identity, for real numbers \\(x,\\)\n\n\\[\n\\bigg[ x + \\frac12 \\bigg] = \\bigg[ 2x \\bigg] - \\bigg[ x \\bigg].\n\\]\n\nHence our sum telescopes:\n\n\\[\n\\sum_{k = 0}^\\infty\\bigg[\\frac{n + 2^k}{2^{k + 1}}\\bigg] = \\sum_{k = 0}^\\infty\\bigg[\\frac{n}{2^{k + 1}} + \\frac12 \\bigg] = \\sum_{k = 0}^\\infty \\left( \\bigg[\\frac{n}{2^{k}} \\bigg] - \\bigg[ \\frac{n}{2^{k+1}} \\bigg] \\right) = \\bigg[ n \\bigg].\n\\]\n\n~Maximilian113\n\n## Resources\n\n- Discussion on AoPS/MathLinks"
] |
IMO-1969-1
|
https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_1
|
Prove that there are infinitely many natural numbers \(a\) with the following property: the number \(z = n^4 + a\) is not prime for any natural number \(n\).
|
[
"Suppose that \\(a = 4k^4\\) for some \\(k\\). We will prove that \\(a\\) satisfies the property outlined above.\n\nThe polynomial \\(n^4 + 4k^4\\) can be factored as follows:\n\n\\[\nn^4 + 4k^4\n\\]\n\n\\[\n= n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2\n\\]\n\n\\[\n= (n^2 + 2k^2)^2 - (2nk)^2\n\\]\n\n\\[\n= (n^2 + 2k^2 - 2nk)(n^2 + 2k^2 + 2nk)\n\\]\n\nBoth factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false.\n\nIt is also simple to prove that \\(n^2 + 2k^2 - 2nk > 1\\) when \\(k > 1\\). Thus, for all \\(k > 1\\), \\(4k^4\\) is a valid value of \\(a\\), completing the proof. \\(\\square\\)\n\n~mathboy100"
] |
IMO-1969-2
|
https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_2
|
Let \(a_1, a_2,\cdots, a_n\) be real constants, \(x\) a real variable, and
\[
f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).
\]
Given that \(f(x_1)=f(x_2)=0,\) prove that \(x_2-x_1=m\pi\) for some integer \(m.\)
|
[
"Because the period of \\(\\cos(x)\\) is \\(2\\pi\\), the period of \\(f(x)\\) is also \\(2\\pi\\).\n\n\\[\nf(x_1)=f(x_2)=f(x_1+x_2-x_1)\n\\]\n\nWe can get \\(x_2-x_1 = 2k\\pi\\) for \\(k\\in N^*\\). Thus, \\(x_2-x_1=m\\pi\\) for some integer \\(m.\\)",
"By the cosine addition formula,\n\n\\[\nf(x)=(\\cos{a_1}+\\frac{1}{2}\\cos{a_2}+\\frac{1}{4}\\cos{a_3}+\\cdots+\\frac{1}{2^{n-1}}\\cos{a_n})\\cos{x}-(\\sin{a_1}+\\frac{1}{2}\\sin{a_2}+\\frac{1}{4}\\sin{a_3}+\\cdots+\\frac{1}{2^{n-1}}\\sin{a_n})\\sin{x}\n\\]\n\nThis implies that if \\(f(x_1)=0\\),\n\n\\[\n\\tan{x_1}=\\frac{\\cos{a_1}+\\frac{1}{2}\\cos{a_2}+\\frac{1}{4}\\cos{a_3}+\\cdots+\\frac{1}{2^{n-1}}\\cos{a_n}}{\\sin{a_1}+\\frac{1}{2}\\sin{a_2}+\\frac{1}{4}\\sin{a_3}+\\cdots+\\frac{1}{2^{n-1}}\\sin{a_n}}\n\\]\n\nSince the period of \\(\\tan{x}\\) is \\(\\pi\\), this means that \\(\\tan{x_1}=\\tan{(x_1+\\pi)}=\\tan{(x_1+m\\pi)}\\) for any natural number \\(m\\). That implies that every value \\(x_1+m\\pi\\) is a zero of \\(f(x)\\).\n\n## Remarks (added by pf02, August 2024)\n\nBoth solutions given above are incorrect.\n\nThe first solution is hopelessly incorrect. It states that (and relies on it) if \\(f(x)\\) has period \\(2\\pi\\) and \\(f(x_1) = f(x_2)\\) then \\(x_2 - x_1 = m\\pi\\) for some integer \\(m\\). This is plainly wrong (think of \\(\\sin{\\pi/3} = \\sin{2\\pi/3}\\)). There is an obvious \"red flag\" as far as solutions go, namely the solution did not use that \\(f(x_1) = 0\\) and \\(f(x_2) = 0\\).\n\nThe second solution starts promising, but then it goes on to (incorrectly) prove the converse of the given problem, namely that if \\(f(x_1) = 0\\) then \\(f(x_1 + m\\pi) = 0\\) for any \\(m\\).\n\nBelow, I will give a solution to the problem.",
"For simplicity of writing, denote \\(b_k = 1/2^{k - 1}\\). \\(f(x) = b_1\\cos{(a_1 + x)} + b_1\\cos{(a_1 + x)} + \\cdots + b_n\\cos{(a_n + x)}\\). First, we want to prove that it is not the case that \\(f(x) = 0\\) for all \\(x\\). To prove this, we will prove that the maximum value of \\(f\\) is at least \\(b_n = 1/2^n\\). This will ensure that \\(f \\ne 0\\).\n\nWe do this by induction. The statement is clear for \\(n = 1\\): \\(f\\) has a maximum value of \\(b_k = 1\\). Assume that we have \\(n - 1\\) terms in \\(f\\), and the maximum value of \\(f\\) is at lease \\(b_{n - 1} = 1/2^{n - 1}\\). Now add the term \\(b_n\\cos{(a_n + x)}\\) (and remember that \\(b_n = 1/2^n\\)). This additional term has values in \\([-1/2^n, 1/2^n]\\), so it can decrease the maximum of \\(f\\) by subtracting at most \\(1/2^n\\) from the previous maximum, which was at least \\(1/2^{n - 1}\\). So, the new maximum is at least \\(1/2^n\\).\n\nNow we will the formula \\(\\cos{(\\alpha + \\beta)} = \\cos{\\alpha}\\cos{\\beta} - \\sin{\\alpha}\\sin{\\beta}\\).\n\nWe get \\(f(x) = (b_1\\cos{a_1} + b_2\\cos{a_2} + \\cdots + b_n\\cos{a_n})\\cos{x} - (b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n})\\sin{x}\\).\n\nIf both \\(b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n} = 0\\) and \\(b_1\\cos{a_1} + b_2\\cos{a_2} + \\cdots + b_n\\cos{a_n} = 0\\) then \\(f(x) = 0\\) for all \\(x\\). So at least one of these sums is \\(\\ne 0\\). I will give the proof for the case \\(b_1\\cos{a_1} + b_2\\cos{a_2} + \\cdots + b_n\\cos{a_n} \\ne 0\\). The other case is proven similarly.\n\nUsing \\(f(x_1) = 0\\), we get \\((b_1\\cos{a_1} + b_2\\cos{a_2} + \\cdots + b_n\\cos{a_n})\\cos{x_1} = (b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n})\\sin{x_1}\\), and similarly for \\(x_2\\).\n\nIf \\(b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n} = 0\\), then \\(\\cos{x_1} = 0\\) and \\(\\cos{x_2} = 0\\). It follows that both \\(x_1\\) and \\(x_2\\) are odd multiples of \\(\\pi/2\\), so they differ by \\(m\\pi\\) for some integer \\(m\\).\n\nIf \\(b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n} \\ne 0\\), then we can divide by this quantity, and we get\n\n\\(\\tan{x_1} = \\tan{x_2} = \\frac{b_1\\cos{a_1} + b_2\\cos{a_2} + \\cdots + b_n\\cos{a_n}} {b_1\\sin{a_1} + b_2\\sin{a_2} + \\cdots + b_n\\sin{a_n}}\\).\n\nThinking of the graph of \\(y = \\tan{x}\\) would be enough for many people to conclude that \\(x_1 - x_2 = m\\pi\\) for some integer \\(m\\). If we want to be more formal, we proceed by writing \\(\\tan{x_1} - \\tan{x_2} = 0\\). Some easy computations yield \\(\\tan{x_1} - \\tan{x_2} = \\frac{\\sin{(x_1 - x_2)}} {\\cos{x_1}\\cos{x_2}} = 0\\). It follows that \\(\\sin{(x_1 - x_2)} = 0\\), which implies that \\(x_1 - x_2 = m\\pi\\) for some integer \\(m\\).\n\n[Solution by pf02, August 2024]"
] |
IMO-1969-3
|
https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_3
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For each of \(k = 1\), \(2\), \(3\), \(4\), \(5\) find necessary and sufficient conditions on \(a > 0\) such that there exists a tetrahedron with \(k\) edges length \(a\) and the remainder length \(1\).
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[
"A plodding question. Take the tetrahedron to be ABCD.\n\nTake k = 1 and AB to have length a, the other edges length 1. Then we can hinge triangles ACD and BCD about CD to vary AB. The extreme values evidently occur with A, B, C, D coplanar. The least value, 0, when A coincides with B, and the greatest value √3, when A and B are on opposite sides of CD. We rule out the extreme values on the grounds that the tetrahedron is degenerate, thus obtaining 0 < a < √3.\n\nFor k = 5, the same argument shows that 0 < 1 < √3 a, and hence a > 1/√3.\n\nFor k = 2, there are two possible configurations: the sides length a adjacent, or not. Consider first the adjacent case. Take the sides length a to be AC and AD. As before, the two extreme cases gave A, B, C, D coplanar. If A and B are on opposite sides of CD, then a = √(2 - √3). If they are on the same side, then a = √(2 + √3). So this configuration allows any a satisfying √(2 - √3) < a < √(2 + √3).\n\nThe other configuration has AB = CD = a. One extreme case has a = 0. We can increase a until we reach the other extreme case with ADBC a square side 1, giving a = √2. So this configuration allows any a satisfying 0 < a < √2. Together, the two configurations allow any a satisfying: 0 < a < √(2 + √3).\n\nThis also solves the case k = 4, and allows any a satisfying: a > 1/√(2 + √3) = √(2 - √3).\n\nFor k = 3, any value of a > 0 is allowed. For a <= 1, we may take the edges length a to form a triangle. For a ≥ 1 we may take a triangle with unit edges and the edges joining the vertices to the fourth vertex to have length a."
] |
IMO-1969-4
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https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_4
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A semicircular arc \(\gamma\) is drawn with \(AB\) as diameter. \(C\) is a point on \(\gamma\) other than \(A\) and \(B\), and \(D\) is the foot of the perpendicular from \(C\) to \(AB\). We consider three circles, \(\gamma_1, \gamma_2, \gamma_3\), all tangent to the line \(AB\). Of these, \(\gamma_1\) is inscribed in \(\triangle ABC\), while \(\gamma_2\) and \(\gamma_3\) are both tangent to \(CD\) and \(\gamma\), one on each side of \(CD\). Prove that \(\gamma_1, \gamma_2\), and \(\gamma_3\) have a second tangent in common.
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[
"Denote the triangle sides \\(a = BC, b = CA, c = AB\\). Let \\(\\omega\\) be the circumcircle of the right angle triangle \\(\\triangle ABC\\) centered at the midpoint \\(O\\) of its hypotenuse \\(c = AB\\). Let \\(R, S, T\\) be the tangency points of the circles \\(K_1, K_2, K_3\\) with the line AB. In an inversion with the center \\(A\\) and positive power \\(r_A^2 = AC^2 = b^2\\) (\\(r_A\\) being the inversion circle radius), the line AB is carried into itself, the circle \\(\\omega\\) is carried into the altitude line \\(CD\\) and the altitude line \\(CD\\) into the circle \\(\\omega\\). This implies that the circle \\(K_3\\) intersecting the inversion circle \\(A\\) is carried into itself, but this is possible only if the circle \\(K_3\\) is perpendicular to the inversion circle \\(A\\). It follows that the tangency point \\(T\\) of the circle \\(K_3\\) is the intersection of the inversion circle \\((A, r_A = b)\\) with the line \\(AB\\). Similarly, in an inversion with the center B and positive power \\(r_B^2 = BC^2 = a^2\\) (\\(r_B\\) being the inversion circle radius), the line AB is carried into itself, the circle \\(\\omega\\) is carried into the altitude line \\(CD\\) and the altitude line \\(CD\\) into the circle \\(\\omega\\). This implies that the circle \\(K_2\\) intersecting the inversion circle \\(B\\) is carried into itself, but this is possible only if the circle \\(K_2\\) is perpendicular to the inversion circle \\(B\\). It follows that the tangency point S of the circle \\(K_2\\) is the intersection of the inversion circle \\((B, r_B = a)\\) with the line \\(AB\\).\n\nThe distance between the tangency points S, T is the equal to ST = AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle \\(K_1\\) of the right angle triangle \\(\\triangle ABC\\) is equal to\n\n\\[\nr = \\frac{|\\triangle ABC|}{s} = \\frac{ab}{a + b + c} = \\frac{a + b - c}{2} = s - c\n\\]\n\nwhere \\(|\\triangle ABC|\\) and s are the area and semiperimeter of the triangle \\(\\triangle ABC\\), for example, because of an obvious identity\n\n\\[\n(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab\n\\]\n\nor just because the angle \\(\\angle C = 90^\\circ\\) is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then\n\n\\[\nAR' = AT - \\frac{ST}{2} = b - r = b - \\frac{a + b - c}{2} = \\frac{c + b - a}{2} = s - a = AR\n\\]\n\nTherefore, the points \\(R' \\equiv R\\) are identical and the midpoint of the segment ST is the tangency point R of the incircle \\(K_1\\) with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles \\(K_2, K_3\\) are tangent to the incircle \\(K_1\\). Radii \\(r_2, r_3\\) of the circles \\(K_2, K_3\\) are now easily calculated:\n\n\\[\nr_2 = SD = BS - BD = a - \\frac{a^2}{c}\n\\]\n\n\\[\nr_3 = TD = AT - AD = b - \\frac{b^2}{c}\n\\]\n\nDenote \\(I, I_2, I_3\\) the centers of the circles \\(K_1, K_2, K_3\\). The line \\(I_2I_3\\) cuts the midline RI of the trapezoid \\(STI_3I_2\\) at the distance from the point R equal to\n\n\\[\n\\frac{SI_2 + TI_3}{2} = \\frac{r_2 + r_3}{2} = \\frac{a + b}{2} - \\frac{a^2 + b^2}{2c} = \\frac{a + b - c}{2} = r = RI\n\\]\n\nAs a result, the centers \\(I_2, I, I_3\\) are collinear (in fact, I is the midpoint of the segment \\(I_2I_3\\)). The common center line \\(I_2I_3\\) and the common external tangent AB of the circles \\(K_1, K_2, K_3\\) meet at their common external homothety center \\(H \\equiv I_2I_3 \\cap AB\\) and the other common external tangent of the circles \\(K_2, K_3\\) from the common homothety center H is a tangent to the circle \\(K_1\\) as well.\n\nThe above solution was posted and copyrighted by yetti. The original thread can be found here: [1]\n\n## Remarks (added by pf02, August 2024)\n\nIt is worth repeating here the note hal9v4ik makes on https://aops.com/community/p376814. This problem is a consequence of a particular case of Thebault's problem III (the Sawayama-Thebault-Streefkerk theorem; see https://en.wikipedia.org/wiki/Th%C3%A9bault%27s_theorem). However, this should not be viewed as a solution to the problem. I believe the point of the problem is proving the Sawayama-Thebault-Streefkerk theorem in a particular case. Also, the theorem is not well known, so it should not be used as a reference.\n\nBelow I will give another solution, based on analytic (coordinate) geometry. It is not elegant, but it is very straightforward and simple (except for the computations which can be tedious at times.)",
"We start by stating a simple proposition. Let \\(L\\) be a line and three circles with centers \\(C_1, C_2, C_3\\) tangent to the line \\(L\\), all on one side. Denote \\(A_1, A_2, A_3\\) the points on the line where the circles touch it. Then there is a second line tangent to the three circles if and only if \\(C_1, C_2, C_3\\) are collinear, and this is the case if and only if \\((\\overline{C_1A_1} - \\overline{C_2A_2})/(\\overline{C_2A_2} - \\overline{C_3A_3}) = (\\overline{A_1A_2})/(\\overline{A_2A_3})\\)\n\n\\[\nProb 1969 4 fig1.png\n\\]\n\nI will not give the full proof of this proposition, and in fact, we need to know only that if the equality of the two fractions holds, then a tangent to two circles is tangent to the third as well. The idea is that if the equality holds then triangles \\(\\triangle{C_1A_1B}, \\triangle{C_2A_2B}, \\triangle{C_3A_3B}\\) are similar, which implies that \\(C_1, C_2, C_3\\) are collinear, which implies that a tangent to two of the circles is tangent to the third as well.\n\nNow let us prove the problem. Look at the following figure:\n\n\\[\nProb 1969 4 fig2.png\n\\]\n\nWe use the notation given in the problem, and we denote \\(O\\) the center of the circle \\(\\gamma\\), \\(C_1, C_2, C_3\\) the centers of the circles \\(\\gamma_1, \\gamma_2,\\gamma_3\\), and \\(A_1, A_2, A_3\\) the points where the circles touch the line \\(AB\\). Let \\(AB\\) be the \\(x\\)-axis, \\(CD\\) be the \\(y\\)-axis, \\(D\\) be the origin, and the orientations of axes as shown in the picture. Let \\(a = \\overline{OD}\\). We think of \\(a\\) as positive, \\(a > 0\\). There is no loss of generality in assuming that the circle is of radius \\(1\\), and that \\(D\\) is to the right of \\(O\\). This implies the following coordinates:\n\n\\[\nD = (0, 0)\n\\]\n\n\\[\nA = (-1 - a, 0)\n\\]\n\n\\[\nB = (1 - a, 0)\n\\]\n\n\\[\nO = (-a, 0)\n\\]\n\nThe circle has the equation \\((x + a)^2 + y^2 = 1\\)\n\nThen \\(C = (0, \\sqrt{1 - a^2})\\)\n\nWe will also make use of the point \\(O_4 = (-a, -1)\\) on the circle, vertically down from \\(O\\).\n\nThe plan is to calculate the coordinates of \\(C_1, C_2, C_3\\) (the centers of the circles \\(\\gamma_1, \\gamma_2,\\gamma_3\\)) as expressions in \\(a\\). Then we will be able to verify that \\((\\overline{C_1A_1} - \\overline{C_2A_2})/(\\overline{C_2A_2} - \\overline{C_3A_3}) = (\\overline{A_1A_2})/(\\overline{A_2A_3})\\), which will have solved the problem.\n\nCalculating the coordinates of \\(C_2 = (x_2, y_2)\\) is easy. Take in account that the distance from \\(C_2\\) to line \\(AB\\) has to be the same as the distance to the circle \\(\\gamma\\). In other words, \\(y_2 = (\\)the radius of \\(\\gamma) - \\overline{C_2O} = 1 - \\sqrt{(x_2 + a)^2 + y_2^2}\\). Also, take in account that \\(C_2\\) is on the bisector of \\(\\angle{ADC}\\), which implies \\(x_2 + y_2 = 0\\).\n\nFrom this system of two equations with two unknowns we get\n\n\\[\n-x_2 = 1 - \\sqrt{(x_2 + a)^2 + x_2^2}\n\\]\n\nSolve this, and get \\(x_2 = 1 - a \\pm \\sqrt{2 - 2a}\\). Since \\(-1 < x_2 \\le 0\\), we see that only \\(x_2 = 1 - a - \\sqrt{2 - 2a}\\) is acceptable. Therefore\n\n\\[\nC_2 = (1 - a - \\sqrt{2 - 2a}, -(1 - a) + \\sqrt{2 - 2a})\n\\]\n\nBy a similar computation, we get\n\n\\[\nC_3 = (-1 - a + \\sqrt{2 + 2a}, -1 - a + \\sqrt{2 + 2a})\n\\]\n\nNow we want to compute the coordinates of \\(C_1 = (x_1, y_1)\\). We could look for the point on \\(CO_4\\) such that the distance to \\(AB\\), which is \\(y_1\\), equals the distance to \\(AC\\). But there is an easier way because we know classical formulas for computing the radius of the incircle of a triangle. Thus, if we denote by \\(A_r\\) and \\(s\\) the area and the semiperimeter of the triangle \\(\\triangle{ABC}\\), we know that\n\n\\(y_1 =\\) the radius of the incircle \\(= A_r/s = \\frac{\\overline{AB} \\ \\overline{DC}}{2} \\ \\frac{2}{\\overline{AC} + \\overline{BC} + \\overline{AB}} = \\cdots = \\frac{\\sqrt{2 + 2a} + \\sqrt{2 - 2a} - 2}{2}\\)\n\n(I skipped some obvious computations at the last equality).\n\nAn easy way to compute \\(x_1\\) is to impose that \\(C, C_1, O_4\\) are collinear.\n\nUse that \\(M = (m_1, m_2), N = (n_1, n_2), P = (p_1, p_2)\\) are collinear if and only if \\(\\frac{n_1 - m_1}{n_2 - m_2} = \\frac{p_1 - m_1}{p_2 - m_2}\\).\n\nWriting down the equality above for \\(C, C_1, O_4\\), we get\n\n\\[\n\\frac{x_1 - 0}{\\frac{\\sqrt{2 + 2a} + \\sqrt{2 - 2a} - 2}{2} - \\sqrt{1 - a^2}} = \\frac{-a - 0}{-1 - \\sqrt{1 - a^2}}\n\\]\n\nSolve for \\(x_1\\) and simplify, after which we get\n\n\\[\nC_1 = \\left( \\frac{\\sqrt{2 + 2a} - \\sqrt{2 - 2a} - 2a}{2}, \\frac{\\sqrt{2 + 2a} + \\sqrt{2 - 2a} - 2}{2} \\right)\n\\]\n\nNow we can verify easily that \\((\\overline{C_1A_1} - \\overline{C_2A_2})/(\\overline{C_2A_2} - \\overline{C_3A_3}) = (\\overline{A_1A_2})/(\\overline{A_2A_3})\\). In fact, we can see that \\(A_1\\) is the midpoint between \\(A_2\\) and \\(A_3\\), and the radius of \\(\\gamma_1\\) is the arithmetic mean of the radii of \\(\\gamma_2\\) and \\(\\gamma_3\\). This concludes the proof.\n\n[Solution by pf02, August 2024]"
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IMO-1969-5
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https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_5
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Given \(n > 4\) points in the plane such that no three are collinear, prove that there are at least \(C(n-3, 2) = {n-3 \choose 2} = \frac{(n-3)(n-4)}{2}\) convex quadrilaterals whose vertices are four of the given points.
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[
"Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they would be collinear. Also no points lie inside the area of ABC because if such a point D did exist, then the angle DAC would be less than BAC - contradiction. So there are none of the other n-3 points inside the area of ABC.\n\nNow we will show that for any two of the remaining n-3 points, you can create a convex quadrilateral containing two of them and two of A,B,C.\n\nNow select two arbitrary points from the remaining n-3. Call them E and F, and draw a line though them; call this line l. If l does not intersect ABC, then ABCEF is convex and you choose the quadrilateral ABEF. The other case is that it separates ABC such that one of A,B,C is one one side of l and the other two are on the other side. Without loss of generality, A is on one side of l while B and C are on the other side. Consider the quadrilateral BCEF (or CBEF, depending on orientation.) We know that BC are on the same side of EF. Also, since no points are in the third quadrant, we know that EF are on the same side of BC. Now we can choose either BCEF or CBEF to be our convex quadrilateral.\n\nSo we now have that for the given ABC, all pairs of the other n points give a distinct convex quadrilateral. So we have n-3 points to choose from, and we must choose two of them. So this gives us at least C(n-3,2) = \\(\\frac{(n-3)(n-4)}{2}\\) convex quadrilaterals.\n\nThe above solution was posted and copyrighted by Philip_Leszczynski. The original thread for this problem can be found here: [1]\n\n## Remarks (added by pf02, August 2024)\n\nThe solution given above is incorrect. This is pointed out very clearly and explicitly by DHu on https://aops.com/community/p364185. I will not repeat the argument here. DHu also gives the idea for a correct proof.\n\nOn the same page (https://aops.com/community/p364185), manuel gives an idea for a proof of a stronger result. His proof is somewhere between vague, incorrect and incomplete.\n\nBelow, I will give two solutions. The first is just an expansion of DHu's idea. The second one is carrying out manuel's idea.",
"We start by proving that given \\(5\\) points, there is at least one convex quadrilateral formed by \\(4\\) of them. Denote the \\(5\\) points by \\(A, B, C, D, E\\). Without loss of generality, we can chose to label the points so that \\(C, D, E\\) are on one side of the line \\(line(AB)\\). Further, we chose to label the point \\(C\\) so that \\(D, E\\) are on one side of \\(line(AC)\\). It follows that \\(D, E\\) are inside \\(\\angle(BAC)\\), with segments \\(AB\\) and \\(AC\\) extended beyond \\(B\\) and \\(C\\) respectively.\n\nNote that from the choice of \\(A, B, C\\) it follows that \\(\\angle(BAC) < 2\\pi\\). Also, we could have chosen \\(A, B, C\\) by considering the convex hull of the set of 5 points. It is either a triangle, or a convex quadrilateral, or a convex pentagon. We choose \\(A\\) to be any of the vertices, and \\(B, C\\) to be the adjacent vertices. (We don't need these notes in the proof, I just wanted to make things easier to visualize.)\n\nThe picture below shows this choice of labels.\n\n\\[\nNo of quads.png\n\\]\n\nThe line \\(line(DE)\\) must intersect at least one of \\(line(AB)\\) and \\(line(AC)\\). Without loss of generality, we can assume it intersects \\(line(AB)\\). In general, it will also intersect \\(line(AC)\\). The argument we are going to give is for the case when \\(line(DE)\\) intersects both \\(line(AB)\\) and \\(line(AC)\\). The argument needed in the limiting case when \\(line(DE) \\parallel line(AC)\\) is identical, and is left as an exercise to the reader.\n\nThere are three cases:\n\nCase 1, when \\(line(DE)\\) intersects \\(segment(AB)\\), but it does not intersect \\(segment(AC)\\). This is shown on the first 3 pictures (one picture would have been enough, but I made all of them, for emphasis). In this case the quadrilateral formed by the points \\(A, D, E, C\\) is convex. (Note that this also applies to the case when \\(line(DE) \\parallel line(AC)\\).)\n\nCase 2, when \\(line(DE)\\) intersects both \\(segment(AB)\\) and \\(segment(AC)\\). This is shown in the 4th picture. In this case the quadrilateral formed by the points \\(B, D, E, C\\) is convex.\n\nCase 3, when \\(line(DE)\\) does not intersect either of \\(segment(AB)\\) or \\(segment(AC)\\). This is shown in the 5th picture. In this case the quadrilateral formed by the points \\(B, D, E, C\\) is convex. (Note that this also applies to the case when \\(line(DE) \\parallel line(AC)\\).)\n\nWe thus proved that in any group of 5 points, there is a subgroup of 4 points which forms a convex quadrilateral. Note that in general, given a group of 5 points, more than one convex quadrilateral can be found by taking subgroups of 4 points.\n\nImportant remark: In fact, we proved more: if \\(D, E\\) are inside the angle \\(\\angle(BAC)\\) formed by \\(line(AB)\\) and \\(line(AC)\\) (with segments \\(AB\\) and \\(AC\\) extended beyond \\(B\\) and \\(C\\) respectively), then we can choose the convex quadrilateral so that \\(D, E\\) are both part of it.\n\nNow the statement of the problem is easy to prove. Assume \\(n\\) points are given. Choose \\(3\\) of them, and label them \\(A, B, C\\), so that all the other points are inside \\(\\angle(BAC)\\). There are \\((n - 3)\\) such points. Now choose all the groups of \\(2\\) points from these \\((n - 3)\\) points. There are \\({n-3 \\choose 2} = \\frac{(n-3)(n-4)}{2}\\) ways of doing this. From what we proved before, we know that for each choice of 2 points, call them \\(D, E\\), there is a group of \\(4\\) points from among \\(A, B, C, D, E\\) which forms a convex quadrilateral, such that \\(D, E\\) are \\(2\\) of the \\(4\\) points. A different choice \\(D_1, E_1\\) will yield a different quadrilateral because \\(\\{D, E\\} \\ne \\{D_1, E_1\\}\\).\n\nThis concludes the proof.",
"As we did in the previous solution, we first prove that given a group of \\(5\\) points, there is a subgroup of \\(4\\) of them which forms a convex quadrilateral. Unlike in the previous solution, we don't care whether \\(2\\) specific points are part of it or not. Note that generally there might be several convex quadrilaterals formed from the group of \\(5\\) points.\n\nNow, given \\(n\\) points, create a list of all the groups of \\(5\\) points chosen from the given \\(n\\) points. This can be done in \\({n \\choose 5}\\) ways, so the list has \\({n \\choose 5}\\) elements. For each group of \\(5\\) points, choose a convex quadrilateral formed by a subgroup of the \\(5\\) points. Thus, create a list \\(M\\) of convex quadrilaterals. This list has \\({n \\choose 5}\\) elements.\n\nHowever, note that some convex quadrilaterals in \\(M\\) are not unique, they might appear several times in \\(M\\). A given quadrilateral could appear up to \\((n - 4)\\) times, as chosen from a group of \\(5\\) points which contains the \\(4\\) points of the quadrilateral, and one point from the remaining \\((n - 4)\\) points. We want to form the list \\(K\\), which is a subset of \\(M\\), such that from each group of duplicates, we keep only one. Let \\(k\\) be the number of elements in \\(K\\).\n\nSince each quadrilateral in \\(M\\) appears at most \\((n - 4)\\) times, it follows that \\(k \\ge {n \\choose 5} / (n-4) = \\frac{C(n, 5)}{n - 4}\\), in other words, there are at least \\({n \\choose 5} / (n-4) = \\frac{C(n, 5)}{n - 4}\\) convex quadrilaterals formed from points in the given \\(n\\) points.\n\nThis is a stronger result than the one in the statement of the problem.\n\nTo prove the statement of the problem, we will prove the assertion I just made above, by showing that \\({n \\choose 5} / (n-4) \\ge {n - 3 \\choose 2}\\). This amounts to proving that\n\n\\[\n\\frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{120(n-4)} \\ge \\frac{(n - 3)(n - 2)}{2}\n\\]\n\nfor \\(n \\ge 5\\). Carry out the computations, and we get that we need to prove that \\(P(n) = n^3 - 3n^2 - 58n + 240 \\ge 0\\) for \\(n \\ge 5\\). The polynomial \\(P(n)\\) has roots \\(-8, 5, 6\\) from which it follows that\n\n\\[\n\\frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{120(n-4)} \\ge \\frac{(n - 3)(n - 2)}{2}\n\\]\n\nis true, and it is an equality for \\(n = 5, 6\\) and a strict inequality for \\(n > 6\\).\n\nThe original thread for this problem can be found here: https://aops.com/community/p364185"
] |
IMO-1969-6
|
https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_6
|
Prove that for all real numbers \(x_1, x_2, y_1, y_2, z_1, z_2\), with \(x_1 > 0, x_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0\), the inequality
\[
\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}
\]
is satisfied. Give necessary and sufficient conditions for equality.
|
[
"Let \\(A=x_1y_1 - z_1^2>0\\) and \\(B=x_2y_2 - z_2^2>0\\)\n\nFrom AM-GM:\n\n\\(\\sqrt{AB} \\le \\frac{A+B}{2}\\) with equality at \\(A=B\\)\n\n\\[\n4AB \\le (A+B)^2\n\\]\n\n\\[\n\\frac{4}{A+B} \\le \\frac{A+B}{AB}\n\\]\n\n\\[\n\\frac{8}{2(A+B)} \\le \\frac{A+B}{AB}\n\\]\n\n\\(\\frac{8}{2(A+B)} \\le \\frac{1}{A}+\\frac{1}{B}\\) [Equation 1]\n\n\\[\n(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2\n\\]\n\n\\[\n(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2=(A+B)+x_1y_2+x_2y_1-2z_1z_2\n\\]\n\nsince \\(x_1y_1>z_1^2\\) and \\(x_2y_2>z_2^2\\), and using the Rearrangement inequality\n\nthen \\(x_1y_1+x_2y_2-z_1z_1-z_2z_2 \\le x_1y_2+x_2y_1-z_1z_2-z_1z_2\\)\n\n\\[\n(A+B) \\le x_1y_2+x_2y_1-2z_1z_2\n\\]\n\n\\[\n2(A+B) \\le x_1y_1+x_2y_2+x_1y_2+x_2y_1-z_1^2-2z_1z_2-z_2^2\n\\]\n\n\\(2(A+B) \\le (x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2\\) [Equation 2]\n\nTherefore, we can can use [Equation 2] into [Equation 1] to get:\n\n\\[\n\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\le \\frac{1}{A}+\\frac{1}{B}\n\\]\n\nThen, from the values of \\(A\\) and \\(B\\) we get:\n\n\\[\n\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\leq \\frac{1}{x_1y_1 - z_1^2} + \\frac{1}{x_2y_2 - z_2^2}\n\\]\n\nWith equality at \\(x_1y_1 - z_1^2=x_2y_2 - z_2^2>0\\) and \\(x_1=x_2, y_1=y_2, z_1=z_2\\)\n\n~Tomas Diaz. [email protected]",
"This solution is actually more difficult but I added it here for fun to see the generalized case as follows:\n\nProve that for all real numbers \\(a_i, b_i\\), for \\(i=1,2,...,n\\) with \\(a_i > 0, b_i > 0\\)\n\nand \\(\\prod_{i=1}^{n-1}a_i-a_n^{n-1}>0, \\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0\\) the inequality\n\n\\[\n\\frac{2^n}{\\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \\leq \\frac{1}{\\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \\frac{1}{\\prod_{i=1}^{n-1}b_i-b_n^{n-1}}\n\\]\n\nis satisfied.\n\nLet \\(A=\\prod_{i=1}^{n-1}a_i-a_n^{n-1}\\) and \\(\\prod_{i=1}^{n-1}b_i-b_n^{n-1}>0\\)\n\nFrom AM-GM:\n\n\\(\\sqrt{AB} \\le \\frac{A+B}{2}\\) with equality at \\(A=B\\)\n\n\\[\n4AB \\le (A+B)^2\n\\]\n\n\\[\n\\frac{4}{A+B} \\le \\frac{A+B}{AB}\n\\]\n\n\\[\n\\frac{2^n}{2^{n-2}(A+B)} \\le \\frac{A+B}{AB}\n\\]\n\n\\(\\frac{2^n}{2^{n-2}(A+B)} \\le \\frac{1}{A}+\\frac{1}{B}\\) [Equation 3]\n\nHere's the difficult part where I'm skipping steps:\n\nwe prove that \\(2^{n-2}(A+B) \\le \\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}\\)\n\nand replace in [Equation 3] to get:\n\n\\[\n\\frac{2^n}{\\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \\le \\frac{1}{A}+\\frac{1}{B}\n\\]\n\nand replace the values of \\(A\\) and \\(B\\) to get:\n\n\\[\n\\frac{2^n}{\\prod_{i=1}^{n-1}(a_i + b_i) - (a_n + b_n)^{n-1}} \\leq \\frac{1}{\\prod_{i=1}^{n-1}a_i-a_n^{n-1}} + \\frac{1}{\\prod_{i=1}^{n-1}b_i-b_n^{n-1}}\n\\]\n\nwith equality at \\(a_i=b_i\\) for all \\(i=1,2,...,n\\)\n\nThen set \\(n=3, a_1=x_1, a_2=y_1, a_3=z_1, b_1=x_2, b_2=y_2, b_3=z_3\\) and substitute in the generalized inequality to get:\n\n\\[\n\\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \\leq \\frac{1}{x_1y_1 - z_1^2} + \\frac{1}{x_2y_2 - z_2^2}\n\\]\n\nwith equality at \\(x_1=x_2, y_1=y_2, z_1=z_2\\)\n\n~Tomas Diaz. [email protected]\n\n## Remarks (added by pf02, July 2024)\n\n1. The solution given above is incorrect. The error is in the incorrect usage of the Rearrangement inequality. The conclusion \\(x_1y_1+x_2y_2-z_1z_1-z_2z_2 \\le x_1y_2+x_2y_1-z_1z_2-z_1z_2\\) is false. For a counterexample take \\(x_1 = y_1 = 2, x_2 = y_2 = 1, z_1 = z_2 = 0.5\\). The left hand side equals \\(5 - 0.25 - 0.25\\) and the right hand side equals \\(4 - 0.25 - 0.25\\).\n\n2. The generalization is reasonable but the idea for a solution is unacceptably vague (at one crucial step, the author says \"Here's the difficult part where I'm skipping steps\"). I don't believe this can be developed into a real proof, since it just follows the idea of the Solution above, which is incorrect.\n\n3. I will give a solution below, which uses calculus. I believe an \"elementary\" solution (i.e. a solution based on elementary algebra and geometry) is possible, but quite difficult.",
"First, remark that given the conditions of the problem, it follows that \\(y_1 > 0, y_2 > 0\\). Also, we can assume \\(z_1 \\ge 0, z_2 \\ge 0\\). Indeed, if \\(z_1 < 0\\), then \\(z_1 < -z_1\\). It follows \\(z_1 + z_2 < -z_1 + z_2\\), so\n\n\\[\n\\frac{1}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} < \\frac{1}{(x_1 + x_2)(y_1 + y_2) - (-z_1 + z_2)^2}\n\\]\n\nSo, if we proved the inequality for positive \\(z_1\\) it is also true for negative \\(z_1\\).\n\nNow consider the function \\(F\\) of three variables\n\n\\[\nF(x, y, z) = \\frac{1}{xy - z^2}\n\\]\n\ndefined on the domain \\(x, y > 0, z \\ge 0, z^2 < xy\\). For simplicity, denote a point \\((x, y, z)\\) in the domain by \\(P\\). The inequality in the problem can be rewritten as\n\n\\[\nF \\left( \\frac{P_1 + P_2}{2} \\right) \\le \\frac{1}{2}[F(P_1) + F(P_2)]\n\\]\n\nThis follows immediately from the inequality expressing the fact that \\(F(x, y, z)\\) is a convex function. (In fact, it is equivalent to the convexity of \\(F\\), but this is not needed for our proof of the given inequality.) Therefore, it is enough to prove that the function \\(F(x, y, z)\\) is convex.\n\nWe will prove that this function is convex. In fact, we will prove that the function is strictly convex. This will imply that equality holds only when \\(P_1 = P_2\\), in other words, \\(x_1 = x_2, y_1 = y_2, z_1 = z_2\\), which will give us the necessary and sufficient conditions for equality.\n\nWe use the theorem which says that a twice differentiable function defined on a convex set is convex if and only if its Hessian is positive definite. Furthermore, if the Hessian is strictly positive definite, then the function is strictly convex.\n\nThe Hessian of the function \\(F(x, y, z)\\) is the 3x3 matrix \\(H\\) formed by the second derivatives of \\(F\\):\n\n\\[\n\\frac{\\partial^2 F}{\\partial x^2} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial x \\partial y} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial x \\partial z} \\\\ \\\\ \\\\ \\frac{\\partial^2 F}{\\partial y \\partial x} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial y^2} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial y \\partial z} \\\\ \\\\ \\\\ \\frac{\\partial^2 F}{\\partial z \\partial x} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial z \\partial y} \\hspace{20pt} \\frac{\\partial^2 F}{\\partial z^2}\n\\]\n\n\\(H\\) is positive definite if for any 3x1 vector \\(v\\) we have \\(v^t H v \\ge 0\\). (\\(v^t\\) is the transpose of \\(v\\), a matrix which is 1x3.) \\(H\\) is strictly positive definite if \\(v^t H v > 0\\) unless \\(v = 0\\).\n\nExplicitly, we need to show that if \\(a, b, c\\) are three numbers, then\n\n\\[\na^2 \\frac{\\partial^2 F}{\\partial x^2} + ab \\frac{\\partial^2 F}{\\partial x \\partial y} + ac \\frac{\\partial^2 F}{\\partial x \\partial z} + ab \\frac{\\partial^2 F}{\\partial y \\partial x} + b^2 \\frac{\\partial^2 F}{\\partial y^2} + bc \\frac{\\partial^2 F}{\\partial y \\partial z} + ac \\frac{\\partial^2 F}{\\partial z \\partial x} + bc \\frac{\\partial^2 F}{\\partial z \\partial y} + c^2 \\frac{\\partial^2 F}{\\partial z^2} \\ge 0\n\\]\n\nand that the expression is \\(= 0\\) only when \\(a = b = c = 0\\).\n\nBefore embarking on the computations proving the inequality expressing the positive definiteness of the Hessian of \\(F\\), let us show that the domain of \\(F\\) is convex. Let \\(P_1 = (x_1, y_1, z_1)\\) and \\(P_2 = (x_2, y_2, z_2)\\) be two points in the domain of \\(F\\). This means that \\(x_1, y_1, x_2, y_2 > 0\\), \\(z_1, z_2 \\ge 0\\), \\(z_1^2 < x_1 y_1\\) and \\(z_2^2 < x_2 y_2\\). We need to verify that the same is true for \\(P = t P_1 + (1 - t) P_2\\) when \\(0 < t < 1\\). The only thing which is not obvious is \\([t z_1 + (1 - t) z_2]^2 < [t x_1 + (1 - t) x_2] [t y_1 + (1 - t) y_2]\\). To see this, work out the multiplications, use what we already know about \\(z_1^2\\) and \\(z_2^2\\), simplify by \\(t(1-t)\\) and we are left with showing that \\(2 z_1 z_2 < x_1 y_2 + x_2 y_1\\). This can be seen as follows: \\(2 z_1 z_2 < 2 \\sqrt{x_1 y_1} \\sqrt{x_2 y_2} = 2 \\sqrt{(x_1 y_2) (x_2 y_1)} \\le x_1 y_2 + x_2 y_1\\).\n\nNow work towards proving the inequality expressing that the Hessian of \\(F\\) is positive definite. Compute all the second derivatives of \\(F\\); notice that they all have \\((xy - z^2)^3\\) at the denominator, so we can simplify with this factor. (The computation is made somewhat shorter by taking advantage of the fact that the order of taking the derivatives makes no difference in mixed derivatives (for example \\(\\frac{\\partial^2 F}{\\partial x \\partial y} = \\frac{\\partial^2 F}{\\partial y \\partial x}\\)).)\n\nWe get (as the thing we have to prove) that\n\n\\(2a^2y^2 + 2ab(xy + z^2) - 8acyz + 2b^2x^2 - 8bcxz + c^2(2xy + 6z^2) \\ge 0\\).\n\nIf \\(a = b = 0\\) the expression equals \\(0\\) if and only if \\(c = 0\\) as well. So, assume that at least one of \\(a, b\\) is \\(\\ne 0\\). We can assume \\(a \\ne 0\\). We will show that in this case, the expression is \\(> 0\\).\n\nSimplify by \\(2\\), and rearrange the terms so we can think of the expression on the left as a polynomial in \\(c\\):\n\n\\((xy + 3z^2)c^2 - 4(xb + ya)zc + [x^2b^2 + y^2a^2 + (xy + z^2)ab] > 0\\).\n\nNotice that this is a polynomial (in \\(c\\)) of degree \\(2\\) of the type \\(AX^2 + BX + C\\), whose leading coefficient is \\(> 0\\). To show that its values are always \\(> 0\\) we need to show that its discriminant is \\(< 0\\), in other words, \\(B^2 - 4AC < 0\\).\n\nTherefore, we need to show that\n\n\\[\n16[(xb + ya)z]^2 - 4(xy + 3z^2)[x^2b^2 + y^2a^2 + (xy + 3z^2)ab] < 0\n\\]\n\nAfter factoring out \\(4\\), and working out all multiplications, this becomes\n\n\\[\n-x^3yb^2 + x^2z^2b^2 - xy^3a^2 + y^2z^2a^2 - x^2y^2ab + 4xyz^2ab - 3z^4ab < 0\n\\]\n\nAfter some rearranging of terms, grouping, and factoring, this becomes\n\n\\[\n-(xy - z^2)[x^2b^2 + (xy - 3z^2)ab + y^2a^2] < 0\n\\]\n\nSince \\(x^2 < xy\\), we need to show\n\n\\[\nx^2b^2 + (xy - 3z^2)ab + y^2a^2 > 0\n\\]\n\nNow, view the expression on the left hand side as a polynomial of degree \\(2\\) in \\(b\\), whose leading coefficient is \\(> 0\\). To conclude that its values are \\(> 0\\) we need to show that its discriminant is \\(< 0\\). That is, we need to show that\n\n\\[\n[(xy - 3z^2)a]^2 - 4x^2y^2a^2 < 0\n\\]\n\nWork out the multiplications, and regroup terms to get\n\n\\(-3a^2(xy - z^2)(xy + 3z^2) < 0\\).\n\nThis is clearly true, which concludes the proof.\n\n[Solution by pf02, July 2024]"
] |
IMO-1970-1
|
https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_1
|
Let \(M\) be a point on the side \(AB\) of \(\triangle ABC\). Let \(r_1, r_2\), and \(r\) be the inscribed circles of triangles \(AMC, BMC\), and \(ABC\). Let \(q_1, q_2\), and \(q\) be the radii of the escribed circles of the same triangles that lie in the angle \(ACB\). Prove that
\[
\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}
\]
.
|
[
"We use the conventional triangle notations.\n\nLet \\(I\\) be the incenter of \\(ABC\\), and let \\(I_{c}\\) be its excenter to side \\(c\\). We observe that\n\n\\[\nr \\left[ \\cot\\left(\\frac{A}{2}\\right) + \\cot\\left(\\frac{B}{2}\\right) \\right] = c\n\\]\n\n,\n\nand likewise,\n\n\\[\n\\begin{matrix} c & = & q \\left[ \\cot\\left(\\frac{\\pi - A}{2}\\right) + \\cot \\left(\\frac{\\pi - B}{2}\\right) \\right] \\\\ \\\\ & = & q \\left[ \\tan\\left(\\frac{A}{2}\\right) + \\tan\\left(\\frac{B}{2}\\right) \\right]\\; . \\end{matrix}\n\\]\n\nSimplifying the quotient of these expressions, we obtain the result\n\n\\[\n\\frac{r}{q} = \\tan (A/2) \\tan (B/2)\n\\]\n\n.\n\nThus we wish to prove that\n\n\\[\n\\tan (A/2) \\tan (B/2) = \\tan (A/2) \\tan (AMC/2) \\tan (B/2) \\tan (CMB/2)\n\\]\n\n.\n\nBut this follows from the fact that the angles \\(AMC\\) and \\(CMB\\) are supplementary.",
"By similar triangles and the fact that both centers lie on the angle bisector of \\(\\angle{C}\\), we have \\(\\frac{r}{q} = \\frac{s-c}{s} = \\frac{a + b - c}{a + b + c}\\), where \\(s\\) is the semi-perimeter of \\(ABC\\). Let \\(ABC\\) have sides \\(a, b, c\\), and let \\(AM = c_1, MB = c_2, MC = d\\). After simple computations, we see that the condition, whose equivalent form is\n\n\\[\n\\frac{b + d - c_1}{b + d + c_1} \\cdot \\frac{a + d - c_2}{a + d + c_2} = \\frac{a + b - c}{a + b + c},\n\\]\n\nis also equivalent to Stewart's Theorem (see Stewart's_theorem or https://en.wikipedia.org/wiki/Stewart's_theorem)\n\n\\[\nd^2 c + c_1 c_2 c = a^2 c_1 + b^2 c_2.\n\\]",
"Let \\(I\\) be the incenter, and \\(E\\) be the excenter relative to C, and let \\(B_1, B_2\\) be the points where the incircle and the excircle relative to \\(C\\) touch \\(AC\\).\n\n\\[\nProb 1970 1.png\n\\]\n\nThe triangles \\(\\triangle CIB_1\\) and \\(\\triangle CEB_2\\) are similar, so \\(\\frac{r}{q} = \\frac{CI}{CE}.\\)\n\nLet \\(I_1\\) be the incenter, and \\(E_1\\) be the excenter relative to C of the triangle \\(\\triangle ACM\\), and \\(I_2\\) be the incenter, and \\(E_2\\) be the excenter relative to C of the triangle \\(\\triangle BCM\\) (\\(E_2\\) is not shown on the picture).\n\nTo solve the problem, we need to prove that \\(\\frac{CI_1}{CE_1} \\cdot \\frac{CI_2}{CE_2} = \\frac{CI}{CE}.\\)\n\nApplying the law of sines (see Law of Sines or https://en.wikipedia.org/wiki/Law_of_sines) in triangle \\(\\triangle ICA\\) we get\n\n\\(\\frac{IC}{\\sin \\angle CAI} = \\frac{AC}{\\sin \\angle CIA}\\), or \\(IC = b\\ \\frac{\\sin \\frac{A}{2}}{\\sin (\\pi - (\\frac{A}{2} + \\frac{C}{2}))}\\). Since \\(C = \\pi - A - B\\) this becomes \\(IC = b\\ \\frac{\\sin \\frac{A}{2}}{\\cos \\frac{B}{2}}.\\)\n\nSimilarly, using the law of sines in triangle \\(\\triangle ECA\\) and replacing some angles, we get \\(EC = b\\ \\frac{\\cos \\frac{A}{2}}{\\sin \\frac{B}{2}}.\\)\n\nIt follows that \\(\\frac{r}{q} = \\tan \\frac{A}{2} \\tan \\frac{B}{2}.\\)\n\nNow, we proceed like in the first solution: Apply the above to triangles \\(\\triangle AMC\\) and \\(\\triangle MBC\\).\n\nWe get that \\(\\frac{r_1}{q_1} = \\tan \\frac{A}{2} \\tan \\frac{\\angle AMC}{2}\\) and \\(\\frac{r_2}{q_2} = \\tan \\frac{\\angle BMC}{2} \\tan \\frac{B}{2}.\\)\n\nThe desired equality follows from \\(\\tan \\frac{\\angle AMC}{2} \\cdot \\tan \\frac{\\angle BMC}{2} = 1\\) (since \\(\\frac{\\angle AMC}{2}\\) and \\(\\frac{\\angle BMC}{2}\\) are complementary).\n\n[Solution by pf02, November 2024]"
] |
IMO-1970-2
|
https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_2
|
Let \(a, b\), and \(n\) be integers greater than 1, and let \(a\) and \(b\) be the bases of two number systems. \(A_{n-1}\) and \(A_{n}\) are numbers in the system with base \(a\) and \(B_{n-1}\) and \(B_{n}\) are numbers in the system with base \(b\); these are related as follows:
\[
A_{n} = x_{n}x_{n-1}\cdots x_{0}, A_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}
\]
,
\[
B_{n} = x_{n}x_{n-1}\cdots x_{0}, B_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}
\]
,
\[
x_{n} \neq 0, x_{n-1} \neq 0.
\]
Prove:
\[
\frac{A_{n-1}}{A_{n}} < \frac{B_{n-1}}{B_{n}}
\]
if and only if
\[
a > b
\]
.
|
[
"Suppose \\(a>b\\). Then for all integers \\(0 \\le k \\le n\\), \\(x_n x_k a^n b^k \\ge x_n x_k b^n a^k\\), with equality only when \\(k=n\\) or \\(x_k = 0\\). (In particular, we have strict inequality for \\(k=n-1\\).) In summation, this becomes\n\n\\[\nx_n a^n \\sum_{k=0}^n x_k b^k > x_n b^n \\sum_{k=0}^n x_k a^k,\n\\]\n\nor\n\n\\[\nx_n a^n \\cdot B_n > x_n b^n \\cdot A_n,\n\\]\n\nwhich is equivalent to\n\n\\[\n\\frac{x_n a^n}{A_n} > \\frac{x_n b^n}{B_n} .\n\\]\n\nThis implies\n\n\\[\n\\frac{A_{n-1}}{A_n} = 1 - \\frac{x_n a^n}{A_n} < 1 - \\frac{x_n b^n}{B_n} = \\frac{B_{n-1}}{B_n} .\n\\]\n\nOn the other hand, if \\(a=b\\), then evidently \\(A_{n-1}/A_n = B_{n-1}/B_n\\), and if \\(a < b\\), then by what we have just shown, \\(A_{n-1}/A_n > B_{n-1}/B_n\\). Hence \\(A_{n-1}/A_n < B_{n-1}/B_n\\) if and only if \\(a>b\\), as desired. \\(\\blacksquare\\)\n\n## Remarks (added by pf02, November 2024)\n\n1. A comment about the problem: it is disappointing. It is surprisingly easy compared to other problems given at an International Mathematical Olympiad, and frankly, uninteresting.\n\n2. It is not stated with the usual rigor typical at this competition. The problem is really the following two statements:\n\nS1: Let \\(a, b, n > 0\\) and \\(x_n, x_{n-1}, \\cdots, x_0\\) be integers, \\(x_k < a, b\\) for all \\(k\\), \\(x_n > 0\\), and \\(x_{n-1}, \\cdots, x_0\\) not all \\(0\\). Define \\(A_n, A_{n-1}, B_n, B_{n-1}\\) as described in the problem. If\n\n\\[\n\\frac{A_{n-1}}{A_{n}} < \\frac{B_{n-1}}{B_{n}}\\ \\ \\ \\ \\ \\ \\ \\ (1)\n\\]\n\nthen \\(a > b\\).\n\nS2: If \\(a > b\\), then (1) is true for any \\(x_n, x_{n-1}, \\cdots, x_0\\).\n\n3. I will give a variation of the solution in which the computations and argumentation are slightly simpler and more direct.",
"Rewrite inequality (1), and give a few steps in which we state equivalent inequalities:\n\n\\[\n\\frac{x_{n-1} a^{n-1} + \\cdots + x_1 a + x_0}{x_n a^n + x_{n-1} a^{n-1} + \\cdots + x_1 a + x_0} < \\frac{x_{n-1} b^{n-1} + \\cdots + x_1 b + x_0}{x_n b^n + x_{n-1} b^{n-1} + \\cdots + x_1 b + x_0}\n\\]\n\n\\[\n\\frac{x_{n-1} a^{n-1} + \\cdots + x_1 a + x_0}{x_n a^n} < \\frac{x_{n-1} b^{n-1} + \\cdots + x_1 b + x_0}{x_n b^n}\n\\]\n\n\\[\n\\frac{x_{n-1} a^{n-1} + \\cdots + x_1 a + x_0}{a^n} < \\frac{x_{n-1} b^{n-1} + \\cdots + x_1 b + x_0}{b^n}\n\\]\n\n\\[\nx_{n-1} \\frac{1}{a} + \\cdots + x_1 \\frac{1}{a^{n-1}} + x_0 \\frac{1}{a^n} < x_{n-1} \\frac{1}{b} + \\cdots + x_1 \\frac{1}{b^{n-1}} + x_0 \\frac{1}{b^n}\\ \\ \\ \\ \\ \\ \\ \\ (2)\n\\]\n\nNow it is clear that if \\(a > b\\) then (2) is true for any \\(x_n, x_{n-1}, \\cdots, x_0\\). This proves statement S2 from the Remarks.\n\n(Note that the problem said \\(x_n, x_{n-1} > 0\\). This is not necessary. All we need is that \\(x_n > 0\\) and at least one of \\(x_{n-1}, \\cdots, x_1, x_0\\) is \\(\\neq 0\\).)\n\nWe prove statement S1 from the Remarks by contradiction. We want to show that if (1) is true for some \\(x_n, \\cdots, x_1, x_0\\), then \\(a > b\\).\n\nAssume \\(a \\le b\\). If we had \\(a = b\\) then (1) would be an equality, not an inequality.\n\nIf \\(a < b\\) then by S1, it would follow that \\(\\frac{A_{n-1}}{A_{n}} > \\frac{B_{n-1}}{B_{n}}\\) for all values of \\(x_n, \\cdots, x_1, x_0\\) which contradicts the hypothesis. Either way, we get a contradiction, which proves S1.\n\n[Solution by pf02, November 2024]\n\n## Resources"
] |
IMO-1970-3
|
https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_3
|
The real numbers \(a_0, a_1, \ldots, a_n, \ldots\) satisfy the condition:
\[
1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots.
\]
The numbers \(b_{1}, b_{2}, \ldots, b_n, \ldots\) are defined by
\[
b_n = \sum_{k=1}^{n} \left( 1 - \frac{a_{k-1}}{a_{k}} \right) \frac{1}{\sqrt{a_k}}
\]
(a) Prove that \(0 \leq b_n < 2\) for all \(n\).
(b) given \(c\) with \(0 \leq c < 2\), prove that there exist numbers \(a_0, a_1, \ldots\) with the above properties such that \(b_n > c\) for large enough \(n\).
|
[
"\\[\nb_n = \\sum_{k=1}^{n} \\frac{1 - \\frac{a_{k-1}}{a_{k}} }{\\sqrt{a_k}} = \\sum_{k=1}^{n} \\frac{a_k - a_{k-1}}{a_k\\sqrt{a_k}} = \\sum_{k=1}^{n} (a_k - a_{k-1})\\left(a_k^{-\\dfrac{3}{2}}\\right)\n\\]\n\nLet \\(X_k\\) be the rectangle with the verticies: \\((a_{k-1},0)\\); \\((a_{k},0)\\); \\((a_{k},a_k^{-\\dfrac{3}{2}})\\); \\((a_{k-1},a_k^{-\\dfrac{3}{2}})\\).\n\n\\[\n[asy] import graph; size(10cm,10cm,IgnoreAspect); Label f; f.p=fontsize(6); xaxis(0,10); yaxis(0,1); real f(real x) { return x^(-3/2); } draw(graph(f,1,10)); draw((1,0)--(1,1)); draw((1,0)--(2,0)--(2,2^(-3/2))--(1,2^(-3/2))--cycle); draw((2,0)--(3,0)--(3,3^(-3/2))--(2,3^(-3/2))--cycle); draw((3,0)--(4,0)--(4,4^(-3/2))--(3,4^(-3/2))--cycle); draw((6,0)--(7,0)--(7,7^(-3/2))--(6,7^(-3/2))--cycle); draw((7,0)--(8,0)--(8,8^(-3/2))--(7,8^(-3/2))--cycle); label(\"$X_1$\",(1.5,0),0.5*N); label(\"$X_2$\",(2.5,0),0.5*N); label(\"$X_3$\",(3.5,0),0.5*N); label(\"$\\cdots$\",(5,0),N); label(\"$X_{n-1}$\",(6.5,0),0.5*N); label(\"$X_n$\",(7.5,0),0.5*N); label(\"$a_0$\",(1,0),0.5*S); label(\"$a_1$\",(2,0),0.5*S); label(\"$a_2$\",(3,0),0.5*S); label(\"$a_3$\",(4,0),0.5*S); label(\"$a_{n-2}$\",(6,0),0.5*S); label(\"$a_{n-1}$\",(7,0),0.5*S); label(\"$a_n$\",(8,0),0.5*S); [/asy]\n\\]\n\nFor all \\(k \\in \\mathbb{N}\\), the area of \\(X_k\\) is \\((a_k - a_{k-1})\\left(a_k^{-\\dfrac{3}{2}}\\right)\\). Therefore, \\(b_n = \\sum_{k=1}^{n} [X_k]\\)\n\nFor all sequences \\(\\{ a_k \\}\\) and all \\(k \\in \\mathbb{N}\\), \\(X_k\\) lies above the \\(x\\)-axis, below the curve \\(f(x) = x^{-\\dfrac{3}{2}}\\), and in between the lines \\(x = 1\\) and \\(x = a_n\\), Also, all such rectangles are disjoint.\n\nThus, \\(b_n = \\sum_{k=1}^{n} [X_k] < \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx = \\left[-\\dfrac{2}{\\sqrt{x}}\\right]_{1}^{a_n} = 2 - \\dfrac{2}{\\sqrt{a_n}} < 2\\) as desired.\n\nBy choosing \\(a_k = 1 + k (\\Delta x)\\), where \\(\\Delta x > 0\\), \\(b_n\\) is a Riemann sum for \\(\\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx\\). Thus, \\(\\lim_{\\Delta x \\to 0^+} b_n = \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx\\).\n\nTherefore, \\(\\lim_{n \\to \\infty} \\left[ \\lim_{\\Delta x \\to 0^+} b_n \\right] = \\lim_{n \\to \\infty} \\int_{1}^{a_n} x^{-\\dfrac{3}{2}} \\,dx\\) \\(= \\lim_{n \\to \\infty} 2 - \\dfrac{2}{\\sqrt{a_n}} = \\lim_{n \\to \\infty} 2 - \\dfrac{2}{\\sqrt{1+n(\\Delta x)}} = 2\\).\n\nSo for any \\(c \\in [0,2)\\), we can always select a small enough \\(\\Delta x > 0\\) to form a sequence \\(\\{ a_n \\}\\)satisfying the above properties such that \\(b_n > c\\) for large enough \\(n\\) as desired.\n\n## Remarks (added by pf02, November 2024)\n\n1. A remark about the problem: it is not a fair problem. It is not a good problem for the International Mathematical Olympiad.\n\nThe reason is that the problem is trivial for most anyone who knows a little bit of integral calculus. Taken out of its natural context, it is not trivial. The young people participating in the International Mathematical Olympiad are from an age group whose mathematical education may not have reached calculus, depending on the country they come from. Hence, the problem is more a comparison between mathematical curricula in different countries rather than problem solving ability of young people.\n\n2. The Solution to question (b) of the problem given above is incorrect. Specifically\n\n\\[\n\\lim_{\\Delta x \\to 0^+} b_n = \\int_{1}^{a_n} x^{-\\frac{3}{2}} dx\n\\]\n\nis incorrect. (To put it bluntly, it makes no sense!) The upper limit of the integral, \\(a_n\\) depends on \\(\\Delta x\\), so it can not be the result of a limit \\(\\lim_{\\Delta x \\to 0^+}\\)! In fact, it is not even clear what is meant by \\(\\lim_{\\Delta x \\to 0^+} b_n\\).\n\nThe errors and inconsistencies just continue accumulating and building up in the remainder of the proof.\n\n3. Below, I will give a corrected proof of part (b) of the problem, similar to some extent to the idea the author of the Solution above might have had in mind. Then, I will give a solution which uses no calculus.",
"For simplicity of writing and visualizing, let us denote \\(\\epsilon = 2 - c\\). So \\(c = 2 - \\epsilon\\). We need to find a sequence \\(\\{a_k\\}\\) so that \\(b_n > 2 - \\epsilon\\) for \\(n\\) large enough.\n\nWe rely on the fact that \\(b_n = \\sum_{k=1}^{n} a_k^{-\\frac{3}{2}} (a_k - a_{k-1})\\) is a Riemann sum for \\(\\int_{1}^{A} x^{-\\frac{3}{2}} dx\\), where \\(A = a_n\\). The denser the points \\(\\{a_k\\}\\), the better the sum approximates the integral.\n\nSince \\(\\int_{1}^{\\infty} x^{-\\frac{3}{2}} dx = 2\\), we can find \\(A\\) so that \\(2 - \\int_{1}^{A} x^{-\\frac{3}{2}} dx < \\frac{\\epsilon}{2}\\).\n\nNow, we can find a set of points \\(1 = a_0 < a_1 < \\cdots < a_n = A\\) in the interval \\([1, A]\\), such that the Riemann sum \\(b_n\\) corresponding to these points is as close to the integral as we want, in particular \\(\\int_{1}^{A} x^{-\\frac{3}{2}} dx - b_n < \\frac{\\epsilon}{2}\\).\n\nThis proves part (b) of the problem.\n\n(Note that we don't have to, but we can take \\(a_k = 1 + k \\alpha\\) as suggested in the Solution. In fact, we can take \\(\\alpha = (A - 1)/N\\) and let \\(N\\) be large enough, to make \\(\\alpha\\) small enough, to make the Riemann sum close enough to the integral on the interval \\([1, A]\\). (Here \\(\\alpha\\) stands for the \\(\\Delta x\\) in the Solution.))\n\n[Solution by pf02, November 2024]",
"This is a solution which does not use calculus. Taking in account that \\(a_{k-1} \\le a_k\\) we have\n\n\\[\n\\left( 1 - \\frac{a_{k-1}}{a_k} \\right) \\frac{1}{\\sqrt{a_k}} = \\frac{a_k - a_{k-1}}{a_k \\sqrt{a_k}} \\le \\frac{a_k - a_{k-1}}{\\sqrt{a_{k-1}}\\ \\sqrt{a_k}\\ \\frac{\\sqrt{a_k} + \\sqrt{a_{k-1}}}{2}} = \\frac{2(\\sqrt{a_k} - \\sqrt{a_{k-1}})}{\\sqrt{a_{k-1}} \\sqrt{a_k}} = 2 \\left( \\frac{1}{\\sqrt{a_{k-1}}} - \\frac{1}{\\sqrt{a_k}} \\right)\n\\]\n\nTaking the sum, we get\n\n\\(b_n = \\sum_{k=1}^n \\left( 1 - \\frac{a_{k-1}}{a_k} \\right) \\frac{1}{\\sqrt{a_k}} \\le \\sum_{k=1}^n 2 \\left( \\frac{1}{\\sqrt{a_{k-1}}} - \\frac{1}{\\sqrt{a_k}} \\right) = 2 \\left( 1 - \\frac{1}{\\sqrt{a_n}} \\right) < 2\\).\n\nThis proves (a).\n\nTo prove (b), let us take \\(a_k = 1 + k \\alpha\\), with \\(\\alpha >0\\). We do similar tricks:\n\n\\[\n\\left( 1 - \\frac{a_{k-1}}{a_k} \\right) \\frac{1}{\\sqrt{a_k}} = \\frac{a_k - a_{k-1}}{a_k \\sqrt{a_k}} = \\frac{\\alpha}{(1 + k \\alpha) \\sqrt{1 + k \\alpha}} \\ge\n\\]\n\n\\[\n\\frac{\\alpha}{\\sqrt{1 + k \\alpha}\\ \\sqrt{1 + (k+1) \\alpha}\\ \\frac{\\sqrt{1 + (k+1) \\alpha} + \\sqrt{1 + k \\alpha}}{2}} = \\frac{2\\alpha (\\sqrt{1 + (k+1) \\alpha} - \\sqrt{1 + k \\alpha})}{\\sqrt{1 + k \\alpha}\\ \\sqrt{1 + (k+1) \\alpha}\\ [(1 + (k+1) \\alpha) - (1 + k \\alpha)]} =\n\\]\n\n\\[\n2 \\left( \\frac{1}{\\sqrt{1 + k \\alpha}} - \\frac{1}{\\sqrt{1 + (k+1) \\alpha}} \\right)\n\\]\n\nNow take the sum for \\(k = 1, \\cdots, n\\):\n\n\\[\nb_n = \\sum_{k=1}^n \\left( 1 - \\frac{a_{k-1}}{a_k} \\right) \\frac{1}{\\sqrt{a_k}} \\ge \\sum_{k=1}^n \\left( \\frac{2}{\\sqrt{1 + k \\alpha}} - \\frac{2}{\\sqrt{1 + (k+1) \\alpha}} \\right) = \\frac{2}{\\sqrt{1 + \\alpha}} - \\frac{2}{\\sqrt{1 + \\alpha + n \\alpha}}\n\\]\n\nFor simplicity of writing and visualizing, let us denote \\(\\epsilon = 2 - c\\). So \\(c = 2 - \\epsilon\\). We need to show that we can take \\(\\alpha\\) small enough and \\(n\\) large enough so that \\(2 - b_n < \\epsilon\\).\n\nStart by choosing a value for \\(\\alpha\\) small enough to make \\(\\frac{2}{\\sqrt{1 + \\alpha}} > 2 - \\frac{\\epsilon}{2}\\).\n\nThis can be done since this condition amounts to \\(\\sqrt{1 + \\alpha} < \\frac{4}{4 - \\epsilon}\\), or in other words to \\(\\alpha < \\left( \\frac{4}{4 - \\epsilon} \\right)^2 - 1\\).\n\nFor the chosen value of \\(\\alpha\\) we now choose \\(n\\) large enough so that \\(\\frac{2}{\\sqrt{1 + \\alpha + n \\alpha}} < \\frac{\\epsilon}{2}\\).\n\nThis can also be done because the condition amounts to \\(\\sqrt{1 + \\alpha + n \\alpha} > \\frac{4}{\\epsilon}\\), or in other words \\(n > \\frac{1}{\\alpha}\\ \\left( \\frac{16}{\\epsilon^2} - 1 - \\alpha \\right)\\).\n\nThis proves part (b) of the problem.\n\n[Solution by pf02, November 2024]"
] |
IMO-1970-4
|
https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_4
|
Find the set of all positive integers \(n\) with the property that the set \(\{ n, n+1, n+2, n+3, n+4, n+5 \}\) can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.
|
[
"The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be \\(\\{ 1, 2, 3, 4, 5, 6 \\}\\), but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.",
"As in the previous solution, none of the six consecutive numbers can be multiples of \\(7\\). This means that together, they take on the values \\(\\{ 1, 2, 3, 4, 5, 6 \\} \\mod 7\\). The product of all the numbers in this set, then, is \\(-1 \\mod 7\\), by Wilson's Theorem. However, \\(-1\\) is not a quadratic residue \\(\\mod 7\\), which means that we cannot partition the original set into two sets of equal product. Thus, no such \\(n\\) exist.\n\n## Remarks (added by pf02, December 2024)\n\n1. I will re-write Solution 2 without making references to Wilson's Theorem or quadratic residue. Beautiful as they are, they are not needed here, so I think it is worth re-writing this elegant proof so that it can be understood by students who know nothing more than the notion of \\(mod\\) (modulo).\n\n2. I will then give another solution, which in fact, proves a more difficult problem: the product of six consecutive integers can not be a perfect square.\n\n3. Just as a matter of general interest, the more difficult problem is the subject of a discussion at https://math.stackexchange.com/questions/90894/product-of-six-consecutive-integers-being-a-perfect-square . As of December 2024 there are several solutions presented in this discussion (I can make sense of two of them), as well as two references to publications where the problem is solved. The references are https://www.jstor.org/stable/2635974?seq=1 (to a journal article of 1874) and http://www.renyi.hu/~p_erdos/1939-03.pdf to a paper from 1939. This second paper proves a more general statement: a product of any number of consecutive integers can not be a perfect square.",
"Let \\(\\{m_1, m_2, m_3\\}\\) and \\(\\{m_4, m_5, m_6\\}\\) be a partition of \\(\\{n, n+1, n+2, n+3, n+4, n+5\\}\\). If a number \\(p \\ge 4\\) is a factor of \\(m_1m_2m_3,\\) then \\(p\\) must also be a factor of \\(m_4m_5m_6\\). In particular, if \\(7\\) is a factor of one of the numbers in \\(\\{n, n+1, n+2, n+3, n+4, n+5\\}\\), then it must be a factor of another, different number in this set. But \\(7\\) can be a factor of at most one of six consecutive integers, so it is a factor of none of them.\n\nSince none of the numbers in \\(\\{n, n+1, n+2, n+3, n+4, n+5\\}\\) is divisible by \\(7\\), the six values of \\(\\{q_i, i = 1, \\dots, 6\\}\\) from \\(n = q_1 \\mod 7, n_1 = q_2 \\mod 7, n_2 = q_3 \\mod 7, n_3 = q_4 \\mod 7, n_4 = q_5 \\mod 7, n_5 = q_6 \\mod 7\\) are \\(1, 2, 3, 4, 5, 6\\).\n\nSo \\(n(n+1)(n+2)(n+3)(n+4)(n+5) = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 = 720 = 6 \\mod 7\\).\n\nOn the other hand, if \\(m\\) is an integer, then \\(m^2 \\mod 7\\) is one of \\(0, 1, 2, 4\\). Indeed let \\(m = 7p + q\\) with \\(q \\in \\{0, 1, 2, 3, 4, 5, 6\\}\\). Calculating \\(m^2\\), we see that the result will be a multiple of \\(7\\) plus \\(0, 1, 4, 9, 16, 25, 36\\), which are \\(0, 1, 4, 2, 2, 4, 1 \\mod 7\\), none of which is \\(6 \\mod 7\\).\n\nSo, \\(n(n+1)(n+2)(n+3)(n+4)(n+5)\\) can not be a perfect square, as it should be if \\(m_1m_2m_3 = m_4m_5m_6\\).",
"If \\(\\{n, n+1, n+2, n+3, n+4, n+5\\}\\) can be partitioned in two sets \\(\\{m_1, m_2, m_3\\}\\) and \\(\\{m_4, m_5, m_6\\}\\) such that \\(m_1m_2m_3 = m_4m_5m_6\\) then \\(N = n(n+1)(n+2)(n+3)(n+4)(n+5)\\) must be a perfect square. To shorten the text, when a number \\(N\\) is a perfect square, we'll write \\(N = \\square\\). Se we want to prove a more difficult problem, namely \\(N = n(n+1)(n+2)(n+3)(n+4)(n+5) \\ne \\square\\). Assume for the sake of reaching a contradiction that \\(N = \\square\\).\n\nRegroup the factors of \\(N\\) and do some multiplications. We get \\((n^2 + 5n)[(n^2 + 5n) + 4][(n^2 + 5n) + 6] = \\square\\). Denote \\(m = n^2 + 5n\\). We have \\(m(m+4)(m+6) = \\square\\). Let \\(m = p^2q\\), such that all the prime factors of \\(q\\) are at power \\(1\\) (in other words, \\(p^2\\) is the product of all the prime factors of \\(m\\) at an even power, and \\(q\\) is the product of all the remaining prime factors of \\(m\\)).\n\nThen, we have \\(q(p^2q + 4)(p^2q + 6) = \\square\\). Multiplying, we get that \\(q(Aq + 24) = \\square\\) for some number \\(A\\) (in fact \\(A = p^4q + 10p^2\\)). Note that if a prime \\(a\\) is a factor of \\(q\\), then it is a factor of the right hand side \\(\\square\\), so \\(a^2\\) is a factor of the right hand side \\(\\square\\), so \\(a^2\\) is a factor of \\(q(Aq + 24)\\), so \\(a\\) is a factor of \\(Aq + 24\\), so \\(a\\) is a factor of \\(24\\). This implies that \\(a = 2\\) or \\(a = 3\\), which in its turn implies that \\(q = 2\\), or \\(q = 3\\) or \\(q = 6\\) (since \\(q\\) has its factors at power \\(1\\)).\n\nWe will continue the proof by contradiction by taking three cases (for \\(q = 2, 3, 6\\)), and showing that none of the following three statements can be true: \\(2(2p^2 + 4)(2p^2 + 6) = \\square\\), \\(3(3p^2 + 4)(3p^2 + 6) = \\square\\), \\(6(6p^2 + 4)(6p^2 + 6) = \\square\\).\n\nCase 1: \\(2(2p^2 + 4)(2p^2 + 6) = \\square\\) becomes \\(2(p^2 + 2)(p^2 + 3) = \\square\\). Note that \\(p^2 + 2\\) and \\(p^2 + 3\\) have no common factors (since they differ by \\(1\\)) so either\n\nCase 1.1: \\(p^2 + 2 = \\square\\) and \\(2(p^2 + 3) = \\square\\)\n\nor\n\nCase 1.2: \\(2(p^2 + 2) = \\square\\) and \\(p^2 + 3 = \\square\\)\n\nCase 1.1 is impossible because \\(p^2 + 2 = \\square\\) is impossible (because two squares differ by more than \\(2\\)).\n\nCase 1.2 implies \\(p = 1\\), because only the squares \\(1\\) and \\(4\\) differ by \\(3\\) (all higher squares differ by more than \\(3\\)). But if \\(p = 1\\) the other condition in case 1.2 becomes \\(2(1 + 2) = \\square\\), which is false.\n\nSo Case 1 is impossible.\n\nCase 2: The condition becomes \\((3p^2 + 4)(p^2 + 2) = \\square\\). We consider two sub-cases corresponding to \\(p =\\) odd or \\(p =\\) even.\n\nCase 2.1: When \\(p =\\) odd, \\(3p^2 + 4\\) and \\(p^2 + 2\\) have no common factors because \\(3p^2 + 4\\) and \\(3(p^2 + 2) = 3p^2 + 6\\) have no common factors (because we have two odd numbers which differ by \\(2\\)). So we must have \\(3p^2 + 4 = \\square\\) and \\(p^2 + 2 = \\square\\). But this is impossible because of the latter condition.\n\nCase 2.2: If \\(p =\\) even, let \\(p = 2r\\). The condition becomes \\(2(3r^2 + 1)(2r^2 + 1) = \\square\\). Note that \\(3r^2 + 1\\) and \\(2r^2 + 1\\) have no common factors (because their multiples by \\(2\\) and \\(3\\) are \\(6r^2 + 2\\) and \\(6r^2 + 3\\) which differ by \\(1\\)).\n\nCase 2.2.1: Take the sub-sub-case \\(2(3r^2 + 1) = \\square\\) and \\(2r^2 + 1 = \\square\\). The first condition implies that we must have \\(3r^2 + 1 = 2a^2\\) for some \\(a\\), or \\(3(r^2 + 1) = 2(a^2 + 1)\\). But this is impossible, since \\(a^2 + 1\\) can not be a multiple of \\(3\\). (Indeed, take \\(a = 3b\\), \\(a = 3b + 1\\), \\(a = 3b + 2\\); \\(a^2 = 0 \\mod 3\\) or \\(a^2 = 1 \\mod 3\\)).\n\nCase 2.2.2: Take the sub-sub-case \\(3r^2 + 1 = \\square\\) and \\(2(2r^2 + 1) = \\square\\). The second condition is impossible because \\(2(2r^2 + 1)\\) has exactly \\(2^1\\) as a factor, so it can not be a square.\n\nCase 3: \\(6(6p^2 + 4)(6p^2 + 6) = \\square\\) becomes \\(2(3p^2 + 2)(p^2 + 1) = \\square\\). We see that \\(3p^2 + 2\\) and \\(p^2 + 1\\) have no common factors since \\(3p^2 + 2\\) and \\(3(p^2 + 1) = 3p^2 + 3\\) differ by \\(1\\). So we have\n\nCase 3.1: \\(2(3p^2 + 2) = \\square\\) and \\(p^2 + 1 = \\square\\). This case is impossible, since the latter condition is impossible (because two squares differ by more than \\(1\\)).\n\nCase 3.2: \\(3p^2 + 2 = \\square\\) and \\(2(p^2 + 1) = \\square\\). The first condition implies \\(3(p^2 + 1) = a^2 + 1\\), but this is impossible since as we saw in Case 2.2.2, \\(a^2 + 1\\) can not be a multiple of \\(3\\).\n\nWe looked at all possible cases, and they were impossible, so \\(N = n(n+1)(n+2)(n+3)(n+4)(n+5)\\) can not be a perfect square.\n\n[Solution by pf02, December 2024]"
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IMO-1970-5
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https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_5
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In the tetrahedron \(ABCD\), angle \(BDC\) is a right angle. Suppose that the foot \(H\) of the perpendicular from \(D\) to the plane \(ABC\) in the tetrahedron is the intersection of the altitudes of \(\triangle ABC\). Prove that
\[
( AB+BC+CA )^2 \leq 6( AD^2 + BD^2 + CD^2 ).
\]
For what tetrahedra does equality hold?
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[
"Let us show first that angles \\(ADB\\) and \\(ADC\\) are also right. Let \\(H\\) be the intersection of the altitudes of \\(ABC\\) and let \\(CH\\) meet \\(AB\\) at \\(E\\). Planes \\(CED\\) and \\(ABC\\) are perpendicular and \\(AB\\) is perpendicular to the line of intersection \\(CE\\). Hence \\(AB\\) is perpendicular to the plane \\(CDE\\) and hence to \\(ED\\). So \\(BD^2 = DE^2 + BE^2.\\) Also \\(CB^2 = CE^2 + BE^2.\\) Therefore \\(CB^2 - BD^2 = CE^2 - DE^2.\\) But \\(CB^2 - BD^2 = CD^2,\\) so \\(CE^2 = CD^2 + DE^2\\), so angle \\(CDE = 90^{\\circ}\\). But angle \\(CDB = 90^{\\circ}\\), so \\(CD\\) is perpendicular to the plane \\(DAB\\), and hence angle \\(CDA\\) = \\(90^{\\circ}\\). Similarly, angle \\(ADB = 90^{\\circ}\\). Hence \\(AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)\\).\n\nBut now we are done, because Cauchy's inequality (applied to vectors \\((AB, BC, CA)\\) and \\((1, 1, 1)\\)) gives \\((AB + BC + CA)^2 \\le 3(AB^2 + BC^2 + CA^2)\\).\n\nWe have equality if and only if we have equality in Cauchy's inequality, which means \\(AB = BC = CA.\\)",
"Let \\(x = DH, a = BC, b = CA, c = AB\\)\n\n\\[\nProb 1970 5.png\n\\]\n\nThe plan of this proof is to compute \\(HA, HB, HC\\) in terms of \\(a, b, c\\), then compute \\(DA^2, DB^2, DC^2\\) in terms of \\(a, b, c, x\\), impose the condition that \\(\\angle BDC = \\pi/2\\) to determine \\(x\\), and calculate \\(DA^2 + DB^2 + DC^2\\) in terms of \\(a, b, c\\). The problem will become a simple inequality in \\(a, b, c\\) which will be easy to prove.\n\nFormulas for the distance from the orthocenter to the vertices are reasonably well known, but to make this solution self contained, we compute them here. From \\(\\triangle ABE_B\\) we have that \\(BE_B = c \\sin A\\). From \\(\\triangle CBE\\) we have \\(BE = a \\cos B\\). From \\(\\triangle ABE_B \\sim \\triangle HBE\\) we have \\(\\frac{HB}{c} = \\frac{BE}{BE_B}\\). It follows that \\(HB = c \\ \\frac{a \\cos B}{c \\sin A} = \\frac{a \\cos B}{\\sin A}\\).\n\nSimilarly, we have \\(HC = \\frac{a \\cos C}{\\sin A}\\) and \\(HA = \\frac{b \\cos A}{\\sin B} = \\frac{a \\cos A}{\\sin A}\\).\n\n(In the last equality we used the Law of Sines: \\(\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}.\\) We want to have \\(\\sin A\\) at the denominator to simplify computations.)\n\nThen \\(DA^2 = x^2 + \\frac{a^2 \\cos^2 A}{\\sin^2 A}, DB^2 = x^2 + \\frac{a^2 \\cos^2 B}{\\sin^2 A}, DC^2 = x^2 + \\frac{a^2 \\cos^2 C}{\\sin^2 A}\\).\n\nSince \\(\\angle BDC = \\pi/2, DB^2 + DC^2 = a^2\\). This yields \\(x^2 = \\frac{a^2}{2 \\sin^2 A} \\ (\\sin^2 A - \\cos^2 B - \\cos^2 C)\\).\n\nUsing this value for \\(x^2\\) we get \\(DA^2 + DB^2 + DC^2 = \\frac{a^2}{2 \\sin^2 A} \\ (3 \\sin^2 A + 2 \\cos^2 A - \\cos^2 B - \\cos^2 C)\\)\n\nAfter some simplifications, and using again the Law of Sines, this becomes\n\n\\[\nDA^2 + DB^2 + DC^2 = \\frac{a^2}{2 \\sin^2 A} \\ (\\sin^2 A + \\sin^2 B + \\sin^2 C) = \\frac{a^2}{2} + \\frac{a^2}{2 \\sin^2 A} \\sin^2 B + \\frac{a^2}{2 \\sin^2 A} \\sin^2 C =\n\\]\n\n\\(\\frac{a^2}{2} + \\frac{b^2}{2 \\sin^2 B} \\sin^2 B + \\frac{c^2}{2 \\sin^2 C} \\sin^2 C = \\frac{1}{2} \\ (a^2 + b^2 + c^2)\\).\n\nWhat we need to prove then is that \\((a + b + c)^2 \\le 3(a^2 + b^2 + c^2)\\). We could invoke the Cauchy-Schwarz Inequality like in the first solution, but we can just as well make the computations, and notice that this is equivalent to\n\n\\(2 a^2 + 2 b^2 + 2 c^2 -2ab - 2bc - 2ca \\ge 0\\) or \\((a - b)^2 + (b - c)^2 + (c - a)^2 \\ge 0\\), which is always true.\n\nWe also see that we have equality if and only if \\(a = b = c\\).\n\n[Solution by pf02, December 2024]"
] |
IMO-1970-6
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https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_6
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In a plane there are \(100\) points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than \(70 \%\) of these triangles are acute-angled.
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[
"At most \\(3\\) of the triangles formed by \\(4\\) points can be acute. It follows that at most \\(7\\) out of the \\(10\\) triangles formed by any \\(5\\) points can be acute. For given \\(10\\) points, the maximum number of acute triangles is: the number of subsets of \\(4\\) points times \\(\\frac{3}{\\text{the number of subsets of 4 points containing 3 given points}}\\). The total number of triangles is the same expression with the first \\(3\\) replaced by \\(4\\). Hence at most \\(\\frac{3}{4}\\) of the \\(10\\), or \\(7.5\\), can be acute, and hence at most \\(7\\) can be acute. The same argument now extends the result to \\(100\\) points. The maximum number of acute triangles formed by \\(100\\) points is: the number of subsets of \\(5\\) points times \\(\\frac{7}{\\text{the number of subsets of 5 points containing 3 given points}}\\). The total number of triangles is the same expression with \\(7\\) replaced by \\(10\\). Hence at most \\(\\frac{7}{10}\\) of the triangles are acute.\n\n## Remarks (added by pf02, December 2024)\n\n1. The solution above contains an error (a typo?) and skips too many steps, which make it very hard to understand. For the benefit of future readers, and as a public service, I re-write the proof below.\n\n2. Note the commonality with 1969 IMO Problems/Problem 5. In fact, the solution to this problem has a strong commonality with Solution 2 of 1969 IMO Problems/Problem 5.",
"Define a \\(n\\)-gon to be a configuration of \\(n\\) points, no three of which are collinear. Given an \\(n\\)-gon draw all the triangles whose vertices are among the \\(n\\) points. We say that a triangle \\(\\triangle ABC\\) is in the \\(n\\)-gon, or that the \\(n\\)-gon has \\(\\triangle ABC\\) in it, if \\(A, B, C\\) are points in the \\(n\\)-gon.\n\nWe start by proving that a \\(4\\)-gon has \\(4\\) triangles, out of which at most \\(3\\) are acute.\n\n\\[\nProb 1970 6.png\n\\]\n\nClearly there are exactly \\(4\\) triangles. We have two cases. If the convex hull of the \\(4\\)-gon is a quadrilateral, then at least one of \\(\\angle A, \\angle B, \\angle C, \\angle D\\) is \\(\\ge \\frac{\\pi}{2}\\) (if they were all \\(< \\frac{\\pi}{2}\\) then \\(\\angle A + \\angle B + \\angle C + \\angle D < 2\\pi\\), which is false). Let us say that \\(\\angle C \\ge \\frac{\\pi}{2}\\). Then \\(\\triangle BCD\\) is obtuse, so at most the other \\(3\\) triangles are acute. (Note that it is possible to have \\(3\\) acute triangles.)\n\nIf the convex hull of the \\(4\\)-gon is a triangle, let us say that \\(C\\) is in the interior of \\(\\triangle ABD\\). Then at least two of \\(\\angle ACB, \\angle BCD, \\angle DCA\\) are \\(\\ge \\frac{2\\pi}{3} > \\frac{\\pi}{2}\\) (if they were all \\(< \\frac{2\\pi}{3}\\) then their sum would be \\(< 2\\pi\\), which is false). So, at least two of \\(\\triangle ACB, \\triangle BCD, \\triangle DCA\\) are obtuse.\n\nNote: We would be tempted to use this result, and the argument which follows below to prove the problem. But that would only yield the result that at most 75% of the triangles in an \\(n\\)-gon (in particular a \\(100\\)-gon) are acute. We need to do better.\n\nNow, we will prove that a \\(5\\)-gon has \\(10\\) triangles, out of which at most \\(7\\) are acute. It is easy to see that there are \\(10\\) triangles in a \\(5\\)-gon, either by counting, or by remembering that the number of triangles \\(= C(5, 3) = {5 \\choose 3} = \\frac{5 \\cdot 4 \\cdot 3}{1 \\cdot 2 \\cdot 3} = 10\\). Now, we could proceed like in the case of a \\(4\\)-gon, by looking at many possible cases and chasing angle sizes. Instead, we will give a different argument, using an idea we will use again later.\n\nA \\(5\\)-gon has \\(5\\) \\(4\\)-gons in it. Each of these \\(4\\)-gons has \\(4\\) triangles in it, of which at most \\(3\\) are acute. Make a list \\(\\mathcal{L}_1\\) of all the triangles, and a sub-list \\(\\mathcal{L}_2\\) of all the acute triangles. The size of \\(\\mathcal{L}_1 = 5 \\cdot 4 = 20\\), and the size of \\(\\mathcal{L}_2 \\le 5 \\cdot 3 = 15\\). These lists contain duplicates. Each triangle is counted exactly twice (corresponding to the two points which are not in the triangle; we can see this as \\(= C(5-3, 4-3) = C(2, 1) = {2 \\choose 1} = 2\\)). Dividing by \\(2\\) to eliminate duplicates, we get that out of the \\(\\frac{20}{2} = 10\\) triangles, at most \\(\\frac{15}{2} = 7.5\\) are acute. We conclude that at most \\(7\\) out of the \\(10\\) are acute.\n\nNow we are ready to prove that given a \\(n\\)-gon with \\(n > 5\\), at most \\(\\frac{7}{10} = 70\\%\\) of its triangles are acute. Consider all the \\(5\\)-gons in the given \\(n\\)-gon, and consider all the triangles in each \\(5\\)-gon. Make a list \\(\\mathcal{L}_1\\) of all these triangles, and a sub-list \\(\\mathcal{L}_2\\) of all the acute triangles. We have that \\(\\frac{\\text{size of } \\mathcal{L}_2}{\\text{size of } \\mathcal{L}_1} \\le \\frac{7}{10}\\) (according to what we showed above). These lists contain duplicates. Each triangle is counted the same number of times, say \\(k\\) times (in fact, \\(k = C(n-3, 5-3) = C(n-3, 2) = {n-3 \\choose 2}\\), but we don't need this). Dividing both the numerator and the denominator by \\(k\\) to eliminate duplicates, we have that at most \\(\\frac{7}{10} = 70\\%\\) of the triangles are acute.\n\n[Re-writing by pf02, December 2024]"
] |
IMO-1971-1
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_1
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Prove that the following assertion is true for \(n=3\) and \(n=5\), and that it is false for every other natural number \(n>2:\)
If \(a_1, a_2,\cdots, a_n\) are arbitrary real numbers, then \((a_1-a_2)(a_1-a_3)\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots (a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots (a_n-a_{n-1})\ge 0.\)
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[
"Denote \\(E_n\\) the expression in the problem, and denote \\(S_n\\) the statement that \\(E_n \\ge 0\\).\n\nTake \\(a_1 < 0\\), and the remaining \\(a_i = 0\\). Then \\(E_n = a_1^{n-1} < 0\\) for \\(n\\) even. So the proposition is false for even \\(n\\).\n\nSuppose \\(n \\ge 7\\) and odd. Take any \\(c > a > b\\), and let \\(a_1 = a\\), \\(a_2 = a_3 = a_4= b\\), and \\(a_5 = a_6 = ... = a_n = c\\). Then \\(E_n = (a - b)^3 (a - c)^{n-4} < 0\\). So the proposition is false for odd \\(n \\ge 7\\).\n\nAssume \\(a_1 \\ge a_2 \\ge a_3\\). Then in \\(E_3\\) the sum of the first two terms is non-negative, because \\(a_1 - a_3 \\ge a_2 - a_3\\). The last term is also non-negative. Hence \\(E_3 \\ge 0\\), and the proposition is true for \\(n = 3\\).\n\nIt remains to prove \\(S_5\\). Suppose \\(a_1 \\ge a_2 \\ge a_3 \\ge a_4 \\ge a_5\\). Then the sum of the first two terms in \\(E_5\\) is \\((a_1 - a_2)[(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)] \\ge 0\\).\n\nThe third term in \\(E_5\\) is non-negative (the first two factors are non-positive and the last two non-negative).\n\nThe sum of the last two terms in \\(E_5\\) is: \\((a_4 - a_5)[(a_1 - a_5)(a_2 - a_5)(a_3 - a_5) - (a_1 - a_4)(a_2 - a_4)(a_3 - a_4)] \\ge 0\\).\n\nHence \\(E_5 \\ge 0\\).\n\nThis solution was posted and copyrighted by e.lopes. The original thread can be found here: [1]\n\n## Remarks (added by pf02, December 2024)\n\n1. As a public service, I fixed a few typos in the solution above.\n\n2. Make the solution a little more complete:\n\n2.1. Let us note that the assumptions \\(a_1 \\ge a_2 \\ge a_3\\) in case \\(n = 3\\) and \\(a_1 \\ge a_2 \\ge a_3 \\ge a_4 \\ge a_5\\) in case \\(n = 5\\) are perfectly legitimate. A different ordering of these numbers could be reduced to this case by a simple change of notation: we would substitute \\(a_i\\) by \\(b_j\\) with the indexes for the \\(b\\)'s chosen in such a way that the inequalities above are true for the \\(b\\)'s.\n\n2.2. The inequality \\((a_1 - a_2)[(a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5)] \\ge 0\\) is true because \\(a_1 - a_2 \\ge 0\\), and \\((a_1 - a_3)(a_1 - a_4)(a_1 - a_5) - (a_2 - a_3)(a_2 - a_4)(a_2 - a_5) \\ge 0\\). To see this latter inequality, just notice that \\(a_1 - a_3 \\ge a_2 - a_3\\), and similarly for the other pairs of factors. The difference of the products is \\(\\ge 0\\) as desired.\n\nThe argument for the sum of the last two terms in \\(E_5\\) is similar.\n\n3. The case \\(n = 3\\) is very easy to prove in a different way. Note that\n\n\\[\n(a_1 - a_2)(a_1 - a_3) + (a_2 - a_1)(a_2 - a_3) + (a_3 - a_1)(a_3 - a_2) = \\frac{1}{2}[(a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_1)^2]\n\\]\n\nI could not find an identity which would give such a simple proof in the case \\(n = 5\\).\n\n4. By looking at the proof above, we can also see that for \\(n = 3\\) we have equality if and only if \\(a_1 = a_2 = a_3\\). For \\(n = 5\\), assuming that \\(a_1 \\ge a_2 \\ge a_3 \\ge a_4 \\ge a_5\\), we have equality if and only if \\(a_1 = a_2\\) and \\(a_3 = a_4 = a_5\\), or \\(a_1 = a_2 = a_3\\) and \\(a_4 = a_5\\).\n\n5. If we denote \\(P(x) = (x - a_1) \\cdots (x - a_n)\\), then the expression in the problem is \\(E_n = P'(a_1) + \\cdots + P'(a_n)\\), where \\(P'(x)\\) is the derivative of \\(P(x)\\). (This is easy to see by calculating \\(P'(x)\\) when \\(P(x)\\) is written as a product rather than as a sum of powers of \\(x\\).)\n\nThe graph of \\(P(x)\\) as \\(x\\) goes from \\(-\\infty\\) to \\(\\infty\\) crosses the \\(x\\)-axis, or is tangent to the \\(x\\)-axis at every root \\(a_k\\). If the graph is tangent to the \\(x\\)-axis, it crosses it, or it stays in the same half plane, depending on the multiplicity of the root. At a simple root \\(a_k, P'(a_k) > 0\\) or \\(P'(a_k) < 0\\) depending on the direction of the graph of \\(P(x)\\) at \\(a_k\\). At a multiple root \\(a_k = \\cdots = a_{k+p}, P'(a_k) = 0\\), and the graph of \\(P(x)\\) crosses the \\(x\\)-axis or not, depending on \\(p\\) being odd or even.\n\nThis way of looking at the problem makes it very easy to find examples which prove the problem for \\(n\\) even, or \\(n \\ge 7\\) and odd, because we would be looking for polynomials whose graph crosses the \\(x\\)-axis once from the upper half plane to the lower half plane at a simple root \\(a_k\\) (thus making \\(P'(a_k) < 0\\)), and is tangent to the \\(x\\)-axis at all the other roots. See the picture below for images showing the graphs of such polynomials.\n\n\\[\nProb 1971 1.png\n\\]\n\n(Wichking makes essentially the same remarks on https://aops.com/community/p366761.)"
] |
IMO-1971-2
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_2
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Consider a convex polyhedron \(P_1\) with nine vertices \(A_1, A_2, \cdots, A_9;\) let \(P_i\) be the polyhedron obtained from \(P_1\) by a translation that moves vertex \(A_1\) to \(A_i(i=2,3,\cdots, 9).\) Prove that at least two of the polyhedra \(P_1, P_2,\cdots, P_9\) have an interior point in common.
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[
"WLOG let \\(A_1\\) be the origin \\(0\\). Take any point \\(A_i\\), then \\(P_i=A_i+P_1\\), lies in \\(2 P_1\\), the polyhedron \\(P_1\\) stretched by the factor \\(2\\) on \\(P_1=0\\). More general: take any \\(p,q\\) in any convex shape \\(S\\). Then \\(p+q \\in 2S\\). Prove: since \\(S\\) is convex, \\(\\frac{p+q}{2} \\in S\\), thus \\(p+q \\in 2S\\).\n\nNow all these nine polyhedrons lie inside \\(2 P_1\\). Let \\(V\\) be the volume of \\(P_1\\). Then some polyhedrons with total sum of volumes \\(9V\\) lie in a shape of volume \\(8V\\), thus they must overlap, meaning that they have an interior point in common.\n\nThe above solution was posted by ZetaX. The original thread for this problem can be found here: [1]"
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IMO-1971-3
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_3
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Prove that the set of integers of the form \(2^k - 3(k = 2; 3; \cdots)\) contains an infinite subset in which every two members are relatively prime.
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[
"Wlog, assume \\(a_{n}\\ge 3\\). Then say \\(p_{1}, p_{2}, \\ldots, p_{k}\\in \\mathbb{N}\\) are all the (pairwise distinct) primes dividing \\(2^{a_{n}}-3\\) and let \\(a_{n+1}= a_{n}\\cdot \\prod_{i=1}^{k}(p_{i}-1)\\). Obviously \\(p_{i}\\) is odd, for any \\(i \\in \\overline{1,k}\\). So \\(\\prod_{i=1}^{k}p_{i}\\) divides \\(2^{a_{n+1}}-1\\), by Fermat's little theorem, and \\(\\gcd(2^{a_{n+1}}-3, 2^{a_{n}}-3) = 1\\). Now, by induction, it follows \\(\\gcd(2^{a_{m}}-3, 2^{a_{n}}-3) = 1\\), for any distinct \\(m, n \\in \\mathbb{N}\\).\n\nThe above solution was posted and copyrighted by s.tringali. The original thread for this problem can be found here: [1]\n\n## Remarks (added by pf02, December 2024)\n\n1. The solution above is incomplete, and somewhat misleading, to such an extent that it can not be called a \"Solution\". The problem is with the statement \"by induction, it follows \\(\\gcd(2^{a_m} - 3, 2^{a_n} - 3) = 1\\), for any distinct \\(m, n \\in \\mathbb{N}\\)\". This statement is not clear at all, it needs a proof. But, as we will see below, we don't need to do this step at all.\n\n2. Below, I will re-phrase the solution, and complete it.",
"Denote \\(A_n = 2^{a_n} - 3\\). The plan is to construct recursively a sequence \\(a_1 < a_2 < \\dots < a_n < \\dots\\) such that \\(A_n\\) and \\(A_k\\) are relatively prime for all \\(k < n\\).\n\nLet \\(a_1 = 3\\). (Note that any other number \\(> 3\\) would serve just as well.)\n\nLet \\(p_{1,1}, \\dots, p_{1,k_1}\\) be all the prime factors of \\(A_1 = 2^{a_1} - 3\\). (Note that we did not specify whether any prime factors are repeated or not. It does not matter for the proof, the important thing is to take all prime factors.) In our case, \\(k_1 = 1\\) and \\(p_{1,1} = 5\\) since we took \\(a_1 = 3\\).\n\nDefine \\(a_2 = a_1 (p_{1,1} - 1) \\dots (p_{1,k_1} - 1)\\). In our case \\(a_2 = 3 (5 - 1) = 12\\).\n\nDefine \\(\\{a_n\\}\\) recursively: let \\(\\{p_{(n-1),1}, \\dots, p_{(n-1),k_{n-1}}\\}\\) be all the prime factors of \\(A_{n-1} = 2^{a_{n-1}} - 3\\). (Note again that we don't care whether factors are repeated or not.) Let us take \\(a_n = a_{n-1} (p_{(n-1),1} - 1) \\dots (p_{(n-1),k_{n-1}} - 1)\\).\n\nAt this point, note for later use that if \\(p\\) is a prime factor of any \\(A_k, k \\in \\{1, \\dots, (n-1)\\}\\) then \\((p - 1)\\) is a factor of \\(a_n\\). Also, note that all such factors \\(p\\) are odd, because all \\(A_k\\) are odd.\n\nLet \\(B = A_n + 2 = 2^{a_n} - 1\\). If \\(p\\) is a prime factor of any of the \\(A_k, k \\in \\{1, \\dots, n-1\\}\\), then \\(p\\) is a factor of \\(B\\). Indeed, note that \\(a_n = b \\cdot (p - 1)\\) for some \\(b\\), so \\(B = 2^{a_n} - 1 = 2^{b \\cdot (p - 1)} - 1 = (2^b)^{p-1} - 1\\), which is divisible by \\(p\\) according to Fermat's_Little_Theorem.\n\nIt follows that \\(A_n = 2^{a_n} - 3\\) is coprime with all \\(A_k, k \\in \\{1, \\dots, (n-1)\\}\\). This can be justified easily, by noting that if there were a common factor \\(b\\) between \\(A_n\\) and an \\(A_k\\), there would exist a prime common factor \\(p\\). But then this \\(p\\) would be a common factor between two consecutive odd integers \\(A_n\\) and \\(B\\), which is impossible.\n\n[Revised, completed solution by pf02, December 2024]"
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IMO-1971-4
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_4
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All the faces of tetrahedron \(ABCD\) are acute-angled triangles. We consider all closed polygonal paths of the form \(XYZTX\) defined as follows: \(X\) is a point on edge \(AB\) distinct from \(A\) and \(B\); similarly, \(Y, Z, T\) are interior points of edges \(BC, CD, DA\), respectively. Prove:
(a) If \(\angle DAB + \angle BCD \neq \angle CDA + \angle ABC\), then among the polygonal paths, there is none of minimal length.
(b) If \(\angle DAB + \angle BCD = \angle CDA + \angle ABC\), then there are infinitely many shortest polygonal paths, their common length being \(2AC \sin(\alpha / 2)\), where \(\alpha = \angle BAC + \angle CAD + \angle DAB\).
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[
"Rotate the triangle \\(BCD\\) around the edge \\(BC\\) until \\(ABCD\\) are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting \\(X\\) and \\(Z\\). Therefore, \\(XYB=ZYC\\). Summing the four equations like this, we get exactly \\(\\angle ABC+\\angle ADC=\\angle BCD+\\angle BAD\\).\n\nNow, draw all four faces in the plane, so that \\(BCD\\) is constructed on the exterior of the edge \\(BC\\) of \\(ABC\\) and so on with edges \\(CD\\) and \\(AD\\).\n\nThe final new edge \\(AB\\) (or rather \\(A'B'\\)) is parallel to the original one (because of the angle equation). Call the direction on \\(AB\\) towards \\(B\\) \"right\" and towards \\(A\\) \"left\". If we choose a vertex \\(X\\) on \\(AB\\) and connect it to the corresponding vertex \\(X'\\) on A'B'. This works for a whole interval of vertices \\(X\\) if \\(C\\) lies to the left of \\(B\\) and \\(D\\) and \\(D\\) lies to the right of \\(A\\). It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.\n\nFinally, regard the sine in half the isosceles triangle \\(ACA'\\) which gives the result with the angles around \\(C\\) instead of \\(A\\), but the role of the vertices is symmetric.\n\n## Remarks (added by pf02, December 2024)\n\nThe solution above is incomplete, and very likely, incorrect. The first part (which claims to prove part (a) of the problem) is incomplete at best, since it is not clear how it leads to the result. (Very likely it is incorrect.) The second part, which claims to prove part (b) of the problem, skips too many steps. Some of the arguments in the proof (e.g. the fact that \\(AB \\parallel A'B'\\)) are true but need proof. I could not make any sense of other arguments in the proof, so I can not judge whether they are correct or not (I suspect not).\n\nI will give a robust solution below. It goes along the same basic idea.",
"The basic idea is to \"fold out\" the tetrahedron into a polygon in the plane. The path \\(XYZTX\\) becomes a collection of connected segments \\(XY, YZ, ZT, TX'\\) with the unconnected ends \\(X, X'\\) on the two line segments \\(AB, A'B'\\).\n\n\\[\nProb 1971 4 1.png\n\\]\n\nSpecifically, we rotate the solid consisting of three faces around \\(BC\\) so \\(\\triangle BCD\\) is in the same plane as \\(\\triangle ABC\\). Then we rotate the solid consisting of two faces around \\(CD\\) so that \\(\\triangle CDA'\\) is in the same plane with the previous triangles. (We denoted \\(A'\\) the new \"copy\" of \\(A\\), since the original \\(A\\) is being used in the picture.) Finally, we rotate \\(\\triangle A'BD\\) around \\(A'D\\), so that it is in the same plane with the other triangles. (We denote \\(B'\\) the new \"copy\" of \\(B\\), and \\(X'\\) the new copy of \\(X\\).)\n\nThe polygonal path \\(XYZTX\\) becomes \\(XYZTX'\\). It is clear that in order to minimize \\(XYZTX\\), we should make \\(XYZTX'\\) be a segment on a straight line. Furthermore, to minimize the segment \\(XX'\\), we want to choose \\(X \\in AB\\) so that when we draw the line segment to its corresponding image \\(X' \\in A'B'\\), the length of \\(XX'\\) is as short as possible.\n\nThis was the \"folding out\" done in the solution above. To continue, we will do a different \"folding out\", which will serve better for solving the problem. Specifically, we rotate \\(\\triangle BCD\\) around \\(BC\\), then we rotate \\(\\triangle ABD\\) around \\(AB\\) to obtain \\(\\triangle ABD'\\), and finally, we rotate \\(\\triangle AD'C\\) around \\(AD'\\) to obtain \\(\\triangle AD'C'\\). Now \\(Z \\in CD\\) becomes \\(Z' \\in C'D'\\). The polygonal path becomes \\(ZYXTZ'\\) going from \\(CD\\) to \\(C'D'\\). We need to make this as small as possible, to find its minimum, if it exists.\n\n\\[\nProb 1971 4 2.png\n\\]\n\nThe idea of the solution to the problem is now easy to explain. First of all, \\(ZYXTZ'\\) needs to be a segment on a straight line, in which case its length is the length of the segment \\(ZZ'\\). Clearly \\(ZYXTZ' = ZZ'\\) has a lower bound (after all, it is \\(> 0\\)), so we can think about its lower limit. If there is a position of \\(Z\\) in which this lower limit is achieved, then this lower limit is a minimum. Otherwise, there is no minimum value for \\(ZYXTZ'\\). (Pay attention to the subtle distinction between lower bound, lower limit, and minimum.)\n\nIn the picture above, \\(CD\\) and \\(C'D'\\) are not parallel. The segment \\(ZZ'\\) would be shortest when \\(Z = D\\) (and \\(Z' = D'\\)). But this is not an acceptable position for \\(ZYXTZ'\\), because the problem stated that \\(Z\\) is between \\(C\\) and \\(D\\), not equal to any of them. So in this picture, there is no minimum for the polygonal path. (In this case there is a lower limit for \\(ZZ'\\), namely \\(DD'\\), which is not a minimum.)\n\nOn the other hand, if \\(CD \\parallel C'D'\\) (see pictures below) then generally there are lots of points \\(Z \\in CD\\) yielding a minimum value for \\(ZYXTZ'\\). Indeed, in this case \\(CC' \\parallel ZZ' \\parallel DD'\\), so the only requirement is for all the points \\(X, Y, Z, T\\) to be inside the respective segments. (In this case, \\(ZZ' = CC' = DD'\\) is a lower limit, and it is a minimum.)\n\nWe will prove that \\(\\angle DAB + \\angle BCD = \\angle CDA + \\angle ABC\\) if and only if \\(CD \\parallel C'D'\\). After this it will be easy to deduce all the statements of the problem.\n\nLet \\(M, N, P\\) be on \\(CA, BD', AC'\\) be such that \\(CD \\parallel MB \\parallel AN \\parallel PD'\\). The equality \\(\\angle DAB + \\angle BCD = \\angle CDA + \\angle ABC\\) becomes \\(\\angle D'AB + \\angle BCD = \\angle C'D'A + \\angle ABC\\), and then it becomes \\(\\angle D'AN + \\angle NAB + \\angle BCD = \\angle C'D'P + \\angle PD'A + \\angle ABM + \\angle MBC\\). Because of the parallelism of \\(CD, MB, AN, PD'\\) we have several equal angles on the two sides of this equality. This equality becomes \\(\\angle C'D'P = 0\\).\n\nSo, the original equality \\(\\angle DAB + \\angle BCD = \\angle CDA + \\angle ABC\\) is true if and only if \\(\\angle C'D'P = 0\\), which is true if and only if \\(CD \\parallel C'D'.\\)\n\nNow it is obvious that when \\(\\angle DAB + \\angle BCD \\neq \\angle CDA + \\angle ABC\\) a lower limit exists (as expected), but it is not a minimum, and a minimum does not exist.\n\nOn the other hand, when \\(\\angle DAB + \\angle BCD = \\angle CDA + \\angle ABC\\), in the typical, general case, \\(ZZ' = DD'\\) is the minimum, and there are infinitely many segments \\(ZZ'\\) of the same size (see Figure 2 below).\n\n\\[\nProb 1971 4 3.png\n\\]\n\nThere is one delicate point we need to worry about: we have to be sure that \\(ZZ'\\) intersects \\(CB, BA, AD'\\). See Figure 3 for an example when this does not happen; note that in this example we have some obtuse angles. Formally, \\(A\\) has to be on the same side of \\(DD'\\) as \\(C, C'\\) and \\(B\\) has to be on the same side of \\(CC'\\) as \\(D, D'\\).\n\nLet us concentrate on the position of \\(A\\) vs. \\(DD'\\) (the statement about \\(B\\) is similar). We want \\(\\angle DAB + \\angle BAD' < \\pi\\). We have \\(\\angle DAB < \\angle CAD\\) and \\(\\angle BAD' = \\text{the original } \\angle BAD \\text{ from the tetrahedron}\\). Since both \\(\\angle CAD, \\angle BAD\\) are acute, it follows that \\(\\angle DAB + \\angle BAD' < \\pi\\).\n\nThe last thing to do is to compute \\(ZZ'\\). We have \\(ZZ' = CC'\\), and from the isosceles triangle \\(\\triangle CAC'\\) we have \\(CM = \\sin \\angle CAM\\), where \\(M\\) is the midpoint of \\(CC'\\). This gives \\(ZZ' = CC' = 2 AC \\sin \\frac{\\alpha}{2}\\), which is the formula we wanted.\n\n[Solution by pf02, December 2024]"
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IMO-1971-5
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_5
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Prove that for every natural number \(m\), there exists a finite set \(S\) of points in a plane with the following property: For every point \(A\) in \(S\), there are exactly \(m\) points in \(S\) which are at unit distance from \(A\).
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[
"I shall prove a more general statement about the unit distance graph(\\(V=\\mathbb{R}^2\\), adjacency iff the Euclidean distance between the points is \\(1\\)) and this will follow as a consequence. if \\(G,H\\) occur as unit distance graphs, then so does \\(G\\times H\\)( here \\(G\\times H\\) is described as \\(V(G\\times H)=V(G)\\times V(H), (v_1,w_1)\\leftrightarrow (v_2,w_2) \\Leftrightarrow v_1=v_2,w_1\\leftrightarrow w_2\\) or \\(v_1\\leftrightarrow v_2,w_1=w_2\\)). this is seen by describing the vertices by complex numbers. suppose there is an embedding of \\(G\\) by the complex numbers \\(v_1,v_2\\ldots,v_n\\) and for \\(H\\) by the numbers \\(w_1,w_2,\\ldots, w_m\\). we claim that for some choice of \\(0\\leq\\theta<2\\pi\\), \\(v_i+e^{i\\theta}w_j\\) will do the job(a suitable rotation). what we need is that \\((v_i+e^{i\\theta}w_j)\\leftrightarrow (v_k+e^{i\\theta}w_l)\\) iff either \\(v_i=v_k,|w_j-w_l|=1\\) or \\(w_j=w_l,|v_i-v_k|=1\\). clearly if the condition holds then the adjacency is satisfied. suppose \\(i\\neq k,j\\neq l\\) and that the corresponding complex numbers are at a distance \\(1\\) from one another. Then this gives a quadratic in \\(e^{i\\theta}\\) and hence \\(\\theta\\) can take only finitely many values here.ruling this out for each such set of \\(i,j,k,l\\) hence rules out finitely many values of \\(\\theta\\) and therefore a suitable choice exists. for the given problem we need a unit distance graph which is regular of degree \\(m\\) for any given \\(m\\). since we can form the graph \\(K_2\\), we can form \\(Q_n=Q_{n-1}\\times K_2\\)(the unit cube) and that solves the problem.\n\nThis solution was posted and copyrighted by seshadri. The original thread for this problem can be found here: [1]",
"Suppose \\(S\\) has the form\n\n\\[\nS_m=\\left\\{\\sum_{\\mathbf v \\in U} \\mathbf v \\;\\middle\\vert\\; U \\subseteq V_m\\right\\}=\\left\\{\\sum_{i=1}^m b_i \\mathbf v_i \\;\\middle\\vert\\; b_i \\in \\{0,1\\}\\right\\}\\!,\n\\]\n\nwhere \\(V_m=\\{\\mathbf v_1,\\mathbf v_2,\\dots,\\mathbf v_m\\}\\) is unknown set of distinct unit vectors in \\(\\mathbb R^2\\). We can build \\(V_m\\) inductively, starting from the empty set and adding vectors to it, one by one. We just need to make sure that two thing are respected: 1. All \\(2^m\\) vectors in \\(S_m\\) are distinct; 2. Two vector sums are at unit distance from one another if and only if they differ in presence of exactly one summand (i.e. one and only one coefficient \\(b_i\\) in the sum changes from \\(0\\) to \\(1\\)). If these two conditions are kept, then each of \\(2^m\\) points at \\(S_m\\) will be at unit distance from exactly \\(m\\) points corresponding to sums at which one and only one of \\(m\\) coefficients differs from coefficients of this point. However, respecting these conditions is not hard because \\(\\left\\|\\sum_{\\mathbf v \\in U_1} \\mathbf v - \\sum_{\\mathbf v \\in U_2} \\mathbf v\\right\\|=\\left\\|\\sum_{\\mathbf v \\in U_1 \\setminus U_2} \\mathbf v \\,-\\! \\sum_{\\mathbf v \\in U_2 \\setminus U_1} \\mathbf v\\right\\|\\) and for each new vector being added to \\(V_m\\) there is at most some finite set of forbidden endpoints given by sums/differences of already determined vectors but the rest of the (infinite) unit circle is permissible.\n\nThis solution was posted and copyrighted by Bandera. The original thread for this problem can be found here: [2]\n\n## Remarks (added by pf02, January 2025)\n\n1. On the original thread at https://artofproblemsolving.com/community/c6h60830p there is a third solution, by Pgm03B. It is simpler than the two solutions above. It has a flaw in the argument, and it is not presented in a nice way, but it is a nice idea, well worth presenting here. As a public service, I will add an edited version of it below.\n\n2. Solutions 1 and 2 above are based on good ideas, but they are presented very poorly. If this page was a reviewed publication, and I were a reviewer, I would reject both of them saying \"rewrite and resubmit\".\n\n2.1. Solution 1 suffers from undefined notation and terminology, from minor errors, and from unacceptable amount of hand waving replacing explanations of details.\n\n2.2. Solution 2 suffers from poor explanation of details and from what seems to me to be an error (starting an inductive proof for a property of \\(m\\) vectors from an empty set).\n\nAs a public service I will rewrite these proofs below, in what I hope is a much more presentable style.\n\n## Definitions and terminology\n\nIt will be helpful (though not necessary) to imagine the set of points \\(S\\) as a graph in the plane. Specifically, the vertices of the graph are the points in \\(S\\), and the edges of the graph are the line segments of length \\(1\\) joining the points. All points at distance \\(1\\) are joined, and only the points at distance \\(1\\) are joined.\n\nWe will call such a graph \\(m\\)-regular iff every \\(A \\in S\\) has exactly \\(m\\) lines of the graph with one end at \\(A\\), in other words, there are exactly \\(m\\) points in \\(S\\) at unit distance from \\(A\\). Using this definition, the problem can be reformulated as \"prove that for any natural number \\(m\\), there exists an \\(m\\)-regular graph.\"\n\nFor the purposes of solution 1, it will be useful to think of the points \\(A \\in S\\) as points in the complex plane. For the purposes of solution 2, it will be useful to think of the plane as having an origin \\(O\\). We will work with vectors in the plane. The points \\(A \\in S\\) will be the end points of vectors. We will try to differentiate between vectors and their end-points, although at times, when there is no danger of confusion, we may be sloppy about it.",
"We will give a proof by induction on \\(m\\). First some notation:\n\nGiven two finite sets of points \\(S, T\\) in the complex plane, and given \\(\\theta \\in [0, 2\\pi)\\), define \\(U = S \\oplus_\\theta T = \\{a + e^{i \\theta} b, \\text{\\ for all\\ } a \\in S, b \\in T\\}\\).\n\n\\(\\mathbf{Proposition}\\): If \\(S\\) is \\(m\\)-regular and \\(T\\) is \\(n\\)-regular, and \\(S \\cap T = \\emptyset\\), then we can choose \\(\\theta\\) so that \\(U = S \\oplus_\\theta T\\) is \\((m + n)\\)-regular.\n\n\\(\\mathbf{Proof}\\): Let \\(a_i, i = 1, \\dots, m\\) be the neighboring points in \\(S\\) at distance \\(1\\) from \\(a\\), and \\(b_j, j = 1, \\dots, n\\) be the neighboring points in \\(T\\) at distance \\(1\\) from \\(b.\\) Clearly \\(a_i + e^{i \\theta} b\\) is at distance \\(1\\) from \\(a + e^{i \\theta} b\\) and \\(a + e^{i \\theta} b_j\\) is at distance \\(1\\) from \\(a + e^{i \\theta} b\\). So \\(a + e^{i \\theta} b\\) has \\((m + n)\\) neighbors at distance \\(1\\) if we choose \\(\\theta\\) such that \\(a_i + e^{i \\theta} b \\neq a + e^{i \\theta} b_j\\) for all \\(i, j\\).\n\nBut the equations \\(a_i + e^{i \\theta} b = a + e^{i \\theta} b_j\\) are just a finite number of linear equations in \\(e^{i \\theta}\\), so we just have to avoid choosing \\(\\theta\\) giving a solution to any of these equations. Thus, with such a choice of \\(\\theta\\) there are definitely at least \\((m + n)\\) points in \\(U\\) at distance \\(1\\) from \\(a + e^{i \\theta} b\\).\n\nWe have to show that we can choose \\(\\theta\\) so that there are no more than \\((m + n)\\) points at distance \\(1\\). If we had more points from \\(U\\) at distance \\(1\\) from \\(a + e^{i \\theta} b\\), we would have \\(\\vert (a + e^{i \\theta} b) - (c + e^{i \\theta} d) \\vert = 1\\) for some \\(c \\in S\\) and \\(d \\in T\\).\n\nThis would imply \\(\\vert (a + e^{i \\theta} b) - (c + e^{i \\theta} d) \\vert^2 = 1\\),\n\nor \\(\\vert (a - c) + e^{i \\theta} (b - d) \\vert^2 = 1\\),\n\nor \\([(a - c) + e^{i \\theta} (b - d)] [\\overline{(a - c)} + e^{-i \\theta} \\overline{(b - d)}] = 1\\),\n\nor \\(\\vert (a - c) \\vert^2 + e^{i \\theta} \\overline{(a - c)}(b - d) + e^{-i \\theta} (a - c) \\overline{(b - d)} + \\vert (b - d) \\vert^2 = 1\\).\n\nThis becomes an equation of degree \\(2\\) in \\(e^{i \\theta}\\). So as long as we choose \\(\\theta\\) not to equal to any of the solutions of these equations, we can be sure that none of the points in \\(U\\) are at distance \\(1\\) from \\(a + e^{i \\theta} b\\).\n\nThis proves the proposition because we have to choose \\(\\theta \\in [0, 2\\pi)\\) different from finite number of values.\n\nNow the problem is very simple to prove by induction.\n\nFor \\(m = 1\\), take \\(S_1 = \\{0, 1\\}\\), which is a \\(1\\)-regular set. For the induction step, assume we have \\(S_{m-1}\\) an \\((m-1)\\)-regular graph. Using the proposition, it follows that we can choose \\(\\theta\\) such that \\(S_m = S_{m-1} \\oplus_{\\theta} T_1\\) is an \\(m\\)-regular graph, where \\(T_1\\) consists of two points in the plane at distance \\(1\\), none of which is in \\(S_{m-1}\\).\n\n(Just for the sake of completeness, let us mention that when we choose \\(\\theta\\), it is a value in \\([0, 2\\pi)\\) which differs from a finite set of values modulo \\(2\\pi\\), dependent on the points in \\(S_{m-1}, T_1\\).)",
"Given a set \\(V = \\{ \\mathbf{v_1}, \\dots, \\mathbf{v_m} \\}\\) of \\(m\\) vectors in the plane, define\n\n\\[\nS(V) = \\left\\{ \\sum_{i=1}^m \\alpha_i \\mathbf{v_i}, \\text{\\ where\\ } \\alpha_i \\in \\{0, 1\\} \\right\\} = \\left\\{ \\sum_{\\mathbf{v} \\in U} \\mathbf{v}, \\text{\\ where\\ } U \\subset V \\right \\}.\n\\]\n\nLet \\(S_V =\\) the set of end points of all \\(\\mathbf{v} \\in S(V)\\).\n\nNote that in the case \\(\\alpha_i = 0\\) for all \\(i = 1, \\dots, m\\), the sum \\(\\sum_{i=1}^m \\alpha_i \\mathbf{v_i} = 0\\), which yields the point \\(O\\) (the origin) in the plane. In the notation \\(\\sum_{\\mathbf{v} \\in U} \\mathbf{v}\\) this corresponds to \\(U = \\emptyset\\). (Just as a curiosity, note that if \\(V\\) contains \\(m\\) vectors, then \\(S_V\\) contains \\(\\leq 2^m\\) points. We are not going to use this fact.)\n\nWe will prove by induction on \\(m\\) that there is a set \\(V\\) of \\(m\\) unit vectors with the starting point at \\(O\\), such that the vector sums in \\(S(V)\\) are all distinct, and \\(S_V\\) is \\(m\\)-regular. If \\(m = 1\\), take \\(V = \\{\\mathbf{v_1}\\}\\), where \\(\\mathbf{v_1}\\) is a unit vector starting at \\(O\\). Then \\(S(V)\\) consists of \\(\\mathbf{v_1}\\), and \\(S_V\\) consists of the origin \\(O\\) and the end point of \\(\\mathbf{v_1}\\), so it is clearly a \\(1\\)-regular set.\n\nNow assume that we have a set of unit vectors starting at \\(O\\), \\(V = \\{ \\mathbf{v_1}, \\dots, \\mathbf{v_m} \\}\\), such that the sums in the corresponding \\(S(V)\\) are all distinct, and \\(S_V\\) is \\(m\\)-regular. We want to show that we can find a unit vector \\(\\mathbf{v_{m+1}}\\) starting at \\(O\\), such that if we take \\(W = V \\cup \\{ \\mathbf{v_{m+1}} \\}\\), then the vector sums in \\(S(W)\\) are all distinct, and \\(S_W\\) is \\((m+1)\\)-regular. Note that \\(S(V) \\subset S(W)\\) and \\(S_V \\subset S_W\\).\n\nProve that we can choose \\(\\mathbf{v_{m+1}}\\) such that the sums in \\(S(W)\\) are all distinct. If \\(s_1 \\neq s_2\\) then \\(s_1 + \\mathbf{v_{m+1}} \\neq s_2 + \\mathbf{v_{m+1}}\\). So we just need to make sure that we choose \\(\\mathbf{v_{m+1}}\\) such that \\(s_1 + \\mathbf{v_{m+1}} \\neq s_2\\) for all \\(s_1, s_2 \\in S(V)\\). This means that \\(\\mathbf{v_{m+1}}\\) has to avoid satisfying a finite number of equations. Since there are infinitely many possibilities for the unit vector \\(\\mathbf{v_{m+1}}\\), we can definitely choose it such that it is not satisfying a finite number of equations.\n\nWe will now prove that we can choose \\(\\mathbf{v_{m+1}}\\) such that \\(S_W\\) is \\((m+1)\\)-regular. First, let \\(A \\in S_V\\) correspond to \\(s_1 \\in S(V)\\). Then, the same \\(A\\) viewed as a point in \\(S_W\\) has at least \\((m + 1)\\) neighbors at unit distance, namely the \\(m\\) neighbors it has in \\(S_V\\) and the point \\(B\\) corresponding to \\(s_1 + \\mathbf{v_{m+1}}\\). It would have more neighbors at unit distance if we there would exist an \\(s_2\\) (where \\(s_2 \\neq s_1\\) and \\(s_2\\) not at distance \\(1\\) from \\(s_1\\)) such that the end points of \\(s_1 + \\mathbf{v_{m+1}}\\) and \\(s_2\\) would be at distance \\(1\\). So, the end point of the unit vector \\(\\mathbf{v_{m+1}}\\) should not be at distance \\(1\\) from the end point of the vector \\(s_2 - s_1\\), or in other words, \\(\\mathbf{v_{m+1}}\\) should not be on the circle of radius \\(1\\) centered at the end of \\(s_2 - s_1\\). This circle intersects the unit circle centered at \\(O\\) (on which the end point of \\(\\mathbf{v_{m+1}}\\) is) in at most \\(2\\) points. This gives yet another finite set of conditions \\(\\mathbf{v_{m+1}}\\) should avoid satisfying.\n\nSecond, take \\(B \\in S_W\\) corresponding to \\(s_1 + \\mathbf{v_{m+1}}\\) for a \\(s_1 \\in S(V)\\). Let \\(t_1, \\dots, t_m \\in S(V)\\) be the vectors which give the points in \\(S_V\\) at unit distance from the point corresponding to \\(s_1\\). Then \\(t_j + \\mathbf{v_{m+1}}, j = 1, \\dots, m\\) and \\(s_1\\) are \\((m + 1)\\) vectors whose end points are at unit distance from \\(B\\). Again, we have to show that we can choose \\(\\mathbf{v_{m+1}}\\) so that there are no more points in \\(S_W\\) at distance \\(1\\) from the end point of \\(s_1 + \\mathbf{v_{m+1}}\\). There would be more points in \\(S_W\\) at distance \\(1\\) from \\(B\\) if we had vectors \\(s_2 \\in S(V)\\) (where \\(s_2 \\neq s_1\\) and \\(s_2\\) not at distance \\(1\\) from \\(s_1\\)) such that the end points of \\(s_1 + \\mathbf{v_{m+1}}\\) and \\(s_2\\), or the end points of \\(s_1 + \\mathbf{v_{m+1}}\\) and \\(s_2 + \\mathbf{v_{m+1}}\\) were at distance \\(1\\). The former is already avoided by the choice of \\(\\mathbf{v_{m+1}}\\) made so far, and the latter is avoided because of the induction hypothesis.\n\nSince there are only finite number of equations the unit vector \\(\\mathbf{v_{m+1}}\\) with starting point at \\(O\\) has to avoid satisfying, it follows that we can find \\(\\mathbf{v_{m+1}}\\) such that \\(S_W\\) is \\((m+1)\\)-regular.",
"We will give a proof by induction on \\(m\\). For \\(m = 1\\) consider the graph \\(S_1\\) consisting of two points at distance \\(1\\) joined by a line segment.\n\nAssume that we have a graph \\(S_{m-1}\\) which is \\((m-1)\\)-regular. Let \\(\\mathbf{v}\\) be a direction along which we make a translation of \\(S_{m-1}\\) of length \\(1\\) (that is, the distance of translation \\(= 1\\)). Denote by \\(T\\) the translated graph, and let \\(U\\) be the graph whose vertices are \\(S_{m-1} \\cup T\\), and whose edges are those of \\(S_{m-1}\\), together with those of \\(T\\), and the edges obtained when we join each \\(A \\in S_{m-1}\\) with the corresponding \\(A' \\in T\\) (where \\(A'\\) is the result of translating \\(A\\)). We will prove that we can choose the direction of translation \\(\\mathbf{v}\\) such that \\(U\\) is \\(m\\)-regular.\n\nFirst, let us make sure we choose \\(\\mathbf{v}\\) such that no point in \\(T\\) is a point of \\(S_{m-1}\\). This means that \\(\\mathbf{v}\\) has to be different from a finite number of values, which is possible since to begin with, we have infinitely many choices for \\(\\mathbf{v}\\).\n\nLet \\(A \\in S_{m-1}\\), and \\(A' \\in T\\) the translated image of \\(A\\). Let \\(B_1, \\dots, B_{m-1}\\) be the points in \\(S_{m-1}\\) at distance \\(1\\) from \\(A\\), and let \\(B'_1, \\dots, B'_{m-1}\\) the coresponding points in \\(T\\). The point \\(A\\) has \\(m\\) neighbors in \\(U\\) at distance \\(1\\), namely \\(B_1, \\dots, B_{m-1}, A'\\), and the point \\(A'\\) has \\(m\\) neighbors in \\(U\\) at distance \\(1\\), namely \\(B'_1, \\dots, B'_{m-1}, A\\). So every point in \\(U\\) has at least \\(m\\) neighbors at distance \\(1\\).\n\nWe have to show that we can choose the direction \\(\\mathbf{v}\\) so that no point has more than \\(m\\)-neighbors at distance \\(1\\). For this, we have to show that for all \\(A, B \\in S_{m-1}\\) and the corresponding \\(A', B' \\in T\\) we can choose \\(\\mathbf{v}\\) such that the distance between \\(A'\\) and \\(B\\) is \\(\\neq 1\\). This means that \\(\\mathbf{v}\\) should be such that \\(A'\\) is not at the intersection of the two circles of radius \\(1\\) centered at \\(A\\) and \\(B\\). Since these two circles intersect in at most \\(2\\) points, and we have finitely many pairs \\(A, B\\), this imposes that the direction \\(\\mathbf{v}\\) does not equal finitely many values. This proves the induction step, and the problem.\n\n(Note that as far as the construction goes, this solution is essentially the same as solution 1, but the formalism and the point of view are so different that I think it should be viewed as a different solution.)\n\nThis solution is based on the idea by Pgm03B from the discussion thread https://artofproblemsolving.com/community/c6h60830p ."
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IMO-1971-6
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https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_6
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Let \(A = (a_{ij})(i, j = 1, 2, \cdots, n)\) be a square matrix whose elements are non-negative integers. Suppose that whenever an element \(a_{ij} = 0\), the sum of the elements in the \(i\)th row and the \(j\)th column is \(\geq n\). Prove that the sum of all the elements of the matrix is \(\geq n^2 / 2\).
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[
"Take the the row or column (without loss of generality, a row) with the minimum possible sum \\(S\\) of its elements. Let \\(z\\) be the number of zeros in this row. We will assume \\(S < \\frac{n}{2}\\) once the other case is obvious. Clearly \\(S \\ge n - z \\Rightarrow z \\ge n - S\\). The total sum \\(T\\) of all elements of the matrix is at least the number of zeros in this row multiplicated by \\(n - S\\) (because the sum of the elements of a row and a column meeting in a zero is \\(\\ge n\\)) plus the number of nonzero elements times \\(S\\) (that is the minimum possible sum), it is,\n\n\\[\nz(n - S) + (n - z)S.\n\\]\n\nNote that, being \\(z \\ge n - S\\) and \\(n - S \\ge S\\), we can put \\(z - 1\\) instead of \\(z\\) and \\(n - z + 1\\) instead of \\(n - z\\) until we get \\(z = n - S\\), what makes the sum become smaller. So we have\n\n\\[\nT \\ge (n - S)^2 + S^2 \\ge 2(\\frac{n - S + S}{2})^2 = \\frac{n^2}{2}\n\\]\n\nby AM - QM inequality.\n\nThe above solution was posted and copyrighted by Renan. The original thread for this problem can be found here: [1]",
"Denote \\(\\ell_i,c_i\\) the \\(i\\)th line and column and \\(s(\\ell_i), s(c_i)\\) the sum of the elements on that row. Let \\(M=\\max_{\\sigma \\in S_n} | \\{ 1\\le i\\le n| a_{i\\sigma(i)}=0\\} |\\). For each permutation \\(\\sigma\\in S_n\\), assign to it the number\n\n\\[\n\\sum\\limits_{a_{i\\sigma(i)=0}} \\left ( s(\\ell_i)+s(c_{\\sigma(i)}) \\right )\n\\]\n\nNow pick a permutation \\(\\sigma\\) such that it generates \\(M\\) and its assigned number is minimal. Suppose wlog that \\(\\{ 1\\le i\\le n| a_{i\\sigma(i)}=0\\}=\\{1,2,..,k\\}\\). If \\(k=n\\), we are done as twice the sum of all elements is\n\n\\[\n\\sum\\limits_{i=1}^{n} s(\\ell_i) +\\sum\\limits_{i=1}^{n} s(c_i)=\\sum\\limits_{i=1}^{n} (s(\\ell_i)+s(c_{\\sigma(i)}))\\ge n^2\n\\]\n\nSo suppose \\(k<n\\) and look at a \\(j\\) such that \\(k+1\\le j\\le n\\). By the maximality of \\(M\\), \\(\\ell_j\\) can have \\(0\\) only on the columns \\(\\sigma(1),...,\\sigma(k)\\). Note that if \\(\\ell_j\\) has a \\(0\\) on \\(c_{\\sigma(r)}\\) with \\(r\\le k\\), i.e. \\(a_{j\\sigma(r)}=0\\), by the minimality of the assigned number of \\(\\sigma\\), we have that \\(s(\\ell_j)\\ge s(\\ell_r)\\). However, if the column \\(c_{\\sigma(j)}\\) has a \\(0\\) on \\(\\ell_r\\), i.e. \\(a_{r\\sigma(j)}=0\\), by the same argument we have \\(s(c_\\sigma(j))\\ge s(c_\\sigma(r))\\), hence \\(s(\\ell_j)+s(c_{\\sigma(j)})\\ge s(r)+s(c_{\\sigma(r)})\\ge n\\).\n\nSuppose that \\(0\\) appears exactly \\(t\\) times on \\(\\ell_j\\). This means that \\(s(\\ell_j)\\ge n-t\\). Suppose that the \\(0\\)s of \\(\\ell_j\\) appear on the columns \\(c_{\\sigma(i_1)},...,c_{\\sigma(i_t)}\\). We have seen earlier that if \\(c_\\sigma(j)\\) has a \\(0\\) on \\(i_1,..,i_t\\) we have that \\(s(\\ell_j)+s(c_{\\sigma(j)})\\ge n\\). So suppose that it doesn't have \\(0\\) on these lines. However, \\(c_{\\sigma(j)}\\) has \\(0\\)s only on the lines \\(\\ell_1,..,\\ell_k\\) by the maximality of \\(M\\), hence \\(c_{\\sigma(j)}\\) has at most \\(k-t\\) zeroes, so \\(s(c_{\\sigma(j)})\\ge n-k+t\\). We thus infer that\n\n\\[\ns(\\ell_j)+s(c_{\\sigma(j)})\\ge (n-t)+(n-k+t)=2n-k\\ge n\n\\]\n\nso \\(s(\\ell_j)+s(c_{\\sigma(j)})\\ge n,\\ \\forall j\\ge k+1\\). By the hypothesis this is also true for \\(j\\le k\\) so summing over all values of \\(j\\) we get that the sum is at least \\(\\dfrac{n^2}{2}\\).\n\nThe above solution was posted and copyrighted by Aiscrim. The original thread for this problem can be found here: [2]",
"If the main diagonal contains all zeroes, we can immediately deduce from the condition that the sum \\(S\\) of the elements is at least \\(n^2/2\\). This motivates us to consider the largest possible subset of elements, all zero, which are pairwise in different rows of columns - let the size of this subset be \\(k\\). (If there is no such subset, \\(S\\ge n^2\\).)\n\nSince the condition is invariant under the operation of switching rows or columns, we may rearrange the matrix such that these \\(k\\) entries are on the main diagonal of the matrix, i.e. \\(a_{11}=a_{22}=\\dots=a_{kk}=0\\). Let \\(X=\\sum_{1\\le i,j\\le k}a_{ij}\\), \\(Y=\\sum_{i\\le k<j}a_{ij}+\\sum_{j\\le k<i}a_{ij}\\) and \\(Z=\\sum_{k<i,j\\le n}a_{ij}\\). Then applying the condition to \\(a_{11},\\dots,a_{kk}\\), we get \\(2X+Y\\ge kn\\).\n\nIf \\(i\\le k<j\\), then if \\(a_{ij}=a_{ji}=0\\), then we may toss \\(a_{ii}\\) out of our set and replace it with \\(a_{ij}\\) and \\(a_{ji}\\), to form a larger feasible set of elements, contradiction. Consequently, \\(a_{ij}+a_{ji}\\ge 1\\) for \\(i\\le k<j\\), which upon summation yields \\(Y\\ge k(n-k)\\). Finally, \\(Z\\ge (n-k)^2\\), trivially. Therefore,\n\n\\[\n2S=2(X+Y+Z)\\ge 2X+Y+Y+Z\\ge kn+k(n-k)+(n-k)^2=k^2+2k(n-k)+(n-k)^2=n^2.\n\\]\n\nThe above solution was posted and copyrighted by randomusername. The original thread for this problem can be found here: [3]\n\n## Remarks (added by pf02, March 2025)\n\nSolutions 1 and 2 above are incorrect, and there is room for better wording in solution 3.\n\nSolution 1 is plainly wrong, but it can be salvaged.\n\nSolution 2 is hopelessly incorrect. An attempt to salvage it would be along the lines of Solution 3, in the context of the unnecessarily complicated setup of Solution 2.\n\nSolution 3 is based on a very good idea, and it is easy to make the necessary improvements to make it robust.\n\nThe discussion page https://artofproblemsolving.com/community/c6h60831s1_sum_of_all_the_elements_in_the_matrix_is_at_least_n22 contains a fourth solution, but it is incomplete. If completed, it would be just a version of the corrected version of solution 1 given below.\n\nBelow I will point out the issues in the three solutions given above, and when I can, I will give the corrected or improved solutions. Then I will give some examples. The examples will show that the inequality \\(\\sum_{i, j = 1}^n a_{ij} \\ge \\frac{n^2}{2}\\) is optimal.\n\nNotation: in all the discussions and proofs given below, I will denote \\(r_i, c_j\\) the \\(i\\)th row and \\(j\\)th column respectively, and \\(sum(r_i), sum(c_j), \\text{or}\\ s(r_i), s(c_j)\\) the sums of their entries.",
"I will use the notation of the solution. We chose a row or column with minimal sum of its elements (we can assume it was a row), \\(S\\) was the sum of the elements in this row, and \\(z\\) was the number of zeros in it.\n\nAt some point in the solution the author writes:\n\n\"Note that, being \\(z \\ge n - S\\) and \\(n - S \\ge S\\), we can put \\(z - 1\\) instead of \\(z\\) and \\(n - z + 1\\) instead of \\(n - z\\) until we get \\(z = n - S\\)\"\n\nThis is wrong! If \\(z\\) was chosen to represent a chosen quantity, we can not change its value in the middle of the proof.\n\nWe could try to make sense of the proof by saying that we replace the given matrix A by a matrix B, where the rows of B are equal to the rows of A, except for the \"minimal\" row we chose, in which we replace one of the zeros by a number \\(> 0\\) (which would replace \\(z\\) by \\(z - 1\\)). This does not work, because if we would show that the sum of the elements of \\(B\\) is \\(\\ge \\frac{n^2}{2}\\), it would not follow that the same is true for \\(A\\).",
"As in the solution, choose a row or column such that the sum of its elements is minimal. We can assume this is a row (either notice that taking the transpose of \\(A\\) changes neither the hypothesis, nor the conclusion; or notice that if we do the proof in the case of a row, the proof in the case of a column is analogous.) Let \\(S\\) be the sum of the elements of this row, and \\(z\\) be the number of zeros in this row.\n\nFor an element \\(a_{ij}\\) of \\(A\\), we denote \\(r_i, c_j\\) the row and the column containing \\(a_{ij}\\). We denote \\(sum(r_i), sum(c_j)\\) the sum of the entries of the matrix in row \\(r_i\\) and \\(c_j\\) respectively.\n\nIf \\(S \\ge \\frac{n}{2}\\), then each row \\(r_i\\) has its sum of elements \\(sum(r_i) \\ge S \\ge \\frac{n}{2}\\), so \\(\\sum_{i, j = 1}^n a_{ij} = \\sum_{i=1}^n sum(r_i) \\ge \\frac{n^2}{2}\\).\n\nSo, let us assume \\(S < \\frac{n}{2}\\). Let \\(r_i = \\{a_{ij}, j = 1, \\dots, n\\}\\) be the row we chose (now \\(i\\) is fixed). For each \\(j\\) consider the sum of the elements of the column \\(c_j\\) containing \\(a_{ij}\\). If \\(a_{ij} = 0\\), we know that \\(sum(r_i) + sum(c_j) \\ge n\\) (by hypothesis), so \\(S + sum(c_j) \\ge n\\), so \\(sum(c_j) \\ge n - S\\). We have \\(z\\) such columns.\n\nIf \\(a_{ij} \\ne 0\\), we know that \\(sum(c_j) \\ge S\\) because \\(S\\) was minimal. We have \\(n - z\\) such columns.\n\nSo, \\(\\sum_{i, j = 1}^n a_{ij} = \\sum_{j = 1}^n sum(c_j) \\ge z(n - S) + (n - z)S\\).\n\nLet \\(S = n - z + E\\). Here we split \\(S\\) in two parts: \\(n - z\\) represents the number of non-zeroes in \\(r_i\\), and \\(E\\) represents the sum of the \"excess\" amounts over \\(1\\) of the non-zero entries. More explicitly, \\(E = \\sum_{\\substack{j = 1 \\\\ a_{ij} \\ne 0}}^n (a_{ij} - 1)\\). (To clarify: if all non-zero entries of \\(r_i\\) would \\(= 1\\), we would have \\(S = n - z\\), and \\(E = 0\\).) In general, \\(E \\ge 0\\).\n\nThen, \\(\\sum_{i, j = 1}^n a_{ij} \\ge z(n - S) + (n - z)S = (n - S + E)(n - S) + (S - E)S = n^2 - 2nS + 2S^2 + nE - 2SE =\\)\n\n\\(\\frac{n^2}{2} + \\frac{1}{2}\\left(n^2 - 4nS + 4S^2\\right) + E(n - 2S) = \\frac{n^2}{2} + \\frac{1}{2}(n - 2S)^2 + E(n - 2S)\\).\n\nThe last term is \\(\\ge 0\\) since \\(E \\ge 0\\) and we assumed we are in the case \\(S < \\frac{n}{2}\\). It follows that \\(\\sum_{i, j = 1}^n a_{ij} > \\frac{n^2}{2}\\).\n\nNote: The strict inequality is not true in general. We obtained it assuming that \\(S < \\frac{n}{2}\\). When \\(S \\ge \\frac{n}{2}\\) we have \\(\\sum_{i, j = 1}^n a_{ij} \\ge \\frac{n^2}{2}\\).\n\nNote: We could have proven directly that \\(z(n - S) + (n - z)S = \\frac{n^2}{2} + \\frac{1}{2}(n - 2S)^2 + (S + z - n)(n - 2S)\\) and note that the last term is the product of two non-negative factors. However, introducing \\(E\\) gives better insight into the inequality.",
"This solution starts by considering all the permutations \\(\\sigma\\) of \\(\\{1, 2, \\dots, n\\}\\). For each \\(\\sigma\\) consider the set \\(S_\\sigma = \\{i, \\text{such that}\\ a_{i\\sigma(i)} = 0\\}\\), and denote \\(M_\\sigma\\) to be the number of elements in \\(S_\\sigma\\). Let \\(M = \\text{maximum of all}\\ M_\\sigma\\). Choose a \\(\\sigma\\) such that \\(M_\\sigma = M\\).\n\nWe can assume that \\(S_\\sigma = \\{1, 2, \\dots, k\\}\\). (Note: this would normally need some justification, but a diligent reader can supply it.)\n\nFor given row \\(r_i\\) and column \\(c_j\\) define \\(s(r_i), s(c_j)\\) the sum of the elements of \\(r_i\\) and \\(c_j\\), respectively.\n\nFor any permutation \\(\\sigma\\) \"assign to it the number\" \\(N_\\sigma = \\sum_{\\substack{i\\ \\text{such that} \\\\ a_{i\\sigma(i)} = 0}} (s(r_i) + s(c_{\\sigma(i)}))\\).\n\nAt this point, the author says:\n\n\"Now pick a permutation \\(\\sigma\\) such that it generates \\(M\\) and its assigned number [\\(N_\\sigma\\)] is minimal.\"\n\nThis is the first problem with this proof. It is not clear whether the author wants to choose \\(\\sigma\\) to minimize \\(N_\\sigma\\) from among the permutations for which \\(M_\\sigma = M\\), or choose a \\(\\sigma\\) to minimize \\(N_\\sigma\\) over all the possible permutations, and at the same time \\(M_\\sigma = M\\). Since the latter may be impossible, we should assume the author meant the former. (Note: it does not make much difference, we could assume that a \\(\\sigma\\) in the sense of the latter interpretation is found, since we will soon see that the proof is flawed anyway.)\n\nUnder the assumption \\(k < n\\), take \\(j \\ge k + 1\\). The author goes on to say:\n\n\"if \\(r_j\\) has a \\(0\\) on \\(c_{\\sigma(p)}\\) with \\(p \\le k\\), i.e. \\(a_{j\\sigma(p)} = 0\\), by the minimality of the assigned number [\\(N_\\sigma\\)] of \\(\\sigma\\), we have that \\(s(r_j) \\ge s(r_p)\\)\"\n\nThis is completely wrong! One flaw is that \\(N_\\sigma\\) is the sum of many terms, one of them being \\(s(r_p)\\). Just because \\(N_\\sigma\\) is in some way minimal, we can not conclude that one of the terms is \\(\\le\\) than a term in some other sum. The other flaw is that it is not clear how the minimality of \\(N_\\sigma\\) is used. The minimality would mean that \\(N_\\sigma \\le N_\\tau\\) for other permutations \\(\\tau\\), but in this solution there is no other permutation defined.\n\nThere is no point in continuing the analysis, since this is a crucial argument in the proof, and it is an unrecoverable error.",
"We could drop all the discussion about the assigned number \\(N_\\sigma\\), and about trying to make it minimal.\n\nThe author wants to prove that for \\(j \\ge k + 1\\) (for which \\(a_{j\\sigma(j)} \\ne 0\\)) we have \\(s(r_j) + s(c_{\\sigma(j)}) \\ge n\\). (This would yield the desired conclusion). I believe this is true, and it can be done with a careful examination of the zeros on \\(r_j\\) and \\(c_{\\sigma(j)}\\). It would use ideas along the lines of the ideas in solution 3. It is not worth the effort of doing this (especially not in the formalism of permutations) since it would use no new ideas. This solution would be just a repetition of solution 3 with a much more complicated formalism.\n\nNote: We know that every permutation is a combination of swaps, so the permutation could be replaced by swapping rows and/or columns. Instead of having \\(a_{i\\sigma(i)} = 0\\) for \\(i = 1, 2, \\dots, k\\), we could make \\(a_{ii} = 0\\) for \\(i = 1, 2, \\dots, k\\), and \\(k\\) be maximal with this property.",
"The author says:\n\n\"consider the largest possible subset of elements, all zero, which are pairwise in different rows of columns - let the size of this subset be \\(k\\)\n\n...\n\nwe may rearrange the matrix such that these \\(k\\) entries are on the main diagonal of the matrix, i.e. \\(a_{11} = a_{22} = \\dots = a_{kk} = 0\\)\"\n\nThis is unclear: the definition of \\(k\\) does not make much sense, but if we insist of giving it some meaning, I would interpret it as counting the zeroes of \\(A\\) which are not on the diagonal. Note that this \\(k\\) could be larger than \\(n\\). Fortunately this flaw is very easy to correct.\n\nSince there are a few more unhappy choices of words in the proof, I will repeat the proof below, making the improvements to wording.",
"Both the hypothesis and the conclusion of the problem stay true if we swap rows, or swap columns, or take the transpose of the given matrix A. If the matrix has no zeroes then the sum of its entries is \\(\\ge n^2\\). So assume there are zeroes in the matrix.\n\nRecursively, swap rows and swap columns to bring more and more zeroes onto the diagonal. Let \\(k\\) be the maximum number of zeroes we can bring onto the diagonal. Swap more rows and columns if needed, so that we will have \\(a_{11} = a_{22} = \\dots = a_{kk} = 0\\).\n\nIf \\(k = n\\) then \\(\\sum_{i = 1}^n (sum(r_i) + sum(c_i)) \\ge n \\cdot n\\) (by hypothesis). On the other hand \\(\\sum_{i = 1}^n (sum(r_i) + sum(c_i)) = 2 \\sum_{i, j = 1}^n a_{ij}\\), so the desired result follows.\n\nAssuming \\(k < n\\), let \\(X = \\sum_{1 \\le i, j \\le k} a_{ij}\\), and let \\(Y = \\sum_{i \\le k < j} a_{ij} + \\sum_{j \\le k < i} a_{ij}\\), and let \\(Z = \\sum_{k < i, j \\le n} a_{ij}\\).\n\nBecause of the hypothesis applied to \\(a_{11}, \\dots, a_{kk}\\), we get \\(2X + Y \\ge kn\\).\n\nAssume we have \\(i \\le k < j\\) and \\(a_{ij} = a_{ji} = 0\\). Now swap columns \\(i\\) and \\(j\\). Then \\(a_{ij}\\) becomes the new entry at \\((i, i)\\) and \\(a_{ji}\\) becomes the new entry at \\((j, j)\\). It follows that we have \\(k + 1\\) zero entries on the diagonal, which contradicts the fact that \\(k\\) was maximal. Consequently, at least one of \\(a_{ij}, a_{ji}\\) is non-zero, so \\(a_{ij} + a_{ji} \\ge 1\\) for \\(i \\le k < j\\). It follows that \\(Y \\ge k(n - k)\\).\n\nIf we had \\(a_{ij} = 0\\) with \\(i, j > k\\) then again, we could swap columns \\(i\\) and \\(j\\). Then \\(a_{ij}\\) becomes the new entry at \\((i, i)\\), and again, we have a contradiction of the fact that \\(k\\) was maximal. Consequently, all elements \\(a_{ij}\\) with \\(i, j > k\\) are non-zero. It follows that \\(Z \\ge (n - k)^2\\).\n\nTherefore, \\(2S = 2(X + Y + Z) \\ge (2X + Y) + Y + Z \\ge kn + k(n - k) + (n - k)^2 = n^2\\).\n\nNote: Another way of finishing off the proof would be to look at \\(\\sum_{i = 1}^n (sum(r_i) + sum(c_i))\\). We know that the terms for \\(1 = 1, \\dots, k\\) are \\(\\ge n\\) by hypothesis. For \\(i > k\\), we know that \\(a_{ij} + a_{ji} \\ge 1\\) for \\(j \\le k\\), and that \\(a_{ij} \\ne 0\\) for \\(j > k\\). It follows that \\(sum(r_i) + sum(c_i) \\ge k + 2(n - k) = 2n - k \\ge n\\). This is the argument solution 2 would have made in its complicated formalism, had it reached this point correctly.\n\n## Examples\n\nWe will say that a matrix is minimal if it satisfies the hypothesis of the problem (i.e. \\(a_{ij} = 0\\) implies \\(sum(r_i) + sum(c_j) \\ge n\\)), and no entry of the matrix could be replaced by a smaller number such that the new matrix still satisfies the hypothesis of the problem.\n\nFor example, in the matrices below, the first, third and fifth are minimal, but the second is not, since one of its entries could be made smaller to yield the third matrix, and similarly the fourth is not, since one of its entries could be made smaller to yield the fifth matrix.\n\n\\begin{matrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} & \\begin{pmatrix} 0 & 1 \\\\ 2 & 0 \\end{pmatrix} & \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} & \\begin{pmatrix} 1 & 0 \\\\ 2 & 2 \\end{pmatrix} & \\begin{pmatrix} 0 & 0 \\\\ 2 & 2 \\end{pmatrix} \\end{matrix}\n\nThe examples I will give are all minimal.\n\nNow consider the following matrices:\n\n\\begin{matrix} \\begin{pmatrix} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 1 \\\\ 1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & 1 \\end{pmatrix} & \\begin{pmatrix} 1 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\end{pmatrix} & \\begin{pmatrix} 0 & 0 & 1 & 1 \\\\ 0 & 0 & 1 & 1 \\\\ 1 & 1 & 0 & 0 \\\\ 1 & 1 & 0 & 0 \\end{pmatrix} & \\begin{pmatrix} 2 & 0 & 0 & 0 \\\\ 0 & 2 & 0 & 0 \\\\ 0 & 0 & 2 & 0 \\\\ 0 & 0 & 0 & 2 \\end{pmatrix} \\end{matrix}\n\nThese examples show that for \\(n\\) even, it is possible to have matrices so that \\(\\sum_{i, j = 1}^n a_{ij} = \\frac{n^2}{2}\\). Note that the first three are essentially the same, they can be obtained from each other by swapping rows and/or columns.\n\nNow let us look at some examples with \\(n\\) odd.\n\n\\begin{matrix} \\begin{pmatrix} 1 & 0 & 1 & 0 & 1 \\\\ 0 & 1 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 1 \\\\ 0 & 1 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 1 \\end{pmatrix} & \\begin{pmatrix} 1 & 1 & 1 & 0 & 0 \\\\ 1 & 1 & 1 & 0 & 0 \\\\ 1 & 1 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & 1 \\end{pmatrix} & \\begin{pmatrix} 0 & 0 & 0 & 0 & 3 \\\\ 0 & 0 & 0 & 3 & 0 \\\\ 0 & 0 & 3 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 0 \\\\ 1 & 1 & 0 & 0 & 0 \\end{pmatrix} & \\begin{pmatrix} 3 & 0 & 0 & 0 & 0 \\\\ 0 & 2 & 1 & 0 & 0 \\\\ 0 & 1 & 2 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & 1 \\end{pmatrix} \\end{matrix}\n\nThe first two are essentially equivalent, since they can be obtained from each other by swapping rows and/or columns. These all show that the minimum sum can be reached. The minimum sum is \\(\\sum_{i, j = 1}^n a_{ij} = \\frac{n^2 + 1}{2}\\), which in this case (\\(n = 5\\)) is \\(13\\).\n\nAnd finally, let us look at a few examples which are minimal, but the sum of the entries is not the minimum (which for \\(n = 5\\) means that the sum is \\(> 13\\)).\n\n\\begin{matrix} \\begin{pmatrix} 3 & 0 & 0 & 0 & 0 \\\\ 0 & 3 & 0 & 0 & 0 \\\\ 0 & 0 & 3 & 0 & 0 \\\\ 0 & 0 & 0 & 3 & 0 \\\\ 0 & 0 & 0 & 0 & 2 \\end{pmatrix} & \\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 5 & 5 & 5 & 5 & 5 \\end{pmatrix} & \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\\\ 0 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 1 & 1 & 1 \\\\ 0 & 1 & 1 & 1 & 1 \\end{pmatrix} \\end{matrix}\n\n\\begin{matrix} \\begin{pmatrix} 1 & 4 & 4 & 4 & 4 \\\\ 1 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 \\\\ 1 & 0 & 0 & 0 & 0 \\end{pmatrix} & \\begin{pmatrix} 1 & 5 & 5 & 5 & 5 \\\\ 4 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix} & \\begin{pmatrix} 0 & 2 & 2 & 0 & 0 \\\\ 2 & 0 & 2 & 0 & 0 \\\\ 2 & 2 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 & 1 \\\\ 0 & 0 & 0 & 1 & 1 \\end{pmatrix} \\end{matrix}"
] |
IMO-1972-1
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https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_1
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Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.
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[
"There are \\(2^{10}-2=1022\\) distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between \\(10\\) and \\(90+91+92+93+94+95+96+97+98+99 < 10 \\cdot 100 = 1000\\). (There are fewer attainable sums.) As \\(1000 < 1022\\), the Pigeonhole Principle implies there are two distinct subsets whose members have the same sum. Let these sets be \\(A\\) and \\(B\\). Now, let \\(A' = A - (A \\cap B)\\) and \\(B' = B - (A \\cap B)\\). Notice \\(A'\\) and \\(B'\\) are disjoint. They are also nonempty because if \\(A = A \\cap B\\) or \\(B = A \\cap B\\), then one of \\(A\\) and \\(B\\) is a subset of the other, so they are either not distinct or have different sums. Therefore \\(A'\\) and \\(B'\\) are disjoint subsets our set of 10 distinct two-digit numbers, which proves the claim. \\(\\square\\)"
] |
IMO-1972-2
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https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_2
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Prove that if \(n \geq 4\), every quadrilateral that can be inscribed in a circle can be dissected into \(n\) quadrilaterals each of which is inscribable in a circle.
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[
"Our initial quadrilateral will be \\(ABCD\\).\n\nFor \\(n=4\\), we do this:\n\nTake \\(E\\in AB,F\\in CD,\\ EF\\|AD\\) with \\(E,F\\) sufficiently close to \\(A,D\\) respectively. Take \\(U\\in AD,V\\in EF\\) such that \\(AEVU\\) is an isosceles trapezoid, with \\(V\\) close enough to \\(F\\) (or \\(U\\) close enough to \\(D\\)) that we can find a circle passing through \\(U,D\\) (or \\(F,V\\)) which cuts the segments \\(UV,DF\\) in \\(X,Y\\). Our four cyclic quadrilaterals are \\(BCFE,\\ AEVU,\\ VFYX,\\ YXUD\\).\n\nFor \\(n\\ge 5\\) we do the exact same thing as above, but now, since we have an isosceles trapezoid, we can add as many trapezoids as we want by dissecting the one trapezoid with lines parallel to its bases.\n\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\n\n## Remarks (added by pf02, March 2025)\n\nThe construction described in the solution above is correct (in the sense that it describes a legitimate way of dissecting an inscribable quadrilateral into four inscribable quadrilaterals). However, the solution is incomplete and sloppily written.\n\nBelow I will discuss and complete the solution given above. Then, I will give a second solution. And finally, I will discuss the cases when \\(n = 2, n = 3\\).",
"The first issue is the fact that a construction is described, but there is no proof, not even a hint, why the quadrilaterals are inscribable. This is not obvious, and it needs a proof. I will give the proof below.\n\nThe second issue is the vagueness of \"close enough\" used twice in the proof. The first time it is used, \"\\(E, F\\) sufficiently close to \\(A, D\\) respectively\" is not needed (indeed, any segment \\(EF\\) parallel to \\(AD\\) would do), so there is no need to make this more precise. The second time it is used, namely \"\\(V\\) close enough to \\(F\\) (or \\(U\\) close enough to \\(D\\)) that we can find a circle passing through \\(U, D\\) (or \\(F, V\\))\" is indeed needed, and it is not at all clear what \"close enough\" should be, or that this is at all possible. I will come back to this shortly.\n\nThe third issue is poor wording. We don't need to \"add as many trapezoids as we want\". We want to dissect the one isosceles trapezoid into as many isosceles trapezoids as we want by lines parallel to its bases.\n\n\\[\nProb 1972 2 1.png\n\\]\n\nBefore giving the missing details, let us remember that a quadrilateral \\(ABCD\\) is inscribable if and only if a pair of opposing angles adds up to \\(\\pi,\\) in other words \\(\\angle A + \\angle C = \\pi,\\) or equivalently, \\(\\angle B + \\angle D = \\pi\\). In particular, any isosceles trapezoid is inscribable.\n\nNow let us show that the four quadrilaterals are inscribable. It is easy to see that the first one, \\(EBCF\\) is inscribable. Indeed, \\(\\angle C + \\angle A = \\pi\\). We know that \\(\\angle A = \\angle FEB\\) because of parallelism, so \\(\\angle C + \\angle FEB = \\pi\\). The second one, \\(AEVU\\) is an isosceles trapezoid by the choice of \\(VU\\), so it is inscribable. The third one, \\(DUXY\\) is inscribable by construction. It remains to be shown that \\(XYFV\\) is inscribable.\n\nWe have \\(\\angle YFV = \\pi - \\angle CFE = \\angle B = \\pi - \\angle D = \\angle YXU = \\pi - \\angle YXV\\). This shows that \\(XYFV\\) is inscribable.\n\nNote that as suggested by the solution, we could have chosen \\(XY\\) so that \\(XYFV\\) is inscribable, in which case a similar argument would have shown that \\(DUXY\\) is inscribable as well.\n\nLet us now make precise what it means that \\(U\\) should be close enough to \\(D\\), or \\(V\\) should be close enough to \\(F\\), so that we can find \\(XY\\), so that \\(DUXY\\) is inscribable.\n\nLet us assume that \\(\\angle D\\) and \\(\\angle DUV = \\angle DAB\\) are acute. One way of constructing the circle \\(DUXY\\) is the following: Pick \\(U\\) on \\(DA\\), pick \\(Y\\) on \\(DF\\), and find the center \\(O\\) of the circle \\(DUY\\) as the intersection of the medians to \\(DY\\) and \\(DU\\). Take \\(X\\) to be the intersection of this circle with \\(UV\\). We want to show that we can choose \\(U\\) and \\(Y\\) so that \\(X\\) will be between \\(U\\) and \\(V\\).\n\nLet us consider the median of \\(DF\\). This median intersects \\(DA\\) someplace on the same side of \\(D\\) as \\(A\\). Let us pick \\(U\\) between \\(D\\) and this point. This is the first condition for \\(U\\) being close enough to \\(D\\). This choice ensures that the midpoint of \\(DU\\) is in the region which is the strip between the median to \\(DF\\) and the perpendicular to \\(DF\\) at \\(D\\).\n\nNow consider \\(UV\\), the median to \\(UV\\), and the perpendicular to \\(UV\\) at \\(U\\). We want to make sure that \\(U\\) is chosen so that the midpoint of \\(DU\\) is in the region which is the strip between these two lines. For this, let us take \\(Z_1\\) to be the midpoint of \\(UV\\) and \\(Z_2\\) be the intersection of the median to \\(UV\\) with \\(DU\\). From the triangle \\(\\triangle UZ_1Z_2\\), we have \\(\\cos (\\angle DUV) = \\frac{UZ_1}{UZ_2}\\), so \\(UZ_2 = \\frac{UV}{2} \\cdot \\frac{1}{\\cos (\\angle DUV)}\\). In order to satisfy the condition about the midpoint of \\(DU\\), it is enough to have \\(\\frac{DU}{2} < UZ_2\\). This translates to \\(DU < \\frac{UV}{\\cos (\\angle DUV)}\\). Note that \\(UV\\) is of fixed slope and length because \\(UVEA\\) has to be an isosceles trapezoid, so this is a second condition expressing how close \\(U\\) has to be to \\(D\\).\n\nWith this choice of \\(U\\), the midpoint of \\(DU\\) is in the region which is the intersection of the two strips. Now pick \\(Y \\in DF\\) close enough to \\(D\\), so that the intersection \\(O\\) of the medians of \\(DY\\) and \\(DU\\) is in the same region. (\\(Y\\) has to be close enough to the foot of the perpendicular from the midpoint of \\(DU\\) to \\(DF\\).) This ensures that the circle of center \\(O\\) and radius \\(OD = OY = OU\\) will intersect \\(UV\\) at a point \\(X\\) between \\(U\\) and \\(V\\).\n\nIn the discussion above, we assumed \\(\\angle D\\) and \\(\\angle DUV\\) to be acute. The cases when either of these angles is not acute can be dealt with in a similar way. We will skip the discussion for these cases, and leave it to the interested reader to work out the details.",
"First, let us remember that a quadrilateral \\(ABCD\\) is inscribable if and only if a pair of opposing angles adds up to \\(\\pi,\\) in other words \\(\\angle A + \\angle C = \\pi,\\) or equivalently, \\(\\angle B + \\angle D = \\pi\\). In particular, any isosceles trapezoid is inscribable.\n\nLet us assume that \\(\\angle A\\) is less or equal than all the other angles. We can assume this because otherwise, we would just need to re-label the vertices.\n\n\\[\nProb 1972 2 2.png\n\\]\n\nChoose a point \\(P\\) inside \\(ABCD\\), and draw the parallel \\(PH\\) to \\(AB\\), with \\(H \\in DA\\). There will be restrictions on the position of \\(P\\), which we will discuss as we do the construction.\n\nChoose \\(E \\in AB\\) so that \\(AEPH\\) is an isosceles trapezoid. In order for this to be possible, \\(P\\) has to be inside the angle \\(\\angle ABB_1\\) formed by drawing the line \\(BB_1\\) such that \\(\\angle ABB_1 = \\angle A\\).\n\nNext, choose \\(F \\in BC\\) so that \\(\\angle PFB = \\angle A\\). This will ensure that \\(EBFP\\) is inscribable because \\(\\angle PEB = \\pi - \\angle A\\). In order for \\(F\\) to exist in \\(BC\\), \\(P\\) has to be inside the angle \\(BCC_1\\) formed by drawing the line \\(CC_1\\) such that \\(\\angle BCC_1 = \\angle A\\).\n\nNext choose \\(G \\in CD\\) such that \\(PG \\parallel BC\\). This will ensure that \\(PFCG\\) is an isosceles trapezoid because \\(\\angle PFC = \\pi - \\angle A = \\angle C\\). In order for \\(G\\) to exist in \\(CD\\), \\(P\\) has to be inside the angle \\(\\angle CDD_1\\) formed by drawing the line \\(DD_1\\) such that \\(\\angle CDD_1 = \\angle A\\).\n\nWith these choices, it is easy to see that \\(GDHP\\) is inscribable. Indeed, \\(\\angle PHD = \\angle A\\), and \\(\\angle DGP = \\pi - \\angle CGP = \\angle C = \\pi - \\angle A\\).\n\nWe now just have to show that the intersection of the interiors of angles \\(\\angle ABB_1, \\angle BCC_1, \\angle CDD_1\\) is not \\(\\O\\). This follows immediately from noticing that \\(DD_1 \\parallel BC\\).\n\nWe now proceed by induction on \\(n\\). For \\(n = 4\\) we dissected the inscribable quadrilateral \\(ABCD\\) in two inscribable rectangles, and \\(2 = 4 - 2\\) isosceles trapezoids. Assume we dissected the inscribable quadrilateral \\(ABCD\\) in two inscribable rectangles, and \\(k - 2\\) isosceles trapezoids. Then we dissect one of the isosceles trapezoids in two isosceles trapezoids by a line parallel to the bases, thus obtaining the statement for \\(k + 1\\).\n\n[Solution by pf02, March 2025]\n\n## Discuss the problem for n = 2, 3\n\nI will only sketch proofs for these cases.\n\n## The case n = 2\n\nIn general, an inscribable quadrilateral can not be dissected into two inscribable quadrilaterals. In fact, the inscribable quadrilateral \\(ABCD\\) can be dissected into two inscribable quadrilaterals if and only if \\(ABCD\\) is an isosceles trapezoid.\n\nTo see this, assume we draw \\(EF\\) as in the figure below, such that \\(ABEF\\) is inscribable.\n\n\\[\nProb 1972 2 3.png\n\\]\n\nIt follows that \\(\\angle FEB = \\pi - \\angle A = \\angle C\\), so \\(EF \\parallel CD\\). Similarly, if \\(EFDC\\) were inscribable, we would have \\(FE \\parallel AB\\). So \\(ABCD\\) would be a trapezoid. Being inscribable, it would have to be isosceles.\n\n## The case n = 3\n\nThis can be done, i.e. an inscribable quadrilateral can be dissected on three inscribable quadrilaterals. To see this, take \\(EF \\parallel CD\\) and \\(HG \\parallel AB\\) (equivalently, these say that \\(ABEF, GHDC\\) are inscribable. It follows that \\(FEGH\\) is inscribable as well. Indeed, \\(\\angle FEG = \\pi - \\angle FEB = \\angle A = \\pi - C = \\angle DHG = \\pi - \\angle FHG\\).\n\n\\(\\mathbf{Note}\\): This splitting does not yield an inductive method for \\(n > 3\\). For that, we still need to prove the case \\(n = 4\\)."
] |
IMO-1972-3
|
https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_3
|
Let \(m\) and \(n\) be arbitrary non-negative integers. Prove that
\[
\frac{(2m)!(2n)!}{m!n!(m+n)!}
\]
is an integer. (\(0! = 1\).)
|
[
"Denote the given expression as \\(f(m,n)\\). We intend to show that \\(f(m,n)\\) is integral for all \\(m,n \\geq 0\\). To start, we would like to find a recurrence relation for \\(f(m,n)\\). First, let's look at \\(f(m,n-1)\\):\n\n\\[\n\\begin{align*} f(m,n-1) &=\\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}\\\\ &=\\frac{(2m)!(2n)!(n-1)(m+n)}{m!n!(m+n)!(2n-1)(2n-2)}\\\\ &=f(m,n) \\cdot \\frac{(n-1)(m+n)}{(2n-1)(2n-2)}\\\\ &=f(m,n) \\cdot \\frac{m+n}{2(2n-1)} \\end{align*}\n\\]\n\nSecond, let's look at \\(f(m+1,n-1)\\):\n\n\\[\n\\begin{align*} f(m+1,n-1) &=\\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!}\\\\ &=\\frac{(2m)!(2n)!(2m+1)(2m+2)(n-1)}{m!n!(m+n)!(m+1)(2n-1)(2n-2)}\\\\ &= f(m,n) \\cdot \\frac{(2m+1)(2m+2)(n-1)}{(m+1)(2n-1)(2n-2)}\\\\ &=f(m,n) \\cdot \\frac{(2m+1)}{2n-1} \\end{align*}\n\\]\n\nCombining,\n\n\\[\n\\begin{align*} 4f(m,n-1)-f(m+1,n-1) &=f(m,n)\\cdot \\Bigg(\\frac{4(m+n)}{2(2n-1)}-\\frac{2m+1}{2n-1}\\Bigg)\\\\ &=f(m,n) \\cdot \\frac{2m+2n-2m-1}{2n-1}\\\\ &=f(m,n). \\end{align*}\n\\]\n\nTherefore, we have found the recurrence relation\n\n\\[\nf(m,n)=4f(m,n-1)-f(m+1,n-1).\n\\]\n\nNote that \\(f(m,0)\\) is just \\(\\binom{2m}{m}\\), which is an integer for all \\(m \\geq 0\\). Then\n\n\\[\nf(m,1)=4f(m,0)-f(m+1,0),\n\\]\n\nso \\(f(m,1)\\) is an integer, and therefore \\(f(m,2)=4f(m,1)-f(m+1,1)\\) must be an integer, etc.\n\nBy induction, \\(f(m,n)\\) is an integer for all \\(m,n \\geq 0\\).\n\nBorrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html",
"Let p be a prime, and n be an integer. Let \\(V_p(n)\\) be the largest positive integer \\(k\\) such that \\(p^k|n\\)\n\nWTS: For all primes \\(p\\), \\(V_p((2m)!)+V_p((2n)!) \\ge V_p(m!)+V_p(n!)+V_p((m+n)!)\\)\n\nWe know\n\n\\[\nV_p(x!)=\\sum_{a=1}^{\\infty} \\left\\lfloor{\\frac{x}{p^a}}\\right\\rfloor\n\\]\n\nLemma 2.1: Let \\(a,b\\) be real numbers. Then \\(\\lfloor{2a}\\rfloor+\\lfloor{2b}\\rfloor\\ge\\lfloor{a}\\rfloor+\\lfloor{b}\\rfloor+\\lfloor{a+b}\\rfloor\\)\n\nProof of Lemma 2.1: Let \\(a_1=\\lfloor{a}\\rfloor\\) and \\(b_1=\\lfloor{b}\\rfloor\\)\n\n\\[\n\\lfloor{2a}\\rfloor+\\lfloor{2b}\\rfloor=2(a_1+b_1)+\\lfloor2\\{a\\}\\rfloor+\\lfloor2\\{b\\}\\rfloor\n\\]\n\nOn the other hand, \\(\\lfloor{a}\\rfloor+\\lfloor{b}\\rfloor+\\lfloor{a+b}\\rfloor=a_1+b_1+(a_1+b_1)+\\lfloor\\{2(a+b)\\}\\rfloor\\)\n\nIt is trivial that \\(\\lfloor2\\{a\\}\\rfloor+\\lfloor2\\{b\\}\\rfloor\\ge\\lfloor\\{2(a+b)\\}\\rfloor\\) (Triangle Inequality)\n\nApply Lemma 2.1 to the problem: and we are pretty much done.\n\nNote: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.",
"The last step seems mistaken. For example [1.5]+[1.5]<[1.5+1.5]. Triangle inequality is about absolute value |x| not floor function [x]. Following is a rectified solution.\n\nLemma: [a+b]<=[a]+[b]+1, a,b>=0.\n\nProof: a<[a]+1, b<[b]+1, so a+b<[a]+[b]+2, [a+b]<=a+b<[a]+[b]+2, so [a+b]<=[a]+[b]+1.\n\nLet (x,2x,…,nx)=n, p is any prime, V2(9!)=V2(1•2•3•4•5•6•7•8•9) =(2,4,6,8)+(4,8)+(8)=[9/2]+[9/4]+[9/8]=7.\n\nGeneralize it and we have, Vp(m!)=Vp(1•2•….•m)=[m/p]+[m/p^2]+…+[m/p^k], k=1~∞, or,\n\nVp(m!)=∑([m/p^k],k=1~∞)\\(\\ \\ \\ \\ (1)\\).\n\nObviously when p^k>m, k>=1, [m/p^k]=0.\n\nNote the target statement is equivalent to:\n\nfor any prime p, Vp((2m)!(2n)!)>=Vp(m!n!(m+n)!), or,\n\nVp((2m)!)+Vp((2n)!)>=Vp(m!)+Vp(n!)+Vp(m+n)! \\(\\ \\ \\ \\ (2)\\).\n\nLet m/p^k=a(k), n/p^k=b(k), k=1~∞. Combine (1), we note that (2) is equivalent to:\n\nfor any k in (1), [2a(k)]+[2b(k)]>=[a(k)]+[b(k)]+[a(k)+b(k)] \\(\\ \\ \\ \\ (3)\\).\n\nFor convenience of viewing, let a(k)=a, b(k)=b, a,b>=0, (3) is then rewritten as:\n\n[2a]+[2b]>=[a]+[b]+[a+b] \\(\\ \\ \\ \\ (4)\\).\n\nWe prove it under different scenarios:\n\n-If a<[a]+1/2, b<[b]+1/2, Then 2[a]<=[2a]<=2a<2[a]+1, [2a]=2[a], likewise, [2b]=2[b], so (4) is equivalent to [a]+[b]>=[a+b]. Per current condition, a+b<[a]+[b]+1, so [a+b]<[[a]+[b]+1]=[a]+[b]+1, also obviously [a+b]>=[a]+[b], so [a+b]=[a]+[b], proved;\n\n-If a<[a]+1/2, b>=[b]+1/2, then [2a]=2[a], [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+1>=[a+b], as per lemma, proved;\n\n-If a>=[a]+1/2, b>=[b]+1/2, so [2a]=2[a]+1, [2b]=2[b]+1, so (4) is equivalent to [a]+[b]+2>=[a+b], as per lemma, proved;\n\nTherefore (4) is proved, so is (2), as every step from (2) to (4) is reversible. Therefore (2m)!(2n)!/(m!n!(m+n)!) is an integer. Proof completed.\n\n## Remarks (added by pf02, April 2025)\n\n1. The first solution has a few errors in the computation, but they cancel out, and the proof is essentially correct. Below I will give the corrections, to make the solution flawless.\n\n2. As pointed out by a contributor, the second solution is incorrect. The correction given is good, but very sloppily written and hard to read (even after I edited the spacing, to make it legible). Below, I will repeat the argument in a simplified form, to make it easier to follow.\n\n3. I will give a third solution below, inspired by the papers mentioned in the next paragraph.\n\n4. The problem is a classical result, attributed to Eugene Catalan, who stated it in a paper published in 1874. The numbers \\(S(m, n) = \\frac{(2m)!(2n)!}{m!n!(m+n)!}\\) are sometimes called the super-Catalan numbers. (They are mentioned at the end of https://en.wikipedia.org/wiki/Catalan_number).\n\nThere are a few papers published about them. See for example the paper \"The Super Catalan Numbers \\(S(m, m + s)\\) for \\(s \\le 3\\) and Some Integer Factorial Ratios\" by Xin Chen & Jane Wang (https://www-users.cse.umn.edu/~reiner/REU/ChenWang2012.pdf), and \"Super Ballot Numbers\" by Ira M. Gessel (especially section 6) (https://www.sciencedirect.com/science/article/pii/0747717192900342/pdf).",
"We have\n\n\\[\nf(m,n-1) = \\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!} = \\frac{(2m)!(2n)! \\cdot (n)(m+n)}{m!n!(m+n)! \\cdot (2n-1)(2n)} = f(m,n) \\cdot \\frac{m+n}{2(2n-1)}\n\\]\n\nand\n\n\\[\nf(m+1,n-1) = \\frac{(2m+2)!(2n-2)!}{(m+1)!(n-1)!(m+n)!} = \\frac{(2m)!(2n)! \\cdot (2m+1)(2m+2)(n)}{m!n!(m+n)! \\cdot (m+1)(2n-1)(2n)} = f(m,n) \\cdot \\frac{2m+1}{2n-1}\n\\]\n\nThe rest of the proof is correct.",
"To prove that \\(S(m, n) = \\frac{(2m)!(2n)!}{m!n!(m+n)!}\\) is an integer, we have to prove that for any prime number \\(p\\), the exponent of the largest power of \\(p\\) in \\((2m)!(2n)!\\) is larger than the exponent of the largest power of \\(p\\) in \\(m!n!(m+n)!\\). (In fact, we have to consider only primes which satisfy \\(p \\le m + n\\)).\n\nThe largest power of \\(p\\) in \\(N\\)! is given by \\(\\nu_p(N!) = \\sum_{i=1}^{\\infty} \\left\\lfloor\\frac{N}{p^i}\\right\\rfloor\\). (The sum is a finite sum, since we have to consider only powers \\(i \\le \\lfloor\\log_p N\\rfloor.)\\) This is easy to see, but it is a reasonably well known fact (see Legendre's Formula or https://en.wikipedia.org/wiki/Legendre's_formula.) Note that \\(\\nu_p(M!N!) = \\nu_p(M!) + \\nu_p(N!)\\).\n\nThus, it is enough to prove that \\(\\left\\lfloor\\frac{2m}{p^i}\\right\\rfloor + \\left\\lfloor\\frac{2n}{p^i}\\right\\rfloor \\ge \\left\\lfloor\\frac{m}{p^i}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^i}\\right\\rfloor + \\left\\lfloor\\frac{m + n}{p^i}\\right\\rfloor\\ \\ \\ \\ \\ \\ \\ \\ (1)\\).\n\nLet \\(m = Q_1 \\cdot p^i + R_1\\) be the integer division of \\(m\\) by \\(p^i\\) (so \\(0 \\le Q_1 < p^i\\)), and let \\(r_1 = \\frac{Q_1}{p^i}\\). In other words, \\(r_1 = \\frac{m}{p^i} - \\left\\lfloor\\frac{m}{p^i}\\right\\rfloor\\). Note that \\(0 \\le r_1 < 1\\). Similarly, obtain \\(r_2\\) from the division of \\(n\\) by \\(p^i\\).\n\nNote that \\(\\left\\lfloor\\frac{2m}{p^i}\\right\\rfloor = 2\\left\\lfloor\\frac{m}{p^i}\\right\\rfloor\\) or \\(\\left\\lfloor\\frac{2m}{p^i}\\right\\rfloor = 2\\left\\lfloor\\frac{m}{p^i}\\right\\rfloor + 1\\) depending on whether \\(r_1 < \\frac{1}{2}\\) or \\(r_1 \\ge \\frac{1}{2}\\) and similarly for \\(n\\) and \\(r_2\\). Also \\(\\left\\lfloor\\frac{m + n}{p^i}\\right\\rfloor = \\left\\lfloor\\frac{m}{p^i}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^i}\\right\\rfloor\\) or \\(\\left\\lfloor\\frac{m + n}{p^i}\\right\\rfloor = \\left\\lfloor\\frac{m}{p^i}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^i}\\right\\rfloor + 1\\) depending on whether \\(r_1 + r_2 < 1\\) or \\(r_1 + r_2 \\ge 1\\).\n\nWe have the following cases:\n\n1. If \\(r_1 < \\frac{1}{2}\\) and \\(r_2 < \\frac{1}{2}\\) then we have equality in (1).\n\n2. If \\(r_1 \\ge \\frac{1}{2}, r_2 < \\frac{1}{2}\\), and \\(r_1 + r_2 < 1\\) then after cancelling terms, the left hand side in (1) becomes 1, and the right hand side becomes 0.\n\n3. If \\(r_1 \\ge \\frac{1}{2}, r_2 < \\frac{1}{2}\\), and \\(r_1 + r_2 \\ge 1\\) then we have equality in (1).\n\n4, 5. The two cases when \\(r_1 < \\frac{1}{2}, r_2 \\ge \\frac{1}{2}\\) are similar to cases 2 and 3.\n\n6. If \\(r_1 \\ge \\frac{1}{2}, r_2 \\ge \\frac{1}{2}\\), then after cancelling terms, the left hand side in (1) becomes 2, and the right hand side becomes 1.\n\nThe above shows that (1) is always true, which proves the problem.",
"This solution uses the fact that \\({n \\choose k} = \\frac{n!}{k!(n - k)!}, k = 0, \\dots n\\) are the coefficients of the powers of \\(x\\) in the expansion of \\((1 + x)^n\\) (see Binomial Theorem or https://en.wikipedia.org/wiki/Binomial_coefficient). Of course these coefficients, hence these expressions in \\(n, k\\) are integers.\n\nWithout loss of generality, we can assume \\(m \\le n\\). We will use the identity\n\n\\(S(m, n) = \\frac{(2m)!(2n)!}{m!n!(m+n)!} = (-1)^m \\sum_{k=0}^{k=2m} (-1)^k {2m \\choose k} {2n \\choose m + n - k}\\).\n\nThis identity is attributed to K. von Szily (1894). Note that we could take the sum over all integers, using the convention that \\({n \\choose k} = 0\\) when \\(k < 0\\) or \\(k > n\\).\n\nThe fact that \\(S(m, n)\\) is integer follows immediately from this identity because the right hand side is integer.\n\nThus, we just need to prove this identity.\n\nStart with \\((1 - x^2)^{m + n} = (1 + x)^{m + n} (1 - x)^{m + n}\\). Expand both sides and look at the coefficient of \\(x^{2m}\\) in both expansions. Equate the two expressions giving the coefficient of \\(x^{2m}\\) in the two sides. We get\n\n\\((-1)^m {m + n \\choose m} = \\sum_{k=0}^{2m} {m + n \\choose k} \\cdot (-1)^{2m - k} {m + n \\choose 2m - k}\\).\n\nWrite out the binomial coefficients in terms of factorials, and multiply both sides by \\((-1)^m \\frac{(2m)!(2n)!}{[(m + n)!]^2}\\). We get\n\n\\[\n\\frac{(m + n)!}{m!n!} \\cdot \\frac{(2m)!(2n)!}{[(m + n)!]^2} = (-1)^m \\sum_{k=0}^{2m} (-1)^k \\frac{(m + n)!}{k!(m + n - k)!} \\cdot \\frac{(m + n)!}{(2m - k)!(n - m + k)!} \\cdot \\frac{(2m)!(2n)!}{[(m + n)!]^2}\n\\]\n\nSimplifying and rearranging factors, we get\n\n\\[\n\\frac{(2m)!(2n)!}{m!n!(m + n)!} = (-1)^m \\sum_{k=0}^{2m} (-1)^k \\frac{(2m)!}{k!(2m - k)!} \\cdot \\frac{(2n)!}{(n + m - k)!(n - m + k)!} = (-1)^m \\sum_{k=0}^{2m} (-1)^k {2m \\choose k} {2n \\choose n + m - k}\n\\]\n\nThis finishes the problem."
] |
IMO-1972-4
|
https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_4
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Find all solutions \((x_1, x_2, x_3, x_4, x_5)\) of the system of inequalities
\[
\begin{align*} (x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \\ (x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \\ (x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \\ (x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \\ (x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0 \end{align*}
\]
where \(x_1, x_2, x_3, x_4, x_5\) are positive real numbers.
|
[
"Add the five inequalities together to get\n\n\\[\n(x_1^2 - x_3 x_5)(x_2^2 - x_3 x_5) + (x_2^2 - x_4 x_1)(x_3^2 - x_4 x_1) + (x_3^2 - x_5 x_2)(x_4^2 - x_5 x_2) +\n\\]\n\n\\[\n(x_4^2 - x_1 x_3)(x_5^2 - x_1 x_3) + (x_5^2 - x_2 x_4)(x_1^2 - x_2 x_4) \\leq 0\n\\]\n\nExpanding, multiplying by \\(2\\), and re-combining terms, we get\n\n\\[\n(x_1 x_2 - x_1 x_4)^2 + (x_2 x_3 - x_2 x_5)^2 + (x_3 x_4 - x_3 x_1)^2 + (x_4 x_5 - x_4 x_2)^2 + (x_5 x_1 - x_5 x_3)^2 +\n\\]\n\n\\[\n(x_1 x_3 - x_1 x_5)^2 + (x_2 x_4 - x_2 x_1)^2 + (x_3 x_5 - x_3 x_2)^2 + (x_4 x_1 - x_4 x_3)^2 + (x_5 x_2 - x_5 x_4)^2 \\leq 0\n\\]\n\nEvery term is \\(\\geq 0\\), so every term must \\(= 0\\).\n\nFrom the first term, we can deduce that \\(x_2 = x_4\\). From the second term, \\(x_3 = x_5\\).\n\nFrom the third term, \\(x_4 = x_1\\). From the fourth term, \\(x_5 = x_2\\).\n\nTherefore, \\(x_1 = x_4 = x_2 = x_5 = x_3\\) is the only solution.\n\nBorrowed from [1]",
"This solution is longer and less elegant than the previous one, but it is more direct, and based on a different idea, so it is worth mentioning.\n\nLooking at the first inequality as an example, we can see that we have either\n\n\\((x_1^2 - x_3x_5) \\le 0\\) and \\((x_2^2 - x_3x_5) \\ge 0\\) \\(\\hspace{35pt}\\) (we will call this a \"direct\" pair of inequalities)\n\nor\n\n\\((x_1^2 - x_3x_5) \\ge 0\\) and \\((x_2^2 - x_3x_5) \\le 0\\) \\(\\hspace{35pt}\\) (we will call this an \"inverted\" pair of inequalities)\n\nWe have a similar conclusion from the other four inequalities. So, we have to look at six cases:\n\n1. all five inequalities imply 5 \"direct\" pairs of inequalities\n\n2. the five inequalities imply 4 \"direct\" pairs and 1 \"inverted\" pair\n\n3. the five inequalities imply 3 \"direct\" pairs and 2 \"inverted\" pairs\n\n4. the five inequalities imply 2 \"direct\" pairs and 3 \"inverted\" pairs\n\n5. the five inequalities imply 1 \"direct\" pair and 4 \"inverted\" pairs\n\n6. all five inequalities imply 5 \"inverted\" pairs of inequalities\n\nSolving the problem in cases 4, 5, 6 is similar to solving it in cases 3, 2, 1 respectively; indeed, all we have to do is switch \\(\\le\\) and \\(\\ge\\). So, we will solve the system of inequalities in cases 1, 2, 3.\n\nIn case 1, we have\n\n\\[\n\\begin{align*} (x_1^2 - x_3x_5) \\le 0 \\hspace{20pt} (1l) \\hspace{40pt} (x_2^2 - x_3x_5) \\ge 0 \\hspace{20pt} (1r) \\\\ (x_2^2 - x_4x_1) \\le 0 \\hspace{20pt} (2l) \\hspace{40pt} (x_3^2 - x_4x_1) \\ge 0 \\hspace{20pt} (2r) \\\\ (x_3^2 - x_5x_2) \\le 0 \\hspace{20pt} (3l) \\hspace{40pt} (x_4^2 - x_5x_2) \\ge 0 \\hspace{20pt} (3r) \\\\ (x_4^2 - x_1x_3) \\le 0 \\hspace{20pt} (4l) \\hspace{40pt} (x_5^2 - x_1x_3) \\ge 0 \\hspace{20pt} (4r) \\\\ (x_5^2 - x_2x_4) \\le 0 \\hspace{20pt} (5l) \\hspace{40pt} (x_1^2 - x_2x_4) \\ge 0 \\hspace{20pt} (5r) \\end{align*}\n\\]\n\n(1l) and (1r) imply \\(x_1^2 \\le x_2^2\\), and we have four similar inequalities from the other four rows. Putting them together, we obtain\n\n\\[\nx_1^2 \\le x_2^2 \\le x_3^2 \\le x_4^2 \\le x_5^2 \\le x_1^2\n\\]\n\nwhich implies \\(x_1 = x_2 = x_3 = x_4 = x_5\\) (remember that \\(x_n\\) are positive).\n\nFor case 2, let us look at the particular situation when the first pair of inequalities is \"inverted\":\n\n\\[\n\\begin{align*} (x_1^2 - x_3x_5) \\ge 0 \\hspace{20pt} (1l) \\hspace{40pt} (x_2^2 - x_3x_5) \\le 0 \\hspace{20pt} (1r) \\\\ (x_2^2 - x_4x_1) \\le 0 \\hspace{20pt} (2l) \\hspace{40pt} (x_3^2 - x_4x_1) \\ge 0 \\hspace{20pt} (2r) \\\\ (x_3^2 - x_5x_2) \\le 0 \\hspace{20pt} (3l) \\hspace{40pt} (x_4^2 - x_5x_2) \\ge 0 \\hspace{20pt} (3r) \\\\ (x_4^2 - x_1x_3) \\le 0 \\hspace{20pt} (4l) \\hspace{40pt} (x_5^2 - x_1x_3) \\ge 0 \\hspace{20pt} (4r) \\\\ (x_5^2 - x_2x_4) \\le 0 \\hspace{20pt} (5l) \\hspace{40pt} (x_1^2 - x_2x_4) \\ge 0 \\hspace{20pt} (5r) \\end{align*}\n\\]\n\nJust like in case 1, this implies\n\n\\[\nx_2^2 \\le x_3^2 \\le x_4^2 \\le x_5^2 \\le x_1^2 \\hspace{30pt} (6)\n\\]\n\n(4r) and (5l) imply \\(x_1 x_3 \\le x_2 x_4\\). Combining this with (6), we obtain \\(x_1 x_2 \\le x_1 x_3 \\le x_2 x_4\\). From this we deduce \\(x_1 \\le x_4\\). Combining this with (6) again, we get \\(x_1 = x_4 = x_5\\).\n\nNow, (2r) and (3l) imply \\(x_4 x_1 \\le x_5 x_2\\), which implies \\(x_1 \\le x_2\\) (because \\(x_4 = x_5\\)). Combining with (6) again, it follows that \\(x_1 = x_2 = x_3 = x_4 = x_5\\).\n\nIf the \"inverted\" pair is the 2nd row, we can make the substitution \\(x_1 = y_5, x_2 = y_1, x_3 = y_2, x_4 = y_3, x_5 = y_4\\). In the new variables, the 1st row would contain the \"inverted\" pair, so our proof applies. If the \"inverted\" pair is in the 3rd, 4th or 5th row, we just repeat this substitution as many times as needed to reduce the system of inequalities to the one we already solved. This finishes the proof in case 2.\n\nFor case 3, let us look at the particular situation when the first two pairs of inequalities are \"inverted\":\n\n\\[\n\\begin{align*} (x_1^2 - x_3x_5) \\ge 0 \\hspace{20pt} (1l) \\hspace{40pt} (x_2^2 - x_3x_5) \\le 0 \\hspace{20pt} (1r) \\\\ (x_2^2 - x_4x_1) \\ge 0 \\hspace{20pt} (2l) \\hspace{40pt} (x_3^2 - x_4x_1) \\le 0 \\hspace{20pt} (2r) \\\\ (x_3^2 - x_5x_2) \\le 0 \\hspace{20pt} (3l) \\hspace{40pt} (x_4^2 - x_5x_2) \\ge 0 \\hspace{20pt} (3r) \\\\ (x_4^2 - x_1x_3) \\le 0 \\hspace{20pt} (4l) \\hspace{40pt} (x_5^2 - x_1x_3) \\ge 0 \\hspace{20pt} (4r) \\\\ (x_5^2 - x_2x_4) \\le 0 \\hspace{20pt} (5l) \\hspace{40pt} (x_1^2 - x_2x_4) \\ge 0 \\hspace{20pt} (5r) \\end{align*}\n\\]\n\nJust like in case 1, this implies\n\n\\[\nx_3^2 \\le x_2^2 \\le x_1^2 \\hspace{55pt} (7)\n\\]\n\n\\[\nx_3^2 \\le x_4^2 \\le x_5^2 \\le x_1^2 \\hspace{30pt} (8)\n\\]\n\nNow, (3r) and (4l) imply \\(x_5 x_2 \\le x_1 x_3\\), and (4r) and (5l) imply \\(x_1 x_3 \\le x_2 x_4\\). This implies \\(x_5 x_2 \\le x_2 x_4\\), so \\(x_5 \\le x_4\\). Using (8), it follows that \\(x_4 = x_5\\).\n\nNow, (2l) and (1r) imply \\(x_4 x_1 \\le x_3 x_5\\). Simplifying with \\(x_4 = x_5\\), we get \\(x_1 \\le x_3\\). Combining this with (7) and (8), we get \\(x_1 = x_2 = x_3 = x_4 = x_5\\).\n\nNote that by applying the substitution described in case 2 as many times as needed, we can always assume that one of the \"inverted\" pairs is in the 1st row. Furthermore, if the \"inverted\" pairs are in the 1st and 5th rows, we can make a substitution which makes the \"inverted\" pairs of inequalities be in the 1st and 2nd rows (1st becomes 2nd, and 5th becomes 1st).\n\nSo we have to consider 2 more situations for case 3: the \"inverted\" pairs are in the 1st and 3rd rows, and the \"inverted\" pairs are in the 1st and 4th rows. The proof in each of these situations goes along the same lines as the one we just gave, and we will not go into these details, but leave them to the diligent reader.\n\n[Solution by pf02, May 2025]"
] |
IMO-1972-5
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https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_5
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Let \(f\) and \(g\) be real-valued functions defined for all real values of \(x\) and \(y\), and satisfying the equation
\[
f(x + y) + f(x - y) = 2f(x)g(y)
\]
for all \(x, y\). Prove that if \(f(x)\) is not identically zero, and if \(|f(x)| \leq 1\) for all \(x\), then \(|g(y)| \leq 1\) for all \(y\).
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[
"Let \\(u>0\\) be the least upper bound for \\(|f(x)|\\) for all \\(x\\). So, \\(|f(x)| \\leq u\\) for all \\(x\\). Then, for all \\(x,y\\),\n\n\\[\n2u \\geq |f(x+y)+f(x-y)| = |2f(x)g(y)|=2|f(x)||g(y)|\n\\]\n\nTherefore, \\(u \\geq |f(x)||g(y)|\\), so \\(|f(x)| \\leq u/|g(y)|\\).\n\nSince \\(u\\) is the least upper bound for \\(|f(x)|\\), \\(u/|g(y)| \\geq u\\). Therefore, \\(|g(y)| \\leq 1\\).\n\nBorrowed from [1]"
] |
IMO-1972-6
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https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_6
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Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.
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[
"Let our planes be \\(\\pi_1,\\pi_2,\\pi_3,\\pi_4\\), which we assume to be parallel to the \\(xy\\)-plane, listed in the increasing order of their \\(z\\)-coordinates. First take a plane \\(\\pi\\) orthogonal to \\(\\pi_i\\), which cuts \\(\\pi_1,\\pi_2,\\pi_3\\) along three lines \\(d_1,d_2,d_3\\). On these three lines, take three vertices \\(A_1,A_2,A_3\\) respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the \\(3\\)-dimensional version of this), and then complete the two regular tetrahedra \\(A_1A_2A_3P_1,A_1A_2A_3P_2\\) having \\(A_1A_2A_3\\) as one of their faces. Both \\(P_i\\) lie below \\(\\pi_4\\).\n\nNow take another plane \\(\\pi\\) and repeat the construction above. If \\(\\pi\\) makes a small enough angle with the \\(\\pi_i\\)'s, one of the \\(P_i\\)'s we get this time must lie above \\(\\pi_4\\). Now, if we move the initial position of \\(\\pi\\) towards the new one continuously and record the \\(z\\)-coordinates of \\(P_1,P_2\\), these will be continuous functions of the angle that \\(\\pi\\) makes with \\(\\pi_i\\), and for one of the points \\(P_1,P_2\\) the \\(z\\)-coordinate will move continuously from being smaller than that of \\(\\pi_4\\) to being larger than it, meaning that at some point, one of the points \\(P_1,P_2\\) will lie on \\(\\pi_4\\), and this is what we want.\n\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]",
"Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If \\(p_1\\), \\(p_2\\), \\(p_3\\), ... is a family of parallel planes in space, then we choose a unit vector \\(\\overrightarrow{v}_p\\) perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the directed distance between two of these planes \\(p_i\\) and \\(p_j\\), we denote the real number k such that the translation with translation vector \\(k\\cdot\\overrightarrow{v}_p\\) maps the plane \\(p_i\\) to the plane \\(p_j\\). Similarly, we define the directed distance between two of a family of parallel lines in a plane.)\n\nThe problem can be rewritten as follows: Given four distinct parallel planes \\(p_1\\), \\(p_2\\), \\(p_3\\), \\(p_4\\) in space, prove that there exists a regular tetrahedron XYZW such that \\(X\\in p_1\\), \\(Y\\in p_2\\), \\(Z\\in p_3\\), \\(W\\in p_4\\).\n\nIn order to do this, it is enough to find a regular tetrahedron ABCD somewhere in space and four parallel planes \\(q_1\\), \\(q_2\\), \\(q_3\\), \\(q_4\\) such that \\(A\\in q_1\\), \\(B\\in q_2\\), \\(C\\in q_3\\), \\(D\\in q_4\\) and \\(d\\left(q_1;\\;q_2\\right): d\\left(q_2;\\;q_3\\right): d\\left(q_3;\\;q_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\). In fact, once we have found such a tetrahedron ABCD and such planes \\(q_1\\), \\(q_2\\), \\(q_3\\), \\(q_4\\), then, because of \\(p_1\\parallel p_2\\parallel p_3\\parallel p_4\\), \\(q_1\\parallel q_2\\parallel q_3\\parallel q_4\\) and \\(d\\left(q_1;\\;q_2\\right): d\\left(q_2;\\;q_3\\right): d\\left(q_3;\\;q_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\), there exists a similitude transformation which maps the planes \\(q_1\\), \\(q_2\\), \\(q_3\\), \\(q_4\\) to the planes \\(p_1\\), \\(p_2\\), \\(p_3\\), \\(p_4\\); this similitude transformation will then obviously map the regular tetrahedron ABCD with \\(A\\in q_1\\), \\(B\\in q_2\\), \\(C\\in q_3\\), \\(D\\in q_4\\) to a regular tetrahedron XYZW with \\(X\\in p_1\\), \\(Y\\in p_2\\), \\(Z\\in p_3\\), \\(W\\in p_4\\); hence, the existence of such a tetrahedron XYZW will be proven, and the problem will be solved.\n\nSo consider a regular tetrahedron ABCD lying arbitrarily in space; we try to find four parallel planes \\(q_1\\), \\(q_2\\), \\(q_3\\), \\(q_4\\) such that \\(A\\in q_1\\), \\(B\\in q_2\\), \\(C\\in q_3\\), \\(D\\in q_4\\) and \\(d\\left(q_1;\\;q_2\\right): d\\left(q_2;\\;q_3\\right): d\\left(q_3;\\;q_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\).\n\nIn fact, we start working in the plane ABC. Let T be the point on the line AC such that \\(AT: TC=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right)\\) (where the segments AT and TC are directed). Let \\(g_2\\) be the line BT, and let \\(g_1\\) and \\(g_3\\) be the parallels to the line \\(g_2=BT\\) through the points A and C, respectively. Then, the lines \\(g_1\\), \\(g_2\\), \\(g_3\\) are parallel and, by Thales, \\(d\\left(g_1;\\;g_2\\right): d\\left(g_2;\\;g_3\\right)=AT: TC\\). Thus, \\(d\\left(g_1;\\;g_2\\right): d\\left(g_2;\\;g_3\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right)\\). Now, denote by \\(g_4\\) the line in the plane ABC which is parallel to the lines \\(g_1\\), \\(g_2\\), \\(g_3\\) and satisfies \\(d\\left(g_1;\\;g_2\\right): d\\left(g_2;\\;g_3\\right): d\\left(g_3;\\;g_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\).\n\nNow, let \\(q_4\\) be the plane passing through the line \\(g_4\\) and the point D. Let \\(q_1\\), \\(q_2\\), \\(q_3\\) be the planes parallel to \\(q_4\\) and passing through the lines \\(g_1\\), \\(g_2\\), \\(g_3\\), respectively (of course, we can construct such planes since the lines \\(g_1\\), \\(g_2\\), \\(g_3\\) are parallel to \\(g_4\\)). Thus, we have found four parallel planes \\(q_1\\), \\(q_2\\), \\(q_3\\), \\(q_4\\) such that \\(A\\in q_1\\), \\(B\\in q_2\\), \\(C\\in q_3\\), \\(D\\in q_4\\), and these planes obviously satisfy \\(d\\left(q_1;\\;q_2\\right): d\\left(q_2;\\;q_3\\right): d\\left(q_3;\\;q_4\\right)=d\\left(g_1;\\;g_2\\right): d\\left(g_2;\\;g_3\\right): d\\left(g_3;\\;g_4\\right)\\). Since \\(d\\left(g_1;\\;g_2\\right): d\\left(g_2;\\;g_3\\right): d\\left(g_3;\\;g_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\), we thus have \\(d\\left(q_1;\\;q_2\\right): d\\left(q_2;\\;q_3\\right): d\\left(q_3;\\;q_4\\right)=d\\left(p_1;\\;p_2\\right): d\\left(p_2;\\;p_3\\right): d\\left(p_3;\\;p_4\\right)\\). Hence, according to the above, the problem is solved.\n\nThe above solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [2]\n\n## Remarks (added by pf02, May 2025)\n\n1. The first \"solution\" is based on a good idea, but it is so incomplete (it contains so mach hand-waiving) that it can not be called a solution. Here are the issues with this \"solution\":\n\na. The first problem (minor) is the statement \"On these three lines, take three vertices \\(A_1, A_2, A_3\\) respectively of an equilateral triangle (it is well-known that this is possible...)\". This is *not* well known, it should be proven.\n\nb. The second problem (a more serious one) is the statement that \"If \\(\\pi\\) makes a small enough angle with the \\(\\pi_i\\)'s, one of the \\(P_i\\)'s we get this time must lie above \\(\\pi_4\\)\". This is probably true, but not obvious at all, and it should have a proof.\n\nc. The third problem (this is so serious that it is a \"show stopper\") is the assumption (not stated explicitly) that the \\(z\\)-coordinates of \\(P_1, P_2\\) as functions of the angle \\(\\theta\\) between \\(\\pi\\) and \\(\\pi_i\\) are continuous functions. This is not even true as stated, it depends on the choice of \\(A_1, A_2, A_3\\).\n\nAs I said, the idea is good, and this could be turned into a solution to the problem. I leave this task to the diligent reader.\n\n2. Below, I will give another solution to the problem. It is similar to solution 2, but different enough to warrant writing it down. Both this solution, and solution 2, show that in fact, the tetrahedron does not have to be regular. The problem could be restated as \"Given four distinct parallel planes and a tetrahedron, prove that there exists a tetrahedron similar to the given one with a vertex on each plane.\"",
"As I remarked, this solution is similar with the previous one. Let \\(P_1, P_2, P_3, P_4\\) be the four planes. We can assume they are horizontal (in some \\(x, y, z\\) coordinate system in which the \\(z\\) axis is vertical) and that their \\(z\\) coordinates are in increasing order. Let \\(d_1\\) be the distance between \\(P_1, P_2\\), \\(d_2\\) be the distance between \\(P_2, P_3\\), and \\(d_3\\) be the distance between \\(P_3, P_4\\).\n\n\\[\nProb 1972 6.png\n\\]\n\nNow consider the tetrahedron \\(ABCD\\). Take \\(C_1 \\in AD, A_1 \\in BC\\) and \\(C_2, A_2 \\in BD\\) such that\n\n\\[\n\\frac{DC_1}{C_1A} = \\frac{d_3}{d_2} \\hspace{52pt} (1)\n\\]\n\n\\[\n\\frac{CA_1}{A_1B} = \\frac{d_2}{d_1} \\hspace{52pt} (2)\n\\]\n\n\\[\n\\frac{DC_2}{C_2B} = \\frac{d_3}{d_2 + d_1} \\hspace{30pt} (3)\n\\]\n\n\\(\\frac{DA_2}{A_2B} = \\frac{d_3 + d_2}{d_1} \\hspace{30pt} (4)\\).\n\nNote that (3) and (4) imply\n\n\\[\n\\frac{DC_2}{C_2A_2} = \\frac{d_3}{d_2} \\hspace{52pt} (5)\n\\]\n\n\\(\\frac{C_2A_2}{A_2B} = \\frac{d_2}{d_1} \\hspace{52pt} (6)\\).\n\n(1) and (5) imply that \\(C_1C_2 \\parallel AA_2\\). Also, (2) and (6) imply that \\(CC_2 \\parallel A_1A_2\\). From these it follows that the planes \\((CC_1C_2)\\) and \\((AA_1A_2)\\) are parallel.\n\nNow take two planes \\(P_B\\) through \\(B\\), and \\(P_D\\) through \\(D\\) parallel to the parallel planes \\((CC_1C_2)\\) and \\((AA_1A_2)\\). If we denote \\(e_1, e_2, e_3\\) the distances between \\(P_B\\) and \\((AA_aA_2)\\), the plane \\((AA_aA_2)\\) and \\((CC_1C_2)\\), and finally, \\((CC_1C_2)\\) and \\(P_D\\), then (5) and (6) imply that\n\n\\(\\frac{d_1}{d_2} = \\frac{e_1}{e_2}\\) and \\(\\frac{d_2}{d_3} = \\frac{e_2}{e_3} \\hspace{50pt} (7)\\).\n\nFrom this it follows that we can find a similar transformation in the 3D space which takes our four planes into the four planes \\(P_1, P_2, P_3, P_4\\). To make this explicit, we first take a rotation followed by a translation which takes \\(P_B\\) into \\(P_1\\). Now we take a scaling (with suitable origin and factor) which keeps \\(P_1\\) in place, and maps the image of \\((AA_1A_2)\\) obtained so far into \\(P_2\\). Because of (7), the image of \\((CC_1C_2)\\) will map into \\(P_3\\) and the image of \\(P_D\\) will map into \\(P_4\\).\n\nThrough these transformations, the image of the tetrahedron \\(ABCD\\) will map into a similar tetrahedron with its vertices on \\(P_1, P_2, P_3, P_4\\).\n\n[Solution by pf02, May 2025]"
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IMO-1973-1
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_1
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Point \(O\) lies on line \(g;\) \(\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}\) are unit vectors such that points \(P_1, P_2, \cdots, P_n\) all lie in a plane containing \(g\) and on one side of \(g.\) Prove that if \(n\) is odd,
\[
\left|\overrightarrow{OP_1}+\overrightarrow{OP_2}+\cdots+ \overrightarrow{OP_n}\right|\ge1.
\]
Here \(\left|\overrightarrow{OM}\right|\) denotes the length of vector \(\overrightarrow{OM}.\)
|
[
"We prove it by induction on the number \\(2n+1\\) of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the \\(2n-1\\) vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm \\(\\ge 1\\) betwen two with norm \\(1\\). The sum of the two vectors of norm \\(1\\) makes an angle of \\(\\le\\frac\\pi 2\\) with the vector of norm \\(\\ge 1\\), so their sum has norm \\(\\ge 1\\), and we're done.\n\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]\n\n## Remarks (added by pf02, May 2025)\n\nThe \"solution\" given above is somewhat incomplete. It simply shoves the difficulty of the problem into the phrase \"... so their sum has norm \\(\\ge 1\\), and we're done.\" Indeed, while true, this is not obvious, and it needs a proof.\n\nBelow, I will fill in the gaps in the first \"solution\". Then, I will give two more solutions.",
"Let \\(n = 2k - 1\\). We give a proof by induction on \\(k\\).\n\nFor \\(k = 0\\), we have one vector \\(\\overrightarrow{OP_1}\\) whose norm (i.e. length) is \\(1\\), so we are done.\n\nAssume the statement is true for \\(0, 1, \\dots, k - 1\\) and prove that it is true for \\(k\\). For ease of visualization, assume the plane to have \\(x, y\\) axes, \\(g\\) to be the line \\(y = 0\\), and \\(O = (0, 0)\\). Let \\(n = 2k + 1\\). Assume the vectors are determined by \\(P_i = (\\cos \\alpha_i, \\sin \\alpha_i)\\), such that \\(0 < \\alpha_1 \\le \\alpha_2 \\le \\dots \\le \\alpha_n < \\pi\\).\n\nLet \\(Q_1 = P_1, Q_3 = P_n\\) and \\(Q_2\\) be determined by \\(\\overrightarrow{OQ_2} = \\overrightarrow{OP_2} + \\dots + \\overrightarrow{OP_{n-1}}\\) (like in the solution above). Since \\(\\overrightarrow{OQ_2}\\) is the sum of \\(n - 2 = 2k - 1\\) unit vectors on one side of \\(y = 0\\), its norm is \\(\\ge 1\\) (because of the induction hypothesis). Let us say that \\(Q_2 = (a \\cos \\beta, a \\sin \\beta)\\). We want to show that the norm \\(\\left| \\overrightarrow{OQ_1} + \\overrightarrow{OQ_2} + \\overrightarrow{OQ_3} \\right| \\ge 1\\).\n\nSince \\(\\overrightarrow{OQ_1}, \\overrightarrow{OQ_3}\\) are unit vectors whose angles with the positive direction of the \\(x\\)-axis are \\(0 < \\alpha_1 \\le \\alpha_n < \\pi\\) the direction of \\(\\overrightarrow{OQ_1} + \\overrightarrow{OQ_3}\\) is \\(\\frac{\\alpha_1 + \\alpha_n}{2}\\). Since \\(\\alpha_1 \\le \\beta \\le \\alpha_n\\), the angle between \\(\\overrightarrow{OQ_1} + \\overrightarrow{OQ_3}\\) and \\(\\overrightarrow{OQ_2}\\) is \\(\\le \\frac{\\pi}{2}\\).\n\nNow use that the norm of \\(\\overrightarrow{OQ_2}\\) is \\(a \\ge 1\\), and the general fact that if the angle between two vectors \\(\\overrightarrow{u}, \\overrightarrow{v}\\) is \\(\\le \\frac{\\pi}{2}\\), then \\(\\left| \\overrightarrow{u} + \\overrightarrow{v} \\right| \\ge \\max \\{ \\left| \\overrightarrow{u} \\right|, \\left| \\overrightarrow{v} \\right| \\}\\).\n\nIt follows that \\(\\left| \\overrightarrow{OQ_1} + \\overrightarrow{OQ_2} + \\overrightarrow{OQ_3} \\right| = \\left| (\\overrightarrow{OQ_1} + \\overrightarrow{OQ_3}) + \\overrightarrow{OQ_2} \\right| \\ge \\max \\left\\{ \\left| \\overrightarrow{OQ_1} + \\overrightarrow{OQ_3} \\right|, a \\right\\} \\ge a \\ge 1\\).\n\nTo finish the proof, let us show this general fact about \\(\\overrightarrow{u}, \\overrightarrow{v}\\) forming an angle \\(\\le \\frac{\\pi}{2}\\). Let \\(\\overrightarrow{u} = \\overrightarrow{OA}, \\overrightarrow{v} = \\overrightarrow{OB}\\) and \\(\\overrightarrow{u} + \\overrightarrow{v} = \\overrightarrow{OC}\\).\n\nLet \\(a, b, c\\) be the lengths of the segments \\(OA, OB, OC\\). Consider the triangle \\(\\triangle OAC\\). The cosine formula in \\(\\triangle OAC\\) yields \\(c^2 = a^2 + b^2 - 2ab\\cos C\\). Since \\(\\angle C\\) is obtuse, we have \\(\\cos C \\le 0\\), so \\(c^2 \\ge a^2 + b^2 \\ge \\max \\{a^2, b^2\\}\\). This implies \\(c \\ge \\max \\{ a, b \\}\\).",
"This solution starts exactly like the previous one, till the line\n\n\\(Q_2 = (a \\cos \\beta, a \\sin \\beta)\\). We want to show that the norm \\(\\left| \\overrightarrow{OQ_1} + \\overrightarrow{OQ_2} + \\overrightarrow{OQ_3} \\right| \\ge 1\\).\n\nThen, we continue as follows:\n\nThe end point of \\(\\overrightarrow{OQ_1} + \\overrightarrow{OQ_2} + \\overrightarrow{OQ_3} = (\\cos \\alpha_1 + a \\cos \\beta + \\cos \\alpha_n, \\sin \\alpha_1 + a \\sin \\beta + \\sin \\alpha_n)\\), and the square of the norm of this vector is \\(S(\\alpha_1, \\alpha_n, \\beta, a) = (\\cos \\alpha_1 + a \\cos \\beta + \\cos \\alpha_n)^2 + (\\sin \\alpha_1 + a \\sin \\beta + \\sin \\alpha_n)^2\\).\n\nAfter some computations, we have \\(S(\\alpha_1, \\alpha_n, \\beta, a) = 2 + a^2 + 2a\\cos (\\alpha_n - \\beta) + 2a\\cos (\\beta - \\alpha_1) + 2\\cos (\\alpha_n - \\alpha_1)\\).\n\nSince \\(\\cos\\) is a decreasing function from \\(1\\) to \\(-1\\) on the interval \\([0, \\pi]\\), it follows that \\(S(\\alpha_1, \\alpha_n, \\beta, a) \\ge S(0, \\pi, \\beta, a)\\).\n\nThus \\(S(\\alpha_1, \\alpha_n, \\beta, a) \\ge 2 + a^2 + 2a\\cos (\\pi - \\beta) + 2a\\cos (\\beta) + 2\\cos (\\pi) = a^2\\).\n\nIt follows that the norm \\(\\left| \\overrightarrow{OQ_1} + \\overrightarrow{OQ_2} + \\overrightarrow{OQ_3} \\right| \\ge a \\ge 1\\).\n\n[Solution by pf02, May 2025]",
"For ease of visualization, assume the plane to have \\(x, y\\) axes, \\(g\\) to be the line \\(y = 0\\), and \\(O = (0, 0)\\). Assume the vectors are determined by \\(P_i = (\\cos \\alpha_i, \\sin \\alpha_i)\\), such that \\(0 < \\alpha_1 \\le \\alpha_2 \\le \\dots \\le \\alpha_n < \\pi\\).\n\nWe are asked to show that \\(\\left| \\sum_{i=1}^n \\overrightarrow{OP_i} \\right| \\ge 1\\). Before embarking on the computations which prove this inequality, let us give the idea of the proof. Let \\(n = 2k + 1\\). We will prove formally that by \"moving\" \\(P_1, P_n\\) to their limiting positions on \\(y = 0\\) (i.e. replacing \\(P_1\\) by \\((1, 0)\\) and \\(P_n\\) by \\((-1, 0)\\)) the norm of the sum of vectors will be decreased. Repeat this argument for \\(P_2, P_{n-1}\\), etc. (for a total of \\(k\\) times). We are left with one vector (the middle vector), which has norm \\(1\\). Now let us turn this idea into a formal proof.\n\nThe point determining the sum of vectors is \\(P = (\\sum_{i=1}^n \\cos \\alpha_i, \\sum_{i=1}^n \\sin \\alpha_i)\\).\n\nThe square of the norm of the sum of vectors is \\(S(\\alpha_1, \\dots, \\alpha_n) = \\left| \\overrightarrow{OP} \\right|^2 = \\left( \\sum_{i=1}^n \\cos \\alpha_i \\right)^2 + \\left( \\sum_{i=1}^n \\sin \\alpha_i \\right)^2\\).\n\nAfter some computations, we get \\(S(\\alpha_1, \\dots, \\alpha_n) = n + 2 \\sum_{1 \\le i < j \\le n} \\cos (\\alpha_j - \\alpha_i)\\).\n\nDenote \\(T(\\alpha_1, \\dots, \\alpha_n) = 2 \\sum_{1 \\le i < j \\le n} \\cos (\\alpha_j - \\alpha_i)\\). When \\(n = 1\\), we define \\(T(\\alpha_1) = 0\\). (This makes sense both formally, and as an accommodation to \\(S(\\alpha_1) = 1\\).)\n\nWe have \\(T(\\alpha_1, \\dots, \\alpha_n) = 2 \\cos (\\alpha_n - \\alpha_1) + 2 \\sum_{1 < i < n} \\cos (\\alpha_n - \\alpha_i) + 2 \\sum_{1 < j < n} \\cos (\\alpha_j - \\alpha_1) + 2 \\sum_{2 \\le i < j \\le n-1} \\cos (\\alpha_j - \\alpha_i)\\).\n\nSince \\(\\cos\\) is a decreasing function from \\(1\\) to \\(-1\\) on the interval \\([0, \\pi]\\), it follows that \\(T(\\alpha_1, \\alpha_2, \\dots, \\alpha_{n-1}, \\alpha_n) \\ge T(0, \\alpha_2, \\dots, \\alpha_{n-1}, \\pi)\\).\n\nBut \\(T(0, \\alpha_2, \\dots, \\alpha_{n-1}, \\pi) = 2 \\cos (\\pi - 0) + 2 \\sum_{1 < i < n} \\cos (\\pi - \\alpha_i) + 2 \\sum_{1 < j < n} \\cos (\\alpha_j - 0) + 2 \\sum_{2 \\le i < j \\le n-1} \\cos (\\alpha_j - \\alpha_i) =\\)\n\n\\[\n= T(\\alpha_2, \\dots, \\alpha_{n-1}) - 2\n\\]\n\nIn short, we showed that \\(T(\\alpha_1, \\dots, \\alpha_n) \\ge T(\\alpha_2, \\dots, \\alpha_{n-1}) - 2\\).\n\nNow, if \\(n = 2k + 1\\), and we apply this argument \\(k\\) times, we get \\(T(\\alpha_1, \\dots, \\alpha_n) \\ge -2k\\).\n\nThus, \\(S(\\alpha_1, \\dots, \\alpha_n) = n + T(\\alpha_1, \\dots, \\alpha_n) \\ge 2k + 1 - 2k = 1\\).\n\n[Solution by pf02, May 2025]",
"Similar to solution \\(3\\), we assume the plane to have the \\(x,y\\) axis, \\(g\\) to be the line \\(y=0\\), \\(O = (0,0)\\), and the vectors \\(\\overrightarrow{OP_k}\\) to be pointed upwards. We will minimize \\(\\Big|\\overrightarrow{OP_1} + \\cdots + \\overrightarrow{OP_n}\\Big|\\). Let\n\n\\[\n\\overrightarrow{O A_1} = \\overrightarrow{OP_2} + \\cdots + \\overrightarrow{OP_n}.\n\\]\n\nBy the law of cosines\n\n\\[\n\\Big|\\overrightarrow{OP_1} + \\overrightarrow{O A_1} \\Big| = 1^2 + \\Big|\\overrightarrow{OA_1} \\Big|^2 - 2 \\Big|\\overrightarrow{OA_1} \\Big| \\cos(\\pi - \\phi_1)\n\\]\n\nwhere \\(\\phi_1\\) is the (positive) angle between \\(\\overrightarrow{OA_1}\\) and \\(\\overrightarrow{OP_1}\\). Apparently \\(A_1\\) lies above the \\(y\\) axis, so that \\(\\phi_1\\) is maximized when \\(\\overrightarrow{OP_1} = (1,0)\\) or \\((-1,0)\\). By the above formula, since \\(\\cos(\\pi-t)\\) is monotonically increasing on \\([0,\\pi]\\), it follows \\(\\Big|\\overrightarrow{OP_1} + \\overrightarrow{O A_1} \\Big|\\) is minimized when \\(\\overrightarrow{OP_1} = (1,0)\\) or \\((-1,0)\\). Since we are trying to minimize\n\n\\[\n\\Big|\\overrightarrow{OP_1} + \\cdots + \\overrightarrow{OP_n}\\Big|\n\\]\n\nwe may now assume WLOG \\(\\overrightarrow{OP_1} = (1,0)\\) or \\((-1,0)\\). Successively repeat this argument for the other \\(n-1\\) vectors (for instance taking \\(\\overrightarrow{OP_1}+\\overrightarrow{OP_3}+\\cdots+\\overrightarrow{OP_n} = \\overrightarrow{OA_2}\\) to show WLOG \\(\\overrightarrow{OP_2} = (1,0)\\) or \\((-1,0)\\)) to get WLOG \\(\\overrightarrow{OP_k} = (1,0)\\) or \\((-1,0)\\) for all \\(k\\). At which point\n\n\\[\n\\Big|\\overrightarrow{OP_1} + \\cdots +\\overrightarrow{OP_n}\\Big| = |m_1-m_2|\n\\]\n\nwhere \\(m_1\\) is the number of vectors equal to \\((1,0)\\) and \\(m_2\\) is the number of vectors equal to \\((-1,0)\\). Also \\(m_1+m_2 = n\\) is odd. So that \\(|m_1-m_2|\\) clearly has minimum value \\(1\\).\n\n~not_detriti"
] |
IMO-1973-2
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_2
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Determine whether or not there exists a finite set \(M\) of points in space not lying in the same plane such that, for any two points \(A\) and \(B\) of \(M\); one can select two other points \(C\) and \(D\) of \(M\) so that lines \(AB\) and \(CD\) are parallel and not coincident.
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[
"In order to solve this problem we can start by finding at least one finite set \\(M\\) that satisfies the condition.\n\nWe start by defining our first set \\(M_{8}\\) with the vertices of a cube of side \\(k\\) as follows: \\(M_{8} = \\{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \\}\\)\n\nSince all the faces of this cube have a parallel face, then any two points on one face will have corresponding 2 points on the opposite face that is parallel. However we have four diagonals on this cube that do not have two points that are parallel to any of these diagonals.\n\nBy doing a reflection of the points on the \\(z=k\\) plane along the \\(xy\\)-plane these four diagonals will have their respective parallel diagonals on the \\(z \\le 0\\) space.\n\nBut now we have four more diagonals on the set of two cubes that do not have a parallel line. That is, diagonal \\((0,0,-k) \\rightarrow (k,k,k)\\) does not have a parallel line and neither do the other three.\n\nBy doing a reflection of the points on the \\(x=k\\) plane along the \\(yz\\)-plane these new four diagonals will have their respective parallel diagonals on the \\(x \\le 0\\) space.\n\nBut now we have four more diagonals on the set of 4 cubes that do not have a parallel line. That is, diagonal \\((-k,0,-k) \\rightarrow (k,k,k)\\) does not have a parallel line and neither do the other three.\n\nBy doing a reflection of the points on the \\(y=k\\) plane along the \\(xz\\)-plane these new four diagonals will have their respective parallel diagonals on the \\(y \\le 0\\) space.\n\nThe new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes.\n\nSo, we found a set at least one finite set \\(M\\) that we can define as \\(M=\\{(x,y,z)\\}\\) where \\(x,y,z \\in \\{-k,0,k\\}\\) giving a total of 27 points. Therefore such a set exists.\n\nAnother way to define this set of points is let \\(M\\) be:\n\nLet \\(V\\) be a solid cube or right angled parallelepiped\n\nLet \\(M_{v}\\) be the set of all 8 vertices of \\(V\\)\n\nLet \\(M_{me}\\) be the set of all 12 midpoints of the edges of \\(V\\)\n\nLet \\(M_{mf}\\) be the set of all 6 midpoints of the faces of \\(V\\)\n\nLet \\(M_{c}\\) be the center of \\(V\\)\n\n\\(M\\)=\\(M_{v} \\cup M_{me} \\cup M_{mf} \\cup M_{c}\\)\n\nIt is possible that one can construct many other sets of \\(M\\) using regular tetrahedrons with some reflections and with less points than 27, or by translating or rotating or skewing all the the points simultaneously of the finite set \\(M\\) that we defined here. But that is not necessary for this problem because it asks to prove whether there exist a set with the described conditions. By showing that at least one set exists with those conditions, the problem is proved.\n\n\\[\nIMO 1973 P2 01.png\n\\]\n\n~Tomas Diaz. [email protected]\n\n## Remarks (added by pf02, June 2025)\n\n1. Very closely related to the solution above, it is easy to verify that if we delete the center (the point \\((0, 0, 0)\\)) from the set \\(M\\), we obtain a set with \\(26\\) points which still satisfies the requirements of the problem. In the second way of defining this set, we don't need \\(M_c\\). In other words, we just take \\(M = M_v \\cup M_{me} \\cup M_{mf}\\).\n\n2. The last paragraph of the solution above ought to be deleted. The solution would gain from deleting it. Stating that maybe one can find a set by \"using regular tetrahedrons with some reflections ... or by translating or rotating or skewing all the the points simultaneously of the finite set \\(M\\) that we defined\" is vague and useless.\n\n3. Below I will give another solution, inspired by rhombic dodecahedrons. Note that there are several non-equivalent rhombic dodecahedrons. See for example https://en.wikipedia.org/wiki/Rhombic_dodecahedron and https://en.wikipedia.org/wiki/Bilinski_dodecahedron (By non-equivalent, I mean that one can not be obtained from the other by an affine transformation.) Not all of them suggest solutions to this problem.",
"Let \\(x, y, z\\) be the coordinate in space, and let us consider the set of \\(14\\) points \\(M = \\{(\\pm 1, \\pm 1, \\pm 1), (\\pm 2, 0, 0), (0, \\pm 2, 0), (0, 0, \\pm 2)\\}\\).\n\nWe will prove that this set satisfies the conditions of the problem. One possible proof (the one I will give below) is purely algebraic. For any two points \\(A, B\\), consider the vectors \\(v_A = \\overrightarrow{OA}, v_B = \\overrightarrow{OB}\\). The segment \\(AB\\) is parallel to the vector \\(v_B - v_A\\). We have to show that we can find two distinct points \\(C, D\\), non-colinear with \\(A, B\\) such that \\(v_B - v_A = \\alpha (v_D - v_C)\\) for some \\(\\alpha\\), where \\(v_C, v_D\\) are the vectors given by \\(C, D\\).\n\nBefore proceeding to do this, let us get a geometric feel for what our set \\(M\\) looks like.\n\n\\[\nProb 1973 2.png\n\\]\n\nThe \\(x, y\\) plane is the plane of the screen of your computer, and the \\(z\\) axis is perpendicular to it, with the positive direction towards you. The red points are the \\(6\\) points \\(\\{(\\pm 2, 0, 0), (0, \\pm 2, 0), (0, 0, \\pm 2)\\}\\) and the blue points are the \\(8\\) points \\(\\{(\\pm 1, \\pm 1, \\pm 1)\\}\\). All the faces are congruent rhombi.\n\nIt is clear that each edge is parallel to several other edges. It is intuitively clear that each diagonal of a face is parallel to a diagonal of another face. And it is intuitively clear that each interior diagonal is parallel to another interior diagonal, or the diagonal of a face, or an edge. But this is not a proof, so we will proceed to give a formal proof.\n\nWe will simply compute all the difference vectors, and show that each difference vector is a scalar multiple of at least one other difference vector. We will do this in a table. We will label all difference vectors with (a), (b), (c), ..., and use the same label when vectors are scalar multiples of each other.\n\nBelow is the table. On the top row (the column headers) and in the left column (the row headers) we list all the points in \\(M\\). In the cells of the table we write the vector \\(\\overrightarrow{OB} - \\overrightarrow{OA}\\), where \\(B\\) is in the top row, and \\(A\\) is in the left-most column. We don't compute the vector for \\(A = B\\) (which would be \\((0, 0, 0)\\)), or, if the index of \\(B\\) is \\(\\le\\) the index of \\(A\\) in the list of points, we don't compute the vector \\(\\overrightarrow{OB} - \\overrightarrow{OA}\\).\n\nNow the proof is done. Indeed, all we have to do is examine the table, and ascertain that each label is there at least twice, and it is attached to proportional vectors. The reader can easily do this task.\n\nFor the sake of making things explicit, and in order to understand the geometry of this rhombic dodecahedron a little better, and to assist the diligent reader who would want to verify that each label appears twice and is attached to proportional vectors, I will give a table showing information about the labels.\n\n[Solution by pf02, June 2025]"
] |
IMO-1973-3
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_3
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Let \(a\) and \(b\) be real numbers for which the equation \(x^4 + ax^3 + bx^2 + ax + 1 = 0\) has at least one real solution. For all such pairs \((a, b)\), find the minimum value of \(a^2 + b^2\).
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[
"Substitute \\(z=x+1/x\\) to change the original equation into \\(z^2+az+b-2=0\\). This equation has solutions \\(z=\\frac{-a \\pm \\sqrt{a^2+8-4b}}{2}\\). We also know that\n\n\\[\n|z|=|x+1/x| \\geq 2.\\ \\ \\ \\ \\ \\ \\ \\ (1)\n\\]\n\nSo,\n\n\\[\n\\left | \\frac{-a \\pm \\sqrt{a^2+8-4b}}{2} \\right | \\geq 2\\ \\ \\ \\ \\ \\ \\ \\ (2)\n\\]\n\n\\[\n\\frac{|a|+\\sqrt{a^2+8-4b}}{2} \\geq 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (3)\n\\]\n\n\\[\n|a|+\\sqrt{a^2+8-4b} \\geq 4\n\\]\n\nRearranging and squaring both sides,\n\n\\[\na^2+8-4b \\geq a^2-16|a|+16\\ \\ \\ \\ \\ \\ \\ \\ (4)\n\\]\n\n\\[\n2|a|-b \\geq 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (5)\n\\]\n\nSo,\n\n\\[\na^2+b^2 \\geq a^2+(2-2|a|)^2\\ \\ \\ \\ \\ \\ \\ \\ (6)\n\\]\n\n\\[\n= 5a^2-8|a|+4 = 5 \\left( |a|-\\frac{4}{5} \\right )^2+\\frac{4}{5}\n\\]\n\n.\n\nTherefore, the smallest possible value of \\(a^2+b^2\\) is \\(\\frac{4}{5}\\), when \\(a=\\pm \\frac{4}{5}\\) and \\(b=\\frac{-2}{5}\\).\n\nBorrowed from [1]\n\n## Remarks (added by pf02, June 2025)\n\n1. The solution above is incomplete and it has some serious errors. The result is correct, but the method for obtaining it is flawed. For the sake of reference to the steps in the solution I added labels, but I did not make any changes to the solution. I will highlight the errors and the missing steps:\n\nA. (1) is true when we know that \\(x\\) is real. By hypothesis the equation has at least one real solution, so (1) is true for this solution. A necessary condition for this is that \\(a^2 + 8 - 4b \\ge 0\\). We have to use this in the proof, or if we obtain a result without having used it, we need to verify that the result satisfies this condition. This is missing from the solution.\n\nB. (2) is equivalent to (3) when \\(\\pm\\) is interpreted as an \\(\\mathbf{\"or\"}\\). (If the \\(\\pm\\) would be an \\(\\mathbf{\"and\"}\\) then (2) and (3) would not be equivalent.) In our problem the \\(\\pm\\) is an \\(\\mathbf{\"or\"}\\) so the proof can proceed. However, while easy, going from (2) to (3) is not obvious, and it is an important step in the proof, so it should be explained.\n\nC. There is a small computational error in (4): we should have \\(8|a|\\), not \\(16|a|\\). This error is in writing, the computation proceeds as if it did not happen.\n\nD. Going from (5) to (6) is a very serious error. Replacing \\(b^2\\) by \\((2 - 2|a|)^2\\) can be done only when certain conditions are satisfied (for example if we know that \\(b \\le 0\\) and \\(2 - 2|a| \\ge 0\\) or if we know that \\(b = 2 - 2|a|\\)). So writing down (6) is completely unjustified.\n\n2. I will not take the task of fixing this solution. Instead, I will give a different solution below.",
"The plan of this solution is the following: we will consider the plane of coordinates \\(a, b\\). We will write the condition(s) for the equation to have at least one real solution. These will be inequalities in \\(a, b\\), which delimit regions in the \\(a, b\\) plane. \\(a^2 + b^2\\) represents the square of the distance from the point of coordinates \\((a, b)\\) to the origin. We will find the point(s) in the region we delimited which are closest to the origin. These are minimizing \\(a^2 + b^2\\).\n\nDividing the given equation by \\(x^2\\) and making the substitution \\(z = x + \\frac{1}{x}\\), we can re-write the given equation as \\(z^2 + az + b - 2 = 0\\). The given equation has at least one real root when this equation in \\(z\\) has real roots. This is true when \\(a^2 - 4b + 8 \\ge 0\\). This represents the parabola \\(b = \\frac{a^2}{4} + 2\\) in the \\(a, b\\) plane. Refer to this parabola as \\(P\\).\n\nThe two solutions of the equation in \\(z\\) are \\(\\frac{-a \\pm \\sqrt{a^2 -4b + 8}}{2}\\). We will take them one at a time, and write down the conditions for its solution \\(x\\) to be real. Start with the \\(+\\) sign, and consider the equation \\(2 x^2 - (-a + \\sqrt{a^2 -4b + 8}) x + 2 = 0\\)\n\nThis equation has real solutions when \\((a - \\sqrt{a^2 - 4b + 8})^2 - 16 \\ge 0\\). Work out the square, rearrange terms and simplify. We get \\(a \\sqrt{a^2 - 4b + 8} \\le a^2 - 2b - 4\\).\n\nNote that at this point we could continue working with inequalities, being vary careful in considering all the cases as we do computations. Instead, we will consider the equation \\(a \\sqrt{a^2 - 4b + 8} = a^2 - 2b - 4\\). This is a curve in the \\(a, b\\) plane, delimiting several regions in the plane. We will study the curve, and decide which are the regions in the \\(a, b\\) plane in which the inequality is true. This will give us a set of points \\((a, b)\\) for which the equation has at least one real root.\n\nSquare both sides and rearrange terms. We get \\(4a^2 = (b + 2)^2\\). As a curve, this is two lines \\(L_1, L_2\\) with equations \\(b = 2a - 2\\) and \\(b = -2a - 2\\). Before proceeding, we have to verify the validity of these solutions since we may have introduced \"fake\" solutions when we squared both sides.\n\nPlug \\(b = 2a - 2\\) into \\(a \\sqrt{a^2 - 4b + 8} = a^2 - 2b - 4\\). We get \\(a \\sqrt{(a - 4)^2} = a (a - 4)\\). This is true when \\(a = 0\\) or \\(a \\ge 4\\). Thus, \\(b = 2a - 2\\) is valid when \\(a \\ge 4, b \\ge 6\\).\n\nSimilarly, plug \\(b = -2a - 2\\) into \\(a \\sqrt{a^2 - 4b + 8} = a^2 - 2b - 4\\), and get that \\(b = -2a - 2\\) is valid when \\(a \\ge -4, b \\le 6\\).\n\nThe parabola \\(P\\), the lines \\(L_1, L_2\\) and the lines \\(a = 4, a = -4, b = 6\\) carve the \\(a, b\\) plane into several regions. We now have to verify which are the regions in which \\((a - \\sqrt{a^2 - 4b + 8})^2 - 16 \\ge 0\\). While mildly tedious, this is very easy: we just have to take a numerical sample in each region and verify the inequality. The results are shown in the image below (the one on the left): if the point \\((a, b)\\) is in the shaded area, then the inequality is satisfied, so equation in the problem has at least one real solution.\n\n\\[\nProb 1973 3.png\n\\]\n\nNow take the \\(-\\) sign from \\(z = \\frac{-a \\pm \\sqrt{a^2 -4b + 8}}{2}\\). We get the equation \\(2 x^2 + (a + \\sqrt{a^2 -4b + 8}) x + 2 = 0\\). This has real solutions when \\((a + \\sqrt{a^2 - 4b + 8})^2 - 16 \\ge 0\\). Consider the regions delimited by \\((a + \\sqrt{a^2 - 4b + 8})^2 - 16 = 0\\), and find the regions when the inequality is true. The computations are very similar to the ones in the preceding case. The regions for \\((a, b)\\) which guarantee that the equation from the problem has at least one real solution are shown in the image above (the one on the right hand side).\n\nIt is now clear that the points \\((a, b)\\) in these regions closest to \\((0, 0)\\) are the ones at the intersection of the line \\(b = 2a - 2\\) with the perpendicular from \\((0, 0)\\), and at the intersection of the line \\(b = -2a - 2\\) with the perpendicular from \\((0, 0)\\). A very simple computation shows that these points are \\((\\frac{4}{5}, -\\frac{2}{5}), (-\\frac{4}{5}, -\\frac{2}{5})\\). Then the square of the distance from \\((0, 0)\\) to these points \\(= a^2 + b^2 = \\frac{4}{5}\\).\n\n[Solution by pf02, June 2025]"
] |
IMO-1973-4
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_4
|
A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path shouid he follow in order to travel the least possible distance and still accomplish his mission?
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[
"Let our triangle be \\(\\triangle ABC\\), let the midpoint of \\(AB\\) be \\(D\\), and let the midpoint of \\(CD\\) be \\(E\\). Let the height of the triangle be \\(h\\). Draw circles around points \\(B\\) and \\(C\\) with radius \\(\\frac{h}{2}\\), and label them \\(c_1\\) and \\(c_2\\). Let the intersection of \\(BE\\) and \\(c_1\\) be \\(F\\).\n\nThe path that is the solution to this problem must go from \\(A\\) to a point on \\(c_2\\) to a point on \\(c_1\\). Let us first find the shortest possible path. We will then prove that this path fits the requirements.\n\nThe shortest path is in fact the path from \\(A\\) to \\(E\\) to \\(F\\). We will prove this as follows:\n\nSuppose that a different point on \\(c_2\\) is the optimal point to go to. Let this point be \\(P\\). Then, the optimal point on \\(c_1\\) would be the intersection of \\(BP\\) and \\(c_1\\) (let this point be \\(R\\)). Let the line through \\(E\\) parallel to \\(AB\\) be \\(l\\), and the intersection of \\(AP\\) and \\(l\\) be \\(Q\\). Then, we have\n\n\\[\nAE + BE < AQ + BQ \\leq AQ + QP + BP = AP + BP.\\textbf{ (1)}\n\\]\n\nThe second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:\n\nSuppose we have a point on \\(l\\), \\(X\\). Reflect \\(B\\) across \\(l\\) to get \\(B'\\). Then, \\(AX + BX = AX + B'X\\), which is obviously minimized at the intersection of the two lines, \\(E\\).\n\nBy \\(\\textbf{(1)}\\), we have\n\n\\[\nAE + EF = AE + BE - \\frac{h}{2} < AP + BP - \\frac{h}{2} = AP + PR,\n\\]\n\nand we have proved the claim.\n\nThis path easily covers the whole triangle. This is because if you draw a line perpendicular to \\(AB\\) from any point in the triangle, this line will hit the path in a distance less than or equal to \\(\\frac{h}{2}\\). We can compute the length of the path to be\n\n\\[\n\\left(\\sqrt{\\frac{7}{3}} - \\frac{1}{2}\\right)h\n\\]\n\n\\[\n= \\left(\\frac{\\sqrt{7}}{2} - \\frac{\\sqrt{3}}{4}\\right)\\cdot\\textrm{the side length of the triangle. }\\square\n\\]\n\n~mathboy100\n\n## Remarks (added by pf02, June 2025)\n\nThe solution given above is so incomplete (and incorrect in a few details) that it can not be called a solution.\n\nBelow I will first point out the shortcomings of the solution. Then I will give a complete solution.\n\n1. The author states: \"The path that is the solution to this problem must go from \\(A\\) to a point on \\(\\mathcal{C}_2\\) to a point on \\(\\mathcal{C}_1\\).\" This needs a proof. I believe this to be true (up to a symmetry), but it is the hardest part of the solution.\n\n2. In the solution above, the author makes the following construction: take \\(P\\) on the arc of \\(\\mathcal{C}_2\\) inside the triangle, and take the point \\(R\\) where \\(PB\\) intersects \\(\\mathcal{C}_1\\). He shows that the length of the curve \\(APR\\) is minimal when \\(P = E\\), where \\(E\\) is the midpoint of the height from \\(C\\).\n\nThen, the author argues that the union of circles of radius \\(\\frac{h}{2}\\) centered at all the points along \\(AEF\\) covers the triangle. (In fact, the argument is incorrect, but the fact is true.)\n\nThis is all good, but the two statements above do not prove that \\(AEF\\) is the shortest curve such that the union of circles of radius \\(\\frac{h}{2}\\) centered at all the points along the curve covers the triangle. In fact, they don't even prove the much easier problem we obtain if we accept that \"the path that is the solution to this problem must go from \\(A\\) to a point on \\(\\mathcal{C}_2\\) to a point on \\(\\mathcal{C}_1\\).\"\n\n3. The author says\n\n\"This path [\\(AEF\\)] easily covers the whole triangle. This is because if you draw a line perpendicular to \\(AB\\) from any point in the triangle, this line will hit the path in a distance less than or equal to \\(\\frac{h}{2}\\).\"\n\nIn fact it is not true that \"if you draw a line perpendicular to \\(AB\\) from any point in the triangle, this line will hit the path in a distance less than or equal to \\(\\frac{h}{2}\\).\" Clearly, there are points \\(T\\) such that a perpendicular from \\(T\\) to \\(AB\\) will not hit the path \\(AEF\\) at all. Besides, this is not what we need in order to conclude that the union of circles of radius \\(\\frac{h}{2}\\) centered at all the points along \\(AEF\\) covers the triangle. And it is hard to believe that something that involves only \\(AB\\) would be sufficient to conclude what we need.",
"Let us say that a curve has the property \\(\\mathcal{P}\\) if one end of it is at \\(A\\), it is inside the triangle, and the union of circles of radius \\(\\frac{h}{2}\\) centered at all the points along the curve covers the triangle. Let \\(\\mathcal{S}\\) be the set of curves having property \\(\\mathcal{P}\\). The problem asks us to find the shortest curve in \\(\\mathcal{S}\\).\n\nIn this solution when we say that a point is in a region, we mean that the point is in the interior, or on the boundary of the region. Also, note that we will use the words curve and path interchangeably.\n\nLet \\(\\mathcal{C}_1\\) and \\(\\mathcal{C}_2\\) be the circles of radius \\(\\frac{h}{2}\\) centered at \\(B, C\\). Let us say that a curve has the property \\(\\mathcal{P}'\\) if one end of it is at \\(A\\), it is inside the triangle, it has a point in the sector of \\(\\mathcal{C}_2\\) inside the triangle, and it has a point in the sector of \\(\\mathcal{C}_1\\) inside the triangle. Let \\(\\mathcal{S}'\\) be the set of curves having property \\(\\mathcal{P}'\\).\n\nWe have \\(\\mathcal{S} \\subset \\mathcal{S}'\\), in other words, if a path has property \\(\\mathcal{P}\\), then it has property \\(\\mathcal{P}'\\). Indeed, if the union of circles of radius \\(\\frac{h}{2}\\) centered at all the points along the path covers the triangle, then this set must cover \\(B\\) and \\(C\\), so there must be points on the path at distance \\(\\le \\frac{h}{2}\\) from \\(B\\) and \\(C.\\) These points will be inside the sectors delimited by the triangle and \\(\\mathcal{C}_1, \\mathcal{C}_2\\).\n\nThe plan of this solution is the following: We will find the shortest path in \\(\\mathcal{S}'\\). (There are two of them, symmetric to each other.) If we show that this path is in \\(\\mathcal{S}\\), then it follows that it is the shortest path in \\(\\mathcal{S}\\). We will verify that the path we found does indeed have property \\(\\mathcal{P}\\).\n\nLet us proceed by taking a path \\(p \\in \\mathcal{S}'\\). Let one end (the \"beginning\") be at \\(A\\), and denote \\(T\\) the other end of the path. Let \\(P\\) and \\(R\\) be points on the curve inside the sectors of \\(\\mathcal{C}_2, \\mathcal{C}_1\\). We can assume that \\(P \\in p\\) comes \"before\" the intersection of \\(p\\) with \\(\\mathcal{C}_1\\).\n\n(Note: if it is the other way, we can just change the notation, or adjust the proof which follows below. This is where we build one of the two solutions, the second solution being a symmetrical transformation of the first.)\n\nLet \\(R \\in p \\cap \\mathcal{C}_1\\) be the \"first\" point of \\(p\\) where it meets \\(\\mathcal{C}_1\\). Define the path \\(p_1\\) to be identical with \\(p\\) up to the point \\(R\\), but then we cut off the rest of the path \\(p\\) (from \\(R\\) to the end \\(T\\)). In other words, the new end point \\(T\\) of \\(p_1\\) is \\(T = R\\).\n\nNow let \\(P \\in p_1 \\cap \\mathcal{C}_2\\) be the \"first\" point of \\(p_1\\) where it meets \\(\\mathcal{C}_2\\). The points in \\(p_1\\) are now the starting point \\(A\\), followed by a set of points inside the triangle, outside of \\(\\mathcal{C}_2, \\mathcal{C}_1\\), then \\(P\\) on \\(\\mathcal{C}_2\\), then a set of points which could be inside or outside of \\(\\mathcal{C}_2\\), and finally \\(R = T\\) on the circle \\(\\mathcal{C}_1\\).\n\nLet \\(p_2\\) be the curve in which we replace the portion of \\(p_1\\) from \\(A\\) to \\(P\\) by a segment on a straight line. Clearly \\(p_2\\) is shorter than \\(p_1\\).\n\nLet \\(p_3\\) be the curve in which we replace the portion of \\(p_2\\) from \\(P\\) to \\(R\\) by a segment on a straight line. Clearly \\(p_3\\) is shorter than \\(p_2\\). Note that \\(p_3\\) consists of the two segments \\(AP\\), \\(PR\\).\n\nLet \\(p_4\\) be the curve identical to \\(p_3\\) from \\(A\\) to \\(P\\), but which replaces the segment \\(PR\\) by a segment \\(PR'\\) along the normal from \\(P\\) to \\(\\mathcal{C}_1\\), where \\(R' \\in \\mathcal{C}_1\\) is the intersection of the normal with the circle. (Note that the normal from \\(P\\) to \\(\\mathcal{C}_1\\) is \\(PB\\).) Clearly \\(p_4\\) is shorter than \\(p_3\\).\n\nFor the sake of ease of notation let us re-label \\(R'\\) to be \\(R\\). Now we are looking at curves which consist of the segment \\(AP\\) (\\(P\\) being a point on \\(\\mathcal{C}_2\\)) followed by the segment \\(PR\\), where \\(R \\in \\mathcal{C}_1\\) is the foot of the normal from \\(P\\) to \\(\\mathcal{C}_1\\). See the figure below.\n\n\\[\nProb 1973 4.png\n\\]\n\nNow, we can proceed as in the previous \"solution\". We will show that \\(APR\\) is shortest when \\(P = E\\), where \\(E\\) is the midpoint of the arc of \\(\\mathcal{C}_2\\) inside the triangle. (It is also the intersection of the arc with the height from \\(C\\), and it is also the midpoint of the height from \\(C\\), and it is also the midpoint of the parallel \\(D_aD_b\\) to \\(AB\\), going through the midpoints \\(D_a, D_b\\) of \\(BC, AC\\)). Denote by \\(F\\) the point corresponding to \\(R\\) when \\(P = E\\), that is \\(F \\in \\mathcal{C}_1\\) is the intersection of the arc with \\(EB\\), or the foot of the normal from \\(E\\) to \\(\\mathcal{C}_1\\).\n\n(Note: we could denote \\(\\alpha = \\angle PCB\\), use coordinates and a little trigonometry to calculate the length of \\(APR = AP + PR\\) in terms of \\(\\alpha\\) and \\(a =\\) the side of the triangle, and then use calculus to find \\(\\alpha \\in \\left[ 0, \\frac{\\pi}{3} \\right]\\) at which this function achieves its minimum. The result would be \\(\\alpha = \\frac{\\pi}{6}\\). But it is simpler to use a little geometry to show that \\(APR\\) is minimum when \\(P = E\\).)\n\nWe want to prove that \\(AE + EF \\le AP + PR\\). Clearly it is enough to prove \\(AE + EB \\le AP + PB\\). Let \\(B'\\) by the point symetrical to \\(B\\) with respect to the line \\(\\mathcal{L} = D_aD_b\\). Let \\(Q = AP \\cap \\mathcal{L}\\). We have \\(AE + EB = AE + EB' \\le AQ + QB' = AQ + QB \\le AQ + QP + PB = AP + PB\\).\n\nThis ends the proof that \\(AEF\\) is the shortest path in \\(\\mathcal{S}'\\) (up to a symmetry). Now we will show that \\(AEF \\in \\mathcal{S}\\), i.e. \\(AEF\\) has property \\(\\mathcal{P}\\).\n\nIn order for a curve \\(p\\) to have property \\(\\mathcal{P}\\) it is sufficient to know that there is a portion of \\(p\\) inside the trapezoid \\(BCD_bD_c\\) which joins a point from the sector of \\(\\mathcal{C}_1\\) inside the triangle to a point from the sector of \\(\\mathcal{C}_2\\) inside the triangle, and similarly for the trapezoids \\(ABD_aD_b\\) and arcs \\(\\mathcal{C}_1, \\mathcal{C}\\), and \\(ACD_aD_c\\) and arcs \\(\\mathcal{C}_2, \\mathcal{C}\\). Obviously \\(AEF\\) satisfies these conditions, and this ends the proof.\n\n(Note: It is interesting to note that in general, \\(APR\\) does not have property \\(\\mathcal{P}\\) because a portion of it is outside the trapezoid \\(ABD_aD_b\\). We would have to extend it by another segment \\(RT\\) for \\(APRT\\) to have the property \\(\\mathcal{P}\\). If we didn't want to extend \\(RT\\) all the way to \\(\\mathcal{C}\\), we would need some other sufficient condition for the curve to satisfy so that it has property \\(\\mathcal{P}\\).)\n\n[Solution by pf02, June 2025]"
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IMO-1973-5
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_5
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\(G\) is a set of non-constant functions of the real variable \(x\) of the form
\[
f(x) = ax + b, a \text{ and } b \text{ are real numbers,}
\]
and \(G\) has the following properties:
(a) If \(f\) and \(g\) are in \(G\), then \(g \circ f\) is in \(G\); here \((g \circ f)(x) = g[f(x)]\).
(b) If \(f\) is in \(G\), then its inverse \(f^{-1}\) is in \(G\); here the inverse of \(f(x) = ax + b\) is \(f^{-1}(x) = (x - b) / a\).
(c) For every \(f\) in \(G\), there exists a real number \(x_f\) such that \(f(x_f) = x_f\).
Prove that there exists a real number \(k\) such that \(f(k) = k\) for all \(f\) in \(G\).
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[
"First, observe that for each function \\(f(x)=ax+b\\) in \\(G\\), if \\(a=1\\) then \\(b=0\\). This is a result of (c); for example, \\(f(x)=x+1\\) could not be in \\(G\\) because it does not have a fixed point. Or if \\(f(x)=x\\), then every point is a fixed point.\n\nAlso, for each function \\(f(x)=ax+b\\) in \\(G\\), if \\(a \\neq 1\\) then the fixed point of \\(f\\) is where \\(y=ax+b\\) intersects \\(y=x\\), namely where \\(x = \\frac{b}{1 - a}\\).\n\nNow, take \\(f_1(x)=a_1 x+b_1\\) and \\(f_2(x)=a_2 x+b_2\\), both in \\(G\\). By (a), \\((f_1 \\circ f_2)(x)=f_1(f_2(x))=f_1(a_2 x+b_2)=a_1 a_2 x + a_1 b_2 + b_1\\) and \\((f_2 \\circ f_1)(x)=f_2(f_1(x))=f_2(a_1 x+b_1)=a_1 a_2 x + a_2 b_1 + b_2\\) must also both be in \\(G\\). By (b), \\((f_1 \\circ f_2)^{-1}(x)=\\frac{x - a_1 b_2 - b_1}{a_1 a_2}\\) must also be in \\(G\\). Finally, by (a),\n\n\\(((f_1 \\circ f_2)^{-1} \\circ (f_2 \\circ f_1))(x)=\\frac{a_1 a_2 x + a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}\\) \\(= x + \\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}\\)\n\nmust also be in \\(G\\).\n\nUsing our first observation, \\(\\frac{a_2 b_1 + b_2 - a_1 b_2 - b_1}{a_1 a_2}=0\\). Rearranging, we get \\(\\frac{b_1}{1 - a_1} = \\frac{b_2}{1 - a_2}\\). Therefore, the fixed point of \\(f_1\\) equals the fixed point of \\(f_2\\). Since we made no assumptions about \\(f_1\\) and \\(f_2\\), this is true for all \\(f\\) in \\(G\\).\n\nBorrowed from [1]"
] |
IMO-1973-6
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https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_6
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Let \(a_1, a_2,\cdots, a_n\) be \(n\) positive numbers, and let \(q\) be a given real number such that \(0<q<1.\) Find \(n\) numbers \(b_1, b_2, \cdots, b_n\) for which
(a) \(a_k<b_k\) for \(k=1,2,\cdots, n,\)
(b) \(q<\dfrac{b_{k+1}}{b_k}<\dfrac{1}{q}\) for \(k=1,2,\cdots,n-1,\)
(c) \(b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n).\)
|
[
"We notice that the constraints are linear, in the sense that if bi is a solution for ai, q, and bi' is a solution for ai', q, then for any k, k' > 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a \"near\" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = qn-h. \"Near\" because the inequalities in (a) and (b) are not strict.\n\nHowever, we might reasonably hope that the inequalities would become strict in the linear combination, and indeed that is true. Define br = qr-1a1 + qr-2a2 + ... + qar-1 + ar + qar+1 + ... + qn-ran. Then we may easily verify that (a) - (c) hold."
] |
IMO-1974-1
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https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_1
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Three players \(A, B\) and \(C\) play the following game: On each of three cards an integer is written. These three numbers \(p, q, r\) satisfy \(0 < p < q < r\). The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.
This process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, \(A\) has 20 counters in all, \(B\) has 10 and \(C\) has 9. At the last round \(B\) received \(r\) counters. Who received \(q\) counters on the first round?
|
[
"Answer: player \\(C\\).\n\nLet \\(n\\) be the number of rounds played, then obviously \\(n (p + q + r) = 20 + 10 + 9 = 39\\). So \\(n\\) must be a divisor of 39, i. e. \\(n \\in \\{ 1, 3, 13, 39 \\}\\). But \\(p \\geq 1,\\) \\(q \\geq p + 1 \\geq 2,\\) \\(r \\geq q + 1 \\geq 3,\\) so \\(p + q + r \\geq 1 + 2 + 3 = 6\\) and \\(n = \\frac{39}{p + q + r} \\leq \\frac{39}{6} < 7\\). Also, by condition \\(n \\geq 2,\\) so we conclude to \\(n = 3\\) and \\(p + q + r = \\frac{39}{n} = 13\\).\n\nAs \\(B\\) received 10 counters including \\(r\\) counters at the last round, \\(10 \\geq p + p + r \\geq 2 + r,\\) so \\(r \\leq 8\\). On the other hand, from number of counters received by \\(A\\) we get \\(r > \\frac{20}{3} > 6\\). So \\(r \\in \\{ 7, 8 \\}\\).\n\nIf \\(r = 7,\\) from number of counters received by \\(B\\) we get \\(3 = 10 - r \\in \\{ 2p, p + q, 2q \\},\\) and since 3 is odd, we get \\(p + q = 3\\). But then \\(p + q + r = 3 + 7 = 10 \\neq 13\\) - a contradiction.\n\nSo \\(r = 8\\) and \\(2 = 10 - r \\in \\{ 2p, p + q, 2q \\}\\). On the other hand, \\(2 \\leq 2p < p + q < 2q,\\) so \\(2 = 2p,\\) \\(p = 1\\) and \\(q = 13 - p - r = 4\\). It is easy to check that the only way to distribute the counters for players is \\(20 = r + r + q,\\) \\(10 = p + p + r\\) and \\(9 = q + q + p,\\) so player \\(C\\) received \\(q\\) counters on the first round."
] |
IMO-1974-2
|
https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_2
|
In the triangle \(ABC\), prove that there is a point \(D\) on side \(AB\) such that \(CD\) is the geometric mean of \(AD\) and \(DB\) if and only if \(\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})\).
|
[
"Let a point \\(D\\) on the side \\(AB\\). Let \\(CF\\) the altitude of the triangle \\(\\triangle ABC\\), and \\(C'\\) the symmetric point of \\(C\\) through \\(F\\). We bring a parallel line \\(L\\) from \\(C'\\) to \\(AB\\). This line intersects the ray \\(CD\\) at the point \\(E\\), and we know that \\(DE=DC\\).\n\nThe distance \\(d(L,AB)\\) between the parallel lines \\(L\\) and \\(AB\\) is \\(CF\\).\n\nLet \\(w = (O,R)\\) the circumscribed circle of \\(\\triangle ABC\\), and \\(MM'\\) the perpendicular diameter to \\(AB\\), such that \\(M,C\\) are on difererent sides of the line \\(AB\\).\n\nIn fact, the problem asks when the line \\(L\\) intersects the circumcircle. Indeed:\n\nSuppose that \\(DC\\) is the geometric mean of \\(DA,DB\\).\n\n\\[\nDA \\cdot DB = DC^{2}\\Rightarrow DA \\cdot DB = DC \\cdot DE\n\\]\n\nThen, from the power of \\(D\\) we can see that \\(E\\) is also a point of the circle \\(w\\). Or else, the line \\(L\\) intersects \\(w \\Leftrightarrow\\)\n\n\\[\nd(L,AB)\\leq d(M,AB) \\Leftrightarrow\n\\]\n\n\\(CF \\leq MN,\\) where \\(MN\\) is the altitude of the isosceles \\(\\triangle MAB\\).\n\n\\(\\Leftrightarrow \\frac{1}{2}CF \\cdot AB \\leq \\frac{1}{2}MN \\cdot AB \\Leftrightarrow\\) \\((ABC) \\leq (MAB) \\Leftrightarrow\\) \\(\\frac{AB \\cdot BC \\cdot AC}{4R}\\leq \\frac{AB \\cdot MA^{2}}{4R}\\Leftrightarrow\\)\n\n\\[\nBC \\cdot AC \\leq MA^{2}\n\\]\n\nWe use the formulas:\n\n\\(BC = 2R \\cdot \\sin A\\) \\(AC = 2R \\cdot \\sin B\\)\n\nand \\(\\angle CMA = \\frac{C}{2}\\Rightarrow MA = 2R \\cdot \\sin\\frac{C}{2}\\)\n\nSo we have \\((2R \\cdot \\sin A)(2R \\cdot \\sin B) \\leq (2R \\cdot \\sin\\frac{C}{2})^{2}\\Leftrightarrow\\) \\(\\sin A \\cdot \\sin B \\leq \\sin^{2}\\frac{C}{2}\\)\n\nFor \\((\\Leftarrow)\\)\n\nSuppose that \\(\\sin A \\cdot \\sin B \\leq \\sin^{2}\\frac{C}{2}\\)\n\nThen we can go inversely and we find that \\(d(L,AB)\\leq d(M,AB) \\Leftrightarrow\\) the line \\(L\\) intersects the circle \\(w\\) (without loss of generality; if \\(d(L,AB)=d(M,AB)\\) then \\(L\\) is tangent to \\(w\\) at \\(M\\))\n\nSo, if \\(E \\in L \\cap w\\) then for the point \\(D = CE \\cap AB\\) we have \\(DC=DE\\) and \\(AD \\cdot AB = CD \\cdot DE \\Rightarrow\\)\n\n\\[\nAD \\cdot AB = CD^{2}\n\\]\n\nThe above solution was posted and copyrighted by pontios. The original thread for this problem can be found here: [1]"
] |
IMO-1974-3
|
https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_3
|
Prove that the number \(\sum^n_{k=0}\binom{2n+1}{2k+1}2^{3k}\) is not divisible by \(5\) for any integer \(n\ge0.\)
|
[
"Everything that follows takes place in \\(\\mathbb F_5(\\sqrt 2)\\), i.e. the field we get by adjoining a root of \\(x^2-2=0\\) to \\(\\mathbb F_5\\), the field with \\(5\\) elements.\n\nWe have \\(\\sum_{k=0}^n\\binom{2n+1}{2k+1}2^{3k}=\\sum_{k=0}^n\\binom{2n+1}{2n-2k}3^k=\\sum_{k=0}^n\\binom{2n+1}{2(n-k)}2^{-k}\\). Now, this is zero iff it's zero when we multiply it by \\(2^n\\), so we may as well prove that \\(\\sum_{k=0}^n\\binom{2n+1}{2(n-k)}\\sqrt 2^{2(n-k)}\\ne 0\\). The LHS is \\(\\alpha\\) from \\((1+\\sqrt 2)^{2n+1}=\\alpha+\\beta\\sqrt 2,\\ \\alpha,\\beta\\in\\mathbb F_5\\). We have \\((1-\\sqrt 2)^{2n+1}=\\alpha-\\beta\\sqrt 2\\), so by multiplying them we get \\(-1=\\alpha^2-2\\beta^2\\). If we were to have \\(\\alpha=0\\), then we would get \\(1=2\\beta^2,\\ \\beta\\in\\mathbb F_5\\), and this is impossible, since it would make \\(3=2^{-1}\\) a square \\(\\beta^2\\) in \\(\\mathbb F_5\\) (i.e. \\(3\\) would be a quadratic residue modulo \\(5\\), and it's not).\n\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]"
] |
IMO-1974-4
|
https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_4
|
Consider decompositions of an \(8\times8\) chessboard into \(p\) non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If \(a_i\) is the number of white squares in the \(i\)-th rectangle, then \(a_1<a_2<\cdots<a_p.\)
Find the maximum value of \(p\) for which such a decomposition is possible. For this value of \(p,\) determine all possible sequences \(a_1, a_2, \cdots, a_p.\)
|
[
"Since each rectangle has the same number of black squares as white squares, \\(a_1+a_2+\\ldots +a_p=\\frac{64}{2}=32\\). Clearly \\(a_i\\ge i\\) for \\(i=1\\) to \\(i=p\\) so \\(32=a_1+a_2+\\ldots + a_p\\ge 1+2+\\ldots +p=\\frac{p(p+1)}{2}\\) so this forces \\(p\\le 7\\). It is possible to decompose the board into \\(7\\) rectangles, as we will show later. But first let us find all such sequences \\(a_i\\). Now \\(32-a_7=a_1+a_2+\\ldots +a_6\\ge 1+2+\\ldots +6=21\\implies 11\\le a_7\\). For a rectangle to have \\(11\\) white squares, it will have an area of \\(22\\) so it's dimensions are either \\(1\\times 22\\) or \\(2\\times 11\\) - neither of which would fit on a \\(8\\times 8\\) board. So \\(a_7\\not= 11\\implies a_7\\le 10\\).\n\nIf \\(a_7=10\\) (which could fit as a \\(4\\times 5\\) rectangle) then \\(a_1+a_2+\\ldots a_6=22\\). Then \\(22-a_6\\ge 1+2+\\ldots +5=15\\) so \\(7\\ge a_6\\). So \\(a_1,a_2,\\ldots ,a_6\\) are 6 numbers among 1-7. If \\(1\\le k\\le 7\\) is the number that is not equal to any \\(a_i\\), then \\(22=a_1+a_2+\\ldots +a_7=1+2+\\ldots +7-k=28-k\\) so \\(k=6\\). Then \\(a_1=1,a_2=2,a_3=3,a_4=4,a_5=5,a_6=7,a_7=10\\). Such a decomposition is possible. Take a \\(4\\times 5\\) rectangle on the top left corner, where there are \\(4\\) squares horizontally and \\(5\\) vertically. Then directly below use a \\(7\\times 2,1\\times 2\\) and a \\(8\\times 1\\) rectangle to cover the 3 rows below it. It's simple from there.\n\nSimilarly, you can find the other possibilities as \\(\\{a_1,a_2,\\ldots ,a_7\\}=\\{1,2,3,4,5,8,9\\}\\) or \\(\\{1,2,3,4,6,7,9\\}\\) or \\(\\{1,2,3,5,6,7,8\\}\\). Tilings are not hard to find.\n\nThe above solution was posted and copyrighted by WakeUp. The original thread for this problem can be found here: [1]"
] |
IMO-1974-5
|
https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_5
|
Determine all possible values of
\[
S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}
\]
where \(a, b, c, d,\) are arbitrary positive numbers.
|
[
"Note that\n\n\\[\n2 = \\frac{a}{a+b}+\\frac{b}{a+b}+\\frac{c}{c+d}+\\frac{d}{c+d} > S > \\frac{a}{a+b+c+d}+\\frac{b}{a+b+c+d}+\\frac{c}{a+b+c+d}+\\frac{d}{a+b+c+d} = 1.\n\\]\n\nWe will now prove that \\(S\\) can reach any range in between \\(1\\) and \\(2\\).\n\nChoose any positive number \\(a\\). For some variables such that \\(k, m, l > 0\\) and \\(k + m + l = 1\\), let \\(b = ak\\), \\(c = am\\), and \\(d = al\\). Plugging this back into the original fraction, we get\n\n\\[\nS = \\frac{a}{a+ak+al}+\\frac{ak}{a+ak+am}+\\frac{am}{ak+am+al}+\\frac{al}{a+am+al} = \\frac{1}{1+k+l}+\\frac{k}{1+k+m}+\\frac{m}{k+m+l}+\\frac{l}{1+m+l}.\n\\]\n\nThe above equation can be further simplified to\n\n\\[\nS = \\frac{1}{2-m}+\\frac{k}{2-l}+m+\\frac{l}{2-k}.\n\\]\n\nNote that \\(S\\) is a continuous function and that \\(f(m) = m + \\frac{1}{2-m}\\) is a strictly increasing function. We can now decrease \\(k\\) and \\(l\\) to make \\(m\\) tend arbitrarily close to \\(1\\). We see \\(\\lim_{m\\to1} m + \\frac{1}{2-m} = 2\\), meaning \\(S\\) can be brought arbitrarily close to \\(2\\). Now, set \\(a = d = x\\) and \\(b = c = y\\) for some positive real numbers \\(x, y\\). Then\n\n\\[\nS = \\frac{2x}{2x+y} + \\frac{2y}{2y+x} = \\frac{2y^2 + 8xy + 2x^2}{2y^2 + 5xy + 2x^2}.\n\\]\n\nNotice that if we treat the numerator and denominator each as a quadratic in \\(y\\), we will get \\(1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2}\\), where \\(g(x)\\) has a degree lower than \\(2\\). This means taking \\(\\lim_{y\\to\\infty} 1 + \\frac{g(x)}{2y^2 + 5xy + 2x^2} = 1\\), which means \\(S\\) can be brought arbitrarily close to \\(1\\). Therefore, we are done.\n\n\\[\n\n\\]\n\n~Imajinary"
] |
IMO-1974-6
|
https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_6
|
Let \(P\) be a non-constant polynomial with integer coefficients. If \(n(P)\) is the number of distinct integers \(k\) such that \((P(k))^2=1,\) prove that \(n(P)-\deg(P)\le2,\) where \(\deg(P)\) denotes the degree of the polynomial \(P.\)
|
[
"Lemma: Let \\(P(x)\\) be a polynomial with integer coefficients which is not constant. Then if \\(P(x)\\) obtains \\(1\\) (or \\(-1\\)) as its values for at least four times then \\(P(x)\\neq -1\\) ( or \\(P(x)\\neq 1\\)) for all \\(x\\). Proof. Assume that \\(P(a)=P(b)=P(c)=P(d)=1\\) for \\(a,b,c,d\\) distince. Then if there's \\(u\\) which \\(P(u)=-1\\) then \\(2=P(a)-P(u)=...\\) so \\(P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)\\) where \\(Q(x)\\) is a polynomial with the integer coefficients! So \\(-2=(u-a)(u-b)(u-c)(u-d)Q(u)\\) which is impossible cause \\(-2\\) can not presents as product of more than three distince numbers! This proved the lemma!\n\nBack to our problem: For convinet put \\(m=n(P)\\) and \\(n=\\deg P\\). Firstly if \\(n\\leq 2\\) then \\(m-n\\leq2\\). Assume \\(n\\geq3\\). If equation \\(P(x)=1\\) with more than three integer points (ie.. at least \\(4\\)) then equation \\((P(x))^2=1\\) implies \\(P(x)=1\\) so \\(m\\leq n\\), ie... \\(m-n\\leq 0\\leq2\\). The same case for equation \\(P(x)=-1\\). So \\(m\\leq 6\\). If \\(n\\geq4\\) then \\(m-n\\leq 6-n\\leq 2\\). Now assume that \\(n=3\\). In this case if \\(m\\leq 5\\) then \\(m-n\\leq 2\\).\n\nSo let us show that \\(m<6\\). In fact if \\(m=6\\) then \\(P(x)-1=0\\) has three integers distince roots, and the same for \\(P(x)+1=0\\). So \\(P(x)-1=k_1(x-a_1)(x-a_2)(x-a_3)\\) and \\(P(x)+1=k_2(x-b_1)(x-b_2)(x-b_3)\\) where \\(a_i\\) distince and \\(b_j\\) distince and all with \\(k_1,k_2\\) are integers! Then \\(k_2(x-b_1)(x-b_2)(x-b_3)-k_1(x-a_1)(x-a_2)(x-a_3)=2\\) for all \\(x\\). So \\(k_1=k_2=k\\). Finally, we have \\(2=k(a_i-b_1)(a_i-b_2)(a_i-b_3)\\) for \\(i=1,2,3\\) and because that \\(\\pm1\\) can not presents as products of three distince numbers so \\(k=\\pm1\\), we may assume \\(k=1\\). Because \\(2=(-2)\\cdot 1\\cdot -1\\) so \\(\\{-2,1,-1\\}=\\{a_i-b_1,a_i-b_2,a_i-b_3\\}\\) This means \\(3a_i-(b_1+b_2+b_3)=-2+1-1=-2\\). So we must have \\(3a_1=3a_2=3a_3\\) which follows \\(a_1=a_2=a_3\\), which contracts!. So \\(m\\leq 5\\) and we're done.\n\nThe above solution was posted and copyrighted by pluricomplex. The original thread for this problem can be found here: [1]"
] |
IMO-1975-1
|
https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_1
|
Let \(x_i, y_i\) \((i=1,2,\cdots,n)\) be real numbers such that
\[
x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.
\]
Prove that, if \(z_1, z_2,\cdots, z_n\) is any permutation of \(y_1, y_2, \cdots, y_n,\) then
\[
\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.
\]
|
[
"We can expand and simplify the inequality a bit, and using the fact that \\(z\\) is a permutation of \\(y\\), we can cancel some terms.\n\n\\[\n\\sum^n_{i=1}x_i^2 + \\sum^n_{i=1}y_i^2 - 2\\sum^n_{i=1}x_iy_i \\leq \\sum^n_{i=1}x_i^2 + \\sum^n_{i=1}z_i^2 - 2\\sum^n_{i=1}x_iz_i\n\\]\n\n\\[\n\\sum^n_{i=1}x_iy_i \\geq \\sum^n_{i=1}x_iz_i\n\\]\n\nConsider the pairing \\(x_1 \\rightarrow y_1\\), \\(x_2 \\rightarrow y_2\\), ... \\(x_n \\rightarrow y_n\\). By switching around some of the \\(y\\) values, we have obtained the pairing \\(x_1 \\rightarrow z_1\\), \\(x_2 \\rightarrow z_2\\), ... \\(x_n \\rightarrow z_n\\). Suppose that we switch around two \\(y\\)-values, \\(y_m\\) and \\(y_n\\), such that \\(y_m > y_n\\). If \\(x_m > x_n\\), call this a type 1 move. Otherwise, call this a type 2 move.\n\nType 2 moves only increase the sum of the products of the pairs. The sum is increased by \\(x_n \\cdot y_m + x_m \\cdot y_n - x_m \\cdot y_m - x_n \\cdot y_n\\). This is equivalent to \\((x_n - x_m)(y_m - y_n)\\), which is clearly nonnegative since \\(y_m > y_n\\) and \\(x_n > x_m\\).\n\nWe will now consider switching from the \\(x\\)-\\(z\\) pairing back to the \\(x\\)-\\(y\\) pairing. We will prove that from any pairing of \\(x\\) and \\(z\\) values, you can use just type 2 moves to navigate back to the pairing of \\(x\\) and \\(y\\) values, which will complete the proof.\n\nSuppose that \\(x_a\\) is the biggest \\(x\\)-value that is not paired with its \\(y\\)-value in the \\(x\\) and \\(z\\) pairing. Then, switch this \\(y\\)-value with the \\(y\\)-value currently paired with \\(x_a\\). This is obviously a type 2 move. Continue this process until you reach back to the \\(x\\)-\\(y\\) pairing. All moves are type 2 moves, so the proof is complete.\n\n~mathboy100",
"We can rewrite the summation as\n\n\\[\n\\sum^n_{i=1} x_i^2 + \\sum^n_{i=1} y_i^2 - \\sum^n_{i=1}x_iy_i \\le \\sum^n_{i=1} x_i^2 + \\sum^n_{i=1} z_i^2 - \\sum^n_{i=1}x_iz_i.\n\\]\n\nSince \\(\\sum^n_{i=1} y_i^2 = \\sum^n_{i=1} z_i^2\\), the above inequality is equivalent to\n\n\\[\n\\sum^n_{i=1}x_iy_i \\ge \\sum^n_{i=1}x_iz_i.\n\\]\n\nWe will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of \\(\\sum^n_{i=1}x_iz_i\\). Obviously, if \\(x_1 = x_2 = \\ldots = x_n\\) or \\(y_1 = y_2 = \\ldots = y_n\\), the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of \\(z_i\\)s that are not ordered in the form \\(z_1 \\ge z_2 \\ge \\ldots \\ge z_n\\) such that \\(\\sum^n_{i=1}x_iz_i\\) is at a maximum out of all possible permutations and is greater than the sum \\(\\sum^n_{i=1}x_iy_i\\). This necessarily means that in the sum \\(\\sum^n_{i=1}x_iz_i\\) there exists two terms \\(x_pz_m\\) and \\(x_qz_n\\) such that \\(x_p > x_q\\) and \\(z_m < z_n\\). Notice that\n\n\\[\nx_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,\n\\]\n\nwhich means if we make the terms \\(x_pz_n\\) and \\(x_qz_m\\) instead of the original \\(x_pz_m\\) and \\(x_qz_n\\), we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality\n\n\\[\n\\sum^n_{i=1}x_iy_i \\ge \\sum^n_{i=1}x_iz_i\n\\]\n\nis proved, which is equivalent to what we wanted to prove.\n\n\\[\n\n\\]\n\n~Imajinary\n\n## Remark\n\nIt is only the most common way of rearrangement inequality after expanding and subtracting same terms.~bluesoul"
] |
IMO-1975-2
|
https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_2
|
Let \(a_1, a_2, a_3, \cdots\) be an infinite increasing sequence of positive integers. Prove that for every \(p \geq 1\) there are infinitely many \(a_m\) which can be written in the form
\[
a_m = xa_p + ya_q
\]
with \(x, y\) positive integers and \(q > p\).
|
[
"If we can find \\(p\\ne q\\) such that \\((a_p,a_q)=1\\), we're done: every sufficiently large positive integer \\(n\\) can be written in the form \\(xa_p+ya_q,\\ x,y\\in\\mathbb N\\). We can thus assume there are no two such \\(p\\ne q\\). We now prove the assertion by induction on the first term of the sequence, \\(a_1\\). The base step is basically proven, since if \\(a_1=1\\) we can take \\(p=1\\) and any \\(q>1\\) we want. There must be a prime divisor \\(u|a_1\\) which divides infinitely many terms of the sequence, which form some subsequence \\((a_{k_n})_{n\\ge 1},\\ k_1=1\\). Now apply the induction hypothesis to the sequence \\(\\left(\\frac{a_{k_n}}u\\right)_{n\\ge 1}\\).\n\nThe above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]"
] |
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