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XLIX OM - I - Problem 6
In triangle $ABC$, where $|AB| > |AC|$, point $D$ is the midpoint of side $BC$, and point $E$ lies on side $AC$. Points $P$ and $Q$ are the orthogonal projections of points $B$ and $E$ onto line $AD$, respectively. Prove that $|BE| = |AE| + |AC|$ if and only if $|AD| = |PQ|$.
|
Note 1: In the official print of the problem texts for the first-level competition of the 49th Mathematical Olympiad, problem 6 was formulated with an error: the assumption that $ |AB| > |AC| $ was missing. (The corrected version of the problem was sent to schools as an Errata a few weeks later.) Without this assumption, the equivalence to be proven is not true. As an exercise, we propose to the Reader to find a counterexample; a hint may be the equivalence (2) obtained in the solution by method III.
The assumption that $ |AB| > |AC| $ guarantees that angle $ ADC $ is acute, and thus the projection $ P $ of point $ B $ onto line $ AD $ lies on the ray $ AD^\to $ outside the triangle $ ABC $; points $ A $, $ C $, $ D $, $ E $, $ Q $ all lie on the same side of line $ BP $; point $ D $ lies between points $ A $ and $ P $.
Through point $ B $, we draw a line parallel to $ AD $; it intersects the ray $ CA^\to $ at a point we will denote by $ F $. The ray $ EQ^\to $ intersects the ray $ BF^\to $ at a point we will call $ S $; it lies on the segment $ BF $ when $ |\measuredangle CAD| \leq 90^\circ $, and outside this segment when $ |\measuredangle CAD| > 90^\circ $ (two variants of figure 2). From the parallelism $ BF \parallel AD $, it follows that point $ A $ is the midpoint of side $ CF $ of triangle $ BCF $. Therefore, $ |AD| = \frac{1}{2}|BF| $.
All the above statements follow from the definitions given in the first two sentences of the problem statement and are true regardless of the truth or falsity of the conditions given in the third sentence, which need to be proven equivalent.
We will now write a sequence of conditions, each of which is obviously equivalent to the previous one:
The equivalence between the first and the last condition is the thesis of the problem.
Note 2: All these equivalences hold regardless of which of the two situations illustrated in the two variants of figure 2 we consider. But it is clear that in the case shown in the right figure, point $ S $ is certainly not the midpoint of segment $ BF $; thus, in this case, the written equivalences bind a chain of statements that are all false. However, the fact that the reasoning is a priori, independent of the case, and at the same time short and transparent, constitutes the elegance of the solution.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,009 |
L OM - II - Task 1
Given is a function $ f: \langle 0,1 \rangle \to \mathbb{R} $ such that $ f\left(\frac{1}{n}\right) = (-1)^n $ for $ n = 1, 2,\ldots $. Prove that there do not exist increasing functions $ g: \langle 0,1 \rangle \to \mathbb{R} $, $ h: \langle 0,1 \rangle \to \mathbb{R} $, such that $ f = g - h $.
|
Suppose there exist functions $ g $ and $ h $ satisfying the conditions of the problem. Since these functions are increasing, for every $ k \in \mathbb{N} $ we have
From the monotonicity of the function $ g $ and the above inequalities, we get
Thus, $ g(1)-g(0) > 2k $ for any $ k \in \mathbb{N} $, which is not possible.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,010 |
XXXI - I - Problem 10
A plane is divided into congruent squares by two families of parallel lines. $ S $ is a set consisting of $ n $ such squares. Prove that there exists a subset of the set $ S $, in which the number of squares is no less than $ \frac{n}{4} $ and no two squares have a common vertex.
|
A set of squares formed on a plane can be divided into two subsets in such a way that two squares sharing a side belong to different subsets (similar to the division of a chessboard into black and white squares). Such a division can be described as follows. We introduce a coordinate system on the plane, whose axes are parallel to the lines defining the division of the plane into squares, with the unit of length equal to the side length of the square, and the origin of the system is the center of a certain square. Thus, the center of each square is a lattice point (i.e., has both coordinates as integers). We include in one subset all squares for which the sum of the coordinates of the center is an even number, and in the other subset, those squares for which the sum of the coordinates of the center is an odd number.
At least one of these subsets contains no fewer than \( n/2 \) squares of the set \( S \). We consider this particular subset \( A \) of the set of squares defined on the plane and divide it into two subsets \( A_1 \) and \( A_2 \) such that two squares sharing a vertex belong to different subsets. We will achieve this division by including in one subset squares whose centers have the first coordinate even, and in the other subset, squares whose centers have the first coordinate odd.
At least one of the subsets \( A_1 \) and \( A_2 \) contains no fewer than half of the squares of the set \( S \cap A \), and thus no fewer than \( n/4 \) squares of the set \( S \). From the construction of the subsets, it follows that no two different squares of this subset have any points in common.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,012 |
XL OM - II - Task 2
For a randomly chosen permutation of the set $ \{1,\ldots, n\} $, denote by $ X(\mathbf{f}) $ the largest number $ k \leq n $ such that $ f_i < f_{i+1} $ for all indices $ i < k $. Prove that the expected value of the random variable $ X $ is $ \sum_{k=1}^n \frac{1}{k!} $.
|
The considered random variable of course depends on $ n $. Let us denote it by $ X_n $. We will find a recursive relationship between the expected values of the variables $ X_{n-1} $ and $ X_n $.
Take any permutation $ f = (f_1, \ldots , f_n) $ of the set $ \{1,\ldots, n\} $. By discarding the term $ f_n $, we obtain a permutation of some $(n -1)$-element set of different numbers. The length of the initial segment forming an increasing sequence is the value of the variable $ X_{n-1} $. (It is not important whether the permuted set is, according to the definition of $ X_{n-1} $, the set $ \{1,\ldots, n-1\} $, or any other set linearly ordered by the relation of being less than.)
We reattach the term $ f_n $. The length of the initial segment forming an increasing sequence will remain unchanged or increase by $ 1 $. This increase will occur only in one case: when the permutation $ f $ is identical to the sequence $ (1,2, \ldots , n) $. The probability of such an event is $ 1/n! $.
We thus obtain the relationship
Hence, we immediately get the recursive formula
Since $ E(X_1) = 1 $, the formula to be proven in the problem is obtained by obvious induction.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,017 |
LIX OM - I -Zadanie 10
Dana jest liczba pierwsza p. Ciąg liczb całkowitych dodatnich $ a_1, a_2, a_3, \dots $ spełnia warunek
Wykazać, że pewien wyraz tego ciągu jest $ p $-tą potęgą liczby całkowitej.
(Uwaga: Symbol $ [x] $ oznacza największą liczbę całkowitą nie przekraczającą $ x $.)
|
Zdefiniujmy ciągi liczb całkowitych nieujemnych $ b_1, b_2, b_3, \dots $ oraz $ r_1, r_2, r_3, \dots $
w następujący sposób:
dla $ n =1,2,3,\dots $. Inaczej mówiąc, $ r_n $ jest różnicą między wyrazem an a największą nie
przekraczającą tego wyrazu $ p $-tą potęgą liczby całkowitej. Warunek (1) przepisujemy teraz w postaci
Zauważmy, ze jeżeli $ r_n = 0 $, to $ a_n = b_n^p $ jest $ p $-tą potęgą liczby całkowitej. Zadanie będzie
więc rozwiązane, jeśli udowodnimy, że $ r_n = 0 $ dla pewnej wartości $ n $.
Przypuśćmy, że wszystkie wyrazy ciągu $ r_1, r_2, r_3, \dots $ są liczbami całkowitymi dodatnimi. Niech
$ r_m $ będzie najmniejszym wśród tych wyrazów. Mamy więc $ a_m = b_m^p +rm $. Ponadto
$ b_{m-1} <b_m $, gdyż w przeciwnym razie mielibyśmy
wbrew określeniu $ m $. Zauważmy dalej, że $ a_{m-1} <(b_{m-1}+1)^p \leqslant b^p_m \leqslant a_m $
i wobec tego zachodzi nierówność
Niech ponadto $ k>m $ będzie najmniejszym wskaźnikiem, dla którego $ b_k >b_m $. Wówczas
$ b_m = b_{m+1} = \dots = b_{k-1} $, więc na mocy (2) otrzymujemy
i sumując stronami dochodzimy do wniosku, że
Wykażemy z kolei, że $ b_k = b_m +1 $. Istotnie, gdyby zachodziła nierówność
$ b_k \geqslant b_m +2= b_{k-1} +2 $, mielibyśmy $ a_{k-1} < (b_{k-1} +1)^p $, $ a_k \geqslant (b_{k-1} +2)^p $
oraz
ale pisząc $ x = b_{k-1} +2 $, $ y = b_{k-1} +1 $ mamy $ x-y =1 $, $ x>y>b_{k-1} $ i widzimy, że
Nierówność (5) jest więc fałszywa, co dowodzi równości $ b_k = b_m +1 $.
Na mocy (4) możemy zatem napisać
czyli
Ze wzoru dwumianowego otrzymujemy
Ponieważ $ p $ jest liczbą pierwszą, więc współczynniki dwumianowe $ \binom{p}{i} $ są dla $ i =1,2,\dots ,p - 1 $
liczbami podzielnymi przez $ p $. Zatem liczba $ (b_m + 1)^p - b_m^p $ daje resztę 1 z dzielenia przez
$ pb_m $. Wobec tego z (6) wynika, że liczba $ r_m -r_k $ daje resztę 1 z dzielenia przez $ pb_m $.
W (3) uzyskaliśmy nierówność $ r_m <pb_{m-1} $ korzystając jedynie z tego, że
$ b_{m-1} <b_m $. Analogicznie wyprowadzamy z nierówności $ b_{k-1} <b_k $, że $ r_k <pb_m $.
Liczby $ r_m $ i $ r_k $ należą więc do przedziału $ (0; pb_m) $, a różnica $ r_m-r_k $ daje resztę 1 z dzielenia
przez $ pb_m $. Wynika stąd, że $ r_k = r_m -1 $, co przeczy założeniu, że $ r_m $ jest najmniejszym wyrazem
w ciągu $ r_1, r_2, r_3, \dots $.
To oznacza, że $ r_n = 0 $ dla pewnego wskaźnika $ n $, skąd wynika teza.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,018 |
XV OM - I - Problem 12
The tetrahedron $ABCD$ is cut into two parts by a plane passing through vertex $D$, point $M$ on edge $AB$, and point $N$ on edge $BC$. Prove that the area of triangle $DMN$ is less than the area of one of the triangles $DAB$, $DBC$, or $DCA$.
|
The theorem we need to prove can be considered the spatial counterpart of the following theorem from plane geometry:
The segment $CM$ connecting vertex $C$ of triangle $ABC$ with an internal point $M$ on the base $AB$ of this triangle is smaller than one of the sides $AC$ and $BC$.
A short proof of this theorem can be found by applying the method of reduction to absurdity. Suppose that $CM \geq AC$ and $CM \geq BC$. In triangles $ACM$ and $BCM$, it then follows that $\measuredangle AMC \leq \measuredangle MAC$, $\measuredangle BMC \leq \measuredangle MBC$, hence
which is contradictory to the fact that $\measuredangle AMC$ and $\measuredangle BMC$ are adjacent angles.
The thesis of this theorem can also be formulated differently. The statement: "the number $x$ is smaller than one of the numbers $a, b, \ldots$" is equivalent to the statement: "the number $x$ is smaller than the largest of the numbers $a, b, \ldots$". Let the symbol
denote the largest of the numbers $a, b, \ldots$; instead of the statement that the segment $CM$ is smaller than one of the segments $AC$ and $BC$, we can write briefly that
Let us note a few obvious properties of the number $\max (a, b, \ldots)$, which we will refer to in the following.
(a) If $d > 0$, then $\max (ad, bd, \ldots) = d \cdot \max (a, b, \ldots)$
(b) If $a_1 > a$, then $\max (a, b, \ldots) \leq \max (a_1, b, \ldots)$
(c) For any $k$, $\max (a, b, \ldots) \leq \max (k, a, b, \ldots)$
(d) $\max [a, b, \ldots, \max (c, d, \ldots)] = \max (a, b, \ldots, c, d, \ldots)$
Note also that the value of $\max (a, b, \ldots)$ does not depend on the order of the numbers $a, b, \ldots$:
We suggest the Reader find other properties of the number $\max (a, b, \ldots)$, as well as the number $\min (a, b, \ldots)$, i.e., the smallest of the numbers $a, b, \ldots$.
Let us proceed to the given problem. Let $M$ be a point on the edge $AB$, and $N$ - a point on the edge $BC$. We need to prove that
1. First, consider the special case where the plane $DMN$ passes through one of the edges of the tetrahedron, for example, through $AD$; in this case, point $M$ coincides with point $A$ (Fig. 10).
Consider the projection of the entire figure onto any plane perpendicular to the line $AD$. Let $A$ be the common projection of points $A$ and $D$, and $B$, $C$, $N$ be the projections of points $B$, $C$, and $N$, respectively.
The lengths of segments $A$, $A$, $A$ are equal to the distances from the line $AA$ to the parallel lines $BB$, $CC$, $NN$, and thus are equal to the distances from points $B$, $C$, $N$ to the line $AA$, i.e., the heights of triangles $DAB$, $DAC$, $DAN$ with the common base $AD$. Since, as we stated above,
then, by (a),
hence, by (c),
2. Suppose next that the plane $DMN$ does not pass through any edge of the tetrahedron; in this case, $M$ and $N$ are internal points of the edges $AB$ and $BC$ (Fig. 11). Draw the plane $DMC$. For the tetrahedron $DMBC$ cut by the plane $DMN$, the case 1 applies; thus,
and since $\text{ area } DMB < \text{ area } DAB$, by (b),
Similarly, in the tetrahedron $DABC$ cut by the plane $DMC$,
From the last two inequalities, by (a) and (d), it follows that
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,020 |
XXI OM - I - Problem 11
Prove that in any division of the plane into three sets, there exist two points in at least one of them that are 1 unit apart.
|
Suppose this theorem is false. Consider equilateral triangles $ABC$ and $ABD$ (Fig. 7) with a common base $\overline{AB}$ and a side length of $1$. From our assumption, it follows that each vertex of triangle $ABC$ belongs to a different one of the three distinct subsets into which we have divided the plane. Similarly, each vertex of triangle $ABD$ belongs to a different one of the three distinct subsets into which we have divided the plane. Therefore, points $C$ and $D$ belong to the same subset. The distance between these points is $\sqrt{3}$.
From the above reasoning, it follows that any two points in the plane that are $\sqrt{3}$ apart belong to the same subset. They can be treated as the vertices of corresponding equilateral triangles with a common base and a side length of $1$.
Now consider triangle $PQR$ (Fig. 8) with sides $\sqrt{3}$, $\sqrt{3}$, and $1$. From the above, it follows that points $P$ and $Q$ as well as $O$ and $R$ belong to the same subset. Therefore, points $P$ and $R$ also belong to the same subset. However, they are $1$ unit apart, which contradicts our initial assumption.
The obtained contradiction proves that the considered theorem is true.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,021 |
XLVIII OM - I - Problem 10
Points $ P $, $ Q $ lie inside an acute-angled triangle $ ABC $, such that $ |\measuredangle ACP| = |\measuredangle BCQ| $ and $ |\measuredangle CAP| = |\measuredangle BAQ| $. Points $ D $, $ E $, $ F $ are the orthogonal projections of point $ P $ onto the sides $ BC $, $ CA $, $ AB $, respectively. Prove that angle $ DEF $ is right if and only if point $ Q $ is the orthocenter of triangle $ BDF $.
|
In the assumptions of equality
they are equivalent to the equalities
Angles $ ACP $ and $ BCQ $, considered as areas of the plane, can be disjoint (as in Figure 6) or they can overlap; this has no significance. A similar remark applies to each of the other three pairs of angles in relations (1) and (2).
Let $ U $, $ V $, $ W $ be the points symmetric to $ P $ with respect to the lines $ BC $, $ CA $, $ AB $, respectively. Thus, $ |CU| = |CP| = |CV| $ and $ |AW| = |AP| = |AV| $, and furthermore
and similarly
Using the first equality (3), the first equality (2), the first equality (4), and the first equality (1), we calculate:
Angles $ UCQ $ and $ VCQ $ are therefore equal. (Each of them has a measure of $ |\measuredangle ACB| $; this information will not be useful, however).
We also know that $ |CU| = |CV| $. It follows that triangle $ UCQ $ is congruent to triangle $ VCQ $, and thus $ |QU| = |QV| $.
Similarly — using the second equalities (3), (2), (4), (1) — we show that angles $ WAQ $ and $ VAQ $ are equal; knowing that $ |AW| = |AV| $, we conclude from this that triangle $ WAQ $ is congruent to triangle $ VAQ $, and thus $ |QW| = |QV| $.
Therefore, point $ Q $ is the center of the circle circumscribed around triangle $ UVW $. A homothety with center $ P $ and scale $ 1/2 $ transforms it into the circle $ \omega $ circumscribed around triangle $ DEF $; the center of circle $ \omega $ is the midpoint $ S $ of segment $ PQ $.
We will now write a sequence of statements, each of which is equivalent to the previous one:
$ Q $ is the intersection point of the altitudes of triangle $ BDF $;
$ QD \perp FB $ and $ QF \perp BD $;
$ QD \parallel PF $ and $ QF \parallel PD $;
quadrilateral $ PDQF $ is a parallelogram;
the midpoint of segment $ DF $ coincides with point $ S $;
segment $ DF $ is a diameter of circle $ \omega $;
angle $ DEF $ is a right angle.
The equivalence between the first and the last statement is the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,022 |
XXIV OM - I - Problem 2
Prove that among 25 different positive numbers, one can choose two such that neither their sum nor their difference is equal to any of the remaining numbers.
|
Let us denote a given set of numbers by $A = \{a_1, a_2, \ldots, a_{25}\}$, where $0 < a_1 < a_2 < \ldots < a_{25}$. Suppose that for any $r$, $s$, where $1 \leq r < s \leq 25$, we have: $a_s + a_r \in A - \{a_r, a_s\}$ or $a_s - a_r \in A - \{a_r, a_s\}$. Of course, $a_r - a_s < 0$ and therefore the number $a_r - a_s$ does not belong to $A$.
Since for $i = 1, 2, \ldots, 24$ we have $a_{25} + a_i > a_{25}$, it follows that $a_{25} + a_i \not \in A$. Therefore, by the assumption, the 24 numbers $a_{25} - a_i$ form a decreasing sequence with terms belonging to the set $A - \{a_{25}\}$, which has 24 elements. Hence
Since for $j = 2, 3, \ldots, 23$ we have $a_{24} + a_j > a_{24} + a_1 = a_{25}$, it follows that $a_{24} + a_i \not \in A$ and hence by the assumption we have $a_{24} - a_j \in A$. Moreover, $a_{24} - a_j \leq a_{24} - a_2 = (a_{25} - a_1) - (a_{25} - a_{23}) = a_{23} - a_1 < a_{23}$. Therefore, the 22 numbers $a_{24} - a_j$ form a decreasing sequence with terms belonging to the set $A - \{a_{23}, a_{24}, a_{25}\}$, which has 22 elements. Hence
In particular, $a_{24} - a_{12} = a_{12}$, and thus $a_{24} - a_{12} \not \in A - \{a_{12}, a_{24}\}$. We also have $a_{24} + a_{12} > a_{24} + a_1 = a_{25}$, and therefore $a_{24} + a_{12} \not \in A$. This contradicts the initial assumption.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,024 |
VII OM - I - Problem 9
Prove that a triangle can be constructed from segments of lengths $ a $, $ b $, $ c $ if and only if
|
From segments of lengths $ a $, $ b $, $ c $, a triangle can be constructed if and only if
If the numbers $ a $, $ b $, $ c $ satisfy inequality (1) and are (as measures of segments) positive, then they also satisfy inequalities (2) and vice versa. To this end, we transform inequality (1). Writing this inequality in the form
we have a quadratic trinomial in terms of $ a^2 $ on the left side; the roots of this trinomial are obtained according to the formula
Therefore, inequalities (3) can be given the form
or
By changing the sign of the second factor on the left side, we finally obtain the inequality
If the numbers $ a $, $ b $, $ c $ satisfy inequality (1), and thus also inequality (4), and are positive, then $ a + b + c > 0 $, so either the remaining 3 factors on the left side of (4) are positive, i.e., inequalities (2) hold, or one of these factors is positive and the other two are negative. However, this last possibility cannot occur, for if, for example, $ b + c - a < 0 $ and $ a + b - c < 0 $, then adding these inequalities side by side we would get $ 2b < 0 $, and thus $ b < 0 $ - contrary to the assumption.
Therefore, from inequality (1) and the condition $ a > 0 $, $ b > 0 $, $ c > 0 $, inequalities (2) follow.
Conversely, if the numbers $ a $, $ b $, $ c $ satisfy inequalities (2), then they also satisfy inequality (1) and are positive. Indeed, adding, for example, the first two inequalities (2) side by side, we get $ 2b > 0 $, i.e., $ b > 0 $, and similarly $ a > 0 $, $ c > 0 $, so $ a + b + c > 0 $. All factors on the left side of inequality (4) are then positive, and inequality (4) is satisfied, hence inequality (1) is also satisfied.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,026 |
XI OM - III - Task 2
A plane passing through the height of a regular tetrahedron intersects the planes of the lateral faces along $ 3 $ lines forming angles $ \alpha $, $ \beta $, $ \gamma $ with the plane of the base of the tetrahedron. Prove that
|
We accept the notation given in Fig. 27, where $HD$ represents the height of the regular tetrahedron $ABCD$, and $MD$, $ND$, $PD$ are the lines of intersection of the plane through $HD$ with the lateral faces $BCD$, $CAD$, and $ABD$, respectively. The angles $\alpha$, $\beta$, $\gamma$ are the angles that the segments $MD$, $ND$, $PD$ form with their projections $MH$, $NH$, and $PH$ on the base plane $ABC$. Therefore,
Denoting the edge length of the tetrahedron by $a$, we have, as it is easy to verify,
Considering formulas (2) and (3), we can give theorem (1) the form
We see that theorem (1) reduces to the planimetric theorem (4) about the equilateral triangle $ABC$ with side $a$, whose sides $BC$, $CA$, $AD$ or their extensions are intersected at points $M$, $N$, $P$ by a line passing through the center of the triangle (Fig. 28).
Since point $H$ is inside the triangle, points $M$, $N$, $P$ lie on different sides of point $H$ on the line $MNP$. Suppose (without loss of generality) that point $M$ lies on one side, and points $N$ and $P$ on the other side of point $H$, with point $N$ on side $CA$ and point $P$ on the extension of side $BA$ beyond point $A$. Let $\lambda$, $\mu$, $\nu$ denote the angles that the segments $HM$, $HN$, $HP$ form with the perpendiculars drawn from point $H$ to the sides $BC$, $CA$, $AB$; these perpendiculars have, as is known, a length of $\frac{a \sqrt{3}}{6}$ and form angles of $120^\circ$ with each other.
The following relationships hold:
where
thus
Therefore,
Using known trigonometric formulas, it is easy to calculate that
Substituting into the previous equation gives
which was to be proved.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,028 |
XVII OM - III - Problem 5
Given a convex hexagon $ABCDEF$, in which each of the diagonals $AD, BE, CF$ divides the hexagon into two parts of equal area. Prove that these three diagonals pass through one point.
|
\spos{1} The area of a polygon $ ABC\ldots $ will be denoted by the symbol $ (ABC\ldots) $. According to the assumption (Fig. 15)
From (1) and (2) we obtain
from which it follows that $ AE \parallel BD $, since the vertices $ A $ and $ E $ of triangles $ ABD $ and $ EDB $ with the common side $ BD $ lie on the same side of the line $ BD $.
Similarly, $ AC \parallel DF $ and $ CE \parallel BF $. Triangles $ ACE $ and $ DFB $ have corresponding sides parallel, so according to the theorem on similar triangles, the lines $ AD $, $ BE $, $ CF $, containing pairs of corresponding vertices of both triangles, pass through one point.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,030 |
XVIII OM - I - Problem 7
On a plane, there are $2n$ points. Prove that there exists a line that does not pass through any of these points and divides the plane into half-planes, each containing $n$ of the given points. Formulate and prove an analogous theorem for space.
|
Let $A_1, A_2, \ldots, A_{2n}$ be given points in the plane and let $a_{ik}$ ($i, k = 1, 2, \ldots, 2n$, $i \ne k$) denote the line passing through points $A_i$ and $A_k$. The set $Z$ of all lines $a_{ik}$ contains at most $\binom{2n}{2} = n(2n-1)$ distinct lines. Therefore, through any point $P$ in the plane, one can draw at most $n(2n-1)$ lines, each of which is perpendicular to some line in the set $Z$. Consequently, through point $P$ there are also lines that are not perpendicular to any of the lines in the set $Z$. Let $p$ be such a line and let $A_i$ be the orthogonal projection of point $A_i$ onto line $p$, and let $U$ be the set of these projections. No two points $A_i$ and $A_k$ ($i \ne k$) in the set $U$ coincide, since the line $A_iA_k$ is not perpendicular to line $p$. Therefore, there are $2n$ distinct points $A_i$ on line $p$. We can arrange their numbering such that in a certain direction of line $p$, point $A_i$ precedes point $A_{i+1}$. The midpoint $M$ of segment $A_n$ divides line $p$ into two rays, each containing $n$ points of the set $U$. The line perpendicular to $p$ through point $M$ divides the plane into two half-planes, each containing $n$ points of the given set.
An analogous theorem for space states: For any set of $2n$ points in space, there exists a plane that does not pass through any of the points in the set, such that each of the half-spaces determined by it contains $n$ points of the given set.
The proof can be conducted similarly to that for the plane.
Namely, through any point $P$ in space, one can draw a plane $\alpha$ that is not perpendicular to any of the lines in the set $Z$ (defined analogously to the plane case).
The orthogonal projections $A_i$ of the given points $A_i$ onto plane $\alpha$ form a set $U$ containing $2n$ distinct points. Let a line $p$ in plane $\alpha$ that does not pass through any point of the set $U$ divide plane $\alpha$ into half-planes, each containing $n$ points of the set $U$. Then the plane $\pi$ passing through line $p$ and perpendicular to $\alpha$ is the plane with the desired property.
Note. In the same way, a more general theorem can be proven. Namely, a set of $n$ points in the plane (space) can be divided into $2$ parts, one containing $k$ points and the other $n - k$ points, by drawing a line (plane) in the plane (space) that does not contain any of these points.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,031 |
XLV OM - II - Task 3
Prove that if a circle can be circumscribed around a hexagonal cross-section of a cube passing through its center, then the cross-section is a regular hexagon.
|
We place a cube in a coordinate system so that its vertices are points with all coordinates equal to $+1$ or $-1$. The center of the cube is then the point $O = (0,0,0)$, the origin of the coordinate system. The plane considered in the problem passes through $O$ and thus has an equation of the form
\[ ax + by + cz = 0 \]
where none of the coefficients $a$, $b$, $c$ are zero (if, for example, the coefficient $a$ were zero, the plane would contain the entire $Ox$ axis, and the intersection of the cube would not be hexagonal).
Let $A$, $B$, $C$ be three consecutive vertices of the hexagon that is the considered intersection. They lie on three edges of the cube forming a broken line with segments parallel to all three coordinate axes. Without loss of generality, we can assume that these are (in order) the edges $PQ$, $QR$, $RP$, where
(see figure 10). All points on the line $PQ$ have the coordinate $y = 1$ and the coordinate $z = -1$. Therefore, the point $A$ has the form $A = (\alpha, 1, -1)$. Similarly, $B = (1, -\beta, -1)$, $C = (1, -1, \gamma)$ (the notation of the second coordinate of point $B$ as $-\beta$ rather than $\beta$ has its purpose). Points $A$, $B$, and $C$ lie on the plane given by equation (1). We thus have the system of equations
\[ a\alpha + b + c(-1) = 0 \]
\[ a + b(-\beta) + c(-1) = 0 \]
\[ a + b(-1) + c\gamma = 0 \]
from which we determine
\[ \alpha = \frac{b - c}{a} \]
\[ \beta = \frac{a - c}{b} \]
\[ \gamma = \frac{a - b}{c} \]
Point $O$ is the center of symmetry of the hexagon (since it is the center of symmetry of the cube). If there is a circle circumscribed around this hexagon, its center is point $O$. This means that points $A$, $B$, and $C$ are at equal distances from $O$. Since $|OA|^2 = 2 + \alpha^2$, $|OB|^2 = 2 + \beta^2$, $|OC|^2 = 2 + \gamma^2$, we conclude that $|\alpha| = |\beta| = |\gamma|$.
If we can show that $\alpha = \beta = \gamma = 0$, it will mean that $A = (0, 1, -1)$, $B = (1, 0, -1)$, $C = (1, -1, 0)$, and thus points $A$, $B$, and $C$ are the midpoints of the corresponding edges, and the hexagon is regular.
We can now forget about the geometric context; the problem has been reduced to showing that if the numbers $\alpha$, $\beta$, $\gamma$ given by formulas (2) have equal magnitudes, then they are all equal to zero.
Since $|\alpha| = |\beta| = |\gamma|$, at least two of the numbers $\alpha$, $\beta$, $\gamma$ are equal; the third either has the same value or differs in sign. We will consider these two cases separately.
If the numbers $\alpha$, $\beta$, $\gamma$ are equal, formulas (2) give the system of equations
\[ \alpha = \frac{b - c}{a} \]
\[ \alpha = \frac{a - c}{b} \]
\[ \alpha = \frac{a - b}{c} \]
Adding them side by side, we get: $\alpha(a + b + c) = 0$. Suppose $\alpha \neq 0$; then $a + b + c = 0$. Substituting $c = -(a + b)$ into the first two equations of system (3), we get the relationships
\[ \alpha = \frac{b + (a + b)}{a} = \frac{a + 2b}{a} \]
\[ \alpha = \frac{a + (a + b)}{b} = \frac{2a + b}{b} \]
which, when multiplied side by side, give the equality $4ab = (1 - \alpha^2)ab$, or $(3 + \alpha^2)ab = 0$. This is a contradiction since $a \neq 0$ and $b \neq 0$. Therefore, in this case, $\alpha = \beta = \gamma = 0$.
The case where two of the numbers $\alpha$, $\beta$, $\gamma$ are equal and the third differs in sign remains to be considered.
The notation system in formulas (2) is cyclic; that is, a cyclic change of variables ($a \mapsto b \mapsto c \mapsto a$) leads to an analogous change in values ($\alpha \mapsto \beta \mapsto \gamma \mapsto \alpha$). We can therefore assume without loss of generality that the numbers $\alpha$ and $\beta$ are equal, and thus $\alpha = \beta = -\gamma$. This time, formulas (2) give the system of equations
\[ \alpha = \frac{b - c}{a} \]
\[ \alpha = \frac{a - c}{b} \]
\[ -\alpha = \frac{a - b}{c} \]
and adding them side by side: $\alpha(a + b - c) = 0$. Reasoning as in the previous case, suppose $\alpha \neq 0$; then $a + b - c = 0$. Substituting $c = a + b$ into the first two equations of system (4), we get the relationships
\[ \alpha = \frac{b - (a + b)}{a} = \frac{-a}{a} = -1 \]
\[ \alpha = \frac{a - (a + b)}{b} = \frac{-b}{b} = -1 \]
which cannot be reconciled (since $a \neq 0$ and $b \neq 0$). Therefore, in this case as well, we conclude that $\alpha = \beta = \gamma = 0$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,033 |
XXVII OM - I - Problem 9
In a circle of radius 1, a triangle with side lengths $a, b, c$ is inscribed. Prove that the triangle is acute if and only if $a^2 + b^2 + c^2 > 8$, right-angled if and only if $a^2 + b^2 + c^2 = 8$, and obtuse if and only if $a^2 + b^2 + c^2 < 8$.
|
From the Law of Sines, we have $ \displaystyle 2 = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} $, where $ \alpha $, $ \beta $, $ \gamma $ are the angles of the considered triangle opposite to the sides of lengths $ a $, $ b $, $ c $, respectively. Therefore,
From the Law of Cosines, we have $ a^2 + b^2 = c^2 + 2ab \cos \gamma $. Thus, by (1) we obtain
Since $ \gamma = \pi - (\alpha + \beta) $, we have $ \cos \gamma = -\cos (\alpha + \beta) = \sin \alpha \sin \beta - \cos \alpha \cos \beta $. It follows that
If the angles $ \alpha $, $ \beta $, $ \gamma $ are acute, then the numbers $ \cos \alpha $, $ \cos \beta $, $ \cos \gamma $ are positive; if one of the angles $ \alpha $, $ \beta $, $ \gamma $ is a right angle, then one of these numbers is zero; if one of the angles $ \alpha $, $ \beta $, $ \gamma $ is obtuse, then one of these numbers is negative. Therefore, the number $ \cos \alpha \cos \beta \cos \gamma $ is positive (zero, negative) if and only if the given triangle is acute (respectively: right or obtuse). Hence, from (2) the thesis of the problem follows.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,034 |
XII OM - II - Task 1
Prove that no number of the form $ 2^n $, where $ n $ is a natural number, is the sum of two or more consecutive natural numbers.
|
Suppose the statement made in the text of the problem is not true and that the equality holds
where $ k $, $ r $, $ n $ denote natural numbers. According to the formula for the sum of terms of an arithmetic progression, we obtain from this
Each of the natural numbers $ 2k + r $ and $ r + 1 $ is greater than $ 1 $. The difference between these numbers $ (2k + r) - (r + 1) = 2k - 1 $ is an odd number, so one of them is odd. The left side of equation (1) is therefore divisible by an odd number greater than one, while the right side of the equation, being a power of $ 2 $, does not have such a divisor. Equation (1) is thus impossible.
Note. Every integer is the sum of consecutive integers, for example :
Please note that the translation preserves the original text's line breaks and formatting.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,035 |
XXX OM - II - Task 5
Prove that among any ten consecutive natural numbers there exists one that is relatively prime to each of the other nine.
|
Every common divisor of two natural numbers is also a divisor of their difference. Therefore, if among ten consecutive natural numbers, some two are not relatively prime, then they have a common divisor which is a prime number less than $10$, i.e., one of the numbers $2$, $3$, $5$, $7$.
Among ten consecutive natural numbers, there are five odd numbers. Among these five odd numbers, at most two are divisible by $3$, one is divisible by $5$, and at most one is divisible by $7$. Therefore, at least one number remains that is not divisible by any of the numbers $2$, $3$, $5$, $7$. Thus, it is relatively prime to each of the other nine.
Note. A similar problem was a contest problem 14(2) at the second stage of the V Mathematical Olympiad.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,037 |
XLIII OM - III - Problem 5
The base of a regular pyramid is a $2n$-sided regular polygon $A_1, A_2, \ldots, A_n$. A sphere passing through the vertex $S$ of the pyramid intersects the lateral edges $SA_i$ at points $B_i$ ($i = 1,2,\ldots, 2n$). Prove that
|
om43_3r_img_12.jpg
Let's adopt the following notation:
(Figure 12). It is tacitly assumed in the problem that $ B_i $ is the point of intersection of the given sphere with the edge $ SA_i $, different from $ S $. Therefore, $ x_i > 0 $.
Let the center of the sphere be denoted by $ O $. The following vector equalities hold:
We square both sides of (2) (in the sense of the dot product, denoted here by a bold dot):
which means
Since points $ S $ and $ B_i $ lie on the sphere with center $ O $, we have $ |OB_i| = |OS| $, which allows us to rewrite equality (3) as
From this, taking into account (1), we have
and dividing both sides by $ x_i $:
Let $ Q $ be the center of the circle circumscribed around the polygon $ A_1A_2 \ldots A_{2n} $. Notice that the polygons $ A_1A_3\ldots A_{2n-1} $ and $ A_2A_4\ldots A_{2n} $ are regular. Therefore,
In this case,
From (4) and the above, we have:
The sums (5) and (6) turned out to be equal to the same number. The equality of these sums is precisely the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,038 |
XXXVII OM - I - Problem 10
Prove that the sequence $ (\sin n) $ does not have a limit.
|
Let's consider sequences of open intervals
($ k = 1, 2, 3, \ldots $). The length of each of them (equal to $ \pi/2 $) is greater than $ 1 $. Therefore, in each interval $ I_k $ there is a natural number $ m_k $ (if there are two, let's choose one of them, for example the smaller one, and denote it by $ m_k $). Similarly, in each interval $ J_k $ we will find a natural number $ n_k $. In this way, we have defined two increasing sequences of natural numbers $ (m_k) $ and $ (n_k) $. On each interval $ I_k $, the sine function takes values less than $ -\sqrt{1/2} $, and on each interval $ J_k $, values greater than $ \sqrt{1/2} $. Therefore
skąd wynika, że nie istnieje granica ciągu $ (\sin n) $.
(Note: The last sentence is in Polish. The English translation of the last sentence is: "It follows that the limit of the sequence $ (\sin n) $ does not exist.")
|
proof
|
Calculus
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,041 |
XXI OM - II - Problem 3
Prove the theorem: There does not exist a natural number $ n > 1 $ such that the number $ 2^n - 1 $ is divisible by $ n $.
|
\spos{1} Suppose that $ n $ is the smallest natural number greater than one such that $ n \mid 2^n - 1 $, and $ k $ is the smallest natural number such that $ n \mid 2^k - 1 $. By the theorem given to prove in problem 9 of the first stage, we have $ k < n $.
Let $ n = km + r $, where $ 0 \leq r < k $. Since $ n \mid 2^n - 1 $ and $ n \mid 2^{km} - 1 $, it follows that $ n \mid 2^n - 2^{km} = 2^{km} (2^r - 1) $, and given the oddness of $ n $, we obtain that $ n \mid 2^r - 1 $. Hence, by the definition of $ k $ and the inequality $ r < k $, it follows that $ r = 0 $. Therefore, $ k \mid n $, $ k \mid 2^k - 1 $, and by the definition of $ n $ and the inequality $ k < n $, we have $ k = 1 $. Thus, $ n \mid 2^1 - 1 $, $ n = 1 $, and we obtain a contradiction.
%Sposób 2. Let $ n $ be the smallest natural number greater than one such that $ n \mid 2^n - 1 $, and $ k $ be any natural number such that $ n \mid 2^k - 1 $ and $ k < n $. Let $ d $ be the greatest common divisor of $ n $ and $ k $. By the theorem given to prove in the preparatory problem B of series II, there exist natural numbers $ a $ and $ b $ such that $ an - bk = d $. Since $ n \mid 2^{an} - 1 $ and $ n \mid 2^{bk} - 1 $, it follows that $ n \mid 2^{an} - 2^{bk} = 2^{bk}(2^{an - bk} - 1) $.
Thus, $ n \mid 2^d - 1 $ and $ n = 1 $. We obtain a contradiction.
%Sposób 3. Let $ n > 1 $ be any natural number such that $ n \mid 2^n - 1 $, and $ p $ be its smallest prime divisor. By the theorem given to prove in problem 9 of the first stage or by Fermat's Little Theorem, we have $ p \mid 2^k - 1 $, where $ k $ is some natural number less than $ p $. From the definition of $ p $, it follows that $ k $ and $ n $ are coprime. Therefore, by the theorem given to prove in the preparatory problem B of series II, there exist natural numbers $ a $ and $ b $ such that $ an - bk = 1 $. Since $ p \mid 2^{an} - 1 $ and $ p \mid 2^{bk} - 1 $, it follows that $ p \mid 2^{an} - 2^{bk} = 2^{bk}(2^{an - bk} - 1) $, which implies $ p \mid 2 - 1 $, so $ p = 2 $. We have obtained a contradiction, since $ n $ is an odd number.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,042 |
XX OM - III - Task 6
Given a set of $ n $ points in the plane not lying on a single line. Prove that there exists a circle passing through at least three of these points, inside which there are no other points of the set.
|
\spos{1} Let us choose from the given set of points $ Z = \{A_1, A_2, \ldots, A_n \} $ two points whose distance is the smallest; suppose these are points $ A_1 $ and $ A_2 $. In the circle $ K $ with diameter $ A_1A_2 $, there is then no other point of the set $ Z $ different from $ A_1 $ and $ A_2 $.
Consider the set $ \Omega $ of all circles, each of which passes through points $ A_1 $ and $ A_2 $ and through some other point of the set $ Z $. Since the points of the set $ Z $ do not lie on a single straight line, the set $ \Omega $ is not empty. It is a finite set, so there exists a circle $ O_S $ with center $ S $ in it, whose radius has the smallest length (Fig. 16).
Inside the circle $ O_S $, there is no point of the set $ Z $. Indeed, if a point $ A_k $ of the set $ Z $ does not lie on the circle $ O_S $ or on the line $ A_1A_2 $ outside $ O_S $, then it lies on some circle $ O_T \in \Omega $ with center $ T $ different from $ S $, whose radius is not smaller than the radius of the circle $ O_S $. The circle $ O_S $ divides the circle $ O_T $ into two arcs with endpoints $ A_1 $, $ A_2 $, of which the larger $ L_1 $ lies (except for the endpoints) outside the circle $ O_S $, and the smaller $ L_2 $ lies inside $ O_S $. The arc $ L_2 $ is smaller than a semicircle, so it lies within the circle $ K $. Therefore, the point $ A_k $ belongs to the arc $ L_1 $.
The circle $ O_S $ is thus the circle whose existence needed to be proven.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,043 |
XLIX OM - III - Problem 3
The convex pentagon $ABCDE$ is the base of the pyramid $ABCDES$. A plane intersects the edges $SA$, $SB$, $SC$, $SD$, $SE$ at points $A'$, $B'$, $C'$, $D'$, $E'$ (different from the vertices of the pyramid). Prove that the points of intersection of the diagonals of the quadrilaterals $ABB'A'$, $BCC'B'$, $CDD'C'$, $DEE'D'$, $EAA'E'$ lie on the same plane.
|
Let $ \pi $ and $ \pi' $ be the planes containing the points $ A $, $ B $, $ C $, $ D $, $ E $ and $ A $, $ B $, $ C $, $ D' $, $ E $, respectively. Furthermore, let $ K $, $ L $, $ M $, $ N $, $ O $ be the points of intersection of the diagonals of the quadrilaterals $ ABB' $, $ BCC' $, $ CDD' $, $ DEE' $, $ EAA' $ (respectively).
First, assume that the planes $ \pi $ and $ \pi' $ are parallel. In this case, the ratio of the distances of these planes from each of the points $ K $, $ L $, $ M $, $ N $, $ O $ is the same. Therefore, these points lie on a single plane, parallel to the planes $ \pi $ and $ \pi' $.
Now, assume that the planes $ \pi $ and $ \pi' $ are not parallel. Let $ \ell $ be their common line. Let $ \sigma $ be the plane containing the line $ \ell $ and passing through the point $ K $. Since the points $ A $, $ C $, $ A' $, $ C' $ are coplanar, the lines $ AC $ and $ A'C' $ are either parallel or intersect at a point $ X $ on the line $ \ell $.
In the first case, both the line $ KL $ (the intersection of the planes $ AB $ and $ A'B' $) and the line $ \ell $ (the intersection of the planes $ ABC $ and $ A'B'C' $) are parallel to the lines $ AC $ and $ A'C' $. Therefore, the lines $ KL $ and $ \ell $ are parallel.
In the second case, the points $ K $, $ L $, $ X $ belong to both planes $ AB $ and $ A'B' $. Hence, these points lie on a single line, meaning that the lines $ KL $ and $ \ell $ have a common point. Thus, in both cases, the point $ L $ lies on the plane $ \sigma $.
Considering the planes $ BC $ and $ B'C' $ and using the fact that $ L \in \sigma $, by reasoning analogous to the above, we prove that the point $ M $ lies on the plane $ \sigma $. Repeating this reasoning twice more, we find that the points $ N $, $ O $ also lie on the plane $ \sigma $. Therefore, the points $ K $, $ L $, $ M $, $ N $, $ O $ lie on the plane $ \sigma $, which completes the proof.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,044 |
XL OM - II - Task 6
In triangle $ABC$, through an internal point $P$, lines $CP$, $AP$, $BP$ are drawn intersecting sides $AB$, $BC$, $CA$ at points $K$, $L$, $M$ respectively. Prove that if circles can be inscribed in quadrilaterals $AKPM$ and $KBLP$, then a circle can also be inscribed in quadrilateral $LCMP$.
|
For a circle inscribed in a quadrilateral (convex or concave), we understand a circle contained within this quadrilateral and tangent to the four lines containing its sides. A necessary and sufficient condition for the existence of such a circle is the equality of the sums of the lengths of opposite sides of the quadrilateral. This fact, well known for convex quadrilaterals, is less commonly known for concave quadrilaterals. However, the proof for the case of a concave quadrilateral is obtained through a simple adaptation of the usual "textbook" proof for the case of a convex quadrilateral. We will provide it at the end of the solution, in the Remark.
The thesis of our problem is a direct conclusion from the given property: the assumption of the problem means that there are circles inscribed in the concave quadrilaterals $PCAB$ and $PABC$. Therefore (Figure 3)
om40_2r_img_3.jpg
Thus
Hence
therefore, there exists a circle inscribed in the concave quadrilateral $PBCA$ - that is, a circle inscribed in the convex quadrilateral $LCMP$.
Remark. Here is the proof of the theorem stating that a circle can be inscribed in a concave quadrilateral if and only if the sums of the lengths of opposite sides are equal.
Let $PABC$ be a concave quadrilateral; assume that point $P$ lies inside triangle $ABC$. The extensions of sides $AP$ and $CP$ intersect sides $BC$ and $AB$ at points $L$ and $K$, respectively.
Suppose there exists a circle tangent to the lines containing the sides of quadrilateral $PABC$. This is, more simply, a circle inscribed in the convex quadrilateral $PKBL$. Let $T$, $U$, $V$, $W$ be the points of tangency of this circle with sides $PK$, $KB$, $BL$, $LP$, respectively (Figure 4).
om40_2r_img_4.jpg
The following pairs of tangent segments to the considered circle have equal lengths:
and
We subtract the two equalities (1) side by side:
We add the equalities (2) side by side:
We then add the obtained equalities side by side and get:
which is the equality we needed to prove.
It remains to prove the converse implication: assuming that equality (3) holds, we need to show that a circle can be inscribed in the concave quadrilateral $PABC$ - that is, that a circle can be inscribed in the convex quadrilateral $PKBL$.
First, we inscribe a circle $\Omega$ in triangle $ABL$. We draw a tangent line from point $C$ to this circle (different from line $CB$), intersecting segment $AB$ at point $K$. Suppose point $K$ does not coincide with $K$. Let $P$ be the intersection of lines $AL$ and $CK$; a (non-degenerate) triangle $CPP$ is formed (Figure 5 shows two possible configurations).
om40_2r_img_5.jpg
Circle $\Omega$ is inscribed in the concave quadrilateral $PABC$. By the already proven part of the theorem, the equality analogous to (3) holds in this quadrilateral:
We subtract this equality from (3) and conclude that
The left side of the last equality is - up to sign - the length of segment $PP$. We thus obtain the following relationship between the lengths of the sides of triangle $CPP$:
In a triangle that does not degenerate into a segment, such an equality cannot hold. The obtained contradiction is a result of the assumption that $K$ does not coincide with $K$. Therefore, point $K$ must coincide with $K$. This means that circle $\Omega$ is inscribed in quadrilateral $PKBL$; the proof is complete.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,045 |
XXVI - II - Problem 5
Prove that if a sphere can be inscribed in a convex polyhedron and each face of the polyhedron can be painted in one of two colors such that any two faces sharing an edge are of different colors, then the sum of the areas of the faces of one color is equal to the sum of the areas of the faces of the other color.
|
Connecting the point of tangency of the sphere with any wall with all vertices belonging to that wall, we obtain a decomposition of that wall into a sum of triangles with disjoint interiors, one of whose vertices is the point of tangency of the sphere, and the others are two vertices of the polyhedron belonging to one of its edges. The triangles $PAB$ and $QAB$, located on different walls of the polyhedron, which have a common side $\overline{AB}$ being an edge of the polyhedron, are painted in different colors. We will prove that they have equal areas.
The plane $\pi$ passing through the center of the sphere and the points of tangency $P$ and $Q$ is perpendicular to the planes containing the considered triangles. Therefore, the plane $\pi$ is perpendicular to the line $AB$.
Let the line $AB$ intersect the plane $\pi$ at point $C$. Then $CP = CQ$, since the segments tangent to the sphere drawn from a fixed point are equal. From the definition of point $C$, it also follows that $\overline{CP} \bot \overline{AB}$ and $\overline{CQ} \bot \overline{AB}$, i.e., segments $\overline{CP}$ and $\overline{CQ}$ are heights of triangles $ABP$ and $ABQ$ respectively. Since these triangles have a common base, it follows that they have equal areas.
All walls of the polyhedron have been represented as a sum of triangles. Each triangle painted in one color has been uniquely associated with a triangle of the same area painted in another color. This leads to the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,047 |
XV OM - I - Problem 3
Prove that the points symmetric to the intersection point of the altitudes of triangle $ABC$ with respect to the lines $AB$, $BC$, and $CA$ lie on the circumcircle of triangle $ABC$.
|
Let $S$ be the intersection point of the altitudes $AH$ and $BK$ of triangle $ABC$, and let $S'$ be the point symmetric to $S$ with respect to the line $AB$. We will consider three cases:
a) Angles $A$ and $B$ of the triangle are acute (Fig. 1). The intersection point $S$ of the lines $AH$ and $BK$ then lies on the same side of the line $AB$ as point $C$. If angle $ACB$ is a right angle, then point $S$ coincides with point $C$, so $\measuredangle ASB = 90^\circ$, and points $S$ and $C$ lie on the circle with diameter $AB$. If $\measuredangle ACB \ne 90^\circ$, then points $C$, $H$, $S$, and $K$ form a quadrilateral in which angles $H$ and $K$ are right angles, so $\measuredangle HSK = 180^\circ - \measuredangle C$, and since $\measuredangle ASB = 180^\circ - \measuredangle C$, we have $\measuredangle ASB = \measuredangle HSK$. Since point $S$ lies on the opposite side of the line $AB$ from points $S$ and $C$, it follows from the last equality that $S$ lies on the circle passing through $A$, $B$, and $C$.
b) One of the angles of the triangle, for example, angle $A$, is obtuse (Fig. 2). The lines $AH$ and $BK$ intersect at point $S$ lying on the opposite side of the line $AB$ from point $C$; point $S$ therefore lies on the same side of the line $AB$ as point $C$.
Right triangles $AKS$ and $AHC$ have equal angles at vertex $A$, so their third angles are also equal, i.e., $\measuredangle ASK = \measuredangle ACB$; but $\measuredangle ASB = 180^\circ - \measuredangle ASK$, so $\measuredangle ASB = 180^\circ - \measuredangle ACB$, and it follows that point $S$ lies on the circle passing through $A$, $B$, and $C$.
c) One of the angles $A$ and $B$, for example, $\measuredangle A$, is a right angle. In this case, points $S$ and $S'$ coincide with point $A$; the theorem is obvious.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,052 |
XXXI - I - Problem 4
Prove that if a polynomial with integer coefficients takes the value 1 for four different integer arguments, then for no integer argument does it take the value -1.
|
Integers $ a $, $ b $, $ c $, $ d $, for which the given polynomial $ w(x) $ takes the value $ 1 $ are roots of the polynomial $ w(x) - 1 $. This latter polynomial is therefore divisible by $ x-a $, $ x-b $, $ x-c $, $ x-d $, and thus $ w(x) - 1 = (x - a) (x - b) (x - c) (x - d) \cdot v (x) $, where $ v (x) $ is some polynomial with integer coefficients. If there were $ w(k) = -1 $ for some integer $ k $, then $ w(k) - 1 = -2 $, so $ (k - a) (k - b) (k - c) (k - d) \cdot v(k) = -2 $. The last equality represents the number $ -2 $ as a product of integers $ k - a $, $ k - b $, $ k - c $, $ k - d $, $ v(k) $, among which the first four are pairwise distinct. Therefore, at most one of the numbers $ k-a $, $ k-b $, $ k-c $, $ k-d $ is equal to $ 1 $, at most one of them is equal to $ -1 $, and at least two have an absolute value greater than $ 1 $. This, however, leads to a contradiction with the fact that $ 2 $ is a prime number. Therefore, there cannot be $ w(k) = -1 $ for any integer $ k $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,054 |
XLIII OM - III - Problem 1
Segments $ AC $ and $ BD $ intersect at point $ P $, such that $ |PA|=|PD| $, $ |PB|=|PC| $. Let $ O $ be the center of the circumcircle of triangle $ PAB $. Prove that the lines $ OP $ and $ CD $ are perpendicular.
|
om43_3r_img_11.jpg
Let line $ l $ be the bisector of angles $ APD $ and $ BPC $, and let line $ m $ be the bisector of angles $ APB $ and $ CPD $. Lines $ l $ and $ m $ are perpendicular. From the given equalities of segments, it follows that triangle $ PCD $ is the image of triangle $ PBA $ under axial symmetry with respect to line $ l $. Let $ Q $ be the point of intersection of line $ m $ with the circumcircle of triangle $ PAB $ (different from $ P $). Since line $ m $ is the bisector of angle $ APB $, the inscribed angles $ BPQ $ and $ APQ $ are equal (congruent). Therefore, the corresponding central angles $ BOQ $ and $ AOQ $ are also equal. Thus, line $ OQ $ is the axis of symmetry of the isosceles triangle $ AOB $; it is therefore perpendicular to the chord $ AB $.
The axis of symmetry of the isosceles triangle $ POQ $ is line $ l $, which is perpendicular to $ m $, so it is parallel to $ l $. Segment $ OP $ is symmetric to $ OQ $ with respect to line $ l $; segment $ CD $ is symmetric to $ AB $ with respect to line $ l $. From the relationships: $ l || l $ and $ OQ \bot AB $, it follows that $ OP \bot CD $ - which is the thesis of the problem. (Figure 11 shows the situation when triangle $ PAB $ is acute, i.e., when point $ O $ lies inside it; the reasoning, however, does not require any changes when one of the angles of triangle $ PAB $ is obtuse.)
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,056 |
LV OM - III - Task 1
Point $ D $ lies on side $ AB $ of triangle $ ABC $. Circles tangent to lines $ AC $ and $ BC $ at points $ A $ and $ B $, respectively, pass through point $ D $ and intersect again at point $ E $. Let $ F $ be the point symmetric to point $ C $ with respect to the perpendicular bisector of segment $ AB $. Prove that points $ D $, $ E $, and $ F $ lie on the same line.
|
Suppose that point $ E $ lies inside triangle $ ABC $. Then
Adding these equalities side by side, we have $ 180^\circ > \measuredangle AEB = 180^\circ + \measuredangle ACB > 180^\circ $. The obtained contradiction proves that points $ C $ and $ E $ lie on opposite sides of line $ AB $ (Fig. 1).
From the equality $ \measuredangle AEB = \measuredangle AED+\measuredangle DEB = \measuredangle BAC+\measuredangle ABC = 180^\circ - \measuredangle ACB $, it follows that points $ A $, $ E $, $ B $, $ C $ lie, in this exact order, on the same circle $ o $. Hence, it follows that point $ F $ also lies on circle $ o $, i.e.,
This equality means that points $ D $, $ E $, and $ F $ lie on the same line.
om55_3r_img_1.jpg
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,057 |
XVIII OM - I - Problem 11
Three circles of the same radius intersect pairwise and have one common point. Prove that the remaining three intersection points of the circles lie on a circle of the same radius.
|
We introduce the notation: $ O_1, O_2, O_3 $ - the centers of the given circles, $ M $ the common point of all three circles, $ A_1, A_2, A_3 $ different from $ M $ are the points of intersection of the circles with centers $ O_2 $ and $ O_3 $, $ O_3 $ and $ O_1 $, $ O_1 $ and $ O_2 $ (Fig. 4).
All sides of the quadrilateral $ O_1A_2O_3M $ are equal (as radii of the given circles), so this quadrilateral is a rhombus. It follows that the vectors $ \overrightarrow{O_1A_2} $ and $ \overrightarrow{MO_3} $ are equal. Similarly, we conclude that $ \overrightarrow{O_2A_1} = \overrightarrow{MO_3} $, hence $ \overrightarrow{O_1A_2} = \overrightarrow{O_2A_1} $. The quadrilateral $ O_1A_2A_1O_2 $ is therefore a parallelogram, so it has opposite sides equal, $ A_1A_2 = O_1O_2 $. Similarly, $ A_2A_3 = O_2O_3 $ and $ A_3A_1 = O_3O_1 $, so the triangles $ A_1A_2A_3 $ and $ O_1O_2O_3 $ are congruent. Since the radius of the circumcircle of triangle $ O_1O_2O_3 $ equals the radius of the given circles, the same applies to triangle $ A_1A_2A_3 $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,058 |
XXX OM - II - Task 6
On the side $ \overline{DC} $ of rectangle $ ABCD $, where $ \frac{AB}{AD} = \sqrt{2} $, a semicircle is constructed externally. Any point $ M $ on the semicircle is connected to $ A $ and $ B $ with segments, intersecting $ \overline{DC} $ at points $ K $ and $ L $, respectively. Prove that $ DL^2 + KC^2 = AB^2 $.
|
Let $ S $ and $ T $ be the points of intersection of the line $ AB $ with the lines $ MD $ and $ MC $, respectively (Fig. 14). Let us choose a point $ P \in AB $ such that $ DP || CT $. Then $ \measuredangle SDP = \measuredangle DMC = \frac{\pi}{2} $ as an inscribed angle in a circle subtended by a diameter. We obviously have $ AP = BT $ and therefore
By the theorem of Thales, under a certain homothety with center at point $ M $, the points $ D $, $ K $, $ L $, $ C $ map to the points $ S $, $ A $, $ B $, $ T $, respectively. Thus
where $ \lambda $ is the ratio of this homothety. From this, we obtain
Therefore, to solve the problem, it suffices to show that
Since $ SB = SA + AB $, $ AT = AB + BT $, $ ST = SA + AB + BT $, equality (3) is equivalent to
or, after transformations, $ AB^2 = 2 \cdot SA \cdot BT $. This last equality follows from (2). Therefore, equality (3) is true.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,059 |
LII OM - III - Problem 5
Points $ K $ and $ L $ lie on sides $ BC $ and $ CD $ of parallelogram $ ABCD $, respectively, such that $ BK \cdot AD = DL \cdot AB $. Segments $ DK $ and $ BL $ intersect at point P. Prove that $ \measuredangle DAP = \measuredangle BAC $.
|
Let $ Y $ be the intersection point of lines $ AP $ and $ CD $, and let $ X $ be the intersection point of lines $ DK $ and $ AB $ (Fig. $ 1 $ and $ 2 $). Then
Hence, and from the equality given in the problem, we obtain:
om52_3r_img_1.jpg
om52_3r_img_2.jpg
This proves that triangles $ ABC $ and $ ADY $ are similar, from which the thesis follows.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,060 |
XXXIV OM - I - Problem 10
Do there exist three points in the plane with coordinates of the form $ a+b\sqrt[3]{2} $, where $ a, b $ are rational numbers, such that at least one of the distances from any point in the plane to each of these points is an irrational number?
|
Such points do exist. We will show that, for example, the triplet of points $ A = (\sqrt[3]{2},0) $, $ B=(-\sqrt[3]{2},0) $, $ C = (0,0) $ has the desired property. Suppose, for instance, that the distance from some point $ P=(x,y) $ to each of the points $ A $, $ B $, $ C $ is a rational number. In this case, the numbers $ PA^2 = (x-\sqrt[3]{2})^2+y^2 $, $ PB^2 = (x+\sqrt[3]{2})^2+y^2 $, $ PC^2 = x^2+y^2 $ are also rational, and thus the number $ PA^2+PB^2-2PC^2 = (x^2-2\sqrt[3]{2}x+\sqrt[3]{4}+y^2) + (x^2+2\sqrt[3]{2}x+\sqrt[3]{4}+y^2)-2(x^2+y^2) = 2\sqrt[3]{4} $ would have to be rational - which is not true. The obtained contradiction proves that at least one of the distances $ PA $, $ PB $, $ PC $ is an irrational number.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,062 |
XLII OM - II - Problem 3
Given are positive integers $ a $, $ b $, $ c $, $ d $, $ e $, $ f $ such that $ a+b = c+d = e+f = 101 $. Prove that the number $ \frac{ace}{bdf} $ cannot be expressed in the form of a fraction $ \frac{m}{n} $ where $ m $, $ n $ are positive integers with a sum less than $ 101 $.
|
From the assumption of the task, the following congruences hold:
We multiply these three relations side by side:
Suppose, contrary to the thesis, that
where $ m $, $ n $ are positive integers such that $ m + n < 101 $. We then have the relationship
which means
This implies that the number $ 101 $ is a divisor of the product $ (m + n)bdf $. As a prime number, it must divide one of the factors. We have reached a contradiction, because each of the numbers $ b $, $ d $, $ f $, $ m + n $ is positive and less than $ 101 $. The contradiction completes the proof.
Note 1. In the proven theorem, the number $ 101 $ can, of course, be replaced by any prime number, and instead of three fractions, we can consider any odd number of fractions.
Note 2. The use of the language and properties of congruences is convenient in the solution, but not necessary; they were useful for justifying the divisibility of the product $ (m + n)bdf $ by $ 101 $.
This conclusion can, without using the concept of congruences, be obtained as follows: according to the assumption, $ a = p - b $, $ c = p - d $, $ e = p - f $ (we reason at once for any prime number $ p $, in accordance with Note 1). Therefore,
Assuming, as before, that $ mbdf = nace $, we have
The right side of this equality is a number divisible by $ p $, so the left side is also.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,063 |
XXVIII - II - Task 2
Let $ X $ be an internal point of triangle $ ABC $. Prove that the product of the distances from point $ X $ to the vertices $ A, B, C $ is at least eight times greater than the product of the distances from this point to the lines $ AB, BC, CA $.
|
Let $ h_a $, $ h_b $, $ h_c $ be the lengths of the altitudes of triangle $ ABC $ drawn from vertices $ A $, $ B $, $ C $ respectively, $ x_a $, $ x_b $, $ x_c $ - the distances from point $ X $ to the lines $ BC $, $ CA $, $ AB $ respectively, and $ S_a $, $ S_b $, $ S_c $ - the areas of triangles $ XBC $, $ XCA $, $ XAB $. Finally, let $ S $ be the area of triangle $ ABC $.
Of course, we have $ XA + x_a \geq h_a $ (Fig. 6) and therefore
Similarly, we prove that
To solve the problem, it is sufficient to prove that
Multiplying both sides of inequality (3) by $ S_aS_bS_c $, and moving all terms to the left side, we arrive at an equivalent inequality
This last inequality is obviously true. Therefore, inequality (3) is also true. Hence, by (1) and (2), we have
Note. It is not difficult to prove that in formula (5), equality holds if and only if $ ABC $ is an equilateral triangle and $ X $ is its centroid.
Indeed, if $ X $ is the centroid of the equilateral triangle $ ABC $, then $ XA = 2x_a $, $ XB = 2x_b $, $ XC = 2x_c $. Therefore, equality holds in formula (5).
Conversely, if equality holds in formula (5), then equality holds in each of formulas (1) - (4). In particular, from (4) it follows that $ S_a = S_b = S_c $, and therefore from (1) and (2) - that $ XA = 2x_a $, $ XB = 2x_b $, $ XC = 2x_c $. From (1) it also follows that $ XA = h_a - x_a $, i.e., point $ X $ lies on the altitude of triangle $ ABC $ drawn from vertex $ A $ and divides it in the ratio $ 2 \colon 1 $.
Let $ A $ be the point of intersection of lines $ AX $ and $ BC $ (Fig. 7). Then
Thus, $ A $, i.e., $ AA $ is a median of triangle $ ABC $.
Similarly, we prove that the other altitudes of triangle $ ABC $ are its medians. Therefore, this triangle is equilateral.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,064 |
VI OM - I - Task 4
Prove that if there exists a sphere tangent to all edges of a tetrahedron, then the sums of the opposite edges of the tetrahedron are equal, and that the converse theorem is also true.
|
a) We will show that if there exists a sphere tangent to all the edges of the tetrahedron $ABCD$, then $AB + CD = AC + BD = AD + BC$. Let $O$ denote the center, and $r$ the radius of such a sphere. The points of tangency of the sphere with the edges of the tetrahedron divide each edge into two segments, with each of the 12 segments emanating from the same vertex of the tetrahedron being equal; for example, each of the three segments emanating from point $A$ has a length of $\sqrt{AO^2 - r^2}$. Let the lengths of the segments emanating from vertices $A$, $B$, $C$, $D$ be denoted by $x$, $y$, $z$, $u$ respectively. Then $AB = x + y$, $CD = z + u$, so $AB + CD = x + y + z + u$. Similarly, $AC + BD = x + z + y + u$, $AD + BC = x + u + y + z$. Indeed, the equality $AB + CD = AC + BD = AD + BC$ holds.
b) We will show that if in the tetrahedron $ABCD$ it is true that $AB + CD = AC + BD = AD + BC$, then there exists a sphere tangent to all the edges of the tetrahedron (Fig. 2). Consider the circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ inscribed in triangles $BCD$ and $ACD$ respectively. Let circle $C_1$ be tangent to edge $CD$ at point $M$, and circle $C_2$ at point $M$. Applying the known formulas for the lengths of segments determined on the sides of a triangle by the inscribed circle to triangles $BCD$ and $ACD$ we have
thus
from which it follows that points $M$ and $M$ coincide, i.e., circles $C_1$ and $C_2$ are tangent to edge $BC$ at the same point $M$. Therefore, there exists a sphere $K$ whose surface passes through circles $C_1$ and $C_2$. Indeed, points $O_1$, $O_2$, and $M$ do not lie on the same line, thus they determine a plane $O_1O_2M$ perpendicular to line $CD$, and therefore perpendicular to planes $BCD$ and $ACD$. In the plane $O_1O_2M$ lie perpendiculars to planes $BCD$ and $ACD$ at points $O_1$ and $O_2$ respectively; these perpendiculars are not parallel (since planes $BCD$ and $ACD$ intersect), so they have a common point $O$. Point $O$ has the same distance, equal to $OM$, from all points of circles $C_1$ and $C_2$, so the surface of sphere $K$ with center $O$ and radius $OM$ passes through circles $C_1$ and $C_2$. Sphere $K$ is tangent to edges $BC$, $BD$, $AC$, $AD$, $CD$; we will show that it is also tangent to edge $AB$. The surface of sphere $K$ intersects plane $ABC$ along a circle $C_3$ tangent to edges $BC$ and $AC$ at the same points as circles $C_1$ and $C_2$. The theorem will be proved when we show that circle $C_3$ is the incircle of triangle $ABC$. Denoting the segments determined on edges $BC$, $BD$, $AC$, $AD$, $CD$ by circles $C_1$ and $C_2$ in the same way as in a), we have $BC = y + z$, $BD = y + u$, $AC = x + z$, $AD = x + u$, $CD = z + u$. According to the assumption $AB + CD = AC + BD$, so $AB = AC + BD - CD$, thus $AB = (x + z) + (y + u) - (z + u) = x + y$. The circle $C_4$ inscribed in triangle $ABC$ determines segments of a certain length $z_1$ on sides $BC$ and $AC$ at vertex $C$, where $z_1 = \frac{1}{2} (BC + AC - AB) = \frac{1}{2} [(y + z) + (x + z) - (x + y)] = z$; hence, circle $C_4$ coincides with circle $C_3$ and sphere $K$ is tangent to all the edges of the tetrahedron $ABCD$, q.e.d.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,065 |
XXII OM - I - Problem 8
Given a cube with edge length 1; let $ S $ denote the surface of this cube. Prove that there exists a point $ A \in S $ such that
1) for every point $ B \in S $, there exists a broken line connecting $ A $ and $ B $ with a length not greater than 2 contained in $ S $;
2) there exists a point $ B \in S $ such that no broken line connecting $ A $ and $ B $ contained in $ S $ has a length less than 2.
|
Let $ A $ be the midpoint of one of the faces of a cube, and $ B $ - the midpoint of the opposite face (Fig. 9). We will prove that point $ A $ satisfies condition 1) of the problem, and points $ A $ and $ B $ satisfy condition 2).
1) Let us cut the surface $ S $ of the cube along the broken lines $ BPC $, $ BQD $, $ BRE $, $ BTF $ and unfold it onto a plane. We will then obtain the figure $ S$ as shown in Fig. 10. Since the distances from point $ A $ to $ B, B $ are equal to $ 2 $, the entire figure $ S $ is contained within a circle centered at point $ A $ with a radius of $ 2 $. It follows that any point of the figure $ S $ can be connected to point $ A $ by a segment no longer than $ 2 $. Such a segment will, of course, be contained in $ S $. Transitioning from the unfolding $ S $ of the surface of the cube to the surface of the cube $ S $, each such segment will define a broken line no longer than $ 2 $ contained in $ S $ connecting point $ A $ with any point belonging to $ S $.
2) Any broken line $ \mathcal{L} $ connecting points $ A $ and $ B $ intersects the quadrilateral $ CDEF $ and the quadrilateral $ PQRT $, because each of these quadrilaterals divides the surface of the cube into two parts, and points $ A $ and $ B $ are in different parts. Let $ K $ be the first intersection point of the broken line $ \mathcal{L} $ with the quadrilateral $ CDEF $, and $ L $ - the last intersection point of the broken line $ \mathcal{L} $ with the quadrilateral $ PQRT $. The length of the broken line $ \mathcal{L} $ is not less than the sum of the lengths of segments $ \overline{AK} $, $ \overline{KL} $, and $ \overline{LB} $, because the length of a segment connecting two points is not greater than the length of any broken line connecting these points. The smallest distance from the center of a square to a point on its side is equal to half the length of the side. Therefore, $ AK \geq \frac{1}{2} $ and $ LB \geq \frac{1}{2} $. Points $ K $ and $ L $ belong to the parallel planes $ CDEF $ and $ PQRT $, which are $ 1 $ unit apart. Thus, the distance between these points is not less than the distance between the planes, i.e., $ KL \geq 1 $. Hence, the length of the broken line $ \mathcal{L} \geq AK + KL + LB \geq \frac{1}{2} + 1 + \frac{1}{2} = 2 $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,068 |
XXII OM - III - Task 6
Given a regular tetrahedron with edge length 1. Prove that:
1) On the surface $ S $ of the tetrahedron, there exist four points such that the distance from any point on the surface $ S $ to one of these four points does not exceed $ \frac{1}{2} $.
2) On the surface $ S $, there do not exist three points with a similar property.
The distance between points on $ S $ is understood as the infimum of the lengths of broken lines contained in $ S $ connecting these points.
|
1) First, let us note that in an equilateral triangle with a side length of $1$, the midpoints of two sides have the property that any point belonging to the triangle is no more than $\frac{1}{2}$ away from one of them. Indeed, by drawing circles with centers at these points (Fig. 15) and radii of length $\frac{1}{2}$, we observe that the triangle is contained within the union of these circles.
In the regular tetrahedron $ABCS$, let us choose points $P$, $Q$, $R$, $T$ as the midpoints of the edges $\overline{AS}$, $\overline{BS}$, $\overline{AC}$, $\overline{BC}$ (Fig. 16). In this way, each face of the tetrahedron contains two of the points $P$, $Q$, $R$, $T$, and they are the midpoints of two edges of that face. From the observation made at the beginning, it follows that any point on any face of the tetrahedron is no more than $\frac{1}{2}$ away from one of the points $P$, $Q$, $R$, $T$.
2) Suppose there exist three points $P$, $Q$, $R$ on the surface of the tetrahedron such that any point on the surface of the tetrahedron is no more than $\frac{1}{2}$ away from one of them. Since the tetrahedron has four vertices, at least two vertices are no more than $\frac{1}{2}$ away from the same point among $P$, $Q$, $R$. Let, for example, points $A$ and $S$ be no more than $\frac{1}{2}$ away from $P$. Since the distance between points $A$ and $S$ is $1$, it follows that point $P$ is the midpoint of the edge $\overline{AS}$.
Notice that any point on the height $\overline{SD}$ of the face $SBC$ (except for point $S$) is more than $\frac{1}{2}$ away from point $P$. Indeed, consider the net of the regular tetrahedron $ABCS$ (Fig. 17). The shaded area consists of points no more than $\frac{1}{2}$ away from $P$. The height $\overline{SD}$ has only point $S$ in common with this area.
If, therefore, point $S$ were more than $\frac{1}{2}$ away from each of the points $Q$ and $R$, then any point in a neighborhood of point $S$ would have a similar property. Thus, points on the segment $\overline{SD}$ sufficiently close to point $S$ would be more than $\frac{1}{2}$ away from each of the points $P$, $Q$, $R$. This, however, contradicts our initial assumption. Therefore, point $S$ is no more than $\frac{1}{2}$ away from one of the points $Q$ and $R$, and similarly, point $A$ is no more than $\frac{1}{2}$ away from one of the points $Q$ and $R$. Since vertices $B$ and $C$ are more than $\frac{1}{2}$ away from point $P$, each of the points $B$ and $C$ is no more than $\frac{1}{2}$ away from one of the points $Q$ and $R$. It follows that at least two of the points $A$, $B$, $C$, $S$ are no more than $\frac{1}{2}$ away from one of the points $Q$ and $R$.
Let, for example, points $B$ and $S$ be no more than $\frac{1}{2}$ away from $Q$ (other cases are considered similarly). Then point $Q$ is the midpoint of the edge $\overline{BS}$, and point $R$ is the midpoint of the edge $\overline{AC}$.
However, reasoning similarly as before, we conclude that points on the height $\overline{BE}$ of the face $ABC$ located sufficiently close to point $B$ are more than $\frac{1}{2}$ away from each of the points $P$, $Q$, $R$. The obtained contradiction proves that there do not exist three points $P$, $Q$, $R$ on the surface of the tetrahedron $ABCS$ such that any point on the surface of the tetrahedron is no more than $\frac{1}{2}$ away from one of them.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,073 |
XXXVIII OM - I - Problem 10
Let $ n = 4^k \cdot 5^m $ ($ k $, $ m $ — natural numbers). Prove that every natural number less than $ n $ is a divisor of $ n $ or a sum of different divisors of $ n $.
|
Note. By a natural number, we mean a positive integer in this task. For $ k = 0 $, the thesis is - as easily noticed - false.
Let $ k \geq 1 $ and $ m \geq 1 $ be fixed and let $ s < 4^k5^m $ be a natural number. We divide $ s $ by $ 5^m $ to get the quotient $ a $ and the remainder $ r $:
Let's write the numbers $ q $ and $ r $ in positional systems with bases $ 2 $ and $ 5 $ (respectively):
and consider the following sets of indices:
Then
All the terms of the obtained sum are divisors of the number $ 4^k5^m $, and they are distinct because the sets $ J_c $ are pairwise disjoint. The thesis is thus proven.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,075 |
XXXII - II - Task 3
Prove that there does not exist a continuous function $ f: \mathbb{R} \to \mathbb{R} $ satisfying the Condition $ f(f(x)) = - x $ for every $ x $.
|
The function $ f $ satisfying the given condition must be injective, because if for some $ x_1 $, $ x_2 $ it were $ f(x_1) = f (x_2) $, then $ f(f(x_1)) = f(f(x_2)) $, which implies $ -x_1 = -x_2 $, so $ x_1 = x_2 $. A continuous and injective function must be monotonic. If $ f $ were an increasing function, then $ f(f) $ would also be an increasing function; if $ f $ were a decreasing function, then $ f(f) $ would be increasing (for $ x_1 < x_2 $, it is $ f(x_1) > f(x_2) $, so $ f(f(x_1)) < f(f(x_2)) $). In either case, we obtain a contradiction, because the function $ g(x) = -x $ is a decreasing function. Therefore, there does not exist a continuous function $ f $ satisfying the condition $ f(f(x)) = -x $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,079 |
LX OM - I - Task 8
The diagonals of the base $ABCD$ of the pyramid $ABCDS$ intersect at a right angle at point $H$, which is the foot of the pyramid's height. Let $K$, $L$, $M$, $N$ be the orthogonal projections of point $H$ onto the faces $ABS$, $BCS$, $CDS$, $DAS$, respectively. Prove that the lines $KL$, $MN$, and $AC$ are either parallel or intersect at a single point.
|
At the beginning, we will show that the lines $KL$ and $AC$ lie in the same plane.
From the assumptions of the problem, it follows that the lines $AC$, $BD$, and $HS$ are mutually perpendicular. This means that any line contained in the plane determined by two of these lines is perpendicular to the third. In particular, the line $SB$ is perpendicular to the line $AC$, and therefore there exists a plane $\pi$ passing through the line $AC$ and perpendicular to the line $SB$.
Since the line $SB$ is contained in the planes $ABS$ and $BCS$, the plane $\pi$ is perpendicular to both of these planes.
It follows that the line $HK$, which is perpendicular to the plane $ABS$, is parallel to the plane $\pi$, and given the relation $H \in \pi$, it is contained in this plane. This means that the point $K$ lies in the plane $\pi$. Similarly, we justify that $L \in \pi$.
Therefore, the lines $KL$ and $AC$ are contained in the plane $\pi$ (Fig. 4).
om60_1r_img_4.jpg
Let $G$ denote the point of intersection of the plane $\pi$ and the line $SB$.
Then
The points $K$ and $L$ thus lie on the circle with diameter $GH$,
tangent to the line $AC$.
Notice now that the lines $KL$ and $AC$ are parallel if and only if $KL \perp GH$. But the points $K$ and $L$ lie on the circle with diameter $GH$, so the condition $KL \perp GH$ is equivalent to the fact that triangles $GKH$ and $GLH$ are symmetric with respect to the line $GH$.
This, in turn, occurs if and only if $\measuredangle AGH = \measuredangle CGH$,
which is equivalent to the isosceles nature of triangle $AGC$, and thus
to the fact that point $H$ is the midpoint of segment $AC$.
Assume then that point $H$ is not the midpoint of segment $AC$.
The further part of the reasoning takes place in the plane $\pi$ (Fig. 5).
We will complete the solution of the problem in two ways.
om60_1r_img_5.jpg
Let $E$ denote the point of intersection of the lines $KL$ and $AC$. From the theorem of the secant and the tangent applied to the circle $o$ we then obtain
On the other hand, the similarity of the right triangles
$GKH$ and $GHA$ gives
while the similarity of the right triangles $GLH$ and $GHC$ yields
By comparing the sides of equations (2) and (3), we conclude that
This proves that the points $A, K, L, C$ lie on the same circle, which implies that
Combining now dependencies (1) and (4) we obtain the condition
The point $E$ on the line $AC$ is uniquely determined by equality (5). By introducing a number line structure on the line $AC$ and assuming that the point $X$ corresponds to the number $x$, we write condition (5) in the form
that is, $(a+c-2h)e = ac-h^2$; since the point $H$ is not the midpoint of the segment $AC$, we have $a+c-2h \neq 0$ and as a result, there is exactly one real number $e$ satisfying equation (6).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,080 |
XXIX OM - I - Problem 9
Prove that from any sequence of natural numbers, one can select a subsequence in which any two terms are coprime, or a subsequence in which all terms have a common divisor greater than 1.
|
Let $ p $ be a fixed prime number. If infinitely many terms of the sequence are divisible by $ p $, then the terms divisible by $ p $ form a subsequence, all of whose terms have a common divisor greater than $ 1 $.
Consider, then, the case where for each prime number $ p $ only a finite number of terms of the sequence are divisible by $ p $. We then define inductively a subsequence $ b_1, b_2, \ldots $ of the given sequence as follows. We take $ b_1 $ to be the first term of the given sequence. If the terms $ b_1, b_2, \ldots, b_n $ of the subsequence have already been chosen so that any two of them are relatively prime, then we take $ b_{n+1} $ to be a term of the given sequence that is not divisible by any prime factor of the number $ b_1b_2 \ldots b_n $. Then, of course, the number $ b_{n+1} $ will be relatively prime to each of the numbers $ b_1, b_2, \ldots, b_n $. Therefore, by the principle of induction, there exists an infinite subsequence $ b_1, b_2, \ldots $, any two terms of which are relatively prime.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,081 |
XXV - I - Problem 8
Given a convex polyhedron and a point $ A $ in its interior. Prove that there exists a face $ S $ of this polyhedron such that the orthogonal projection of point $ A $ onto its plane belongs to $ S $.
|
Let $ S $ be such a face of a given polyhedron that the distance $ d $ from point $ A $ to the plane of face $ S $ is not greater than the distance from this point to the plane of any other face of the polyhedron. Let $ k $ be the ray with origin at point $ A $ perpendicular to the plane of face $ S $. If the ray $ k $ does not intersect face $ S $, then it intersects some other face $ T $ of the given polyhedron at point $ P $. From the convexity of the given polyhedron, it follows that points $ A $ and $ P $ belong to the same half-space determined by the plane of face $ S $. Therefore, $ AP < d $, and the distance from point $ A $ to the plane of face $ T $ is even smaller than $ d $. The obtained contradiction proves that the ray $ k $ intersects face $ S $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,085 |
XXIV OM - III - Task 5
Prove that every positive proper fraction $ \frac{m}{n} $ can be represented as a finite sum of the reciprocals of different natural numbers.
|
\spos{1} Let us choose a natural number $ k $ such that $ n \leq 2^k $. Let $ q $ and $ r $ be the quotient and remainder, respectively, of the division of $ 2^km $ by $ n $, i.e., let $ 2^km = qn + r $, where $ 0 \leq r < n $.
Then
We have $ qn \leq gn + r = 2^km < 2^kn $, since $ \displaystyle \frac{m}{n} < 1 $. Hence $ q < 2^k $. Therefore, the binary representation of the number $ q $ has the form
where $ q_i = 0 $ or $ 1 $. Hence
From the definition of the numbers $ r $ and $ k $, we have $ r < n \leq 2^k $. Therefore, the binary representation of the number $ r $ has the form
where $ r_j = 0 $ or $ 1 $.
Hence
For $ j = 0, 1, \ldots, k-1 $, we have
since $ r < n $. Therefore, each non-zero term in sum (3) is smaller than any non-zero term in sum (2). The terms are thus distinct.
From (1), (2), and (3), we obtain a representation of the number $ \displaystyle \frac{m}{n} $ as a sum of distinct fractions of the form $ \displaystyle \frac{1}{t} $, where $ t $ is a natural number.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,086 |
XXXVIII OM - I - Problem 8
The sequence of points $ A_1, A_2, \ldots $ in the plane satisfies for every $ n > 1 $ the following conditions: $ |A_{n+1}A_n| = n $ and the directed angle $ A_{n+1}A_nA_1 $ is $ 90^{\circ} $. Prove that every ray starting at $ A_1 $ intersects infinitely many segments $ A_nA_{n+1} $ ($ n = 1, 2, \ldots $).
|
Let's denote: $ |A_1A_n| = r_n $, $ |\measuredangle A_nA_1A_{n+1}| = a_n $ for $ n = 2, 3, 4, \ldots $, and moreover $ a = r_2^2 - 1 $ (the problem statement does not specify the length of the segment $ A_1A_2 $, so $ a $ can be any number greater than $ -1 $). From the relationships in the right triangle $ A_1A_nA_{n+1} $ we obtain
and
By taking the last formula for $ n = 2, 3, 4, \ldots $ we get:
and generally (by induction)
which means
Let $ P $ be any point on the plane different from $ A_1 $. The position of point $ P $ is fully determined by two quantities: the distance from point $ A_1 $ and the measure of the directed angle $ \measuredangle (\overrightarrow{A_1A_2}, \overrightarrow{A_1P}) $; we will call this measure the angular coordinate of point $ P $. We adopt an unlimited scale of angles; two numbers differing by a multiple of 360° will be considered as measures of the same directed angle.
om38_1r_img_7.jpg
The segments $ A_2A_3 $, $ A_3A_4 $, etc., form the line shown in Figure 7. If $ P $ is a point moving along this line in the direction determined by the numbering of points $ A_n $, then its distance from $ A_1 $ increases (because the hypotenuse of any right triangle is longer than the leg). When $ P $ is one of the points $ A_n $ ($ n \geq 3 $), the sum
is its angular coordinate (more precisely: one of the possible values of this coordinate). When $ P $ is a point on the segment $ A_nA_{n+1} $, we choose the value of its angular coordinate that lies in the interval $ \langle \phi_n; \phi_{n+1} \rangle $ and denote it by $ \phi(P) $. Thus, $ \phi(A_n) = \phi_n $. As the point $ P $ moves (along the considered line), $ \phi(P) $ increases (because during the traversal of the segment $ A_nA_{n+1} $, the velocity vector is perpendicular to the position vector at the initial moment). To prove the thesis of the problem, it suffices to show that
Indeed: suppose we have proven this and take any ray with origin at $ A_1 $. Let $ \theta $ be the angular coordinate of the points of this ray, taken from the interval $ \langle 0^\circ; 360^\circ \rangle $. Then any larger measure of the same angle (i.e., any of the values $ \theta + k \cdot 360^\circ $, $ k = 1, 2, 3, \ldots $) belongs to some interval $ \langle \phi_n; \phi_{n+1} \rangle $, and is thus the angular coordinate of some point on the segment $ A_nA_{n+1} $. Different values of $ k $ correspond to different values of $ n $ (because $ \phi_{n+1} - \phi_n = a_n < 90^\circ $). This means that the considered ray intersects infinitely many segments $ A_nA_{n+1} $.
(The intuitive sense of the above considerations is as follows: the considered line spirals around the point $ A_1 $; we want to prove that it spirals around it infinitely many times, and this is precisely what the limiting relation (3) expresses.)
It remains to prove that $ \phi_n \to \infty $.
Suppose (3) does not hold. The sequence ($ \phi_n $) is increasing. Therefore, there exists a finite limit $ \lim \phi_n = \psi $.
Take a number $ m $ such that $ \phi_m > \psi - 90^\circ $ (we have returned to degree measure of angles). The ray with origin at $ A_1 $, whose points have the angular coordinate $ \psi $, intersects the ray $ A_mA_{m+1}^\to $ at a point we will denote by $ B $ (Figure 8).
om38_1r_img_8.jpg
All points $ A_n $ with numbers $ n \geq m $ lie within the convex angle $ A_mA_1B $. We will show that they must lie within the right triangle $ A_mA_1B $. The proof is by induction. Denote this triangle by $ \Delta $. Clearly, $ A_m \in \Delta $. Suppose that $ A_n $ is (for some $ n \geq m $) a point of the triangle $ \Delta $. The point $ A_{n+1} $ lies between the rays $ A_1A_n^\to $ and $ A_1B^\to $. If it were outside the triangle $ \Delta $, the angle $ A_1A_nA_{n+1} $ would be obtuse - contrary to the condition of the problem, which states that this angle is right. Therefore, $ A_{n+1} \in \Delta $. By the principle of induction, all points $ A_n $ ($ n \geq m $) lie within the triangle $ \Delta $.
Their distances from $ A_1 $ must therefore form a bounded sequence. However, from formula (2) it is clear that $ r_n \to \infty $.
The obtained contradiction proves the validity of condition (3), and thus the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,087 |
V OM - II - Task 2
Prove that among ten consecutive natural numbers there is always at least one, and at most four numbers that are not divisible by any of the numbers $ 2 $, $ 3 $, $ 5 $, $ 7 $.
|
In a sequence consisting of ten consecutive natural numbers, there are five even numbers and five odd numbers. Therefore, the problem can be reduced to the following:
Prove that among five consecutive odd numbers, there is at least one, and at most four numbers not divisible by any of the numbers $3$, $5$, $7$.
Notice that in the sequence of odd numbers, every third number is divisible by $3$, every fifth by $5$, and every seventh by $7$. Therefore:
a) Among five consecutive odd numbers, at least one is divisible by $3$; hence, the number of numbers not divisible by $3$, and consequently, the number of numbers not divisible by $3$, $5$, or $7$ cannot be more than four. The example of the numbers $11$, $13$, $15$, $17$, $19$ shows that among five consecutive odd numbers, there can be four numbers not divisible by any of the numbers $3$, $5$, $7$.
b) Among five consecutive odd numbers, at most two are divisible by $3$, at most one is divisible by $5$, and at most one is divisible by $7$. Therefore, at least one of these five numbers is not divisible by any of $3$, $5$, or $7$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,089 |
XLVIII OM - III - Problem 6
On a circle of radius $ 1 $, there are $ n $ different points ($ n \geq 2 $). Let $ q $ be the number of segments with endpoints at these points and length greater than $ \sqrt{2} $. Prove that $ 3q \leq n^2 $.
|
We will call a segment with endpoints in the considered set long if its length is greater than $ \sqrt{2} $, and short otherwise. Thus, $ q $ is the number of long segments. We inscribe a square $ ABCD $ in a given circle so that none of the given $ n $ points is a vertex of the square. Let us assume that on the arcs $ AB $, $ BC $, $ CD $, $ DA $ there are respectively $ a $ points, $ b $ points, $ c $ points, $ d $ points. Consider the set of all quadrilaterals having vertices at the given points, one on each of the four arcs. There are $ abcd $ such quadrilaterals. Each quadrilateral inscribed in a given circle has at least one short side.
Let us associate with each quadrilateral (from the considered set) one of its short sides and paint this chosen side green. A fixed green segment with endpoints on arcs $ AB $ and $ BC $ could have been associated with at most $ c \cdot d $ quadrilaterals, and thus at most $ m $ quadrilaterals, where $ m $ is the largest of the numbers $ ab $, $ bc $, $ cd $, $ da $. Similarly, we can estimate the number of quadrilaterals to which a fixed green segment (with endpoints on any two adjacent arcs) has been associated. Hence, the total number of green segments is no less than $ (abcd)/m $.
All green segments are short. Of course, all such segments with both endpoints on the same arc are short. Therefore, the number of short segments, equal to $ {n \choose 2} - q $, satisfies the estimate:
We can assume (by shifting the notation if necessary) that $ m = ab $. The obtained inequality, after transformation, takes the form:
(we used the fact that $ a + b + c + d = n $). Therefore,
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,090 |
IX OM - III - Task 2
Each side of the convex quadrilateral $ABCD$ is divided into three equal parts; a line is drawn through the division points on sides $AB$ and $AD$ that are closer to vertex $A$, and similarly for vertices $B$, $C$, and $D$. Prove that the centroid of the quadrilateral formed by the drawn lines coincides with the centroid of quadrilateral $ABCD$.
|
We will adopt the notations shown in Fig. 27. We need to prove that the center of gravity of quadrilateral $ABCD$ coincides with the center of gravity of quadrilateral $MNPQ$, which is a parallelogram, since lines $MN$ and $QP$ (as well as lines $14$ and $85$) are parallel to line $AC$, and lines $MQ$ and $NP$ are parallel to line $BD$.
First, we will determine the centers of gravity of triangles $ABC$ and $ADC$, into which diagonal $AC$ divides quadrilateral $ABCD$. The center of gravity $S$ of triangle $ABC$ lies on the median $BK$ of this triangle, with $BS = \frac{2}{3} BK$; since triangle $1B4$ is similar to triangle $ABC$ in a ratio of $2:3$, point $S$ lies at the midpoint of segment $14$. Similarly, the center of gravity $T$ of triangle $ADC$ lies at the midpoint of segment $85$.
It is known that if a figure consists of two parts (not overlapping), then the center of gravity of this figure lies on the segment connecting the centers of gravity of both parts. The center of gravity of quadrilateral $ABCD$ thus lies on segment $ST$, and therefore on the line passing through the midpoints of sides $MN$ and $QP$ of parallelogram $MNPQ$. Similarly, it must lie on the line passing through the midpoints of sides $MQ$ and $NP$ of this parallelogram. Both mentioned lines intersect at the center of parallelogram $MNPQ$, which is, of course, the center of gravity of this parallelogram. Indeed, then, the centers of gravity of quadrilateral $ABCD$ and parallelogram $MNPQ$ coincide.
Note 1. The theorem is also true for a concave quadrilateral; the proof proceeds in the same way as for a convex quadrilateral, with the only difference being that if $AC$ is the diagonal that lies outside quadrilateral $ABCD$ (Fig. 28), then the center of gravity of quadrilateral $ABCD$ does not lie on segment $TS$, but on its extension beyond point $S$.
Note 2. The previous proof can be modified as follows. If $S$ and $T$ are the centers of gravity of triangles $ABC$ and $ADC$, then the center of gravity of quadrilateral $ABCD$ is the point on line $TS$ that divides segment $TS$ (internally in the case of Fig. 27, and externally in the case of Fig. 28) inversely proportional to the areas of triangles $ADC$ and $ABC$. Points $S$ and $T$ are also the centers of gravity of parallelograms $MNVU$ and $UVPQ$, so the center of gravity of parallelogram $MNPQ$ divides segment $TS$ inversely proportional to the areas of $UVPQ$ and $MNVU$. However, $\text{area of } UVPQ = \text{area of } AC67 = \frac{8}{9}$, $\text{area of } ADC$, and similarly $\text{area of } MNVU = \text{area of } A23C = \frac{8}{9} \text{area of } ABC$, so the ratio of the areas of triangles $ADC$ and $ABC$ equals the ratio of the areas of parallelograms $UVPQ$ and $MNVU$. Hence, the centers of gravity of quadrilaterals $ABCD$ and $MNPQ$ coincide.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,091 |
L OM - II - Task 3
Convex quadrilateral $ABCD$ is inscribed in a circle. Points $E$ and $F$ lie on sides $AB$ and $CD$ respectively, such that $AE: EB = CF: FD$. Point $P$ lies on segment $EF$ and satisfies the condition $EP: PF = AB: CD$. Prove that the ratio of the areas of triangles $APD$ and $BPC$ does not depend on the choice of points $E$ and $F$.
|
First, assume that lines $AD$ and $BC$ are not parallel and intersect at point $S$. Then triangles $ASB$ and $CSD$ are similar. Since $AE : EB = CF : FD$, triangles $ASE$ and $CSF$ are also similar. Therefore, $\measuredangle DSE = \measuredangle CSF$. Moreover,
from which we get $\measuredangle ESP = \measuredangle FSP$. Thus, point $P$ is equidistant from lines $AD$ and $BC$. The ratio of the areas of triangles $APD$ and $BPC$ is therefore equal to the ratio of the lengths of segments $AD$ and $BC$, i.e., it does not depend on the choice of points $E$ and $F$.
Now assume that lines $AD$ and $BC$ are parallel. Then $ABCD$ is an isosceles trapezoid, where $AB = CD$. Point $P$ is the midpoint of segment $EF$ and $BE = DF$. Points $E$ and $F$ are therefore equally distant from lines $BC$ and $AD$, respectively. Additionally, point $P$ is equidistant from lines passing through points $E$, $F$ and parallel to the bases of the trapezoid $ABCD$. Thus, in this case as well, the distances from point $P$ to lines $AD$ and $BC$ are the same. Repeating the conclusion from the previous case, we obtain the thesis.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,092 |
LVI OM - III - Task 3
In a square table of dimensions $ 2n \times 2n $, where $ n $ is a natural number, there are $ 4n^2 $ real numbers with a total sum of 0 (one number on each cell of the table). The absolute value of each of these numbers is no greater than 1. Prove that the absolute value of the sum of all numbers in some row (horizontal or vertical) does not exceed $ n $.
|
Let's number the rows and columns of the given table with the numbers $1, 2, \ldots, 2n$.
Let $r_1, r_2, \ldots, r_{2n}$ be the sums of the numbers written in the rows numbered $1, 2, \ldots, 2n$, and $c_1, c_2, \ldots, c_{2n}$ be the sums of the numbers in the columns numbered $1, 2, \ldots, 2n$. By changing the numbering of the rows and columns, we can assume that $r_1 \geq r_2 \geq \ldots \geq r_{2n}$ and $c_1 \geq c_2 \geq \ldots \geq c_{2n}$.
Let $A$, $B$, $C$, $D$ be the sums of the numbers in the four quarters of the table, as shown in Figure 3.
Then $A + B + C + D = 0$, which implies that in the sequence $(A, B, C, D, A)$ there are two consecutive numbers, one of which is non-negative and the other non-positive. For example, let's assume $AB \leq 0$. The reasoning in the other cases is analogous.
om56_3r_img_3.jpg
The numbers $A$ and $B$ are the result of summing $n^2$ real numbers with absolute values not greater than 1, so $|A| \leq n^2$ and $|B| \leq n^2$. One of the numbers $A$ and $B$ is non-negative, and the other is non-positive, so $|A + B| \leq \max(|A|, |B|) \leq n^2$. Therefore, we also have $|C + D| = |A + B| \leq n^2$.
If $r_n \geq 0$, then the numbers $r_1, r_2, \ldots, r_n$ are non-negative. Then their sum $A + B$ is also non-negative, so we get $r_1 + r_2 + \ldots + r_n = A + B \leq n^2$. Therefore, at least one of the (non-negative) numbers $r_1, r_2, \ldots, r_n$ does not exceed $n$.
If, however, $r_n < 0$, then the numbers $r_{n+1}, r_{n+2}, \ldots, r_{2n}$ are negative and their sum is $C + D$. Then $|r_{n+1}| + |r_{n+2}| + \ldots + |r_{2n}| = |C + D| \leq n^2$. This implies that at least one of the numbers $|r_{n+1}|, |r_{n+2}|, \ldots, |r_{2n}|$ does not exceed $n$.
This proves the problem.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,096 |
XXXII - II - Problem 5
In the plane, there are two disjoint sets $ A $ and $ B $, each consisting of $ n $ points, such that no three points of the set $ A \cup B $ lie on the same line. Prove that there exists a set of $ n $ disjoint closed segments, each of which has one endpoint in set $ A $ and the other in set $ B $.
|
For each one-to-one assignment of points $B_j$ from set $B$ to points $A_i$ from set $A$, we can calculate the sum of distances determined by this assignment for pairs of points $(A_i, B_j)$. Suppose that by assigning point $A_i$ to point $B_i$ for $i = 1,2,\ldots,n$, we obtain the minimum value of this sum. Then the segments $\overline{A_iB_i}$ do not have common points (Fig. 12), because if $C$ were a common point of segments $\overline{A_iB_i}$ and $\overline{A_jB_j}$ for $i \ne j$, then in view of the one-to-one assignment, $C$ is not the endpoint of any of these segments, and in triangles $A_iB_jC$ and $A_jB_iC$ it would be $A_iB_j < A_iC + B_jC$ and $A_jB_i < A_jC + B_iC$, so $A_iB_j + A_jB_i < A_iC + CB_i + B_jC + CA_j$, and thus $A_iB_j + A_jB_i < A_iB_i + A_jB_j$, which contradicts the assumption of the minimum sum of distances.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,101 |
LVIII OM - III - Problem 3
The plane has been divided by horizontal and vertical lines into unit squares.
A positive integer must be written in each square so that each
positive integer appears on the plane exactly once.
Determine whether this can be done in such a way that each written number
is a divisor of the sum of the numbers written in the four adjacent squares.
|
We will prove that the postulated method of writing numbers exists.
Let us shade some unit squares and draw a path passing through each unit square exactly once as shown in Fig. 15. Then, in the marked squares of the path, we write numbers according to the following rules:
We start from the square marked with a thick dot.
om58_3r_img_15.jpg
om58_3r_img_16.jpg
A) If a given square is not shaded, we write the smallest positive integer that has not yet been written on the plane.
B) If a given square $K$ is shaded, it has a common side with exactly one square $H$, in which a number has already been written, and for which the three remaining neighboring squares (different from $K$) already have numbers written in them. In this case, we write in $K$ the smallest positive integer $m$ that has not yet appeared on the plane, such that the number written in square $H$ is a divisor of the sum of the four numbers in the squares neighboring $H$.
(See Fig. 16, which shows the result of executing steps A) and B) for several dozen initial squares on the path. For example, for the square $K$ in which the number 37 is written, the corresponding square $H$ contains the number 13, and the remaining three neighboring squares have the numbers 27, 12, and 15 written in them. Therefore, the number $m$ must be chosen so that the sum $27+12+15+m=54+m$ is divisible by $13$. The numbers $m=11$ and $m=24$ have already been written on the plane, so we choose the value $m=37$.)
First, note that a number $m$ consistent with the requirements of condition B) always exists. Indeed, if the numbers $a$, $b$, and $c$ are already written in the three squares neighboring square $H$, and the number $d$ is in square $H$, then we need to choose $m$ such that the number $a+b+c+m$ is divisible by $d$. This is possible because only a finite number of numbers have been written on the plane so far.
Thus, proceeding in the indicated manner, we will write a positive integer in each square on the plane. It remains to show that this method of writing satisfies the conditions of the problem.
Since an infinite number of squares are unshaded, step A) ensures that every positive integer will appear on the plane at some point. Furthermore, in steps A) and B), we always choose a number that has not yet been written. Therefore, every positive integer will be written in exactly one square. Finally, for any square $H$, among the four squares neighboring $H$, the one appearing last on the marked path is shaded, and thus the writing of a number in this square follows procedure B). From the content of this procedure, we conclude that the number in square $H$ is a divisor of the sum of the four numbers written in the squares neighboring $H$.
We have thus shown that the described method of placing numbers in squares satisfies the conditions of the problem.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,104 |
XLII OM - II - Problem 1
The numbers $ a_i $, $ b_i $, $ c_i $, $ d_i $ satisfy the conditions $ 0\leq c_i \leq a_i \leq b_i \leq d_i $ and $ a_i+b_i = c_i+d_i $ for $ i=1,2,\ldots,n $. Prove that
|
Induction. For $ n = 1 $, the inequality to be proven holds (with an equality sign; since $ a_1 + b_1 = c_1 + d_1 $, in accordance with the assumption).
Let us fix a natural number $ n \geq 1 $ and assume the validity of the given theorem for this very number $ n $. We need to prove its validity for $ n + 1 $.
Let then the numbers $ a_i $, $ b_i $, $ c_i $, $ d_i $ ($ i = 1, \ldots, n + 1 $) satisfy the conditions $ 0 \leq c_i \leq a_i \leq b_i \leq d_i $ and $ a_i + b_i = c_i + d_i $. Denote
From the induction hypothesis, we have the inequality $ A + B \leq C + D $, that is,
we need to prove that
By the conditions satisfied by the numbers $ a_i $, $ b_i $, $ c_i $, $ d_i $, the following equalities and inequalities hold:
where the expressions on both sides of each of the relations (1), (3), (4), (5) are non-negative. Inequalities directed consistently, binding non-negative numbers, can be multiplied side by side. Therefore, we multiply (1) by (4) and (3) by (5):
We add the obtained inequalities:
We have obtained the inductive thesis (2).
By the principle of induction, the theorem holds for any natural number $ n $.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,105 |
LVII OM - I - Problem 6
In an acute triangle $ABC$, the altitudes intersect at point $H$. A line passing through $H$ intersects segments $AC$ and $BC$ at points $D$ and $E$, respectively. A line passing through $H$ and perpendicular to line $DE$ intersects line $AB$ at point $F$. Prove that
|
Let $ K $ be the intersection point of the line $ AH $ with the line passing through point $ B $ and parallel to the line
$ FH $ (Fig. 2). Denote by $ X $ the orthogonal projection of point $ A $ onto the line $ BC $, and by $ Y $ the intersection point of the
perpendiculars $ DE $ and $ BK $.
om57_1r_img_2.jpg
The segments $ BX $ and $ HY $ are altitudes in the triangle $ BHK $, and thus the line $ KE $ contains the third altitude of this triangle.
Therefore, the line $ KE $ is perpendicular to the line $ BH $, i.e., parallel to the line $ AC $. Hence
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,108 |
XLIV OM - I - Problem 5
Given is a half-plane and points $ A $ and $ C $ on its edge. For each point $ B $ of this half-plane, we consider squares $ ABKL $ and $ BCMN $ lying outside the triangle $ ABC $. They determine a line $ LM $ corresponding to the point $ B $. Prove that all lines corresponding to different positions of point $ B $ pass through one point.
|
Let's choose point $ B $ in the given half-plane. Due to the symmetry of the roles of the considered squares, we can assume, without loss of generality, that $ |AB| \geq |BC| $. Angle $ BAC $ is therefore acute; angle $ ACB $ can be acute or not (Figure 2). Let $ D $, $ P $, $ Q $ be the orthogonal projections of points $ B $, $ L $, $ M $ onto line $ AC $, and let $ R $ be the projection of point $ M $ onto line $ LP $.
From the definition of points $ L $ and $ M $, it follows that
From these equalities, we see that the right triangle $ APL $ is congruent to triangle $ BDA $, and triangle $ CQM $ is congruent to $ BDC $. Therefore,
Angles $ CAL $ and $ ACM $ are obtuse. Thus, points $ A $ and $ C $ lie inside segment $ PQ $. From the obtained equality $ |AP| = |CQ| $, we conclude that the midpoint $ S $ of segment $ AC $ is also the midpoint of segment $ PQ $.
om44_1r_img_2.jpg
Let $ l $ be the common perpendicular bisector of these segments; it intersects lines $ ML $ and $ MR $ at points $ X $ and $ Y $, respectively, which, according to Thales' theorem, are the midpoints of segments $ ML $ and $ MR $. Therefore, the equality $ |XY| = \frac{1}{2}|LR| $ holds.
We will calculate the length of segment $ SX $; in the transformations, the sign $ \pm $ (plus-minus) should be read as "plus" when angle $ ACB $ is acute, and as "minus" when angle $ ACB $ is obtuse (see the two variants of Figure 2):
From this, in view of equality (1),
- which means that the position of point $ X $ on line $ l $ does not depend on the choice of point $ B $. Point $ X $ is (in each case) the center of the square, located in the considered half-plane, one of whose sides is segment $ AC $. This is the sought common point of all possible lines $ LM $. (As follows from earlier observations, this point not only lies on each such line $ LM $, but is the midpoint of segment $ LM $.)
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,110 |
XXVI - II - Problem 2
In a convex quadrilateral $ABCD$, points $M$ and $N$ are chosen on the adjacent sides $\overline{AB}$ and $\overline{BC}$, respectively, and point $O$ is the intersection of segments $AN$ and $GM$. Prove that if circles can be inscribed in quadrilaterals $AOCD$ and $BMON$, then a circle can also be inscribed in quadrilateral $ABCD$.
|
Let $ P $, $ Q $, $ R $, $ S $ be the points of tangency of the inscribed circle in quadrilateral $ AOCD $ with sides $ \overline{AO} $, $ \overline{OC} $, $ \overline{CD} $, $ \overline{DA} $ respectively, and similarly $ E $, $ F $, $ G $, $ H $ - the points of tangency of the inscribed circle in quadrilateral $ BMON $ with sides $ \overline{MB} $, $ \overline{BN} $, $ \overline{NO} $, $ \overline{OM} $ respectively (Fig. 9). Using the theorem: segments of tangents drawn from a point outside a circle have equal lengths, we obtain successively:
thus $ AS + CF = AE + CR $. Similarly, we prove that $ DS + BF = DR + BE $. Adding the last two equalities side by side, we get $ AD + BC = AB + CD $. It follows from this, as is known, that a circle can be inscribed in quadrilateral $ ABCD $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,111 |
LIII OM - III - Task 3
Three non-negative integers are written on the board. We choose two of these numbers $ k $, $ m $ and replace them with the numbers $ k + m $ and $ |k - m| $, while the third number remains unchanged. We proceed in the same way with the resulting triplet. Determine whether, from any initial triplet of non-negative integers, continuing this procedure, one can obtain a triplet in which at least two of the numbers are zeros.
|
Every triple of non-negative integers can be represented in the form
\[
(2^pa, 2^qb, 2^rc)
\]
where \( p \), \( q \), \( r \) are non-negative integers, and each of the numbers \( a \), \( b \), \( c \) is odd or equal to \( 0 \). The weight of a triple represented in form (1) will be the quantity \( a + b + c \).
We will show that from any triple of non-negative integers with at least two non-zero numbers, by performing the operations described in the problem, we can obtain a triple with a smaller weight. This will prove that we can always obtain a triple in which at least two numbers are zeros.
Assume, therefore, that in the triple \( (2^pa, 2^qb, 2^rc) \) at least two numbers are different from \( 0 \). Without loss of generality, assume that \( b, c \neq 0 \) and \( q \leq r \).
By performing the operation \( (k, l, m) \to (k + l, |k - l|, m) \) \( 2(r-q) \) times, we transform the triple \( (2^pa, 2^qb, 2^rc) \) into the triple \( (2^{p+r-q}a, 2^rb, 2^rc) \), which we then transform into \( (2^{p+r-q}a, 2^r|b - c|, 2^r(b + c)) \). The numbers \( b \) and \( c \) are odd, so the weight of the last triple does not exceed
\[
2^{p+r-q}a + 2^r|b - c| + 2^r(b + c)
\]
which is less than \( a + b + c \), i.e., less than the weight of the triple \( (2^pa, 2^qb, 2^rc) \).
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,113 |
XXXVII OM - III - Problem 6
In triangle $ABC$, points $K$ and $L$ are the orthogonal projections of vertices $B$ and $C$ onto the angle bisector of $\angle BAC$, point $M$ is the foot of the altitude from vertex $A$ in triangle $ABC$, and point $N$ is the midpoint of side $BC$. Prove that points $K$, $L$, $M$, and $N$ lie on the same circle.
|
When $ |AB| = |AC| $, points $ K $, $ L $, $ M $, $ N $ coincide. We will exclude this trivial case from further considerations. We will present three methods of solving.
\spos{I} Without loss of generality, we can assume that $ |AB| < |AC| $. Let us first assume that angle $ ABC $ is not a right angle - it is therefore either acute or obtuse. Figures 12a and 12b illustrate these two situations.
om37_3r_img_12a.jpg
om37_3r_img_12b.jpg
Consider the circle whose diameter is segment $ AB $. Since $ |\measuredangle AKB| = | \measuredangle AMB| = 90^\circ $, points $ K $ and $ M $ lie on this circle. Inscribed angles $ BAK $ and $ KMB $ are subtended either by the same arc (when $ |\measuredangle B| > 90^\circ $ - Fig. 12b), or by arcs that complete each other to a full circle (when $ |\measuredangle B | < 90^\circ $ - Fig. 12a). Thus,
In both cases, the expression obtained on the right side is equal to the measure of angle $ KMN $. Therefore,
When angle $ ABC $ is a right angle, point $ M $ coincides with $ B $, and angles $ BAK $ and $ KMN $ have sides that are perpendicular to each other. Thus, equality (1) holds in this case as well.
Extend side $ AB $ to intersect line $ CL $ at a point we will denote by $ E $. Segment $ AL $ is both the angle bisector and the altitude in triangle $ ACE $; it is therefore an isosceles triangle and point $ L $ is the midpoint of side $ CE $. Thus, segment $ NL $, as connecting the midpoints of sides $ BC $ and $ CE $ of triangle $ BCE $, is parallel to side $ BE $, i.e., to line $ AB $. Therefore, $ |\measuredangle BAK| = |\measuredangle KLN| $. Combined with equality (1), this gives the equality $ |\measuredangle KMN| = |\measuredangle KLN| $, from which it follows that points $ K $, $ N $, $ L $, $ M $ lie on the same circle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,114 |
LIV OM - I - Problem 8
In the tetrahedron $ABCD$, points $M$ and $N$ are the midpoints of edges $AB$ and $CD$, respectively. Point $P$ lies on segment $MN$, such that $MP = CN$ and $NP = AM$. Point $O$ is the center of the sphere circumscribed around the tetrahedron $ABCD$. Prove that if $O \neq P$, then $OP \perp MN$.
|
Let's introduce the following notations:
Then, by the Law of Cosines, we have
Furthermore, using the Pythagorean theorem, we obtain
By subtracting equation (1) and utilizing the relationship (2), we get $ 2(a + b)x \cos \alpha = 0 $. Therefore, $ \alpha = 90^\circ $, which means $ OP \perp MN $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,115 |
XLIX OM - II - Problem 6
Given is a tetrahedron $ABCD$. Prove that the edges $AB$ and $CD$ are perpendicular if and only if there exists a parallelogram $CDPQ$ in space such that $PA = PB = PD$ and $QA = QB = QC$.
|
Let $ k $ be the line perpendicular to the plane of triangle $ ABC $ and passing through the center of the circle circumscribed around triangle $ ABC $. Similarly, let $ \ell $ be the line perpendicular to the plane of triangle $ ABD $ and passing through the center of the circle circumscribed around triangle $ ABD $. The lines $ k $, $ \ell $ intersect in space (at the center of the sphere circumscribed around the tetrahedron $ ABCD $) and both are perpendicular to the edge $ AB $. Therefore, the plane $ \pi $, containing the lines $ k $ and $ \ell $, is perpendicular to the edge $ AB $ of the tetrahedron $ ABCD $.
Assume first that there exists a parallelogram $ CDPQ $ such that $ PA = PB = PD $ and $ QA = QB = QC $. Then the point $ P $ lies on the line $ \ell $, and the point $ Q $ lies on the line $ k $. The line $ PQ $ lies in the plane $ \pi $, which means it is perpendicular to the edge $ AB $. Since $ CD \parallel PQ $, then $ CD \perp AB $.
Now assume that the edges $ AB $ and $ CD $ are perpendicular. Then the line $ CD $ is parallel to the plane $ \pi $. By translating the line $ k $ by the vector $ \overrightarrow{CD} $, we obtain a line $ k' $ also lying in the plane $ \pi $ (and not parallel to $ \ell $). Therefore, there is a common point of the lines $ k' $ and $ \ell $, which we will denote by $ P $. Let $ Q $ be the image of the point $ P $ under the translation by the vector $ \overrightarrow{DC} $. Then the point $ Q $ lies on the line $ k $. The point $ P $ lies on the line $ \ell $. The quadrilateral $ CDPQ $ is a parallelogram, in which $ PA = PB = PD $ and $ QA = QB = QC $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,116 |
XLVII OM - I - Problem 9
A polynomial with integer coefficients, when divided by the polynomial $ x^2 - 12x + 11 $, gives a remainder of $ 990x - 889 $. Prove that this polynomial does not have integer roots.
|
According to the assumption, the considered polynomial has the form
where $ Q(x) $ is some polynomial.
Further reasoning is based on the following observation: for every pair of different integers $ k, m $
[Justification: if $ P(x) = a_0 + a_1x + \ldots +a_nx^n $ ($ a_i $ integers), then the difference $ P(k) - P(m) $ is the sum of terms $ a_i(k^i-m^i) $ for $ i = 1,\ldots,n $, and each of these terms is divisible by $ k - m $.]
Suppose that the polynomial $ P(x) $ has an integer root $ x_0 $. By taking $ k = 1, m = x_0 $, and then $ k = 11, m = x_0 $ in (2), we conclude that
However, from the representation (1), it is clear that $ P(1) = 101 $, and $ P(11) = 10001 $. The difference $ 1 - x_0 $, as a divisor of the prime number $ P(1) = 101 $, must be one of the numbers: $ -101 $, $ -1 $, $ 1 $, $ 101 $. For these values of $ x_0 $, the difference $ 11 - x_0 $ takes the respective values: $ -91 $, $ 9 $, $ 11 $, $ 111 $. However, none of these values is a divisor of the number $ P(11) = 10001 ( = 73 \cdot 137) $.
Simultaneous satisfaction of conditions (3) is therefore impossible. The contradiction proves that the polynomial $ P(x) $ does not have an integer root.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,118 |
XXXV OM - III - Task 6
The towns $ P_1, \ldots, P_{1025} $ are served by the airlines $ A_1, \ldots, A_{10} $, such that for any towns $ P_k $ and $ P_m $ ($ k \neq m $) there exists an airline whose planes fly directly from $ P_k $ to $ P_m $ and directly from $ P_m $ to $ P_k $. Prove that one of these airlines can offer a trip with an odd number of landings starting and ending in the same town.
|
Suppose that none of the lines $ A_1, \ldots, A_{10} $ can offer a journey with an odd number of landings that starts and ends in the same town. It follows that if there is a connection from town $ P_k $ to town $ P_m $ by planes of a certain line, with an even number of landings, then in any other connection from $ P_k $ to $ P_m $ by planes of the same line, there is also an even number of landings.
For each line $ A_j $, we can divide the set of towns into two subsets $ M_j $ and $ N_j $. We choose an arbitrary town, for example, $ P_1 $, and include it in the subset $ M_j $. Then, every town that can be reached from $ P_1 $ by planes of line $ A_j $ with an even number of landings is included in $ M_j $, while every town that can be reached from $ P_1 $ by planes of line $ A_j $ with an odd number of landings is included in $ N_j $. If there are still some towns left after this division, we arbitrarily choose one of them, for example, $ P_k $, and include it in $ M_j $ along with all those towns that can be reached from $ P_k $ by planes of line $ A_j $ with an even number of landings. Those towns that can be reached from $ P_k $ by planes of line $ A_j $ with an odd number of landings are included in $ N_j $. We continue this process until all towns are exhausted. One of the sets $ M_1 $ or $ N_1 $ has no fewer than $ 513 $ elements. These are towns that do not have direct connections by planes of line $ A_1 $ (two towns, between which there is a connection by planes of line $ A_1 $ with an odd number of landings, and thus in particular with one landing, belong to different sets). Among these $ 513 $ towns, no fewer than $ 257 $ belong to one of the sets $ M_2 $ or $ N_2 $. No two of these $ 257 $ towns have a direct connection by planes of lines $ A_1 $ and $ A_2 $. No fewer than half of them, i.e., $ 129 $ towns, belong to one of the sets $ M_3 $ or $ N_3 $. Between no two of these $ 129 $ towns is there a direct connection by planes of lines $ A_1 $, $ A_2 $, and $ A_3 $. Reasoning in this way, we find that there are $ 65 $ towns between which there is no direct connection by planes of lines $ A_1 $, $ A_2 $, $ A_3 $, and $ A_4 $, then $ 33 $ towns without a direct connection by planes of lines $ A_1, \ldots, A_5 $, followed by $ 17 $ towns without a direct connection by planes of lines $ A_1, \ldots, A_6 $. 9 of these towns have no direct connection by planes of lines $ A_1, \ldots, A_7 $, 5 of them have no direct connection by planes of lines $ A_1, \ldots, A_8 $, 3 towns have no direct connection by any plane of lines $ A_1, \ldots, A_9 $, and finally, two towns have no direct connection by any plane of the ten lines considered. The last statement, however, contradicts the assumption. This contradiction was caused by the assumption that no line can offer a journey with an odd number of landings starting and ending in the same town. Therefore, some line can offer such a journey.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,120 |
XXXIX OM - III - Problem 4
Let $ d $ be a positive integer, and $ f \: \langle 0; d\rangle \to \mathbb{R} $ a continuous function such that $ f(0) = f(d) $. Prove that there exists $ x \in \langle 0; d-1\rangle $ such that $ f(x) = -f(x + 1) $.
|
We consider the function $ g(x) = f(x+1)-f(x) $ defined and continuous in the interval $ \langle 0; d-1 \rangle $. The sum
of the terms, by assumption, is zero, which means either all its terms are zero, or some two terms are of different signs. In any case, there exist integers $ a, b \in \langle 0; d-1 \rangle $ such that $ g(a) \leq 0 \leq g(b) $. From the continuity of the function $ g $, it follows that there is a point $ x_0 $ in the interval with endpoints $ a $, $ b $ where it takes the value $ 0 $ (Darboux property). The equality $ g(x_0) = 0 $ is equivalent to $ f(x_0) = f(x_0+1) $. We have thus found a point $ x_0 \in \langle 0; d-1 \rangle $, which is what we were looking for in the problem.
Note: It is not difficult to show that the given theorem is not true for any number $ d > 1 $ that is not an integer. It suffices to take any function $ h \colon \mathbb{R} \to \mathbb{R} $ that is periodic with period $ 1 $, continuous, and such that $ h(0) = 0 $, $ h(d) = d $, and set $ f(x) = h(x)-x $. The function thus defined satisfies the condition $ f(0) = f(d) $, but does not satisfy the equality $ f(x) = f (x +1) $ for any $ x $; such an equality would imply that $ h(x+1) = h(x) + 1 $, which contradicts the periodicity of the function $ h $.
|
proof
|
Calculus
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,121 |
VI OM - I - Task 2
A factory sends goods in packages of $ 3 $ kg and $ 5 $ kg. Prove that in this way, any whole number of kilograms greater than $ 7 $ can be sent. Can the numbers in this task be replaced by other numbers?
|
Every integer greater than $7$ can be represented in one of the forms $3k-1$, $3k$, $3k+1$, where $k$ is an integer greater than $2$. Since $3k - 1 = 3 (k - 2) + 5$, and $3k + 1 = 3 (k - 3) + 2 \cdot 5$, we conclude from this that every integer greater than $7$ can be represented in the form $3x + 5y$, where $x$ and $y$ are non-negative integers. Indeed, it is therefore possible to send any integer amount of kilograms of goods greater than $7$ using only three- and five-kilogram packages. It is easy to verify that it is not possible to send $7$ kg in this way.
This theorem can be generalized as follows. Suppose that goods are sent in packages of $a$ and $b$ kilograms. If $a$ and $b$ have a common divisor $d > 1$, then of course the number of kilograms of each package of goods sent must be divisible by $d$. Let us assume, then, that the natural numbers $a$ and $b$ are relatively prime, with neither equal to $1$. We will prove that in packages of $a$ kg and $b$ kg, any integer amount $c$ of kilograms greater than $ab - a - b$ can be sent, but $ab - a - b$ kilograms cannot be sent.
To prove this, consider the equation $ax + by = c$,
Let us assign the variable $x$ the values $x_0 = 0, x_1 = 1, x_2 = 2, \ldots, x_{b-1} = b - 1$; the corresponding values of the variable $y$ we denote by the letters $y_0, y_1, y_2, \ldots, y_{b-1}$; these values form a decreasing sequence. It is easy to prove that one and only one of them is an integer. Indeed, each of the numbers $c - ax_i$ ($i = 0, 1, \ldots, b - 1$) gives a different remainder when divided by $b$; if, for example, two of them, say $c - ax_i$ and $c - ax_k$, gave the same remainder, then the difference
would be divisible by $b$, which is impossible, since the number $a$ is relatively prime to $b$, and the number $x_k - x_i$ is absolutely less than $b$.
Hence, one and only one of these $b$ remainders is equal to zero, and thus one and only one of the numbers $y_0, y_1, \ldots, y_{b-1}$ is an integer. For this number to be non-negative, i.e., greater than $-1$, it suffices that the smallest of the numbers $y_0, y_1, \ldots$, i.e., the number $y_{b-1}$, be greater than $-1$, which gives the condition
Therefore, if this condition is satisfied, the equation $ax + by = c$ has a solution in non-negative integers $(x, y)$.
If $c = ab - a - b$, then
The equation $ax + by = c$ then has no non-negative integer solutions, since, given that $y_{b-1}$ is an integer, none of the numbers $y_0, y_1, \ldots, y_{b-2}$ is an integer, and if $x > b - 1$, then $y < -1$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,122 |
XIX OM - II - Problem 6
On a plane, $ n \geq 3 $ points not lying on a single straight line are chosen. By drawing lines through every pair of these points, $ k $ lines are obtained. Prove that $ k \geq n $.
|
We will first prove the theorem:
If $ n \geq 3 $ and $ A_1, A_2, \ldots, A_n $ are points in the plane not lying on a single line, then there exists a line that contains exactly two of these points.
Proof. From the assumption, it follows that some triples of points are non-collinear. If $ A_i $, $ A_k $, $ A_l $ is such a triple, let $ d_{ikl} $ denote the distance from point $ A_i $ to the line $ A_kA_l $. The set of positive numbers $ d_{ikl} $ is finite, so there is a number in it that is not smaller than any other number in this set. We can assume that this number is, for example, $ d_{123} $, as this can be achieved by appropriately labeling the points (Fig. 15).
The line $ A_2A_3 $ does not then contain any of the remaining points $ A_i $. If, for example, point $ A_4 $ lay on it, then two of the points $ A_2 $, $ A_3 $, $ A_4 $, for example, points $ A_3 $ and $ A_4 $, would lie on the same side of the perpendicular from point $ A_1 $ to the line $ A_2A_3 $, and then one of the inequalities would hold
which would contradict the definition of the number $ d_{123} $.
Based on the above theorem, the proof of the theorem stated in the problem is easily obtained by induction.
When $ n = 3 $, the thesis of the theorem is true, as in this case $ k = 3 $. Suppose that the thesis of the theorem is true for some natural number $ n \geq 3 $. Let $ Z= \{ A_1,A_2,\ldots,A_n, A_{n + 1}\} $ be a set of points in the plane, not lying on a single line, and let $ A_1A_{n+1} $ be a line that contains exactly two of the points in the set $ Z $. From the assumption, it follows that at least one of the $ n $-point sets $ Z_1 = \{A_1, A_2, \ldots, A_n\} $ and $ Z_2 = \{ A_2, A_3,\ldots, A_{n + 1}\} $ contains non-collinear points. Suppose, for example, that the points of set $ Z $ do not lie on a single line. According to the induction hypothesis, the number of lines passing through any two points of the set $ Z_1 $ is at least $ n $. Since the line $ A_1A_{n+1} $ does not belong to these lines, as it passes through only one point of the set $ Z_1 $, the number of lines passing through any two points of the set $ Z $ is at least $ n + 1 $. The inductive proof of the theorem has been completed.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,123 |
XXX OM - III - Task 2
Prove that four lines connecting the vertices of a tetrahedron with the centers of the circles inscribed in the opposite faces have a common point if and only if the three products of the lengths of opposite edges of the tetrahedron are equal.
|
Let $ABCD$ be a given tetrahedron, and let points $P$ and $Q$ be the centers of the circles inscribed in the faces $ABC$ and $ABD$ (Fig. 15).
om30_3r_img_15.jpg
Let $R$ and $R'$ be the points of intersection of the line $AB$ with the lines $CP$ and $DQ$, respectively. Since the center of the circle inscribed in a triangle is the point of intersection of the angle bisectors of its angles, $\overline{CR}$ is the bisector of $\measuredangle ACB$, and $\overline{DR'}$ is the bisector of $\measuredangle ADB$. The angle bisector of a triangle divides the opposite side into segments proportional to the other sides. Therefore,
Thus, $R$ and $R'$ coincide if and only if $\frac{AC}{BC} = \frac{AD}{BD}$.
The lines $DP$ and $CQ$ have a common point if and only if the points $C$, $D$, $P$, $Q$ lie in the same plane, which means that the lines $CP$ and $DQ$ have a common point, and thus, $R$ and $R'$ coincide.
We have thus proved that the lines $m_C$ and $m_D$ connecting the vertices $C$ and $D$ with the centers of the circles inscribed in the opposite faces have a common point if and only if $AC \cdot BD = BC \cdot AD$.
Similarly, we prove that the lines $m_B$ and $m_C$ drawn from the vertices $B$ and $C$, respectively, have a common point if and only if $AB \cdot DC = DB \cdot AC$; also, the lines $m_A$ and $m_E$ have a common point if and only if $AC \cdot BD = BC \cdot AD$, etc.
If, therefore, the lines $m_A$, $m_B$, $m_C$, $m_D$ have a common point, then
Conversely, if the equalities (1) hold, then each pair of lines among $m_A$, $m_B$, $m_C$, $m_D$ has a common point. For example, if the points of intersection of the line $m_C$ with the lines $m_A$ and $m_B$ were different, then the line $m_C$ would be contained in the plane determined by the intersecting lines $m_A$ and $m_B$. This is impossible, since the plane containing points $A$, $B$, and $C$ does not contain any of these lines. Therefore, the line $m_C$ intersects each of the lines $m_A$, $m_B$, $m_D$ at the same point. Thus, all four lines have a common point.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,124 |
XLV OM - II - Problem 5
The circle inscribed in triangle $ABC$ is tangent to sides $AB$ and $BC$ of this triangle at points $P$ and $Q$, respectively. The line $PQ$ intersects the angle bisector of $\angle BAC$ at point $S$. Prove that this angle bisector is perpendicular to the line $SC$.
|
Let's denote the center of the inscribed circle in triangle $ABC$ by $I$, and the measures of angles $CAB$, $ABC$, and $BCA$ by $\alpha$, $\beta$, and $\gamma$ respectively. When $\beta = \gamma$, the theorem is obvious. Therefore, let's assume that $\beta \neq \gamma$. There are two possible cases: $\beta < \gamma$ or $\beta > \gamma$; Figure 11 illustrates these two situations. In the first case, point $S$ lies on segment $PQ$, and in the second case, point $Q$ lies on segment $PS$. In both cases, point $I$ lies on segment $AS$. Looking at the rays $IA^\rightarrow$, $IP^\rightarrow$, $IQ^\rightarrow$, $IS^\rightarrow$, we see that
Angles $IPB$ and $IQB$ are right angles; therefore, a circle can be circumscribed around quadrilateral $IPBQ$. Consequently, the following angle equalities hold:
Of course, $|\measuredangle AIP| = 90^\circ - |\measuredangle IAP| = 90^\circ - \frac{1}{2} \alpha$. Substituting this equality along with equality (2) into formula (1), we conclude that
From the arrangement of points $P$, $Q$, $S$ on line $PQ$ and from the relationship (3), we have
The sum of the angles in triangle $IQS$ is $180^\circ$. Therefore,
Since $\frac{1}{2}\gamma$ is the measure of angle $ICQ$, from the obtained relationship, we infer that points $I$, $C$, $Q$, $S$ lie on the same circle. Angle $IQC$ is a right angle. Therefore, segment $IC$ is the diameter of this circle, and consequently, angle $ISC$ is also a right angle. This is the thesis of the problem.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,125 |
LV OM - I - Task 10
Given is a convex polygon with an even number of sides. Each side of the polygon has a length of 2 or 3, and the number of sides of each of these lengths is even. Prove that there exist two vertices of the polygon that divide its perimeter into two parts of equal length.
|
Let the number of sides of the polygon be $2n$. Denote the vertices of the polygon in order as $A_1, A_2, A_3, \ldots, A_{2n}$. For $i = 1, 2, \ldots, 2n$ let
\[
f(i) = \text{the difference in the lengths of the parts into which the perimeter of the polygon is divided by the points } A_i \text{ and } A_{i+n}
\]
where $A_{k+2n} = A_k$ for $k = 1, 2, \ldots, 2n-1$. In other words, $f(i)$ is the difference in the lengths of the parts into which the perimeter of the polygon is divided by the points $A_i$ and $A_{i+n}$.
For $i = 1, 2, \ldots, 2n$, the number $f(i)$ is an even integer, and
\[
f(i) = -f(i + n).
\]
Therefore, the sequence of numbers
\[
f(1), f(2), \ldots, f(2n)
\]
consists of even numbers, and the difference between consecutive terms is no more than 2. Hence, there exists an $i$ such that $f(i) = 0$, which is what we needed to prove.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,126 |
III OM - I - Problem 12
Prove that the area $ S $ of a quadrilateral inscribed in a circle with sides $ a $, $ b $, $ c $, $ d $ is given by the formula
where $ 2p = a + b + c + d $.
|
Marking the sides and angles of the quadrilateral as in Fig. 25, we have
and since in a cyclic quadrilateral $ A + C = 180^\circ $, therefore
The relationship between angle $ A $ and the sides of the quadrilateral can be found by calculating the length of $ BD $ from triangle $ ABD $ and from triangle $ BCD $:
from which
and since $ \cos C = - \cos A $, therefore
From equations (2) and (4), we can eliminate angle $ A $. For this purpose, we multiply both sides of equation (2) by $ 2 $, square equations (2) and (4), and add them; as a result, we obtain
Hence
and finally
We introduce the notation
and transform the factors on the right side of equation (5) in a known way:
Similarly,
and
Substituting these expressions into equation (5), we get
from which
which was to be proved.
Note 1. The formula (6) for the area of a cyclic quadrilateral is a special case of the formula concerning the area of any quadrilateral (convex or concave).
Equations (1) and (3) apply, as is easily verified, to any quadrilateral. Squaring these equations on both sides, we get:
Adding these equations side by side:
and substituting $ \cos (A + C) = 2 \cos^2 \frac{A+C}{2} - 1 $, we get
Applying the same transformation to the first two terms on the right side of the last equation, we get
from which we finally have the formula
When $ A+C = 180^\circ $, i.e., when the quadrilateral is cyclic, then $ \cos \frac{A+C}{2} = 0 $, so the second term under the square root disappears and formula (7) takes the form (6).
Note 2. From formula (7), we can immediately infer the converse of the theorem stated in problem 27:
If the area of a quadrilateral is expressed by formula (6), then a circle can be circumscribed around the quadrilateral.
Indeed, from formulas (6) and (7), it follows that $ \cos \frac{A+C}{2} = 0 $, so $ A + C = 180^\circ $.
Let us also note the following interesting conclusion from formula (7): when the sides $ a $, $ b $, $ c $, $ d $ are constant, formula (7) gives the maximum value of the area $ S $ when $ A + C = 180^\circ $. Therefore:
A cyclic quadrilateral has a larger area than any other quadrilateral with the same sides.
We can state a stronger theorem (theorem $ A $ is stronger than theorem $ B $, i.e., $ B $ is weaker than $ A $, if theorem $ B $ follows from theorem $ A $, but not vice versa):
Among all quadrilaterals that can be constructed from given sides $ a $, $ b $, $ c $, $ d $, the one with the largest area is the cyclic quadrilateral.
It might seem at first glance that both theorems express the same thing. However, it is easy to see the difference in the content of the theorems when we formulate them as follows:
a) If quadrilaterals $ A $ and $ B $ have the same sides and a circle can be circumscribed around quadrilateral $ A $, but not around quadrilateral $ B $, then the area of quadrilateral $ A $ is greater than the area of quadrilateral $ B $.
b) Among all quadrilaterals with the same sides, there exists a quadrilateral with the largest area; it is the one on which a circle can be circumscribed.
We have proved theorem a) above; to prove theorem b), we need to show that among quadrilaterals with given sides $ a $, $ b $, $ c $, $ d $, there exists a cyclic quadrilateral. We can do this as follows:
Let $ ABCD $ be a quadrilateral with sides $ a $, $ b $, $ c $, $ d $ (Fig. 26), in which $ A + C < 180^\circ $. If we choose a point $ D_1 $ arbitrarily on the extension of diagonal $ BD $—such that the segment $ x = BD_1 $ is shorter than $ a + d $ and $ b + c $—then we can construct (as indicated in Fig. 26) a specified quadrilateral $ A_1BC_1D_1 $ with sides $ a $, $ b $, $ c $, $ d $. In this case, $ A_1 > A $ and $ C_1 > C $.
As point $ D_1 $ moves away from point $ D $, angles $ A_1 $ and $ C_1 $ increase, and we can find a position of point $ D_1 $ such that $ A_1 + C_1 > 180^\circ $. If, for example, $ a + d < b + c $, it suffices to construct a triangle $ A_1BD_1 $ in which $ A_1 = 180^\circ - C $; then $ A_1 + C_1 = 180^\circ - C + C_1 > 180^\circ $.
The sum $ A_1 + C_1 $ is a continuous function of the variable $ x = BD_1 $, which can be easily verified by calculating angles $ A_1 $ and $ C_1 $ from triangles $ A_1BD_1 $ and $ BC_1D_1 $ in terms of $ x $ and $ a $, $ b $, $ c $, $ d $. As point $ D_1 $ moves along the extension of segment $ BD $ starting from point $ D $, this function transitions from values less than $ 180^\circ $ to values greater than $ 180^\circ $, so it must pass through the value $ 180^\circ $, i.e., there exists a point $ D_1 $ such that quadrilateral $ A_1BC_1D_1 $ is cyclic, which was to be proved.
The above theorem b) is a special case of Cramer's theorem (Swiss mathematician, 1704 - 1752):
Among all polygons with given sides $ a_1, a_2, \ldots, a_n $ (where $ n \geq 3 $), the one with the largest area is the one that can be inscribed in a circle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,129 |
XLI OM - I - Problem 4
From a chessboard with dimensions $8 \times 80$ squares, colored in the usual way, two squares of different colors have been removed. Prove that the remaining $62$ squares can be covered by 31 rectangles of dimensions $2 \times 1$.
|
Let $ R $ be the smallest rectangle containing both removed fields $ A $ and $ B $. Since these fields are of different colors and lie in opposite corners of the rectangle $ R $, the lengths of the sides of $ R $ are natural numbers of different parity. Without loss of generality, we can assume that the rectangle $ R $ consists of $ k $ vertical rows and $ l $ horizontal rows, where $ k $ is an even number and $ l $ is an odd number (see Figure 2).
We will show that the rectangle $ R $ without fields $ A $ and $ B $ can be covered with tiles of dimensions $ 2 \times 1 $. When $ l = 1 $, we have to cover a strip of length $ k - 2 $ and width $ 1 $; the way to cover it is obvious. When $ l > 1 $, then each of the extreme vertical rows (without field $ A $, respectively $ B $) has dimensions $ (l-1) \times 1 $ and, due to the oddness of $ l $, can be covered with $ 2 \times 1 $ tiles, placed vertically. The remaining part of the rectangle $ R $ is a rectangle of dimensions $ (k-2) \times l $ (or an empty set if $ k = 2 $) and can be covered with tiles placed horizontally.
It remains to cover the rest of the chessboard. The lines containing the vertical sides of the rectangle $ R $ cut out two rectangular strips running the entire height of the chessboard; in special cases, one of them (or both) may be an empty set. Each of these strips is covered with vertical $ 2 \times 1 $ tiles; this is possible because $ 8 $ is an even number.
The part of the chessboard between the mentioned lines, after removing the rectangle $ R $, splits into two rectangles (upper and lower), each of width $ k $ (again: one of them - or both - may be an empty set). Since $ k $ is an even number, we can cover these areas with $ 2 \times 1 $ tiles, placed horizontally.
In this way, we have covered the entire chessboard without fields $ A $ and $ B $ with $ 2 \times 1 $ tiles. This is not the only possible way; there are many others. Note also that the method described here is effective for any chessboard with even dimensions.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,131 |
VII OM - I - Problem 11
In a triangle with sides $a$, $b$, $c$, segments $m$, $n$, $p$ tangent to the inscribed circle of the triangle have been drawn, with endpoints on the sides of the triangle and parallel to the sides $a$, $b$, $c$ respectively. Prove that
|
Let $2s$ denote the perimeter of the triangle, $P$ - its area, $r$ - the radius of the inscribed circle (Fig. 11).
The triangle cut off from the given triangle by the segment $m$ is similar to the given triangle, and its height relative to the side $m$ is equal to $h_a - 2r$, where $h_a$ denotes the height of the given triangle relative to the side $a$. Due to the similarity of the triangles we have
and since, as is known, $P = r \cdot s$, we have
Similarly
Therefore
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,133 |
XXXVIII OM - I - Problem 7
Given are non-negative real numbers $ a_1, a_2, \ldots , a_n $, $ x_1, x_2, \ldots , x_n $.
We assume $ a = \sum_{i=1}^n a_i $, $ c = \sum_{i=1}^n a_ix_i $. Prove that
|
The inequality $ \sqrt{1+x^2} \leq 1+x $ is satisfied for all $ x > 0 $ (verification by squaring, which, when performed within the realm of non-negative numbers, yields an equivalent inequality to the original). When we substitute $ x = x_i $, multiply both sides by $ a_i $ ($ i = 1, \ldots , n $), and sum up, we obtain the right-hand side of the inequality to be proven. It remains to prove the left-hand side of the inequality:
By writing $ c_i = a_i x_i $, we transform inequality (1) into an equivalent form:
Let the left and right sides of (2) be denoted by $ L $ and $ P $. We transform:
from which
Since $ \sqrt{(s^2+t^2)(u^2+v^2)} \geq su+tv $ for any quadruple of non-negative numbers $ s $, $ t $, $ u $, $ v $ (verification by squaring), all the terms of the resulting sum are non-negative. Therefore, $ L^2 \leq P^2 $, and hence $ L \leq P $ (since $ L, P \geq 0 $).
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,134 |
XXXV OM - I - Problem 5
Prove that there exists a multiple of the number $ 5^n $, whose decimal representation consists of $ n $ digits different from zero.
|
The number $5^n$ has in its decimal representation no more than $n$ digits: $5^n = c_0 + c_1 \cdot 10 + c_2 \cdot 10^2 + \ldots + c_{n-1} \cdot 10^{n-1}$, where some of the digits $c_i$ may be zeros. Suppose that $k$ is the smallest natural number for which $c_k = 0$. Consider the number $5^n + 10^k \cdot 5^{n-k}$. This number has in its decimal representation the form $c_0 + c_1 \cdot 10 + \ldots + c_{k-1} \cdot 10^{k-1} + 5 \cdot 10^k + \ldots$, where $c_i \ne 0$ for $i \leq k$. If any of the subsequent digits of this number is zero, e.g., $c_r = 0$ for some $r > k$, but $c_j \ne 0$ for $j < r$, then we consider the number $5^n + 10^k \cdot 5^{n-k} + 10^r \cdot 5^{n-r}$, which in its decimal representation has $r + 1$ consecutive digits different from zero.
As a result of such operations, we obtain a number in which the non-zero digits in its decimal representation correspond to the orders $1$, $10$, $10^2$, $10^{n-1}$. Moreover, this number does not exceed $5^n + 10 \cdot 5^{n-t} + 10^2 \cdot 5^{n-2} + \ldots + 10^{n-t} \cdot 5 = 5^n + 2 \cdot 5^n + 2^2 \cdot 5^n + \ldots + 2^{n-1} \cdot 5^n = 5^n(1 + 2 + 2^2 + \ldots + 2^{n-1}) = 5^n(2^n - 1) < 10^n$.
This is therefore a number that has in its decimal representation exactly $n$ digits different from zero. It is also a multiple of the number $5^n$, as it is formed by adding to $5^n$ terms of the form $10^k \cdot 5^{n-k} = 5^n \cdot 2^k$, which are multiples of the number $5^n$.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,136 |
XXV - I - Problem 5
Prove that for every $ k $, the $ k $-th digit from the end in the decimal representation of the numbers $ 5, 5^2, 5^3, \ldots $ forms a periodic sequence from some point onwards.
|
Let $ a_n $ be the number formed by the last $ k $ digits of the number $ 5^n $. It suffices to prove that the sequence $ \{a_n\} $ is periodic. The number $ a_n $ is the remainder of the division of the number $ 5^n $ by $ 10^k $, i.e., $ 5^n = 10^k \cdot b_n + a_n $, where $ b_n \geq 0 $ and $ 0 \leq a_n < 10^k $. Therefore, $ 5^{n+1} = 5 \cdot 10^k \cdot b_n + 5 \cdot a_n $ and $ 5^{n+1} = 10^k \cdot b_{n+1} + a_{n+1} $. Thus, the numbers $ 5^{n+1} $ and $ 5 \cdot a_n $ give the same remainder $ a_{n+1} $ when divided by $ 10^k $. Hence, $ a_{n+1} $ is the number formed by the last $ k $ digits of the number $ 5 \cdot a_n $. It follows that each term of the sequence $ \{a_n\} $ uniquely determines the next term. Since the sequence $ \{a_n\} $ can take only a finite number of values (no more than $ 10^k $), some term of the sequence must repeat: $ a_r = a_{r+t} $ for some natural numbers $ r $ and $ t $. Therefore, $ a_{r+1} = a_{r+t+1} $ and generally $ a_{r+s} = a_{r+t+s} $ for $ s = 1, 2, \ldots $. The sequence $ \{a_n\} $ is thus periodic from some point onward.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,137 |
XXXIV OM - I - Problem 12
A sphere is circumscribed around a tetrahedron $ABCD$. $\alpha, \beta, \gamma, \delta$ are planes tangent to this sphere at vertices $A, B, C, D$ respectively, where $\alpha \cap \beta = p$, $\gamma \cap \delta = q$, and $p \cap CD \neq \emptyset$. Prove that the lines $q$ and $AB$ lie in the same plane.
|
Let $ S $ and $ E $ be the sphere circumscribed around the tetrahedron $ ABCD $ and the point of intersection of the lines $ p $ and $ CD $, respectively. Point $ E $ lies outside the segment $ CD $; it can be assumed that it belongs to the ray $ CD^\to $, as shown in Fig. 4. Since $ E \subset \alpha $, the line $ AE $ is tangent to $ S $, and is therefore also tangent to the circle circumscribed around the triangle $ ACD $. If $ O $ is the center of the circle, then since $ 2 \measuredangle DAO+ \measuredangle AOD = 180^\circ $, the angle $ AOD $ is a central angle subtended by the arc $ \overline{AD} $, and the angle $ ACD $ is an inscribed angle subtended by the same arc, so $ \measuredangle ACD = \frac{1}{2} \measuredangle AOD $. Therefore,
Triangles $ EAD $ and $ ECA $ share the angle at vertex $ E $, and are therefore similar by the above equality. It follows that
Since $ E \in \beta $, the line $ BE $ is tangent to $ S $ and to the circle circumscribed around the triangle $ BCD $. Reasoning analogously, we obtain the proportion
From the last two equalities, it follows that
since $ EA = EB $ as the lengths of the segments tangent to the sphere $ S $ drawn from point $ E $.
Suppose that one of the planes $ \gamma $ and $ \delta $ is parallel to the line $ AB $. Assume that the relation $ \gamma || AB $ holds. Then $ AC = BC $ and from (1) it follows that $ AD = BD $, and thus $ \sigma || AB $. From this, we obtain that $ q =\gamma \cap \delta $ is a line parallel to $ AB $. Therefore, the lines $ q $ and $ AB $ lie in the same plane.
Assume that the line $ AB $ is not parallel to either of the planes $ \gamma $ and $ \delta $ and $ F $, $ G $ are the respective points of intersection. Reasoning analogously to the beginning of the solution, we obtain the proportions
The equalities $ FC = FB $, $ GD = GB $ hold, since the segments $ \overline{FC} $, $ \overline{FB} $, $ \overline{GD} $, and $ \overline{GB} $ drawn from $ F $ and $ G $, respectively, are tangent to $ S $. Therefore, by (1), we obtain the following equality
Points $ F $ and $ G $ lie outside the sphere $ S $, so they do not belong to the segment $ AB $, and since they belong to the line $ AB $, it follows from the above equality that $ F = G $. Therefore, $ F \in \gamma \cap \delta = q $. Thus, $ F $ is the point of intersection of the lines $ q $ and $ AB $, which implies that these lines lie in the same plane.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,138 |
XLVIII OM - III - Problem 3
The medians of the lateral faces $ABD$, $ACD$, $BCD$ of the pyramid $ABCD$ drawn from vertex $D$ form equal angles with the edges to which they are drawn. Prove that the area of each lateral face is less than the sum of the areas of the other two lateral faces.
|
We will prove the inequality
(the other two inequalities can be obtained analogously). Let $ K $, $ L $, $ M $ be the midpoints of the edges $ BC $, $ CA $, $ AB $, respectively. The segments $ DK $, $ DL $, $ DM $ form equal angles with the edges $ BC $, $ CA $, $ AB $, respectively; let this common angle be denoted by $ \alpha $. Let $ DH $ be the altitude of triangle $ ABD $. We have:
Similarly,
Since $ \sin\alpha \neq 0 $, using equations (2) and (3), we can rewrite inequality (1) in the following equivalent form:
Consider the tetrahedron $ KLMD $. Let $ N $ be a point in the plane $ KLM $ such that $ KN = KD $, $ LN = LD $, and points $ M $ and $ N $ lie on opposite sides of the line $ KL $. Let $ P $ be the intersection point of the line $ KL $ with the segment $ MN $. Since triangles $ KLD $ and $ KLN $ are congruent, $ DP = NP $. For the quadrilateral $ MKNL $, the Ptolemy's inequality holds:
From this, we obtain
which is inequality (4), equivalent to inequality (1).
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,141 |
LII OM - III - Task 3
We consider the sequence $ (x_n) $ defined recursively by the formulas
where $ a $ and $ b $ are real numbers.
We will call the number $ c $ a multiple value of the sequence $ (x_n) $ if there exist at least two different positive integers $ k $ and $ l $ such that $ x_k = x_l = c $. Prove that one can choose the numbers $ a $ and $ b $ such that the sequence $ (x_n) $ has more than $ 2000 $ multiple values, but it is not possible to choose $ a $ and $ b $ such that the sequence has infinitely many multiple values.
|
The sequence $ (x_n) $ satisfies the same recurrence relation as the Fibonacci sequence.
Let additionally $ F_0 = 0 $ and $ F_{-n} = (-1)^{n+1}F_n $ for $ n = 1,2,3,\ldots $. Then the equation $ F_{n+2} = F_{n+1} + F_n $ holds for any integer $ n $.
Let $ a = F_{-4001} $ and $ b = F_{-4000} $. Then $ x_n = F_{n-4002} $. For $ n = 1,3,5,7,\ldots,4001 $, we get $ x_n = F_{n-4002} = F_{4002-n} = x_{8004-n} $. Therefore, the numbers $ F_1,F_3,F_5,\ldots,F_{4001} $ are multiple values of the sequence $ (x_n) $; they are distinct and there are more than $ 2000 $ of them.
For any sequence $ (x_n) $ satisfying the condition
there exist real numbers $ A $ and $ B $ such that
where $ \alpha=\frac{1}{2}(1+\sqrt{5}) $, $ \beta=\frac{1}{2}(1-\sqrt{5}) $ are the roots of the polynomial $ x^2-x - 1 $.
If $ A = B = 0 $, then all terms of the sequence $ (x_n) $ are zeros and the number $ 0 $ is the only multiple value of the sequence $ (x_n) $.
If $ A = 0 $ and $ B \neq 0 $, then all terms of the sequence $ (x_n) $ are distinct and the sequence $ (x_n) $ has no multiple values.
If, however, $ A \ne 0 $, then
Thus, there exists a natural number $ N $ such that for all $ n \geq N $, $ x_n \ne 0 $. Moreover,
hence $ \lim_{n\to \infty} \frac{x_{n+1}}{x_n} =\alpha>1 $.
This implies that the sequence $ (x_n) $ is increasing from some point onward and therefore cannot have infinitely many multiple values.
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,142 |
LIV OM - III - Task 6
Let n be a positive even integer. Prove that there exists a permutation $ (x_1,x_2, \ldots ,x_n) $ of the set $ \{1,2,\ldots,n\} $, satisfying for every $ i \in \{1,2,\ldots ,n\} $ the condition:
$ x_{i+1} $ is one of the numbers $ 2x_i $, $ 2x_i-1 $, $ 2x_i-n $, $ 2x_i-n-1 $,
where $ x_{n+1} = x_1 $.
|
Let's adopt the notation:
We consider a directed graph with $ n $ vertices numbered $ 1,2,\ldots,n $, where from each vertex $ x $, two directed edges emerge: to vertex $ f(x) $ and to vertex $ g(x) $. The thesis reduces to the existence of a closed path that passes through each vertex exactly once (a Hamiltonian cycle).
Starting from vertex $ 1 $ and applying the functions $ f $, $ g $, we can obtain: in one step - points $ 1 $ and $ 2 $; in two steps - points $ 1 $, $ 2 $, $ 3 $, $ 4 $; in three steps - points from $ 1 $ to $ 8 $; and so on. It is clear that every point is reachable from vertex $ 1 $.
Let $ S: x_1 \to x_2 \to \ldots \to x_m $ be the longest path passing through distinct vertices, among which is vertex $ 1 $. Both edges emerging from vertex $ x_m $ must lead to some points on path $ S $ (otherwise, the path could be extended). Thus, we have edges $ x_m \to x_k $, $ x_m \to x_\ell $ ($ k,\ell \leq m $, $ k \neq \ell $); let, for example, $ f (x_m) = x_k $, $ g(x_m) = x_\ell $.
Suppose that $ k,\ell > 1 $. Then
The function $ f $ takes only even values, and $ g $ takes only odd values. Moreover, each of the equalities $ f (x) = f (y) $ and $ g(x) = g(y) $ implies $ x \equiv y (\mathrm{mod}\ n/2) $. Therefore,
we have $ x_m \equiv x_{k-1} \equiv x_{\ell-1} (\mathrm{mod}\ n/2) $ - a contradiction, because $ x_m, x_{k-1}, x_{\ell- 1} $ are three different numbers from the set $ \{1, 2, \ldots , n\} $.
Thus, $ k = 1 $ or $ \ell = 1 $, which means that $ x_m \to x_1 $ is an edge of the graph. This edge completes path $ S $ into a closed cycle. Assume that there are still some vertices of the graph left outside this cycle. They are reachable from vertex $ 1 $ (which lies on path $ S $). Therefore, from some point on this path, $ x_p $, there must be an edge to a point $ y $ outside the path. Then $ x_{p+1} \to \ldots \to x_m \to x_1 \to \ldots \to x_p \to y $ is a path passing through $ m+1 $ distinct vertices (among which is point $ 1 $), contradicting the maximality of $ m $.
Conclusion: the path $ S $, completed by the edge $ x_m \to x_1 $, is a cycle passing through all vertices of the graph.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,143 |
XI OM - II - Problem 5
Given three distinct points $ A $, $ B $, $ C $ on a line and a point $ S $ outside this line; perpendiculars drawn from points $ A $, $ B $, $ C $ to the lines $ SA $, $ SB $, $ SC $ intersect at points $ M $, $ N $, $ P $. Prove that the points $ M $, $ N $, $ P $, $ S $ lie on a circle.
|
We will adopt the notation such that point $B$ lies between points $A$ and $C$, point $M$ lies on the perpendiculars to $SB$ and $SC$, point $N$ - on the perpendiculars to $SC$ and $SA$, and point $P$ - on the perpendiculars to $SA$ and $SB$.
If any of the lines $SA$, $SB$, $SC$ is perpendicular to a given line, then two of the points $M$, $N$, $P$, for example, points $N$ and $P$, coincide with two of the points $ABC$, in which case the angles $SNM$ and $SPM$ are right angles. The points $M$, $N$, $P$, $S$ lie on a circle with diameter $SM$ (Fig. 21).
If none of the lines $SA$, $SB$, $SC$ is perpendicular to the given line, then one of two cases occurs:
a) the oblique segments $SA$, $SB$, $SC$ lie on different sides of the perpendicular from point $S$ to the given line, for example, $SA$ on one side and $SB$ and $SC$ on the other (Fig. 22).
In this case, points $P$ and $N$ lie on the opposite side of the given line from point $S$, and point $M$ lies on the same side as point $S$. Since angles $SAP$ and $SBP$ are right angles, points $S$, $A$, $P$, $B$ lie on a circle with diameter $SP$, and the equality of inscribed angles holds:
Similarly, since angles $SAN$ and $SCN$ are right angles, points $S$, $A$, $N$, $C$ lie on a circle with diameter $SN$, from which we infer, as above, that
Since $\measuredangle SAB = \measuredangle SAC$, because point $B$ lies on segment $AC$, it follows from the above equalities that
But $\measuredangle SPB = \measuredangle SPM$, and $\measuredangle SNC = \measuredangle SNM$, because point $B$ lies on segment $PM$, and point $C$ on segment $NM$, hence
It follows that points $P$ and $N$ lie on a certain circle passing through points $S$ and $M$.
b) All three segments $SA$, $SB$, $SC$ lie on the same side of the perpendicular from point $S$ to the given line (Fig. 23). In this case, all three points $M$, $N$, $P$ lie on the same side of the given line as point $S$ (The reader will wish to justify this fact rigorously; it is best to use the theorem that two lines intersected by a third line intersect on the side of that third line where the sum of the interior angles on the same side is less than $180^\circ$). Since $\measuredangle SAP = \measuredangle SBP = 90^\circ$, points $S$, $A$, $B$, $P$ lie on a circle with diameter $SP$, with points $S$ and $B$ separating points $A$ and $P$, so the angles at vertices $A$ and $P$ inscribed in this circle sum to $180^\circ$, i.e.,
Similarly, $\measuredangle SAN = \measuredangle SCN = 90^\circ$, so points $S$, $A$, $C$, $N$ lie on a circle with diameter $SN$, with pairs of points $S$, $C$ and $A$, $N$ separating each other, so as above
But $\measuredangle SAB = \measuredangle SAC$, so from the above equalities it follows that
Since point $B$ lies between points $A$ and $C$, point $P$ lies between points $B$ and $M$ (since if point $A$ moves in the direction of $AC$, point $P$ moves away from point $B$), and point $N$ between points $C$ and $M$. Therefore, $\measuredangle SPB = 180^\circ - \measuredangle SPM$, $\measuredangle SNC = 180^\circ - \measuredangle SNM$, and from the previous equality we obtain
It follows, as before, that points $P$ and $N$ lie on a circle passing through points $S$ and $M$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,145 |
XXXVII OM - III - Problem 5
In a chess tournament, $2n$ ($n > 1$) players participate, and any two of them play at most one game against each other. Prove that such a tournament, in which no three players play three games among themselves, is possible if and only if the total number of games played in the tournament does not exceed $n^2$.
|
We need to prove the equivalence of the following two statements $A$ and $B$ (for given natural numbers $m$ and $n$, $n > 1$):
A. It is possible to organize a tournament with $2n$ participants such that
$A_1$) every player plays at most one game with each other player,
$A_2$) no trio of players plays three games among themselves,
$A_3$) the total number of games in the tournament is $m$.
B. The numbers $m$ and $n$ satisfy the inequality $m \leq n^2$.
Proof of the implication $B \Longrightarrow A$. We divide the participants into two groups of $n$ players each. The number of pairs where the partners belong to different groups is $n^2$. We select any $m$ of these pairs and conduct the games in the selected pairs. The satisfaction of conditions $A_1$, $A_2$, and $A_3$ is obvious.
Proof of the implication $A \Longrightarrow B$. We use induction on $n$. When $n = 2$, we have $4$ players and $6$ possible pairs. If the number of games played exceeds $4$, it means that all possible matches are played except for at most one; this would then form a "triangle of matches" (and even two such triangles).
Assume the statement is true for some $n \geq 2$ and suppose a tournament with $2(n + 1)$ participants has been organized such that conditions $A_1$, $A_2$, and $A_3$ are satisfied. We need to show that $m \leq (n + 1)^2$. Choose any two players who play a game against each other and name them $X$ and $Y$. The number of games played between the remaining $2n$ players does not exceed $n^2$, by the induction hypothesis. Let $x$ be the number of games played by player $X$ (excluding the match with $Y$), and $y$ be the number of games played by player $Y$ (excluding the match with $X$). Then $x + y \leq 2n$; otherwise, $X$ and $Y$ would have a common opponent and a "triangle of matches" would form, contradicting condition $A_2$. Considering the match between $X$ and $Y$ as well, we see that the total number of matches in the tournament does not exceed $n^2 + 2n + 1$, which is $(n + 1)^2$. This completes the inductive proof.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,147 |
LII OM - II - Problem 2
Points $ A $, $ B $, $ C $ lie on a straight line in that order, with $ AB < BC $. Points $ D $, $ E $ are vertices of the square $ ABDE $. The circle with diameter $ AC $ intersects the line $ DE $ at points $ P $ and $ Q $, where point $ P $ lies on the segment $ DE $. The lines $ AQ $ and $ BD $ intersect at point $ R $. Prove that $ DP = DR $.
|
Angle $ APC $ is a right angle as an inscribed angle based on the diameter (Fig. 1). Hence
Furthermore, from the equality of arcs $ AP $ and $ CQ $, the equality of angles $ ACP $ and $ CAQ $ follows, hence
om52_2r_img_1.jpg
Triangles $ PAE $ and $ RBA $ are therefore congruent (angle-side-angle criterion), which proves that $ EP = BR $. Hence $ DP = DR $.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,148 |
XXIII OM - I - Problem 10
Given six points in space, not all in the same plane. Prove that among the lines determined by these points, there is a line that is not parallel to any of the others.
|
Let points $A_1, A_3, \ldots, A_6$ not lie in the same plane, and let $p_{ij}$ be the line containing points $A_i$ and $A_j$, where $\{i, j\} \ne \{s, t\}$. Suppose that each of the lines $p_{ij}$ is parallel to some line $p_{s,t}$. Since the number of lines $p_{ij}$ is $\binom{6}{2} = 15$, which is odd, there exists a triplet of parallel lines.
1) If the three parallel lines are distinct, then they do not lie in the same plane, because the given set of points is not contained in a single plane, and each line contains two given points. Therefore, the given points are the vertices of a triangular prism. The diagonal of any side face of the prism is not parallel to any other line determined by its vertices. This contradicts the initial assumption.
2) If the three parallel lines are not distinct, then there is a triplet of collinear points in the given set. Let, for example, points $A_1$, $A_2$, $A_3$ be collinear.
2a) If $A_4 \in A_1A_2$, then there exists a plane $\pi$ containing $A_1, A_2, A_3, A_4, A_5$, and from the conditions of the problem, $A_6 \not \in \pi$. Then, for example, the line $p_{16}$ is not parallel to any of the other lines $p_{ij}$. We have $p_{16} \nparallel p_{i6}$ for $i = 2, 3, 4, 5$, because these lines intersect at point $A_6$. Similarly, $p_{16} \nparallel p_{ij}$ for $i < j \leq 5$, because $p_{ij} \subset \pi$, and the line $p_{16}$ intersects $\pi$. Thus, we have a contradiction with the initial assumption in this case as well.
2b) If, finally, $A_4 \not \in A_1A_2$, then points $A_1, A_2, A_4$ determine some plane $\sigma$. We can also assume that $A_5, A_6 \not \in \sigma$. Then no two of the lines $p_{12}, p_{14}, p_{24}, p_{34}$ are parallel (Fig. 10). Therefore, the lines $p_{14}$ and $p_{24}$ are not parallel to any other line $p_{ij}$ contained in the plane $\sigma$, nor to any line intersecting $\sigma$. Thus, the lines $p_{14}$ and $p_{24}$ can only be parallel to the line $p_{56}$. However, both cannot be parallel to it, because $p_{14} \nparallel p_{24}$. Therefore, one of the lines $p_{14}, p_{24}$ is not parallel to any of the other lines $p_{ij}$. Thus, we have a contradiction with the initial assumption in this case as well. The initial assumption is therefore false.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,153 |
XLVII OM - I - Zadanie 6
Dane są dwa ciągi liczb całkowitych dodatnich: ciąg arytmetyczny o różnicy $ r > 0 $ i ciąg geometryczny o ilorazie $ q > 1 $; liczby naturalne $ r $, $ q $ są względnie pierwsze. Udowodnić, że jeśli te ciągi mają jeden wspólny wyraz, to mają nieskończenie wiele wspólnych wyrazów.
|
Oznaczmy przez $ r_i $ resztę z dzielenia liczby $ q^i $ przez $ r\ (i = 1,2,3,\ldots) $. W nieskończonym ciągu $ (r_1,r_2,r_3,\ldots) $, którego wyrazy przybierają tylko skończenie wiele wartości, muszą wystąpić powtórzenia. Istnieją zatem różne numery $ i $, $ j $ takie, że $ r_i = r_j $. Różnica $ q^i-q^j $ jest wówczas podzielna przez $ r $. Przyjmując, że $ i >j $, możemy tę różnicę zapisać tak:
Liczba po prawej stronie ma się dzielić przez $ r $ (bo lewa strona dzieli się przez $ r $). Czynnik $ q^j $ jest względnie pierwszy z $ r $ (warunek zadania). Zatem czynnik $ (q^{i-j}-1) $ musi dzielić się przez $ r $. Istnieje więc liczba całkowita $ l $ taka, że $ q^{i-j}-1=lr $. Oznaczając różnicę $ i-j $ przez $ k $ mamy równość $ q^k=1+lr $.
Niech $ (a_1,a_2,\ldots) $ będzie rozważanym ciągiem arytmetycznym, a $ (b_1,b_2, \ldots) $ rozważanym ciągiem geometrycznym. Zakładamy, że mają one wspólny wyraz $ a_m = b_n $. Wówczas
gdzie $ s=la_m $. Znaleźliśmy w ten sposób nową parę wspólnych wyrazów rozpatrywanych ciągów (o numerach większych od $ m $ i $ n $). Powtarzając to rozuwanie znajdujemy nieskończenie wiele wspólnych wyrazów.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,154 |
III OM - II - Task 2
Prove that if $ a $, $ b $, $ c $, $ d $ are the sides of a quadrilateral that can have a circle circumscribed around it and a circle inscribed in it, then the area $ S $ of the quadrilateral is given by the formula
|
If a quadrilateral with sides $a$, $b$, $c$, $d$ is inscribed in a circle, then the area $S$ is given by (see problem 27) the formula
in which
If such a quadrilateral is also circumscribed about a circle, then
From equation (2), it follows that in such a quadrilateral
Substituting these values into formula (1), we obtain
which was to be shown.
Note. Let us ask whether the converse theorem holds, i.e., whether from the assumption that the area of the quadrilateral is given by formula (3), it follows that a circle can be circumscribed about the quadrilateral and that a circle can be inscribed in the quadrilateral. It is not difficult to see that this is not the case. For example, if the quadrilateral is a rectangle with sides $a$, $b$, $c = a$, $d = b$, where $a \ne b$, then $S = ab = \sqrt{(ab)^2} = \sqrt{abcd}$, but a circle cannot be inscribed in this quadrilateral, although a circle can be circumscribed about it. This example can be modified to obtain a quadrilateral that satisfies condition (3) but is neither circumscribed about a circle nor inscribed in a circle. For this purpose, instead of a rectangle, take a quadrilateral $ABCD$ inscribed in a circle, where $AB = a$, $BC = b$, $CD = c$, $DA = d$, with $a = b$, $c \ne d$, and angles $B$ and $D$ are right angles (Fig. 27).
The area of quadrilateral $ABCD$ is given by the formula
The right-angled isosceles triangle $ABC$ has the same base as the right-angled triangle $ADC$, but a greater height, so $\text{area of } ABC > \text{area of } ADC$, i.e., $ab > cd$; since the arithmetic mean of the unequal numbers $ab$ and $cd$ is greater than their geometric mean $\sqrt{abcd}$, we have
If, keeping the lengths of the sides $a$, $b$, $c$, $d$ unchanged, we increase the diagonal $AC$, we obtain a quadrilateral $AB$ (Fig. 27) with an area $S$ smaller than $S$. The goal is to choose point $C$ in such a way that the equality $S$ holds. For this purpose, the obtuse angles $B$ and $D$ must be chosen so that two conditions are satisfied:
There is no need to conduct a general discussion of these equations; for our purpose, it suffices to provide a solution in a particular case. Suppose that the quadrilateral $ABCD$ is inscribed in a circle of radius $1$ and that $CD$ is a side of a regular hexagon inscribed in this circle, so $a = b = \sqrt{2}$, $c = 1$, $d = \sqrt{3}$. Substituting these values into the above equations, we obtain the equations
Squaring the first equation and then expressing the cosines of the angles in terms of their sines, we get the equation
Dividing this equation by the second of the previous equations, we have
Thus
The values obtained are, as is easily verified, less than $1$, so they determine the obtuse angles $B$ and $D$.
The quadrilateral $AB$ thus chosen has an area $S$, but it is neither inscribed in a circle (since it has two opposite obtuse angles) nor circumscribed about a circle (since the sums of the opposite sides are not equal).
We also note that a partially converse theorem to the theorem from problem 29 holds:
If the area $S$ of a quadrilateral with sides $a$, $b$, $c$, $d$ is given by the formula $S = \sqrt{abcd}$ and a circle can be inscribed in this quadrilateral, then a circle can also be circumscribed about it.
Indeed, the area of quadrilateral $ABCD$ with sides $a$, $b$, $c$, $d$ satisfies, according to formula (6) in problem 27, the condition
From the assumption that the quadrilateral is circumscribed about a circle, as we previously stated (based on formulas (2), p. 94), the relationship follows
From formulas (3), (4), and (5), we obtain
thus
from which
which proves that a circle can be circumscribed about quadrilateral $ABCD$.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,155 |
LIX OM - II - Task 4
In each field of a square table of size $ n \times n $, an integer is written. We can perform the following operation multiple times: We select any field of the table and decrease the number written in it by the number of adjacent fields (having a common side with the selected field), while increasing each of the numbers written in the adjacent fields by 1.
For each integer $ n \geqslant 2 $, determine whether from any initial table, in which the sum of all $ n^2 $ numbers is zero, one can obtain a table consisting entirely of zeros.
|
Let $ P $ denote the diagonal of the table connecting its top-left corner with the bottom-right corner. We will prove that as a result of performing the described operation, the parity of the sum of the numbers located on the diagonal $ P $ (the fields lying on the diagonal $ P $ are shaded in Fig. 2) does not change.
om59_2r_img_3.jpg
For each field on the diagonal $ P $, the number of adjacent fields is 2 or 4, and none of them lie on the diagonal $ P $. Each field adjacent to the diagonal $ P $ (marked with crosses in Fig. 2) is adjacent to exactly two fields lying on the diagonal $ P $. The remaining fields of the table do not lie on the diagonal $ P $ and do not adjoin it. Therefore, if the sum of the numbers on the diagonal $ P $ before performing the operation was $ s $, then as a result of its execution, this sum will be equal to $ s-4 $, $ s-2 $, $ s $, or $ s+2 $, i.e., its parity will not change.
Thus, if at the beginning we enter the number 1 in the top-left corner, the number -1 in the field located in the top-right corner, and the number 0 in each of the remaining fields, then the sum of all the numbers in the table will be zero, but after performing any number of operations, the sum of the numbers lying on the diagonal $ P $ will be odd. Therefore, it is not possible for the diagonal $ P $, let alone the entire table, to consist of all zeros.
Answer: For every $ n \geqslant 2 $, there exists an initial table with a sum of numbers equal to zero, from which it is not possible to obtain a table composed entirely of zeros.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,160 |
XV OM - II - Task 2
A circle is divided into four non-overlapping arcs $ AB $, $ BC $, $ CD $, and $ DA $. Prove that the segment connecting the midpoints of arcs $ AB $ and $ CD $ is perpendicular to the segment connecting the midpoints of arcs $ BC $ and $ DA $.
|
Let $ M $, $ N $, $ P $, $ Q $ denote the midpoints of arcs $ AB $, $ BC $, $ CD $, $ DA $ (Fig. 12).
The theorem will be proven when we show that the sum of the inscribed angles $ MNQ $ and $ PMN $ is $ 90^\circ $. The angle $ MNQ $ is subtended by the arc $ MAQ $ equal to the sum of the arcs $ MA $ and $ AQ $, and the inscribed angle $ PMN $ is subtended by the arc $ PCN $ equal to the sum of the arcs $ PC $ and $ CN $. Since inscribed angles in the same circle and subtended by equal arcs are equal, the sum of the inscribed angles $ MNQ $ and $ PMN $ equals the inscribed angle in the given circle subtended by an arc equal to the sum of the arcs $ MA $, $ AQ $, $ PC $, $ CN $. Now, arc $ MA $ = arc $ MB $, arc $ AQ $ = arc $ QD $, arc $ PC $ = arc $ DP $, arc $ CN $ = arc $ NB $, so the sum of the arcs $ MA $, $ AQ $, $ PC $, $ CN $ equals the sum of the arcs $ MB $, $ QD $, $ DP $, $ NB $, and each of these sums equals a semicircle. The sum of the inscribed angles $ MNQ $ and $ PMN $ equals the inscribed angle subtended by a semicircle, i.e., a right angle.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,163 |
LVI OM - I - Problem 5
Quadrilateral $ABCD$ is inscribed in a circle, and the circles inscribed in triangles $ABC$ and $BCD$ have equal radii. Determine whether it follows from these assumptions that the circles inscribed in triangles $CDA$ and $DAB$ also have equal radii.
|
The answer given to the question in the task is affirmative.
Let $ I $, $ J $ be the centers of the incircles of triangles $ ABC $ and $ BCD $ (Fig. 2).
om56_1r_img_2.jpg
om56_1r_img_3.jpg
From the equality of the radii of the incircles of triangles $ ABC $ and $ BCD $, it follows that lines $ BC $ and $ IJ $ are parallel. Furthermore,
Similarly, we prove that $ \measuredangle BJC = 90^\circ + \frac{1}{2} \measuredangle BDC $. From the obtained dependencies and the equality $ \measuredangle BAC = \measuredangle BDC $, we conclude that a circle can be circumscribed around quadrilateral $ BCJI $. Therefore, $ BCJI $ is an isosceles trapezoid. From this, we obtain the equalities
As a result, we get that quadrilateral $ ABCD $ is an isosceles trapezoid with bases $ BC $ and $ AD $. Hence, triangles $ CDA $ and $ DAB $ are congruent, and thus the incircles of these triangles have equal radii.
Note: In the solution, we have shown that points $ B $, $ I $, $ J $, $ C $ lie on the same circle. It is not difficult to determine the center of this circle: it is the point $ M $ of intersection of lines $ AI $ and $ DJ $, i.e., the midpoint of arc $ BC $ of the circumcircle of quadrilateral $ ABCD $ (Fig. 3). The proof of this fact can be found in the brochure LI Mathematical Olympiad, Report of the Main Committee, Warsaw 2001, Appendix E, p. 113.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,164 |
XL OM - II - Task 4
Given are integers $ a_1, a_2, \ldots , a_{11} $. Prove that there exists a non-zero sequence $ x_1, x_2, \ldots, x_{11} $ with terms from the set $ \{-1,0,1\} $, such that the number $ x_1a_1 + \ldots x_{11}a_{11} $ is divisible by 1989.
|
To each $11$-element sequence $(u_1, \ldots, u_{11})$ with terms equal to $0$ or $1$, we assign the number $u_1a_1 + \ldots + u_{11}a_{11}$. There are $2^{11}$ such sequences, which is more than $1989$. Therefore, there exists a pair of sequences $(u_1, \ldots, u_{11}) \neq (v_1, \ldots, v_{11})$, $u_i, v_i \in \{0,1\}$, such that the numbers assigned to these sequences
give the same remainder when divided by $1989$. This means that the number
is divisible by $1989$.
It suffices to set $x_i = u_i - v_i$ for $i = 1, \ldots, 11$. The terms of the sequence $(x_1, \ldots, x_{11})$ are equal to $0$, $1$, or $-1$, and not all of them are zeros.
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,165 |
5. From the equation $2 \cos ^{2} a_{n}=\cos a_{n+1}$, it follows that for all $n$, the value of $\cos a_{n+1} \geq 0$ and $\left|\cos a_{n}\right|=\sqrt{\frac{\cos a_{n+1}}{2}} \leq \frac{1}{\sqrt{2}}$, i.e., on the unit circle, all values of $a_{n}$ fall within the interval $[\pi / 4 ; \pi / 2]$.
From the fact that the numbers $a_{n}$ form an arithmetic progression, it follows that the common difference $d$ of this progression must be a multiple of the circumference of the circle $2 \pi$. Indeed, if the value of $d$ is greater than the nearest multiple of $2 \pi$ to the number $a_{n}$, then one of the subsequent terms of the sequence, located on the unit circle counterclockwise from $a_{n}$ and moving from it by a constant value along the arc, will exit the arc segment $[\pi / 4 ; \pi / 2]$. Similarly, if the value of $d$ is less than the nearest multiple of $2 \pi$ to the number $a_{n}$, then one of the subsequent terms of the sequence, located on the unit circle clockwise from $a_{n}$ and moving from it by a constant value along the arc, will also exit the arc segment $[\pi / 4 ; \pi / 2]$. Thus,
$$
d=2 \pi k, k \in Z
$$
Here, $k \neq 0$ because by the problem's condition, $d \neq 0$.
From the proven statement, it follows that for all $n$,
$$
\cos a_{n}=\cos \left(a_{1}+(n-1) d\right)=\cos \left(a_{1}+(n-1) \cdot 2 \pi k\right)=\cos a_{1}
$$
This fact gives an equation for determining the first term of the progression
$$
2 \cos ^{2} a_{1}=\cos a_{1}
$$
Solving the equation, we find
$$
\left[\begin{array} { l }
{ \operatorname { c o s } a _ { 1 } = 0 } \\
{ \operatorname { c o s } a _ { 1 } = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
a_{1}=\frac{\pi}{2}+2 \pi m, m \in Z \\
a_{1}= \pm \frac{\pi}{3}+2 \pi m, m \in Z
\end{array}\right.\right.
$$
The first series gives
$$
a_{n}=\frac{\pi}{2}+2 \pi m+(n-1) 2 \pi k \Rightarrow b_{n}=a_{n}
$$
and the two remaining series give
$$
a_{n}= \pm \frac{\pi}{3}+2 \pi m+(n-1) 2 \pi k \Rightarrow b_{n}= \pm \frac{\sqrt{3}}{2} a_{n}
$$
All series satisfy the conditions of the problem.
$$
\text { 1) } a_{1}=\frac{\pi}{2}+\pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0
$$
|
Answer: 2$) a_{1}=\frac{\pi}{3}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0$;
$$
\text { 3) } a_{1}=-\frac{\pi}{3}+2 \pi m, m \in Z ; d=2 \pi k, k \in Z, k \neq 0 \text {. }
$$
|
proof
|
Algebra
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,175 |
Problem 3. A line passing through any point $D$ on the median $B M$ of triangle $A B C$ and parallel to side $A B$, intersects the line passing through vertex $C$ and parallel to $B M$, at point $E$. Prove that $B E=A D$.
|
Solution. Additional construction: through point $M$ we draw line $MN$, parallel to $DE$. Figure $MDEN$ is a parallelogram (by construction).
Triangles $ABM$ and $MNC$ are equal (by two equal angles adjacent to equal sides $AM$ and $MC$). Then $AB=MN=DE$ and quadrilateral $ABED$ is a parallelogram. $AD$ and $BE$ are its opposite sides.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,179 |
Problem 2. Prove that for any positive numbers $a$ and $b$ the following inequality holds
$$
\sqrt{\frac{a^{2}}{b}}+\sqrt{\frac{b^{2}}{a}} \geq \sqrt{a}+\sqrt{b}
$$
|
Solution. Let $b \geq a$. Divide the inequality by $\sqrt{a}$:
$$
\begin{aligned}
& \sqrt{\frac{a}{b}}+\sqrt{\frac{b^{2}}{a^{2}}} \geq 1+\sqrt{\frac{b}{a}} \rightarrow \sqrt{\frac{b}{a}}\left(\sqrt{\frac{b}{a}}-1\right) \geq 1-\sqrt{\frac{a}{b}} \rightarrow \sqrt{\frac{b}{a}} \cdot \frac{1}{\sqrt{a}}(\sqrt{b}-\sqrt{a}) \geq \frac{(\sqrt{b}-\sqrt{a})}{\sqrt{b}} \rightarrow \\
& \rightarrow \frac{b}{a}(\sqrt{b}-\sqrt{a}) \geq(\sqrt{b}-\sqrt{a}) \rightarrow(\sqrt{b}-\sqrt{a})\left(\frac{b}{a}-1\right) \geq 0
\end{aligned}
$$
The last inequality is true. The case $a \geq b$ is analogous.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,181 |
11.8. Given real numbers $a_{1}, a_{2}$, ?, $a_{7}$ such that $a_{1}=a_{7}=0$. Is it true that one can always choose an index $k \leq 5$ such that the inequality $a_{k}+a_{k+2} \leq a_{k+1} \sqrt{3}$ holds?
|
Answer. Correct.
Solution. Suppose the opposite, that is, for any $k=1,2,3,4,5$ the inequality $a_{k}+a_{k+2}>a_{k+1} \sqrt{3}$ holds.
Then $\quad\left(a_{1}+a_{3}\right)+\left(a_{3}+a_{5}\right)>a_{2} \sqrt{3}+a_{4} \sqrt{3}=\left(a_{2}+a_{4}\right) \sqrt{3}>\left(a_{3} \sqrt{3}\right) \sqrt{3}=3 a_{3}$ $\left(a_{3}+a_{5}\right)+\left(a_{5}+a_{7}\right)>a_{4} \sqrt{3}+a_{6} \sqrt{3}=\left(a_{4}+a_{6}\right) \sqrt{3}>\left(a_{5} \sqrt{3}\right) \sqrt{3}=3 a_{5}$.
Adding these inequalities, we get
$$
\left(a_{1}+a_{3}\right)+\left(a_{3}+a_{5}\right)+\left(a_{3}+a_{5}\right)+\left(a_{5}+a_{7}\right)>3 a_{3}+3 a_{5}
$$
Considering that $a_{1}=a_{7}=0$, we obtain a contradiction $3 a_{3}+3 a_{5}>3 a_{3}+3 a_{5}$.
Comment. A correct answer without justification - 0 points.
|
proof
|
Inequalities
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,183 |
Problem 3. On the side $BC$ of an acute-angled triangle $ABC$, a point $D$ is chosen such that $AB + BD = DC$. Prove that $\angle ADC = 90^{\circ}$, given that $\angle B = 2 \angle C$.
|
Solution. Mark point $M$ on the extension of side $B C$ beyond point $B$ such that $A B = B M$. In the isosceles triangle $A B M$, the external angle at vertex $B$ is equal to the sum of the two equal angles at the base $A M$, so $\angle A M B = \frac{\angle B}{2} = \angle C$. Therefore, triangle $A M C$ is isosceles and $A M = A C$. Since $M D = M B + B D = A B + B D = D C$, segment $A D$ is the median of this isosceles triangle, but then it is also its height, i.e., $\angle A D C = 90^{\circ}$.
Remark. A similar solution is obtained if point $N$ is marked on segment $D C$ such that $B D = D N$.
## Criteria
The largest suitable criterion is used:
7 p. Any complete solution of the problem.
2 6. Either point $M$ or point $N$ is constructed, but there is no further progress.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,196 |
Problem 3. In an acute-angled triangle $ABC$, the altitude $AD$ is drawn. It turns out that $AB + BD = DC$. Prove that $\angle B = 2 \angle C$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solution. Mark point $M$ on the extension of side $B C$ beyond point $B$ such that $A B = B M$. Then, in the isosceles triangle $A B M$, the external angle at vertex $B$ is equal to the sum of the two equal angles at the base $A M$, so $\angle A B C = 2 \angle A M B$.
The triangle $A M C$ is also isosceles, since segment $A D$ is both its height and median (since $D C = A B + B D = B M + B D = M D$). Therefore, $\angle A M B = \angle A C B$ - its angles at the base.
Thus, $\angle A B C = 2 \angle A M B = 2 \angle A C B$, as required.
Remark. A similar solution is obtained if a point $N$ is marked on segment $D C$ such that $B D = D N$.
## Criteria
The largest suitable criterion is used:
7 points. Any complete solution to the problem.
5-6 points. It is correctly proven that triangles $A B M$ and $A M C$ are isosceles, but there is no further progress.
2 points. Point $M$ or point $N$ (or an analogous construction) is constructed, and there are further developments, but they do not lead to the correct answer.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,197 |
Problem 5. A grid rectangle $42 \times 44$ was cut along the grid lines into rectangles $1 \times 8$, one square $2 \times 2$, and one tetromino. Prove that this tetromino is also a square. (All possible tetrominoes are shown in the figure below, they can be rotated and flipped.)
|
Solution. Suppose the remaining tetromino is not a square. Then there exists a row (horizontal or vertical) that intersects this tetromino at exactly one cell. Now let's paint black the entire row, as well as all rows that are at a distance from it that is a multiple of four (for example, if the 11th row is chosen, then the rows with numbers $3, 7, 11, 15$, etc., are painted black). The remaining cells will be painted white. Since each black row contains an even number of black cells, the total number of black cells in the $42 \times 44$ rectangle is even.
Notice that each $1 \times 8$ rectangle contains an even number of black cells (0, 2, or 8). Each $2 \times 2$ square also contains an even number of black cells (0 or 2). Together with the single black cell in the remaining tetromino, we get an odd number of black cells. However, the total number of black cells in the $42 \times 44$ rectangle is even, which is a contradiction.
## Criteria
The following criteria are used, with the highest applicable criterion selected:
7 points. Any complete solution to the problem.
0 points. One or several possible dissections are provided where the tetromino is a square, but it is not proven that this tetromino cannot be non-square.
|
proof
|
Logic and Puzzles
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,198 |
6.5. 30 students are walking in pairs, with students of different heights in each pair. Prove that they can stand in a circle so that the height of each differs from the height of their neighbors.
|
Solution. Let one of the students be called Petya, and all other students of Petya's height be called Vasya. We will build a row from left to right, adding pairs. First, we place Petya with his partner, with Petya on the left. If there is a pair with Vasya, we place it next, with Vasya also on the left. Then to the left of Vasya will be Petya's partner, who differs in height from Vasya. We add all other pairs with Vasyas in the same way. When the Vasyas run out, we add the remaining pairs in any order. The students in a pair are of different heights, so at least one of them differs in height from the rightmost in the row. We place the one who differs on the left, and his partner on the right. As a result, any pair of neighbors in the row will be of different heights. Moreover, the rightmost in the row will differ in height from Petya, since all Vasyas were placed on the left and cannot be the rightmost. Therefore, the row can be closed into a circle.
Criterion. 3 points if the algorithm allows arranging everyone in an open row.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,199 |
9.3. Given an isosceles triangle \(ABC\). On the lateral side \(AB\), a point \(M\) is marked such that \(CM = AC\). Then, on the lateral side \(BC\), a point \(N\) is marked such that \(BN = MN\), and the angle bisector \(NH\) of triangle \(CNM\) is drawn. Prove that \(H\) lies on the median \(BK\) of triangle \(ABC\).
|
Solution. In isosceles triangles $A C M$ and $M N B$, $\angle A M C = \angle A$ and $\angle B M N = \angle B$. Therefore, $\angle C M N = 180^{\circ} - \angle A M C - \angle B M N = 180^{\circ} - \angle A - \angle B = \angle C = \angle A = \angle A M C$. Thus, $H$ is the intersection point of the angle bisectors of $\angle A M N$ and $\angle M N C$. Therefore, the distance from $H$ to line $M N$ is equal to the distances from $H$ to lines $A B$ and $B C$, respectively. Hence, point $H$ is equidistant from lines $A B$ and $B C$, which means that $B H$ is the angle bisector of $\angle A B C$. Since the triangle is isosceles, the angle bisector coincides with the median.
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,202 |
9.5. There are 64 checkers of three colors, divided into pairs such that the colors of the checkers in each pair are different. Prove that all the checkers can be arranged on a chessboard so that the checkers in each two-square rectangle are of different colors.
|
Solution. Let's distribute the checkers into three monochromatic piles. In each pile, there will be no more than 32 checkers. Now, let's distribute the largest pile on white squares, filling the rows starting from the top. The second largest pile will be distributed on black squares, filling the rows starting from the bottom.
We will call a row complete if it contains 4 checkers from the 1st pile or 4 checkers from the second pile.
From the first pile, no more than 3 checkers will not fall into a complete row, and from the second pile - also no more than 3. But in these two piles together, there are more than $2 / 3$ of all the checkers, that is, no less than 43. This means that no less than 43-6=37 checkers are in complete rows. Therefore, there are no less than 9 complete rows. But there are only 8 rows in total, so there must be a special row where 4 checkers from both piles have landed. The free squares of different colors lie on opposite sides of the special row and do not touch each other. We will distribute the checkers from the third pile on them.
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,203 |
Problem 2. In a popular intellectual game "Clash of Minds," 10 people participate. By the end of the game, each player scores an integer number of points. It turned out that in the semifinal, all the quantities of points scored by the players have different last digits. Prove that in the final of the game, where the players will collectively score twice as many points as in the semifinal, such a situation cannot occur.
|
2. In a popular intellectual game "Clash of Minds," 10 people participate. By the end of the game, each player scores an integer number of points. It turned out that in the semifinal, all the quantities of points scored by the players have different last digits. Prove that in the final of the game, where the players collectively score twice as many points as in the semifinal, such a situation cannot occur.
Assume the opposite - by the end of the final, the players scored such quantities of points that their last digits are different, i.e., they are 0, 1, 2, ..., 9 in some order. Then the last digit of the sum of all points coincides with the last digit of the sum \(0 + 1 + 2 + \ldots + 9 = 45\). On the other hand, by the condition, this sum is an even number. Contradiction.
- The solution involves the idea of calculating the last digit of the sum without further progress. 2 points
- The problem is solved under the assumption that each participant in the final will score twice as many points as in the semifinal. 0 points
|
proof
|
Number Theory
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,204 |
Problem 3. In an acute-angled triangle $A B C$, the altitude $A H$ is drawn. Let $P$ and $Q$ be the feet of the perpendiculars dropped from point $H$ to sides $A B$ and $A C$ respectively. Prove that $\angle B Q H=\angle C P H$.
|
3. In an acute triangle $ABC$, the altitude $AH$ is drawn. Let $P$ and $Q$ be the feet of the perpendiculars dropped from point $H$ to sides $AB$ and $AC$, respectively. Prove that $\angle B Q H = \angle C P H$.
Since points $P$ and $Q$ lie on the circle constructed with $AH$ as the diameter, the angles $\angle P Q A = \angle P H A$ are equal as inscribed angles. On the other hand, angles $\angle P H A$ and $\angle H B A$ are equal, as they both complement angle $B A H$ to a right angle. Angles $\angle A Q P$ and $\angle A B C$ are equal, so quadrilateral $B P Q C$ can be inscribed in a circle. Then the angles $\angle B P C$ and $\angle B Q C$ are equal as inscribed angles. The desired equality is obtained by subtracting $90^{\circ}$ from each of them.
$\mp$ The inscribed nature of quadrilateral $B P Q C$ is proven, but the solution is not complete. 3 points
- The inscribed nature of quadrilateral $A P H Q$ is proven. 1 point
|
proof
|
Geometry
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,205 |
Problem 5. In a physics competition, 17 schoolchildren are participating. The participants were offered 12 problems. As a result, each problem was solved correctly by more than half of the participants. Prove that there will definitely be three schoolchildren who, together, have solved all the problems.
http://olimpiadakurchatov.ru
|
5. In a physics competition, 17 schoolchildren are participating. The participants were given 12 problems. As a result, each problem was solved correctly by more than half of the participants. Prove that there will definitely be three schoolchildren who, together, have solved all the problems.
Let's estimate the number of triples of schoolchildren who did not solve the first problem. This problem was not solved by 8 schoolchildren or fewer, so the number of such triples does not exceed $8 \cdot 7 \cdot 6 / 6=56$. Since there are 12 problems in total, the number of triples of schoolchildren who did not solve some problem does not exceed $12 \times 56=672$. On the other hand, there are $17 \cdot 16 \cdot 15 / 6=680$ triples of schoolchildren in total, which is more. Therefore, there will be three schoolchildren who, together, have solved all the problems.
$\mp$ The solution involves the idea of counting the number of triples of participants who did not solve a specific problem. No less than 3 points
- Several specific cases were considered. 0 points
|
proof
|
Combinatorics
|
proof
|
olympiads
| false |
nlile/NuminaMath-1.5-proofs-only
|
train
| 1,206 |
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