8. (ROM 2) $)^{\mathrm{IMO} 3}$ In a plane two different points $O$ and $A$ are given. For each point $X \neq O$ of the plane denote by $\alpha(X)$ the angle $A O X$ measured in radians $(0 \leq \alpha(X)<2 \pi)$ and by $C(X)$ the circle with center $O$ and radius $O X+\frac{\alpha(X)}{O X}$. Suppose each point of the plane is colored by one of a finite number of colors. Show that there exists a point $X$ with $\alpha(X)>0$ such that its color appears somewhere on the circle $C(X)$.

Solution: 8. Suppose that the statement of the problem is false. Consider two arbitrary circles $R=(O, r)$ and $S=(O, s)$ with $0<r<s<1$. The point $X \in R$ with $\alpha(X)=r(s-r)<2 \pi$ satisfies that $C(X)=S$. It follows that the color of the point $X$ does not appear on $S$. Consequently, the set of colors that appear on $R$ is not the same as the set of colors that appear on $S$. Hence any two distinct circles with center at $O$ and radii less than 1 have distinct sets of colors. This is a contradiction, since there are infinitely many such circles but only finitely many possible sets of colors.


9. (POL 2) Let $a, b, c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{\sqrt{3}}{2}$. Prove that the system of equations  $$ \begin{aligned} & \sqrt{y-a}+\sqrt{z-a}=1, \\ & \sqrt{z-b}+\sqrt{x-b}=1 \\ & \sqrt{x-c}+\sqrt{y-c}=1 \end{aligned} $$  has exactly one solution $(x, y, z)$ in real numbers.

Solution: 9. Let us show first that the system has at most one solution. Suppose that $(x, y, z)$ and $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ are two distinct solutions and that w.l.o.g. $x<x^{\prime}$. Then the second and third equation imply that $y>y^{\prime}$ and $z>z^{\prime}$, but then $\sqrt{y-a}+\sqrt{z-a}>\sqrt{y^{\prime}-a}+\sqrt{z^{\prime}-a}$, which is a contradiction. We shall now prove the existence of at least one solution. Let $P$ be an arbitrary point in the plane and $K, L, M$ points such that $P K=\sqrt{a}$, $P L=\sqrt{b}, P M=\sqrt{c}$, and $\angle K P L=\angle L P M=\angle M P K=120^{\circ}$. The lines through $K, L, M$ perpendicular respectively to $P K, P L, P M$ form an equilateral triangle $A B C$, where $K \in B C, L \in A C$, and $M \in A B$. Since its area equals $A B^{2} \sqrt{3} / 4=S_{\triangle B P C}+S_{\triangle A P C}+$ $S_{\triangle A P B}=A B(\sqrt{a}+\sqrt{b}+\sqrt{c}) / 2$, it follows that $A B=1$. Therefore $x=P A^{2}, y=P B^{2}$, and $z=P C^{2}$ is a solution of the system (indeed, $\sqrt{y-a}+\sqrt{z-a}=\sqrt{P B^{2}-P K^{2}}+\sqrt{P C^{2}-P K^{2}}=B K+C K=1$, etc.).


1. (MON 1) ${ }^{\mathrm{IMO} 4}$ Given a set $M$ of 1985 positive integers, none of which has a prime divisor larger than 26 , prove that the set has four distinct elements whose geometric mean is an integer.

Solution: 1. Since there are 9 primes ( $p_{1}=2<p_{2}=3<\cdots<p_{9}=23$ ) less than 26 , each number $x_{j} \in M$ is of the form $\prod_{i=1}^{9} p_{i}^{a_{i j}}$, where $0 \leq a_{i j}$. Now, $x_{j} x_{k}$ is a square if $a_{i j}+a_{i k} \equiv 0(\bmod 2)$ for $i=1, \ldots, 9$. Since the number of distinct ninetuples modulo 2 is $2^{9}$, any subset of $M$ with at least 513 elements contains two elements with square product. Starting from $M$ and eliminating such pairs, one obtains $(1985-513) / 2=736>513$ distinct two-element subsets of $M$ each having a square as the product of elements. Reasoning as above, we find at least one (in fact many) pair of such squares whose product is a fourth power.


10. 2b.(VIE 1) Prove that for every point $M$ on the surface of a regular tetrahedron there exists a point $M^{\prime}$ such that there are at least three different curves on the surface joining $M$ to $M^{\prime}$ with the smallest possible length among all curves on the surface joining $M$ to $M^{\prime}$.

Solution: 10. If $M$ is at a vertex of the regular tetrahedron $A B C D(A B=1)$, then one can take $M^{\prime}$ at the center of the opposite face of the tetrahedron. Let $M$ be on the face $(A B C)$ of the tetrahedron, excluding the vertices. Consider a continuous mapping $f$ of $\mathbb{C}$ onto the surface $S$ of $A B C D$ that maps $m+n e^{2 \pi / 3}$ for $m, n \in$ $\mathbb{Z}$ onto $A, B, C, D$ if $(m, n) \equiv$ $(1,1),(1,0),(0,1),(0,0)(\bmod 2)$ re- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-484.jpg?height=464&width=446&top_left_y=1356&top_left_x=857) spectively, and maps each unit equilateral triangle with vertices of the form $m+n e^{2 \pi / 3}$ isometrically onto the corresponding face of $A B C D$. The point $M$ then has one preimage $M_{j}, j=1,2, \ldots, 6$, in each of the six preimages of $\triangle A B C$ having two vertices on the unit circle. The $M_{j}$ 's form a convex centrally symmetric (possibly degenerate) hexagon. Of the triangles formed by two adjacent sides of this hexagon consider the one, say $M_{1} M_{2} M_{3}$, with the smallest radius of circumcircle and denote by $\widehat{M^{\prime}}$ its circumcenter. Then we can choose $M^{\prime}=f\left(\widehat{M^{\prime}}\right)$. Indeed, the images of the segments $M_{1} \widehat{M^{\prime}}, M_{2} \widehat{M^{\prime}}, M_{3} \widehat{M^{\prime}}$ are three different shortest paths on $S$ from $M$ to $M^{\prime}$.


11. 3a.(USS 3) Find a method by which one can compute the coefficients of $P(x)=x^{6}+a_{1} x^{5}+\cdots+a_{6}$ from the roots of $P(x)=0$ by performing not more than 15 additions and 15 multiplications.

Solution: 11. Let $-x_{1}, \ldots,-x_{6}$ be the roots of the polynomial. Let $s_{k, i}(k \leq i \leq 6)$ denote the sum of all products of $k$ of the numbers $x_{1}, \ldots, x_{i}$. By Vieta's formula we have $a_{k}=s_{k, 6}$ for $k=1, \ldots, 6$. Since $s_{k, i}=s_{k-1, i-1} x_{i}+$ $s_{k, i-1}$, one can compute the $a_{k}$ by the following scheme (the horizontal and vertical arrows denote multiplications and additions respectively): ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-485.jpg?height=457&width=609&top_left_y=646&top_left_x=486)


12. 3b.(GBR 4) A sequence of polynomials $P_{m}(x, y, z), m=0,1,2, \ldots$, in $x, y$, and $z$ is defined by $P_{0}(x, y, z)=1$ and by  $$ P_{m}(x, y, z)=(x+z)(y+z) P_{m-1}(x, y, z+1)-z^{2} P_{m-1}(x, y, z) $$  for $m>0$. Prove that each $P_{m}(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z$.

Solution: 12. We shall prove by induction on $m$ that $P_{m}(x, y, z)$ is symmetric and that  $$ (x+y) P_{m}(x, z, y+1)-(x+z) P_{m}(x, y, z+1)=(y-z) P_{m}(x, y, z) $$  holds for all $x, y, z$. This is trivial for $m=0$. Assume now that it holds for $m=n-1$. Since obviously $P_{n}(x, y, z)=P_{n}(y, x, z)$, the symmetry of $P_{n}$ will follow if we prove that $P_{n}(x, y, z)=P_{n}(x, z, y)$. Using (1) we have $P_{n}(x, z, y)-$ $P_{n}(x, y, z)=(y+z)\left[(x+y) P_{n-1}(x, z, y+1)-(x+z) P_{n-1}(x, y, z+1)\right]-\left(y^{2}-\right.$ $\left.z^{2}\right) P_{n-1}(x, y, z)=(y+z)(y-z) P_{n-1}(x, y, z)-\left(y^{2}-z^{2}\right) P_{n-1}(x, y, z)=0$. It remains to prove (1) for $m=n$. Using the already established symmetry we have  $$ \begin{aligned} & (x+y) P_{n}(x, z, y+1)-(x+z) P_{n}(x, y, z+1) \\ & =(x+y) P_{n}(y+1, z, x)-(x+z) P_{n}(z+1, y, x) \\ & =(x+y)\left[(y+x+1)(z+x) P_{n-1}(y+1, z, x+1)-x^{2} P_{n-1}(y+1, z, x)\right] \\ & \quad-(x+z)\left[(z+x+1)(y+x) P_{n-1}(z+1, y, x+1)-x^{2} P_{n-1}(z+1, y, x)\right] \\ & =(x+y)(x+z)(y-z) P_{n-1}(x+1, y, z)-x^{2}(y-z) P_{n-1}(x, y, z) \\ & =(y-z) P_{n}(z, y, x)=(y-z) P_{n}(x, y, z), \end{aligned} $$  as claimed.


13. 4a.(BUL 1) Let $m$ boxes be given, with some balls in each box. Let $n<m$ be a given integer. The following operation is performed: choose $n$ of the boxes and put 1 ball in each of them. Prove: (a) If $m$ and $n$ are relatively prime, then it is possible, by performing the operation a finite number of times, to arrive at the situation that all the boxes contain an equal number of balls. (b) If $m$ and $n$ are not relatively prime, there exist initial distributions of balls in the boxes such that an equal distribution is not possible to achieve.

Solution: 13. If $m$ and $n$ are relatively prime, there exist positive integers $p, q$ such that $p m=q n+1$. Thus by putting $m$ balls in some boxes $p$ times we can achieve that one box receives $q+1$ balls while all others receive $q$ balls. Repeating this process sufficiently many times, we can obtain an equal distribution of the balls. Now assume $\operatorname{gcd}(m, n)>1$. If initially there is only one ball in the boxes, then after $k$ operations the number of balls will be $1+k m$, which is never divisible by $n$. Hence the task cannot be done.


14. 4b.(IRE 4) A set of 1985 points is distributed around the circumference of a circle and each of the points is marked with 1 or -1 . A point is called "good" if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with -1 is less than 662 , there must be at least one good point.

Solution: 14. It suffices to prove the existence of a good point in the case of exactly 661 -1 's. We prove by induction on $k$ that in any arrangement with $3 k+2$ points $k$ of which are -1 's a good point exists. For $k=1$ this is clear by inspection. Assume that the assertion holds for all arrangements of $3 n+2$ points and consider an arrangement of $3(n+1)+2$ points. Now there exists a sequence of consecutive -1 's surrounded by two +1 's. There is a point $P$ which is good for the arrangement obtained by removing the two +1 's bordering the sequence of -1 's and one of these -1 's. Since $P$ is out of this sequence, clearly the removal either leaves a partial sum as it was or diminishes it by 1 , so $P$ is good for the original arrangement. Second solution. Denote the number on an arbitrary point by $a_{1}$, and the numbers on successive points going in the positive direction by $a_{2}, a_{3}, \ldots$ (in particular, $a_{k+1985}=a_{k}$ ). We define the partial sums $s_{0}=0, s_{n}=$ $a_{1}+a_{2}+\cdots+a_{n}$ for all positive integers $n$; then $s_{k+1985}=s_{k}+s_{1985}$ and $s_{1985} \geq 663$. Since $s_{1985 m} \geq 663 m$ and $3 \cdot 663 m>1985(m+2)+1$ for large $m$, not all values $0,1,2, \ldots 663 m$ can appear thrice among the $1985(m+2)+1$ sums $s_{-1985}, s_{-1984}, \ldots, s_{1985(m+1)}$ (and none of them appears out of this set). Thus there is an integral value $s>0$ that appears at most twice as a partial sum, say $s_{k}=s_{l}=s, k<l$. Then either $a_{k}$ or $a_{l}$ is a good point. Actually, $s_{i}>s$ must hold for all $i>l$, and $s_{i}<s$ for all $i<k$ (otherwise, the sum $s$ would appear more than twice). Also, for the same reason there cannot exist indices $p, q$ between $k$ and $l$ such that $s_{p}>s$ and $s_{q}<s$; i.e., for $k<p<l, s_{p}$ 's are either all greater than or equal to $s$, or smaller than or equal to $s$. In the former case $a_{k}$ is good, while in the latter $a_{l}$ is good.


15. 5a.(FRA 3) Let $K$ and $K^{\prime}$ be two squares in the same plane, their sides of equal length. Is it possible to decompose $K$ into a finite number of triangles $T_{1}, T_{2}, \ldots, T_{p}$ with mutually disjoint interiors and find translations $t_{1}, t_{2}, \ldots, t_{p}$ such that  $$ K^{\prime}=\bigcup_{i=1}^{p} t_{i}\left(T_{i}\right) ? $$

Solution: 15. There is no loss of generality if we assume $K=A B C D, K^{\prime}=$ $A B^{\prime} C^{\prime} D^{\prime}$, and that $K^{\prime}$ is obtained from $K$ bya clockwise rotation around $A$ by $\phi, 0 \leq \phi \leq \pi / 4$. Let $C^{\prime} D^{\prime}, B^{\prime} C^{\prime}$, and the parallel to $A B$ through $D^{\prime}$ meet the line $B C$ at $E$, $F$, and $G$ respectively. Let us now choose points $E^{\prime} \in A B^{\prime}, G^{\prime} \in A B$, $C^{\prime \prime} \in A D^{\prime}$, and $E^{\prime \prime} \in A D$ such that ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-486.jpg?height=421&width=516&top_left_y=1572&top_left_x=831) the triangles $A E^{\prime} G^{\prime}$ and $A C^{\prime \prime} E^{\prime \prime}$ are translates of the triangles $D^{\prime} E G$ and $F C^{\prime} E$ respectively. Since $A E^{\prime}=D^{\prime} E$ and $A C^{\prime \prime}=F C^{\prime}$, we have $C^{\prime \prime} E^{\prime \prime}=C^{\prime} E=B^{\prime} E^{\prime}$ and $C^{\prime \prime} D^{\prime}=B^{\prime} F$, which imply that $\triangle E^{\prime \prime} C^{\prime \prime} D^{\prime}$ is a translate of $\triangle E^{\prime} B^{\prime} F$, and consequently $E^{\prime \prime} D^{\prime}=E^{\prime} F$ and $E^{\prime \prime} D^{\prime} \| E^{\prime} F$. It follows that there exist points $H \in C D, H^{\prime} \in B F$, and $D^{\prime \prime} \in E^{\prime} G^{\prime}$ such that $E^{\prime \prime} D^{\prime} H D$ is a translate of $E^{\prime} F H^{\prime} D^{\prime \prime}$. The remaining parts of $K$ and $K^{\prime}$ are the rectangles $D^{\prime} G C H$ and $D^{\prime \prime} H^{\prime} B G^{\prime}$ of equal area. We shall now show that two rectangles with parallel sides and equal areas can be decomposed into translation invariant parts. Let the sides of the rectangles $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}\left(X Y \| X^{\prime} Y^{\prime}\right)$ satisfy $X^{\prime} Y^{\prime}<X Y$, $Y^{\prime} Z^{\prime}>Y Z$, and $X^{\prime} Y^{\prime} \cdot Y^{\prime} Z^{\prime}=X Y \cdot Y Z$. Suppose that $2 X^{\prime} Y^{\prime}>X Y$ (otherwise, we may cut off congruent rectangles from both the original ones until we reduce them to the case of $2 X^{\prime} Y^{\prime}>X Y$ ). Let $U \in X Y$ and $V \in Z T$ be points such that $Y U=T V=X^{\prime} Y^{\prime}$ and $W \in X V$ be a point such that $U W \| X T$. Then translating $\triangle X U W$ to a triangle $V Z R$ and $\triangle X V T$ to a triangle $W R S$ results in a rectangle $U Y R S$ congruent to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$. Thus we have partitioned $K$ and $K^{\prime}$ into translation-invariant parts. Although not all the parts are triangles, we may simply triangulate them.


16. 5b.(BEL 2) If possible, construct an equilateral triangle whose three vertices are on three given circles.

Solution: 16. Let the three circles be $\alpha(A, a), \beta(B, b)$, and $\gamma(C, c)$, and assume $c \leq a, b$. We denote by $\mathcal{R}_{X, \varphi}$ the rotation around $X$ through an angle $\varphi$. Let $P Q R$ be an equilateral triangle, say of positive orientation (the case of negatively oriented $\triangle P Q R$ is analogous), with $P \in \alpha, Q \in \beta$, and $R \in \gamma$. Then $Q=\mathcal{R}_{P,-60^{\circ}}(R) \in \mathcal{R}_{P,-60^{\circ}}(\gamma) \cap \beta$. Since the center of $\mathcal{R}_{P,-60^{\circ}}(\gamma)$ is $\mathcal{R}_{P,-60^{\circ}}(C)=\mathcal{R}_{C, 60^{\circ}}(P)$ and it lies on $\mathcal{R}_{C, 60^{\circ}}(\alpha)$, the union of circles $\mathcal{R}_{P,-60^{\circ}}(\gamma)$ as $P$ varies on $\alpha$ is the annulus $\mathcal{U}$ with center $A^{\prime}=\mathcal{R}_{C, 60^{\circ}}(A)$ and radii $a-c$ and $a+c$. Hence there is a solution if and only if $\mathcal{U} \cap \beta$ is nonempty.


17. 6a. (SWE 3) ${ }^{\mathrm{IMO} O}$ The sequence $f_{1}, f_{2}, \ldots, f_{n}, \ldots$ of functions is defined for $x>0$ recursively by  $$ f_{1}(x)=x, \quad f_{n+1}(x)=f_{n}(x)\left(f_{n}(x)+\frac{1}{n}\right) . $$  Prove that there exists one and only one positive number $a$ such that $0<f_{n}(a)<f_{n+1}(a)<1$ for all integers $n \geq 1$.

Solution: 17. The statement of the problem is equivalent to the statement that there is one and only one $a$ such that $1-1 / n<f_{n}(a)<1$ for all $n$. We note that each $f_{n}$ is a polynomial with positive coefficients, and therefore increasing and convex in $\mathbb{R}^{+}$. Define $x_{n}$ and $y_{n}$ by $f_{n}\left(x_{n}\right)=1-1 / n$ and $f_{n}\left(y_{n}\right)=1$. Since  $$ f_{n+1}\left(x_{n}\right)=\left(1-\frac{1}{n}\right)^{2}+\left(1-\frac{1}{n}\right) \frac{1}{n}=1-\frac{1}{n} $$  and $f_{n+1}\left(y_{n}\right)=1+1 / n$, it follows that $x_{n}<x_{n+1}<y_{n+1}<y_{n}$. Moreover, the convexity of $f_{n}$ together with the fact that $f_{n}(x)>x$ for all $x>0$ implies that $y_{n}-x_{n}<f_{n}\left(y_{n}\right)-f_{n}\left(x_{n}\right)=1 / n$. Therefore the sequences have a common limit $a$, which is the only number lying between $x_{n}$ and $y_{n}$ for all $n$. By the definition of $x_{n}$ and $y_{n}$, the statement immediately follows.


Solution: 18. Set $y_{i}=\frac{x_{i}^{2}}{x_{i+1} x_{i+2}}$, where $x_{n+i}=x_{i}$. Then $\prod_{i=1}^{n} y_{i}=1$ and the inequality to be proved becomes $\sum_{i=1}^{n} \frac{y_{i}}{1+y_{i}} \leq n-1$, or equivalently  $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq 1 $$  We prove this inequality by induction on $n$. Since $\frac{1}{1+y}+\frac{1}{1+y^{-1}}=1$, the inequality is true for $n=2$. Assume that it is true for $n-1$, and let there be given $y_{1}, \ldots, y_{n}>0$ with $\prod_{i=1}^{n} y_{i}=1$. Then $\frac{1}{1+y_{n-1}}+\frac{1}{1+y_{n}}>\frac{1}{1+y_{n-1} y_{n}}$, which is equivalent to $1+y_{n} y_{n-1}(1+$ $\left.y_{n}+y_{n-1}\right)>0$. Hence by the inductive hypothesis  $$ \sum_{i=1}^{n} \frac{1}{1+y_{i}} \geq \sum_{i=1}^{n-2} \frac{1}{1+y_{i}}+\frac{1}{1+y_{n-1} y_{n}} \geq 1 $$  Remark. The constant $n-1$ is best possible (take for example $x_{i}=a^{i}$ with $a$ arbitrarily large).


19. (ISR 3) For which integers $n \geq 3$ does there exist a regular $n$-gon in the plane such that all its vertices have integer coordinates in a rectangular coordinate system?

Solution: 19. Suppose that for some $n>6$ there is a regular $n$-gon with vertices having integer coordinates, and that $A_{1} A_{2} \ldots A_{n}$ is the smallest such $n$-gon, of side length $a$. If $O$ is the origin and $B_{i}$ the point such that $\overrightarrow{O B_{i}}=\overrightarrow{A_{i-1} A_{i}}$, $i=1,2, \ldots, n$ (where $A_{0}=A_{n}$ ), then $B_{i}$ has integer coordinates and $B_{1} B_{2} \ldots B_{n}$ is a regular polygon of side length $2 a \sin (\pi / n)<a$, which is impossible. It remains to analyze the cases $n \leq 6$. If $\mathcal{P}$ is a regular $n$-gon with $n=$ $3,5,6$, then its center $C$ has rational coordinates. We may suppose that $C$ also has integer coordinates and then rotate $\mathcal{P}$ around $C$ thrice through $90^{\circ}$, thus obtaining a regular 12 -gon or 20 -gon, which is impossible. Hence we must have $n=4$ which is indeed a solution.


2. (BRA 3) A polyhedron has 12 faces and is such that: (i) all faces are isosceles triangles, (ii) all edges have length either $x$ or $y$, (iii) at each vertex either 3 or 6 edges meet, and (iv) all dihedral angles are equal.  Find the ratio $x / y$.

Solution: 2. The polyhedron has $3 \cdot 12 / 2=18$ edges, and by Euler's formula, 8 vertices. Let $v_{1}$ and $v_{2}$ be the numbers of vertices at which respectively 3 and 6 edges meet. Then $v_{1}+v_{2}=8$ and $3 v_{1}+6 v_{2}=2 \cdot 18$, implying that $v_{1}=4$. Let $A, B, C, D$ be the vertices at which three edges meet. Since the dihedral angles are equal, all the edges meeting at $A$, say $A E, A F, A G$, must have equal length, say $x$. (If $x=A E=A F \neq A G=y$, and $A E F$, $A F G$, and $A G E$ are isosceles, $\angle E A F \neq \angle F A G$, in contradiction to the equality of the dihedral angles.) It is easy to see that at $E, F$, and $G$ six edges meet. One proceeds to conclude that if $H$ is the fourth vertex of this kind, $E F G H$ must be a regular tetrahedron of edge length $y$, and the other vertices $A, B, C$, and $D$ are tops of isosceles pyramids based on $E F G, E F H, F G H$, and $G E H$. Let the plane through $A, B, C$ meet $E F$, $H F$, and $G F$, at $E^{\prime}, H^{\prime}$, and $G^{\prime}$. Then $A E^{\prime} B H^{\prime} C G^{\prime}$ is a regular hexagon, and since $x=F A=F E^{\prime}$, we have $E^{\prime} G^{\prime}=x$ and $A E^{\prime}=x / \sqrt{3}$. From the isosceles triangles $A E F$ and $F A E^{\prime}$ we obtain finally, with $\measuredangle E F A=\alpha$, $\frac{y}{2 x}=\cos \alpha=1-2 \sin ^{2}(\alpha / 2), x /(2 x \sqrt{3})=\sin (\alpha / 2)$, and $y / x=5 / 3$.


20. $(\mathbf{G B R} 5)^{\mathrm{IMO}} \mathrm{A}$ circle whose center is on the side $E D$ of the cyclic quadrilateral $B C D E$ touches the other three sides. Prove that $E B+C D=$ $E D$.

Solution: 20. Let $O$ be the center of the circle touching the three sides of $B C D E$ and let $F \in(E D)$ be the point such that $E F=E B$. Then $\angle E F B=90^{\circ}-$ $\angle E / 2=\angle C / 2=\angle O C B$, which implies that $B, C, F, O$ lie on a circle. It follows that $\angle D F C=\angle O B C=\angle B / 2=90^{\circ}-\angle D / 2$ and consequently $\angle D C F=\angle D F C$. Hence $E D=E F+F D=E B+C D$. Second solution. Let $r$ be the radius of the small circle and let $M, N$ be the points of tangency of the circle with $B E$ and $C D$ respectively. Then $E M=r \cot E, D N=r \cot D, M B=r \cot (\angle B / 2)=r \tan (\angle D / 2)$, $N C=r \tan (\angle E / 2)$, and $E D=E O+O D=r / \sin D+r / \sin E$. The statement follows from the identity $\cot x+\tan (x / 2)=1 / \sin x$.


21. (IRE 1) The tangents at $B$ and $C$ to the circumcircle of the acute-angled triangle $A B C$ meet at $X$. Let $M$ be the midpoint of $B C$. Prove that (a) $\angle B A M=\angle C A X$, and (b) $\frac{A M}{A X}=\cos \angle B A C$.

Solution: 21. Let $B_{1}$ and $C_{1}$ be the points on the rays $A C$ and $A B$ respectively such that $X B_{1}=X C=X B=X C_{1}$. Then $\angle X B_{1} C=\angle X C B_{1}=\angle A B C$ and $\angle X C_{1} B=\angle X B C_{1}=\angle A C B$, which imply that $B_{1}, X, C_{1}$ are collinear and $\triangle A B_{1} C_{1} \sim \triangle A B C$. Moreover, $X$ is the midpoint of $B_{1} C_{1}$ because $X B_{1}=X C=X B=X C_{1}$, from which we conclude that $\triangle A X C_{1} \sim$ $\triangle A M C$. Therefore $\angle B A X=\angle C A M$ and  $$ \frac{A M}{A X}=\frac{C M}{X C_{1}}=\frac{C M}{X C}=\cos \alpha $$


22. (USS 5) ${ }^{\mathrm{IMO}} \mathrm{A}$ circle with center $O$ passes through points $A$ and $C$ and intersects the sides $A B$ and $B C$ of the triangle $A B C$ at points $K$ and $N$, respectively. The circumscribed circles of the triangles $A B C$ and $K B N$ intersect at two distinct points $B$ and $M$. Prove that $\angle O M B=90^{\circ}$.

Solution: 22. Assume that $\triangle A B C$ is acute (the case of an obtuse $\triangle A B C$ is similar). Let $S$ and $R$ be the centers of the circumcircles of $\triangle A B C$ and $\triangle K B N$, respectively. Since $\angle B N K=\angle B A C$, the triangles $B N K$ and $B A C$ are similar. Now we have $\angle C B R=\angle A B S=90^{\circ}-\angle A C B$, which gives us $B R \perp A C$ and consequently $B R \| O S$. Similarly $B S \perp K N$ implies that $B S \| O R$. Hence $B R O S$ is a parallelogram. Let $L$ be the point symmetric to $B$ with respect to $R$. Then $R L O S$ is also a parallelogram, and since $S R \perp B M$, we obtain $O L \perp B M$. However, we also have $L M \perp B M$, from which we conclude that $O, L, M$ are collinear and $O M \perp B M$. Second solution. The lines $B M, N K$, and $C A$ are the radical axes of pairs of the three circles, and hence they intersect at a single point $P$. Also, the quadrilateral $M N C P$ is cyclic. Let $O A=O C=O K=O N=r$. We then have $B M \cdot B P=B N \cdot B C=O B^{2}-r^{2}$, $P M \cdot P B=P N \cdot P K=O P^{2}-r^{2}$. It follows that $O B^{2}-O P^{2}=$ $B P(B M-P M)=B M^{2}-P M^{2}$, which implies that $O M \perp M B$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-489.jpg?height=360&width=532&top_left_y=760&top_left_x=800)


3. (NET 3) ${ }^{\mathrm{IMO} 3}$ The weight $w(p)$ of a polynomial $p, p(x)=\sum_{i=0}^{n} a_{i} x^{i}$, with integer coefficients $a_{i}$ is defined as the number of its odd coefficients. For $i=0,1,2, \ldots$, let $q_{i}(x)=(1+x)^{i}$. Prove that for any finite sequence $0 \leq i_{1}<i_{2}<\cdots<i_{n}$, the inequality  $$ w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right) \geq w\left(q_{i_{1}}\right) $$  holds.

Solution: 3. We shall write $P \equiv Q$ for two polynomials $P$ and $Q$ if $P(x)-Q(x)$ has even coefficients. We observe that $(1+x)^{2^{m}} \equiv 1+x^{2^{m}}$ for every $m \in \mathbb{N}$. Consequently, for every polynomial $p$ with degree less than $k=2^{m}, w\left(p \cdot q_{k}\right)=2 w(p)$. Now we prove the inequality from the problem by induction on $i_{n}$. If $i_{n} \leq 1$, the inequality is trivial. Assume it is true for any sequence with $i_{1}<\cdots<i_{n}<2^{m}(m \geq 1)$, and let there be given a sequence with $k=2^{m} \leq i_{n}<2^{m+1}$. Consider two cases. (i) $i_{1} \geq k$. Then $w\left(q_{i_{1}}+\cdots+q_{i_{n}}\right)=2 w\left(q_{i_{1}-k}+\cdots+q_{i_{n}-k}\right) \geq 2 w\left(q_{i_{1}-k}\right)=$ $w\left(q_{i_{1}}\right)$. (ii) $i_{1}<k$. Then the polynomial $p=q_{i_{1}}+\cdots+q_{i_{n}}$ has the form  $$ p=\sum_{i=0}^{k-1} a_{i} x^{i}+(1+x)^{k} \sum_{i=0}^{k-1} b_{i} x^{i} \equiv \sum_{i=0}^{k-1}\left[\left(a_{i}+b_{i}\right) x^{i}+b_{i} x^{i+k}\right] $$  Whenever some $a_{i}$ is odd, either $a_{i}+b_{i}$ or $b_{i}$ in the above sum will be odd. It follows that $w(p) \geq w\left(q_{i_{1}}\right)$, as claimed.  The proof is complete.


4. (AUS 1) ${ }^{\mathrm{IMO} 2}$ Each of the numbers in the set $N=\{1,2,3, \ldots, n-1\}$, where $n \geq 3$, is colored with one of two colors, say red or black, so that: (i) $i$ and $n-i$ always receive the same color, and (ii) for some $j \in N$, relatively prime to $n, i$ and $|j-i|$ receive the same color for all $i \in N, i \neq j$. Prove that all numbers in $N$ must receive the same color.

Solution: 4. Let $\langle x\rangle$ denote the residue of an integer $x$ modulo $n$. Also, we write $a \sim b$ if $a$ and $b$ receive the same color. We claim that all the numbers $\langle i j\rangle$, $i=1,2, \ldots, n-1$, are of the same color. Since $j$ and $n$ are coprime, this will imply the desired result. We use induction on $i$. For $i=1$ the statement is trivial. Assume now that the statement is true for $i=1, \ldots, k-1$. For $1<k<n$ we have $\langle k j\rangle \neq j$. If $\langle k j\rangle>j$, then by (ii), $\langle k j\rangle \sim\langle k j\rangle-j=\langle(k-1) j\rangle$. If otherwise $\langle k j\rangle<j$, then by (ii) and (i), $\langle k j\rangle \sim j-\langle k j\rangle \sim n-j+\langle k j\rangle=\langle(k-1) j\rangle$. This completes the induction.


5. (ROM 1) Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f: D \rightarrow \mathbb{R}, f(M)=\frac{|M A|}{\left|M M^{\prime}\right|}$, where $M^{\prime}=(A M \cap C$, is strictly convex; i.e., $f(P)<\frac{f\left(M_{1}\right)+f\left(M_{2}\right)}{2}, \forall M_{1}, M_{2} \in D, M_{1} \neq M_{2}$, where $P$ is the midpoint of the segment $M_{1} M_{2}$.

Solution: 5. Let w.l.o.g. circle $C$ have unit radius. For each $m \in \mathbb{R}$, the locus of points $M$ such that $f(M)=m$ is the circle $C_{m}$ with radius $r_{m}=m /(m+1)$, that is tangent to $C$ at $A$. Let $O_{m}$ be the center of $C_{m}$. We have to show that if $M \in C_{m}$ and $N \in C_{n}$, where $m, n>0$, then the midpoint $P$ of $M N$ lies inside the circle $C_{(m+n) / 2}$. This is trivial if $m=n$, so let $m \neq n$. For fixed $M, P$ is in the image $C_{n}^{\prime}$ of $C_{n}$ under the homothety with center $M$ and coefficient $1 / 2$. The center of the circle $C_{n}^{\prime}$ is at the midpoint of $O_{n} M$. If we let both $M$ and $N$ vary, $P$ will be on the union of circles with radius $r_{n} / 2$ and centers in the image of $C_{m}$ under the homothety with center $O_{n}$ and coefficient $1 / 2$. Hence $P$ is not outside the circle centered at the midpoint $O_{m} O_{n}$ and with radius $\left(r_{m}+r_{n}\right) / 2$. It remains to show that $r_{(m+n) / 2}>\left(r_{m}+r_{n}\right) / 2$. But this inequality is easily reduced to $(m-n)^{2}>0$, which is true.


6. (POL 1) Let $x_{n}=\sqrt[2]{2+\sqrt[3]{3+\ldots+\sqrt[n]{n}}}$. Prove that  $$ x_{n+1}-x_{n}<\frac{1}{n!}, \quad n=2,3, \ldots $$  ## Alternatives

Solution: 6. Let us set  $$ \begin{aligned} & x_{n, i}=\sqrt[i]{i+\sqrt[i+1]{i+1+\cdots+\sqrt[n]{n}}} \\ & y_{n, i}=x_{n+1, i}^{i-1}+x_{n+1, i}^{i-2} x_{n, i}+\cdots+x_{n, i}^{i-1} \end{aligned} $$  In particular, $x_{n, 2}=x_{n}$ and $x_{n, i}=0$ for $i>n$. We observe that for $n>i>2$,  $$ x_{n+1, i}-x_{n, i}=\frac{x_{n+1, i}^{i}-x_{n, i}^{i}}{y_{n, i}}=\frac{x_{n+1, i+1}-x_{n, i+1}}{y_{n, i}} . $$  Since $y_{n, i}>i x_{n, i}^{i-1} \geq i^{1+(i-1) / i} \geq i^{3 / 2}$ and $x_{n+1, n+1}-x_{n, n+1}=\sqrt[n+1]{n+1}$, simple induction gives  $$ x_{n+1}-x_{n} \leq \frac{\sqrt[n+1]{n+1}}{(n!)^{3 / 2}}<\frac{1}{n!} \quad \text { for } n>2 $$  The inequality for $n=2$ is directly verified.


7. 1a.(CZS 3) The positive integers $x_{1}, \ldots, x_{n}, n \geq 3$, satisfy $x_{1}<x_{2}<$ $\cdots<x_{n}<2 x_{1}$. Set $P=x_{1} x_{2} \cdots x_{n}$. Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $p^{k}$, then $\frac{P}{p^{k}} \geq n!$.

Solution: 7. Let $k_{i} \geq 0$ be the largest integer such that $p^{k_{i}} \mid x_{i}, i=1, \ldots, n$, and $y_{i}=x_{i} / p^{k_{i}}$. We may assume that $k=k_{1}+\cdots+k_{n}$. All the $y_{i}$ must be distinct. Indeed, if $y_{i}=y_{j}$ and $k_{i}>k_{j}$, then $x_{i} \geq p x_{j} \geq 2 x_{i} \geq 2 x_{1}$, which is impossible. Thus $y_{1} y_{2} \ldots y_{n}=P / p^{k} \geq n!$. If equality holds, we must have $y_{i}=1, y_{j}=2$ and $y_{k}=3$ for some $i, j, k$. Thus $p \geq 5$, which implies that either $y_{i} / y_{j} \leq 1 / 2$ or $y_{i} / y_{j} \geq 5 / 2$, which is impossible. Hence the inequality is strict.


8. 1b.(TUR 5) Find the smallest positive integer $n$ such that (i) $n$ has exactly 144 distinct positive divisors, and (ii) there are ten consecutive integers among the positive divisors of $n$.

Solution: 8. Among ten consecutive integers that divide $n$, there must exist numbers divisible by $2^{3}, 3^{2}, 5$, and 7 . Thus the desired number has the form $n=$ $2^{\alpha_{1}} 3^{\alpha_{2}} 5^{\alpha_{3}} 7^{\alpha_{4}} 11^{\alpha_{5}} \cdots$, where $\alpha_{1} \geq 3, \alpha_{2} \geq 2, \alpha_{3} \geq 1, \alpha_{4} \geq 1$. Since $n$ has $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) \cdots$ distinct factors, and $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+\right.$ $1)\left(\alpha_{4}+1\right) \geq 48$, we must have $\left(\alpha_{5}+1\right) \cdots \leq 3$. Hence at most one $\alpha_{j}$, $j>4$, is positive, and in the minimal $n$ this must be $\alpha_{5}$. Checking through the possible combinations satisfying $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{5}+1\right)=144$ one finds that the minimal $n$ is $2^{5} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11=110880$.


9. 2a.(USA 3) Determine the radius of a sphere $S$ that passes through the centroids of each face of a given tetrahedron $T$ inscribed in a unit sphere with center $O$. Also, determine the distance from $O$ to the center of $S$ as a function of the edges of $T$.

Solution: 9. Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ denote the vectors $\overrightarrow{O A}, \overrightarrow{O B}, \overrightarrow{O C}, \overrightarrow{O D}$ respectively. Then $|\vec{a}|=|\vec{b}|=|\vec{c}|=|\vec{d}|=1$. The centroids of the faces are $(\vec{b}+\vec{c}+\vec{d}) / 3$, $(\vec{a}+\vec{c}+\vec{d}) / 3$, etc., and each of these is at distance $1 / 3$ from $P=$ $(\vec{a}+\vec{b}+\vec{c}+\vec{d}) / 3$; hence the required radius is $1 / 3$. To compute $|P|$ as a function of the edges of $A B C D$, observe that $A B^{2}=(\vec{b}-\vec{a})^{2}=$ $2-2 \vec{a} \cdot \vec{b}$ etc. Now  $$ \begin{aligned} P^{2} & =\frac{|\vec{a}+\vec{b}+\vec{c}+\vec{d}|^{2}}{9} \\ & =\frac{16-2\left(A B^{2}+B C^{2}+A C^{2}+A D^{2}+B D^{2}+C D^{2}\right)}{9} \end{aligned} $$


1. (GBR 3) ${ }^{\text {IMO5 }}$ Find, with proof, all functions $f$ defined on the nonnegative real numbers and taking nonnegative real values such that (i) $f[x f(y)] f(y)=f(x+y)$, (ii) $f(2)=0$ but $f(x) \neq 0$ for $0 \leq x<2$.

Solution: 1. If $w>2$, then setting in (i) $x=w-2, y=2$, we get $f(w)=f((w-$ 2) $f(w)) f(2)=0$. Thus  $$ f(x)=0 \quad \text { if and only if } \quad x \geq 2 $$  Now let $0 \leq y<2$ and $x \geq 0$. The LHS in (i) is zero if and only if $x f(y) \geq 2$, while the RHS is zero if and only if $x+y \geq 2$. It follows that $x \geq 2 / f(y)$ if and only if $x \geq 2-y$. Therefore  $$ f(y)=\left\{\begin{array}{cl} \frac{2}{2-y} & \text { for } 0 \leq y<2 \\ 0 & \text { for } y \geq 2 \end{array}\right. $$  The confirmation that $f$ satisfies the given conditions is straightforward.


10. (HUN 2) Three persons $A, B, C$, are playing the following game: A $k$ element subset of the set $\{1, \ldots, 1986\}$ is randomly chosen, with an equal probability of each choice, where $k$ is a fixed positive integer less than or equal to 1986. The winner is $A, B$ or $C$, respectively, if the sum of the chosen numbers leaves a remainder of 0,1 , or 2 when divided by 3 . For what values of $k$ is this game a fair one? (A game is fair if the three outcomes are equally probable.)

Solution: 10. The set $X=\{1, \ldots, 1986\}$ splits into triads $T_{1}, \ldots, T_{662}$, where $T_{j}=$ $\{3 j-2,3 j-1,3 j\}$. Let $\mathcal{F}$ be the family of all $k$-element subsets $P$ such that $\left|P \cap T_{j}\right|=1$ or 2 for some index $j$. If $j_{0}$ is the smallest such $j_{0}$, we define $P^{\prime}$ to be the $k$-element set obtained from $P$ by replacing the elements of $P \cap T_{j_{0}}$ by the ones following cyclically inside $T_{j_{0}}$. Let $s(P)$ denote the remainder modulo 3 of the sum of elements of $P$. Then $s(P), s\left(P^{\prime}\right), s\left(P^{\prime \prime}\right)$ are distinct, and $P^{\prime \prime \prime}=P$. Thus the operator ${ }^{\prime}$ gives us a bijective correspondence between the sets $X \in \mathcal{F}$ with $s(P)=0$, those with $s(P)=1$, and those with $s(P)=2$. If $3 \nmid k$ is not divisible by 3 , then each $k$-element subset of $X$ belongs to $\mathcal{F}$, and the game is fair. If $3 \mid k$, then $k$-element subsets not belonging to $\mathcal{F}$ are those that are unions of several triads. Since every such subset has the sum of elements divisible by 3 , it follows that player $A$ has the advantage.


11. (BUL 1) Let $f(n)$ be the least number of distinct points in the plane such that for each $k=1,2, \ldots, n$ there exists a straight line containing exactly $k$ of these points. Find an explicit expression for $f(n)$. Simplified version. Show that $f(n)=\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]$ ( $[x]$ denoting the greatest integer not exceeding $x$ ).

Solution: 11. Let $X$ be a finite set in the plane and $l_{k}$ a line containing exactly $k$ points of $X(k=1, \ldots, n)$. Then $l_{n}$ contains $n$ points, $l_{n-1}$ contains at least $n-2$ points not lying on $l_{n}, l_{n-2}$ contains at least $n-4$ points not lying on $l_{n}$ or $l_{n-1}$, etc. It follows that  $$ |X| \geq g(n)=n+(n-2)+(n-4)+\cdots+\left(n-2\left[\frac{n}{2}\right]\right) . $$  Hence $f(n) \geq g(n)=\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]$, where the last equality is easily proved by induction. We claim that $f(n)=g(n)$. To prove this, we shall inductively construct a set $X_{n}$ of cardinality $g(n)$ with the required property. For $n \leq 2$ a one-point and two-point set satisfy the requirements. Assume that $X_{n}$ is a set of $g(n)$ points and that $l_{k}$ is a line containing exactly $k$ points of $X_{n}, k=1, \ldots, n$. Consider any line $l$ not parallel to any of the $l_{k}$ 's and not containing any point of $X_{n}$ or any intersection point of the $l_{k}$. Let $l$ intersect $l_{k}$ in a point $P_{k}, k=1, \ldots, n$, and let $P_{n+1}, P_{n+2}$ be two points on $l$ other than $P_{1}, \ldots, P_{n}$. We define $X_{n+2}=X_{n} \cup\left\{P_{1}, \ldots, P_{n+2}\right\}$. The set $X_{n+2}$ consists of $g(n)+(n+2)=g(n+2)$ points. Since the lines $l, l_{n}, \ldots, l_{2}, l_{1}$ meet $X_{n}$ in $n+2, n+1, \ldots, 3,2$ points respectively (and there clearly exists a line containing only one point of $X_{n+2}$ ), this set also meets the demands.


12. (GDR 3) ${ }^{\mathrm{IMO} 3}$ To each vertex $P_{i}(i=1, \ldots, 5)$ of a pentagon an integer $x_{i}$ is assigned, the sum $s=\sum x_{i}$ being positive. The following operation is allowed, provided at least one of the $x_{i}$ 's is negative: Choose a negative $x_{i}$, replace it by $-x_{i}$, and add the former value of $x_{i}$ to the integers assigned to the two neighboring vertices of $P_{i}$ (the remaining two integers are left unchanged). This operation is to be performed repeatedly until all negative integers disappear. Decide whether this procedure must eventually terminate.

Solution: 12. We define $f\left(x_{1}, \ldots, x_{5}\right)=\sum_{i=1}^{5}\left(x_{i+1}-x_{i-1}\right)^{2}\left(x_{0}=x_{5}, x_{6}=x_{1}\right)$. Assuming that $x_{3}<0$, according to the rules the lattice vector $X=$ $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)$ changes into $Y=\left(x_{1}, x_{2}+x_{3},-x_{3}, x_{4}+x_{3}, x_{5}\right)$. Then  $$ \begin{aligned} f(Y)-f(X)= & \left(x_{2}+x_{3}-x_{5}\right)^{2}+\left(x_{1}+x_{3}\right)^{2}+\left(x_{2}-x_{4}\right)^{2} \\ & +\left(x_{3}+x_{5}\right)^{2}+\left(x_{1}-x_{3}-x_{4}\right)^{2}-\left(x_{2}-x_{5}\right)^{2} \\ & -\left(x_{3}-x_{1}\right)^{2}-\left(x_{4}-x_{2}\right)^{2}-\left(x_{5}-x_{3}\right)^{2}-\left(x_{1}-x_{4}\right)^{2} \\ = & 2 x_{3}\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=2 x_{3} S<0 \end{aligned} $$  Thus $f$ strictly decreases after each step, and since it takes only positive integer values, the number of steps must be finite. Remark. One could inspect the behavior of $g(x)=\sum_{i=1}^{5} \sum_{j=1}^{5} \mid x_{i}+x_{i+1}+$ $\cdots+x_{j-1} \mid$ instead. Then $g(Y)-g(X)=\left|S+x_{3}\right|-\left|S-x_{3}\right|>0$.


13. (FRG 3) A particle moves from $(0,0)$ to $(n, n)$ directed by a fair coin. For each head it moves one step east and for each tail it moves one step north. At $(n, y), y<n$, it stays there if a head comes up and at $(x, n)$, $x<n$, it stays there if a tail comes up. Let $k$ be a fixed positive integer. Find the probability that the particle needs exactly $2 n+k$ tosses to reach $(n, n)$.

Solution: 13. Let us consider the infinite integer lattice and assume that having reached a point $(x, n)$ or $(n, y)$, the particle continues moving east and north following the rules of the game. The required probability $p_{k}$ is equal to the probability of getting to one of the points $E_{1}(n, n+k), E_{2}(n+k, n)$, but without passing through $(n, n+k-1)$ or $(n+k-1, n)$. Thus $p$ is equal to the probability $p_{1}$ of getting to $E_{1}(n, n+k)$ via $D_{1}(n-1, n+k)$ plus the probability $p_{2}$ of getting to $E_{2}(n+k, n)$ via $D_{2}(n+k, n-1)$. Both $p_{1}$ and $p_{2}$ are easily seen to be equal to $\binom{2 n+k-1}{n-1} 2^{-2 n-k}$, and therefore $p=\binom{2 n+k-1}{n-1} 2^{-2 n-k+1}$.


14. (IRE 1) The circle inscribed in a triangle $A B C$ touches the sides $B C, C A, A B$ in $D, E, F$, respectively, and $X, Y, Z$ are the midpoints of $E F, F D, D E$, respectively. Prove that the centers of the inscribed circle and of the circles around $X Y Z$ and $A B C$ are collinear.

Solution: 14. We shall use the following simple fact.  Lemma. If $\widehat{k}$ is the image of a circle $k$ under an inversion centered at a point $Z$, and $O_{1}, O_{2}$ are centers of $k$ and $\widehat{k}$, then $O_{1}, O_{2}$, and $Z$ are collinear. Proof. The result follows immediately from the symmetry with respect to the line $Z O_{1}$. Let $I$ be the center of the inscribed circle $i$. Since $I X \cdot I A=I E^{2}$, the inversion with respect to $i$ takes points $A$ into $X$, and analogously $B, C$ into $Y, Z$ respectively. It follows from the lemma that the center of circle $A B C$, the center of circle $X Y Z$, and point $I$ are collinear.


15. (NET 2) Let $A B C D$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a quadrangle such that $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ are the centers of the circumcircles of triangles $B C D, A C D, A B D$, and $A B C$. We write $T(A B C D)=A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. Let us define $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}=$ $T\left(A^{\prime} B^{\prime} C^{\prime} D^{\prime}\right)=T(T(A B C D))$. (a) Prove that $A B C D$ and $A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}$ are similar. (b) The ratio of similitude depends on the size of the angles of $A B C D$. Determine this ratio.

Solution: 15. (a) This is the same problem as $S L 82$-14. (b) If $S$ is the midpoint of $A C$, we have $B^{\prime} S=A C \frac{\cos \angle D}{2 \sin \angle D}, D^{\prime} S=$ $A C \frac{\cos \angle B}{2 \sin \angle B}, B^{\prime} D^{\prime}=A C\left|\frac{\sin (\angle B+\angle D)}{2 \sin \angle B \sin \angle D}\right|$. These formulas are true also if $\angle B>90^{\circ}$ or $\angle D>90^{\circ}$. We similarly obtain that $A^{\prime \prime} C^{\prime \prime}=$ $B^{\prime} D^{\prime}\left|\frac{\sin \left(\angle A^{\prime}+\angle C^{\prime}\right)}{2 \sin \angle A^{\prime} \sin \angle C^{\prime}}\right|$. Therefore  $$ A^{\prime \prime} C^{\prime \prime}=A C \frac{\sin ^{2}(\angle A+\angle C)}{4 \sin \angle A \sin \angle B \sin \angle C \sin \angle D} $$


16. (ISR 1) ${ }^{\mathrm{IMO}}$ Let $A, B$ be adjacent vertices of a regular $n$-gon in the plane and let $O$ be its center. Now let the triangle $A B O$ glide around the polygon in such a way that the points $A$ and $B$ move along the whole circumference of the polygon. Describe the figure traced by the vertex $O$.

Solution: 16. Let $Z$ be the center of the polygon.  Suppose that at some moment we have $A \in P_{i-1} P_{i}$ and $B \in$ $P_{i} P_{i+1}$, where $P_{i-1}, P_{i}, P_{i+1}$ are adjacent vertices of the polygon. Since $\angle A O B=180^{\circ}-\angle P_{i-1} P_{i} P_{i+1}$, the quadrilateral $A P_{i} B O$ is cyclic. Hence $\angle A P_{i} O=\angle A B O=\angle A P_{i} Z$, which means that $O \in P_{i} Z$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-495.jpg?height=329&width=390&top_left_y=251&top_left_x=903)  Moreover, from $O P_{i}=2 r \sin \angle P_{i} A O$, where $r$ is the radius of circle $A P_{i} B O$, we obtain that $Z P_{i} \leq O P_{i} \leq Z P_{i} / \cos (\pi / n)$. Thus $O$ traces a segment $Z Z_{i}$ as $A$ and $B$ move along $P_{i-1} P_{i}$ and $P_{i} P_{i+1}$ respectively, where $Z_{i}$ is a point on the ray $P_{i} Z$ with $P_{i} Z_{i} \cos (\pi / n)=P_{i} Z$. When $A, B$ move along the whole circumference of the polygon, $O$ traces an asterisk consisting of $n$ segments of equal length emanating from $Z$ and pointing away from the vertices.


17. ( $\mathbf{C H N} 3)^{\mathrm{IMO} 2}$ Let $A, B, C$ be fixed points in the plane. A man starts from a certain point $P_{0}$ and walks directly to $A$. At $A$ he turns his direction by $60^{\circ}$ to the left and walks to $P_{1}$ such that $P_{0} A=A P_{1}$. After he does the same action 1986 times successively around the points $A, B, C, A, B, C, \ldots$, he returns to the starting point. Prove that $\triangle A B C$ is equilateral and that the vertices $A, B, C$ are arranged counterclockwise.

Solution: 17. We use complex numbers to represent the position of a point in the plane. For convenience, let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, \ldots$ be $A, B, C, A, B, \ldots$ respectively, and let $P_{0}$ be the origin. After the $k$ th step, the position of $P_{k}$ will be $P_{k}=A_{k}+\left(P_{k-1}-A_{k}\right) u, k=1,2,3, \ldots$, where $u=e^{4 \pi \tau / 3}$. We easily obtain  $$ P_{k}=(1-u)\left(A_{k}+u A_{k-1}+u^{2} A_{k-2}+\cdots+u^{k-1} A_{1}\right) $$  The condition $P_{0} \equiv P_{1986}$ is equivalent to $A_{1986}+u A_{1985}+\cdots+u^{1984} A_{2}+$ $u^{1985} A_{1}=0$, which, having in mind that $A_{1}=A_{4}=A_{7}=\cdots, A_{2}=A_{5}=$ $A_{8}=\cdots, A_{3}=A_{6}=A_{9}=\cdots$, reduces to  $$ 662\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=\left(1+u^{3}+\cdots+u^{1983}\right)\left(A_{3}+u A_{2}+u^{2} A_{1}\right)=0 $$  It follows that $A_{3}-A_{1}=u\left(A_{1}-A_{2}\right)$, and the assertion follows. Second solution. Let $f_{P}$ denote the rotation with center $P$ through $120^{\circ}$ clockwise. Let $f_{1}=f_{A}$. Then $f_{1}\left(P_{0}\right)=P_{1}$. Let $B^{\prime}=f_{1}(B), C^{\prime}=f_{1}(C)$, and $f_{2}=f_{B^{\prime}}$. Then $f_{2}\left(P_{1}\right)=P_{2}$ and $f_{2}\left(A B^{\prime} C^{\prime}\right)=A^{\prime} B^{\prime} C^{\prime \prime}$. Finally, let $f_{3}=f_{C^{\prime \prime}}$ and $f_{3}\left(A^{\prime} B^{\prime} C^{\prime \prime}\right)=A^{\prime \prime} B^{\prime \prime} C^{\prime \prime}$. Then $g=f_{3} f_{2} f_{1}$ is a translation sending $P_{0}$ to $P_{3}$ and $C$ to $C^{\prime \prime}$. Now $P_{1986}=P_{0}$ implies that $g^{662}$ is the identity, and thus $C=C^{\prime \prime}$. Let $K$ be such that $A B K$ is equilateral and positively oriented. We observe that $f_{2} f_{1}(K)=K$; therefore the rotation $f_{2} f_{1}$ satisfies $f_{2} f_{1}(P) \neq P$ for $P \neq K$. Hence $f_{2} f_{1}(C)=C^{\prime \prime}=C$ implies $K=C$.


18. (TUR 1) Let $A X, B Y, C Z$ be three cevians concurrent at an interior point $D$ of a triangle $A B C$. Prove that if two of the quadrangles $D Y A Z, D Z B X, D X C Y$ are circumscribable, so is the third.

Solution: 18. We shall use the following criterion for a quadrangle to be circumscribable. Lemma. The quadrangle $A Y D Z$ is circumscribable if and only if $D B-$ $D C=A B-A C$. Proof. Suppose that $A Y D Z$ is circumscribable and that the incircle is tangent to $A Z, Z D, D Y, Y A$ at $M, N, P, Q$ respectively. Then $D B-D C=P B-N C=M B-Q C=A B-A C$. Conversely, assume that $D B-D C=A B-A C$ and let a tangent from $D$ to the incircle of the triangle $A C Z$ meet $C Z$ and $C A$ at $D^{\prime} \neq Z$ and $Y^{\prime} \neq A$ respectively. According to the first part we have $D^{\prime} B-D^{\prime} C=A B-A C$. It follows that $\left|D^{\prime} B-D B\right|=$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-496.jpg?height=315&width=387&top_left_y=236&top_left_x=900) $\left|D^{\prime} C-D C\right|=D D^{\prime}$, implying that $D^{\prime} \equiv D$. Let us assume that $D Z B X$ and $D X C Y$ are circumscribable. Using the lemma we obtain $D C-D A=B C-B A$ and $D A-D B=C A-C B$. Adding these two inequalities yields $D C-D B=A C-A B$, and the statement follows from the lemma.


19. (BUL 2) A tetrahedron $A B C D$ is given such that $A D=B C=a$; $A C=B D=b ; A B \cdot C D=c^{2}$. Let $f(P)=A P+B P+C P+D P$, where $P$ is an arbitrary point in space. Compute the least value of $f(P)$.

Solution: 19. Let $M$ and $N$ be the midpoints of segments $A B$ and $C D$, respectively. The given conditions imply that $\triangle A B D \cong \triangle B A C$ and $\triangle C D A \cong \triangle D C B$; hence $M C=M D$ and $N A=N B$. It follows that $M$ and $N$ both lie on the perpendicular bisectors of $A B$ and $C D$, and consequently $M N$ is the common perpendicular bisector of $A B$ and $C D$. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to $M N$. Now if $P$ is a point in space and $P^{\prime}$ the point symmetric to $P$ with respect to $M N$, we have $B P=A P^{\prime}, C P=D P^{\prime}$, and thus $f(P)=A P+A P^{\prime}+D P+D P^{\prime}$. Let $P P^{\prime}$ intersect $M N$ in $Q$. Then $A P+A P^{\prime} \geq 2 A Q$ and $D P+D P^{\prime} \geq 2 D Q$, from which it follows that $f(P) \geq 2(A Q+D Q)=f(Q)$. It remains to minimize $f(Q)$ with $Q$ moving along the line $M N$. Let us rotate point $D$ around $M N$ to a point $D^{\prime}$ that belongs to the plane $A M N$, on the side of $M N$ opposite to $A$. Then $f(Q)=2\left(A Q+D^{\prime} Q\right) \geq$ $A D^{\prime}$, and equality occurs when $Q$ is the intersection of $A D^{\prime}$ and $M N$. Thus $\min f(Q)=A D^{\prime}$. We note that $4 M D^{2}=2 A D^{2}+2 B D^{2}-A B^{2}=$ $2 a^{2}+2 b^{2}-A B^{2}$ and $4 M N^{2}=4 M D^{2}-C D^{2}=2 a^{2}+2 b^{2}-A B^{2}-C D^{2}$. Now, $A D^{\prime 2}=\left(A M+D^{\prime} N\right)^{2}+M N^{2}$, which together with $A M+D^{\prime} N=$ $(a+b) / 2$ gives us  $$ A D^{\prime 2}=\frac{a^{2}+b^{2}+A B \cdot C D}{2}=\frac{a^{2}+b^{2}+c^{2}}{2} $$  We conclude that $\min f(Q)=\sqrt{\left(a^{2}+b^{2}+c^{2}\right) / 2}$.


2. (SWE 2) Let $f(x)=x^{n}$ where $n$ is a fixed positive integer and $x=$ $1,2, \ldots$. Is the decimal expansion $a=0 . f(1) f(2) f(3) \ldots$ rational for any value of $n$ ? The decimal expansion of $a$ is defined as follows: If $f(x)=d_{1}(x) d_{2}(x) \ldots$ $\ldots d_{r(x)}(x)$ is the decimal expansion of $f(x)$, then $a=0.1 d_{1}(2) d_{2}(2) \ldots$ $\ldots d_{r(2)}(2) d_{1}(3) \ldots d_{r(3)}(3) d_{1}(4) \ldots$.

Solution: 2. No. If $a$ were rational, its decimal expansion would be periodic from some point. Let $p$ be the number of decimals in the period. Since $f\left(10^{2 p}\right)$ has $2 n p$ zeros, it contains a full periodic part; hence the period would consist only of zeros, which is impossible.


20. (CAN 3) Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.

Solution: 20. If the faces of the tetrahedron $A B C D$ are congruent triangles, we must have $A B=C D, A C=B D$, and $A D=B C$. Then the sum of angles at $A$ is $\angle B A C+\angle C A D+\angle D A B=\angle B D C+\angle C B D+\angle D C B=180^{\circ}$. We now assume that the sum of angles at each vertex is $180^{\circ}$. Let us construct triangles $B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ in the plane $A B C$, exterior to $\triangle A B C$, such that $\triangle B C D^{\prime} \cong \triangle B C D, \triangle C A D^{\prime \prime} \cong \triangle C A D$, and $\triangle A B D^{\prime \prime \prime} \cong \triangle A B D$. Then by the assumption, $A \in D^{\prime \prime} D^{\prime \prime \prime}, B \in D^{\prime \prime \prime} D^{\prime}$, and $C \in D^{\prime} D^{\prime \prime}$. Since also $D^{\prime \prime} A=D^{\prime \prime \prime} A=D A$, etc., $A, B, C$ are the mid- points of segments $D^{\prime \prime} D^{\prime \prime \prime}, D^{\prime \prime \prime} D^{\prime}, D^{\prime} D^{\prime \prime}$ respectively. Thus the triangles $A B C, B C D^{\prime}, C A D^{\prime \prime}, A B D^{\prime \prime \prime}$ are congruent, and the statement follows.


21. (TUR 2) Let $A B C D$ be a tetrahedron having each sum of opposite sides equal to 1. Prove that  $$ r_{A}+r_{B}+r_{C}+r_{D} \leq \frac{\sqrt{3}}{3} $$  where $r_{A}, r_{B}, r_{C}, r_{D}$ are the inradii of the faces, equality holding only if $A B C D$ is regular.

Solution: 21. Since the sum of all edges of $A B C D$ is 3 , the statement of the problem is an immediate consequence of the following statement: Lemma. Let $r$ be the inradius of a triangle with sides $a, b, c$. Then $a+$ $b+c \geq 6 \sqrt{3} \cdot r$, with equality if and only if the triangle is equilateral. Proof. If $S$ and $p$ denotes the area and semiperimeter of the triangle, by Heron's formula and the AM-GM inequality we have  $$ \begin{aligned} p r & =S=\sqrt{p(p-a)(p-b)(p-c)} \\ & \leq \sqrt{p\left(\frac{(p-a)+(p-b)+(p-c)}{3}\right)^{3}}=\sqrt{\frac{p^{4}}{27}}=\frac{p^{2}}{3 \sqrt{3}}, \end{aligned} $$  i.e., $p \geq 3 \sqrt{3} \cdot r$, which is equivalent to the claim.


3. (USA 3) Let $A, B$, and $C$ be three points on the edge of a circular chord such that $B$ is due west of $C$ and $A B C$ is an equilateral triangle whose side is 86 meters long. A boy swam from $A$ directly toward $B$. After covering a distance of $x$ meters, he turned and swam westward, reaching the shore after covering a distance of $y$ meters. If $x$ and $y$ are both positive integers, determine $y$.

Solution: 3. Let $E$ be the point where the boy turned westward, reaching the shore at $D$. Let the ray $D E$ cut $A C$ at $F$ and the shore again at $G$. Then $E F=$ $A E=x$ (because $A E F$ is an equilateral triangle) and $F G=D E=y$. From $A E \cdot E B=D E \cdot E G$ we obtain $x(86-x)=y(x+y)$. If $x$ is odd, then $x(86-x)$ is odd, while $y(x+y)$ is even. Hence $x$ is even, and so $y$ must also be even. Let $y=2 y_{1}$. The above equation can be rewritten as  $$ \left(x+y_{1}-43\right)^{2}+\left(2 y_{1}\right)^{2}=\left(43-y_{1}\right)^{2} \text {. } $$  Since $y_{1}<43$, we have $\left(2 y_{1}, 43-y_{1}\right)=1$, and thus $\left(\left|x+y_{1}-43\right|, 2 y_{1}, 43-\right.$ $\left.y_{1}\right)$ is a primitive Pythagorean triple. Consequently there exist integers $a>b>0$ such that $y_{1}=a b$ and $43-y_{1}=a^{2}+b^{2}$. We obtain that $a^{2}+b^{2}+a b=43$, which has the unique solution $a=6, b=1$. Hence $y=12$ and $x=2$ or $x=72$. Remark. The Diophantine equation $x(86-x)=y(x+y)$ can be also solved directly. Namely, we have that $x(344-3 x)=(2 y+x)^{2}$ is a square, and since $x$ is even, we have $(x, 344-3 x)=2$ or 4 . Consequently $x, 344-3 x$ are either both squares or both two times squares. The rest is easy.


4. (CZS 3) Let $n$ be a positive integer and let $p$ be a prime number, $p>3$. Find at least $3(n+1)$ [easier version: $2(n+1)]$ sequences of positive integers $x, y, z$ satisfying  $$ x y z=p^{n}(x+y+z) $$  that do not differ only by permutation.

Solution: 4. Let $x=p^{\alpha} x^{\prime}, y=p^{\beta} y^{\prime}, z=p^{\gamma} z^{\prime}$ with $p \nmid x^{\prime} y^{\prime} z^{\prime}$ and $\alpha \geq \beta \geq \gamma$. From the given equation it follows that $p^{n}(x+y)=z\left(x y-p^{n}\right)$ and consequently $z^{\prime} \mid x+y$. Since also $p^{\gamma} \mid x+y$, we have $z \mid x+y$, i.e., $x+y=q z$. The given equation together with the last condition gives us  $$ x y=p^{n}(q+1) \quad \text { and } \quad x+y=q z . $$  Conversely, every solution of (1) gives a solution of the given equation.  For $q=1$ and $q=2$ we obtain the following classes of $n+1$ solutions each:  $$ \begin{array}{ll} q=1:(x, y, z)=\left(2 p^{i}, p^{n-i}, 2 p^{i}+p^{n-i}\right) & \text { for } i=0,1,2, \ldots, n \\ q=2:(x, y, z)=\left(3 p^{j}, p^{n-j}, \frac{3 p^{j}+p^{n-j}}{2}\right) & \text { for } j=0,1,2, \ldots, n \end{array} $$  For $n=2 k$ these two classes have a common solution $\left(2 p^{k}, p^{k}, 3 p^{k}\right)$; otherwise, all these solutions are distinct. One further solution is given by $(x, y, z)=\left(1, p^{n}\left(p^{n}+3\right) / 2, p^{2}+2\right)$, not included in the above classes for $p>3$. Thus we have found $2(n+1)$ solutions. Another type of solution is obtained if we put $q=p^{k}+p^{n-k}$. This yields the solutions  $$ (x, y, z)=\left(p^{k}, p^{n}+p^{n-k}+p^{2 n-2 k}, p^{n-k}+1\right) \quad \text { for } k=0,1, \ldots, n $$  For $k<n$ these are indeed new solutions. So far, we have found $3(n+1)-1$ or $3(n+1)$ solutions. One more solution is given by $(x, y, z)=\left(p, p^{n}+\right.$ $\left.p^{n-1}, p^{n-1}+p^{n-2}+1\right)$.


5. (FRG 1) ${ }^{\mathrm{IMO1}}$ The set $S=\{2,5,13\}$ has the property that for every $a, b \in S, a \neq b$, the number $a b-1$ is a perfect square. Show that for every positive integer $d$ not in $S$, the set $S \cup\{d\}$ does not have the above property.

Solution: 5. Suppose that for every $a, b \in\{2,5,13, d\}, a \neq b$, the number $a b-1$ is a perfect square. In particular, for some integers $x, y, z$ we have  $$ 2 d-1=x^{2}, \quad 5 d-1=y^{2}, \quad 13 d-1=z^{2} $$  Since $x$ is clearly odd, $d=\left(x^{2}+1\right) / 2$ is also odd because $4 \nmid x^{2}+1$. It follows that $y$ and $z$ are even, say $y=2 y_{1}$ and $z=2 z_{1}$. Hence $\left(z_{1}-\right.$ $\left.y_{1}\right)\left(z_{1}+y_{1}\right)=\left(z^{2}-y^{2}\right) / 4=2 d$. But in this case one of the factors $z_{1}-y_{1}$, $z_{1}+y_{1}$ is odd and the other one is even, which is impossible.


6. (NET 1) Find four positive integers each not exceeding 70000 and each having more than 100 divisors.

Solution: 6 . There are five such numbers:  $$ \begin{aligned} 69300 & =2^{2} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 11: & 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 & =108 \text { divisors; } \\ 50400 & =2^{5} \cdot 3^{2} \cdot 5^{2} \cdot 7: & 6 \cdot 3 \cdot 3 \cdot 2 & =108 \text { divisors; } \\ 60480 & =2^{6} \cdot 3^{3} \cdot 5 \cdot 7: & 7 \cdot 4 \cdot 2 \cdot 2 & =112 \text { divisors; } \\ 55440 & =2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 11: & 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 & =120 \text { divisors; } \\ 65520 & =2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 13: & 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 & =120 \text { divisors. } \end{aligned} $$


7. (FRA 5) Let real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy $0<x_{1}<x_{2}<\cdots<$ $x_{n}<1$ and set $x_{0}=0, x_{n+1}=1$. Suppose that these numbers satisfy the following system of equations:  $$ \sum_{j=0, j \neq i}^{n+1} \frac{1}{x_{i}-x_{j}}=0 \quad \text { where } i=1,2, \ldots, n $$  Prove that $x_{n+1-i}=1-x_{i}$ for $i=1,2, \ldots, n$.

Solution: 7. Let $P(x)=\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right)\left(x-x_{n+1}\right)$. Then  $$ P^{\prime}(x)=\sum_{j=0}^{n+1} \frac{P(x)}{x-x_{j}} \quad \text { and } \quad P^{\prime \prime}(x)=\sum_{j=0}^{n+1} \sum_{k \neq j} \frac{P(x)}{\left(x-x_{j}\right)\left(x-x_{k}\right)} . $$  Therefore  $$ P^{\prime \prime}\left(x_{i}\right)=2 P^{\prime}\left(x_{i}\right) \sum_{j \neq i} \frac{1}{\left(x_{i}-x_{j}\right)} $$  for $i=0,1, \ldots, n+1$, and the given condition implies $P^{\prime \prime}\left(x_{i}\right)=0$ for $i=1,2, \ldots, n$. Consequently,  $$ x(x-1) P^{\prime \prime}(x)=(n+2)(n+1) P(x) . $$  It is easy to observe that there is a unique monic polynomial of degree $n+2$ satisfying differential equation (1). On the other hand, the polynomial $Q(x)=(-1)^{n} P(1-x)$ also satisfies this equation, is monic, and $\operatorname{deg} Q=$ $n+2$. Therefore $(-1)^{n} P(1-x)=P(x)$, and the result follows.


8. (USA 1) From a collection of $n$ persons $q$ distinct two-member teams are selected and ranked $1, \ldots, q$ (no ties). Let $m$ be the least integer larger than or equal to $2 q / n$. Show that there are $m$ distinct teams that may be listed so that (i) each pair of consecutive teams on the list have one member in common and (ii) the chain of teams on the list are in rank order. Alternative formulation. Given a graph with $n$ vertices and $q$ edges numbered $1, \ldots, q$, show that there exists a chain of $m$ edges, $m \geq \frac{2 q}{n}$, each two consecutive edges having a common vertex, arranged monotonically with respect to the numbering.

Solution: 8. We shall solve the problem in the alternative formulation. Let $L_{G}(v)$ denote the length of the longest directed chain of edges in the given graph $G$ that begins in a vertex $v$ and is arranged decreasingly relative to the numbering. By the pigeonhole principle it suffices to show that $\sum_{v} L(v) \geq 2 q$ in every such graph. We do this by induction on $q$. For $q=1$ the claim is obvious. We assume that it is true for $q-1$ and consider a graph $G$ with $q$ edges numbered $1, \ldots, q$. Let the edge number $q$ connect vertices $u$ and $w$. Removing this edge, we get a graph $G^{\prime}$ with $q-1$ edges. We then have  $$ L_{G}(u) \geq L_{G^{\prime}}(w)+1, L_{G}(w) \geq L_{G^{\prime}}(u)+1, L_{G}(v) \geq L_{G^{\prime}}(v) \text { for other } v $$  Since $\sum L_{G^{\prime}}(v) \geq 2(q-1)$ by inductive assumption, it follows that $\sum L_{G}(v) \geq 2(q-1)+2=2 q$ as desired. Second solution. Let us place a spider at each vertex of the graph. Let us now interchange the positions of the two spiders at the endpoints of each edge, listing the edges increasingly with respect to the numbering. This way we will move spiders exactly $2 q$ times (two for each edge). Hence there is a spider that will be moved at least $2 q / n$ times. All that remains is to notice that the path of each spider consists of edges numbered in increasing order. Remark. A chain of the stated length having all vertices distinct does not necessarily exist. An example is $n=4, q=6$ with the numbering following the order $a b, c d, a c, b d, a d, b c$.


9. (GDR 1) ${ }^{\mathrm{IMO} 6}$ Prove or disprove: Given a finite set of points with integer coordinates in the plane, it is possible to color some of these points red and the remaining ones white in such a way that for any straight line $L$ parallel to one of the coordinate axes, the number of red colored points and the number of white colored points on $L$ differ by at most 1.

Solution: 9. We shall use induction on the number $n$ of points. The case $n=1$ is trivial. Let us suppose that the statement is true for all $1,2, \ldots, n-1$, and that we are given a set $T$ of $n$ points. If there exists a point $P \in T$ and a line $l$ that is parallel to an axis and contains $P$ and no other points of $T$, then by the inductive hypothesis we can color the set $T \backslash\{P\}$ and then use a suitable color for $P$. Let us now suppose that whenever a line parallel to an axis contains a point of $T$, it contains another point of $T$. It follows that for an arbitrary point $P_{0} \in T$ we can choose points $P_{1}, P_{2}, \ldots$ such that $P_{k} P_{k+1}$ is parallel to the $x$-axis for $k$ even, and to the $y$-axis for $k$ odd. We eventually come to a pair of integers $(r, s)$ of the same parity, $0 \leq r<s$, such that lines $P_{r} P_{r+1}$ and $P_{s} P_{s+1}$ coincide. Hence the closed polygonal line $P_{r+1} P_{r+2} \ldots P_{s} P_{r+1}$ is of even length. Thus we may color the points of this polygonal line alternately and then apply the inductive assumption for the rest of the set $T$. The induction is complete.  Second solution. Let $P_{1}, P_{2}, \ldots, P_{k}$ be the points lying on a line $l$ parallel to an axis, going from left to right or from up to down. We draw segments joining $P_{1}$ with $P_{2}, P_{3}$ with $P_{4}$, and generally $P_{2 i-1}$ with $P_{2 i}$. Having this done for every such line $l$, we obtain a set of segments forming certain polygonal lines. If one of these polygonal lines is closed, then it must have an even number of vertices. Thus, we can color the vertices on each of the polygonal lines alternately (a point not lying on any of the polygonal lines may be colored arbitrarily). The obtained coloring satisfies the conditions.


1. (AUS 6) Let $f$ be a function that satisfies the following conditions: (i) If $x>y$ and $f(y)-y \geq v \geq f(x)-x$, then $f(z)=v+z$, for some number $z$ between $x$ and $y$. (ii) The equation $f(x)=0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; (iii) $f(0)=1$. (iv) $f(1987) \leq 1988$. (v) $f(x) f(y)=f(x f(y)+y f(x)-x y)$.  Find $f(1987)$.

Solution: 1. By (ii), $f(x)=0$ has at least one solution, and there is the greatest among them, say $x_{0}$. Then by (v), for any $x$,  $$ 0=f(x) f\left(x_{0}\right)=f\left(x f\left(x_{0}\right)+x_{0} f(x)-x_{0} x\right)=f\left(x_{0}(f(x)-x)\right) $$  It follows that $x_{0} \geq x_{0}(f(x)-x)$. Suppose $x_{0}>0$. By (i) and (iii), since $f\left(x_{0}\right)-x_{0}<0<f(0)-0$, there is a number $z$ between 0 and $x_{0}$ such that $f(z)=z$. By (1), $0=f\left(x_{0}(f(z)-\right.$ $z))=f(0)=1$, a contradiction. Hence, $x_{0}<0$. Now the inequality $x_{0} \geq x_{0}(f(x)-x)$ gives $f(x)-x \geq 1$ for all $x$; so, $f(1987) \geq 1988$. Therefore $f(1987)=1988$.


10. (ICE 3) Let $S_{1}$ and $S_{2}$ be two spheres with distinct radii that touch externally. The spheres lie inside a cone $C$, and each sphere touches the cone in a full circle. Inside the cone there are $n$ additional solid spheres arranged in a ring in such a way that each solid sphere touches the cone $C$, both of the spheres $S_{1}$ and $S_{2}$ externally, as well as the two neighboring solid spheres. What are the possible values of $n$ ?

Solution: 10. Denote by $r, R$ (take w.l.o.g. $r<R$ ) the radii and by $A, B$ the centers of the spheres $S_{1}, S_{2}$ respectively. Let $s$ be the common radius of the spheres in the ring, $C$ the center of one of them, say $S$, and $D$ the foot of the perpendicular from $C$ to $A B$. The centers of the spheres in the ring form a regular $n$-gon with center $D$, and thus $\sin (\pi / n)=s / C D$. Using Heron's formula on the triangle $A B C$, we obtain $(r+R)^{2} C D^{2}=$ $4 r R s(r+R+s)$, and hence ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-502.jpg?height=565&width=438&top_left_y=1118&top_left_x=865)  $$ \sin ^{2} \frac{\pi}{n}=\frac{s^{2}}{C D^{2}}=\frac{(r+R)^{2} s}{4(r+R+s) r R} $$  Choosing the unit of length so that $r+R=2$, for simplicity of writing, we write (1) as $1 / \sin ^{2}(\pi / n)=r R(1+2 / s)$. Let now $v$ be half the angle at the top of the cone. Then clearly $R-r=(R+r) \sin v=2 \sin v$, giving us $R=1+\sin v, r=1-\sin v$. It follows that  $$ \frac{1}{\sin ^{2} \frac{\pi}{n}}=\left(1+\frac{2}{s}\right) \cos ^{2} v $$  We need to express $s$ as a function of $R$ and $r$. Let $E_{1}, E_{2}, E$ be collinear points of tangency of $S_{1}, S_{2}$, and $S$ with the cone. Obviously, $E_{1} E_{2}=$ $E_{1} E+E_{2} E$, i.e., $2 \sqrt{r s}+2 \sqrt{R s}=2 \sqrt{R r}=(R+r) \cos v=2 \cos v$. Hence,  $$ \cos ^{2} v=s(\sqrt{R}+\sqrt{r})^{2}=s(R+r+2 \sqrt{R r})=s(2+2 \cos v) . $$  Substituting this into (2), we obtain $2+\cos v=1 / \sin (\pi / n)$. Therefore $1 / 3<\sin (\pi / n)<1 / 2$, and we conclude that the possible values for $n$ are 7,8 , and 9 .


11. (POL 1) Find the number of partitions of the set $\{1,2, \ldots, n\}$ into three subsets $A_{1}, A_{2}, A_{3}$, some of which may be empty, such that the following conditions are satisfied: (i) After the elements of every subset have been put in ascending order, every two consecutive elements of any subset have different parity. (ii) If $A_{1}, A_{2}, A_{3}$ are all nonempty, then in exactly one of them the minimal number is even.

Solution: 11. Let $A_{1}$ be the set that contains 1 , and let the minimal element of $A_{2}$ be less than that of $A_{3}$. We shall construct the partitions with required properties by allocating successively numbers to the subsets that always obey the rules. The number 1 must go to $A_{1}$; we show that for every subsequent number we have exactly two possibilities. Actually, while $A_{2}$ and $A_{3}$ are both empty, every successive number can enter either $A_{1}$ or $A_{2}$. Further, when $A_{2}$ is no longer empty, we use induction on the number to be placed, denote it by $m$ : if $m$ can enter $A_{i}$ or $A_{j}$ but not $A_{k}$, and it enters $A_{i}$, then $m+1$ can be placed in $A_{i}$ or $A_{k}$, but not in $A_{j}$. The induction step is finished. This immediately gives us that the final answer is $2^{n-1}$.


12. (POL 5) Given a nonequilateral triangle $A B C$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A^{\prime} B^{\prime} C^{\prime}$ (the vertices listed counterclockwise) for which the triples of points $A, B^{\prime}, C^{\prime} ; A^{\prime}, B, C^{\prime}$; and $A^{\prime}, B^{\prime}, C$ are collinear.

Solution: 12. Here all angles will be oriented and measured counterclockwise.  Note that $\measuredangle C A^{\prime} B=\measuredangle A B^{\prime} C=$ $\measuredangle B C^{\prime} A=\pi / 3$. Let $a^{\prime}, b^{\prime}, c^{\prime}$ denote respectively the inner bisectors of angles $A^{\prime}, B^{\prime}, C^{\prime}$ in triangle $A^{\prime} B^{\prime} C^{\prime}$. The lines $a^{\prime}, b^{\prime}, c^{\prime}$ meet at the centroid $X$ of $A^{\prime} B^{\prime} C^{\prime}$, and $\measuredangle\left(a^{\prime}, b^{\prime}\right)=$ $\measuredangle\left(b^{\prime}, c^{\prime}\right)=\measuredangle\left(c^{\prime}, a^{\prime}\right)=2 \pi / 3$. Now let $K, L, M$ be the points such that $K B=K C, L C=L A, M A=M B$, and $\measuredangle B K C=\measuredangle C L A=\measuredangle A M B=$ $2 \pi / 3$, and let $C_{1}, C_{2}, C_{3}$ be the circles circumscribed about triangles ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-503.jpg?height=511&width=491&top_left_y=1136&top_left_x=834) $B K C, C L A$, and $A M B$ respectively. These circles are characterized by $C_{1}=\{Z \mid \measuredangle B Z C=2 \pi / 3\}$, etc.; hence we deduce that they meet at a point $P$ such that $\measuredangle B P C=\measuredangle C P A=\measuredangle A P B=2 \pi / 3$ (Torricelli's point). Points $A^{\prime}, B^{\prime}, C^{\prime}$ run over $C_{1} \backslash\{P\}, C_{2} \backslash\{P\}, C_{3} \backslash\{P\}$ respectively. As for $a^{\prime}, b^{\prime}, c^{\prime}$, we see that $K \in a^{\prime}, L \in b^{\prime}, M \in c^{\prime}$, and also that they can take all possible directions except $K P, L P, M P$ respectively (if $K=P, K P$ is assumed to be the corresponding tangent at $K$ ). Then, since $\measuredangle K X L=$ $2 \pi / 3, X$ runs over the circle defined by $\{Z \mid \measuredangle K Z L=2 \pi / 3\}$, without $P$. But analogously, $X$ runs over the circle $\{Z \mid \measuredangle L Z M=2 \pi / 3\}$, from which we can conclude that these two circles are the same, both equal to the circumcircle of $K L M$, and consequently also that triangle $K L M$ is equilateral (which is, anyway, a well-known fact). Therefore, the locus of the points $X$ is the circumcircle of $K L M$ minus point $P$.


13. (GDR 2) ${ }^{\mathrm{IMO} 5}$ Is it possible to put 1987 points in the Euclidean plane such that the distance between each pair of points is irrational and each three points determine a nondegenerate triangle with rational area?

Solution: 13. We claim that the points $P_{i}\left(i, i^{2}\right), i=1,2, \ldots, 1987$, satisfy the conditions. In fact: (i) $\overline{P_{i} P_{j}}=\sqrt{(i-j)^{2}+\left(i^{2}-j^{2}\right)^{2}}=|i-j| \sqrt{1+(i+j)^{2}}$.  It is known that for each positive integer $n, \sqrt{n}$ is either an integer or an irrational number. Since $i+j<\sqrt{1+(i+j)^{2}}<i+j+1$, $\sqrt{1+(i+j)^{2}}$ is not an integer, it is irrational, and so is $\overline{P_{i} P_{j}}$. (ii) The area $A$ of the triangle $P_{i} P_{j} P_{k}$, for distinct $i, j, k$, is given by  $$ \begin{aligned} A & =\left|\frac{i^{2}+j^{2}}{2}(i-j)+\frac{j^{2}+k^{2}}{2}(j-k)+\frac{k^{2}+i^{2}}{2}(k-i)\right| \\ & =\left|\frac{(i-j)(j-k)(k-i)}{2}\right| \in \mathbb{Q} \backslash\{0\}, \end{aligned} $$  also showing that this triangle is nondegenerate.


14. (FRG 1) How many words with $n$ digits can be formed from the alphabet $\{0,1,2,3,4\}$, if neighboring digits must differ by exactly one?

Solution: 14. Let $x_{n}$ be the total number of counted words of length $n$, and $y_{n}, z_{n}, u_{n}$, $z_{n}, y_{n}$ the numbers of counted words of length $n$ starting with $0,1,2,3,4$, respectively (indeed, by symmetry, words starting with 0 are equally numbered as those starting with 4 , etc.). We have the clear relations  $$ \begin{array}{ll} \text { (1) } y_{n}=z_{n-1} ; & \text { (2) } z_{n}=y_{n-1}+u_{n-1} \\ \text { (3) } u_{n}=2 z_{n-1} ; & \text { (4) } x_{n}=2 y_{n}+2 z_{n}+u_{n} \end{array} $$  From (1), (2), and (3) we get $z_{n}=z_{n-2}+2 z_{n-2}=3 z_{n-2}$, with $z_{1}=1$, $z_{2}=2$, which gives  $$ z_{2 n}=2 \cdot 3^{n-1}, \quad z_{2 n+1}=3^{n} $$  Then (1), (3), and (4) obviously imply  $$ \begin{array}{ll} y_{2 n}=3^{n-1}, & y_{2 n+1}=2 \cdot 3^{n-1} \\ u_{2 n}=2 \cdot 3^{n-1}, & u_{2 n+1}=4 \cdot 3^{n-1} \\ x_{2 n}=8 \cdot 3^{n-1}, & x_{2 n+1}=14 \cdot 3^{n-1} \end{array} $$  with the initial number $x_{1}=5$.


15. (FRG 2) $)^{\mathrm{IMO} 3}$ Suppose $x_{1}, x_{2}, \ldots, x_{n}$ are real numbers with $x_{1}^{2}+x_{2}^{2}+$ $\cdots+x_{n}^{2}=1$. Prove that for any integer $k>1$ there are integers $e_{i}$ not all 0 and with $\left|e_{i}\right|<k$ such that  $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$

Solution: 15. Since $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, we get by the Cauchy-Schwarz inequality  $$ \left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right| \leq \sqrt{n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)}=\sqrt{n} $$  Hence all $k^{n}$ sums of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$, with $e_{i} \in$ $\{0,1,2, \ldots, k-1\}$, must lie in some closed interval $\Im$ of length $(k-1) \sqrt{n}$. This interval can be covered with $k^{n}-1$ closed subintervals of length $\frac{k-1}{k^{n}-1} \sqrt{n}$. By the pigeonhole principle there must be two of these sums lying in the same subinterval. Their difference, which is of the form $e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}$ where $e_{i} \in\{0, \pm 1, \ldots, \pm(k-1)\}$, satisfies  $$ \left|e_{1} x_{1}+e_{2} x_{2}+\cdots+e_{n} x_{n}\right| \leq \frac{(k-1) \sqrt{n}}{k^{n}-1} $$


16. (FRG 3) ${ }^{\mathrm{IMO1}}$ Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_{n}(k)$. Prove: (a) $\sum_{k=0}^{n} k p_{n}(k)=n$ !; (b) $\sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=n$ !.

Solution: 16. We assume that $S=\{1,2, \ldots, n\}$, and use the obvious fact  $$ \sum_{k=0}^{n} p_{n}(k)=n! $$  (a) To each permutation $\pi$ of $S$ we assign an $n$-vector $\left(e_{1}, e_{2}, \ldots, e_{n}\right)$, where $e_{i}$ is 1 if $i$ is a fixed point of $\pi$, and 0 otherwise. Since exactly $p_{n}(k)$ of the assigned vectors contain exactly $k$ " 1 "s, the considered sum $\sum_{k=0}^{n} k p_{n}(k)$ counts all the " 1 "s occurring in all the $n$ ! assigned vectors. But for each $i, 1 \leq i \leq n$, there are exactly $(n-1)$ ! permutations that fix $i$; i.e., exactly $(n-1)$ ! of the vectors have $e_{i}=1$. Therefore the total number of " 1 "s is $n \cdot(n-1)$ ! = n!, implying  $$ \sum_{k=0}^{n} k p_{n}(k)=n! $$  (b) In this case, to each permutation $\pi$ of $S$ we assign a vector $\left(d_{1}, \ldots, d_{n}\right)$ instead, with $d_{i}=k$ if $i$ is a fixed point of $\pi$, and $d_{i}=0$ otherwise, where $k$ is the number of fixed points of $\pi$. Let us count the sum $Z$ of all components $d_{i}$ for all the $n$ ! permutations. There are $p_{n}(k)$ such vectors with exactly $k$ components equal to $k$, and sums of components equal to $k^{2}$. Thus, $Z=\sum_{k=0}^{n} k^{2} p_{n}(k)$. On the other hand, we may first calculate the sum of all components $d_{i}$ for fixed $i$. In fact, the value $d_{i}=k>0$ will occur exactly $p_{n-1}(k-1)$ times, so that the sum of the $d_{i}$ 's is $\sum_{k=1}^{n} k p_{n-1}(k-1)=\sum_{k=0}^{n-1}(k+$ 1) $p_{n-1}(k)=2(n-1)$ !. Summation over $i$ yields  $$ Z=\sum_{k=0}^{n} k^{2} p_{n}(k)=2 n!. $$  From (0), (1), and (2), we conclude that  $$ \sum_{k=0}^{n}(k-1)^{2} p_{n}(k)=\sum_{k=0}^{n} k^{2} p_{n}(k)-2 \sum_{k=0}^{n} k p_{n}(k)+\sum_{k=0}^{n} p_{n}(k)=n! $$  Remark. Only the first part of this problem was given on the IMO.


17. (ROM 1) Prove that there exists a four-coloring of the set $M=$ $\{1,2, \ldots, 1987\}$ such that any arithmetic progression with 10 terms in the set $M$ is not monochromatic. Alternative formulation. Let $M=\{1,2, \ldots, 1987\}$. Prove that there is a function $f: M \rightarrow\{1,2,3,4\}$ that is not constant on every set of 10 terms from $M$ that form an arithmetic progression.

Solution: 17. The number of 4 -colorings of the set $M$ is equal to $4^{1987}$. Let $A$ be the number of arithmetic progressions in $M$ with 10 terms. The number of colorings containing a monochromatic arithmetic progression with 10 terms is less than $4 A \cdot 4^{1977}$. So, if $A<4^{9}$, then there exist 4 -colorings with the required property.  Now we estimate the value of $A$. If the first term of a 10-term progression is $k$ and the difference is $d$, then $1 \leq k \leq 1978$ and $d \leq\left[\frac{1987-k}{9}\right]$; hence  $$ A=\sum_{k=1}^{1978}\left[\frac{1987-k}{9}\right]<\frac{1986+1985+\cdots+9}{9}=\frac{1995 \cdot 1978}{18}<4^{9} $$


18. (ROM 4) For any integer $r \geq 1$, determine the smallest integer $h(r) \geq 1$ such that for any partition of the set $\{1,2, \ldots, h(r)\}$ into $r$ classes, there are integers $a \geq 0,1 \leq x \leq y$, such that $a+x, a+y, a+x+y$ belong to the same class.

Solution: 18. Note first that the statement that some $a+x, a+y, a+x+y$ belong to a class $C$ is equivalent to the following statement: (1) There are positive integers $p, q \in C$ such that $p<q \leq 2 p$.  Indeed, given $p, q$, take simply $x=y=q-p, a=2 p-q$; conversely, if $a, x, y(x \leq y)$ exist such that $a+x, a+y, a+x+y \in C$, take $p=a+y$, $q=a+x+y$ : clearly, $p<q \leq 2 p$. We will show that $h(r)=2 r$. Let $\{1,2, \ldots, 2 r\}=C_{1} \cup C_{2} \cup \cdots \cup C_{r}$ be an arbitrary partition into $r$ classes. By the pigeonhole principle, two among the $r+1$ numbers $r, r+1, \ldots, 2 r$ belong to the same class, say $i, j \in C_{k}$. If w.l.o.g. $i<j$, then obviously $i<j \leq 2 i$, and so by (1) this $C_{k}$ has the required property. On the other hand, we consider the partition  $$ \{1,2, \ldots, 2 r-t\}=\bigcup_{k=1}^{r-t}\{k, k+r\} \cup\{r-t+1\} \cup \cdots \cup\{r\} $$  and prove that (1), and thus also the required property, does not hold. In fact, none of the classes in the partition contains $p$ and $q$ with $p<q \leq 2 p$, because $k+r>2 k$.


19. (USS 2) Let $\alpha, \beta, \gamma$ be positive real numbers such that $\alpha+\beta+\gamma<\pi$, $\alpha+\beta>\gamma, \beta+\gamma>\alpha, \gamma+\alpha>\beta$. Prove that with the segments of lengths $\sin \alpha, \sin \beta, \sin \gamma$ we can construct a triangle and that its area is not greater than  $$ \frac{1}{8}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) $$

Solution: 19. The facts given in the problem allow us to draw a triangular pyramid with angles $2 \alpha, 2 \beta, 2 \gamma$ at the top and lateral edges of length $1 / 2$. At the base there is a triangle whose side lengths are exactly $\sin \alpha, \sin \beta, \sin \gamma$. The area of this triangle does not exceed the sum of areas of the lateral sides, which equals $(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) / 8$.


2. (USA 3) At a party attended by $n$ married couples, each person talks to everyone else at the party except his or her spouse. The conversations involve sets of persons or cliques $C_{1}, C_{2}, \ldots, C_{k}$ with the following property: no couple are members of the same clique, but for every other pair of persons there is exactly one clique to which both members belong. Prove that if $n \geq 4$, then $k \geq 2 n$.

Solution: 2. Let $d_{i}$ denote the number of cliques of which person $i$ is a member. Clearly $d_{i} \geq 2$. We now distinguish two cases: (i) For some $i, d_{i}=2$. Suppose that $i$ is a member of two cliques, $C_{p}$ and $C_{q}$. Then $\left|C_{p}\right|=\left|C_{q}\right|=n$, since for each couple other than $i$ and his/her spouse, one member is in $C_{p}$ and one in $C_{q}$. There are thus $(n-1)(n-2)$ pairs $(r, s)$ of nonspouse persons distinct from $i$, where $r \in C_{p}, s \in C_{q}$. We observe that each such pair accounts for a different clique. Otherwise, we find two members of $C_{p}$ or $C_{q}$ who belong to one other clique. It follows that $k \geq 2+(n-1)(n-2) \geq 2 n$ for $n \geq 4$. (ii) For every $i, d_{i} \geq 3$. Suppose that $k<2 n$. For $i=1,2, \ldots, 2 n$ assign to person $i$ an indeterminant $x_{i}$, and for $j=1,2, \ldots, k$ set $y=\sum_{i \in C_{j}} x_{i}$. From linear algebra, we know that if $k<2 n$, then there exist $x_{1}, x_{2}, \ldots, x_{2 n}$, not all zero, such that $y_{1}=y_{2}=\cdots=y_{k}=0$. On the other hand, suppose that $y_{1}=y_{2}=\cdots=y_{k}=0$. Let $M$ be the set of the couples and $M^{\prime}$ the set of all other pairs of persons. Then  $$ \begin{aligned} 0 & =\sum_{j=1}^{k} y_{j}^{2}=\sum_{i=1}^{2 n} d_{i} x_{i}^{2}+2 \sum_{(i, j) \in M^{\prime}} x_{i} x_{j} \\ & =\sum_{i=1}^{2 n}\left(d_{i}-2\right) x_{i}^{2}+\left(x_{1}+x_{2}+\cdots+x_{2 n}\right)^{2}+\sum_{(i, j) \in M}\left(x_{i}-x_{j}\right)^{2} \\ & \geq \sum_{i=1}^{2 n} x_{i}^{2}>0 \end{aligned} $$  if not all $x_{1}, x_{2}, \ldots, x_{2 n}$ are zero, which is a contradiction. Hence $k \geq$ $2 n$. Remark. The condition $n \geq 4$ is essential. For a party attended by 3 couples $\{(1,4),(2,5),(3,6)\}$, there is a collection of 4 cliques satisfying the conditions: $\{(1,2,3),(3,4,5),(5,6,1),(2,4,6)\}$.


20. (USS 3) ${ }^{\text {IMO6 }}$ Let $f(x)=x^{2}+x+p, p \in \mathbb{N}$. Prove that if the numbers $f(0), f(1), \ldots, f([\sqrt{p / 3}])$ are primes, then all the numbers $f(0), f(1), \ldots$, $f(p-2)$ are primes.

Solution: 20. Let $y$ be the smallest nonnegative integer with $y \leq p-2$ for which $f(y)$ is a composite number. Denote by $q$ the smallest prime divisor of $f(y)$. We claim that $y<q$. Suppose the contrary, that $y \geq q$. Let $r$ be a positive integer such that $y \equiv r(\bmod q)$. Then $f(y) \equiv f(r) \equiv 0(\bmod q)$, and since $q \leq y \leq p-2 \leq$ $f(r)$, we conclude that $q \mid f(r)$, which is a contradiction to the minimality of $y$. Now, we will prove that $q>2 y$. Suppose the contrary, that $q \leq 2 y$. Since  $$ f(y)-f(x)=(y-x)(y+x+1) $$  we observe that $f(y)-f(q-1-y)=(2 y-q+1) q$, from which it follows that $f(q-1-y)$ is divisible by $q$. But by the assumptions, $q-1-y<y$, implying that $f(q-1-y)$ is prime and therefore equal to $q$. This is impossible, because  $$ f(q-1-y)=(q-1-y)^{2}+(q-1-y)+p>q+p-y-1 \geq q . $$  Therefore $q \geq 2 y+1$. Now, since $f(y)$, being composite, cannot be equal to $q$, and $q$ is its smallest prime divisor, we obtain that $f(y) \geq q^{2}$. Consequently,  $$ y^{2}+y+p \geq q^{2} \geq(2 y+1)^{2}=4 y^{2}+4 y+1 \Rightarrow 3\left(y^{2}+y\right) \leq p-1 $$  and from this we easily conclude that $y<\sqrt{p / 3}$, which contradicts the condition of the problem. In this way, all the numbers  $$ f(0), f(1), \ldots, f(p-2) $$  must be prime.


21. (USS 4) ${ }^{\mathrm{IMO} 2}$ The prolongation of the bisector $A L(L \in B C)$ in the acuteangled triangle $A B C$ intersects the circumscribed circle at point $N$. From point $L$ to the sides $A B$ and $A C$ are drawn the perpendiculars $L K$ and $L M$ respectively. Prove that the area of the triangle $A B C$ is equal to the area of the quadrilateral $A K N M$.

Solution: 21. Let $P$ be the second point of intersection of segment $B C$ and the circle circumscribed about quadrilateral $A K L M$. Denote by $E$ the intersection point of the lines $K N$ and $B C$ and by $F$ the intersection point of the lines $M N$ and $B C$. Then $\angle B C N=\angle B A N$ and $\angle M A L=$ $\angle M P L$, as angles on the same arc. Since $A L$ is a bisector, $\angle B C N=$ $\angle B A L=\angle M A L=\angle M P L$, and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-507.jpg?height=452&width=414&top_left_y=813&top_left_x=866) consequently $P M \| N C$. Similarly we prove $K P \| B N$. Then the quadrilaterals $B K P N$ and $N P M C$ are trapezoids; hence  $$ S_{B K E}=S_{N P E} \quad \text { and } \quad S_{N P F}=S_{C M F} $$  Therefore $S_{A B C}=S_{A K N M}$.


22. (VIE 3) ${ }^{\mathrm{IMO} 4}$ Does there exist a function $f: \mathbb{N} \rightarrow \mathbb{N}$, such that $f(f(n))=$ $n+1987$ for every natural number $n$ ?

Solution: 22. Suppose that there exists such function $f$. Then we obtain  $$ f(n+1987)=f(f(f(n)))=f(n)+1987 \quad \text { for all } n \in \mathbb{N} $$  and from here, by induction, $f(n+1987 t)=f(n)+1987 t$ for all $n, t \in \mathbb{N}$. Further, for any $r \in\{0,1, \ldots, 1986\}$, let $f(r)=1987 k+l, k, l \in \mathbb{N}$, $l \leq 1986$. We have  $$ r+1987=f(f(r))=f(l+1987 k)=f(l)+1987 k, $$  and consequently there are two possibilities: (i) $k=1 \Rightarrow f(r)=l+1987$ and $f(l)=r$; (ii) $k=0 \Rightarrow f(r)=l$ and $f(l)=r+1987$; in both cases, $r \neq l$. In this way, the set $\{0,1, \ldots, 1986\}$ decomposes to pairs $\{a, b\}$ such that  $$ f(a)=b \text { and } f(b)=a+1987, \quad \text { or } \quad f(b)=a \text { and } f(a)=b+1987 . $$  But the set $\{0,1, \ldots, 1986\}$ has an odd number of elements, and cannot be decomposed into pairs. Contradiction.


23. (YUG 2) Prove that for every natural number $k(k \geq 2)$ there exists an irrational number $r$ such that for every natural number $m$,  $$ \left[r^{m}\right] \equiv-1 \quad(\bmod k) $$  Remark. An easier variant: Find $r$ as a root of a polynomial of second degree with integer coefficients.

Solution: 23. If we prove the existence of $p, q \in \mathbb{N}$ such that the roots $r, s$ of  $$ f(x)=x^{2}-k p \cdot x+k q=0 $$  are irrational real numbers with $0<s<1$ (and consequently $r>1$ ), then we are done, because from $r+s, r s \equiv 0(\bmod k)$ we get $r^{m}+s^{m} \equiv 0$ $(\bmod k)$, and $0<s^{m}<1$ yields the assertion. To prove the existence of such natural numbers $p$ and $q$, we can take them such that $f(0)>0>f(1)$, i.e.,  $$ k q>0>k(q-p)+1 \quad \Rightarrow \quad p>q>0 $$  The irrationality of $r$ can be obtained by taking $q=p-1$, because the discriminant $D=(k p)^{2}-4 k p+4 k$, for $(k p-2)^{2}<D<(k p-1)^{2}$, is not a perfect square for $p \geq 2$.


3. (FIN 3) Does there exist a second-degree polynomial $p(x, y)$ in two variables such that every nonnegative integer $n$ equals $p(k, m)$ for one and only one ordered pair $(k, m)$ of nonnegative integers?

Solution: 3. The answer: yes. Set  $$ p(k, m)=k+[1+2+\cdots+(k+m)]=\frac{(k+m)^{2}+3 k+m}{2} . $$  It is obviously of the desired type.


4. (FRA 5) Let $A B C D E F G H$ be a parallelepiped with $A E\|B F\| C G \| D H$. Prove the inequality  $$ A F+A H+A C \leq A B+A D+A E+A G $$  In what cases does equality hold?

Solution: 4. Setting $x_{1}=\overrightarrow{A B}, x_{2}=\overrightarrow{A D}, x_{3}=\overrightarrow{A E}$, we have to prove that  $$ \left\|x_{1}+x_{2}\right\|+\left\|x_{2}+x_{3}\right\|+\left\|x_{3}+x_{1}\right\| \leq\left\|x_{1}\right\|+\left\|x_{2}\right\|+\left\|x_{3}\right\|+\left\|x_{1}+x_{2}+x_{3}\right\| . $$  We have  $$ \begin{aligned} & \left(\left\|x_{1}\right\|+\left\|x_{2}\right\|+\left\|x_{3}\right\|\right)^{2}-\left\|x_{1}+x_{2}+x_{3}\right\|^{2} \\ & \quad=2 \sum_{1 \leq i<j \leq 3}\left(\left\|x_{i}\right\|\left\|x_{j}\right\|-\left\langle x_{i}, x_{j}\right\rangle\right)=\sum_{1 \leq i<j \leq 3}\left[\left(\left\|x_{i}\right\|+\left\|x_{j}\right\|\right)^{2}-\left\|x_{i}+x_{j}\right\|^{2}\right] \\ & \quad=\sum_{1 \leq i<j \leq 3}\left(\left\|x_{i}\right\|+\left\|x_{j}\right\|+\left\|x_{i}+x_{j}\right\|\right)\left(\left\|x_{i}\right\|+\left\|x_{j}\right\|-\left\|x_{i}+x_{j}\right\|\right) . \end{aligned} $$  The following two inequalities are obvious:  $$ \begin{array}{r} \left\|x_{i}\right\|+\left\|x_{j}\right\|-\left\|x_{i}+x_{j}\right\| \geq 0 \\ \left\|x_{i}\right\|+\left\|x_{j}\right\|+\left\|x_{i}+x_{j}\right\| \leq\left\|x_{1}\right\|+\left\|x_{2}\right\|+\left\|x_{3}\right\|+\left\|x_{1}+x_{2}+x_{3}\right\| \end{array} $$  It follows that  $$ \begin{aligned} & \left(\left\|x_{1}\right\|+\left\|x_{2}\right\|+\left\|x_{3}\right\|\right)^{2}-\left\|x_{1}+x_{2}+x_{3}\right\|^{2} \\ & \quad \leq\left(\sum_{i=1}^{3}\left\|x_{i}\right\|+\left\|\sum_{i=1}^{3} x_{i}\right\|\right)\left(2 \sum_{i=1}^{3}\left\|x_{i}\right\|-\sum_{1 \leq i<j \leq 3}\left\|x_{i}+x_{j}\right\|\right) \end{aligned} $$  and dividing by the positive number $\sum_{i=1}^{3}\left\|x_{i}\right\|+\left\|\sum_{i=1}^{3} x_{i}\right\|$ we obtain  $$ \sum_{i=1}^{3}\left\|x_{i}\right\|-\left\|\sum_{i=1}^{3} x_{i}\right\| \leq 2 \sum_{i=1}^{3}\left\|x_{i}\right\|-\sum_{1 \leq i<j \leq 3}\left\|x_{i}+x_{j}\right\| $$  The inequality is proven. Let us analyze the cases of equality. If one of the vectors is null, then equality obviously holds. Suppose that $x_{i} \neq 0$, $i=1,2,3$. For every $i, j$, at least one of (1) and (2) is equality. Equality in (1) holds if and only if $x_{i}$ and $x_{j}$ are collinear with the same direction, while in (2) it holds if and only if $-x_{k}$ and $x_{1}+x_{2}+x_{3}$ are collinear with the same direction. If not all the vectors are collinear, then there are at least two distinct pairs $x_{i}, x_{j}, i<j$, for which (2) is an equality, so at least two of $x_{i}$ are collinear with $x_{1}+x_{2}+x_{3}$, but then so is the third; hence, the sum $x_{1}+x_{2}+x_{3}$ must be 0 . Thus the cases of equality are (a) the vectors are collinear with the same direction; (b) the vectors are collinear, two of them have the same direction, say $x_{i}, x_{j}$, and $\left\|x_{k}\right\| \geq\left\|x_{i}\right\|+\left\|x_{j}\right\|$; (c) one of the vectors is 0 ; (d) their sum is 0 .  Second solution. The following technique, although not quite elementary, is often used to effectively reduce geometric inequalities of first degree, like this one, to the one-dimensional case. Let $\sigma$ be a fixed sphere with center $O$. For an arbitrary segment $d$ in space, and any line $l$, we denote by $\pi_{l}(d)$ the length of the projection of $d$ onto $l$. Consider the integral of lengths of these projections on all possible directions of $O P$, with $P$ moving on the sphere: $\int_{\sigma} \pi_{O P}(d) d \sigma$. It is clear that this value depends only on the length of $d$ (because of symmetry); hence  $$ \int_{\sigma} \pi_{O P} d \sigma=c \cdot|d| \quad \text { for some constant } c \neq 0 $$  Notice that by the one-dimensional case, for any point $P \in \sigma$,  $$ \begin{gathered} \pi_{O P}\left(x_{1}\right)+\pi_{O P}\left(x_{2}\right)+\pi_{O P}\left(x_{3}\right)+\pi_{O P}\left(x_{1}+x_{2}+x_{3}\right) \\ \geq \pi_{O P}\left(x_{1}+x_{2}\right)+\pi_{O P}\left(x_{1}+x_{3}\right)+\pi_{O P}\left(x_{2}+x_{3}\right) . \end{gathered} $$  By integration on $\sigma$, using (1), we obtain $c\left(\left\|x_{1}\right\|+\left\|x_{2}\right\|+\left\|x_{3}\right\|+\left\|x_{1}+x_{2}+x_{3}\right\|\right) \geq c\left(\left\|x_{1}+x_{2}\right\|+\left\|x_{1}+x_{3}\right\|+\left\|x_{2}+x_{3}\right\|\right)$.


5. (GBR 1) Find, with proof, the point $P$ in the interior of an acute-angled triangle $A B C$ for which $B L^{2}+C M^{2}+A N^{2}$ is a minimum, where $L, M, N$ are the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively.

Solution: 5. Assuming the notation $a=\overline{B C}, b=\overline{A C}, c=\overline{A B} ; x=\overline{B L}, y=\overline{C M}$, $z=\overline{A N}$, from the Pythagorean theorem we obtain  $$ \begin{aligned} (a-x)^{2}+(b-y)^{2} & +(c-z)^{2}=x^{2}+y^{2}+z^{2} \\ & =\frac{x^{2}+(a-x)^{2}+y^{2}+(b-y)^{2}+z^{2}+(c-z)^{2}}{2} \end{aligned} $$  Since $x^{2}+(a-x)^{2}=a^{2} / 2+(a-2 x)^{2} / 2 \geq a^{2} / 2$ and similarly $y^{2}+(b-y)^{2} \geq$ $b^{2} / 2$ and $z^{2}+(c-z)^{2} \geq c^{2} / 2$, we get  $$ x^{2}+y^{2}+z^{2} \geq \frac{a^{2}+b^{2}+c^{2}}{4} $$  Equality holds if and only if $P$ is the circumcenter of the triangle $A B C$, i.e., when $x=a / 2, y=b / 2, z=c / 2$.


6. (GRE 4) Show that if $a, b, c$ are the lengths of the sides of a triangle and if $2 S=a+b+c$, then  $$ \frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geq\left(\frac{2}{3}\right)^{n-2} S^{n-1}, \quad n \geq 1 $$

Solution: 6. Suppose w.l.o.g. that $a \geq b \geq c$. Then $1 /(b+c) \geq 1 /(a+c) \geq 1 /(a+b)$. Chebyshev's inequality yields  $$ \frac{a^{n}}{b+c}+\frac{b^{n}}{a+c}+\frac{c^{n}}{a+b} \geq \frac{1}{3}\left(a^{n}+b^{n}+c^{n}\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right) . $$  By the Cauchy-Schwarz inequality we have  $$ 2(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right) \geq 9 $$  and the mean inequality yields $\left(a^{n}+b^{n}+c^{n}\right) / 3 \geq[(a+b+c) / 3]^{n}$. We obtain from (1) that  $$ \begin{aligned} \frac{a^{n}}{b+c}+\frac{b^{n}}{a+c}+\frac{c^{n}}{a+b} & \geq\left(\frac{a+b+c}{3}\right)^{n}\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right) \\ & \geq \frac{3}{2}\left(\frac{a+b+c}{3}\right)^{n-1}=\left(\frac{2}{3}\right)^{n-2} S^{n-1} . \end{aligned} $$


7. (NET 1) Given five real numbers $u_{0}, u_{1}, u_{2}, u_{3}, u_{4}$, prove that it is always possible to find five real numbers $v_{0}, v_{1}, v_{2}, v_{3}, v_{4}$ that satisfy the following conditions: (i) $u_{i}-v_{i} \in \mathbb{N}$. (ii) $\sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2}<4$.

Solution: 7. For all real numbers $v$ the following inequality holds:  $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} $$  Indeed,  $$ \begin{aligned} \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} & =\sum_{0 \leq i<j \leq 4}\left[\left(v_{i}-v\right)-\left(v_{j}-v\right)\right]^{2} \\ & =5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2}-\left(\sum_{i=0}^{4}\left(v_{i}-v\right)\right)^{2} \leq 5 \sum_{i=0}^{4}\left(v_{i}-v\right)^{2} . \end{aligned} $$  Let us first take $v_{i}$ 's, satisfying condition (1), so that w.l.o.g. $v_{0} \leq v_{1} \leq$ $v_{2} \leq v_{3} \leq v_{4} \leq 1+v_{0}$. Defining $v_{5}=1+v_{0}$, we see that one of the differences $v_{j+1}-v_{j}, j=0, \ldots, 4$, is at most $1 / 5$. Take $v=\left(v_{j+1}+v_{j}\right) / 2$, and then place the other three $v_{j}$ 's in the segment $[v-1 / 2, v+1 / 2]$. Now we have $\left|v-v_{j}\right| \leq 1 / 10,\left|v-v_{j+1}\right| \leq 1 / 10$, and $\left|v-v_{k}\right| \leq 1 / 2$, for any $k$ different from $j, j+1$. The $v_{i}$ 's thus obtained have the required property. In fact, using the inequality (1), we obtain  $$ \sum_{0 \leq i<j \leq 4}\left(v_{i}-v_{j}\right)^{2} \leq 5\left(2\left(\frac{1}{10}\right)^{2}+3\left(\frac{1}{2}\right)^{2}\right)=3.85<4 $$  Remark. The best possible estimate for the right-hand side is 2 .


8. (HUN 1) (a) Let $(m, k)=1$. Prove that there exist integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}, \ldots, b_{k}$ such that each product $a_{i} b_{j}(i=1,2, \ldots, m ; j=$ $1,2, \ldots, k)$ gives a different residue when divided by $m k$. (b) Let $(m, k)>1$. Prove that for any integers $a_{1}, a_{2}, \ldots, a_{m}$ and $b_{1}, b_{2}$, $\ldots, b_{k}$ there must be two products $a_{i} b_{j}$ and $a_{s} b_{t}((i, j) \neq(s, t))$ that give the same residue when divided by $m k$.

Solution: 8. (a) Consider  $$ a_{i}=i k+1, \quad i=1,2, \ldots, m ; \quad b_{j}=j m+1, \quad j=1,2, \ldots, k $$  Assume that $m k \mid a_{i} b_{j}-a_{s} b_{t}=(i k+1)(j m+1)-(s k+1)(t m+1)=$ $k m(i j-s t)+m(j-t)+k(i-s)$. Since $m$ divides this sum, we get that $m \mid k(i-s)$, or, together with $\operatorname{gcd}(k, m)=1$, that $i=s$. Similarly $j=t$, which proves part (a). (b) Suppose the opposite, i.e., that all the residues are distinct. Then the residue 0 must also occur, say at $a_{1} b_{1}: m k \mid a_{1} b_{1}$; so, for some $a^{\prime}$ and $b^{\prime}, a^{\prime}\left|a_{1}, b^{\prime}\right| b_{1}$, and $a^{\prime} b^{\prime}=m k$. Assuming that for some $i, s \neq i$, $a^{\prime} \mid a_{i}-a_{s}$, we obtain $m k=a^{\prime} b^{\prime} \mid a_{i} b_{1}-a_{s} b_{1}$, a contradiction. This shows that $a^{\prime} \geq m$ and similarly $b^{\prime} \geq k$, and thus from $a^{\prime} b^{\prime}=m k$ we have $a^{\prime}=m, b^{\prime}=k$. We also get (1): all $a_{i}$ 's give distinct residues modulo $m=a^{\prime}$, and all $b_{j}$ 's give distinct residues modulo $k=b^{\prime}$. Now let $p$ be a common prime divisor of $m$ and $k$. By $(*)$, exactly $\frac{p-1}{p} m$ of $a_{i}$ 's and exactly $\frac{p-1}{p} k$ of $b_{j}$ 's are not divisible by $p$. Therefore there are precisely $\frac{(p-1)^{2}}{p^{2}} m k$ products $a_{i} b_{j}$ that are not divisible by $p$, although from the assumption that they all give distinct residues it follows that the number of such products is $\frac{p-1}{p} m k \neq \frac{(p-1)^{2}}{p^{2}} m k$. We have arrived at a contradiction, thus proving (b).


9. (HUN 2) Does there exist a set $M$ in usual Euclidean space such that for every plane $\lambda$ the intersection $M \cap \lambda$ is finite and nonempty?

Solution: 9. The answer is yes. Consider the curve  $$ C=\left\{(x, y, z) \mid x=t, y=t^{3}, z=t^{5}, \quad t \in \mathbb{R}\right\} $$  Any plane defined by an equation of the form $a x+b y+c z+d=0$ intersects the curve $C$ at points $\left(t, t^{3}, t^{5}\right)$ with $t$ satisfying $c t^{5}+b t^{3}+a t+d=0$. This last equation has at least one but only finitely many solutions.


1. (BUL 1) An integer sequence is defined by  $$ a_{n}=2 a_{n-1}+a_{n-2} \quad(n>1), \quad a_{0}=0, \quad a_{1}=1 $$  Prove that $2^{k}$ divides $a_{n}$ if and only if $2^{k}$ divides $n$.

Solution: 1. Assume that $p$ and $q$ are real and $b_{0}, b_{1}, b_{2}, \ldots$ is a sequence such that $b_{n}=p b_{n-1}+q b_{n-2}$ for all $n>1$. From the equalities $b_{n}=p b_{n-1}+q b_{n-2}$, $b_{n+1}=p b_{n}+q b_{n-1}, b_{n+2}=p b_{n+1}+q b_{n}$, eliminating $b_{n+1}$ and $b_{n-1}$ we obtain that $b_{n+2}=\left(p^{2}+2 q\right) b_{n}-q^{2} b_{n-2}$. So the sequence $b_{0}, b_{2}, b_{4}, \ldots$ has the property  $$ b_{2 n}=P b_{2 n-2}+Q b_{2 n-4}, \quad P=p^{2}+2 q, \quad Q=-q^{2} . $$  We shall solve the problem by induction. The sequence $a_{n}$ has $p=2$, $q=1$, and hence $P=6, Q=-1$. Let $k=1$. Then $a_{0}=0, a_{1}=1$, and $a_{n}$ is of the same parity as $a_{n-2}$; i.e., it is even if and only if $n$ is even. Let $k \geq 1$. We assume that for $n=2^{k} m$, the numbers $a_{n}$ are divisible by $2^{k}$, but divisible by $2^{k+1}$ if and only if $m$ is even. We assume also that the sequence $c_{0}, c_{1}, \ldots$, with $c_{m}=a_{m \cdot 2^{k}}$, satisfies the condition $c_{n}=$ $p c_{n-1}-c_{n-2}$, where $p \equiv 2(\bmod 4)($ for $k=1$ it is true). We shall prove the same statement for $k+1$. According to (1), $c_{2 n}=P c_{2 n-2}-c_{2 n-4}$, where $P=p^{2}-2$. Obviously $P \equiv 2(\bmod 4)$. Since $P=4 s+2$ for some integer $s$, and $c_{2 n}=2^{k+1} d_{2 n}, c_{0}=0, c_{1} \equiv 2^{k}\left(\bmod 2^{k+1}\right)$, and $c_{2}=p c_{1} \equiv 2^{k+1}$ $\left(\bmod 2^{k+2}\right)$, we have  $$ c_{2 n}=(4 s+2) 2^{k+1} d_{2 n-2}-c_{2 n-4} \equiv c_{2 n-4}\left(\bmod 2^{k+2}\right) $$  i.e., $0 \equiv c_{0} \equiv c_{4} \equiv c_{8} \equiv \cdots$ and $2^{k+1} \equiv c_{2} \equiv c_{6} \equiv \cdots\left(\bmod 2^{k+2}\right)$, which proves the statement. Second solution. The recursion is solved by  $$ a_{n}=\frac{1}{2 \sqrt{2}}\left((1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right)=\binom{n}{1}+2\binom{n}{3}+2^{2}\binom{n}{5}+\cdots . $$  Let $n=2^{k} m$ with $m$ odd; then for $p>0$ the summand  $$ 2^{p}\binom{n}{2 p+1}=2^{k+p} m \frac{(n-1) \ldots(n-2 p)}{(2 p+1)!}=2^{k+p} \frac{m}{2 p+1}\binom{n-1}{2 p} $$  is divisible by $2^{k+p}$, because the denominator $2 p+1$ is odd. Hence  $$ a_{n}=n+\sum_{p>0} 2^{p}\binom{n}{2 p+1}=2^{k} m+2^{k+1} N $$  for some integer $N$, so that $a_{n}$ is exactly divisible by $2^{k}$. Third solution. It can be proven by induction that $a_{2 n}=2 a_{n}\left(a_{n}+a_{n+1}\right)$. The required result follows easily, again by induction on $k$.