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IMOSL-1974-2
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Prove that the squares with sides \(1 / 1, 1 / 2, 1 / 3, \ldots\) may be put into the square with side \(3 / 2\) in such a way that no two of them have any interior point in common.
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We denote by \(q_{i}\) the square with side \(\frac{1}{i}\) . Let us divide the big square into rectangles \(r_{i}\) by parallel lines, where the size of \(r_{i}\) is \(\frac{3}{2} \times \frac{1}{2^{i}}\) for \(i = 2,3,\ldots\) and \(\frac{3}{2} \times 1\) for \(i = 1\) (this can be done because \(1 + \sum_{i = 2}^{\infty} \frac{1}{2^{i}} = \frac{3}{2}\) ). In rectangle \(r_{1}\) , one can put the squares \(q_{1}, q_{2}, q_{3}\) , as is done on the figure. Also, since \(\frac{1}{2^{i}} + \cdots + \frac{1}{2^{i+1}-1} < 2^{i} \cdot \frac{1}{2^{i}} = 1 < \frac{3}{2}\) , in each \(r_{i}\) , \(i \geq 2\) , one can put \(q_{2^{i}}, \ldots , q_{2^{i+1}-1}\) . This completes the proof.
Remark. It can be shown that the squares \(q_{1}, q_{2}\) cannot fit in any square of side less than \(\frac{3}{2}\) .
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IMOSL-1974-3
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Let \(P(x)\) be a polynomial with integer coefficients. If \(n(P)\) is the number of (distinct) integers \(k\) such that \(P^{2}(k) = 1\) , prove that
\[n(P) - \deg (P)\leq 2,\]
where \(\deg (P)\) denotes the degree of the polynomial \(P\) .
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For \(\deg (P) \leq 2\) the statement is obvious, since \(n(P) \leq \deg (P^{2}) = 2 \deg (P) \leq \deg (P) + 2\) .
Suppose now that \(\deg (P) \geq 3\) and \(n(P) > \deg (P) + 2\) . Then there is at least one integer \(b\) for which \(P(b) = - 1\) , and at least one \(x\) with \(P(x) = 1\) . We may assume w.l.o.g. that \(b = 0\) (if necessary, we consider the polynomial \(P(x + b)\) instead). If \(k_{1}, \ldots , k_{m}\) are all integers for which \(P(k_{i}) = 1\) , then \(P(x) = Q(x)(x - k_{1}) \cdots (x - k_{m}) + 1\) for some polynomial \(Q(x)\) with integer coefficients. Setting \(x = 0\) we obtain \((- 1)^{m} Q(0) k_{1} \cdots k_{m} = 1 - P(0) = 2\) . It follows that \(k_{1} \cdots k_{m} \mid 2\) , and hence \(m\) is at most 3. The same holds for the polynomial \(- P(x)\) , and thus \(P(x) = - 1\) also has at most 3 integer solutions. This counts for 6 solutions of \(P^{2}(x) = 1\) in total, implying the statement for \(\deg (P) \geq 4\) .
It remains to verify the statement for \(n = 3\) . If \(\deg (P) = 3\) and \(n(P) = 6\) , then it follows from the above consideration that \(P(x)\) is either \(- (x^{2} - 1)(x - 2) + 1\) or \((x^{2} - 1)(x + 2) + 1\) . It is directly checked that \(n(P)\) equals only 4 in both cases.
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IMOSL-1974-4
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The sum of the squares of five real numbers \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\) equals 1. Prove that the least of the numbers \((a_{i} - a_{j})^{2}\) , where \(i, j = 1, 2, 3, 4, 5\) and \(i \neq j\) , does not exceed \(1 / 10\) .
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Assume w.l.o.g. that \(a_{1} \leq a_{2} \leq a_{3} \leq a_{4} \leq a_{5}\) . If \(m\) is the least value of \(|a_{i} - a_{j}|\) , \(i \neq j\) , then \(a_{i + 1} - a_{i} \geq m\) for \(i = 1,2,\ldots ,5\) , and consequently \(a_{i} - a_{j} \geq (i - j)m\) for any \(i,j \in \{1,\ldots ,5\}\) , \(i > j\) . Then it follows that
\[\sum_{i > j}(a_{i} - a_{j})^{2} \geq m^{2} \sum_{i > j}(i - j)^{2} = 50m^{2}.\]
On the other hand, by the condition of the problem,
\[\sum_{i > j}(a_{i} - a_{j})^{2} = 5 \sum_{i = 1}^{5} a_{i}^{2} - (a_{1} + \dots + a_{5})^{2} \leq 5.\]
Therefore \(50m^{2} \leq 5\) ; i.e., \(m^{2} \leq \frac{1}{10}\) .
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IMOSL-1974-5
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Let \(A_{r},B_{r},C_{r}\) be points on the circumference of a given circle \(S\) . From the triangle \(A_{r}B_{r}C_{r}\) , called \(\triangle_{r}\) , the triangle \(\triangle_{r + 1}\) is obtained by constructing the points \(A_{r + 1},B_{r + 1},C_{r + 1}\) on \(S\) such that \(A_{r + 1}A_{r}\) is parallel to \(B_{r}C_{r}\) , \(B_{r + 1}B_{r}\) is parallel to \(C_{r}A_{r}\) , and \(C_{r + 1}C_{r}\) is parallel to \(A_{r}B_{r}\) . Each angle of \(\triangle_{1}\) is an integer number of degrees and those integers are not multiples of 45. Prove that at least two of the triangles \(\triangle_{1},\triangle_{2},\ldots ,\triangle_{15}\) are congruent.
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All the angles are assumed to be oriented and measured modulo \(180^{\circ}\) . Denote by \(\alpha_{i}, \beta_{i}, \gamma_{i}\) the angles of triangle \(\triangle_{i}\) , at \(A_{i}, B_{i}, C_{i}\) respectively. Let us determine the angles of \(\triangle_{i + 1}\) . If \(D_{i}\) is the intersection of lines \(B_{i}B_{i + 1}\) and \(C_{i}C_{i + 1}\) , we have \(\angle B_{i + 1}A_{i + 1}C_{i + 1} = \angle D_{i}B_{i}C_{i + 1} = \angle B_{i}D_{i}C_{i + 1} + \angle D_{i}C_{i + 1}B_{i} = \angle B_{i}D_{i}C_{i} - \angle B_{i}C_{i + 1}C_{i} = -2\angle B_{i}A_{i}C_{i}\) . We conclude that
\[\alpha_{i + 1} = -2\alpha_{i},\quad \mathrm{and~analogously}\quad \beta_{i + 1} = -2\beta_{i},\quad \gamma_{i + 1} = -2\gamma_{i}.\]
Therefore \(\alpha_{r + t} = (- 2)^{t}\alpha_{r}\) . However, since \((- 2)^{12} \equiv 1\) (mod 45) and consequently \((- 2)^{14} \equiv (- 2)^{2}\) (mod 180), it follows that \(\alpha_{15} = \alpha_{3}\) , since all values are modulo \(180^{\circ}\) . Analogously, \(\beta_{15} = \beta_{3}\) and \(\gamma_{15} = \gamma_{3}\) , and moreover, \(\triangle_{3}\) and \(\triangle_{15}\) are inscribed in the same circle; hence \(\triangle_{3} \cong \triangle_{15}\) .
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IMOSL-1974-6
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Does there exist a natural number \(n\) for which the number
\[\sum_{k = 0}^{n}{\binom{2n + 1}{2k + 1}}2^{3k}\]
is divisible by 5?
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We set
\[x = \sum_{k = 0}^{n}\binom{2n + 1}{2k + 1}2^{3k} = \frac{1}{\sqrt{8}}\sum_{k = 0}^{n}\binom{2n + 1}{2k + 1}\sqrt{8}^{2k + 1},\] \[y = \sum_{k = 0}^{n}\binom{2n + 1}{2k}2^{3k} = \sum_{k = 0}^{n}\binom{2n + 1}{{2k}}\sqrt{8}^{2k}.\]
Both \(x\) and \(y\) are positive integers. Also, from the binomial formula we obtain
\[y + x\sqrt{8} = \sum_{i = 0}^{2n + 1}\binom{2n + 1}{i}\sqrt{8}^{i} = (1 + \sqrt{8})^{2n + 1},\]
and similarly \(y - x\sqrt{8} = (1 - \sqrt{8})^{2n + 1}\) .
Multiplying these equalities, we get \(y^{2} - 8x^{2} = (1 + \sqrt{8})^{2n + 1}(1 - \sqrt{8})^{2n + 1} = - 7^{2n + 1}\) . Reducing modulo 5 gives us
\[3x^{2} - y^{2} \equiv 2^{2n + 1} \equiv 2 \cdot (-1)^{n}.\]
Now we see that if \(x\) is divisible by 5, then \(y^{2} \equiv \pm 2\) (mod 5), which is impossible. Therefore \(x\) is never divisible by 5.
Second solution. Another standard way is considering recurrent formulas. If we set
\[x_{m} = \sum_{k}\binom{m}{2k + 1}8^{k},\qquad y_{m} = \sum_{k}\binom{m}{2k}8^{k},\]
then since \(\binom{a}{b}=\binom{a- 1}{b}+\binom{a- 1}{b- 1}\) , it follows that \(x_{m + 1} = x_{m} + y_{m}\) and \(y_{m + 1} =\) \(8x_{m} + y_{m}\) ; therefore \(x_{m + 1} = 2x_{m} + 7x_{m - 1}\) . We need to show that none of \(x_{2n + 1}\) are divisible by 5. Considering the sequence \(\{x_{m}\}\) modulo 5, we get that \(x_{m} =\) 0,1,2,1,1,4,0,3,1,3,3,2,0,4,3,4,4,1,....Zeros occur in the initial position of blocks of length 6, where each subsequent block is obtained by multiplying the previous one by 3 (modulo 5). Consequently, \(x_{m}\) is divisible by 5 if and only if \(m\) is a multiple of 6, which cannot happen if \(m = 2n + 1\)
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IMOSL-1974-7
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Let \(a_{i},b_{i}\) be coprime positive integers for \(i = 1,2,\ldots ,k\) , and \(m\) the least common multiple of \(b_{1},\ldots ,b_{k}\) . Prove that the greatest common divisor of \(a_{1}\frac{m}{b_{1}},\ldots ,a_{k}\frac{m}{b_{k}}\) equals the greatest common divisor of \(a_{1},\ldots ,a_{k}\) .
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Consider an arbitrary prime number \(p\) . If \(p \mid m\) , then there exists \(b_{i}\) that is divisible by the same power of \(p\) as \(m\) . Then \(p\) divides neither \(a_{i} \frac{m}{b_{i}}\) nor \(a_{i}\) , because \((a_{i}, b_{i}) = 1\) . If otherwise \(p \nmid m\) , then \(\frac{m}{b_{i}}\) is not divisible by \(p\) for any \(i\) , hence \(p\) divides \(a_{i}\) and \(a_{i} \frac{m}{b_{i}}\) to the same power. Therefore \((a_{1}, \ldots , a_{k})\) and \(\left(a_{1} \frac{m}{b_{1}}, \ldots , a_{k} \frac{m}{b_{k}}\right)\) have the same factorization; hence they are equal.
Second solution. For \(k = 2\) we can easily verify the formula
\[\left(m\frac{a_{1}}{b_{1}},m\frac{a_{2}}{b_{2}}\right) = \frac{m}{b_{1}b_{2}} (a_{1}b_{2},a_{2}b_{1}) = \frac{1}{b_{1}b_{2}} [b_{1},b_{2}](a_{1},a_{2})(b_{1},b_{2}) = (a_{1},a_{2}),\]
since \([b_{1},b_{2}]\cdot (b_{1},b_{2}) = b_{1}b_{2}\) . We proceed by induction:
\[\left(a_{1}\frac{m}{b_{1}},\ldots ,a_{k}\frac{m}{b_{k}},a_{k + 1}\frac{m}{b_{k + 1}}\right) = \left(\frac{m}{[b_{1},\ldots,b_{k}]} (a_{1},\ldots ,a_{k}),a_{k + 1}\frac{m}{b_{k + 1}}\right)\] \[= (a_{1},\ldots ,a_{k},a_{k + 1}).\]
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IMOSL-1974-8
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If \(a,b,c,d\) are arbitrary positive real numbers, find all possible values of
\[S = \frac{a}{a + b + d} +\frac{b}{a + b + c} +\frac{c}{b + c + d} +\frac{d}{a + c + d}.\]
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It is clear that
\[\frac{a}{a + b + c + d} +\frac{b}{a + b + c + d} +\frac{c}{a + b + c + d} +\frac{d}{a + b + c + d} < S\] \[\mathrm{and} S< \frac{a}{a + b} +\frac{b}{a + b} +\frac{c}{c + d} +\frac{d}{c + d},\]
or equivalently, \(1 < S < 2\) .
On the other hand, all values from (1, 2) are attained. Since \(S = 1\) for \((a, b, c, d) = (0, 0, 1, 1)\) and \(S = 2\) for \((a, b, c, d) = (0, 1, 0, 1)\) , due to continuity all the values from (1, 2) are obtained, for example, for \((a, b, c, d) = (x(1 - x), x, 1 - x, 1)\) , where \(x\) goes through \((0, 1)\) .
Second solution. Set
\[S_{1} = \frac{a}{a + b + d} +\frac{c}{b + c + d}\qquad \mathrm{and}\qquad S_{2} = \frac{b}{a + b + c} +\frac{d}{a + c + d}.\]
We may assume without loss of generality that \(a + b + c + d = 1\) . Putting \(a + c = x\) and \(b + d = y\) (then \(x + y = 1\) ), we obtain that the set of values of
\[S_{1} = \frac{a}{1 - c} +\frac{c}{1 - a} = \frac{2ac + x - x^{2}}{ac + 1 - x}\]
is \(\left(x,\frac{2x}{2 - x}\right]\) . Having the analogous result for \(S_{2}\) in mind, we conclude that the values that \(S = S_{1} + S_{2}\) can take are \(\left(x + y,\frac{2x}{2 - x} +\frac{2y}{2 - y}\right]\) . Since \(x + y = 1\) and
\[\frac{2x}{2 - x} +\frac{2y}{2 - y} = \frac{4 - 4xy}{2 + xy}\leq 2\]
with equality for \(xy = 0\) , the desired set of values for \(S\) is \((1,2)\) .
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IMOSL-1974-9
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Let \(x,y,z\) be real numbers each of whose absolute value is different from \(1 / \sqrt{3}\) such that \(x + y + z = xyz\) . Prove that
\[\frac{3x - x^{3}}{1 - 3x^{2}} +\frac{3y - y^{3}}{1 - 3y^{2}} +\frac{3z - z^{3}}{1 - 3z^{2}} = \frac{3x - x^{3}}{1 - 3x^{2}}\cdot \frac{3y - y^{3}}{1 - 3y^{2}}\cdot \frac{3z - z^{3}}{1 - 3z^{2}}.\]
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There exist real numbers \(a,b,c\) with \(\tan a = x\) , \(\tan b = y\) , \(\tan c = z\) . Then using the additive formula for tangents we obtain
\[\tan (a + b + c) = \frac{x + y + z - xyz}{1 - xy - xz - yz}.\]
We are given that \(xyz = x + y + z\) . In this case \(xy + yz + zx = 1\) is impossible; otherwise, \(x,y,z\) would be the zeros of a cubic polynomial \(t^{3} - \lambda t^{2} + t - \lambda = (t^{2} + 1)(t - \lambda)\) (where \(\lambda = xyz\) ), which has only one real root. It follows that
\[x + y + z = xyz\iff \tan (a + b + c) = 0. \quad (1)\]
Hence \(a + b + c = k\pi\) for some \(k\in \mathbb{Z}\) . We note that \(\frac{3x - x^{3}}{1 - 3x^{2}}\) actually expresses \(\tan 3a\) . Since \(3a + 3b + 3c = 3k\pi\) , the result follows from (1) for the numbers \(\frac{3x - x^{3}}{1 - 3x^{2}}\) , \(\frac{3y - y^{3}}{1 - 3y^{2}}\) , \(\frac{3z - z^{3}}{1 - 3z^{2}}\) .
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IMOSL-1974-10
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Let \(\triangle ABC\) be a triangle. Prove that there exists a point \(D\) on the side \(AB\) such that \(CD\) is the geometric mean of \(AD\) and \(BD\) if and only if \(\sqrt{\sin A\sin B}\leq \sin \frac{C}{2}\) .
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If we set \(\angle ACD = \gamma_{1}\) and \(\angle BCD = \gamma_{2}\) for a point \(D\) on the segment \(AB\) , then by the sine theorem,
\[f(D) = \frac{CD^{2}}{AD\cdot BD} = \frac{CD}{AD}\cdot \frac{CD}{BD} = \frac{\sin\alpha\sin\beta}{\sin\gamma_{1}\sin\gamma_{2}}.\]
The denominator of the last fraction is
\[\sin \gamma_{1}\sin \gamma_{2} = \frac{1}{2} (\cos (\gamma_{1} - \gamma_{2}) - \cos (\gamma_{1} + \gamma_{2}))\] \[\qquad = \frac{1}{2} (\cos (\gamma_{1} - \gamma_{2}) - \cos \gamma)\leq \frac{1 - \cos\gamma}{2} = \sin^{2}\frac{\gamma}{2}.\]
Now we deduce that the set of values of \(f(D)\) is the interval \(\left[\frac{\sin\alpha\sin\beta}{\sin^{2}\frac{\gamma}{2}}, + \infty\right)\) . Hence \(f(D) = 1\) (equivalently, \(CD^{2} = AD\cdot BD\) ) is possible if and only if \(\sin \alpha \sin \beta \leq \sin^{2}\frac{\gamma}{2}\) , i.e.,
\[\sqrt{\sin\alpha\sin\beta}\leq \sin \frac{\gamma}{2}.\]
Second solution. Let \(E\) be the second point of intersection of the line \(CD\) with the circumcircle \(k\) of \(ABC\) . Since \(AD\cdot BD = CD\cdot ED\) (power of \(D\) with respect to \(k\) ),
\(C D^{2} = A D\cdot B D\) ie equivalent to \(E D = C D\) . Clearly the ratio \(\frac{E D}{C D} (D\in A B)\) takes a maximal value when \(E\) is the midpoint of the arc \(A B\) not containing \(C\) . (This follows from \(E D:C D = E^{\prime}D:C^{\prime}D\) when \(C^{\prime}\) and \(E^{\prime}\) are respectively projections from \(C\) and \(E\) onto \(A B\) .) On the other hand, it is directly shown that in this case
\[\frac{E D}{C D} = \frac{\sin^{2}\frac{\gamma}{2}}{\sin\alpha\sin\beta},\]
and the assertion follows.
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IMOSL-1974-11
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Consider a partition of an \(8\times 8\) chessboard into \(p\) rectangles whose interiors are disjoint such that each of them has an equal number of white and black cells. Assume that \(a_{1}< a_{2}< \dots < a_{p}\) , where \(a_{i}\) denotes the number of white cells in the \(i\) th rectangle. Find the maximal \(p\) for which such a partition is possible and for that \(p\) determine all possible corresponding sequences \(a_{1},a_{2},\dots,a_{p}\) .
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First, we notice that \(a_{1} + a_{2} + \dots +a_{p} = 32\) . The numbers \(a_{i}\) are distinct, and consequently \(a_{i}\geq i\) and \(a_{1} + \dots +a_{p}\geq p(p + 1) / 2\) . Therefore \(p\leq 7\) . The number 32 can be represented as a sum of 7 mutually distinct positive integers in the following ways:
\[32 = 1 + 2 + 3 + 4 + 5 + 6 + 11;\] \[32 = 1 + 2 + 3 + 4 + 5 + 7 + 10;\] \[32 = 1 + 2 + 3 + 4 + 5 - 8 + 9;\] \[32 = 1 + 2 + 3 + 4 + 6 + 7 + 9;\] \[32 = 1 + 2 + 3 + 5 + 6 + 7 + 8.\]
The case (1) is eliminated because there is no rectangle with 22 cells on an \(8\times 8\) chessboard. In the other cases the partitions are realized as below.
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IMOSL-1974-12
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In a certain language words are formed using an alphabet of three letters. Some words of two or more letters are not allowed, and any two such distinct words are of different lengths. Prove that one can form a word of arbitrary length that does not contain any nonallowed word.
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We say that a word is good if it doesn't contain any nonallowed word. Let \(a_{n}\) be the number of good words of length \(n\) . If we prolong any good word of length \(n\) by adding one letter to its end (there are \(3a_{n}\) words that can be so obtained), we get either
(i) a good word of length \(n + 1\) , or (ii) an \((n + 1)\) -letter word of the form \(X Y\) , where \(X\) is a good word and \(Y\) a nonallowed word.
The number of words of type (ii) with word \(Y\) of length \(k\) is exactly \(a_{n + 1 - k}\) ; hence the total number of words of kind (ii) doesn't exceed \(a_{n - 1} + \dots +a_{1} + a_{0}\) (where \(a_{0} = 1\) ). Hence
\[a_{n + 1}\geq 3a_{n} - (a_{n - 1} + \dots +a_{1} + a_{0}),\qquad a_{0} = 1,a_{1} = 3. \quad (1)\]
We prove by induction that \(a_{n + 1} > 2a_{n}\) for all \(n\) . For \(n = 1\) the claim is trivial. If it holds for \(i\leq n\) , then \(a_{i}\leq 2^{i - n}a_{n}\) ; thus we obtain from (1)
\[a_{n + 1} > a_{n}\left(3 - \frac{1}{2} -\frac{1}{2^{2}} -\dots -\frac{1}{2^{n}}\right) > 2a_{n}.\]
Therefore \(a_{n} \geq 2^{n}\) for all \(n\) (moreover, one can show from (1) that \(a_{n} \geq (n + 2)2^{n - 1}\) ); hence there exist good words of length \(n\) .
Remark. If there are two nonallowed words (instead of one) of each length greater than 1, the statement of the problem need not remain true.
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IMOSL-1975-1
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There are six ports on a lake. Is it possible to organize a series of routes satisfying the following conditions:
(i) Every route includes exactly three ports;
(ii) No two routes contain the same three ports;
(iii) The series offers exactly two routes to each tourist who desires to visit two different arbitrary ports?
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First, we observe that there cannot exist three routes of the form \((A,B,C)\) \((A,B,D)\) \((A,C,D)\) , for if \(E,F\) are the remaining two ports, there can be only one route covering \(A,E\) , namely, \((A,E,F)\) . Thus if \((A,B,C)\) \((A,B,D)\) are two routes, the one covering \(A,C\) must be w.l.o.g. \((A,C,E)\) . The other roots are uniquely determined: These are \((A,D,F)\) \((A,E,F)\) \((B,D,E)\) \((B,E,F)\) \((B,C,F)\) \((C,D,E)\) \((C,D,F)\)
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IMOSL-1975-2
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Let \(x_{1}\geq x_{2}\geq \dots \geq x_{n}\) and \(y_{1}\geq y_{2}\geq \dots \geq y_{n}\) be two \(n\) -tuples of numbers. Prove that
\[\sum_{i = 1}^{n}(x_{i} - y_{i})^{2}\leq \sum_{i = 1}^{n}(x_{i} - z_{i})^{2}\]
is true when \(z_{1},z_{2},\ldots ,z_{n}\) denote \(y_{1},y_{2},\ldots ,y_{n}\) taken in another order.
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Since there are finitely many arrangements of the \(z_{i}\) 's, assume that \(z_{1},\ldots ,z_{n}\) is the one for which \(\textstyle \sum_{i = 1}^{n}(x_{i} - z_{i})^{2}\) is minimal. We claim that in this case if \(i< j\) and \(x_{i}\neq x_{j}\) then \(z_{i}\geq z_{j}\) , from which the claim of the problem directly follows. Indeed, otherwise we would have
\[(x_{i} - z_{i})^{2} + (x_{j} - z_{i})^{2} = (x_{i} - z_{i})^{2} + (x_{j} - z_{\bar{i}})^{2}\] \[\qquad +2(x_{i}z_{i} + x_{j}z_{j} - x_{i}z_{j} - x_{j}z_{i})\] \[\qquad = (x_{i} - z_{i})^{2} + (x_{j} - z_{j})^{2} + 2(x_{i} - x_{j})(z_{i} - z_{j})\] \[\qquad \leq (x_{i} - z_{i})^{2} + (x_{j} - z_{j})^{2},\]
contradicting the assumption.
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IMOSL-1975-3
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Find the integer represented by \(\left[\sum_{n = 1}^{10^{9}}n^{-2 / 3}\right]\) . Here \([x]\) denotes the greatest integer less than or equal to \(x\) (e.g. \([\sqrt{2} ] = 1\) ).
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From \(\left((k + 1)^{2 / 3} + (k + 1)^{1 / 3}k^{1 / 3} + k^{2 / 3}\right)\left((k + 1)^{1 / 3} - k^{1 / 3}\right) = 1\) and \(3k^{2 / 3}<\) \((k + 1)^{2 / 3} + (k + 1)^{1 / 3}k^{1 / 3} + k^{2 / 3}< 3(k + 1)^{2 / 3}\) we obtain
\[3\left((k + 1)^{1 / 3} - k^{1 / 3}\right)< k^{-2 / 3}< 3\left(k^{1 / 3} - (k - 1)^{1 / 3}\right).\]
Summing from 1 to \(n\) we get
\[1 + 3\left((n + 1)^{1 / 3} - 2^{1 / 3}\right)< \sum_{k = 1}^{n}k^{-2 / 3}< 1 + 3(n^{1 / 3} - 1).\]
In particular, for \(n = 10^{9}\) this inequality gives
\[2997< 1 + 3\left((10^{9} + 1)^{1 / 3} - 2^{1 / 3}\right)< \sum_{k = 1}^{10^{9}}k^{-2 / 3}< 2998.\]
Therefore \(\left[\sum_{k = 1}^{10^{9}}k^{- 2 / 3}\right] = 2997\)
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IMOSL-1975-4
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Let \(a_{1},a_{2},\ldots ,a_{n},\ldots\) be a sequence of real numbers such that \(0\leq a_{n}\leq 1\) and \(a_{n} - 2a_{n + 1} + a_{n + 2}\geq 0\) for \(n = 1,2,3,\ldots\) . Prove that
\[0\leq (n + 1)(a_{n} - a_{n + 1})\leq 2\qquad \mathrm{for} n = 1,2,3,\ldots\]
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Put \(\Delta a_{n} = a_{n} - a_{n + 1}\) . By the imposed condition, \(\Delta a_{n} > \Delta a_{n + 1}\) . Suppose that for some \(n\) , \(\Delta a_{n}< 0\) : Then for each \(k\geq n\) , \(\Delta a_{k}< \Delta a_{n}\) ; hence \(a_{n} - a_{n + m} = \Delta a_{n}+\) \(\dots +\Delta a_{n + m - 1}< m\Delta a_{n}\) . Thus for sufficiently large \(m\) it holds that \(a_{n} - a_{n + m}<\) \(- 1\) , which is impossible. This proves the first part of the inequality.
Next one observes that
\[n\geq \sum_{k = 1}^{n}a_{k} = na_{n + 1} + \sum_{k = 1}^{n}k\Delta a_{k}\geq (1 + 2 + \dots +n)\Delta a_{n} = \frac{n(n + 1)}{2}\Delta a_{n}.\]
Hence \((n + 1)\Delta a_{n}\leq 2\)
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IMOSL-1975-5
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Let \(M\) be the set of all positive integers that do not contain the digit 9 (base 10). If \(x_{1},\ldots ,x_{n}\) are arbitrary but distinct elements in \(M\) , prove that
\[\sum_{j = 1}^{n}\frac{1}{x_{j}} < 80.\]
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There are exactly \(8 \cdot 9^{k - 1} k\) -digit numbers in \(M\) (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is \(10^{k}\) , and hence
\[\sum_{x_{j}< 10^{k}}\frac{1}{x_{j}} = \sum_{i = 1}^{k}\sum_{10^{i - 1}\leq x_{j}< 10^{i}}\frac{1}{x_{j}} < \sum_{i = 1}^{k}\sum_{10^{i - 1}\leq x_{j}< 10^{i}}\frac{1}{10^{i - 1}}\] \[\qquad = \sum_{i = 1}^{k}\frac{8\cdot 9^{i - 1}}{10^{i - 1}} = 80\left(1 - \frac{9^{k}}{10^{k}}\right)< 80.\]
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IMOSL-1975-6
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Let \(A\) be the sum of the digits of the number \(16^{16}\) and \(B\) the sum of the digits of the number \(A\) . Find the sum of the digits of the number \(B\) without calculating \(16^{16}\) .
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Let us denote by \(C\) the sum of digits of \(B\) . We know that \(16^{16} \equiv A \equiv B \equiv C \pmod{9}\) . Since \(16^{16} = 2^{64} = 2^{6 \cdot 10 + 4} \equiv 2^{4} \equiv 7\) , we get \(C \equiv 7 \pmod{9}\) . Moreover, \(16^{16} < 100^{16} = 10^{32}\) , hence \(A\) cannot exceed \(9 \cdot 32 = 288\) ; consequently, \(B\) cannot exceed 19 and \(C\) is at most 10. Therefore \(C = 7\) .
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IMOSL-1975-7
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Prove that from \(x + y = 1\) \((x,y\in \mathbb{R})\) it follows that
\[x^{m + 1}\sum_{j = 0}^{n}\binom{m + j}{j} y^{j} + y^{n + 1}\sum_{i = 0}^{m}\binom{n + i}{i} x^{i} = 1\qquad (m,n = 0,1,2,\ldots).\]
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We use induction on \(m\) . Denote by \(S_{m}\) the left-hand side of the equality to be proved. First \(S_{0} = (1 - y)(1 + y + \dots + y^{n}) + y^{n + 1} = 1\) , since \(x = 1 - y\) . Furthermore,
\[S_{m + 1} - S_{m\] \[= \binom{m + n + 1}{m + 1} x^{m + 1}y^{n + 1} + x^{m + 1}\sum_{j = 0}^{n}\left(\binom{m + 1 + j}{j}xy^{j} - \binom{m + j}{j}y^{j}\right)\] \[= \binom{m + n + 1}{m + 1} x^{m + 1}y^{n + 1\] \[+x^{m + 1}\sum_{j = 0}^{n}\left(\binom{m + 1 + j}{j}y^{j} - \binom{m + j}{j}y^{j} - \binom{m + 1 + j}{j}y^{j + 1}\right)\] \[= x^{m + 1}\left[\binom{m + n + 1}{n}y^{n + 1} + \sum_{j = 0}^{n}\left(\binom{m + j}{j - 1}y^{j} - \binom{m + j + 1}{j}y^{j + 1}\right)\right]\] \[= 0;\]
i.e., \(S_{m + 1} = S_{m} = 1\) for every \(m\) .
Second solution. Let us be given an unfair coin that, when tossed, shows heads with probability \(x\) and tails with probability \(y\) . Note that \(x^{m + 1} \binom{m + j}{j} y^{j}\) is the probability that until the moment when the \((m + 1)\) th head appears, exactly \(j\) tails \((j < n + 1)\) have appeared. Similarly, \(y^{n + 1} \binom{n + i}{i} x^{i}\) is the probability that exactly \(i\) heads will appear before the \((n + 1)\) th tail occurs. Therefore, the above sum is the probability that either \(m + 1\) heads will appear before \(n + 1\) tails, or vice versa, and this probability is clearly 1.
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IMOSL-1975-8
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On the sides of an arbitrary triangle \(ABC\) , triangles \(BPC\) , \(CQA\) , and \(ARB\) are externally erected such that
\[\angle PBC = \angle CAQ = 45^{\circ},\] \[\angle BCP = \angle QCA = 30^{\circ},\] \[\angle ABR = \angle BAR = 15^{\circ}.\]
Prove that \(\angle QRP = 90^{\circ}\) and \(QR = RP\) .
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Denote by \(K\) and \(L\) the feet of perpendiculars from the points \(P\) and \(Q\) to the lines \(BC\) and \(AC\) respectively.
Let \(M\) and \(N\) be the points on \(AB\) (ordered \(A - N - M - B\) ) such that the triangle \(RMN\) is isosceles with \(\angle R = 90^{\circ}\) . By sine theorem we have \(\frac{BM}{BA} = \frac{BM}{BR}\) . \(\frac{BR}{BA} = \frac{\sin 15^{\circ}}{\sin 45^{\circ}}\) . Since \(\frac{BK}{BC} = \frac{\sin 45^{\circ} \sin 30^{\circ}}{\cos 15^{\circ}} = \frac{\sin 15^{\circ}}{\sin 45^{\circ}}\) , we deduce that \(MK \parallel AC\) and \(MK = AL\) . Similarly, \(NL \parallel BC\) and \(NL = BK\) . It follows that the vectors \(\overrightarrow{RN}\) , \(\overrightarrow{NL}\) , and \(\overrightarrow{LQ}\) are the images of \(\overrightarrow{RM}\) , \(\overrightarrow{KP}\) , and \(\overrightarrow{MK}\) respectively under a rotation of \(90^{\circ}\) , and consequently the same holds for their sums \(\overrightarrow{RQ}\) and \(\overrightarrow{RP}\) . Therefore, \(QR = RP\) and \(\angle QRP = 90^{\circ}\) .
and \(NL = BK\) . It follows that the vectors \(\overrightarrow{RN}\) , \(\overrightarrow{NL}\) , and \(\overrightarrow{LQ}\) are the images of \(\overrightarrow{RM}\) , \(\overrightarrow{KP}\) , and \(\overrightarrow{MK}\) respectively under a rotation of \(90^{\circ}\) , and consequently the same holds for their sums \(\overrightarrow{RQ}\) and \(\overrightarrow{RP}\) . Therefore, \(QR = RP\) and \(\angle QRP = 90^{\circ}\) .
Second solution. Let \(ABS\) be the equilateral triangle constructed in the exterior of \(\triangle ABC\) . Obviously, the triangles \(BPC\) , \(BRS\) , \(ARS\) , \(AQC\) are similar. Let \(f\) be the rotational homothety centered at \(B\) that maps \(P\) onto \(C\) , and let \(g\) be the rotational homothety about \(A\) that maps \(C\) onto \(Q\) . The composition \(h = g \circ f\) is also a rotational homothety; its angle is \(\angle PBC + \angle CAQ = 90^{\circ}\) , and the coefficient is \(\frac{BC}{BP} \cdot \frac{AQ}{AC} = 1\) . Moreover, \(R\) is a fixed point of \(h\) because \(f(R) = S\) and \(g(S) = R\) . Hence \(R\) is the center of \(h\) , and the statement follows from \(h(P) = Q\) .
Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of \(RP, RQ, PQ\) by the cosine theorem.
Second remark. The problem allows a generalization: Given that \(\angle CBP = \angle CAQ = \alpha\) , \(\angle BCP = \angle ACQ = \beta\) , and \(\angle RAB = \angle RBA = 90^{\circ} - \alpha - \beta\) , show that \(RP = RQ\) and \(\angle PRQ = 2\alpha\) .
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IMOSL-1975-9
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Let \(f(x)\) be a continuous function defined on the closed interval \(0\leq x\leq\) 1. Let \(G(f)\) denote the graph of \(f(x)\) : \(G(f) = \{(x,y)\in \mathbb{R}^{2}\mid 0\leq x\leq 1,y = f(x)\}\) Let \(G_{a}(f)\) denote the graph of the translated function \(f(x - a)\) (translated over a distance \(a\) ), defined by \(G_{a}(f) = \{(x,y)\in \mathbb{R}^{2}\mid a\leq x\leq a + 1,y = f(x - a)\}\) Is it possible to find for every \(a\) , \(0< a< 1\) , a continuous function \(f(x)\) , defined on \(0\leq x\leq 1\) , such that \(f(0) = f(1) = 0\) and \(G(f)\) and \(G_{a}(f)\) are disjoint point sets?
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Suppose \(n\) is the natural number with \(na \leq 1 < (n + 1)a\) . Let \(f_a(x) = f(x - a)\) . If a function \(f\) with the desired properties exists, then \(f_a(a) = 0\) and let w.l.o.g. \(f(a) > 0\) , or equivalently, let the graph of \(f_a\) lie below the graph of \(f\) . In this case also \(f(2a) > f(a)\) , since otherwise, the graphs of \(f\) and \(f_a\) would intersect between \(a\) and \(2a\) . Continuing in this way we are led to \(0 = f(0) < f(a) < f(2a) < \dots < f(na)\) . Thus if \(na = 1\) , i.e., \(a = 1/n\) , such an \(f\) does not exist. On the other hand, if \(a \neq 1/n\) , then we similarly obtain \(f(1) > f(1 - a) > f(1 - 2a) > \dots > f(1 - na)\) . Choosing values of \(f\) at \(ia, 1 - ia, i = 1, \dots, n\) , so that they satisfy \(f(1 - na) < \dots < f(1 - a) < 0 < f(a) < \dots < f(na)\) , we can extend \(f\) to other values of \([0, 1]\) by linear interpolation. A function obtained this way has the desired property.
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IMOSL-1975-10
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The function \(f(x,y)\) is a homogeneous polynomial of the \(n\) th degree in \(x\) and \(y\) . If \(f(1,0) = 1\) and for all \(a,b,c\) ,
\[f(a + b,c) + f(b + c,a) + f(c + a,b) = 0,\]
prove that \(f(x,y) = (x - 2y)(x + y)^{n - 1}\) .
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We shall prove that for all \(x, y\) with \(x + y = 1\) it holds that \(f(x, y) = x - 2y\) . In this case \(f(x, y) = f(x, 1 - x)\) can be regarded as a polynomial in \(z = x - 2y = 3x - 2\) , say \(f(x, 1 - x) = F(z)\) . Putting in the given relation \(a = b = x/2\) , \(c = 1 - x\) , we obtain \(f(x, 1 - x) + 2f(1 - x/2, x/2) = 0\) ; hence \(F(z) + 2F(- z/2) = 0\) . Now \(F(1) = 1\) , and we get that for all \(k\) , \(F((- 2)^k) = (- 2)^k\) . Thus \(F(z) = z\) for infinitely many values of \(z\) ; hence \(F(z) \equiv z\) . Consequently \(f(x, y) = x - 2y\) if \(x + y = 1\) .
For general \(x,y\) with \(x + y\neq 0\) , since \(f\) is homogeneous, we have \(f(x,y) = (x+\) \(y)^{n}f\left(\frac{x}{x + y},\frac{y}{x + y}\right) = (x + y)^{n}\left(\frac{x}{x + y} - 2\frac{y}{x + y}\right) = (x + y)^{n - 1}(x - 2y)\) . The same is true for \(x + y = 0\) , because \(f\) is a polynomial.
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IMOSL-1975-11
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Let \(a_1, a_2, a_3, \ldots\) be any infinite increasing sequence of positive integers. (For every integer \(i > 0\) , \(a_{i+1} > a_i\).) Prove that there are infinitely many \(m\) for which positive integers \(x, y, h, k\) can be found such that \(0 < h < k < m\) and \(a_m = xa_h + ya_k\) .
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Let \((a_{k_{i}})\) be the subsequence of \((a_{k})\) consisting of all \(a_{k}\) 's that give remainder \(r\) upon division by \(a_{1}\) . For every \(i > 1\) , \(a_{k_{i}}\equiv a_{k_{1}}\) (mod \(a_{1}\) ); hence \(a_{k_{i}} = a_{k_{1}} + ya_{1}\) for some integer \(y > 0\) . It follows that for every \(r = 0,1,\ldots ,a_{1} - 1\) there is exactly one member of the corresponding \((a_{k_{i}})_{i\geq 1}\) that cannot be represented as \(x a_{l} +\) \(y a_{m}\) , and hence at most \(a_{1} + 1\) members of \((a_{k})\) in total are not representable in the given form.
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IMOSL-1975-12
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Consider on the first quadrant of the trigonometric circle the arcs \(AM_1 = x_1, AM_2 = x_2, AM_3 = x_3, \ldots, AM_v = x_v\) , such that \(x_1 < x_2 < x_3 < \cdots < x_v\) . Prove that
\[\sum_{i = 0}^{\nu -1}\sin 2x_i - \sum_{i = 0}^{\nu -1}\sin (x_i - x_{i + 1})< \frac{\pi}{2} +\sum_{i = 0}^{\nu -1}\sin (x_i + x_{i + 1}).\]
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Since \(\sin 2x_{i} = 2\sin x_{i}\cos x_{i}\) and \(\sin (x_{i} + x_{i + 1}) + \sin (x_{i} - x_{i + 1}) = 2\sin x_{i}\cos x_{i + 1}\) , the inequality from the problem is equivalent to
\[\begin{array}{c}{{(\cos x_{1}-\cos x_{2})\sin x_{1}+(\cos x_{2}-\cos x_{3})\sin x_{2}+\cdots}}\\ {{\cdots+(\cos x_{\nu-1}-\cos x_{\nu})\sin x_{\nu-1}< \frac{\pi}{4}.}}\end{array} \quad (1)\]
Consider the unit circle with center at \(O(0,0)\) and points \(M_{i}(\cos x_{i},\sin x_{i})\) on it. Also, choose the points \(N_{i}(\cos x_{i},0)\) and \(M_{i}^{\prime}(\cos x_{i + 1},\sin x_{i})\) . It is clear that \((\cos x_{i} - \cos x_{i + 1})\sin x_{i}\) is equal to the area of the rectangle \(M_{i}N_{i}N_{i + 1}M_{i}^{\prime}\) . Since all these rectangles are disjoint and lie inside the quarter circle in the first quadrant whose area is \(\frac{\pi}{4}\) , inequality (1) follows.
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IMOSL-1975-13
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Let \(A_0, A_1, \ldots, A_n\) be points in a plane such that
(i) \(A_0A_1 \leq \frac{1}{2} A_1A_2 \leq \dots \leq \frac{1}{2^{n-1}} A_{n-1}A_n\) and
(ii) \(0 < \angle A_0A_1A_2 < \angle A_1A_2A_3 < \dots < \angle A_{n-2}A_{n-1}A_n < 180^\circ\)
where all these angles have the same orientation. Prove that the segments \(A_kA_{k+1}, A_mA_{m+1}\) do not intersect for each \(k\) and \(m\) such that \(0 \leq k \leq m - 2 < n - 2\) .
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Suppose that \(A_{k}A_{k + 1}\cap A_{m}A_{m + 1}\neq \emptyset\) for some \(k,m > k + 1\) . Without loss of generality we may suppose that \(k = 0\) \(m = n - 1\) and that no two segments \(A_{k}A_{k + 1}\) and \(A_{m}A_{m + 1}\) intersect for \(0\leq k< m - 1< n - 1\) except for \(k = 0\) \(m = n - 1\) Also, shortening \(A_{0}A_{1}\) , we may suppose that \(A_{0}\in A_{n - 1}A_{n}\) . Finally, we may reduce the problem to the case that \(A_{0}\ldots A_{n - 1}\) is convex: Otherwise, the segment \(A_{n - 1}A_{n}\) can be prolonged so that it intersects some \(A_{k}A_{k + 1}\) \(0< k< n - 2\)
If \(n = 3\) , then \(A_{1}A_{2}\geq 2A_{0}A_{1}\) implies \(A_{0}A_{2} > A_{0}A_{1}\) , hence \(\angle A_{0}A_{1}A_{2} > \angle A_{1}A_{2}A_{3}\) a contradiction.
Let \(n = 4\) . From \(A_{3}A_{2} > A_{1}A_{2}\) we conclude that \(\angle A_{3}A_{1}A_{2} > \angle A_{1}A_{3}A_{2}\) . Using the inequality \(\angle A_{0}A_{3}A_{2} > \angle A_{0}A_{1}A_{2}\) we obtain that \(\angle A_{0}A_{3}A_{1} > \angle A_{0}A_{1}A_{3}\) implying \(A_{0}A_{1} > A_{0}A_{3}\) . Now we have \(A_{2}A_{3}< A_{3}A_{0} + A_{0}A_{1} + A_{1}A_{2}< 2A_{0}A_{1} + A_{1}A_{2}\leq\) \(2A_{1}A_{2}\leq A_{2}A_{3}\) , which is not possible.
Now suppose \(n\geq 5\) . If \(\alpha_{i}\) is the exterior angle at \(A_{i}\) , then \(\alpha_{1} > \dots >\alpha_{n - 1}\) ; hence \(\alpha_{n - 1}< \frac{360^{\circ}}{n - 1}\leq 90^{\circ}\) . Consequently \(\angle A_{n - 2}A_{n - 1}A_{0}\geq 90^{\circ}\) and \(A_{0}A_{n - 2} > A_{n - 1}A_{n - 2}\) . On the other hand, \(A_{0}A_{n - 2}< A_{0}A_{1} + A_{1}A_{2} + \dots +A_{n - 3}A_{n - 2}< \left(\frac{1}{2^{n - 2}} +\frac{1}{2^{n - 3}} +\dots +\frac{1}{2}\right)A_{n - 1}A_{n - 2}< A_{n - 1}A_{n - 2}\) , which contradicts the previous relation.
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IMOSL-1975-14
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Let \(x_0 = 5\) and \(x_{n+1} = x_n + \frac{1}{x_n}\) \((n = 0, 1, 2, \ldots)\) . Prove that \(45 < x_{1000} < 45.1\) .
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We shall prove that for every \(n\in \mathbb{N}\) , \(\sqrt{2n + 25}\leq x_{n}\leq \sqrt{2n + 25} +0.1\) . Note that for \(n = 1000\) this gives us exactly the desired inequalities.
First, notice that the recurrent relation is equivalent to
\[2x_{k}(x_{k + 1} - x_{k}) = 2. \quad (1)\]
Since \(x_{0}< x_{1}< \dots < x_{k}< \dots\) , from (1) we get \(x_{k + 1}^{2} - x_{k}^{2} = (x_{k + 1} + x_{k})(x_{k + 1} - x_{k}) > 2\) . Adding these up we obtain \(x_{n}^{2}\geq x_{0}^{2} + 2n\) , which proves the first inequality.
On the other hand, \(x_{k + 1} = x_{k} + \frac{1}{x_{k}}\leq x_{k} + 0.2\) (for \(x_{k}\geq 5\) ), and one also deduces from (1) that \(x_{k + 1}^{2} - x_{k}^{2} - 0.2(x_{k + 1} - x_{k}) = (x_{k + 1} + x_{k} - 0.2)(x_{k + 1} - x_{k})\leq 2\) . Again, adding these inequalities up, \((k = 0,\ldots ,n - 1)\) yields
\[x_{n}^{2}\leq 2n + x_{0}^{2} + 0.2(x_{n} - x_{0}) = 2n + 24 + 0.2x_{n}.\]
Solving the corresponding quadratic equation, we obtain
\[x_{n}< 0.1 + \sqrt{2n + 24.01} < 0.1 + \sqrt{2n + 25}.\]
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IMOSL-1975-15
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Is it possible to plot 1975 points on a circle with radius 1 so that the distance between any two of them is a rational number (distances have to be measured by chords)?
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Assume that the center of the circle is at the origin \(O(0,0)\) , and that the points \(A_{1},A_{2},\ldots ,A_{1975}\) are arranged on the upper half-circle so that \(\angle A_{1}OA_{1} = \alpha_{i}\) ( \(\alpha_{1} =\) 0). The distance \(A_{i}A_{j}\) equals \(2\sin {\frac{\alpha_{j} - \alpha_{i}}{2}} = 2\sin {\frac{\alpha_{j}}{2}}\cos {\frac{\alpha_{j}}{2}} - \cos {\frac{\alpha_{j}}{2}}\sin {\frac{\alpha_{j}}{2}}\) , and it will be rational if all \(\sin {\frac{\alpha_{k}}{2}},\cos {\frac{\alpha_{k}}{2}}\) are rational.
Finally, observe that there exist infinitely many angles \(\alpha\) such that both \(\sin \alpha\) , \(\cos \alpha\) are rational, and that such \(\alpha\) can be arbitrarily small. For example, take \(\alpha\) so that \(\sin \alpha = \frac{2t}{t^{2} + 1}\) and \(\cos \alpha = \frac{t^{2} - 1}{t^{2} + 1}\) for any \(t \in \mathbb{Q}\) .
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IMOSL-1976-1
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Let \(ABC\) be a triangle with bisectors \(AA_{1}, BB_{1}, CC_{1}\) \((A_{1} \in BC\) , etc.) and \(M\) their common point. Consider the triangles \(MB_{1}A, MC_{1}A, MC_{1}B, MA_{1}B, MA_{1}C, MB_{1}C\) , and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then \(AB = BC = CA\) .
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Let \(r\) denote the common inradius. Some two of the four triangles with the inradii \(\rho\) have cross angles at \(M\) : Suppose these are \(\triangle A M B_{1}\) and \(\triangle B M A_{1}\) . We shall show that \(\triangle A M B_{1}\cong \triangle B M A_{1}\) . Indeed, the altitudes of these two triangles are both equal to \(r\) , the inradius of \(\triangle A B C\) , and their interior angles at \(M\) are equal to some angle \(\phi\) . If \(P\) is the point of tangency of the incircle of \(\triangle A_{1}M B\) with \(M B\) , then \(\frac{r}{\rho} = \frac{A_{1}M + B M + A_{1}B}{A_{1}B}\) , which also implies \(\frac{r - 2\rho}{\rho} = \frac{A_{1}M + B M - A_{1}B}{A_{1}B} = \frac{2M P}{A_{1}B} = \frac{2r\cot(\phi / 2)}{A_{1}B}\) . Since similarly \(\frac{r - 2\rho}{\rho} = \frac{2r\cot(\phi / 2)}{B_{1}A}\) , we obtain \(A_{1}B = B_{1}A\) and consequently \(\triangle A M B_{1}\cong \triangle B M A_{1}\) . Thus \(\angle B A C = \angle A B C\) and \(C C_{1}\perp A B\) . There are two alternatives for the other two incircles:
(i) If the inradii of \(A M C_{1}\) and \(A M B_{1}\) are equal to \(r\) , it is easy to obtain that \(\triangle A M C_{1}\cong \triangle A M B_{1}\) . Hence \(\angle A B_{1}M = \angle A C_{1}M = 90^{\circ}\) , and \(\triangle A B C\) is equilateral.
(ii) The inradii of \(A M B_{1}\) and \(C M B_{1}\) are equal to \(r\) . Put \(x = \angle M A C_{1} = \angle M B C_{1}\) . In this case \(\phi = 2x\) and \(\angle B_{1}M C = 90^{\circ} - x\) . Now we have \(\frac{A B_{1}}{C B_{1}} = \frac{S_{A M B_{1}}}{S_{C M B_{1}}} = \frac{A M + M B_{1} + A B_{1}}{C M + M B_{1} + C B_{1}} = \frac{A M + M B_{1} - A B_{1}}{C M + M B_{1} - C B_{1}} = \frac{\cot x}{\cot(45^{\circ} - x / 2)}\) . On the other hand, we have \(\frac{A B_{1}}{C B_{1}} = \frac{A B}{B C} = 2\cos 2x\) . Thus we have an equation for \(x\) : \(\tan (45^{\circ} - x / 2) = 2\cos 2x\tan x\) , or equivalently
\[2\tan \left(45^{\circ} - \frac{x}{2}\right)\sin \left(45^{\circ} - \frac{x}{2}\right)\cos \left(45^{\circ} - \frac{x}{2}\right) = 2\cos 2x\sin x.\]
Hence \(\sin 3x - \sin x = 2\sin^{2}\left(45^{\circ} - \frac{x}{2}\right) = 1 - \sin x\) , implying \(\sin 3x = 1\) , i.e., \(x = 30^{\circ}\) . Therefore \(\triangle A B C\) is equilateral.
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IMOSL-1976-2
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Let \(a_{0}, a_{1}, \ldots , a_{n}, a_{n + 1}\) be a sequence of real numbers satisfying the following conditions:
\[a_{0} = a_{n + 1} = 0,\] \[|a_{k - 1} - 2a_{k} + a_{k + 1}| \leq 1 \quad (k = 1,2,\ldots ,n).\]
Prove that \(|a_{k}| \leq \frac{k(n + 1 - k)}{2} (k = 0,1,\ldots ,n + 1)\) .
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Let us put \(b_{i} = i(n + 1 - i) / 2\) , and let \(c_{i} = a_{i} - b_{i}\) , \(i = 0,1,\ldots ,n + 1\) . It is easy to verify that \(b_{0} = b_{n + 1} = 0\) and \(b_{i - 1} - 2b_{i} + b_{i + 1} = -1\) . Subtracting this inequality from \(a_{i - 1} - 2a_{i} + a_{i + 1}\geq -1\) , we obtain \(c_{i - 1} - 2c_{i} + c_{i + 1}\geq 0\) , i.e., \(2c_{i}\leq c_{i - 1}+\) \(c_{i + 1}\) . We also have \(c_{0} = c_{n + 1} = 0\) .
Suppose that there exists \(i\in \{1,\ldots ,n\}\) for which \(c_{i} > 0\) , and let \(c_{k}\) be the maximal such \(c_{i}\) . Assuming w.l.o.g. that \(c_{k - 1}< c_{k}\) , we obtain \(c_{k - 1} + c_{k + 1}< 2c_{k}\) which is a contradiction. Hence \(c_{i}\leq 0\) for all \(i\) ; i.e., \(a_{i}\leq b_{i}\) .
Similarly, considering the sequence \(c_{i}^{\prime} = a_{i} + b_{i}\) one can show that \(c_{i}^{\prime}\geq 0\) , i.e., \(a_{i}\geq - b_{i}\) for all \(i\) . This completes the proof.
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IMOSL-1976-3
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In a convex quadrangle with area \(32 \mathrm{cm}^{2}\) , the sum of the lengths of two nonadjacent edges and of the length of one diagonal is equal to \(16 \mathrm{cm}\) .
(a) What is the length of the other diagonal?
(b) What are the lengths of the edges of the quadrangle if the perimeter is a minimum?
(c) Is it possible to choose the edges in such a way that the perimeter is a maximum?
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(a) Let \(A B C D\) be a quadrangle with \(16 = d = A B + C D + A C\) , and let \(S\) be its area. Then \(S\leq (A C\cdot A B + A C\cdot C D) / 2 = A C(d - A C) / 2\leq d^{2} / 8 = 32\) , where equality occurs if and only if \(A B\perp A C\perp C D\) and \(A C = A B + C D = 8\) . In this case \(B D = 8\sqrt{2}\) .
(b) Let \(A^{\prime}\) be the point with \(\overrightarrow{D A^{\prime}} = \overrightarrow{A C}\) . The triangular inequality implies \(A D +\) \(B C\geq A A^{\prime} = 8\sqrt{5}\) . Thus the perimeter attains its minimum for \(A B = C D =\) 4.
(c) Let us assume w.l.o.g. that \(C D\leq A B\) . Then \(C\) lies inside \(\triangle B D A^{\prime}\) and hence \(B C + A D = B C + C A^{\prime}< B D + D A^{\prime}\) . The maximal value \(B D + D A^{\prime}\) of \(B C+\) \(A D\) is attained when \(C\) approaches \(D\) , making a degenerate quadrangle.
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IMOSL-1976-4
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For all positive integral \(n\) , \(u_{n + 1} = u_{n}(u_{n - 1}^{2} - 2) - u_{1}\) , \(u_{0} = 2\) , and \(u_{1} = 5 / 2\) . Prove that \(3 \log_{2}[u_{n}] = 2^{n} - (-1)^{n}\) , where \([x]\) is the integral part of \(x\) .
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The first few values are easily verified to be \(2^{r_{n}} + 2^{-r_{n}}\) , where \(r_{0} = 0\) , \(r_{1} = r_{2} = 1\) , \(r_{3} = 3\) , \(r_{4} = 5\) , \(r_{5} = 11\) , ... . Let us put \(u_{n} = 2^{r_{n}} + 2^{-r_{n}}\) (we will show that \(r_{n}\) exists and is integer for each \(n\) ). A simple calculation gives us \(u_{n}(u_{n - 1}^{2} - 2) = 2^{r_{n} + 2r_{n - 1}} + 2^{-r_{n} - 2r_{n - 1}} + 2^{r_{n} - 2r_{n - 1}} + 2^{-r_{n} + 2r_{n - 1}}\) . If an array \(q_{n}\) , with \(q_{0} = 0\) and \(q_{1} = 1\) , is set so as to satisfy the linear recurrence \(q_{n + 1} = q_{n} + 2q_{n - 1}\) , then it also satisfies \(q_{n} - 2q_{n - 1} = -(q_{n - 1} - 2q_{n - 2}) = \dots = (-1)^{n - 1}(q_{1} - 2q_{0}) = (-1)^{n - 1}\) . Assuming inductively up to \(n\) \(r_{i} = q_{i}\) , the expression for \(u_{n}(u_{n - 1}^{2} - 2) = u_{n + 1} + u_{1}\) reduces to \(2^{q_{n + 1}} + 2^{- q_{n + 1}} + u_{1}\) . Therefore, \(r_{n + 1} = q_{n + 1}\) . The solution to this linear recurrence with \(r_{0} = 0\) , \(r_{1} = 1\) is \(r_{n} = q_{n} = \frac{2^{n} - (-1)^{n}}{3}\) , and since \([u_{n}] = 2^{r_{n}}\) for \(n \geq 0\) , the result follows.
Remark. One could simply guess that \(u_{n} = 2^{r_{n}} + 2^{- r_{n}}\) for \(r_{n} = \frac{2^{n} - (-1)^{n}}{3}\) , and then prove this result by induction.
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IMOSL-1976-5
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Let a set of \(p\) equations be given,
\[a_{11}x_{1} + \dots +a_{1q}x_{q} = 0,\] \[a_{21}x_{1} + \dots +a_{2q}x_{q} = 0,\] \[\qquad \vdots\] \[a_{p1}x_{1} + \dots +a_{pq}x_{q} = 0,\]
with coefficients \(a_{ij}\) satisfying \(a_{ij} = - 1\) , 0, or \(+1\) for all \(i = 1,\ldots ,p\) and \(j = 1,\ldots ,q\) . Prove that if \(q = 2p\) , there exists a solution \(x_{1},\ldots ,x_{q}\) of this system such that all \(x_{j}\) \((j = 1,\ldots ,q)\) are integers satisfying \(|x_{j}|\leq q\) and \(x_{j}\neq 0\) for at least one value of \(j\) .
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If one substitutes an integer \(q\) -tuple \((x_{1},\ldots ,x_{q})\) satisfying \(|x_{i}|\leq p\) for all \(i\) in an equation of the given system, the absolute value of the right-hand member never exceeds \(p q\) . So for the right-hand member of the system there are \((2p q + 1)^{p}\) possibilities There are \((2p + 1)^{q}\) possible \(q\) -tuples \((x_{1},\ldots ,x_{q})\) . Since \((2p + 1)^{q} > (2p q + 1)^{p}\) , there are at least two \(q\) -tuples \((y_{1},\ldots ,y_{q})\) and \((z_{1},\ldots ,z_{q})\) giving the same right-hand members in the given system. The difference \((x_{1},\ldots ,x_{q}) = (y_{1} - z_{1},\ldots ,y_{q} - z_{q})\) thus satisfies all the requirements of the problem.
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IMOSL-1976-6
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A rectangular box can be filled completely with unit cubes. If one places the maximal number of cubes with volume 2 in the box such that their edges are parallel to the edges of the box, one can fill exactly \(40\%\) of the box. Determine all possible (interior) sizes of the box.
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Suppose \(a_{1} \leq a_{2} \leq a_{3}\) are the dimensions of the box. If we set \(b_{i} = [a_{i} / \sqrt[3]{2}]\) , the condition of the problem is equivalent to \(\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} = 5\) . We list some values of \(a, b = [a / \sqrt[3]{2}]\) and \(a / b\) :
<table><tr><td>a</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>b</td><td>1</td><td>2</td><td>3</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>7</td></tr><tr><td>a/b</td><td>2</td><td>1.5</td><td>1.33</td><td>1.67</td><td>1.5</td><td>1.4</td><td>1.33</td><td>1.29</td><td>1.43</td></tr></table>
We note that if \(a > 2\) , then \(a / b \leq 5 / 3\) , and if \(a > 5\) , then \(a / b \leq 3 / 2\) . If \(a_{1} > 2\) , then \(\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} < (5 / 3)^{3} < 5\) , a contradiction. Hence \(a_{1} = 2\) . If also \(a_{2} = 2\) , then \(a_{3} / b_{3} = 5 / 4 < \sqrt[3]{2}\) , which is impossible. Also, if \(a_{2} \geq 6\) , then \(\frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} \leq (1.5)^{2} < 2.5\) , again a contradiction. We thus have the following cases:
(i) \(a_{1} = 2, a_{2} = 3\) , then \(a_{3} / b_{3} = 5 / 3\) , which holds only if \(a_{3} = 5\) ;
(ii) \(a_{1} = 2, a_{2} = 4\) , then \(a_{3} / b_{3} = 15 / 8\) , which is impossible;
(iii) \(a_{1} = 2, a_{2} = 5\) , then \(a_{3} / b_{3} = 3 / 2\) , which holds only if \(a_{3} = 6\) .
The only possible sizes of the box are therefore \((2,3,5)\) and \((2,5,6)\) .
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IMOSL-1976-7
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Let \(I = (0,1]\) be the unit interval of the real line. For a given number \(a\in (0,1)\) we define a map \(T:I\to I\) by the formula
\[
T(x, y) =
\begin{cases}
x + (1 - a), & \text{if } 0 < x \le a, \\
x - a, & \text{if } a < x \le 1.
\end{cases}
\]
Show that for every interval \(J\subset I\) there exists an integer \(n > 0\) such that \(T^{n}(J)\cap\) \(J\neq \emptyset\)
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The map \(T\) transforms the interval \((0, a]\) onto \((1 - a, 1]\) and the interval \((a, 1]\) onto \((0, 1 - a]\) . Clearly \(T\) preserves the measure. Since the measure of the interval \([0, 1]\) is finite, there exist two positive integers \(k, l > k\) such that \(T^{k}(J)\) and \(T^{l}(J)\) are not disjoint. But the map \(T\) is bijective; hence \(T^{l - k}(J)\) and \(J\) are not disjoint.
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IMOSL-1976-8
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Let \(P\) be a polynomial with real coefficients such that \(P(x) > 0\) if \(x > 0\) . Prove that there exist polynomials \(Q\) and \(R\) with nonnegative coefficients such that \(P(x) = \frac{Q(x)}{R(x)}\) if \(x > 0\) .
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Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result only for a quadratic polynomial \(P(x) = x^{2} - 2ax + b^{2}\) , with \(a > 0\) and \(b^{2} > a^{2}\) . Using the identity
\[(x^{2} + b^{2})^{2n} - (2ax)^{2n} = (x^{2} - 2ax + b^{2})\sum_{k = 0}^{2n - 1}(x^{2} + b^{2})^{k}(2ax)^{2n - k - 1}\]
we have solved the problem if we can choose \(n\) such that \(b^{2n}\binom{2n}{n} >2^{2n}a^{2n}\) However, it is is easy to show that \(2n\binom{2n}{n} < 2^{2n}\) ; hence it is enough to take \(n\) such that \(\left(b / a\right)^{2n} > 2n\) . Since \(\begin{array}{r}{\lim_{n\to \infty}(2n)^{1 / (2n)} = 1< b / a} \end{array}\) , such an \(n\) always exists.
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IMOSL-1976-9
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Let \(P_{1}(x) = x^{2} - 2\) , \(P_{j}(x) = P_{1}(P_{j - 1}(x))\) , \(j = 2,3,\ldots\) . Show that for arbitrary \(n\) the roots of the equation \(P_{n}(x) = x\) are real and different from one another.
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The equation \(P_{n}(x) = x\) is of degree \(2^{n}\) , and has at most \(2^{n}\) distinct roots. If \(x > 2\) , then by simple induction \(P_{n}(x) > x\) for all \(n\) . Similarly, if \(x< -1\) , then \(P_{1}(x) > 2\) , which implies \(P_{n}(x) > 2\) for all \(n\) . It follows that all real roots of the equation \(P_{n}(x) = x\) lie in the interval \([-2,2]\) , and thus have the form \(x = 2\cos t\) . Observe that \(P_{1}(2\cos t) = 4\cos^{2}t - 2 = 2\cos 2t\) , and in general \(P_{n}(2\cos t) = 2\cos 2^{n}t\) . Our equation becomes
\[\cos 2^{n}t = \cos t,\]
which indeed has \(2^{n}\) different solutions \(t = \frac{2\pi m}{2^{n} - 1}\) \((m = 0,1,\ldots ,2^{n - 1} - 1)\) and \(t = \frac{2\pi m}{2^{n} + 1}\) \((m = 1,2,\ldots ,2^{n - 1})\)
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IMOSL-1976-10
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Find the largest number obtainable as the product of positive integers whose sum is 1976.
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Let \(a_{1}\leq a_{2}\leq \dots \leq a_{n}\) be positive integers whose sum is 1976. Let \(M\) denote the maximal value of \(a_{1}a_{2}\dots a_{n}\) . We make the following observations:
(1) \(a_{1} = 1\) does not yield the maximum, since replacing \(1,a_{2}\) by \(1 + a_{2}\) increases the product.
(2) \(a_{j} - a_{i}\geq 2\) does not yield the maximal value, since replacing \(a_{i},a_{j}\) by \(a_{i}+\) \(1,a_{j} - 1\) increases the product.
(3) \(a_{i}\geq 5\) does not yield the maximal value, since \(2(a_{i} - 2) = 2a_{i} - 4 > a_{i}\) Since \(4 = 2^{2}\) , we may assume that all \(a_{i}\) are either 2 or 3, and \(M = 2^{k}3^{l}\) , where \(2k + 3l = 1976\) (4) \(k\geq 3\) does not yield the maximal value, since \(2\cdot 2\cdot 2< 3\cdot 3\) Hence \(k\leq 2\) and \(2k\equiv 1976\) (mod 3) gives us \(k = 1\) \(l = 658\) and \(M = 2\cdot 3^{658}\)
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IMOSL-1976-11
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Prove that there exist infinitely many positive integers \(n\) such that the decimal representation of \(5^{n}\) contains a block of 1976 consecutive zeros.
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We shall show by induction that \(5^{2^{k}} - 1 = 2^{k + 2}q_{k}\) for each \(k = 0,1,\ldots\) , where \(q_{k}\in \mathbb{N}\) . Indeed, the statement is true for \(k = 0\) , and if it holds for some \(k\) then \(5^{2^{k + 1}} - 1 = \left(5^{2^{k}} + 1\right)\left(5^{2^{k}} - 1\right) = 2^{k + 3}d_{k + 1}\) where \(d_{k + 1} = \left(5^{2^{k}} + 1\right)d_{k} / 2\) is an integer by the inductive hypothesis.
Let us now choose \(n = 2^{k} + k + 2\) . We have \(5^{n} = 10^{k + 2}q_{k} + 5^{k + 2}\) . It follows from \(5^{4}< 10^{3}\) that \(5^{k + 2}\) has at most \([3(k + 2) / 4] + 2\) nonzero digits, while \(10^{k + 2}q_{k}\) ends in \(k + 2\) zeros. Hence the decimal representation of \(5^{n}\) contains at least \([(k + 2) / 4] - 2\) consecutive zeros. Now it suffices to take \(k > 4\cdot 1978\)
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IMOSL-1976-12
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The polynomial \(1976(x + x^{2} + \dots +x^{n})\) is decomposed into a sum of polynomials of the form \(a_{1}x + a_{2}x^{2} + \ldots +a_{n}x^{n}\) , where \(a_{1},a_{2},\dots ,a_{n}\) are distinct positive integers not greater than \(n\) . Find all values of \(n\) for which such a decomposition is possible.
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Suppose the decomposition into \(k\) polynomials is possible. The sum of coefficients of each polynomial \(a_{1}x + a_{2}x^{2} + \dots +a_{n}x^{n}\) equals \(1 + \dots +n = n(n + 1) / 2\) while the sum of coefficients of \(1976(x + x^{2} + \dots +x^{n})\) is \(1976n\) . Hence we must have \(1976n = kn(n + 1) / 2\) , which reduces to \((n + 1)\mid 3952 = 2^{4}\cdot 13\cdot 19\) . In other words, \(n\) is of the form \(n = 2^{\alpha}13^{\beta}19^{\gamma} - 1\) , with \(0\leq \alpha \leq 4\) , \(0\leq \beta \leq 1\) , \(0\leq \gamma \leq 1\) .
We can immediately eliminate the values \(n = 0\) and \(n = 3951\) that correspond to \(\alpha = \beta = \gamma = 0\) and \(\alpha = 4\) , \(\beta = \gamma = 1\) .
We claim that all other values \(n\) are permitted. There are two cases.
\(\alpha \leq 3\) . In this case \(k = 3952 / (n + 1)\) is even. The simple choice of the polynomials \(P = x + 2x^{2} + \dots +nx^{n}\) and \(P^{\prime} = nx + (n - 1)x^{2} + \dots +x^{n}\) suffices, since \(k(P + P^{\prime}) / 2 = 1976(x + x^{2} + \dots +x^{n})\) .
\(\alpha = 4\) . Then \(k\) is odd. Consider \((k - 3) / 2\) pairs \((P,P^{\prime})\) of the former case and
\[P_{1} = \left[nx + (n - 1)x^{3} + \dots +\frac{n + 1}{2} x^{n}\right]\] \[+\left[\frac{n - 1}{2} x^{2} + \frac{n - 3}{2} x^{4} + \dots +x^{n - 1}\right];\] \[P_{2} = \left[\frac{n + 1}{2} x + \frac{n - 1}{2} x^{3} + \dots +x^{n}\right]\] \[+\left[nx^{2} + (n - 1)x^{4} + \dots +\frac{n + 3}{2} x^{n - 1}\right].\]
Then \(P + P_{1} + P_{2} = 3(n + 1)(x + x^{2} + \dots +x^{n}) / 2\) and therefore \((k - 3)(P + P^{\prime}) / 2 + (P + P_{1} + P_{2}) = 1976(x + x^{2} + \dots +x^{n})\) .
It follows that the desired decomposition is possible if and only if \(1 < n < 3951\) and \(n + 1 \mid 2 \cdot 1976\) .
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IMOLL-1977-1
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A pentagon ABCDE inscribed in a circle for which \(BC< CD\) and \(AB< DE\) is the base of a pyramid with vertex \(S\) . If \(AS\) is the longest edge starting from \(S\) , prove that \(BS > CS\) .
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Let \(P\) be the projection of \(S\) onto the plane \(ABCDE\) . Obviously \(BS > CS\) is equivalent to \(BP > CP\) . The conditions of the problem imply that \(PA > PB\) and \(PA > PE\) . The locus of such points \(P\) is the region of the plane that is determined by the perpendicular bisectors of segments \(AB\) and \(AE\) and that contains the point diametrically opposite \(A\) . But since \(AB < DE\) , the whole of this region lies on one side of the perpendicular bisector of \(BC\) . The result follows immediately.
Remark. The assumption \(BC < CD\) is redundant.
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IMOLL-1977-2
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Let \(f:\mathbb{N}\to \mathbb{N}\) be a function that satisfies the inequality \(f(n + 1)>\) \(f(f(n))\) for all \(n\in \mathbb{N}\) . Prove that \(f(n) = n\) for all natural numbers \(n\)
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We shall prove by induction on \(n\) that \(f(x) > f(n)\) whenever \(x > n\) . The case \(n = 0\) is trivial. Suppose that \(n \geq 1\) and that \(x > k\) implies \(f(x) > f(k)\) for all \(k < n\) . It follows that \(f(x) \geq n\) holds for all \(x \geq n\) . Let \(f(m) = \min_{x \geq n} f(x)\) . If we suppose that \(m > n\) , then \(m - 1 \geq n\) and consequently \(f(m - 1) \geq n\) . But in this case the inequality \(f(m) > f(f(m - 1))\) contradicts the minimality property of \(m\) . The inductive proof is thus completed.
It follows that \(f\) is strictly increasing, so \(f(n + 1) > f(f(n))\) implies that \(n + 1 > f(n)\) . But since \(f(n) \geq n\) we must have \(f(n) = n\) .
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IMOLL-1977-3
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In a company of \(n\) persons, each person has no more than \(d\) acquaintances, and in that company there exists a group of \(k\) persons, \(k \geq d\) , who are not acquainted with each other. Prove that the number of acquainted pairs is not greater than \([n^2 /4]\) .
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Let \(\nu_{1}, \nu_{2}, \ldots , \nu_{k}\) be \(k\) persons who are not acquainted with each other. Let us denote by \(m\) the number of acquainted couples and by \(d_{j}\) the number of acquaintances of person \(\nu_{j}\) . Then
\[m\leq d_{k + 1} + d_{k + 2} + \dots +d_{n}\leq d(n - k)\leq k(n - k)\leq \left(\frac{k + (n - k)}{2}\right)^{2} = \frac{n^{2}}{4}.\]
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IMOLL-1977-4
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We are given \(n\) points in space. Some pairs of these points are connected by line segments so that the number of segments equals \([n^2 /4]\) , and a connected triangle exists. Prove that any point from which the maximal number of segments starts is a vertex of a connected triangle.
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Consider any vertex \(\nu_{n}\) from which the maximal number \(d\) of segments start, and suppose it is not a vertex of a triangle. Let \(\mathcal{A} = \{\nu_{1}, \nu_{2}, \ldots , \nu_{d}\}\) be the set of points that are connected to \(\nu_{n}\) , and let \(\mathcal{B} = \{\nu_{d + 1}, \nu_{d + 2}, \ldots , \nu_{n}\}\) be the set of the other points. Since \(\nu_{n}\) is not a vertex of a triangle, there is no segment both of whose vertices lie in \(\mathcal{A}\) ; i.e., each segment has an end in \(\mathcal{B}\) . Thus, if \(d_{j}\) denotes the number of segments at \(\nu_{j}\) and \(m\) denotes the total number of segments, we have \(m \leq d_{d + 1} + d_{d + 2} + \dots + d_{n} \leq d(n - d) \leq \left\lfloor n^{2} / 4 \right\rfloor = m\) . This means that each inequality must be equality, implying that each point in \(\mathcal{B}\) is a vertex of \(d\) segments, and each of these segments has the other end in \(\mathcal{A}\) . Then there is no triangle at all, which is a contradiction.
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IMOLL-1977-5
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A lattice point in the plane is a point both of whose coordinates are integers. Each lattice point has four neighboring points: upper, lower, left, and right. Let \(k\) be a circle with radius \(r\geq 2\) , that does not pass through any lattice point. An interior boundary point is a lattice point lying inside the circle \(k\) that has a neighboring point lying outside \(k\) . Similarly, an exterior boundary point is a lattice point lying outside the circle \(k\) that has a neighboring point lying inside \(k\) . Prove that there are four more exterior boundary points than interior boundary points.
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Let us denote by \(I\) and \(E\) the sets of interior boundary points and exterior boundary points. Let \(ABCD\) be the square inscribed in the circle \(k\) with sides parallel to the coordinate axes. Lines \(AB, BC, CD, DA\) divide the plane into 9 regions: \(\mathcal{R}, \mathcal{R}_{A}, \mathcal{R}_{B}, \mathcal{R}_{C}, \mathcal{R}_{D}, \mathcal{R}_{AB}, \mathcal{R}_{BC}, \mathcal{R}_{CD}, \mathcal{R}_{DA}\) . There is a unique pair of lattice points \(A_{I} \in \mathcal{R}, A_{E} \in \mathcal{R}_{A}\) that are opposite vertices of a
unit square. We similarly define \(B_{I}, C_{I}, D_{I}, B_{E}, C_{E}, D_{E}\) . Let us form a graph \(G\) by connecting each point from \(E\) lying in \(\mathcal{R}_{AB}\) (respectively \(\mathcal{R}_{BC}, \mathcal{R}_{CD}, \mathcal{R}_{DA}\) ) to its upper (respectively left, lower, right) neighbor point (which clearly belongs to \(I\) ). It is easy to see that:
(i) All vertices from \(I\) other than \(A_{I}, B_{I}, C_{I}, D_{I}\) have degree 1.
(ii) \(A_{E}\) is not in \(E\) if and only if \(A_{I} \in I\) and \(\deg A_{I} = 2\) .
(iii) No other lattice points inside \(\mathcal{R}_{A}\) belong to \(E\) .
Thus if \(m\) is the number of edges of the graph \(G\) and \(s\) is the number of points among \(A_{E}, B_{E}, C_{E}\) , and \(D_{E}\) that are in \(E\) , using (i)–(iii) we easily obtain \(|E| = m + s\) and \(|I| = m - (4 - s) = |E| + 4\) .
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IMOLL-1977-6
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Let \(x_{1},x_{2},\ldots ,x_{n}\) \((n\geq 1)\) be real numbers such that \(0\leq x_{j}\leq \pi\) \(j =\) \(1,2,\ldots ,n\) . Prove that if \(\textstyle \sum_{j = 1}^{n}(\cos x_{j} + 1)\) is an odd integer, then \(\textstyle \sum_{j = 1}^{n}\sin x_{j}\geq 1\)
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Let \(\langle y \rangle\) denote the distance from \(y \in \mathbb{R}\) to the closest even integer. We claim that
\[\langle 1 + \cos x \rangle \leq \sin x \quad \text{for all} x \in [0, \pi ].\]
Indeed, if \(\cos x \geq 0\) , then \(\langle 1 + \cos x \rangle = 1 - \cos x \leq 1 - \cos^2 x = \sin^2 x \leq \sin x\) ; the proof is similar if \(\cos x < 0\) .
We note that \(\langle x + y \rangle \leq \langle x \rangle + \langle y \rangle\) holds for all \(x, y \in \mathbb{R}\) . Therefore
\[\sum_{j = 1}^{n} \sin x_{j} \geq \sum_{j = 1}^{n} \langle 1 + \cos x_{j} \rangle \geq \left\langle \sum_{j = 1}^{n} (1 + \cos x_{j}) \right\rangle = 1.\]
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IMOLL-1977-7
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Prove the following assertion: If \(c_{1},c_{2},\ldots ,c_{n}(n\geq 2)\) are real numbers such that
\[(n - 1)(c_{1}^{2} + c_{2}^{2} + \dots +c_{n}^{2}) = (c_{1} + c_{2} + \dots +c_{n})^{2},\]
then either all these numbers are nonnegative or all these numbers are nonpositive.
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Let us suppose that \(c_{1} \leq c_{2} \leq \dots \leq c_{n}\) and that \(c_{1} < 0 < c_{n}\) . There exists \(k\) , \(1 \leq k < n\) , such that \(c_{k} \leq 0 < c_{k + 1}\) . Then we have
\[(n - 1)(c_{1}^{2} + c_{2}^{2} + \dots +c_{n}^{2}) \geq k(c_{1}^{2} + \dots +c_{k}^{2}) + (n - k)(c_{k + 1}^{2} + \dots +c_{n}^{2})\] \[\qquad \geq (c_{1} + \dots +c_{k})^{2} + (c_{k + 1} + \dots +c_{n})^{2}\] \[\qquad = (c_{1} + \dots +c_{n})^{2}\] \[\qquad -2(c_{1} + \dots +c_{k})(c_{k + 1} + \dots +c_{n}),\]
from which we obtain \((c_{1} + \dots +c_{k})(c_{k + 1} + \dots +c_{n}) \geq 0\) , a contradiction.
Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have
\[(c_{1} + \dots +c_{n})^{2} = (n - 1)(c_{1}^{2} + \dots +c_{n - 1}^{2}) + (n - 1)c_{n}^{2}\] \[\qquad \geq (c_{1} + \dots +c_{n - 1})^{2} + (n - 1)c_{n}^{2},\]
which is equivalent to \(2(c_{1} + c_{2} + \dots +c_{n})c_{n} \geq nc_{n}^{2}\) . Similarly, \(2(c_{1} + c_{2} + \dots +c_{n})c_{i} \geq nc_{i}^{2}\) for all \(i = 1, \ldots , n\) . Hence all \(c_{i}\) are of the same sign.
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IMOLL-1977-8
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A hexahedron \(ABCDE\) is made of two regular congruent tetrahedra \(ABCD\) and \(ABCE\) . Prove that there exists only one isometry \(\mathcal{L}\) that maps points \(A,B,C,D,E\) onto \(B,C,A,E,D\) , respectively. Find all points \(X\) on the surface of hexahedron whose distance from \(\mathcal{L}(X)\) is minimal.
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There is exactly one point satisfying the given condition on each face of the hexahedron. Namely, on the face \(ABD\) it is the point that divides the median from \(D\) in the ratio \(32:3\) .
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IMOLL-1977-9
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Let \(ABCD\) be a regular tetrahedron and \(\mathcal{L}\) an isometry mapping \(A,B\) , \(C,D\) into \(B,C,D,A\) , respectively. Find the set \(\mathcal{M}\) of all points \(X\) of the face \(ABC\) whose distance from \(\mathcal{L}(X)\) is equal to a given number \(t\) . Find necessary and sufficient conditions for the set \(\mathcal{M}\) to be nonempty.
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A necessary and sufficient condition for \(\mathcal{M}\) to be nonempty is that \(1 / \sqrt{10} \leq t \leq 1\) .
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IMOLL-1977-10
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Let \(a\) and \(b\) be natural numbers and let \(q\) and \(r\) be the quotient and remainder respectively when \(a^2 +b^2\) is divided by \(a + b\) . Determine the numbers \(a\) and \(b\) if \(q^2 +r = 1977\) .
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Integers \(a, b, q, r\) satisfy
\[a^{2} + b^{2} = (a + b)q + r,\quad 0\leq r< a + b,\quad q^{2} + r = 1977.\]
From \(q^{2} \leq 1977\) it follows that \(q \leq 44\) , and consequently \(a^{2} + b^{2} < 45(a + b)\) . Having in mind the inequality \((a + b)^{2} \leq 2(a^{2} + b^{2})\) , we get \((a + b)^{2} < 90(a + b)\) , i.e., \(a + b < 90\) and consequently \(r < 90\) . Now from \(q^{2} = 1977 - r > 1977 - 90 = 1887\) it follows that \(q > 43\) ; hence \(q = 44\) and \(r = 41\) . It remains to find positive integers \(a\) and \(b\) satisfying \(a^{2} + b^{2} = 44(a + b) + 41\) , or equivalently
\[(a - 22)^{2} + (b - 22)^{2} = 1009.\]
The Diophantine equation \(A^{2} + B^{2} = 1009\) has only two pairs of positive solutions: \((15,28)\) and \((28,15)\) . Hence \((|a - 22|, |b - 22|) \in \{(15,28), (28,15)\}\) , which implies \((a, b) \in \{(7,50), (37,50), (50,7), (50,37)\}\) .
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IMOLL-1977-11
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Let \(n\) and \(z\) be integers greater than 1 and \((n,z) = 1\) . Prove:
(a) At least one of the numbers \(z_{i} = 1 + z + z^{2} + \dots +z^{i}\) , \(i = 0,1,\ldots ,n - 1\) , is divisible by \(n\) .
(b) If \((z - 1,n) = 1\) , then at least one of the numbers \(z_{i}\) , \(i = 0,1,\ldots ,n - 2\) , is divisible by \(n\) .
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(a) Suppose to the contrary that none of the numbers \(z_{0}, z_{1}, \ldots , z_{n - 1}\) is divisible by \(n\) . Then two of these numbers, say \(z_{k}\) and \(z_{l}\) \((0 \leq k < l \leq n - 1)\) , are congruent modulo \(n\) , and thus \(n \mid z_{l} - z_{k} = z^{k + 1}z_{l - k - 1}\) . But since \((n, z) = 1\) , this implies \(n \mid z_{l - k - 1}\) , which is a contradiction. (b) Again suppose the contrary, that none of \(z_{0}, z_{1}, \ldots , z_{n - 2}\) is divisible by \(n\) . Since \((z - 1, n) = 1\) , this is equivalent to \(n \nmid (z - 1)z_{j}\) , i.e., \(z^{k} \neq 1\) (mod \(n\) ) for all \(k = 1, 2, \ldots , n - 1\) . But since \((z, n) = 1\) , we also have that \(z^{k} \neq 0\) (mod \(n\) ). It follows that there exist \(k, l\) , \(1 \leq k < l \leq n - 1\) such that \(z^{k} \equiv z^{l}\) , i.e., \(z^{l - k} \equiv 1\) (mod \(n\) ), which is a contradiction.
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IMOLL-1977-12
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Let \(z\) be an integer \(>1\) and let \(M\) be the set of all numbers of the form \(z_{k} = 1 + z + \dots +z^{k}\) , \(k = 0,1,\ldots\) . Determine the set \(T\) of divisors of at least one of the numbers \(z_{k}\) from \(M\) .
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According to part (a) of the previous problem we can conclude that \(T = \{n \in \mathbb{N} \mid (n, z) = 1\}\) .
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IMOLL-1977-13
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Describe all closed bounded figures \(\Phi\) in the plane any two points of which are connectable by a semicircle lying in \(\Phi\) .
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The figure \(\Phi\) contains two points \(A\) and \(B\) having maximum distance. Let \(h\) be the semicircle with diameter \(AB\) that lies in \(\Phi\) , and let \(k\) be the circle containing \(h\) . Consider any point \(M\) inside \(k\) . The line passing through \(M\) that is orthogonal to \(AM\) meets \(h\) in some point \(P\) (because \(\angle AMB > 90^{\circ}\) ). Let \(h'\) and \(\overline{h''}\) be the two semicircles with diameter \(AP\) , where \(M \in h'\) . Since \(\overline{h''}\) contains a point \(C\) such that \(BC > AB\) , it cannot be contained in \(\Phi\) , implying that \(h' \subset \Phi\) . Hence \(M\) belongs to \(\Phi\) . Since \(\Phi\) contains no points outside the circle \(k\) , it must coincide with the disk determined by \(k\) . On the other hand, any disk has the required property.
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IMOLL-1977-14
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There are \(2^n\) words of length \(n\) over the alphabet \(\{0,1\}\) . Prove that the following algorithm generates the sequence \(w_0, w_1, \ldots , w_{2^n - 1}\) of all these words such that any two consecutive words differ in exactly one digit.
(1) \(w_0 = 00\ldots 0\) ( \(n\) zeros).
(2) Suppose \(w_{m - 1} = a_1a_2\ldots a_n\) , \(a_i\in \{0,1\}\) . Let \(e(m)\) be the exponent of 2 in the representation of \(n\) as a product of primes, and let \(j = 1 + e(m)\) . Replace the digit \(a_j\) in the word \(w_{m - 1}\) by \(1 - a_j\) . The obtained word is \(w_m\) .
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We prove by induction on \(n\) that independently of the word \(w_{0}\) , the given algorithm generates all words of length \(n\) . This is clear for \(n = 1\) . Suppose now the statement is true for \(n - 1\) , and that we are given a word \(w_{0} = c_{1}c_{2}\ldots c_{n}\) of length \(n\) . Obviously, the words \(w_{0},w_{1},\ldots ,w_{2^{n - 1} - 1}\) all have the \(n\) th digit \(c_{n}\) , and by the inductive hypothesis these are all words whose \(n\) th digit is \(c_{n}\) . Similarly, by the inductive hypothesis \(w_{2^{n - 1}},\ldots ,w_{2^{n} - 1}\) are all words whose \(n\) th digit is \(1 - c_{n}\) , and the induction is complete.
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IMOLL-1977-15
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Let \(n\) be an integer greater than 1. In the Cartesian coordinate system we consider all squares with integer vertices \((x,y)\) such that \(1\leq x,y\leq n\) . Denote by \(p_{k}(k = 0,1,2,\ldots)\) the number of pairs of points that are vertices of exactly \(k\) such squares. Prove that \(\textstyle \sum_{k}(k - 1)p_{k} = 0\)
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Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore \(p_{k} = 0\) for \(k > 3\) , and we have to prove that
\[p_{0} = p_{2} + 2p_{3}. \quad (1)\]
Let us calculate the number \(q(n)\) of considered squares. Each of these squares is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are \((n - s)^{2}\) squares of side \(s\) with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly \(s\) of the considered squares. It follows that \(q(n) = \sum_{s = 1}^{n - 1}(n - s)^{2}s = n^{2}(n^{2} - 1) / 12\) . Computing the number of edges and diagonals of the considered squares in two ways, we obtain that
\[p_{1} + 2p_{2} + 3p_{3} = 6q(n). \quad (2)\]
On the other hand, the total number of segments with endpoints in the considered integer points is given by
\[p_{0} + p_{1} + p_{2} + p_{3} = \binom{n^{2}}{2} = \frac{n^{2}(n^{2} - 1)}{2} = 6q(n). \quad (3)\]
Now (1) follows immediately from (2) and (3).
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IMOLL-1977-16
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Let \(n\) be a positive integer. How many integer solutions \((i,j,k,l)\) , \(1 \leq i,j,k,l \leq n\) , does the following system of inequalities have:
\[1\leq -j + k + l\leq n\] \[1\leq i - k + l\leq n\] \[1\leq i - j + l\leq n\] \[1\leq i + j - k\leq n?\]
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For \(i = k\) and \(j = l\) the system is reduced to \(1 \leq i, j \leq n\) , and has exactly \(n^{2}\) solutions. Let us assume that \(i \neq k\) or \(j \neq l\) . The points \(A(i, j)\) , \(B(k, l)\) , \(C(-j + k + l, i - k + l)\) , \(D(i - j + l, i + j - k)\) are vertices of a negatively oriented square with integer vertices lying inside the square \([1, n] \times [1, n]\) , and each of these squares corresponds to exactly 4 solutions to the system. By the previous problem there are exactly \(q(n) = n^{2}(n^{2} - 1) / 12\) such squares. Hence the number of solutions is equal to \(n^{2} + 4q(n) = n^{2}(n^{2} + 2) / 3\) .
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IMOLL-1977-17
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A ball \(K\) of radius \(r\) is touched from the outside by mutually equal balls of radius \(R\) . Two of these balls are tangent to each other. Moreover, for two balls \(K_{1}\) and \(K_{2}\) tangent to \(K\) and tangent to each other there exist two other balls tangent to \(K_{1}\) , \(K_{2}\) and also to \(K\) . How many balls are tangent to \(K\) ? For a given \(r\) determine \(R\) .
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Centers of the balls that are tangent to \(K\) are vertices of a regular polyhedron with triangular faces, with edge length \(2R\) and radius of circumscribed sphere \(r + R\) . Therefore the number \(n\) of these balls is 4, 6, or 20. It is straightforward to obtain that:
(i) If \(n = 4\) , then \(r + R = 2R(\sqrt{6} /4)\) , whence \(R = r(2 + \sqrt{6})\) . (ii) If \(n = 6\) , then \(r + R = 2R(\sqrt{2} /2)\) , whence \(R = r(1 + \sqrt{2})\) . (iii) If \(n = 20\) , then \(r + R = 2R\sqrt{5 + \sqrt{5}} /8\) . In this case we can conclude that \(R = r\left[\sqrt{5 - 2\sqrt{5}} + (3 - \sqrt{5}) / 2\right]\) .
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IMOLL-1977-18
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Given an isosceles triangle \(ABC\) with a right angle at \(C\) , construct the center \(M\) and radius \(r\) of a circle cutting on segments \(AB\) , \(BC\) , \(CA\) the segments \(DE\) , \(FG\) , and \(HK\) , respectively, such that \(\angle DME + \angle FMG + \angle HMK = 180^\circ\) and \(DE : FG : HK = AB : BC : CA\) .
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Let \(U\) be the midpoint of the segment \(AB\) . The point \(M\) belongs to \(CU\) and \(CM = (\sqrt{5} - 1)CU / 2\) , \(r = CU\sqrt{\sqrt{5} - 2}\) .
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IMOLL-1977-19
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Given any integer \(m > 1\) prove that there exist infinitely many positive integers \(n\) such that the last \(m\) digits of \(5^{n}\) are a sequence \(a_{m}, a_{m-1}, \ldots , a_{1} = 5\) \((0 \leq a_{j} < 10)\) in which each digit except the last is of opposite parity to its successor (i.e., if \(a_{i}\) is even, then \(a_{i-1}\) is odd, and if \(a_{i}\) is odd, then \(a_{i-1}\) is even).
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We shall prove the statement by induction on \(m\) . For \(m = 2\) it is trivial, since each power of 5 greater than 5 ends in 25. Suppose that the statement is true for some \(m \geq 2\) , and that the last \(m\) digits of \(5^{n}\) alternate in parity. It can be shown by induction that the maximum power of 2 that divides \(5^{2m - 2} - 1\) is \(2^{m}\) , and consequently the difference \(5^{n + 2m - 2} - 5^{n}\) is divisible by \(10^{m}\) but not by \(2 \cdot 10^{m}\) . It follows that the last \(m\) digits of the numbers \(5^{n + 2m - 2}\) and \(5^{n}\) coincide, but the digits at the position \(m + 1\) have opposite parity. Hence the last \(m + 1\) digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.
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IMOLL-1977-20
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Let \(a, b, A, B\) be given constant real numbers and
\[f(x) = 1 - a\cos x - b\sin x - A\cos 2x - B\sin 2x.\]
Prove that if \(f(x) \geq 0\) for all real \(x\) , then
\[a^{2} + b^{2} \leq 2 \quad \text{and} \quad A^{2} + B^{2} \leq 1.\]
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There exist \(u,v\) such that \(a\cos x + b\sin x = r\cos (x - u)\) and \(A\cos 2x + B\sin 2x =\) \(R\cos 2(x - v)\) , where \(r = \sqrt{a^{2} + b^{2}}\) and \(R = \sqrt{A^{2} + B^{2}}\) . Then \(1 - f(x) = r\cos (x-\) \(u) + R\cos 2(x - v)\leq 1\) holds for all \(x\in \mathbb{R}\)
There exists \(x\in \mathbb{R}\) such that \(\cos (x - u)\geq 0\) and \(\cos 2(x - v) = 1\) (indeed, either \(x = v\) or \(x = v + \pi\) works). It follows that \(R\leq 1\) . Similarly, there exists \(x\in \mathbb{R}\) such that \(\cos (x - u) = 1 / \sqrt{2}\) and \(\cos 2(x - v)\geq 0\) (either \(x = u - \pi /4\) or \(x = u + \pi /4\) works). It follows that \(r\leq \sqrt{2}\) .
Remark. The proposition of this problem contained as an addendum the following, more difficult, inequality:
\[\sqrt{a^{2} + b^{2}} +\sqrt{A^{2} + B^{2}}\leq 2.\]
The proof follows from the existence of \(x\in \mathbb{R}\) such that \(\cos (x - u)\geq 1 / 2\) and \(\cos 2(x - v)\geq 1 / 2\) .
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IMOLL-1977-21
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Given that \(x_{1} + x_{2} + x_{3} = y_{1} + y_{2} + y_{3} = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = 0\) , prove that
\[\frac{x_{1}^{2}}{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} +\frac{y_{1}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} = \frac{2}{3}.\]
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Let us consider the vectors \(v_{1} = (x_{1},x_{2},x_{3})\) , \(v_{2} = (y_{1},y_{2},y_{3})\) , \(v_{3} = (1,1,1)\) in space. The given equalities express the condition that these three vectors are mutually perpendicular. Also, \(\frac{x_{1}^{2}}{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}\) , \(\frac{y_{1}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}\) , and \(1 / 3\) are the squares of the projections of the vector \((1,0,0)\) onto the directions of \(v_{1},v_{2},v_{3}\) , respectively. The result follows from the fact that the sum of squares of projections of a unit vector on three mutually perpendicular directions is 1.
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IMOLL-1977-22
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Let \(S\) be a convex quadrilateral \(ABCD\) and \(O\) a point inside it. The feet of the perpendiculars from \(O\) to \(AB\) , \(BC\) , \(CD\) , \(DA\) are \(A_{1}\) , \(B_{1}\) , \(C_{1}\) , \(D_{1}\) respectively. The feet of the perpendiculars from \(O\) to the sides of \(S_{i}\) , the quadrilateral \(A_{i}B_{i}C_{i}D_{i}\) , are \(A_{i + 1}B_{i + 1}C_{i + 1}D_{i + 1}\) , where \(i = 1,2,3\) . Prove that \(S_{4}\) is similar to \(S\) .
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Since the quadrilateral \(OA_{1}BB_{1}\) is cyclic, \(\angle OA_{1}B_{1} = \angle OBC\) . By using the analogous equalities we obtain \(\angle OA_{4}B_{4} = \angle OB_{3}C_{3} = \angle OC_{2}D_{2} = \angle OD_{1}A_{1} = \angle OAB\) , and similarly \(\angle OB_{4}A_{4} = \angle OBA\) . Hence \(\triangle OA_{4}B_{4} \sim \triangle OAB\) . Analogously, we have for the other three pairs of triangles \(\triangle OB_{4}C_{4} \sim \triangle OBC\) , \(\triangle OC_{4}D_{4} \sim \triangle OCD\) , \(\triangle OD_{4}A_{4} \sim \triangle ODA\) , and consequently \(ABCD \sim A_{4}B_{4}C_{4}D_{4}\) .
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IMOLL-1977-23
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For which positive integers \(n\) do there exist two polynomials \(f\) and \(g\) with integer coefficients of \(n\) variables \(x_{1}, x_{2}, \ldots , x_{n}\) such that the following equality is satisfied:
\[\left(\sum_{i = 1}^{n} x_{i}\right) f(x_{1}, x_{2}, \ldots , x_{n}) = g(x_{1}^{2}, x_{2}^{2}, \ldots , x_{n}^{2})?\]
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Every polynomial \(q(x_{1},\ldots ,x_{n})\) with integer coefficients can be expressed in the form \(q = r_{1} + x_{1}r_{2}\) , where \(r_{1},r_{2}\) are polynomials in \(x_{1},\ldots ,x_{n}\) with integer coefficients in which the variable \(x_{1}\) occurs only with even exponents. Thus if \(q_{1} = r_{1} - x_{1}r_{2}\) , the polynomial \(qq_{1} = r_{1}^{2} - x_{1}^{2}r_{2}^{2}\) contains \(x_{1}\) only with even exponents. We can continue inductively constructing polynomials \(q_{j}\) , \(j = 2,3,\ldots ,n\) , such that \(qq_{1}q_{2}\cdot \cdot \cdot q_{j}\) contains each of the variables \(x_{1},x_{2},\ldots ,x_{j}\) only with even exponents. Thus the polynomial \(qq_{1}\cdot \cdot \cdot q_{n}\) is a polynomial in \(x_{1}^{2},\ldots ,x_{n}^{2}\) . The polynomials \(f\) and \(g\) exist for every \(n\in \mathbb{N}\) . In fact, it suffices to construct \(q_{1},\ldots ,q_{n}\) for the polynomial \(q = x_{1} + \cdot \cdot \cdot +x_{n}\) and take \(f = q_{1}q_{2}\cdot \cdot \cdot q_{n}\) .
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IMOLL-1977-24
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Determine all real functions \(f(x)\) that are defined and continuous on the interval \((-1, 1)\) and that satisfy the functional equation
\[f(x + y) = \frac{f(x) + f(y)}{1 - f(x)f(y)}\qquad (x,y,x + y\in (-1,1)).\]
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Setting \(x = y = 0\) gives us \(f(0) = 0\) . Let us put \(g(x) = \arctan f(x)\) . The given functional equation becomes \(\tan g(x + y) = \tan (g(x) + g(y))\) ; hence
\[g(x + y) = g(x) + g(y) + k(x,y)\pi ,\]
where \(k(x,y)\) is an integer function. But \(k(x,y)\) is continuous and \(k(0,0) = 0\) , therefore \(k(x,y) = 0\) . Thus we obtain the classical Cauchy's functional equation \(g(x + y) = g(x) + g(y)\) on the interval \((- 1,1)\) , all of whose continuous solutions are of the form \(g(x) = ax\) for some real \(a\) . Moreover, \(g(x) \in (-\pi ,\pi)\) implies \(|a| \leq \pi /2\) .
Therefore \(f(x) = \tan ax\) for some \(|a| \leq \pi /2\) , and this is indeed a solution to the given equation.
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IMOLL-1977-25
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Prove the identity
\[(z + a)^{n} = z^{n} + a\sum_{k = 1}^{n}\binom{n}{k} (a - k b)^{k - 1}(z + k b)^{n - k}.\]
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Let
\[f_{n}(z) = z^{n} + a\sum_{k = 1}^{n}\binom{n}{k} (a - k b)^{k - 1}(z + k b)^{n - k}.\]
We shall prove by induction on \(n\) that \(f_{n}(z) = (z + a)^{n}\) . This is trivial for \(n = 1\) . Suppose that the statement is true for some positive integer \(n - 1\) . Then
\[f_{n}^{\prime}(z) = n z^{n - 1} + a\sum_{k = 1}^{n - 1}\binom{n}{k} (n - k)(a - k b)^{k - 1}(z + k b)^{n - k - 1}\] \[\qquad = n z^{n - 1} + n a\sum_{k = 1}^{n - 1}\binom{n - 1}{k} (a - k b)^{k - 1}(z + k b)^{n - k - 1}\] \[\qquad = n f_{n - 1}(z) = n(z + a)^{n - 1}.\]
It remains to prove that \(f_{n}(- a) = 0\) . For \(z = - a\) we have by the lemma of (SL81- 13),
\[f_{n}(-a) = (-a)^{n} + a\sum_{k = 1}^{n}\binom{n}{k} (-1)^{n - k}(a - k b)^{n - 1}\] \[\qquad = a\sum_{k = 0}^{n}\binom{n}{k} (-1)^{n - k}(a - k b)^{n - 1} = 0.\]
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IMOLL-1977-26
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Let \(p\) be a prime number greater than 5. Let \(V\) be the collection of all positive integers \(n\) that can be written in the form \(n = kp + 1\) or \(n = kp - 1\) \((k = 1, 2, \ldots)\) . A number \(n \in V\) is called indecomposable in \(V\) if it is impossible to find \(k, l \in V\) such that \(n = kl\) . Prove that there exists a number \(N \in V\) that can be factorized into indecomposable factors in \(V\) in more than one way.
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The result is an immediate consequence (for \(G = \{-1,1\}\) ) of the following generalization.
(1) Let \(G\) be a proper subgroup of \(\mathbb{Z}_{n}^{*}\) (the multiplicative group of residue classes modulo \(n\) coprime to \(n\) ), and let \(V\) be the union of elements of \(G\) . A number \(m \in V\) is called indecomposable in \(V\) if there do not exist numbers \(p, q \in V\) , \(p, q \notin \{-1,1\}\) , such that \(pq = m\) . There exists a number \(r \in V\) that can be expressed as a product of elements indecomposable in \(V\) in more than one way.
First proof. We shall start by proving the following lemma.
Lemma. There are infinitely many primes not in \(V\) that do not divide \(n\) .
Proof. There is at least one such prime: In fact, any number other than \(\pm 1\) not in \(V\) must have a prime factor not in \(V\) , since \(V\) is closed under multiplication. If there were a finite number of such primes, say \(p_{1}, p_{2}, \ldots , p_{k}\) , then one of the numbers \(p_{1} p_{2} \cdots p_{k} + n\) , \(p_{1}^{2} p_{2} \cdots p_{k} + n\) is not in \(V\) and is coprime to \(n\) and \(p_{1}, \ldots , p_{k}\) , which is a contradiction.
[This lemma is actually a direct consequence of Dirichlet's theorem.]
Let us consider two such primes \(p, q\) that are congruent modulo \(n\) . Let \(p^{k}\) be the least power of \(p\) that is in \(V\) . Then \(p^{k}, q^{k}, p^{k - 1} q, p q^{k - 1}\) belong to \(V\) and are indecomposable in \(V\) . It follows that
\[r = p^{k}\cdot q^{k} = p^{k - 1}q\cdot p q^{k - 1}\]
has the desired property.
Second proof. Let \(p\) be any prime not in \(V\) that does not divide \(n\) , and let \(p^k\) be the least power of \(p\) that is in \(V\) . Obviously \(p^k\) is indecomposable in \(V\) . Then the number
\[r = p^{k}\cdot (p^{k - 1} + n)(p + n) = p(p^{k - 1} + n)\cdot p^{k - 1}(p + n)\]
has at least two different factorizations into indecomposable factors.
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IMOLL-1977-27
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Let \(n\) be an integer greater than 2. Define \(V = \{1 + kn \mid k = 1,2, \ldots \}\) . A number \(p \in V\) is called indecomposable in \(V\) if it is not possible to find numbers \(q_{1}, q_{2} \in V\) such that \(q_{1} q_{2} = p\) . Prove that there exists a number \(N \in V\) that can be factorized into indecomposable factors in \(V\) in more than one way.
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The result is a consequence of the generalization from the previous problem for \(G = \{1\}\) .
Remark. There is an explicit example: \(r = (n - 1)^2 \cdot (2n - 1)^2 = [(n - 1)(2n - 1)]^2\) .
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IMOLL-1977-28
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Let \(n\) be an integer greater than 1. Define
\[x_{1} = n, y_{1} = 1, \quad x_{i + 1} = \left[\frac{x_{i} + y_{i}}{2}\right], y_{i + 1} = \left[\frac{n}{x_{i + 1}}\right] \quad \text{for} i = 1,2, \ldots ,\]
where \([z]\) denotes the largest integer less than or equal to \(z\) . Prove that
\[\min \{x_{1}, x_{2}, \ldots , x_{n}\} = \lfloor \sqrt{n} \rfloor .\]
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The recurrent relations give us that
\[x_{i + 1} = \left[\frac{x_i + [n / x_i]}{2}\right] = \left[\frac{x_i + n / x_i}{2}\right]\geq [\sqrt{n} ].\]
On the other hand, if \(x_{i} > [\sqrt{n} ]\) for some \(i\) , then we have \(x_{i + 1}< x_{i}\) . This follows from the fact that \(x_{i + 1}< x_{i}\) is equivalent to \(x_{i} > (x_{i} + n / x_{i}) / 2\) , i.e., to \(x_{i}^{2} > n\) . Therefore \(x_{i} = [\sqrt{n} ]\) holds for at least one \(i\leq n - [\sqrt{n} ] + 1\) .
Remark. If \(n + 1\) is a perfect square, then \(x_{i} = [\sqrt{n} ]\) implies \(x_{i + 1} = [\sqrt{n} ] + 1\) . Otherwise, \(x_{i} = [\sqrt{n} ]\) implies \(x_{i + 1} = [\sqrt{n} ]\) .
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IMOLL-1977-29
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On the sides of a square \(ABCD\) one constructs inwardly equilateral triangles \(ABK\) , \(BCL\) , \(CDM\) , \(DAN\) . Prove that the midpoints of the four segments \(KL\) , \(LM\) , \(MN\) , \(NK\) , together with the midpoints of the eight segments \(AK\) , \(BK\) , \(BL\) , \(CL\) , \(CM\) , \(DM\) , \(DN\) , \(AN\) , are the 12 vertices of a regular dodecagon.
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Let us denote the midpoints of segments \(LM\) , \(AN\) , \(BL\) , \(MN\) , \(BK\) , \(CM\) , \(NK\) , \(CL\) ,
\(DN\) , \(KL\) , \(DM\) , \(AK\) by \(P_{1}\) , \(P_{2}\) , \(P_{3}\) , \(P_{4}\) , \(P_{5}\) , \(P_{6}\) , \(P_{7}\) , \(P_{8}\) , \(P_{9}\) , \(P_{10}\) , \(P_{11}\) , \(P_{12}\) , respectively.
We shall prove that the dodecagon \(P_{1}P_{2}P_{3}\ldots P_{11}P_{12}\) is regular. From \(BL = BA\) and \(\angle ABL = 30^{\circ}\) it follows that \(\angle BAL = 75^{\circ}\) . Similarly \(\angle DAM = 75^{\circ}\) , and therefore \(\angle LAM = 60^{\circ}\) , which together with the fact \(AL = AM\) implies that \(\triangle ALM\) is equilateral. Now, from the triangles \(OLM\) and \(ALN\) , we deduce \(OP_{1} = LM / 2\) , \(OP_{2} = AL / 2\) and \(OP_{2} \parallel AL\) .
Hence \(OP_{1} = OP_{2}\) , \(\angle P_{1}OP_{2} = \angle P_{1}AL = 30^{\circ}\) and \(\angle P_{2}OM = \angle LAD = 15^{\circ}\) . The desired result follows from symmetry.
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IMOLL-1977-30
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A triangle \(ABC\) with \(\angle A = 30^{\circ}\) and \(\angle C = 54^{\circ}\) is given. On \(BC\) a point \(D\) is chosen such that \(\angle CAD = 12^{\circ}\) . On \(AB\) a point \(E\) is chosen such that \(\angle ACE = 6^{\circ}\) . Let \(S\) be the point of intersection of \(AD\) and \(CE\) . Prove that \(BS = BC\) .
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Suppose \(\angle SBA = x\) . By the trigonometric form of Ceva's theorem we have
\[\frac{\sin(96^{\circ} - x)}{\sin x} \frac{\sin 18^{\circ}}{\sin 12^{\circ}} \frac{\sin 6^{\circ}}{\sin 48^{\circ}} = 1. \quad (1)\]
We claim that \(x = 12^{\circ}\) is a solution of this equation. To prove this, it is enough to show that \(\sin 84^{\circ} \sin 6^{\circ} \sin 18^{\circ} = \sin 48^{\circ} \sin 12^{\circ} \sin 12^{\circ}\) , which is equivalent to \(\sin 18^{\circ} = 2 \sin 48^{\circ} \sin 12^{\circ} = \cos 36^{\circ} - \cos 60^{\circ}\) . The last equality can be checked directly.
Since the equation is equivalent to
\[(\sin 96^{\circ}\cot x - \cos 96^{\circ})\sin 6^{\circ}\sin 18^{\circ} = \sin 48^{\circ}\sin 12^{\circ},\]
the solution \(x\in [0,\pi)\) is unique. Hence \(x = 12^{\circ}\)
Second solution. We know that if \(a,b,c,a^{\prime},b^{\prime},c^{\prime}\) are points on the unit circle in the complex plane, the lines \(a a^{\prime}\) , \(b b^{\prime}\) , \(c c^{\prime}\) are concurrent if and only if
\[(a - b^{\prime})(b - c^{\prime})(c - a^{\prime}) = (a - c^{\prime})(b - a^{\prime})(c - b^{\prime}). \quad (1)\]
We shall prove that \(x = 12^{\circ}\) . We may suppose that \(ABC\) is the triangle in the complex plane with vertices \(a = 1\) , \(b = \epsilon^{9}\) , \(c = \epsilon^{14}\) , where \(\epsilon = \cos \frac{\pi}{15} +i\sin \frac{\pi}{15}\) . If \(a^{\prime} = \epsilon^{12}\) , \(b^{\prime} = \epsilon^{28}\) , \(c^{\prime} = \epsilon\) , our task is the same as proving that lines \(aa^{\prime}\) , \(bb^{\prime}\) , \(cc^{\prime}\) are concurrent, or by (1) that
\[(1 - \epsilon^{28})(\epsilon^{9} - \epsilon)(\epsilon^{14} - \epsilon^{12}) - (1 - \epsilon)(\epsilon^{9} - \epsilon^{12})(\epsilon^{14} - \epsilon^{28}) = 0.\]
The last equality holds, since the left- hand side is divisible by the minimum polynomial of \(\epsilon \colon z^{8} + z^{7} - z^{5} - z^{4} - z^{3} + z + 1\) .
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IMOLL-1977-31
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Let \(f\) be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that \(f\) satisfies the following conditions:
(1) \(f(ab,c) = f(a,c)f(b,c),f(c,ab) = f(c,a)f(c,b);\) (2) \(f(a,1 - a) = 1\)
Prove that \(f(a,a) = f(a, - a) = 1,f(a,b)f(b,a) = 1.\)
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We obtain from (1) that \(f(1,c) = f(1,c)f(1,c)\) ; hence \(f(1,c) = 1\) and consequently \(f(-1,c)f(-1,c) = f(1,c) = 1\) , i.e. \(f(-1,c) = 1\) . Analogously, \(f(c,1) = f(c, - 1) = 1\) .
Clearly \(f(1,1) = f(- 1,1) = f(1, - 1) = 1\) . Now let us assume that \(a \neq 1\) . Observe that \(f(x^{- 1},y) = f(x,y^{- 1}) = f(x,y)^{- 1}\) . Thus by (1) and (2) we get
\[1 = f(a,1 - a)f(1 / a,1 - 1 / a)\] \[= f(a,1 - a)f\left(a,\frac{1}{1 - 1 / a}\right) = f\left(a,\frac{1 - a}{1 - 1 / a}\right) = f(a, - a).\]
We now have \(f(a,a) = f(a, - 1)f(a, - a) = 1\cdot 1 = 1\) and \(1 = f(ab,ab) =\) \(f(a,ab)f(b,ab) = f(a,a)f(a,b)f(b,a)f(b,b) = f(a,b)f(b,a).\)
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IMOLL-1977-32
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In a room there are nine men. Among every three of them there are two mutually acquainted. Prove that some four of them are mutually acquainted.
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It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.
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IMOLL-1977-33
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A circle \(K\) centered at \((0,0)\) is given. Prove that for every vector \((a_{1},a_{2})\) there is a positive integer \(n\) such that the circle \(K\) translated by the vector \(n(a_{1},a_{2})\) contains a lattice point (i.e., a point both of whose coordinates are integers).
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Let \(r\) be the radius of \(K\) and \(s > \sqrt{2} /r\) an integer. Consider the points \(A_{k}(ka_{1} - [ka_{1}],ka_{2} - [ka_{2}])\) , where \(k = 0,1,2,\ldots ,s^{2}\) . Since all these points are in the unit square, two of them, say \(A_{p},A_{q}\) , \(q > p\) , are in a small square with side \(1 / s\) , and consequently \(A_{p}A_{q} \leq \sqrt{2} /s < r\) . Therefore, for \(n = q - p\) , \(m_{1} = [qa_{1}] - [pa_{1}]\) and \(m_{2} = [qa_{2}] - [pa_{2}]\) the distance between the points \(n(a_{1},a_{2})\) and \((m_{1},m_{2})\) is less then \(r\) , i.e., the point \((m_{1},m_{2})\) is in the circle \(K + n(a_{1},a_{2})\) .
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IMOLL-1977-34
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Let \(B\) be a set of \(k\) sequences each having \(n\) terms equal to 1 or \(-1\) . The product of two such sequences \((a_{1}, a_{2}, \ldots , a_{n})\) and \((b_{1}, b_{2}, \ldots , b_{n})\) is defined as \((a_{1} b_{1}, a_{2} b_{2}, \ldots , a_{n} b_{n})\) . Prove that there exists a sequence \((c_{1}, c_{2}, \ldots , c_{n})\) such that the intersection of \(B\) and the set containing all sequences from \(B\) multiplied by \((c_{1}, c_{2}, \ldots , c_{n})\) contains at most \(k^{2} / 2^{n}\) sequences.
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Let \(A\) be the set of the \(2^{n}\) sequences of \(n\) terms equal to \(\pm 1\) . Since there are \(k^{2}\) products \(ab\) with \(a,b\in B\) , by the pigeonhole principle there exists \(c\in A\) such that \(ab = c\) holds for at most \(k^{2} / 2^{n}\) pairs \((a,b)\in B\times B\) . Then \(cb\in B\) holds for at most \(k^{2} / 2^{n}\) values \(b\in B\) , which means that \(|B\cap cB|\leq k^{2} / 2^{n}\) .
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IMOLL-1977-35
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Find all numbers \(N = \overline{a_{1}a_{2}\ldots a_{n}}\) for which \(9\times \overline{a_{1}a_{2}\ldots a_{n}} = \overline{a_{n}\ldots a_{2}a_{1}}\) such that at most one of the digits \(a_{1},a_{2},\ldots ,a_{n}\) is zero.
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The solutions are 0 and \(N_{k} = 1099\underbrace{..}_{k}989\) , where \(k = 0,1,2,\ldots\) .
Remark. If we omit the condition that at most one of the digits is zero, the solutions are numbers of the form \(N_{k_{1}}N_{k_{2}}\ldots N_{k_{r}}\) , where \(k_{1} = k_{r}\) , \(k_{2} = k_{r - 1}\) etc. The more general problem \(k\cdot \overline{a_{1}a_{2}\ldots a_{n}} = \overline{a_{n}\ldots a_{2}a_{1}}\) has solutions only for \(k = 9\) and for \(k = 4\) (namely 0, 2199. . . 978 and combinations as above).
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IMOLL-1977-36
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Consider a sequence of numbers \((a_{1},a_{2},\ldots ,a_{2^{n}})\) . Define the operation
\[S((a_{1},a_{2},\ldots ,a_{2^{n}})) = (a_{1}a_{2},a_{2}a_{3},\ldots ,a_{2^{n} - 1}a_{2^{n}},a_{2^{n}a_{1}}).\]
Prove that whatever the sequence \((a_{1},a_{2},\ldots ,a_{2^{n}})\) is, with \(a_{i}\in \{- 1,1\}\) for \(i =\) \(1,2,\ldots ,2^{n}\) , after finitely many applications of the operation we get the sequence \((1,1,\ldots ,1)\)
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It can be shown by simple induction that \(S^{m}(a_{1},\ldots ,a_{2^{n}}) = (b_{1},\ldots ,b_{2^{n}})\) , where
\[b_{k} = \prod_{i = 0}^{m}a_{k + i}^{\binom{n}{i}} \text{(assuming that} a_{k + 2^{n}} = a_{k}).\]
If we take \(m = 2^{n}\) all the binomial coefficients \(\binom{n}{i}\) apart from \(i = 0\) and \(i = m\) will be even, and thus \(b_{k} = a_{k}a_{k + m} = 1\) for all \(k\) .
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IMOLL-1977-37
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Let \(A_{1},A_{2},\ldots ,A_{n + 1}\) be positive integers such that \((A_{i},A_{n + 1}) = 1\) for every \(i = 1,2,\ldots ,n\) . Show that the equation
\[x_{1}^{A_{1}} + x_{2}^{A_{2}} + \dots +x_{n}^{A_{n}} = x_{n + 1}^{A_{n + 1}}\]
has an infinite set of solutions \((x_{1},x_{2},\ldots ,x_{n + 1})\) in positive integers.
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We look for a solution with \(x_{1}^{A_{1}} = \dots = x_{n}^{A_{n}} = n^{A_{1}A_{2}\dots A_{n}x}\) and \(x_{n + 1} = n^{y}\) . In order for this to be a solution we must have \(A_{1}A_{2}\dots A_{n}x + 1 = A_{n + 1}y\) . This equation has infinitely many solutions \((x,y)\) in \(\mathbb{N}\) , since \(A_{1}A_{2}\dots A_{n}\) and \(A_{n + 1}\) are coprime.
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IMOLL-1977-38
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Let \(m_{j} > 0\) for \(j = 1,2,\ldots ,n\) and \(a_{1}\leq \dots \leq a_{n}< b_{1}\leq \dots \leq b_{n}< c_{1}\leq\) \(\dots \leq c_{n}\) be real numbers. Prove:
\[\left[\sum_{j = 1}^{n}m_{j}(a_{j} + b_{j} + c_{j})\right]^{2} > 3\left(\sum_{j = 1}^{n}m_{j}\right)\left[\sum_{j = 1}^{n}m_{j}(a_{j}b_{j} + b_{j}c_{j} + c_{j}a_{j})\right].\]
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The condition says that the quadratic equation \(f(x) = 0\) has distinct real solutions, where
\[f(x) = 3x^{2}\sum_{j = 1}^{n}m_{j} - 2x\sum_{j = 1}^{n}m_{j}(a_{j} + b_{j} + c_{j}) + \sum_{j = 1}^{n}m_{j}(a_{j}b_{j} + b_{j}c_{j} + c_{j}a_{j}).\]
It is easy to verify that the function \(f\) is the derivative of
\[F(x) = \sum_{j = 1}^{n}m_{j}(x - a_{j})(x - b_{j})(x - c_{j}).\]
Since \(F(a_{1})\leq 0\leq F(a_{n})\) , \(F(b_{1})\leq 0\leq F(b_{n})\) and \(F(c_{1})\leq 0\leq F(c_{n})\) , \(F(x)\) has three distinct real roots, and hence by Rolle's theorem its derivative \(f(x)\) has two distinct real roots.
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IMOLL-1977-39
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Consider 37 distinct points in space, all with integer coordinates. Prove that we may find among them three distinct points such that their barycenter has integer coordinates.
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By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their \(x\) -coordinates are congruent and their \(y\) -coordinates are congruent modulo 3. Now among these 5 points either there exist three with \(z\) -coordinates congruent modulo 3, or there exist three whose \(z\) -coordinates are congruent to 0, 1, 2 modulo 3. These three points are the desired ones.
Remark. The minimum number \(n\) such that among any \(n\) integer points in space one can find three points whose barycenter is an integer point is \(n = 19\) . Each proof of this result seems to consist in studying a great number of cases.
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IMOLL-1977-40
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The numbers \(1,2,3,\ldots ,64\) are placed on a chessboard, one number in each square. Consider all squares on the chessboard of size \(2\times 2\) . Prove that
there are at least three such squares for which the sum of the 4 numbers contained exceeds 100.
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Let us divide the chessboard into 16 squares \(Q_{1}, Q_{2}, \ldots , Q_{16}\) of size \(2 \times 2\) . Let \(s_{k}\) be the sum of numbers in \(Q_{k}\) , and let us assume that \(s_{1} \geq s_{2} \geq \dots \geq s_{16}\) . Since \(s_{4} + s_{5} + \dots + s_{16} \geq 1 + 2 + \dots + 52 = 1378\) , we must have \(s_{4} \geq 100\) and hence \(s_{1}, s_{2}, s_{3} \geq 100\) as well.
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IMOLL-1977-41
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A wheel consists of a fixed circular disk and a mobile circular ring. On the disk the numbers \(1,2,3,\ldots ,N\) are marked, and on the ring \(N\) integers
\(a_{1},a_{2},\ldots ,a_{N}\) of sum 1 are marked (see the figure). The ring can be turned into \(N\) different positions in which the numbers on the disk and on the ring match each other. Multiply every number on the ring with the corresponding number on the disk and form the sum of \(N\) products. In this way a sum is obtained for every position of the ring. Prove that the \(N\) sums are different.
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The considered sums are congruent modulo \(N\) to \(S_{k} = \sum_{i = 1}^{N}(i + k)a_{i}\) , \(k = 0, 1, \ldots , N - 1\) . Since \(S_{k} = S_{0} + k(a_{1} + \dots + a_{n}) = S_{0} + k\) , all these sums give distinct residues modulo \(N\) and therefore are distinct.
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IMOLL-1977-42
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The sequence \(a_{n,k},k = 1,2,3,\ldots ,2^{n},n = 0,1,2,\ldots\) , is defined by the following recurrence formula:
\[a_{1} = 2,\qquad a_{n,k} = 2a_{n - 1,k}^{3},\qquad a_{n,k + 2^{n - 1}} = \frac{1}{2} a_{n - 1,k}^{3}\]
\[\mathrm{for} k = 1,2,3,\ldots ,2^{n - 1},n = 0,1,2,\ldots .\]
Prove that the numbers \(a_{n,k}\) are all different.
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It can be proved by induction on \(n\) that
\[\{a_{n,k} \mid 1 \leq k \leq 2^{n}\} = \{2^{m} \mid m = 3^{n} + 3^{n - 1}s_{1} + \dots + 3^{1}s_{n - 1} + s_{n} (s_{i} = \pm 1)\} .\]
Thus the result is an immediate consequence of the following lemma.
Lemma. Each positive integer \(s\) can be uniquely represented in the form
\[s = 3^{n} + 3^{n - 1}s_{1} + \dots + 3^{1}s_{n - 1} + s_{n},\quad \mathrm{where} s_{i}\in \{-1,0,1\} . \quad (1)\]
Proof. Both the existence and the uniqueness can be shown by simple induction on \(s\) . The statement is trivial for \(s = 1\) , while for \(s > 1\) there exist \(q \in \mathbb{N}\) , \(r \in \{- 1,0,1\}\) such that \(s = 3q + r\) , and \(q\) has a unique representation of the form (1).
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IMOLL-1977-43
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Evaluate
\[S = \sum_{k = 1}^{n}k(k + 1)\dots (k + p),\]
where \(n\) and \(p\) are positive integers.
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Since \(k(k + 1) \dots (k + p) = (p + 1)! \binom{k + p}{p + 1} = (p + 1)! \left[ \binom{k + p + 1}{p + 2} - \binom{k + p}{p + 2} \right]\) , it follows that
\[\sum_{k = 1}^{n}k(k + 1)\dots (k + p) = (p + 1)!\binom{n + p + 1}{p + 2} = \frac{n(n + 1)\dots(n + p + 1)}{p + 2}.\]
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IMOLL-1977-44
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Let \(E\) be a finite set of points in space such that \(E\) is not contained in a plane and no three points of \(E\) are collinear. Show that \(E\) contains the vertices of a tetrahedron \(T = ABCD\) such that \(T\cap E = \{A,B,C,D\}\) (including interior points of \(T\) ) and such that the projection of \(A\) onto the plane \(BCD\) is inside a triangle that is similar to the triangle \(BCD\) and whose sides have midpoints \(B,C,D\) .
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Let \(d(X, \sigma)\) denote the distance from a point \(X\) to a plane \(\sigma\) . Let us consider the pair \((A, \pi)\) where \(A \in E\) and \(\pi\) is a plane containing some three points \(B, C, D \in E\) such that \(d(A, \pi)\) is the smallest possible. We may suppose that \(B, C, D\) are selected such that \(\triangle BCD\) contains no other points of \(E\) . Let \(A'\) be the projection of \(A\) on \(\pi\) , and let \(l_b, l_c, l_d\) be lines through \(B, C, D\) parallel to \(CD, DB, BC\) respectively. If \(A'\) is in the half-plane determined by \(l_d\) not containing \(BC\) , then \(d(D, ABC) \leq d(A', ABC) < d(A, BCD)\) , which is impossible. Similarly, \(A'\) lies in the half-planes determined by \(l_b, l_c\) that contain \(D\) , and hence \(A'\) is inside the triangle bordered by \(l_b, l_c, l_d\) . The minimality property of \((A, \pi)\) and the way in which \(BCD\) was selected guarantee that \(E \cap T = \{A, B, C, D\}\) .
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IMOLL-1977-45
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Let \(E\) be a finite set of points such that \(E\) is not contained in a plane and no three points of \(E\) are collinear. Show that at least one of the following alternatives holds:
(i) \(E\) contains five points that are vertices of a convex pyramid having no other points in common with \(E\) ;
(ii) some plane contains exactly three points from \(E\) .
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As in the previous problem, let us choose the pair \((A, \pi)\) such that \(d(A, \pi)\) is minimal. If \(\pi\) contains only three points of \(E\) , we are done. If not, there are four points in \(E \cap P\) , say \(A_1, A_2, A_3, A_4\) , such that the quadrilateral \(Q = A_1A_2A_3A_4\) contains no other points of \(E\) . Suppose \(Q\) is not convex, and that w.l.o.g. \(A_1\) is inside the triangle \(A_2A_3A_4\) . If \(A_0\) is the projection of \(A\) on \(P\) , the point \(A_1\) belongs to one of the triangles \(A_0A_2A_3\) , \(A_0A_3A_4\) , \(A_0A_4A_2\) , say \(A_0A_2A_3\) . Then \(d(A_1, AA_2A_3) \leq d(A_0, AA_2A_3) < AA_0\) , which is impossible. Hence \(Q\) is convex. Also, by the minimality property of \((A, \pi)\) the pyramid \(AA_1A_2A_3A_4\) contains no other points of \(E\) .
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IMOLL-1977-46
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Let \(f\) be a strictly increasing function defined on the set of real numbers. For \(x\) real and \(t\) positive, set
\[g(x,t) = \frac{f(x + t) - f(x)}{f(x) - f(x - t)}.\]
Assume that the inequalities
\[2^{-1}< g(x,t)< 2\]
hold for all positive \(t\) if \(x = 0\) , and for all \(t\leq |x|\) otherwise.
Show that
\[14^{-1}< g(x,t)< 14\]
for all real \(x\) and positive \(t\) .
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We need to consider only the case \(t > |x|\) . There is no loss of generality in assuming \(x > 0\) .
To obtain the estimate from below, set
\[a_{1} = f\left(-\frac{x + t}{2}\right) - f(-(x + t)),\qquad a_{2} = f(0) - f\left(-\frac{x + t}{2}\right),\] \[a_{3} = f\left(\frac{x + t}{2}\right) - f(0),\qquad a_{4} = f(x + t) - f\left(\frac{x + t}{2}\right).\]
Since \(- (x + t)< x - t\) and \(x< (x + t) / 2\) , we have \(f(x) - f(x - t)\leq a_{1} + a_{2} + a_{3}\) . Since \(2^{- 1}< a_{j + 1} / a_{j}< 2\) , it follows that
\[g(x,t) > \frac{a_{4}}{a_{1} + a_{2} + a_{3}} >\frac{a_{3} / 2}{4a_{3} + 2a_{3} + a_{3}} = 14^{-1}.\]
To obtain the estimate from above, set
\[b_{1} = f(0) - f\left(-\frac{x + t}{3}\right),\qquad b_{2} = f\left(\frac{x + t}{3}\right) - f(0),\] \[b_{3} = f\left(\frac{2(x + t)}{3}\right) - f\left(\frac{x + t}{3}\right),\qquad b_{4} = f(x + t) - f\left(\frac{2(x + t)}{3}\right).\]
If \(t< 2x\) , then \(x - t< -(x + t) / 3\) and therefore \(f(x) - f(x - t)\geq b_{1}\) . If \(t\geq 2x\) then \((x + t) / 3\leq x\) and therefore \(f(x) - f(x - t)\geq b_{2}\) . Since \(2^{- 1}< b_{j + 1} / b_{j}< 2\) we get
\[g(x,t)< \frac{b_{2} + b_{3} + b_{4}}{\min \{b_{1},b_{2}\}} < \frac{b_{2} + 2b_{2} + 4b_{2}}{b_{2} / 2} = 14.\]
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IMOLL-1977-47
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A square \(ABCD\) is given. A line passing through \(A\) intersects \(CD\) at \(Q\) . Draw a line parallel to \(AQ\) that intersects the boundary of the square at points \(M\) and \(N\) such that the area of the quadrilateral \(AMNQ\) is maximal.
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\(M\) lies on \(AB\) and \(N\) lies on \(BC\) . If \(CQ\leq 2CD / 3\) , then \(BM = CQ / 2\) . If \(CQ > 2CD / 3\) , then \(N\) coincides with \(C\) .
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IMOLL-1977-48
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The intersection of a plane with a regular tetrahedron with edge \(a\) is a quadrilateral with perimeter \(P\) . Prove that \(2a \leq P \leq 3a\) .
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Let a plane cut the edges \(AB,BC,CD,DA\) at points \(K,L,M,N\) respectively.
Let \(D^{\prime},A^{\prime},B^{\prime}\) be distinct points in the plane \(ABC\) such that the triangles \(BCD^{\prime}\)
\(CD^{\prime}A^{\prime}\) , \(D^{\prime}A^{\prime}B^{\prime}\) are equilateral, and \(M^{\prime}\in [CD^{\prime}]\) , \(N^{\prime}\in [D^{\prime}A^{\prime}]\) , and \(K^{\prime}\in\) \([A^{\prime}B^{\prime}]\) such that \(CM^{\prime} = CM\) , \(A^{\prime}N^{\prime} =\) \(AN\) , and \(A^{\prime}K^{\prime} = AK\) . The perimeter \(P\) of the quadrilateral \(KLMN\) is equal to the length of the polygonal line \(KLM^{\prime}N^{\prime}K^{\prime}\) , which is not less than \(KK^{\prime}\) .
It follows that \(P\geq 2a\) .
Let us consider all quadrilaterals \(KLMN\) that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane \(\alpha\) . The lengths of the segments \(KL,LM,MN,NK\) are linear functions in \(AK\) , and so is \(P\) . Thus \(P\) takes its maximum at an endpoint of the interval, i.e., when the plane \(KLMN\) passes through one of the vertices \(A,B,C,D\) , and it is easy to see that in this case \(P\leq 3a\) .
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IMOLL-1977-49
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Find all pairs of integers \((p, q)\) for which all roots of the trinomials \(x^2 + px + q\) and \(x^2 + qx + p\) are integers.
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If one of \(p,q\) , say \(p\) , is zero, then \(-q\) is a perfect square. Conversely, \((p,q) = (0, - t^2)\) and \((p,q) = (- t^2,0)\) satisfy the conditions for \(t \in \mathbb{Z}\) .
We now assume that \(p,q\) are nonzero. If the trinomial \(x^{2} + px + q\) has two integer roots \(x_{1},x_{2}\) , then \(|q| = |x_{1}x_{2}|\geq |x_{1}| + |x_{2}| - 1\geq |p| - 1\) . Similarly, if \(x^{2} + qx + p\) has integer roots, then \(|p|\geq |q| - 1\) and \(q^{2} - 4p\) is a square. Thus we have two cases to investigate:
(i) \(|p| = |q|\) . Then \(p^{2} - 4q = p^{2}\pm 4p\) is a square, so \((p,q) = (4,4)\) .
(ii) \(|p| = |q|\pm 1\) . The solutions for \((p,q)\) are \((t, - 1 - t)\) for \(t \in \mathbb{Z}\) and \((5,6)\) , \((6,5)\) .
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IMOLL-1977-50
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Determine all positive integers \(n\) for which there exists a polynomial \(P_n(x)\) of degree \(n\) with integer coefficients that is equal to \(n\) at \(n\) different integer points and that equals zero at zero.
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Suppose that \(P_{n}(x) = n\) for \(x \in \{x_{1}, x_{2}, \ldots , x_{n}\}\) . Then
\[P_{n}(x) = (x - x_{1})(x - x_{2})\dots (x - x_{n}) + n.\]
From \(P_{n}(0) = 0\) we obtain \(n = |x_{1}x_{2}\dots x_{n}|\geq 2^{n - 2}\) (because at least \(n - 2\) factors are different from \(\pm 1\) ) and therefore \(n\geq 2^{n - 2}\) . It follows that \(n\leq 4\) .
For each positive integer \(n\leq 4\) there exists a polynomial \(P_{n}\) . Here is the list of such polynomials:
\[n = 1:\pm x, n = 2:2x^{2}, x^{2}\pm x, -x^{2}\pm 3x,\] \[n = 3:\pm (x^{3} - x) + 3x^{2}, n = 4:-x^{4} + 5x^{2}.\]
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IMOLL-1977-51
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Several segments, which we shall call white, are given, and the sum of their lengths is 1. Several other segments, which we shall call black, are given, and the sum of their lengths is 1. Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the same color are disjoint, and segments of different colors are either disjoint or one is inside the other. Prove that there exists a system that cannot be distributed in that way on the segment that is 1.49 long.
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We shall use the following algorithm:
Choose a segment of maximum length ("basic" segment) and put on it unused segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used.
Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one- half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1.5 and can be distributed on a segment of length 1.51.
On the other hand, if we are given two white segments of lengths 0.5 and two black segments of lengths 0.999 and 0.001, we cannot distribute them on a segment of length less than 1.499.
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IMOLL-1977-52
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Two perpendicular chords are drawn through a given interior point \(P\) of a circle with radius \(R\) . Determine, with proof, the maximum and the minimum of the sum of the lengths of these two chords if the distance from \(P\) to the center of the circle is \(kR\) .
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The maximum and minimum are \(2R\sqrt{4 - 2k^{2}}\) and \(2R\left(1 + \sqrt{1 - k^{2}}\right)\) respectively.
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IMOLL-1977-53
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Find all pairs of integers \(a\) and \(b\) for which
\[7a + 14b = 5a^2 +5ab + 5b^2.\]
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The discriminant of the given equation considered as a quadratic equation in \(b\) is \(196 - 75a^{2}\) . Thus \(75a^{2}\leq 196\) and hence \(-1\leq a\leq 1\) . Now the integer solutions of the given equation are easily found: \((-1,3)\) , \((0,0)\) , \((1,2)\) .
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IMOLL-1977-54
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If \(0 \leq a \leq b \leq c \leq d\) , prove that
\[d^b b^c c^d d^a \geq b^a c^b d^c a^d.\]
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We shall use the following lemma.
Lemma. If a real function \(f\) is convex on the interval \(I\) and \(x,y,z\in I,x\leq y\leq z\) then
\[(y - z)f(x) + (z - x)f(y) + (x - y)f(z)\leq 0.\]
Proof. The inequality is obvious for \(x = y = z\) . If \(x< z\) , then there exist \(p,r\) such that \(p + r = 1\) and \(y = px + rz\) . Then by Jensen's inequality \(f(px + rz)\leq\) \(pf(x) + rf(z)\) , which is equivalent to the statement of the lemma.
By applying the lemma to the convex function \(-\ln x\) we obtain \(x^{y}y^{z}z^{x}\geq y^{x}z^{y}x^{z}\) for any \(0< x\leq y\leq z\) . Multiplying the inequalities \(a^{b}b^{c}c^{a}\geq b^{a}c^{b}a^{c}\) and \(a^{c}c^{d}d^{a}\geq\) \(c^{d}d^{c}a^{d}\) we get the desired inequality.
Remark. Similarly, for \(0< a_{1}\leq a_{2}\leq \dots \leq a_{n}\) it holds that \(a_{1}^{a_{2}}a_{2}^{a_{3}}\dots a_{n}^{a_{1}}\geq\) \(a_{2}^{a_{1}}a_{3}^{a_{2}}\dots a_{1}^{a_{n}}\)
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IMOLL-1977-55
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Through a point \(O\) on the diagonal \(BD\) of a parallelogram \(ABCD\) , segments \(MN\) parallel to \(AB\) , and \(PQ\) parallel to \(AD\) , are drawn, with \(M\) on \(AD\) , and \(Q\) on \(AB\) . Prove that diagonals \(AO\) , \(BP\) , \(DN\) (extended if necessary) will be concurrent.
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The statement is true without the assumption that \(O\in BD\) . Let \(BP\cap DN = \{K\}\) . If we denote \(\overrightarrow{AB} = a,\overrightarrow{AD} = b\) and \(\overrightarrow{AO} = \alpha a + \beta b\) for some \(\alpha ,\beta \in \mathbb{R}\) , \(1 / \alpha +1 / \beta \neq 1\) , by straightforward calculation we obtain that
\[\overrightarrow{AK} = \frac{\alpha}{\alpha + \beta - \alpha\beta} a + \frac{\beta}{\alpha + \beta - \alpha\beta} b = \frac{1}{\alpha + \beta - \alpha\beta}\overrightarrow{AO}.\]
Hence \(A,K,O\) are collinear.
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IMOLL-1977-56
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The four circumcircles of the four faces of a tetrahedron have equal radii. Prove that the four faces of the tetrahedron are congruent triangles.
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Suppose that the skew edges of the tetrahedron \(ABCD\) are equal. Let \(K, L, M, P, Q, R\) be the midpoints of edges \(AB, AC, AD, CD, DB, BC\) respectively. Segments \(KP, LQ, MR\) have the common midpoint \(T\) .
We claim that the lines \(KP\) , \(LQ\) and \(MR\) are axes of symmetry of the tetrahedron \(ABCD\) . From \(LM \parallel CD \parallel RQ\) and similarly \(LR \parallel MQ\) and \(LM = CD / 2 = AB / 2 = LR\) it follows that \(LMQR\) is a rhombus and therefore \(LQ \perp MR\) . We similarly show that \(KP\) is perpendicular to \(LQ\) and \(MR\) , and thus it is perpendicular to the plane \(LMQR\) . Since the lines \(AB\) and \(CD\) are parallel to the plane \(LMQR\) , they are perpendicular to \(KP\) . Hence the points \(A\) and \(C\) are symmetric to \(B\) and \(D\) with respect to the line \(KP\) , which means that \(KP\) is an axis of symmetry of the tetrahedron \(ABCD\) . Similarly, so are the lines \(LQ\) and \(MR\) .
\(LMQR\) . Since the lines \(AB\) and \(CD\) are parallel to the plane \(LMQR\) , they are perpendicular to \(KP\) . Hence the points \(A\) and \(C\) are symmetric to \(B\) and \(D\) with respect to the line \(KP\) , which means that \(KP\) is an axis of symmetry of the tetrahedron \(ABCD\) . Similarly, so are the lines \(LQ\) and \(MR\) .
The centers of circumscribed and inscribed spheres of tetrahedron \(ABCD\) must lie on every axis of symmetry of the tetrahedron, and hence both must coincide with \(T\) .
Conversely, suppose that the centers of circumscribed and inscribed spheres of the tetrahedron \(ABCD\) coincide with some point \(T\) . Then the orthogonal projections of \(T\) onto the faces \(ABC\) and \(ABD\) are the circumcenters \(O_1\) and \(O_2\) of these two triangles, and moreover, \(TO_1 = TO_2\) . Pythagoras's theorem gives \(AO_1 = AO_2\) , which by the law of sines implies \(\angle ACB = \angle ADB\) . Now it easily follows that the sum of the angles at one vertex of the tetrahedron is equal to \(180^\circ\) . Let \(D'\) , \(D''\) , and \(D'''\) be the points in the plane \(ABC\) lying outside \(\triangle ABC\) such that \(\triangle D'BC \cong \triangle DBC\) , \(\triangle D'CA \cong \triangle DCA\) , and \(\triangle D''AB \cong \triangle DAB\) . The angle \(D''AD'''\) is then straight, and hence \(A, B, C\) are midpoints of the segments \(D''D''', D''D', D'D''\) respectively. Hence \(AD = D''D''/2 = BC\) , and analogously \(AB = CD\) and \(AC = BD\) .
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IMOLL-1977-57
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The length of a finite sequence is defined as the number of terms of this sequence. Determine the maximal possible length of a finite sequence that satisfies the following condition: The sum of each seven successive terms is negative, and the sum of each eleven successive terms is positive.
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Suppose that there exists a sequence of 17 terms \(a_{1},a_{2},\ldots ,a_{17}\) satisfying the required conditions. Then the sum of terms in each row of the rectangular array below is positive, while the sum of terms in each column is negative, which is a contradiction.
\[a_{1} a_{2} \dots a_{11}\] \[a_{2} a_{3} \dots a_{12}\] \[\vdots \vdots \vdots\] \[a_{7} a_{8} \dots a_{17}\]
On the other hand, there exist 16- term sequences with the required property. An example is \(5,5, - 13,5,5,5, - 13,5,5, - 13,5,5,5, - 13,5\) which can be obtained by solving the system of equations \(\sum_{i = k}^{k + 10}a_{i} = 1\) \((k = 1,2,\ldots ,6)\) and \(\sum_{i = l}^{l + 6}a_{i} = - 1\) \((l = 1,2,\ldots ,10)\) .
Second solution. We shall prove a stronger statement: If 7 and 11 in the question are replaced by any positive integers \(m,n\) , then the maximum number of terms is \(m + n - (m,n) - 1\) .
Let \(a_{1},a_{2},\ldots ,a_{l}\) be a sequence of real numbers, and let us define \(s_{0} = 0\) and \(s_{k} = a_{1} + \dots +a_{k}\) \((k = 1,\ldots ,l)\) . The given conditions are equivalent to \(s_{k} > s_{k + m}\) for \(0\leq k\leq l - m\) and \(s_{k}< s_{k + n}\) for \(0\leq k\leq l - n\) .
Let \(d = (m,n)\) and \(m = m^{\prime}d\) , \(n = n^{\prime}d\) . Suppose that there exists a sequence \((a_{k})\) of length greater than or equal to \(l = m + n - d\) satisfying the required conditions. Then the \(m^{\prime} + n^{\prime}\) numbers \(s_{0},s_{d},\ldots ,s_{(m^{\prime} + n^{\prime} - 1)d}\) satisfy \(n^{\prime}\) inequalities \(s_{k + m}< s_{k}\) and \(m^{\prime}\) inequalities \(s_{k}< s_{k + n}\) . Moreover, each term \(s_{kd}\) appears twice in these inequalities: once on the left- hand and once on the right- hand side. It follows that there exists a ring of inequalities \(s_{i_{1}}< s_{i_{2}}< \dots < s_{i_{k}}< s_{i_{1}}\) , giving a contradiction.
On the other hand, suppose that such a ring of inequalities can be made also for \(l = m + n - d - 1\) , say \(s_{i_{1}}< s_{i_{2}}< \dots < s_{i_{k}}< s_{i_{1}}\) . If there are \(p\) inequalities of
the form \(a_{k + m}< a_{k}\) and \(q\) inequalities of the form \(a_{k + n} > a_{k}\) in the ring, then \(q n = r m\) , which implies \(m^{\prime}\mid q,n^{\prime}\mid p\) and thus \(k = p + q\geq m^{\prime} + n^{\prime}\) . But since all \(i_{1},i_{2},\ldots ,i_{k}\) are congruent modulo \(d\) , we have \(k\leq m^{\prime} + n^{\prime} - 1\) , a contradiction. Hence there exists a sequence of length \(m + n - d - 1\) with the required property.
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IMOLL-1977-58
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Prove that for every triangle the following inequality holds:
\[\frac{ab + bc + ca}{4S} \geq \cot \frac{\pi}{6},\]
where \(a, b, c\) are lengths of the sides and \(S\) is the area of the triangle.
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The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove:
\[2a b + 2b c + 2c a - a^{2} - b^{2} - c^{2}\geq 4S\sqrt{3}. \quad (1)\]
First proof. Let us set \(2x = b + c - a\) , \(2y = c + a - b\) , \(2z = a + b - c\) . Then \(x,y,z > 0\) and the inequality (1) becomes
\[y^{2}z^{2} + z^{2}x^{2} + x^{2}y^{2}\geq xy z(x + y + z),\]
which is equivalent to the obvious inequality \((xy - yz)^{2} + (yz - zx)^{2} + (zx - xy)^{2}\geq 0\)
Second proof. Using the known relations for a triangle
\[a^{2} + b^{2} + c^{2} = 2s^{2} - 2r^{2} - 8rR,\] \[ab + bc + ca = s^{2} + r^{2} + 4rR,\] \[S = rs,\]
where \(r\) and \(R\) are the radii of the incircle and the circumcircle, \(s\) the semiperimeter and \(S\) the area, we can transform (1) into
\[s\sqrt{3}\leq 4R + r.\]
The last inequality is a consequence of the inequalities \(2r\leq R\) and \(s^{2}\leq 4R^{2}+\) \(4R r + 3r^{2}\) , where the last one follows from the equality \(H I^{2} = 4R^{2} + 4R r + 3r^{2}-\) \(s^{2}\) ( \(H\) and \(I\) being the orthocenter and the incenter of the triangle).
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IMOLL-1977-59
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Let \(E\) be a set of \(n\) points in the plane \((n \geq 3)\) whose coordinates are integers such that any three points from \(E\) are vertices of a nondegenerate triangle whose centroid doesn't have both coordinates integers. Determine the maximal \(n\) .
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Let us consider the set \(R\) of pairs of coordinates of the points from \(E\) reduced modulo 3. If some element of \(R\) occurs thrice, then the corresponding points are vertices of a triangle with integer barycenter. Also, no three elements from \(E\) can have distinct \(x\) -coordinates and distinct \(y\) -coordinates. By an easy discussion we can conclude that the set \(R\) contains at most four elements. Hence \(|E|\leq 8\) . An example of a set \(E\) consisting of 8 points that satisfies the required condition is
\[E = \{(0,0),(1,0),(0,1),(1,1),(3,6),(4,6),(3,7),(4,7)\} .\]
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IMOLL-1977-60
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Suppose \(x_0, x_1, \ldots , x_n\) are integers and \(x_0 > x_1 > \dots > x_n\) . Prove that at least one of the numbers \(|F(x_0)|, |F(x_1)|, |F(x_2)|, \ldots , |F(x_n)|\) , where
\[F(x) = x^n +a_1x^{n - 1} + \dots +a_n,\qquad a_i\in \mathbb{R}, i = 1,\ldots ,n,\]
is greater than \(\frac{n!}{2^n}\) .
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By Lagrange's interpolation formula we have
\[F(x) = \sum_{j = 0}^{n}F(x_{j})\frac{\prod_{i\neq j}(x - x_{j})}{\prod_{i\neq j}(x_{i} - x_{j})}.\]
Since the leading coefficient in \(F(x)\) is 1, it follows that
\[1 = \sum_{j = 0}^{n}\frac{F(x_{j})}{\prod_{i\neq j}(x_{i} - x_{j})}.\]
Since
\[\left|\prod_{i\neq j}(x_{i} - x_{j})\right| = \prod_{i = 0}^{j - 1}|x_{i} - x_{j}|\prod_{i = j + 1}^{n}|x_{i} - x_{j}|\geq j!(n - j)!,\]
we have
\[1\leq \sum_{j = 0}^{n}\frac{|F(x_{j})|}{|\prod_{i\neq j}(x_{i} - x_{j})|}\leq \frac{1}{n!}\sum_{j = 0}^{n}\binom{n}{j}|F(x_{j})|\leq \frac{2^{n}}{n!}\max |F(x_{j})|.\]
Now the required inequality follows immediately.
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IMOSL-1978-1
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The set \(M = \{1, 2, \ldots , 2n\}\) is partitioned into \(k\) nonintersecting subsets \(M_1, M_2, \ldots , M_k\) , where \(n \geq k^3 + k\) . Prove that there exist even numbers \(2j_1, 2j_2, \ldots , 2j_{k+1}\) in \(M\) that are in one and the same subset \(M_i\) ( \(1 \leq i \leq k\) ) such that the numbers \(2j_1 - 1, 2j_2 - 1, \ldots , 2j_{k+1} - 1\) are also in one and the same subset \(M_j\) ( \(1 \leq j \leq k\) ).
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There exists an \(M_{s}\) that contains at least \(2n / k = 2(k^{2} + 1)\) elements. It follows that \(M_{s}\) contains either at least \(k^{2} + 1\) even numbers or at least \(k^{2} + 1\) odd numbers. In the former case, consider the predecessors of those \(k^{2} + 1\) numbers: among them, at least \(\frac{k^{2} + 1}{k + 1} >k\) , i.e., at least \(k + 1\) , belong to the same subset, say \(M_{t}\) . Then we choose \(s,t\) . The latter case is similar.
Second solution. For all \(i,j\in \{1,2,\ldots ,k\}\) , consider the set \(N_{ij} = \{r\mid 2r\in\) \(M_{i}\) \(2r - 1\in M_{j}\}\) . Then \(\{N_{ij}\mid i,j\}\) is a partition of \(\{1,2,\ldots ,n\}\) into \(k^{2}\) subsets. For \(n\geq k^{3} + 1\) one of these subsets contains at least \(k + 1\) elements, and the statement follows.
Remark. The statement is not necessarily true when \(n = k^{3}\)
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IMOSL-1978-2
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Two identically oriented equilateral triangles, \(ABC\) with center \(S\) and \(A'B'C\) , are given in the plane. We also have \(A' \neq S\) and \(B' \neq S\) . If \(M\) is the midpoint of \(A'B\) and \(N\) the midpoint of \(AB'\) , prove that the triangles \(SB'M\) and \(SA'N\) are similar.
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Consider the transformation \(\phi\) of the plane defined as the homothety \(\mathcal{H}\) with center \(B\) and coefficient 2 followed by the rotation \(\mathcal{R}\) about the center \(O\) through
an angle of \(60^{\circ}\) . Being direct, this mapping must be a rotational homothety. We also see that \(\mathcal{H}\) maps \(S\) into the point symmetric to \(S\) with respect to \(OA\) , and \(\mathcal{R}\) takes it back to \(S\) . Hence \(S\) is a fixed point, and is consequently also the center of \(\phi\) . Therefore \(\phi\) is the rotational homothety about \(S\) with the angle \(60^{\circ}\)
and coefficient 2. (In fact, this could also be seen from the fact that \(\phi\) preserves angles of triangles and maps the segment \(SR\) onto \(SB\) , where \(R\) is the midpoint of \(AB\) .)
Since \(\phi (M) = B'\) , we conclude that \(\angle MSB' = 60^{\circ}\) and \(SB'/SM = 2\) . Similarly, \(\angle NSA' = 60^{\circ}\) and \(SA'/SN = 2\) , so triangles \(MSB'\) and \(NSA'\) are indeed similar.
Second solution. Probably the simplest way here is using complex numbers. Put the origin at \(O\) and complex numbers \(a,a'\) at points \(A,A'\) , and denote the primitive sixth root of 1 by \(\omega\) . Then the numbers at \(B\) , \(B'\) , \(S\) and \(N\) are \(\omega a\) , \(\omega a'\) , \((a + \omega a) / 3\) , and \((a + \omega a') / 2\) respectively. Now it is easy to verify that \((n - s) = \omega (a' - s) / 2\) , i.e., that \(\angle NSA' = 60^{\circ}\) and \(SA'/SN = 2\) .
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