8. $(\mathbf{C Z S} 2)^{\mathrm{IMO1}} A B C D$ is a parallelogram; $A B=a, A D=1, \alpha$ is the size of $\angle D A B$, and the three angles of the triangle $A B D$ are acute. Prove that the four circles $K_{A}, K_{B}, K_{C}, K_{D}$, each of radius 1, whose centers are the vertices $A, B, C, D$, cover the parallelogram if and only if $a \leq$ $\cos \alpha+\sqrt{3} \sin \alpha$.

Solution: 8. The circles $K_{A}, K_{B}, K_{C}, K_{D}$ cover the parallelogram if and only if for every point $X$ inside the parallelogram, the length of one of the segments $X A, X B, X C, X D$ does not exceed 1. Let $O$ and $r$ be the center and radius of the circumcircle of $\triangle A B D$. For every point $X$ inside $\triangle A B D$, it holds that $X A \leq r$ or $X B \leq r$ or $X D \leq r$. Similarly, for $X$ inside $\triangle B C D, X B \leq r$ or $X C \leq r$ or $X D \leq r$. Hence $K_{A}, K_{B}, K_{C}, K_{D}$ cover the parallelogram if and only if $r \leq 1$, which is equivalent to $\angle A B D \geq 30^{\circ}$. However, this last is exactly equivalent to $a=A B=2 r \sin \angle A D B \leq 2 \sin \left(\alpha+30^{\circ}\right)=\sqrt{3} \sin \alpha+\cos \alpha$.


9. (CZS 3) The circle $k$ and its diameter $A B$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $A B$ and two other vertices on $k$.

Solution: 9. The incenter of any such triangle lies inside the circle $k$. We shall show that every point $S$ interior to the circle $S$ is the incenter of one such triangle. If $S$ lies on the segment $A B$, then it is obviously the incenter of an isosceles triangle inscribed in $k$ that has $A B$ as an axis of symmetry. Let us now suppose $S$ does not lie on $A B$. Let $X$ and $Y$ be the intersection points of lines $A S$ and $B S$ with $k$, and let $Z$ be the foot of the perpendicular from $S$ to $A B$. Since the quadrilateral $B Z S X$ is cyclic, we have $\angle Z X S=$ $\angle A B S=\angle S X Y$ and analogously $\angle Z Y S=\angle S Y X$, which implies that $S$ is the incenter of $\triangle X Y Z$.


1. (SWE 2) Two ships sail on the sea with constant speeds and fixed directions. It is known that at 9:00 the distance between them was 20 miles; at 9:35, 15 miles; and at 9:55, 13 miles. At what moment were the ships the smallest distance from each other, and what was that distance?

Solution: 1. Since the ships are sailing with constant speeds and directions, the second ship is sailing at a constant speed and direction in reference to the first ship. Let $A$ be the constant position of the first ship in this frame. Let $B_{1}$, $B_{2}, B_{3}$, and $B$ on line $b$ defining the trajectory of the ship be positions of the second ship with respect to the first ship at 9:00, 9:35, 9:55, and at the moment the two ships were closest. Then we have the following equations for distances (in miles):  $$ \begin{gathered} A B_{1}=20, \quad A B_{2}=15, \quad A B_{3}=13 \\ B_{1} B_{2}: B_{2} B_{3}=7: 4, \quad A B_{i}^{2}=A B^{2}+B B_{i}^{2} \end{gathered} $$  Since $B B_{1}>B B_{2}>B B_{3}$, it follows that $\mathcal{B}\left(B_{3}, B, B_{2}, B_{1}\right)$ or $\mathcal{B}\left(B, B_{3}, B_{2}\right.$, $B_{1}$ ). We get a system of three quadratic equations with three unknowns: $A B, B B_{3}$ and $B_{3} B_{2}$ ( $B B_{3}$ being negative if $\mathcal{B}\left(B_{3}, B, B_{1}, B_{2}\right)$, positive otherwise). This can be solved by eliminating $A B$ and then $B B_{3}$. The unique solution ends up being  $$ A B=12, \quad B B_{3}=5, \quad B_{3} B_{2}=4 $$  and consequently, the two ships are closest at 10:20 when they are at a distance of 12 miles.


10. (ROM 4) Consider two segments of length $a, b(a>b)$ and a segment of length $c=\sqrt{a b}$. (a) For what values of $a / b$ can these segments be sides of a triangle? (b) For what values of $a / b$ is this triangle right-angled, obtuse-angled, or acute-angled?

Solution: 10. (a) Let us set $k=a / b>1$. Then $a=k b$ and $c=\sqrt{k} b$, and $a>c>b$. The segments $a, b, c$ form a triangle if and only if $k<\sqrt{k}+1$, which holds if and only if $1<k<\frac{3+\sqrt{5}}{2}$. (b) The triangle is right-angled if and only if $a^{2}=b^{2}+c^{2} \Leftrightarrow k^{2}=k+1 \Leftrightarrow$ $k=\frac{1+\sqrt{5}}{2}$. Also, it is acute-angled if and only if $k^{2}<k+1 \Leftrightarrow 1<$ $k<\frac{1+\sqrt{5}}{2}$ and obtuse-angled if $\frac{1+\sqrt{5}}{2}<k<\frac{3+\sqrt{5}}{2}$.


Solution: 11. Introducing $y_{i}=\frac{1}{x_{i}}$, we transform our equation to  $$ \begin{aligned} 0 & =1+y_{1}+\left(1+y_{1}\right) y_{2}+\cdots+\left(1+y_{1}\right) \cdots\left(1+y_{n-1}\right) y_{n} \\ & =\left(1+y_{1}\right)\left(1+y_{2}\right) \cdots\left(1+y_{n}\right) \end{aligned} $$  The solutions are $n$-tuples $\left(y_{1}, \ldots, y_{n}\right)$ with $y_{i} \neq 0$ for all $i$ and $y_{j}=-1$ for at least one index $j$. Returning to $x_{i}$, we conclude that the solutions are all the $n$-tuples $\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \neq 0$ for all $i$, and $x_{j}=-1$ for at least one index $j$.


12. (POL 1) If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that  $$ \left(1+\frac{a}{b}\right)^{m}+\left(1+\frac{b}{a}\right)^{m} \geq 2^{m+1} $$

Solution: 12. The given inequality is equivalent to $(a+b)^{m} / b^{m}+(a+b)^{m} / a^{m} \geq 2^{m+1}$, which can be rewritten as  $$ \frac{1}{2}\left(\frac{1}{a^{m}}+\frac{1}{b^{m}}\right) \geq\left(\frac{2}{a+b}\right)^{m} $$  Since $f(x)=1 / x^{m}$ is a convex function for every $m \in \mathbb{Z}$, the last inequality immediately follows from Jensen's inequality $(f(a)+f(b)) / 2 \geq$ $f((a+b) / 2)$.


13. (POL 5) Given two congruent triangles $A_{1} A_{2} A_{3}$ and $B_{1} B_{2} B_{3}\left(A_{i} A_{k}=\right.$ $B_{i} B_{k}$ ), prove that there exists a plane such that the orthogonal projections of these triangles onto it are congruent and equally oriented.

Solution: 13. Translating one of the triangles if necessary, we may assume w.l.o.g. that $B_{1} \equiv A_{1}$. We also assume that $B_{2} \not \equiv A_{2}$ and $B_{3} \not \equiv A_{3}$, since the result is obvious otherwise. There exists a plane $\pi$ through $A_{1}$ that is parallel to both $A_{2} B_{2}$ and $A_{3} B_{3}$. Let $A_{2}^{\prime}, A_{3}^{\prime}, B_{2}^{\prime}, B_{3}^{\prime}$ denote the orthogonal projections of $A_{2}, A_{3}, B_{2}, B_{3}$ onto $\pi$, and let $h_{2}, h_{3}$ denote the distances of $A_{2}, B_{2}$ and of $A_{3}, B_{3}$ from $\pi$. By the Pythagorean theorem, $A_{2}^{\prime} A_{3}^{\prime 2}=A_{2} A_{3}^{2}-\left(h_{2}+h_{3}\right)^{2}=B_{2} B_{3}^{2}-$ $\left(h_{2}+h_{3}\right)^{2}=B_{2}^{\prime}{B_{3}^{\prime}}^{2}$, and similarly $A_{1} A_{2}^{\prime}=A_{1} B_{2}^{\prime}$ and $A_{1} A_{3}^{\prime}=A_{1} B_{3}^{\prime}$; hence $\triangle A_{1} A_{2}^{\prime} A_{3}^{\prime}$ and $\triangle A_{1} B_{2}^{\prime} B_{3}^{\prime}$ are congruent. If these two triangles are equally oriented, then we have finished. Otherwise, they are symmetric with respect to some line $a$ passing through $A_{1}$, and consequently the projections of the triangles $A_{1} A_{2} A_{3}$ and $A_{1} B_{2} B_{3}$ onto the plane through $a$ perpendicular to $\pi$ coincide.


14. (BUL 5) A line in the plane of a triangle $A B C$ intersects the sides $A B$ and $A C$ respectively at points $X$ and $Y$ such that $B X=C Y$. Find the locus of the center of the circumcircle of triangle $X A Y$.

Solution: 14. Let $O, D, E$ be the circumcenter of $\triangle A B C$ and the midpoints of $A B$ and $A C$, and given arbitrary $X \in A B$ and $Y \in A C$ such that $B X=C Y$, let $O_{1}, D_{1}, E_{1}$ be the circumcenter of $\triangle A X Y$ and the midpoints of $A X$ and $A Y$, respectively. Since $A D=A B / 2$ and $A D_{1}=A X / 2$, it follows that $D D_{1}=B X / 2$ and similarly $E E_{1}=C Y / 2$. Hence $O_{1}$ is at the same distance $B X / 2=C Y / 2$ from the lines $O D$ and $O E$ and lies on the halfline bisector $l$ of $\angle D O E$. If we let $X, Y$ vary along the segments $A B$ and $A C$, we obtain that the locus of $O_{1}$ is the segment $O P$, where $P \in l$ is a point at distance $\min (A B, A C) / 2$ from $O D$ and $O E$.


15. (GBR 1) ${ }^{\mathrm{IMO} 6}$ Let $[x]$ denote the integer part of $x$, i.e., the greatest integer not exceeding $x$. If $n$ is a positive integer, express as a simple function of $n$ the sum  $$ \left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\cdots+\left[\frac{n+2^{i}}{2^{i+1}}\right]+\cdots $$

Solution: 15. Set  $$ f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\cdots+\left[\frac{n+2^{i}}{2^{i+1}}\right]+\ldots $$  We prove by induction that $f(n)=n$. This obviously holds for $n=1$. Let us assume that $f(n-1)=n-1$. Define  $$ g(i, n)=\left[\frac{n+2^{i}}{2^{i+1}}\right]-\left[\frac{n-1+2^{i}}{2^{i+1}}\right] . $$  We have that $f(n)-f(n+1)=\sum_{i=0}^{\infty} g(i, n)$. We also note that $g(i, n)=1$ if and only if $2^{i+1} \mid n+2^{i}$; otherwise, $g(i, n)=0$. The divisibility $2^{i+1} \mid$ $n+2^{i}$ is equivalent to $2^{i} \mid n$ and $2^{i+1} \nmid n$, which for a given $n$ holds for exactly one $i \in \mathbb{N}_{0}$. Thus it follows that $f(n)-f(n-1)=1 \Rightarrow f(n)=n$. The proof by induction is now complete. Second solution. It is easy to show that $[x+1 / 2]=[2 x]-[x]$ for $x \in \mathbb{R}$. Now $f(x)=([x]-[x / 2])+([x / 2]-[x / 4])+\cdots=[x]$. Hence, $f(n)=n$ for all $n \in \mathbb{N}$.


16. (GBR 3) A polynomial $p(x)=a_{0} x^{k}+a_{1} x^{k-1}+\cdots+a_{k}$ with integer coefficients is said to be divisible by an integer $m$ if $p(x)$ is divisible by $m$ for all integers $x$. Prove that if $p(x)$ is divisible by $m$, then $k!a_{0}$ is also divisible by $m$. Also prove that if $a_{0}, k, m$ are nonnegative integers for which $k!a_{0}$ is divisible by $m$, there exists a polynomial $p(x)=a_{0} x^{k}+\cdots+$ $a_{k}$ divisible by $m$.

Solution: 16. We shall prove the result by induction on $k$. It trivially holds for $k=0$. Assume that the statement is true for some $k-1$, and let $p(x)$ be a polynomial of degree $k$. Let us set $p_{1}(x)=p(x+1)-p(x)$. Then $p_{1}(x)$ is a polynomial of degree $k-1$ with leading coefficient $k a_{0}$. Also, $m \mid p_{1}(x)$ for all $x \in \mathbb{Z}$ and hence by the inductive assumption $m \mid(k-1)!\cdot k a_{0}=k!a_{0}$, which completes the induction. On the other hand, for any $a_{0}, k$ and $m \mid k!a_{0}, p(x)=k!a_{0}\binom{x}{k}$ is a polynomial with leading coefficient $a_{0}$ that is divisible by $m$.


17. (GBR 4) Given a point $O$ and lengths $x, y, z$, prove that there exists an equilateral triangle $A B C$ for which $O A=x, O B=y, O C=z$, if and only if $x+y \geq z, y+z \geq x, z+x \geq y$ (the points $O, A, B, C$ are coplanar).

Solution: 17. Let there be given an equilateral triangle $A B C$ and a point $O$ such that $O A=x, O B=y, O C=z$. Let $X$ be the point in the plane such that $\triangle C X B$ and $\triangle C O A$ are congruent and equally oriented. Then $B X=x$ and the triangle $X O C$ is equilateral, which implies $O X=z$. Thus we have a triangle $O B X$ with $B X=x, B O=y$, and $O X=z$. Conversely, given a triangle $O B X$ with $B X=x, B O=y$ and $O X=z$ it is easy to construct the triangle $A B C$.


18. (ITA 2) If an acute-angled triangle $A B C$ is given, construct an equilateral triangle $A^{\prime} B^{\prime} C^{\prime}$ in space such that lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ pass through a given point.

Solution: 18. The required construction is not feasible. In fact, let us consider the special case $\angle B O C=135^{\circ}, \angle A O C=120^{\circ}, \angle A O B=90^{\circ}$, where $A A^{\prime} \cap B B^{\prime} \cap$ $C C^{\prime}=\{O\}$. Denoting $O A^{\prime}, O B^{\prime}, O C^{\prime}$ by $a, b, c$ respectively we obtain the system of equations $a^{2}+b^{2}=a^{2}+c^{2}+a c=b^{2}+c^{2}+\sqrt{2} b c$. Assuming w.l.o.g. $c=1$ we easily obtain $a^{3}-a^{2}-a-1=0$, which is an irreducible equation of third degree. By a known theorem, its solution $a$ is not constructible by ruler and compass.


19. (ITA 5) We are given a fixed point on the circle of radius 1, and going from this point along the circumference in the positive direction on curved distances $0,1,2, \ldots$ from it we obtain points with abscisas $n=0,1,2, \ldots$ respectively. How many points among them should we take to ensure that some two of them are less than the distance $1 / 5$ apart?

Solution: 19. We shall denote by $d_{n}$ the shortest curved distance from the initial point to the $n$th point in the positive direction. The sequence $d_{n}$ goes as follows: $0,1,2,3,4,5,6,0.72,1.72, \ldots, 5.72,0.43,1.43, \ldots, 5.43,0.15=d_{19}$. Hence the required number of points is 20 .


2. (ROM 5) ${ }^{\mathrm{IMO} 1}$ Prove that there exists a unique triangle whose side lengths are consecutive natural numbers and one of whose angles is twice the measure of one of the others.

Solution: 2. The sides $a, b, c$ of a triangle $A B C$ with $\angle A B C=2 \angle B A C$ satisfy $b^{2}=$ $a(a+c)$ (this statement is the lemma in (SL98-7)). Taking into account the remaining condition that $a, b, c$ are consecutive integers with $a<b$, we obtain three cases: (i) $a=n, b=n+1, c=n+2$. We get the equation $(n+1)^{2}=n(2 n+2)$, giving us $(a, b, c)=(1,2,3)$, which is not a valid triangle. (ii) $a=n, b=n+2, c=n+1$. We get $(n+2)^{2}=n(2 n+1) \Rightarrow$ $(n-4)(n+1)=0$, giving us the triangle $(a, b, c)=(4,6,5)$. (iii) $a=n+1, b=n+2, c=n$. We get $(n+2)^{2}=(n+1)(2 n+1) \Rightarrow$ $n^{2}-n-3=0$, which has no positive integer solutions for $n$. Hence, the only solution is the triangle with sides of lengths 4,5 , and 6 .


20. (CZS 1) Given $n(n \geq 3)$ points in space such that every three of them form a triangle with one angle greater than or equal to $120^{\circ}$, prove that these points can be denoted by $A_{1}, A_{2}, \ldots, A_{n}$ in such a way that for each $i, j, k, 1 \leq i<j<k \leq n$, angle $A_{i} A_{j} A_{k}$ is greater than or equal to $120^{\circ}$.

Solution: 20. Let us denote the points $A_{1}, A_{2}, \ldots, A_{n}$ in such a manner that $A_{1} A_{n}$ is a diameter of the set of given points, and $A_{1} A_{2} \leq A_{1} A_{3} \leq \cdots \leq A_{1} A_{n}$. Since for each $1<i<n$ it holds that $A_{1} A_{i}<A_{1} A_{n}$, we have $\angle A_{i} A_{1} A_{n}<120^{\circ}$ and hence $\angle A_{i} A_{1} A_{n}<60^{\circ}$ (otherwise, all angles in $\triangle A_{1} A_{i} A_{n}$ are less than $120^{\circ}$ ). It follows that for all $1<i<j \leq n$, $\angle A_{i} A_{1} A_{j}<120^{\circ}$. Consequently, the angle in the triangle $A_{1} A_{i} A_{j}$ that is at least $120^{\circ}$ must be $\angle A_{1} A_{i} A_{j}$. Moreover, for any $1<i<j<k \leq n$ it holds that $\angle A_{i} A_{j} A_{k} \geq \angle A_{1} A_{j} A_{k}-\angle A_{1} A_{j} A_{i}>120^{\circ}-60^{\circ}=60^{\circ}$ (because $\angle A_{1} A_{j} A_{i}<60^{\circ}$ ); hence $\angle A_{i} A_{j} A_{k} \geq 120^{\circ}$. This proves that the denotation is correct. Remark. It is easy to show that the diameter is unique. Hence the denotation is also unique.


21. (CZS 2) Let $a_{0}, a_{1}, \ldots, a_{k}(k \geq 1)$ be positive integers. Find all positive integers $y$ such that  $$ a_{0}\left|y ; \quad\left(a_{0}+a_{1}\right)\right|\left(y+a_{1}\right) ; \ldots ; \quad\left(a_{0}+a_{n}\right) \mid\left(y+a_{n}\right) . $$

Solution: 21. The given conditions are equivalent to $y-a_{0}$ being divisible by $a_{0}, a_{0}+$ $a_{1}, a_{0}+a_{2}, \ldots, a_{0}+a_{n}$, i.e., to $y=k\left[a_{0}, a_{0}+a_{1}, \ldots, a_{0}+a_{n}\right]+a_{0}, k \in \mathbb{N}_{0}$.


22. (CZS 3) ${ }^{\mathrm{IMO} 2}$ Find all positive integers $x$ for which $p(x)=x^{2}-10 x-22$, where $p(x)$ denotes the product of the digits of $x$.

Solution: 22. It can be shown by induction on the number of digits of $x$ that $p(x) \leq x$ for all $x \in \mathbb{N}$. It follows that $x^{2}-10 x-22 \leq x$, which implies $x \leq 12$.  Since $0<x^{2}-10 x-22=(x-12)(x+2)+2$, one easily obtains $x \geq 12$. Now one can directly check that $x=12$ is indeed a solution, and thus the only one.


23. (CZS 4) Find all complex numbers $m$ such that polynomial  $$ x^{3}+y^{3}+z^{3}+m x y z $$  can be represented as the product of three linear trinomials.

Solution: 23. We may assume w.l.o.g. that in all the factors the coefficient of $x$ is 1 . Suppose that $x+a y+b z$ is one of the linear factors of $p(x, y, z)=x^{3}+$ $y^{3}+z^{3}+m x y z$. Then $p(x)$ is 0 at every point $(x, y, z)$ with $z=-a x-b y$. Hence $x^{3}+y^{3}+(-a x-b y)^{3}+m x y(-a x-b y)=\left(1-a^{3}\right) x^{3}-(3 a b+$ $m)(a x+b y) x y+\left(1-b^{3}\right) y^{3} \equiv 0$. This is obviously equivalent to $a^{3}=b^{3}=1$ and $m=-3 a b$, from which it follows that $m \in\left\{-3,-3 \omega,-3 \omega^{2}\right\}$, where $\omega=\frac{-1+i \sqrt{3}}{2}$. Conversely, for each of the three possible values for $m$ there are exactly three possibilities $(a, b)$. Hence $-3,-3 \omega,-3 \omega^{2}$ are the desired values.


24. (MON 1) Find the number of all $n$-digit numbers for which some fixed digit stands only in the $i$ th $(1<i<n)$ place and the last $j$ digits are distinct. ${ }^{3}$

Solution: 24. If the $i$ th digit is 0 , then the result is $9^{k-j} 9$ !/ $(10-j)$ ! if $i>k-j$ and $9^{k-j-1} 9$ !/ $(9-j)$ ! otherwise. If the $i$ th digit is not 0 , then the above results are multiplied by 8 .


25. (MON 2) Given $k$ parallel lines and a few points on each of them, find the number of all possible triangles with vertices at these given points. ${ }^{4}$

Solution: 25. The answer is  $$ \sum_{1 \leq p<q<r \leq k} n_{p} n_{q} n_{r}+\sum_{1 \leq p<q \leq k}\left[n_{p}\binom{n_{q}}{2}+n_{q}\binom{n_{p}}{2}\right] . $$


26. (GDR) ${ }^{\text {IMO5 }}$ Let $a>0$ be a real number and $f(x)$ a real function defined on all of $\mathbb{R}$, satisfying for all $x \in \mathbb{R}$,  $$ f(x+a)=\frac{1}{2}+\sqrt{f(x)-f(x)^{2}} . $$  (a) Prove that the function $f$ is periodic; i.e., there exists $b>0$ such that for all $x, f(x+b)=f(x)$. (b) Give an example of such a nonconstant function for $a=1$.  [^2]

Solution: 26. (a) We shall show that the period of $f$ is $2 a$. From $(f(x+a)-1 / 2)^{2}=$ $f(x)-f(x)^{2}$ we obtain  $$ \left(f(x)-f(x)^{2}\right)+\left(f(x+a)-f(x+a)^{2}\right)=\frac{1}{4} $$  Subtracting the above relation for $x+a$ in place of $x$ we get $f(x)-$ $f(x)^{2}=f(x+2 a)-f(x+2 a)^{2}$, which implies $(f(x)-1 / 2)^{2}=$ $(f(x+2 a)-1 / 2)^{2}$. Since $f(x) \geq 1 / 2$ holds for all $x$ by the condition of the problem, we conclude that $f(x+2 a)=f(x)$. (b) The following function, as is directly verified, satisfies the conditions:  $$ f(x)=\left\{\begin{array}{cl} 1 / 2 & \text { if } 2 n \leq x<2 n+1, \\ 1 & \text { if } 2 n+1 \leq x<2 n+2, \end{array} \text { for } n=0,1,2, \ldots\right. $$


3. (POL 4) ${ }^{\mathrm{IMO} 4}$ Prove that in any tetrahedron there is a vertex such that the lengths of its sides through that vertex are sides of a triangle.

Solution: 3. A triangle cannot be formed out of three lengths if and only if one of them is larger than the sum of the other two. Let us assume this is the case for all triplets of edges out of each vertex in a tetrahedron $A B C D$. Let w.l.o.g. $A B$ be the largest edge of the tetrahedron. Then $A B \geq A C+A D$ and $A B \geq B C+B D$, from which it follows that $2 A B \geq A C+A D+B C+B D$. This implies that either $A B \geq A C+B C$ or $A B \geq A D+B D$, contradicting the triangle inequality. Hence the three edges coming out of at least one of the vertices $A$ and $B$ form a triangle. Remark. The proof can be generalized to prove that in a polyhedron with only triangular surfaces there is a vertex such that the edges coming out of this vertex form a triangle.


4. (BUL 2) ${ }^{\mathrm{IMO} 3}$ Let $a, b, c$ be real numbers. Prove that the system of equations  $$ \left\{\begin{array}{r} a x_{1}^{2}+b x_{1}+c=x_{2} \\ a x_{2}^{2}+b x_{2}+c=x_{3} \\ \cdots \cdots \cdots \cdots \\ a x_{n-1}^{2}+b x_{n-1}+c=x_{n} \\ a x_{n}^{2}+b x_{n}+c=x_{1} \end{array}\right. $$  has a unique real solution if and only if $(b-1)^{2}-4 a c=0$. Remark. It is assumed that $a \neq 0$.

Solution: 4. We will prove the equivalence in the two directions separately: $(\Rightarrow)$ Suppose $\left\{x_{1}, \ldots, x_{n}\right\}$ is the unique solution of the equation. Since $\left\{x_{n}, x_{1}, x_{2} \ldots, x_{n-1}\right\}$ is also a solution, it follows that $x_{1}=x_{2}=\cdots=$ $x_{n}=x$ and the system of equations reduces to a single equation $a x^{2}+$ $(b-1) x+c=0$. For the solution for $x$ to be unique the discriminant $(b-1)^{2}-4 a c$ of this quadratic equation must be 0 . $(\Leftarrow)$ Assume $(b-1)^{2}-4 a c=0$. Adding up the equations, we get  $$ \sum_{i=1}^{n} f\left(x_{i}\right)=0, \quad \text { where } \quad f(x)=a x^{2}+(b-1) x+c $$  But by the assumed condition, $f(x)=a\left(x+\frac{b-1}{2 a}\right)^{2}$. Hence we must have $f\left(x_{i}\right)=0$ for all $i$, and $x_{i}=-\frac{b-1}{2 a}$, which is indeed a solution.


5. (BUL 5) Let $h_{n}$ be the apothem (distance from the center to one of the sides) of a regular $n$-gon $(n \geq 3)$ inscribed in a circle of radius $r$. Prove the inequality  $$ (n+1) h_{n+1}-n h_{n}>r . $$  Also prove that if $r$ on the right side is replaced with a greater number, the inequality will not remain true for all $n \geq 3$.

Solution: 5. We have $h_{k}=r \cos (\pi / k)$ for all $k \in \mathbb{N}$. Using $\cos x=1-2 \sin ^{2}(x / 2)$ and $\cos x=2 /\left(1+\tan ^{2}(x / 2)\right)-1$ and $\tan x>x>\sin x$ for all $0<x<\pi / 2$, it suffices to prove  $$ \begin{aligned} & (n+1)\left(1-2 \frac{\pi^{2}}{4(n+1)^{2}}\right)-n\left(\frac{2}{1+\pi^{2} /\left(4 n^{2}\right)}-1\right)>1 \\ \Leftrightarrow & 1+2 n\left(1-\frac{1}{1+\pi^{2} /\left(4 n^{2}\right)}\right)-\frac{\pi^{2}}{2(n+1)}>1 \\ \Leftrightarrow & 1+\frac{\pi^{2}}{2}\left(\frac{1}{n+\pi^{2} /(4 n)}-\frac{1}{n+1}\right)>1, \end{aligned} $$  where the last inequality holds because $\pi^{2}<4 n$. It is also apparent that as $n$ tends to infinity the term in parentheses tends to 0 , and hence it is not possible to strengthen the bound. This completes the proof.


6. (HUN 1) If $a_{i}(i=1,2, \ldots, n)$ are distinct non-zero real numbers, prove that the equation  $$ \frac{a_{1}}{a_{1}-x}+\frac{a_{2}}{a_{2}-x}+\cdots+\frac{a_{n}}{a_{n}-x}=n $$  has at least $n-1$ real roots.

Solution: 6. We define $f(x)=\frac{a_{1}}{a_{1}-x}+\frac{a_{2}}{a_{2}-x}+\cdots+\frac{a_{n}}{a_{n}-x}$. Let us assume w.l.o.g. $a_{1}<a_{2}<\cdots<a_{n}$. We note that for all $1 \leq i<n$ the function $f$ is continuous in the interval $\left(a_{i}, a_{i+1}\right)$ and satisfies $\lim _{x \rightarrow a_{i}} f(x)=-\infty$ and $\lim _{x \rightarrow a_{i+1}} f(x)=\infty$. Hence the equation $f(x)=n$ will have a real solution in each of the $n-1$ intervals $\left(a_{i}, a_{i+1}\right)$. Remark. In fact, this equation has exactly $n$ solutions, and hence they are all real. Moreover, the solutions are distinct if all $a_{i}$ are of the same sign, since $x=0$ is an evident solution.


7. (HUN 5) Prove that the product of the radii of three circles exscribed to a given triangle does not exceed $\frac{3 \sqrt{3}}{8}$ times the product of the side lengths of the triangle. When does equality hold?

Solution: 7. Let $r_{a}, r_{b}, r_{c}$ denote the radii of the exscribed circles corresponding to the sides of lengths $a, b, c$ respectively, and $R, p$ and $S$ denote the circumradius, semiperimeter, and area of the given triangle. It is well-known that $r_{a}(p-a)=r_{b}(p-b)=r_{c}(p-c)=S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{a b c}{4 R}$. Hence, the desired inequality $r_{a} r_{b} r_{c} \leq \frac{3 \sqrt{3}}{8} a b c$ reduces to $p \leq \frac{3 \sqrt{3}}{2} R$, which is by the law of sines equivalent to  $$ \sin \alpha+\sin \beta+\sin \gamma \leq \frac{3 \sqrt{3}}{2} $$  This inequality immediately follows from Jensen's inequality, since the sine is concave on $[0, \pi]$. Equality holds if and only if the triangle is equilateral.


8. (ROM 2) Given an oriented line $\Delta$ and a fixed point $A$ on it, consider all trapezoids $A B C D$ one of whose bases $A B$ lies on $\Delta$, in the positive direction. Let $E, F$ be the midpoints of $A B$ and $C D$ respectively. Find the loci of vertices $B, C, D$ of trapezoids that satisfy the following: (i) $|A B| \leq a \quad(a$ fixed); (ii) $|E F|=l \quad(l$ fixed); (iii) the sum of squares of the nonparallel sides of the trapezoid is constant. Remark. The constants are chosen so that such trapezoids exist.

Solution: 8. Let $G$ be the point such that $B C D G$ is a parallelogram and let $H$ be the midpoint of $A G$. Obviously $H E F D$ is also a parallelogram, and thus $D H=E F=l$. If $A D^{2}+B C^{2}=m^{2}$ is fixed, then from the Stewart theorem we have  $$ D H^{2}=\frac{2 D A^{2}+2 D G^{2}-A G^{2}}{4}=\frac{2 m^{2}-A G^{2}}{4} $$  which is fixed. Thus $G$ and $H$ are fixed points, and from here the locus of $D$ is a circle with center $H$ and radius $l$. The locus of $B$ is the segment (GI], where $I \in \Delta$ is a point in the positive direction such that $A I=a$. Finally, the locus of $C$ is a region of the plane consisting of a rectangle sandwiched between two semicircles of radius $l$ centered at points $H$ and $H^{\prime}$, where $H^{\prime}$ is a point such that $\overrightarrow{I H^{\prime}}=\overrightarrow{G H}$.


9. (ROM 3) Let $A B C$ be an arbitrary triangle and $M$ a point inside it. Let $d_{a}, d_{b}, d_{c}$ be the distances from $M$ to sides $B C, C A, A B ; a, b, c$ the lengths of the sides respectively, and $S$ the area of the triangle $A B C$. Prove the inequality  $$ a b d_{a} d_{b}+b c d_{b} d_{c}+c a d_{c} d_{a} \leq \frac{4 S^{2}}{3} $$  Prove that the left-hand side attains its maximum when $M$ is the centroid of the triangle.

Solution: 9. We note that $S_{a}=a d_{a} / 2, S_{b}=b d_{b} / 2$, and $S_{c}=c d_{c} / 2$ are the areas of the triangles $M B C, M C A$, and $M A B$ respectively. The desired inequality now follows from  $$ S_{a} S_{b}+S_{b} S_{c}+S_{c} S_{a} \leq \frac{1}{3}\left(S_{a}+S_{b}+S_{c}\right)^{2}=\frac{S^{2}}{3} $$  Equality holds if and only if $S_{a}=S_{b}=S_{c}$, which is equivalent to $M$ being the centroid of the triangle.


1. Prove that there exist infinitely many natural numbers $a$ with the following property: the number $z=n^{4}+a$ is not prime for any natural number $n$.

Solution: 1. Set $a=4 m^{4}$, where $m \in \mathbb{N}$ and $m>1$. We then have $z=n^{4}+4 m^{4}=$ $\left(n^{2}+2 m^{2}\right)^{2}-(2 m n)^{2}=\left(n^{2}+2 m^{2}+2 m n\right)\left(n^{2}+2 m^{2}-2 m n\right)$. Since $n^{2}+2 m^{2}-2 m n=(n-m)^{2}+m^{2} \geq m^{2}>1$, it follows that $z$ must be composite. Thus we have found infinitely many $a$ that satisfy the condition of the problem.


2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real constants and  $$ y(x)=\cos \left(a_{1}+x\right)+\frac{\cos \left(a_{2}+x\right)}{2}+\frac{\cos \left(a_{3}+x\right)}{2^{2}}+\cdots+\frac{\cos \left(a_{n}+x\right)}{2^{n-1}} . $$  If $x_{1}, x_{2}$ are real and $y\left(x_{1}\right)=y\left(x_{2}\right)=0$, prove that $x_{1}-x_{2}=m \pi$ for some integer $m$.

Solution: 2. Using $\cos (a+x)=\cos a \cos x-\sin a \sin x$, we obtain $f(x)=A \sin x+$ $B \cos x$ where $A=-\sin a_{1}-\sin a_{2} / 2-\cdots-\sin a_{n} / 2^{n-1}$ and $B=\cos a_{1}+$ $\cos a_{2} / 2+\cdots+\cos a_{n} / 2^{n-1}$. Numbers $A$ and $B$ cannot both be equal to 0 , for otherwise $f$ would be identically equal to 0 , while on the other hand, we have $f\left(-a_{1}\right)=\cos \left(a_{1}-a_{1}\right)+\cos \left(a_{2}-a_{1}\right) / 2+\cdots+\cos \left(a_{n}-a_{1}\right) / 2^{n-1} \geq$ $1-1 / 2-\cdots-1 / 2^{n-1}=1 / 2^{n-1}>0$. Setting $A=C \cos \phi$ and $B=C \sin \phi$, where $C \neq 0$ (such $C$ and $\phi$ always exist), we get $f(x)=C \sin (x+\phi)$. It follows that the zeros of $f$ are of the form $x_{0} \in-\phi+\pi \mathbb{Z}$, from which $f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}-x_{2}=m \pi$ immediately follows.


3. Find conditions on the positive real number $a$ such that there exists a tetrahedron $k$ of whose edges $(k=1,2,3,4,5)$ have length $a$, and the other $6-k$ edges have length 1 .  Second Day (July 11)

Solution: 3. We have several cases: $1^{\circ} k=1$. W.l.o.g. let $A B=a$ and the remaining segments have length 1. Let $M$ be the midpoint of $C D$. Then $A M=B M=\sqrt{3} / 2(\triangle C D A$ and $\triangle C D B$ are equilateral) and $0<A B<A M+B M=\sqrt{3}$, i.e., $0<a<\sqrt{3}$. It is evident that all values of $a$ within this interval are realizable. $2^{\circ} k=2$. We have two subcases. First, let $A C=A D=a$. Let $M$ be the midpoint of $C D$. We have $C D=1, A M=\sqrt{a^{2}-1 / 4}$, and $B M=\sqrt{3} / 2$. Then we have $1-$ $\sqrt{3} / 2=A B-B M<A M<A B+B M=1+\sqrt{3} / 2$, which gives us $\sqrt{2-\sqrt{3}}<a<\sqrt{2+\sqrt{3}}$. Second, let $A B=C D=a$. Let $M$ be the midpoint of $C D$. From $\triangle M A B$ we get $a<\sqrt{2}$. Thus, from $\sqrt{2-\sqrt{3}}<\sqrt{2}<\sqrt{2+\sqrt{3}}$ it follows that the required condition in this case is $0<a<\sqrt{2+\sqrt{3}}$. All values for $a$ in this range are realizable. $3^{\circ} k=3$. We show that such a tetrahedron exists for all $a$. Assume $a>1$. Assume $A B=A C=A D=a$. Varying $A$ along the line perpendicular to the plane $B C D$ and through the center of $\triangle B C D$ we achieve all values of $a>1 / \sqrt{3}$. For $a<1 / \sqrt{3}$ we can observe a similar tetrahedron with three edges of length $1 / a$ and three of length 1 and proceed as before. $4^{\circ} k=4$. By observing the similar tetrahedron we reduce this case to $k=2$ with length $1 / a$ instead of $a$. Thus we get $a>\sqrt{2-\sqrt{3}}$. $5^{\circ} k=5$. We reduce to $k=1$ and get $a>1 / \sqrt{3}$.


4. Let $A B$ be a diameter of a circle $\gamma$. A point $C$ different from $A$ and $B$ is on the circle $\gamma$. Let $D$ be the projection of the point $C$ onto the line $A B$. Consider three other circles $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ with the common tangent $A B: \gamma_{1}$ inscribed in the triangle $A B C$, and $\gamma_{2}$ and $\gamma_{3}$ tangent to both (the segment) $C D$ and $\gamma$. Prove that $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ have two common tangents.

Solution: 4. Let $O$ be the midpoint of $A B$, i.e., the center of $\gamma$. Let $O_{1}, O_{2}$, and $O_{3}$ respectively be the centers of $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ and let $r_{1}, r_{2}, r_{3}$ respectively be the radii of $\gamma_{1}, \gamma_{2}$ and $\gamma_{3}$. Let $C_{1}, C_{2}$, and $C_{3}$ respectively be the points of tangency of $\gamma_{1}, \gamma_{2}$ and $\gamma_{3}$ with $A B$. Let $D_{2}$ and $D_{3}$ respectively be the points of tangency of $\gamma_{2}$ and $\gamma_{3}$ with $C D$. Finally, let $G_{2}$ and $G_{3}$ respectively be the points of tangency of $\gamma_{2}$ and $\gamma_{3}$ with $\gamma$. We have $\mathcal{B}\left(G_{2}, O_{2}, O\right)$, $G_{2} O_{2}=O_{2} D_{2}$, and $G_{2} O=O B$. Hence, $G_{2}, D_{2}, B$ are collinear. Similarly, $G_{3}, D_{3}, A$ are collinear. It follows that $A G_{2} D_{2} D$ and $B G_{3} D_{3} D$ are cyclic, since $\angle A G_{2} D_{2}=\angle D_{2} D A=\angle D_{3} D B=\angle B G_{3} D_{3}=90^{\circ}$. Hence $B C_{2}^{2}=B D_{2} \cdot B G_{2}=B D \cdot B A=B C^{2} \Rightarrow B C_{2}=B C$ and hence $A C_{2}=A B-B C$. Similarly, $A C_{3}=A C$. We thus have $A C_{1}=(A C+A B-B C) / 2=\left(A C_{3}+A C_{2}\right) / 2$. Hence, $C_{1}$ is the midpoint of $C_{2} C_{3}$. We also have $r_{2}+r_{3}=C_{2} C_{3}=A C+B C-A B=2 r_{1}$, from which it follows that $O_{1}, O_{2}, O_{3}$ are collinear. Second solution. We shall prove the statement for arbitrary points $A, B, C$ on $\gamma$. Let us apply the inversion $\psi$ with respect to the circle $\gamma_{1}$. We denote by $\widehat{X}$ the image of an object $X$ under $\psi$. Also, $\psi$ maps lines $B C, C A, A B$ onto circles $\widehat{a}, \widehat{b}, \widehat{c}$, respectively. Circles $\widehat{a}, \widehat{b}, \widehat{c}$ pass through the center $O_{1}$ of $\gamma_{1}$ and have radii equal to the radius of $\widehat{\gamma}$. Let $P, Q, R$ be the centers of $\widehat{a}, \widehat{b}, \widehat{c}$ respectively. The line $C D$ maps onto a circle $k$ through $\widehat{C}$ and $O_{1}$ that is perpendicular to $\widehat{c}$. Therefore its center $K$ lies in the intersection of the tangent $t$ to $\widehat{c}$ and the line $P Q$ (which bisects $\left.\widehat{C} O_{1}\right)$. Let $O$ be a point such that $R O_{1} K O$ is a parallelogram and $\gamma_{2}^{\prime}, \gamma_{3}^{\prime}$ the circles centered at $O$ tangent to $k$. It is easy to see that $\gamma_{2}^{\prime}$ and $\gamma_{3}^{\prime}$ are also tangent to $\widehat{c}$, since $O R$ and $O K$ have lengths equal to the radii of $k$ and $\widehat{c}$. Hence $\gamma_{2}^{\prime}$ and $\gamma_{3}^{\prime}$ are the images of $\gamma_{2}$ and $\gamma_{3}$ under $\psi$. Moreover, since $Q \widehat{A} O K$ and $P \widehat{B} O K$ are parallelograms and $Q, P, K$ are collinear, it follows that $\widehat{A}, \widehat{B}, O$ are also collinear. Hence the centers of $\gamma_{1}, \gamma_{2}, \gamma_{3}$ are collinear, lying on the line $O_{1} O$, and the statement follows.  Third solution. Moreover, the statement holds for an arbitrary point $D \in B C$. Let $E, F, G, H$ be the points of tangency of $\gamma_{2}$ with $A B, C D$ and of $\gamma_{3}$ with $A B, C D$, respectively. Let $O_{i}$ be the center of $\gamma_{i}, i=1,2,3$. As is shown in the third solution of (SL93-3), $E F$ and $G H$ meet at $O_{1}$. Hence the problem of proving the collinearity of $O_{1}, O_{2}, O_{3}$ reduces to the following simple problem:  Let $D, E, F, G, H$ be points such that $D \in E G, F \in D H$ and $D E=D F, D G=D H$. Let $O_{1}, O_{2}, O_{3}$ be points such that $\angle O_{2} E D=$ $\angle O_{2} F D=90^{\circ}, \angle O_{3} G D=\angle O_{3} H D=90^{\circ}$, and $O_{1}=E F \cap G H$. Then $O_{1}, O_{2}, O_{3}$ are collinear. Let $K_{2}=D O_{2} \cap E F$ and $K_{3}=D O_{3} \cap G H$. Then $O_{2} K_{2} / O_{2} D=$ $D K_{3} / D O_{3}=K_{2} O_{1} / D O_{3}$ and hence by Thales' theorem $O_{1} \in O_{2} O_{3}$.


5. Given $n$ points in the plane such that no three of them are collinear, prove that one can find at least $\binom{n-3}{2}$ convex quadrilaterals with their vertices at these points.

Solution: 5. We first prove the following lemma.  Lemma. If of five points in a plane no three belong to a single line, then there exist four that are the vertices of a convex quadrilateral.  Proof. If the convex hull of the five points $A, B, C, D, E$ is a pentagon or a quadrilateral, the statement automatically holds. If the convex hull is a triangle, then w.l.o.g. let $\triangle A B C$ be that triangle and $D, E$ points in its interior. Let the line $D E$ w.l.o.g. intersect $[A B]$ and $[A C]$. Then $B, C, D, E$ form the desired quadrilateral. We now observe each quintuplet of points within the set. There are $\binom{n}{5}$ such quintuplets, and for each of them there is at least one quadruplet of points forming a convex quadrilateral. Each quadruplet, however, will be counted up to $n-4$ times. Hence we have found at least $\frac{1}{n-4}\binom{n}{5}$ quadruplets. Since $\frac{1}{n-4}\binom{n}{5} \geq\binom{ n-3}{2} \Leftrightarrow(n-5)(n-6)(n+8) \geq 0$, which always holds, it follows that we have found at least $\binom{n-3}{2}$ desired quadruplets of points.


Solution: 6. Define $u_{1}=\sqrt{x_{1} y_{1}}+z_{1}, u_{2}=\sqrt{x_{2} y_{2}}+z_{2}, v_{1}=\sqrt{x_{1} y_{1}}-z_{1}$, and $v_{2}=$ $\sqrt{x_{2} y_{2}}-z_{2}$. By expanding both sides of the equation we can easily verify $\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}=\left(u_{1}+u_{2}\right)\left(v_{1}+v_{2}\right)+\left(\sqrt{x_{1} y_{2}}-\sqrt{x_{2} y_{1}}\right)^{2} \geq$ $\left(u_{1}+u_{2}\right)\left(v_{1}+v_{2}\right)$. Since $x_{i} y_{i}-z_{i}^{2}=u_{i} v_{i}$ for $i=1,2$, it suffices to prove  $$ \begin{aligned} & \frac{8}{\left(u_{1}+u_{2}\right)\left(v_{1}+v_{2}\right)} \leq \frac{1}{u_{1} v_{1}}+\frac{1}{u_{2} v_{2}} \\ \Leftrightarrow & 8 u_{1} u_{2} v_{1} v_{2} \leq\left(u_{1}+u_{2}\right)\left(v_{1}+v_{2}\right)\left(u_{1} v_{1}+u_{2} v_{2}\right) \end{aligned} $$  which trivially follows from the AM-GM inequalities $2 \sqrt{u_{1} u_{2}} \leq u_{1}+u_{2}$, $2 \sqrt{v_{1} v_{2}} \leq v_{1}+v_{2}$ and $2 \sqrt{u_{1} v_{1} u_{2} v_{2}} \leq u_{1} v_{1}+u_{2} v_{2}$. Equality holds if and only if $x_{1} y_{2}=x_{2} y_{1}, u_{1}=u_{2}$ and $v_{1}=v_{2}$, i.e. if and only if $x_{1}=x_{2}, y_{1}=y_{2}$ and $z_{1}=z_{2}$. Second solution. Let us define $f(x, y, z)=1 /\left(x y-z^{2}\right)$. The problem actually states that  $$ 2 f\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right) \leq f\left(x_{1}, y_{1}, z_{1}\right)+f\left(x_{2}, y_{2}, z_{2}\right) $$  i.e., that the function $f$ is convex on the set $D=\left\{(x, y, z) \in \mathbb{R}^{2} \mid x y-\right.$ $\left.z^{2}>0\right\}$. It is known that a twice continuously differentiable function $f\left(t_{1}, t_{2}, \ldots, t_{n}\right)$ is convex if and only if its Hessian $\left[f_{i j}^{\prime \prime}\right]_{i, j=1}^{n}$ is positive semidefinite, or equivalently, if its principal minors $D_{k}=\operatorname{det}\left[f_{i j}^{\prime \prime}\right]_{i, j=1}^{k}, k=$ $1,2, \ldots, n$, are nonnegative. In the case of our $f$ this is directly verified: $D_{1}=2 y^{2} /\left(x y-z^{2}\right)^{3}, D_{2}=3 x y+z^{2} /\left(x y-z^{2}\right)^{5}, D_{3}=6 /\left(x y-z^{2}\right)^{6}$ are obviously positive.


1. (BEL 3) Consider a regular $2 n$-gon and the $n$ diagonals of it that pass through its center. Let $P$ be a point of the inscribed circle and let $a_{1}, a_{2}, \ldots, a_{n}$ be the angles in which the diagonals mentioned are visible from the point $P$. Prove that  $$ \sum_{i=1}^{n} \tan ^{2} a_{i}=2 n \frac{\cos ^{2} \frac{\pi}{2 n}}{\sin ^{4} \frac{\pi}{2 n}} $$

Solution: 1. Denote respectively by $R$ and $r$ the radii of the circumcircle and incircle, by $A_{1}, \ldots, A_{n}, B_{1}, \ldots, B_{n}$, the vertices of the $2 n$-gon and by $O$ its center. Let $P^{\prime}$ be the point symmetric to $P$ with respect to $O$. Then $A_{i} P^{\prime} B_{i} P$ is a parallelogram, and applying cosine theorem on triangles $A_{i} B_{i} P$ and $P P^{\prime} B_{i}$ yields  $$ \begin{aligned} 4 R^{2} & =P A_{i}^{2}+P B_{i}^{2}-2 P A_{i} \cdot P B_{i} \cos a_{i} \\ 4 r^{2} & =P B_{i}^{2}+P^{\prime} B_{i}^{2}-2 P B_{i} \cdot P^{\prime} B_{i} \cos \angle P B_{i} P^{\prime} \end{aligned} $$  Since $A_{i} P^{\prime} B_{i} P$ is a parallelogram, we have that $P^{\prime} B_{i}=P A_{i}$ and $\angle P B_{i} P^{\prime}=\pi-a_{i}$. Subtracting the expression for $4 r^{2}$ from the one for $4 R^{2}$ yields $4\left(R^{2}-r^{2}\right)=-4 P A_{i} \cdot P B_{i} \cos a_{i}=-8 S_{\triangle A_{i} B_{i} P} \cot a_{i}$, hence we conclude that  $$ \tan ^{2} a_{i}=\frac{4 S_{\triangle A_{i} B_{i} P}^{2}}{\left(R^{2}-r^{2}\right)^{2}} $$  Denote by $M_{i}$ the foot of the perpendicular from $P$ to $A_{i} B_{i}$ and let $m_{i}=$ $P M_{i}$. Then $S_{\triangle A_{i} B_{i} P}=R m_{i}$. Substituting this into (1) and adding up these relations for $i=1,2, \ldots, n$, we obtain  $$ \sum_{i=1}^{n} \tan ^{2} a_{i}=\frac{4 R^{2}}{\left(R^{2}-r^{2}\right)^{2}}\left(\sum_{i=1}^{n} m_{i}^{2}\right) $$  Note that all the points $M_{i}$ lie on a circle with diameter $O P$ and form a regular $n$-gon. Denote its center by $F$. We have that $m_{i}^{2}=\left\|\overrightarrow{P M_{i}}\right\|^{2}=$ $\left\|\overrightarrow{F M_{i}}-\overrightarrow{F P}\right\|^{2}=\left\|\overrightarrow{F M}_{i}^{2}\right\|+\left\|\overrightarrow{F P}^{2}\right\|-2\left\langle\overrightarrow{F M_{i}}, \overrightarrow{F P}\right\rangle=r^{2} / 2-2\left\langle\overrightarrow{F M_{i}}, \overrightarrow{F P}\right\rangle$. From this it follows that $\sum_{i=1}^{n} m_{i}^{2}=2 n(r / 2)^{2}-2 \sum_{i=1}^{n}\left\langle\overrightarrow{F M_{i}}, \overrightarrow{F P}\right\rangle=$ $2 n(r / 2)^{2}-2\left\langle\sum_{i=1}^{n} \overrightarrow{F M_{i}}, \overrightarrow{F P}\right\rangle=2 n(r / 2)^{2}$, because $\sum_{i=1}^{n} \overrightarrow{F M_{i}}=\overrightarrow{0}$. Thus  $$ \sum_{i=1}^{n} \tan ^{2} a_{i}=\frac{4 R^{2}}{\left(R^{2}-r^{2}\right)^{2}} 2 n\left(\frac{r}{2}\right)^{2}=2 n \frac{(r / R)^{2}}{\left(1-(r / R)^{2}\right)^{2}}=2 n \frac{\cos ^{2} \frac{\pi}{2 n}}{\sin ^{4} \frac{\pi}{2 n}} $$  Remark. For $n=1$ there is no regular 2-gon. However, if we think of a 2-gon as a line segment, the statement will remain true.


10. (SWE 4) ${ }^{\mathrm{IMO} 3}$ Let $1=a_{0} \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots$ be a sequence of real numbers. Consider the sequence $b_{1}, b_{2}, \ldots$ defined by:  $$ b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}} $$  Prove that: (a) For all natural numbers $n, 0 \leq b_{n}<2$. (b) Given an arbitrary $0 \leq b<2$, there is a sequence $a_{0}, a_{1}, \ldots, a_{n}, \ldots$ of the above type such that $b_{n}>b$ is true for infinitely many natural numbers $n$.

Solution: 10. (a) Since $a_{n-1}<a_{n}$, we have  $$ \begin{aligned} \left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}} & =\frac{a_{k}-a_{k-1}}{a_{k}^{3 / 2}} \\ & \leq \frac{2\left(\sqrt{a_{k}}-\sqrt{a_{k-1}}\right) \sqrt{a_{k}}}{a_{k} \sqrt{a_{k-1}}}=2\left(\frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_{k}}}\right) . \end{aligned} $$  Summing up all these inequalities for $k=1,2, \ldots, n$ we obtain  $$ b_{n} \leq 2\left(\frac{1}{\sqrt{a_{0}}}-\frac{1}{\sqrt{a_{n}}}\right)<2 $$  (b) Choose a real number $q>1$, and let $a_{k}=q^{k}, k=1,2, \ldots$. Then $\left(1-a_{k-1} / a_{k}\right) / \sqrt{a_{k}}=(1-1 / q) / q^{k / 2}$, and consequently  $$ b_{n}=\left(1-\frac{1}{q}\right) \sum_{k=1}^{n} \frac{1}{q^{k / 2}}=\frac{\sqrt{q}+1}{q}\left(1-\frac{1}{q^{n / 2}}\right) . $$  Since $(\sqrt{q}+1) / q$ can be arbitrarily close to 2 , one can set $q$ such that $(\sqrt{q}+1) / q>b$. Then $b_{n} \geq b$ for all sufficiently large $n$. Second solution. (a) Note that  $$ b_{n}=\sum_{k=1}^{n}\left(1-\frac{a_{k-1}}{a_{k}}\right) \frac{1}{\sqrt{a_{k}}}=\sum_{k=1}^{n}\left(a_{k}-a_{k-1}\right) \cdot \frac{1}{a_{k}^{3 / 2}} $$  hence $b_{n}$ represents exactly the lower Darboux sum for the function $f(x)=x^{-3 / 2}$ on the interval $\left[a_{0}, a_{n}\right]$. Then $b_{n} \leq \int_{a_{0}}^{a_{n}} x^{-3 / 2} d x<$ $\int_{1}^{+\infty} x^{-3 / 2} d x=2$. (b) For each $b<2$ there exists a number $\alpha>1$ such that $\int_{1}^{\alpha} x^{-3 / 2} d x>$ $b+(2-b) / 2$. Now, by Darboux's theorem, there exists an array $1=$ $a_{0} \leq a_{1} \leq \cdots \leq a_{n}=\alpha$ such that the corresponding Darboux sums are arbitrarily close to the value of the integral. In particular, there is an array $a_{0}, \ldots, a_{n}$ with $b_{n}>b$.


11. (SWE 6) Let $P, Q, R$ be polynomials and let $S(x)=P\left(x^{3}\right)+x Q\left(x^{3}\right)+$ $x^{2} R\left(x^{3}\right)$ be a polynomial of degree $n$ whose roots $x_{1}, \ldots, x_{n}$ are distinct. Construct with the aid of the polynomials $P, Q, R$ a polynomial $T$ of degree $n$ that has the roots $x_{1}^{3}, x_{2}^{3}, \ldots, x_{n}^{3}$.

Solution: 11. Let $S(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)$. We have $x^{3}-x_{i}^{3}=\left(x-x_{i}\right)(\omega x-$ $\left.x_{i}\right)\left(\omega^{2} x-x_{i}\right)$, where $\omega$ is a primitive third root of 1 . Multiplying these equalities for $i=1, \ldots, n$ we obtain  $$ T\left(x^{3}\right)=\left(x^{3}-x_{1}^{3}\right)\left(x^{3}-x_{2}^{3}\right) \cdots\left(x^{3}-x_{n}^{3}\right)=S(x) S(\omega x) S\left(\omega^{2} x\right) $$  Since $S(\omega x)=P\left(x^{3}\right)+\omega x Q\left(x^{3}\right)+\omega^{2} x^{2} R\left(x^{3}\right)$ and $S\left(\omega^{2} x\right)=P\left(x^{3}\right)+$ $\omega^{2} x Q\left(x^{3}\right)+\omega x^{2} R\left(x^{3}\right)$, the above expression reduces to  $$ T\left(x^{3}\right)=P^{3}\left(x^{3}\right)+x^{3} Q^{3}\left(x^{3}\right)+x^{6} R^{3}\left(x^{3}\right)-3 P\left(x^{3}\right) Q\left(x^{3}\right) R\left(x^{3}\right) $$  Therefore the zeros of the polynomial  $$ T(x)=P^{3}(x)+x Q^{3}(x)+x^{2} R^{3}(x)-3 P(x) Q(x) R(x) $$  are exactly $x_{1}^{3}, \ldots, x_{n}^{3}$. It is easily verified that $\operatorname{deg} T=\operatorname{deg} S=n$, and hence $T$ is the desired polynomial.


12. (USS 4) ${ }^{\mathrm{IMO} 6}$ We are given 100 points in the plane, no three of which are on the same line. Consider all triangles that have all vertices chosen from the 100 given points. Prove that at most $70 \%$ of these triangles are acute angled.

Solution: 12. Lemma. Five points are given in the plane such that no three of them are collinear. Then there are at least three triangles with vertices at these points that are not acute-angled. Proof. We consider three cases, according to whether the convex hull of these points is a triangle, quadrilateral, or pentagon. (i) Let a triangle $A B C$ be the convex hull and two other points $D$ and $E$ lie inside the triangle. At least two of the triangles $A D B, B D C$ and $C D A$ have obtuse angles at the point $D$. Similarly, at least two of the triangles $A E B, B E C$ and $C E A$ are obtuse-angled. Thus there are at least four non-acute-angled triangles. (ii) Suppose that $A B C D$ is the convex hull and that $E$ is a point of its interior. At least one angle of the quadrilateral is not acute, determining one non-acute-angled triangle. Also, the point $E$ lies in the interior of either $\triangle A B C$ or $\triangle C D A$ hence, as in the previous case, it determines another two obtuse-angled triangles. (iii) It is easy to see that at least two of the angles of the pentagon are not acute. We may assume that these two angles are among the angles corresponding to vertices $A, B$, and $C$. Now consider the quadrilateral $A C D E$. At least one its angles is not acute. Hence, there are at least three triangles that are not acute-angled.  Now we consider all combinations of 5 points chosen from the given 100. There are $\binom{100}{5}$ such combinations, and for each of them there are at least three non-acute-angled triangles with vertices in it. On the other hand, vertices of each of the triangles are counted $\binom{97}{2}$ times. Hence there are at least $3\binom{100}{5} /\binom{97}{2}$ non-acute-angled triangles with vertices in the given 100 points. Since the number of all triangles with vertices in the given points is $\binom{100}{3}$, the ratio between the number of acute-angled triangles and the number of all triangles cannot be greater than  $$ 1-\frac{3\binom{100}{5}}{\binom{97}{2}\binom{100}{3}}=0.7 $$


2. (ROM 1) ${ }^{\mathrm{IMO} 2}$ Let $a$ and $b$ be the bases of two number systems and let  $$ \begin{array}{ll} A_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(a)}, & A_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(a)}, \\ B_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(b)}, & B_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(b)}, \end{array} $$  be numbers in the number systems with respective bases $a$ and $b$, so that $x_{0}, x_{1}, x_{2}, \ldots, x_{n}$ denote digits in the number system with base $a$ as well as in the number system with base $b$. Suppose that neither $x_{0}$ nor $x_{1}$ is zero. Prove that $a>b$ if and only if  $$ \frac{A_{n}}{A_{n+1}}<\frac{B_{n}}{B_{n+1}} . $$

Solution: 2. Suppose that $a>b$. Consider the polynomial $P(X)=x_{1} X^{n-1}+x_{2} X^{n-2}+$ $\cdots+x_{n-1} X+x_{n}$. We have $A_{n}=P(a), B_{n}=P(b), A_{n+1}=x_{0} a^{n}+$ $P(a)$, and $B_{n+1}=x_{0} b^{n}+P(b)$. Now $A_{n} / A_{n+1}<B_{n} / B_{n+1}$ becomes $P(a) /\left(x_{0} a^{n}+P(a)\right)<P(b) /\left(x_{0} b^{n}+P(b)\right)$, i.e.,  $$ b^{n} P(a)<a^{n} P(b) $$  Since $a>b$, we have that $a^{i}>b^{i}$ and hence $x_{i} a^{n} b^{n-i} \geq x_{i} b^{n} a^{n-i}$ (also, for $i \geq 1$ the inequality is strict). Summing up all these inequalities for $i=1, \ldots, n$ we get $a^{n} P(b)>b^{n} P(a)$, which completes the proof for $a>b$.  On the other hand, for $a<b$ we analogously obtain the opposite inequality $A_{n} / A_{n+1}>B_{n} / B_{n+1}$, while for $a=b$ we have equality. Thus $A_{n} / A_{n+1}<$ $B_{n} / B_{n+1} \Leftrightarrow a>b$.


3. (BUL 6) ${ }^{\mathrm{IMO}}$ In the tetrahedron $S A B C$ the angle $B S C$ is a right angle, and the projection of the vertex $S$ to the plane $A B C$ is the intersection of the altitudes of the triangle $A B C$. Let $z$ be the radius of the inscribed circle of the triangle $A B C$. Prove that  $$ S A^{2}+S B^{2}+S C^{2} \geq 18 z^{2} $$

Solution: 3 . We shall use the following lemma Lemma. If an altitude of a tetrahedron passes through the orthocenter of the opposite side, then each of the other altitudes possesses the same property. Proof. Denote the tetrahedron by $S A B C$ and let $a=B C, b=C A$, $c=A B, m=S A, n=S B, p=S C$. It is enough to prove that an altitude passes through the orthocenter of the opposite side if and only if $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$. Suppose that the foot $S^{\prime}$ of the altitude from $S$ is the orthocenter of $A B C$. Then $S S^{\prime} \perp A B C \Rightarrow S B^{2}-S C^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. But from $A S^{\prime} \perp B C$ it follows that $A B^{2}-A C^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. From these two equalities it can be concluded that $n^{2}-p^{2}=c^{2}-b^{2}$, or equivalently, $n^{2}+b^{2}=c^{2}+p^{2}$. Analogously, $a^{2}+m^{2}=n^{2}+b^{2}$, so we have proved the first part of the equivalence. Now suppose that $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$. Defining $S^{\prime}$ as before, we get $n^{2}-p^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. From the condition $n^{2}-p^{2}=c^{2}-b^{2}$ $\left(\Leftrightarrow b^{2}+n^{2}=c^{2}+p^{2}\right.$ ) we conclude that $A S^{\prime} \perp B C$. In the same way $C S^{\prime} \perp A B$, which proves that $S^{\prime}$ is the orthocenter of $\triangle A B C$. The lemma is thus proven. Now using the lemma it is easy to see that if one of the angles at $S$ is right, than so are the others. Indeed, suppose that $\angle A S B=\pi / 2$. From the lemma we have that the altitude from $C$ passes through the orthocenter of $\triangle A S B$, which is $S$, so $C S \perp A S B$ and $\angle C S A=\angle C S B=\pi / 2$. Therefore $m^{2}+n^{2}=c^{2}, n^{2}+p^{2}=a^{2}$, and $p^{2}+m^{2}=b^{2}$, so it follows that $m^{2}+n^{2}+p^{2}=\left(a^{2}+b^{2}+c^{2}\right) / 2$. By the inequality between the arithmetic and quadric means, we have that $\left(a^{2}+b^{2}+c^{2}\right) / 2 \geq 2 s^{2} / 3$, where $s$ denotes the semiperimeter of $\triangle A B C$. It remains to be shown that $2 s^{2} / 3 \geq 18 r^{2}$. Since $S_{\triangle A B C}=s r$, this is equivalent to $2 s^{4} / 3 \geq$ $18 S_{A B C}^{2}=18 s(s-a)(s-b)(s-c)$ by Heron's formula. This reduces to $s^{3} \geq 27(s-a)(s-b)(s-c)$, which is an obvious consequence of the AM-GM mean inequality. Remark. In the place of the lemma one could prove that the opposite edges of the tetrahedron are mutually perpendicular and proceed in the same way.


4. (CZS 1) $)^{\mathrm{IMO} 4}$ For what natural numbers $n$ can the product of some of the numbers $n, n+1, n+2, n+3, n+4, n+5$ be equal to the product of the remaining ones?

Solution: 4. Suppose that $n$ is such a natural number. If a prime number $p$ divides any of the numbers $n, n+1, \ldots, n+5$, then it must divide another one of them, so the only possibilities are $p=2,3,5$. Moreover, $n+1, n+2, n+3, n+4$ have no prime divisors other than 2 and 3 (if some prime number greater than 3 divides one of them, then none of the remaining numbers can have that divisor). Since two of these numbers are odd, they must be powers of  3 (greater than 1). However, there are no two powers of 3 whose difference is 2 . Therefore there is no such natural number $n$. Second solution. Obviously, none of $n, n+1, \ldots, n+5$ is divisible by 7; hence they form a reduced system of residues. We deduce that $n(n+$ 1) $\cdots(n+5) \equiv 1 \cdot 2 \cdots 6 \equiv-1(\bmod 7)$. If $\{n, \ldots, n+5\}$ can be partitioned into two subsets with the same products, both congruent to, say, $p$ modulo 7 , then $p^{2} \equiv-1(\bmod 7)$, which is impossible. Remark. Erdős has proved that a set $n, n+1, \ldots, n+m$ of consecutive natural numbers can never be partitioned into two subsets with equal products of elements.


5. (CZS 3) Let $M$ be an interior point of the tetrahedron $A B C D$. Prove that  $$ \begin{aligned} & \overrightarrow{M A} \operatorname{vol}(M B C D)+\overrightarrow{M B} \operatorname{vol}(M A C D) \\ & \quad+\overrightarrow{M C} \operatorname{vol}(M A B D)+\overrightarrow{M D} \operatorname{vol}(M A B C)=0 \end{aligned} $$  ( $\operatorname{vol}(P Q R S)$ denotes the volume of the tetrahedron $P Q R S)$.

Solution: 5. Denote respectively by $A_{1}, B_{1}, C_{1}$ and $D_{1}$ the points of intersection of the lines $A M, B M, C M$, and $D M$ with the opposite sides of the tetrahedron. Since $\operatorname{vol}(M B C D)=\operatorname{vol}(A B C D) \overrightarrow{M A_{1}} / \overrightarrow{A A_{1}}$, the relation we have to prove is equivalent to  $$ \overrightarrow{M A} \cdot \frac{\overrightarrow{M A_{1}}}{\overrightarrow{A A_{1}}}+\overrightarrow{M B} \cdot \frac{\overrightarrow{M B_{1}}}{\overrightarrow{B B_{1}}}+\overrightarrow{M C} \cdot \frac{\overrightarrow{M C_{1}}}{\overrightarrow{C C_{1}}}+\overrightarrow{M D} \cdot \frac{\overrightarrow{M D_{1}}}{\overrightarrow{D D_{1}}}=0 $$  There exist unique real numbers $\alpha, \beta, \gamma$, and $\delta$ such that $\alpha+\beta+\gamma+\delta=1$ and for every point $O$ in space  $$ \overrightarrow{O M}=\alpha \overrightarrow{O A}+\beta \overrightarrow{O B}+\gamma \overrightarrow{O C}+\delta \overrightarrow{O D} $$  (This follows easily from $\overrightarrow{O M}=\overrightarrow{O A}+\overrightarrow{A M}=\overrightarrow{O A}+k \overrightarrow{A B}+l \overrightarrow{A C}+m \overrightarrow{A D}=$ $\overrightarrow{A B}+k(\overrightarrow{O B}-\overrightarrow{O A})+l(\overrightarrow{O C}-\overrightarrow{O A})+m(\overrightarrow{O D}-\overrightarrow{O A})$ for some $k, l, m \in \mathbb{R}$.) Further, from the condition that $A_{1}$ belongs to the plane $B C D$ we obtain for every $O$ in space the following equality for some $\beta^{\prime}, \gamma^{\prime}, \delta^{\prime}$ :  $$ \overrightarrow{O A_{1}}=\beta^{\prime} \overrightarrow{O B}+\gamma^{\prime} \overrightarrow{O C}+\delta^{\prime} \overrightarrow{O D} $$  However, for $\lambda=\overrightarrow{M A_{1}} / \overrightarrow{A A_{1}}, \overrightarrow{O M}=\lambda \overrightarrow{O A}+(1-\lambda) \overrightarrow{O A_{1}}$; hence substituting (2) and (3) in this expression and equating coefficients for $\overrightarrow{O A}$ we obtain $\lambda=\overrightarrow{M A_{1}} / \overrightarrow{A A_{1}}=\alpha$. Analogously, $\beta=\overrightarrow{M B_{1}} / \overrightarrow{B B_{1}}, \gamma=\overrightarrow{M C_{1}} / \overrightarrow{C C_{1}}$, and $\delta=\overrightarrow{M D_{1}} / \overrightarrow{D D_{1}}$; hence (1) follows immediately for $O=M$. Remark. The statement of the problem actually follows from the fact that $M$ is the center of mass of the system with masses $\operatorname{vol}(M B C D)$, $\operatorname{vol}(M A C D), \operatorname{vol}(M A B D), \operatorname{vol}(M A B C)$ at $A, B, C, D$ respectively. Our proof is actually a formal verification of this fact.


6. (FRA 1) In the triangle $A B C$ let $B^{\prime}$ and $C^{\prime}$ be the midpoints of the sides $A C$ and $A B$ respectively and $H$ the foot of the altitude passing through the vertex $A$. Prove that the circumcircles of the triangles $A B^{\prime} C^{\prime}, B C^{\prime} H$, and $B^{\prime} C H$ have a common point $I$ and that the line $H I$ passes through the midpoint of the segment $B^{\prime} C^{\prime}$.

Solution: 6. Let $F$ be the midpoint of $B^{\prime} C^{\prime}, A^{\prime}$ the midpoint of $B C$, and $I$ the intersection point of the line $H F$ and the circle circumscribed about $\triangle B H C^{\prime}$. Denote by $M$ the intersection point of the line $A A^{\prime}$ with the circumscribed circle about the triangle $A B C$. Triangles $H B^{\prime} C^{\prime}$ and $A B C$ are similar. Since $\angle C^{\prime} I F=\angle A B C=\angle A^{\prime} M C, \angle C^{\prime} F I=\angle A A^{\prime} B=\angle M A^{\prime} C$, $2 C^{\prime} F=C^{\prime} B^{\prime}$, and $2 A^{\prime} C=C B$, it follows that $\triangle C^{\prime} I B^{\prime} \sim \triangle C M B$, hence $\angle F I B^{\prime}=\angle A^{\prime} M B=\angle A C B$. Now one concludes that $I$ belongs to the circumscribed circles of $\triangle A B^{\prime} C^{\prime}$ (since $\left.\angle C^{\prime} I B^{\prime}=180^{\circ}-\angle C^{\prime} A B^{\prime}\right)$ and $\triangle H C B^{\prime}$. Second Solution. We denote the angles of $\triangle A B C$ by $\alpha, \beta, \gamma$. Evidently $\triangle A B C \sim \triangle H C^{\prime} B^{\prime}$. Within $\triangle H C^{\prime} B^{\prime}$ there exists a unique point $I$ such that $\angle H I B^{\prime}=180^{\circ}-\gamma, \angle H I C^{\prime}=180^{\circ}-\beta$, and $\angle C^{\prime} I B^{\prime}=180^{\circ}-\alpha$, and all three circles must contain this point. Let $H I$ and $B^{\prime} C^{\prime}$ intersect in $F$. It remains to show that $F B^{\prime}=F C^{\prime}$. From $\angle H I B^{\prime}+\angle H B^{\prime} F=180^{\circ}$ we obtain $\angle I H B^{\prime}=\angle I B^{\prime} F$. Similarly, $\angle I H C^{\prime}=\angle I C^{\prime} F$. Thus circles around $\triangle I H C^{\prime}$ and $\triangle I H B^{\prime}$ are both tangent to $B^{\prime} C^{\prime}$, giving us $F B^{\prime 2}=$ $F I \cdot F H=F C^{\prime 2}$.


7. (USS 5) For which digits $a$ do exist integers $n \geq 4$ such that each digit of $\frac{n(n+1)}{2}$ equals $a$ ?

Solution: 7. For $a=5$ one can take $n=10$, while for $a=6$ one takes $n=11$. Now assume $a \notin\{5,6\}$. If there exists an integer $n$ such that each digit of $n(n+1) / 2$ is equal to $a$, then there is an integer $k$ such that $n(n+1) / 2=\left(10^{k}-1\right) a / 9$. After multiplying both sides of the equation by 72 , one obtains $36 n^{2}+36 n=$ $8 a \cdot 10^{k}-8 a$, which is equivalent to  $$ 9(2 n+1)^{2}=8 a \cdot 10^{k}-8 a+9 $$  So $8 a \cdot 10^{k}-8 a+9$ is the square of some odd integer. This means that its last digit is 1,5 , or 9 . Therefore $a \in\{1,3,5,6,8\}$. If $a=3$ or $a=8$, the number on the RHS of (1) is divisible by 5 , but not by 25 (for $k \geq 2$ ), and thus cannot be a square. It remains to check the case $a=1$. In that case, (1) becomes $9(2 n+1)^{2}=8 \cdot 10^{k}+1$, or equivalently $[3(2 n+1)-1][3(2 n+1)+1]=8 \cdot 10^{k} \Rightarrow(3 n+1)(3 n+2)=2 \cdot 10^{k}$. Since the factors $3 n+1,3 n+2$ are relatively prime, this implies that one of them is $2^{k+1}$ and the other one is $5^{k}$. It is directly checked that their difference really equals 1 only for $k=1$ and $n=1$, which is excluded. Hence, the desired $n$ exists only for $a \in\{5,6\}$.


8. (POL 2) ${ }^{\mathrm{IMO} 1}$ Given a point $M$ on the side $A B$ of the triangle $A B C$, let $r_{1}$ and $r_{2}$ be the radii of the inscribed circles of the triangles $A C M$ and $B C M$ respectively and let $\rho_{1}$ and $\rho_{2}$ be the radii of the excircles of the triangles $A C M$ and $B C M$ at the sides $A M$ and $B M$ respectively. Let $r$ and $\rho$ denote the radii of the inscribed circle and the excircle at the side $A B$ of the triangle $A B C$ respectively. Prove that  $$ \frac{r_{1}}{\rho_{1}} \frac{r_{2}}{\rho_{2}}=\frac{r}{\rho} $$

Solution: 8. Let $A C=b, B C=a, A M=x, B M=y, C M=l$. Denote by $I_{1}$ the incenter and by $S_{1}$ the center of the excircle of $\triangle A M C$. Suppose that $P_{1}$ and $Q_{1}$ are feet of perpendiculars from $I_{1}$ and $S_{1}$, respectively, to the line $A C$. Then $\triangle I_{1} C P_{1} \sim \triangle S_{1} C Q_{1}$, hence $r_{1} / \rho_{1}=C P_{1} / C Q_{1}$. We have $C P_{1}=(A C+M C-A M) / 2=(b+l-x) / 2$ and $C Q_{1}=$ $(A C+M C+A M) / 2=(b+l+x) / 2$. Hence  $$ \frac{r_{1}}{\rho_{1}}=\frac{b+l-x}{b+l+x} $$  We similarly obtain  $$ \frac{r_{2}}{\rho_{2}}=\frac{b+l-y}{b+l+y} \text { and } \frac{r}{\rho}=\frac{a+b-x-y}{a+b+x+y} $$  What we have to prove is now equivalent to  $$ \frac{(b+l-x)(a+l-y)}{(b+l+x)(a+l+y)}=\frac{a+b-x-y}{a+b+x+y} $$  Multiplying both sides of (1) by $(a+l+y)(b+l+x)(a+b+x+y)$ we obtain an expression that reduces to $l^{2} x+l^{2} y+x^{2} y+x y^{2}=b^{2} y+a^{2} x$. Dividing both sides by $c=x+y$, we get that (1) is equivalent to $l^{2}=$ $b^{2} y /(x+y)+a^{2} x /(x+y)-x y$, which is exactly Stewart's theorem for $l$. This finally proves the desired result.


9. (GDR 3) Let $u_{1}, u_{2}, \ldots, u_{n}, v_{1}, v_{2}, \ldots, v_{n}$ be real numbers. Prove that  $$ 1+\sum_{i=1}^{n}\left(u_{i}+v_{i}\right)^{2} \leq \frac{4}{3}\left(1+\sum_{i=1}^{n} u_{i}^{2}\right)\left(1+\sum_{i=1}^{n} v_{i}^{2}\right) $$  In what case does equality hold?

Solution: 9. Let us set $a=\sqrt{\sum_{i=1}^{n} u_{i}^{2}}$ and $b=\sqrt{\sum_{i=1}^{n} v_{i}^{2}}$. By Minkowski's inequality (for $p=2$ ) we have $\sum_{i=1}^{n}\left(u_{i}+v_{i}\right)^{2} \leq(a+b)^{2}$. Hence the LHS of the desired inequality is not greater than $1+(a+b)^{2}$, while the RHS is equal to $4\left(1+a^{2}\right)\left(1+b^{2}\right) / 3$. Now it is sufficient to prove that  $$ 3+3(a+b)^{2} \leq 4\left(1+a^{2}\right)\left(1+b^{2}\right) $$  The last inequality can be reduced to the trivial $0 \leq(a-b)^{2}+(2 a b-1)^{2}$. The equality in the initial inequality holds if and only if $u_{i} / v_{i}=c$ for some $c \in \mathbb{R}$ and $a=b=1 / \sqrt{2}$.


1. (BUL 2) Consider a sequence of polynomials $P_{0}(x), P_{1}(x), P_{2}(x), \ldots$, $P_{n}(x), \ldots$, where $P_{0}(x)=2, P_{1}(x)=x$ and for every $n \geq 1$ the following equality holds:  $$ P_{n+1}(x)+P_{n-1}(x)=x P_{n}(x) $$  Prove that there exist three real numbers $a, b, c$ such that for all $n \geq 1$,  $$ \left(x^{2}-4\right)\left[P_{n}^{2}(x)-4\right]=\left[a P_{n+1}(x)+b P_{n}(x)+c P_{n-1}(x)\right]^{2} $$

Solution: 1. Assuming that $a, b, c$ in (1) exist, let us find what their values should be. Since $P_{2}(x)=x^{2}-2$, equation (1) for $n=1$ becomes $\left(x^{2}-4\right)^{2}=$ $\left[a\left(x^{2}-2\right)+b x+2 c\right]^{2}$. Therefore, there are two possibilities for $(a, b, c)$ : $(1,0,-1)$ and $(-1,0,1)$. In both cases we must prove that  $$ \left(x^{2}-4\right)\left[P_{n}(x)^{2}-4\right]=\left[P_{n+1}(x)-P_{n-1}(x)\right]^{2} $$  It suffices to prove (2) for all $x$ in the interval $[-2,2]$. In this interval we can set $x=2 \cos t$ for some real $t$. We prove by induction that  $$ P_{n}(x)=2 \cos n t \quad \text { for all } n $$  This is trivial for $n=0,1$. Assume (3) holds for some $n-1$ and $n$. Then $P_{n+1}(x)=4 \cos t \cos n t-2 \cos (n-1) t=2 \cos (n+1) t$ by the additive formula for the cosine. This completes the induction. Now (2) reduces to the obviously correct equality  $$ 16 \sin ^{2} t \sin ^{2} n t=(2 \cos (n+1) t-2 \cos (n-1) t)^{2} $$  Second solution. If $x$ is fixed, the linear recurrence relation $P_{n+1}(x)+$ $P_{n-1}(x)=x P_{n}(x)$ can be solved in the standard way. The characteristic polynomial $t^{2}-x t+1$ has zeros $t_{1,2}$ with $t_{1}+t_{2}=x$ and $t_{1} t_{2}=1$; hence, the general $P_{n}(x)$ has the form $a t_{1}^{n}+b t_{2}^{n}$ for some constants $a$, $b$. From $P_{0}=2$ and $P_{1}=x$ we obtain that  $$ P_{n}(x)=t_{1}^{n}+t_{2}^{n} . $$  Plugging in these values and using $t_{1} t_{2}=1$ one easily verifies (2).


10. (POL 2) ${ }^{\mathrm{IMO} 3}$ Prove that the sequence $2^{n}-3(n>1)$ contains a subsequence of numbers relatively prime in pairs.

Solution: 10. We use induction. Suppose that every two of the numbers $a_{1}=2^{n_{1}}-$ $3, a_{2}=2^{n_{2}}-3, \ldots, a_{k}=2^{n_{k}}-3$, where $2=n_{1}<n_{2}<\cdots<n_{k}$, are coprime. Then one can construct $a_{k+1}=2^{n_{k+1}}-3$ in the following way:  Set $s=a_{1} a_{2} \ldots a_{k}$. Among the numbers $2^{0}, 2^{1}, \ldots, 2^{s}$, two give the same residue upon division by $s$, say $s \mid 2^{\alpha}-2^{\beta}$. Since $s$ is odd, it can be assumed w.l.o.g. that $\beta=0$ (this is actually a direct consequence of Euler's theorem). Let $2^{\alpha}-1=q s, q \in \mathbb{N}$. Since $2^{\alpha+2}-3=4 q s+1$ is then coprime to $s$, it is enough to take $n_{k+1}=\alpha+2$. We obviously have $n_{k+1}>n_{k}$.


11. (POL 3) The matrix  $$ \left(\begin{array}{ccc} a_{11} & \ldots & a_{1 n} \\ \vdots & \ldots & \vdots \\ a_{n 1} & \ldots & a_{n n} \end{array}\right) $$  satisfies the inequality $\sum_{j=1}^{n}\left|a_{j 1} x_{1}+\cdots+a_{j n} x_{n}\right| \leq M$ for each choice of numbers $x_{i}$ equal to $\pm 1$. Show that  $$ \left|a_{11}+a_{22}+\cdots+a_{n n}\right| \leq M $$

Solution: 11. We use induction. The statement for $n=1$ is trivial. Suppose that it holds for $n=k$ and consider $n=k+1$. From the given condition, we have  $$ \begin{gathered} \sum_{j=1}^{k}\left|a_{j, 1} x_{1}+\cdots+a_{j, k} x_{k}+a_{j, k+1}\right| \\ +\left|a_{k+1,1} x_{1}+\cdots+a_{k+1, k} x_{k}+a_{k+1, k+1}\right| \leq M \\ \sum_{j=1}^{k}\left|a_{j, 1} x_{1}+\cdots+a_{j, k} x_{k}-a_{j, k+1}\right| \\ +\left|a_{k+1,1} x_{1}+\cdots+a_{k+1, k} x_{k}-a_{k+1, k+1}\right| \leq M \end{gathered} $$  for each choice of $x_{i}= \pm 1$. Since $|a+b|+|a-b| \geq 2|a|$ for all $a, b$, we obtain  $$ \begin{aligned} 2 \sum_{j=1}^{k}\left|a_{j 1} x_{1}+\cdots+a_{j k} x_{k}\right|+2\left|a_{k+1, k+1}\right| & \leq 2 M, \text { that is, } \\ \sum_{j=1}^{k}\left|a_{j 1} x_{1}+\cdots+a_{j k} x_{k}\right| & \leq M-\left|a_{k+1, k+1}\right| \end{aligned} $$  Now by the inductive assumption $\sum_{j=1}^{k}\left|a_{j j}\right| \leq M-\left|a_{k+1, k+1}\right|$, which is equivalent to the desired inequality.


12. (POL 6) Two congruent equilateral triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ in the plane are given. Show that the midpoints of the segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$ either are collinear or form an equilateral triangle.

Solution: 12. Let us start with the case $A=A^{\prime}$. If the triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ are oppositely oriented, then they are symmetric with respect to some axis, and the statement is true. Suppose that they are equally oriented. There is a rotation around $A$ by $60^{\circ}$ that maps $A B B^{\prime}$ onto $A C C^{\prime}$. This rotation also maps the midpoint $B_{0}$ of $B B^{\prime}$ onto the midpoint $C_{0}$ of $C C^{\prime}$, hence the triangle $A B_{0} C_{0}$ is equilateral. In the general case, when $A \neq A^{\prime}$, let us denote by $T$ the translation that maps $A$ onto $A^{\prime}$. Let $X^{\prime}$ be the image of a point $X$ under the (unique) isometry mapping $A B C$ onto $A^{\prime} B^{\prime} C^{\prime}$, and $X^{\prime \prime}$ the image of $X$ under $T$. Furthermore, let $X_{0}, X_{0}^{\prime}$ be the midpoints of segments $X X^{\prime}, X^{\prime} X^{\prime \prime}$. Then $X_{0}$ is the image of $X_{0}^{\prime}$ under the translation $-(1 / 2) T$. However, since it has already been proven that the triangle $A_{0}^{\prime} B_{0}^{\prime} C_{0}^{\prime}$ is equilateral, its image $A_{0} B_{0} C_{0}$ under $(1 / 2) T$ is also equilateral. The statement of the problem is thus proven.


13. (SWE 5) ${ }^{\mathrm{IMO} 6}$ Consider the $n \times n$ array of nonnegative integers  $$ \left(\begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1 n} \\ a_{21} & a_{22} & \ldots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \ldots & a_{n n} \end{array}\right) $$  with the following property: If an element $a_{i j}$ is zero, then the sum of the elements of the $i$ th row and the $j$ th column is greater than or equal to $n$. Prove that the sum of all the elements is greater than or equal to $\frac{1}{2} n^{2}$.

Solution: 13. Let $p$ be the least of all the sums of elements in one row or column. If $p \geq n / 2$, then the sum of all elements of the array is $s \geq n p \geq n^{2} / 2$.  Now suppose that $p<n / 2$. Without loss of generality, one can assume that the sum of elements in the first row is $p$, and that exactly the first $q$ elements of it are different from zero. Then the sum of elements in the last $n-q$ columns is greater than or equal to $(n-p)(n-q)$. Furthermore, the sum of elements in the first $q$ columns is greater than or equal to $p q$. This implies that the sum of all elements in the array is  $$ s \geq(n-p)(n-q)+p q=\frac{1}{2} n^{2}+\frac{1}{2}(n-2 p)(n-2 q) \geq \frac{1}{2} n^{2} $$  since $n \geq 2 p \geq 2 q$.


14. (USS 1) A broken line $A_{1} A_{2} \ldots A_{n}$ is drawn in a $50 \times 50$ square, so that the distance from any point of the square to the broken line is less than 1. Prove that its total length is greater than 1248.

Solution: 14. Denote by $V$ the figure made by a circle of radius 1 whose center moves along the broken line. From the condition of the problem, $V$ contains the whole $50 \times 50$ square, and thus the area $S(V)$ of $V$ is not less than 2500 . Let $L$ be the length of the broken line. We shall show that $S(V) \leq 2 L+\pi$, from which it will follow that $L \geq 1250-\pi / 2>1248$. For each segment $l_{i}=A_{i} A_{i+1}$ of the broken line, consider the figure $V_{i}$ obtained by a circle of radius 1 whose center moves along it, and let $\overline{V_{i}}$ be obtained by cutting off the circle of radius 1 with center at the starting point of $l_{i}$. The area of $\overline{V_{i}}$ is equal to $2 A_{i} A_{i+1}$. It is clear that the union of all the figures $\overline{V_{i}}$ together with a semicircle with center in $A_{1}$ and a semicircle with center in $A_{n}$ contains $V$ completely. Therefore  $$ S(V) \leq \pi+2 A_{1} A_{2}+2 A_{2} A_{3}+\cdots+2 A_{n-1} A_{n}=\pi+2 L . $$  This completes the proof.


15. (USS 2) Natural numbers from 1 to 99 (not necessarily distinct) are written on 99 cards. It is given that the sum of the numbers on any subset of cards (including the set of all cards) is not divisible by 100. Show that all the cards contain the same number.

Solution: 15. Assume the opposite. Then one can numerate the cards 1 to 99, with a number $n_{i}$ written on the card $i$, so that $n_{98} \neq n_{99}$. Denote by $x_{i}$ the remainder of $n_{1}+n_{2}+\cdots+n_{i}$ upon division by 100 , for $i=1,2, \ldots, 99$. All $x_{i}$ must be distinct: Indeed, if $x_{i}=x_{j}, i<j$, then $n_{i+1}+\cdots+n_{j}$ is divisible by 100, which is impossible. Also, no $x_{i}$ can be equal to 0 . Thus, the numbers $x_{1}, x_{2}, \ldots, x_{99}$ take exactly the values $1,2, \ldots, 99$ in some order. Let $x$ be the remainder of $n_{1}+n_{2}+\cdots+n_{97}+n_{99}$ upon division by 100 . It is not zero; hence it must be equal to $x_{k}$ for some $k \in\{1,2, \ldots, 99\}$. There are three cases: (i) $x=x_{k}, k \leq 97$. Then $n_{k+1}+n_{k+2}+\cdots+n_{97}+n_{99}$ is divisible by 100, a contradiction; (ii) $x=x_{98}$. Then $n_{98}=n_{99}$, a contradiction; (iii) $x=x_{99}$. Then $n_{98}$ is divisible by 100, a contradiction.  Therefore, all the cards contain the same number.


16. (USS 5) ${ }^{\mathrm{IMO} 2}$ Given a convex polyhedron $P_{1}$ with 9 vertices $A_{1}, \ldots, A_{9}$, let us denote by $P_{2}, P_{3}, \ldots, P_{9}$ the images of $P_{1}$ under the translations mapping the vertex $A_{1}$ to $A_{2}, A_{3}, \ldots, A_{9}$ respectively. Prove that among the polyhedra $P_{1}, \ldots, P_{9}$ at least two have a common interior point.

Solution: 16. Denote by $P^{\prime}$ the polyhedron defined as the image of $P$ under the homothety with center at $A_{1}$ and coefficient of similarity 2 . It is easy to see that all $P_{i}, i=1, \ldots, 9$, are contained in $P^{\prime}$ (indeed, if $M \in P_{k}$, then $\frac{1}{2} \overrightarrow{A_{1} M}=\frac{1}{2}\left(\overrightarrow{A_{1} A_{k}}+\overrightarrow{A_{1} M^{\prime}}\right)$ for some $M^{\prime} \in P$, and the claim follows from the convexity of $P$ ). But the volume of $P^{\prime}$ is exactly 8 times the volume of $P$, while the volumes of $P_{i}$ add up to 9 times that volume. We conclude that not all $P_{i}$ have disjoint interiors.


Solution: 17. We use the following obvious consequences of $(a+b)^{2} \geq 4 a b$ :  $$ \begin{aligned} & \frac{1}{\left(a_{1}+a_{2}\right)\left(a_{3}+a_{4}\right)} \geq \frac{4}{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}} \\ & \frac{1}{\left(a_{1}+a_{4}\right)\left(a_{2}+a_{3}\right)} \geq \frac{4}{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}} \end{aligned} $$  Now we have  $$ \begin{aligned} & \frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\frac{a_{4}+a_{2}}{a_{4}+a_{1}} \\ = & \frac{\left(a_{1}+a_{3}\right)\left(a_{1}+a_{2}+a_{3}+a_{4}\right)}{\left(a_{1}+a_{2}\right)\left(a_{3}+a_{4}\right)}+\frac{\left(a_{2}+a_{4}\right)\left(a_{1}+a_{2}+a_{3}+a_{4}\right)}{\left(a_{1}+a_{4}\right)\left(a_{2}+a_{3}\right)} \\ \geq & \frac{4\left(a_{1}+a_{3}\right)}{a_{1}+a_{2}+a_{3}+a_{4}}+\frac{4\left(a_{2}+a_{4}\right)}{a_{1}+a_{2}+a_{3}+a_{4}}=4 . \end{aligned} $$


2. (BUL 5) ${ }^{\mathrm{IMO} 5}$ Prove that for every natural number $m \geq 1$ there exists a finite set $S_{m}$ of points in the plane satisfying the following condition: If $A$ is any point in $S_{m}$, then there are exactly $m$ points in $S_{m}$ whose distance to $A$ equals 1.

Solution: 2 . We will construct such a set $S_{m}$ of $2^{m}$ points. Take vectors $u_{1}, \ldots, u_{m}$ in a given plane, such that $\left|u_{i}\right|=1 / 2$ and $0 \neq\left|c_{1} u_{1}+c_{2} u_{2}+\cdots+c_{n} u_{n}\right| \neq 1 / 2$ for any choice of numbers $c_{i}$ equal to 0 or $\pm 1$. Such vectors are easily constructed by induction on $m$ : For $u_{1}, \ldots, u_{m-1}$ fixed, there are only finitely many vector values $u_{m}$ that violate the upper condition, and we may set $u_{m}$ to be any other vector of length $1 / 2$. Let $S_{m}$ be the set of all points $M_{0}+\varepsilon_{1} u_{1}+\varepsilon_{2} u_{2}+\cdots+\varepsilon_{m} u_{m}$, where $M_{0}$ is any fixed point in the plane and $\varepsilon_{i}= \pm 1$ for $i=1, \ldots, m$. Then $S_{m}$ obviously satisfies the condition of the problem.


3. (GDR 1) Knowing that the system  $$ \begin{aligned} x+y+z & =3, \\ x^{3}+y^{3}+z^{3} & =15, \\ x^{4}+y^{4}+z^{4} & =35, \end{aligned} $$  has a real solution $x, y, z$ for which $x^{2}+y^{2}+z^{2}<10$, find the value of $x^{5}+y^{5}+z^{5}$ for that solution.

Solution: 3. Let $x, y, z$ be a solution of the given system with $x^{2}+y^{2}+z^{2}=\alpha<10$. Then  $$ x y+y z+z x=\frac{(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)}{2}=\frac{9-\alpha}{2} . $$  Furthermore, $3 x y z=x^{3}+y^{3}+z^{3}-(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$, which gives us $x y z=3(9-\alpha) / 2-4$. We now have  $$ \begin{aligned} 35= & x^{4}+y^{4}+z^{4}=\left(x^{3}+y^{3}+z^{3}\right)(x+y+z) \\ & -\left(x^{2}+y^{2}+z^{2}\right)(x y+y z+z x)+x y z(x+y+z) \\ = & 45-\frac{\alpha(9-\alpha)}{2}+\frac{9(9-\alpha)}{2}-12 . \end{aligned} $$  The solutions in $\alpha$ are $\alpha=7$ and $\alpha=11$. Therefore $\alpha=7$, xyz $=-1$, $x y+x z+y z=1$, and  $$ \begin{aligned} x^{5}+y^{5}+z^{5}= & \left(x^{4}+y^{4}+z^{4}\right)(x+y+z) \\ & -\left(x^{3}+y^{3}+z^{3}\right)(x y+x z+y z)+x y z\left(x^{2}+y^{2}+z^{2}\right) \\ = & 35 \cdot 3-15 \cdot 1+7 \cdot(-1)=83 . \end{aligned} $$


4. (GBR 3) We are given two mutually tangent circles in the plane, with radii $r_{1}, r_{2}$. A line intersects these circles in four points, determining three segments of equal length. Find this length as a function of $r_{1}$ and $r_{2}$ and the condition for the solvability of the problem.

Solution: 4. In the coordinate system in which the $x$-axis passes through the centers of the circles and the $y$-axis is their common tangent, the circles have equations  $$ x^{2}+y^{2}+2 r_{1} x=0, \quad x^{2}+y^{2}-2 r_{2} x=0 . $$  Let $p$ be the desired line with equation $y=a x+b$. The abscissas of points of intersection of $p$ with both circles satisfy one of  $$ \left(1+a^{2}\right) x^{2}+2\left(a b+r_{1}\right) x+b^{2}=0, \quad\left(1+a^{2}\right) x^{2}+2\left(a b-r_{2}\right) x+b^{2}=0 $$  Let us denote the lengths of the chords and their projections onto the $x$-axis by $d$ and $d_{1}$, respectively. From these equations it follows that  $$ d_{1}^{2}=\frac{4\left(a b+r_{1}\right)^{2}}{\left(1+a^{2}\right)^{2}}-\frac{4 b^{2}}{1+a^{2}}=\frac{4\left(a b-r_{2}\right)^{2}}{\left(1+a^{2}\right)^{2}}-\frac{4 b^{2}}{1+a^{2}} $$  Consider the point of intersection of $p$ with the $y$-axis. This point has equal powers with respect to both circles. Hence, if that point divides the segment determined on $p$ by the two circles on two segments of lengths $x$ and $y$, this power equals $x(x+d)=y(y+d)$, which implies $x=y=d / 2$. Thus each of the equations in (1) has two roots, one of which is thrice the other. This fact gives us $\left(a b+r_{1}\right)^{2}=4\left(1+a^{2}\right) b^{2} / 3$. From (1) and this we obtain  $$ \begin{gathered} a b=\frac{r_{2}-r_{1}}{2}, \quad 4 b^{2}+a^{2} b^{2}=3\left[\left(a b+r_{1}\right)^{2}-a^{2} b^{2}\right]=3 r_{1} r_{2} \\ a^{2}=\frac{4\left(r_{2}-r_{1}\right)^{2}}{14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}}, \quad b^{2}=\frac{14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}}{16} \\ d_{1}^{2}=\frac{\left(14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}\right)^{2}}{36\left(r_{1}+r_{2}\right)^{2}} \end{gathered} $$  Finally, since $d^{2}=d_{1}^{2}\left(1+a^{2}\right)$, we conclude that  $$ d^{2}=\frac{1}{12}\left(14 r_{1} r_{2}-r_{1}^{2}-r_{2}^{2}\right), $$  and that the problem is solvable if and only if $7-4 \sqrt{3} \leq \frac{r_{1}}{r_{2}} \leq 7+4 \sqrt{3}$.


5. (HUN 1) ${ }^{\mathrm{IMO}}$ Let $a, b, c, d, e$ be real numbers. Prove that the expression  $$ \begin{gathered} (a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e)+(c-a)(c-b)(c-d)(c-e) \\ +(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d) \end{gathered} $$  is nonnegative.

Solution: 5. Without loss of generality, we may assume that $a \geq b \geq c \geq d \geq e$. Then $a-b=-(b-a) \geq 0, a-c \geq b-c \geq 0, a-d \geq b-d \geq 0$ and $a-e \geq b-e \geq 0$, and hence  $$ (a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e) \geq 0 $$  Analogously, $(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d) \geq 0$. Finally, $(c-a)(c-b)(c-d)(c-e) \geq 0$ as a product of two nonnegative numbers, from which the inequality stated in the problem follows. Remark. The problem in an alternative formulation, accepted for the IMO, asked to prove that the analogous inequality  $$ \begin{gathered} \left(a_{1}-a_{2}\right)\left(a_{1}-a_{2}\right) \cdots\left(a_{1}-a_{n}\right)+\left(a_{2}-a_{1}\right)\left(a_{2}-a_{3}\right) \cdots\left(a_{2}-a_{n}\right)+\cdots \\ +\left(a_{n}-a_{1}\right)\left(a_{n}-a_{2}\right) \cdots\left(a_{n}-a_{n-1}\right) \geq 0 \end{gathered} $$  holds for arbitrary real numbers $a_{i}$ if and only if $n=3$ or $n=5$. The case $n=3$ is analogous to $n=5$. For $n=4$, a counterexample is $a_{1}=0, a_{2}=a_{3}=a_{4}=1$, while for $n>5$ one can take $a_{1}=a_{2}=\cdots=$ $a_{n-4}=0, a_{n-3}=a_{n-2}=a_{n-1}=2, a_{n}=1$ as a counterexample.


6. (HUN 7) Let $n \geq 2$ be a natural number. Find a way to assign natural numbers to the vertices of a regular $2^{n}$-gon such that the following conditions are satisfied: (1) only digits 1 and 2 are used; (2) each number consists of exactly $n$ digits; (3) different numbers are assigned to different vertices; (4) the numbers assigned to two neighboring vertices differ at exactly one digit.

Solution: 6. The proof goes by induction on $n$. For $n=2$, the following numeration satisfies the conditions (a)-(d): $C_{1}=11, C_{2}=12, C_{3}=22, C_{4}=21$. Suppose that $n>2$, and that the numeration $C_{1}, C_{2}, \ldots, C_{2^{n-1}}$ of a regular $2^{n-1}$-gon, in cyclical order, satisfies (i)-(iv). Then one can assign to the vertices of a $2^{n}$-gon cyclically the following numbers:  $$ \overline{1 C_{1}}, \overline{1 C_{2}}, \ldots, \overline{1 C_{2^{n-1}}}, \overline{2 C_{2^{n-1}}}, \ldots, \overline{2 C_{2}}, \overline{2 C_{1}} $$  The conditions (i), (ii) obviously hold, while (iii) and (iv) follow from the inductive assumption.


7. (NET 1) ${ }^{\mathrm{IMO} /}$ Given a tetrahedron $A B C D$ whose all faces are acuteangled triangles, set  $$ \sigma=\measuredangle D A B+\measuredangle B C D-\measuredangle A B C-\measuredangle C D A $$  Consider all closed broken lines $X Y Z T X$ whose vertices $X, Y, Z, T$ lie in the interior of segments $A B, B C, C D, D A$ respectively. Prove that: (a) if $\sigma \neq 0$, then there is no broken line $X Y Z T$ of minimal length; (b) if $\sigma=0$, then there are infinitely many such broken lines of minimal length. That length equals $2 A C \sin (\alpha / 2)$, where  $$ \alpha=\measuredangle B A C+\measuredangle C A D+\measuredangle D A B $$

Solution: 7. (a) Suppose that $X, Y, Z$ are fixed on segments $A B, B C, C D$. It is proven in a standard way that if $\angle A T X \neq \angle Z T D$, then $Z T+T X$ can be reduced. It follows that if there exists a broken line $X Y Z T X$ of minimal length, then the following conditions hold:  $$ \begin{aligned} & \angle D A B=\pi-\angle A T X-\angle A X T \\ & \angle A B C=\pi-\angle B X Y-\angle B Y X=\pi-\angle A X T-\angle C Y Z \\ & \angle B C D=\pi-\angle C Y Z-\angle C Z Y \\ & \angle C D A=\pi-\angle D T Z-\angle D Z T=\pi-\angle A T X-\angle C Z Y . \end{aligned} $$  Thus $\sigma=0$. (b) Now let $\sigma=0$. Let us cut the surface of the tetrahedron along the edges $A C, C D$, and $D B$ and set it down into a plane. Consider the plane figure $\mathcal{S}=A C D^{\prime} B D^{\prime \prime} C^{\prime}$ thus obtained made up of triangles $B C D^{\prime}, A B C, A B D^{\prime \prime}$, and $A C^{\prime} D^{\prime \prime}$, with $Z^{\prime}, T^{\prime}, Z^{\prime \prime}$ respectively on $C D^{\prime}, A D^{\prime \prime}, C^{\prime} D^{\prime \prime}$ (here $C^{\prime}$ corresponds to $C$, etc.). Since $\angle C^{\prime} D^{\prime \prime} A+\angle D^{\prime \prime} A B+\angle A B C+\angle B C D^{\prime}=0$ as an oriented angle (because $\sigma=0$ ), the lines $C D^{\prime}$ and $C^{\prime} D^{\prime \prime}$ are parallel and equally oriented; i.e., $C D^{\prime} D^{\prime \prime} C^{\prime}$ is a parallelogram. The broken line $X Y Z T X$ has minimal length if and only if $Z^{\prime \prime}, T^{\prime}, X$, $Y, Z^{\prime}$ are collinear (where $Z^{\prime} Z^{\prime \prime} \|$ $C C^{\prime}$ ), and then this length equals $Z^{\prime} Z^{\prime \prime}=C C^{\prime}=2 A C \sin (\alpha / 2)$. There is an infinity of such lines, one for every line $Z^{\prime} Z^{\prime \prime}$ parallel to $C C^{\prime}$ that meets the interiors of all the segments $C B, B A, A D^{\prime \prime}$. Such ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-389.jpg?height=300&width=516&top_left_y=408&top_left_x=835) $Z^{\prime} Z^{\prime \prime}$ exist. Indeed, the triangles $C A B$ and $D^{\prime \prime} A B$ are acute-angled, and thus the segment $A B$ has a common interior point with the parallelogram $C D^{\prime} D^{\prime \prime} C^{\prime}$. Therefore the desired result follows.


8. (NET 4) Determine whether there exist distinct real numbers $a, b, c, t$ for which: (i) the equation $a x^{2}+b t x+c=0$ has two distinct real roots $x_{1}, x_{2}$, (ii) the equation $b x^{2}+c t x+a=0$ has two distinct real roots $x_{2}, x_{3}$, (iii) the equation $c x^{2}+a t x+b=0$ has two distinct real roots $x_{3}, x_{1}$.

Solution: 8. Suppose that $a, b, c, t$ satisfy all the conditions. Then $a b c \neq 0$ and  $$ x_{1} x_{2}=\frac{c}{a}, \quad x_{2} x_{3}=\frac{a}{b}, \quad x_{3} x_{1}=\frac{b}{c} . $$  Multiplying these equations, we obtain $x_{1}^{2} x_{2}^{2} x_{3}^{2}=1$, and hence $x_{1} x_{2} x_{3}=$ $\varepsilon= \pm 1$. From (1) we get $x_{1}=\varepsilon b / a, x_{2}=\varepsilon c / b, x_{3}=\varepsilon a / c$. Substituting $x_{1}$ in the first equation, we get $a b^{2} / a^{2}+t \varepsilon b^{2} / a+c=0$, which gives us  $$ b^{2}(1+t \varepsilon)=-a c . $$  Analogously, $c^{2}(1+t \varepsilon)=-a b$ and $a^{2}(1+t \varepsilon)=-b c$, and therefore $(1+$ $t \varepsilon)^{3}=-1$; i.e., $1+t \varepsilon=-1$, since it is real. This also implies together with (1) that $b^{2}=a c, c^{2}=a b$, and $a^{2}=b c$, and consequently  $$ a=b=c \text {. } $$  Thus the three equations in the problem are equal, which is impossible. Hence, such $a, b, c, t$ do not exist.


9. (POL 1) Let $T_{k}=k-1$ for $k=1,2,3,4$ and  $$ T_{2 k-1}=T_{2 k-2}+2^{k-2}, \quad T_{2 k}=T_{2 k-5}+2^{k} \quad(k \geq 3) $$  Show that for all $k$,  $$ 1+T_{2 n-1}=\left[\frac{12}{7} 2^{n-1}\right] \quad \text { and } \quad 1+T_{2 n}=\left[\frac{17}{7} 2^{n-1}\right] $$  where $[x]$ denotes the greatest integer not exceeding $x$.

Solution: 9. We use induction. Since $T_{1}=0, T_{2}=1, T_{3}=2, T_{4}=3, T_{5}=5, T_{6}=8$, the statement is true for $n=1,2,3$. Suppose that both formulas from the problem hold for some $n \geq 3$. Then  $$ \begin{aligned} & T_{2 n+1}=1+T_{2 n}+2^{n-1}=\left[\frac{17}{7} 2^{n-1}+2^{n-1}\right]=\left[\frac{12}{7} 2^{n}\right] \\ & T_{2 n+2}=1+T_{2 n-3}+2^{n+1}=\left[\frac{12}{7} 2^{n-2}+2^{n+1}\right]=\left[\frac{17}{7} 2^{n}\right] . \end{aligned} $$  Therefore the formulas hold for $n+1$, which completes the proof.


1. (BUL 7) ${ }^{\mathrm{IMO} 5}$ Let $f$ and $\varphi$ be real functions defined on the set $\mathbb{R}$ satisfying the functional equation  $$ f(x+y)+f(x-y)=2 \varphi(y) f(x) $$  for arbitrary real $x, y$ (give examples of such functions). Prove that if $f(x)$ is not identically 0 and $|f(x)| \leq 1$ for all $x$, then $|\varphi(x)| \leq 1$ for all $x$.

Solution: 1. Suppose that $f\left(x_{0}\right) \neq 0$ and for a given $y$ define the sequence $x_{k}$ by the formula  $$ x_{k+1}= \begin{cases}x_{k}+y, & \text { if }\left|f\left(x_{k}+y\right)\right| \geq\left|f\left(x_{k}-y\right)\right| \\ x_{k}-y, & \text { otherwise. }\end{cases} $$  It follows from (1) that $\left|f\left(x_{k+1}\right)\right| \geq|\varphi(y)|\left|f\left(x_{k}\right)\right|$; hence by induction, $\left|f\left(x_{k}\right)\right| \geq|\varphi(y)|^{k}\left|f\left(x_{0}\right)\right|$. Since $\left|f\left(x_{k}\right)\right| \leq 1$ for all $k$, we obtain $|\varphi(y)| \leq 1$. Second solution. Let $M=\sup f(x) \leq 1$, and $x_{k}$ any sequence, possibly constant, such that $f\left(x_{k}\right) \rightarrow M, k \rightarrow \infty$. Then for all $k$,  $$ |\varphi(y)|=\frac{\left|f\left(x_{k}+y\right)+f\left(x_{k}-y\right)\right|}{2\left|f\left(x_{k}\right)\right|} \leq \frac{2 M}{2\left|f\left(x_{k}\right)\right|} \rightarrow 1, \quad k \rightarrow \infty $$


10. (NET 3) ${ }^{\text {IMO2 }}$ Prove that for each $n \geq 4$ every cyclic quadrilateral can be decomposed into $n$ cyclic quadrilaterals.

Solution: 10. Consider first a triangle. It can be decomposed into $k=3$ cyclic quadrilaterals by perpendiculars from some interior point of it to the sides; also, it can be decomposed into a cyclic quadrilateral and a triangle, and it follows by induction that this decomposition is possible for every $k$. Since every triangle can be cut into two triangles, the required decomposition is possible for each $n \geq 6$. It remains to treat the cases $n=4$ and $n=5$. $n=4$. If the center $O$ of the circumcircle is inside a cyclic quadrilateral $A B C D$, then the required decomposition is effected by perpendiculars from $O$ to the four sides. Otherwise, let $C$ and $D$ be the vertices of the obtuse angles of the quadrilateral. Draw the perpendiculars at $C$ and $D$ to the lines $B C$ and $A D$ respectively, and choose points $P$ and $Q$ on them such that $P Q \| A B$. Then the required decomposition is effected by $C P, P Q, Q D$ and the perpendiculars from $P$ and $Q$ to $A B$. $n=5$. If $A B C D$ is an isosceles trapezoid with $A B \| C D$ and $A D=B C$, then it is trivially decomposed by lines parallel to $A B$. Otherwise, $A B C D$ can be decomposed into a cyclic quadrilateral and a trapezoid; this trapezoid can be cut into an isosceles trapezoid and a triangle, which can further be cut into three cyclic quadrilaterals and an isosceles trapezoid.  Remark. It can be shown that the assertion is not true for $n=2$ and $n=3$.


11. (NET 6) Consider a sequence of circles $K_{1}, K_{2}, K_{3}, K_{4}, \ldots$ of radii $r_{1}, r_{2}, r_{3}, r_{4}, \ldots$, respectively, situated inside a triangle $A B C$. The circle $K_{1}$ is tangent to $A B$ and $A C ; K_{2}$ is tangent to $K_{1}, B A$, and $B C ; K_{3}$ is tangent to $K_{2}, C A$, and $C B ; K_{4}$ is tangent to $K_{3}, A B$, and $A C$; etc. (a) Prove the relation  $$ r_{1} \cot \frac{1}{2} A+2 \sqrt{r_{1} r_{2}}+r_{2} \cot \frac{1}{2} B=r\left(\cot \frac{1}{2} A+\cot \frac{1}{2} B\right), $$  where $r$ is the radius of the incircle of the triangle $A B C$. Deduce the existence of a $t_{1}$ such that  $$ r_{1}=r \cot \frac{1}{2} B \cot \frac{1}{2} C \sin ^{2} t_{1} . $$  (b) Prove that the sequence of circles $K_{1}, K_{2}, \ldots$ is periodic.

Solution: 11. Let $\angle A=2 x, \angle B=2 y, \angle C=2 z$. (a) Denote by $M_{i}$ the center of $K_{i}, i=1,2, \ldots$ If $N_{1}, N_{2}$ are the projections of $M_{1}, M_{2}$ onto $A B$, we have $A N_{1}=r_{1} \cot x, N_{2} B=r_{2} \cot y$, and $N_{1} N_{2}=\sqrt{\left(r_{1}+r_{2}\right)^{2}-\left(r_{1}-r_{2}\right)^{2}}=2 \sqrt{r_{1} r_{2}}$. The required relation between $r_{1}, r_{2}$ follows from $A B=A N_{1}+N_{1} N_{2}+N_{2} B$. If this relation is further considered as a quadratic equation in $\sqrt{r_{2}}$, then its discriminant, which equals  $$ \Delta=4\left(r(\cot x+\cot y) \cot y-r_{1}(\cot x \cot y-1)\right), $$  must be nonnegative, and therefore $r_{1} \leq r \cot y \cot z$. Then $t_{1}, t_{2}, \ldots$ exist, and we can assume that $t_{i} \in[0, \pi / 2]$. (b) Substituting $r_{1}=r \cot y \cot z \sin ^{2} t_{1}, r_{2}=r \cot z \cot x \sin ^{2} t_{2}$ in the relation of (a) we obtain that $\sin ^{2} t_{1}+\sin ^{2} t_{2}+k^{2}+2 k \sin t_{1} \sin t_{2}=1$, where we set $k=\sqrt{\tan x \tan y}$. It follows that $\left(k+\sin t_{1} \sin t_{2}\right)^{2}=$ $\left(1-\sin ^{2} t_{1}\right)\left(1-\sin ^{2} t_{2}\right)=\cos ^{2} t_{1} \cos ^{2} t_{2}$, and hence  $$ \cos \left(t_{1}+t_{2}\right)=\cos t_{1} \cos t_{2}-\sin t_{1} \sin t_{2}=k=\sqrt{\tan x \tan y} $$  which is constant. Writing the analogous relations for each $t_{i}, t_{i+1}$ we conclude that $t_{1}+t_{2}=t_{4}+t_{5}, t_{2}+t_{3}=t_{5}+t_{6}$, and $t_{3}+t_{4}=t_{6}+t_{7}$. It follows that $t_{1}=t_{7}$, i.e., $K_{1}=K_{7}$.


12. (USS 2) ${ }^{\mathrm{IMO} 1} \mathrm{~A}$ set of 10 positive integers is given such that the decimal expansion of each of them has two digits. Prove that there are two disjoint subsets of the set with equal sums of their elements.

Solution: 12. First we observe that it is not essential to require the subsets to be disjoint (if they aren't, one simply excludes their intersection). There are $2^{10}-1=$ 1023 different subsets and at most 990 different sums. By the pigeonhole principle there are two different subsets with equal sums.


2. (CZS 1) We are given $3 n$ points $A_{1}, A_{2}, \ldots, A_{3 n}$ in the plane, no three of them collinear. Prove that one can construct $n$ disjoint triangles with vertices at the points $A_{i}$.

Solution: 2. We use induction. For $n=1$ the assertion is obvious. Assume that it is true for a positive integer $n$. Let $A_{1}, A_{2}, \ldots, A_{3 n+3}$ be given $3 n+3$ points, and let w.l.o.g. $A_{1} A_{2} \ldots A_{m}$ be their convex hull. Among all the points $A_{i}$ distinct from $A_{1}, A_{2}$, we choose the one, say $A_{k}$, for which the angle $\angle A_{k} A_{1} A_{2}$ is minimal (this point is uniquely determined, since no three points are collinear). The line $A_{1} A_{k}$ separates the plane into two half-planes, one of which contains $A_{2}$ only, and the other one all the remaining $3 n$ points. By the inductive hypothesis, one can construct $n$ disjoint triangles with vertices in these $3 n$ points. Together with the triangle $A_{1} A_{2} A_{k}$, they form the required system of disjoint triangles.


3. (CZS 4) Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\cdots+x_{n}=$ 0 . Let $m$ be the least and $M$ the greatest among them. Prove that  $$ x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leq-n m M $$

Solution: 3. We have for each $k=1,2, \ldots, n$ that $m \leq x_{k} \leq M$, which gives $(M-$ $\left.x_{k}\right)\left(m-x_{k}\right) \leq 0$. It follows directly that  $$ 0 \geq \sum_{k=1}^{n}\left(M-x_{k}\right)\left(m-x_{k}\right)=n m M-(m+M) \sum_{k=1}^{n} x_{k}+\sum_{k=1}^{n} x_{k}^{2} $$  But $\sum_{k=1}^{n} x_{k}=0$, implying the required inequality.


4. (GDR 1) Let $n_{1}, n_{2}$ be positive integers. Consider in a plane $E$ two disjoint sets of points $M_{1}$ and $M_{2}$ consisting of $2 n_{1}$ and $2 n_{2}$ points, respectively, and such that no three points of the union $M_{1} \cup M_{2}$ are collinear. Prove that there exists a straightline $g$ with the following property: Each of the two half-planes determined by $g$ on $E$ ( $g$ not being included in either) contains exactly half of the points of $M_{1}$ and exactly half of the points of $M_{2}$.

Solution: 4. Choose in $E$ a half-line $s$ beginning at a point $O$. For every $\alpha$ in the interval $\left[0,180^{\circ}\right]$, denote by $s(\alpha)$ the line obtained by rotation of $s$ about $O$ by $\alpha$, and by $g(\alpha)$ the oriented line containing $s(\alpha)$ on which $s(\alpha)$ defines the positive direction. For each $P$ in $M_{i}, i=1,2$, let $P(\alpha)$ be the foot of the perpendicular from $P$ to $g(\alpha)$, and $l_{P}(\alpha)$ the oriented (positive, negative or zero) distance of $P(\alpha)$ from $O$. Then for $i=1,2$ one can arrange the $l_{P}(\alpha)\left(P \in M_{i}\right)$ in ascending order, as $l_{1}(\alpha), l_{2}(\alpha), \ldots, l_{2 n_{i}}(\alpha)$. Call $J_{i}(\alpha)$ the interval $\left[l_{n_{i}}(\alpha), l_{n_{i}+1}(\alpha)\right]$. It is easy to see that any line perpendicular to $g(\alpha)$ and passing through the point with the distance $l$ in the interior of $J_{i}(\alpha)$ from $O$, will divide the set $M_{i}$ into two subsets of equal cardinality. Therefore it remains to show that for some $\alpha$, the interiors of intervals $J_{1}(\alpha)$ and $J_{2}(\alpha)$ have a common point. If this holds for $\alpha=0$, then we have finished. Suppose w.l.o.g. that $J_{1}(0)$ lies on $g(0)$ to the left of $J_{2}(0)$; then $J_{1}\left(180^{\circ}\right)$ lies to the right of $J_{2}\left(180^{\circ}\right)$. Note that $J_{1}$ and $J_{2}$ cannot simultaneously degenerate to a point (otherwise, we would have four collinear points in $M_{1} \cup M_{2}$ ); also, each of them degenerates to a point for only finitely many values of $\alpha$. Since $J_{1}(\alpha)$ and $J_{2}(\alpha)$ move continuously, there exists a subinterval $I$ of $\left[0,180^{\circ}\right]$ on which they are not disjoint. Thus, at some point of $I$, they are both nondegenerate and have a common interior point, as desired.


5. (GDR 2) Prove the following assertion: The four altitudes of a tetrahedron $A B C D$ intersect in a point if and only if  $$ A B^{2}+C D^{2}=B C^{2}+A D^{2}=C A^{2}+B D^{2} $$

Solution: 5. Lemma. If $X, Y, Z, T$ are points in space, then the lines $X Z$ and $Y T$ are perpendicular if and only if $X Y^{2}+Z T^{2}=Y Z^{2}+T X^{2}$. Proof. Consider the plane $\pi$ through $X Z$ parallel to $Y T$. If $Y^{\prime}, T^{\prime}$ are the feet of the perpendiculars to $\pi$ from $Y, T$ respectively, then  $$ \begin{aligned} & \\ & X Y^{2}+Z T^{2}=X Y^{\prime 2}+Z T^{\prime 2}+2 Y Y^{\prime 2} \\ & \text { and } \quad Y Z^{2}+T X^{2}=Y^{\prime} Z^{2}+T^{\prime} X^{2}+2 Y Y^{\prime 2} \end{aligned} $$  Since by the Pythagorean theorem $X Y^{\prime 2}+Z T^{\prime 2}=Y^{\prime} Z^{2}+T^{\prime} X^{2}$, i.e., $X Y^{\prime 2}-Y^{\prime} Z^{2}=X T^{\prime 2}-T^{\prime} Z^{2}$, if and only if $Y^{\prime} T^{\prime} \perp X Z$, the statement follows. Assume that the four altitudes intersect in a point $P$. Then we have $D P \perp$ $A B C \Rightarrow D P \perp A B$ and $C P \perp A B D \Rightarrow C P \perp A B$, which implies that $C D P \perp A B$, and $C D \perp A B$. By the lemma, $A C^{2}+B D^{2}=A D^{2}+B C^{2}$. Using the same procedure we obtain the relation $A D^{2}+B C^{2}=A B^{2}+$ $C D^{2}$ 。 Conversely, assume that $A B^{2}+C D^{2}=A C^{2}+B D^{2}=A D^{2}+B C^{2}$. The lemma implies that $A B \perp C D, A C \perp B D, A D \perp B C$. Let $\pi$ be the plane containing $C D$ that is perpendicular to $A B$, and let $h_{D}$ be the altitude from $D$ to $A B C$. Since $\pi \perp A B$, we have $\pi \perp A B C \Rightarrow h_{D} \subset \pi$ and $\pi \perp A B D \Rightarrow h_{C} \subset \pi$. The altitudes $h_{D}$ and $h_{C}$ are not parallel; thus they have an intersection point $P_{C D}$. Analogously, $h_{B} \cap h_{C}=\left\{P_{B C}\right\}$ and $h_{B} \cap h_{D}=\left\{P_{B D}\right\}$, where both these points belong to $\pi$. On the other hand, $h_{B}$ doesn't belong to $\pi$; otherwise, it would be perpendicular to both $A C D$ and $A B \subset \pi$, i.e. $A B \subset A C D$, which is impossible. Hence, $h_{B}$ can have at most one common point with $\pi$, implying $P_{B D}=P_{C D}$. Analogously, $P_{A B}=P_{B D}=P_{C D}=P_{A B C D}$.


6. (GDR 3) Show that for any $n \not \equiv 0(\bmod 10)$ there exists a multiple of $n$ not containing the digit 0 in its decimal expansion.

Solution: 6. Let $n=2^{\alpha} 5^{\beta} m$, where $\alpha=0$ or $\beta=0$. These two cases are analogous, and we treat only $\alpha=0, n=5^{\beta} m$. The case $m=1$ is settled by the following lemma. Lemma. For any integer $\beta \geq 1$ there exists a multiple $M_{\beta}$ of $5^{\beta}$ with $\beta$ digits in decimal expansion, all different from 0. Proof. For $\beta=1, M_{1}=5$ works. Assume that the lemma is true for $\beta=k$. There is a positive integer $C_{k} \leq 5$ such that $C_{k} 2^{k}+m_{k} \equiv$ $0(\bmod 5)$, where $5^{k} m_{k}=M_{k}$, i.e. $C_{k} 10^{k}+M_{k} \equiv 0\left(\bmod 5^{k+1}\right)$. Then $M_{k+1}=C_{k} 10^{k}+M_{k}$ satisfies the conditions, and proves the lemma.  In the general case, consider, the sequence $1,10^{\beta}, 10^{2 \beta}, \ldots$ It contains two numbers congruent modulo $\left(10^{\beta}-1\right) m$, and therefore for some $k>0$, $10^{k \beta} \equiv 1\left(\bmod \left(10^{\beta}-1\right) m\right)$ (this is in fact a consequence of Fermat's theorem). The number  $$ \frac{10^{k \beta}-1}{10^{\beta}-1} M_{\beta}=10^{(k-1) \beta} M_{\beta}+10^{(k-2) \beta} M_{\beta}+\cdots+M_{\beta} $$  is a multiple of $n=5^{\beta} m$ with the required property.