7. $(\mathbf{G B R} 1)^{\mathrm{IMO} 6}$ (a) A plane $\pi$ passes through the vertex $O$ of the regular tetrahedron $O P Q R$. We define $p, q, r$ to be the signed distances of $P, Q, R$ from $\pi$ measured along a directed normal to $\pi$. Prove that  $$ p^{2}+q^{2}+r^{2}+(q-r)^{2}+(r-p)^{2}+(p-q)^{2}=2 a^{2} $$  where $a$ is the length of an edge of a tetrahedron. (b) Given four parallel planes not all of which are coincident, show that a regular tetrahedron exists with a vertex on each plane.

Solution: 7. (i) Consider the circumscribing cube $O Q_{1} P R_{1} O_{1} Q P_{1} R$ (that is, the cube in which the edges of the tetrahedron are small diagonals), of side $b=a \sqrt{2} / 2$. The left-hand side is the sum of squares of the projections of the edges of the tetrahedron onto a perpendicular $l$ to $\pi$. On the other hand, if $l$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-395.jpg?height=363&width=333&top_left_y=609&top_left_x=936) forms angles $\varphi_{1}, \varphi_{2}, \varphi_{3}$ with $O O_{1}, O Q_{1}, O R_{1}$ respectively, then the projections of $O P$ and $Q R$ onto $l$ have lengths $b\left(\cos \varphi_{2}+\cos \varphi_{3}\right)$ and $b\left|\cos \varphi_{2}-\cos \varphi_{3}\right|$. Summing up all these expressions, we obtain  $$ 4 b^{2}\left(\cos ^{2} \varphi_{1}+\cos ^{2} \varphi_{2}+\cos ^{2} \varphi_{3}\right)=4 b^{2}=2 a^{2} $$  (ii) We construct a required tetrahedron of edge length $a$ given in (i). Take $O$ arbitrarily on $\pi_{0}$, and let $p, q, r$ be the distances of $O$ from $\pi_{1}, \pi_{2}, \pi_{3}$. Since $a>p, q, r,|p-q|$, we can choose $P$ on $\pi_{1}$ anywhere at distance $a$ from $O$, and $Q$ at one of the two points on $\pi_{2}$ at distance $a$ from both $O$ and $P$. Consider the fourth vertex of the tetrahedron: its distance from $\pi_{0}$ will satisfy the equation from (i); i.e., there are two values for this distance; clearly, one of them is $r$, putting $R$ on $\pi_{3}$.


8. (GBR 2) ${ }^{\mathrm{IMO} 3}$ Let $m$ and $n$ be nonnegative integers. Prove that $m!n!(m+$ $n)$ ! divides $(2 m)!(2 n)!$.

Solution: 8. Let $f(m, n)=\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$. Then it is directly shown that  $$ f(m, n)=4 f(m, n-1)-f(m+1, n-1), $$  and thus $n$ may be successively reduced until one obtains $f(m, n)=$ $\sum_{r} c_{r} f(r, 0)$. Now $f(r, 0)$ is a simple binomial coefficient, and the $c_{r}$ 's are integers. Second solution. For each prime $p$, the greatest exponents of $p$ that divide the numerator $(2 m)!(2 n)$ ! and denominator $m!n!(m+n)$ ! are respectively  $$ \sum_{k>0}\left(\left[\frac{2 m}{p^{k}}\right]+\left[\frac{2 n}{p^{k}}\right]\right) \quad \text { and } \quad \sum_{k>0}\left(\left[\frac{m}{p^{k}}\right]+\left[\frac{n}{p^{k}}\right]+\left[\frac{m+n}{p^{k}}\right]\right) $$  hence it suffices to show that the first exponent is not less than the second one for every $p$. This follows from the fact that for each real $x,[2 x]+[2 y] \geq$ $[x]+[y]+[x+y]$, which is straightforward to prove (for example, using $[2 x]=[x]+[x+1 / 2])$.


Solution: 9. Clearly $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$ is a solution. We shall show that this describes all solutions. Suppose that not all $x_{i}$ are equal. Then among $x_{3}, x_{5}, x_{2}, x_{4}, x_{1}$ two consecutive are distinct: Assume w.l.o.g. that $x_{3} \neq x_{5}$. Moreover, since $\left(1 / x_{1}, \ldots, 1 / x_{5}\right)$ is a solution whenever $\left(x_{1}, \ldots, x_{5}\right)$ is, we may assume that $x_{3}<x_{5}$. Consider first the case $x_{1} \leq x_{2}$. We infer from (i) that $x_{1} \leq \sqrt{x_{3} x_{5}}<x_{5}$ and $x_{2} \geq \sqrt{x_{3} x_{5}}>x_{3}$. Then $x_{5}^{2}>x_{1} x_{3}$, which together with (iv) gives $x_{4}^{2} \leq x_{1} x_{3}<x_{3} x_{5}$; but we also have $x_{3}^{2} \leq x_{5} x_{2}$; hence by (iii), $x_{4}^{2} \geq$ $x_{5} x_{2}>x_{5} x_{3}$, a contradiction. Consider next the case $x_{1}>x_{2}$. We infer from (i) that $x_{1} \geq \sqrt{x_{3} x_{5}}>x_{3}$ and $x_{2} \leq \sqrt{x_{3} x_{5}}<x_{5}$. Then by (ii) and (v),  $$ x_{1} x_{4} \leq \max \left(x_{2}^{2}, x_{3}^{2}\right) \leq x_{3} x_{5} \quad \text { and } \quad x_{2} x_{4} \geq \min \left(x_{1}^{2}, x_{5}^{2}\right) \geq x_{3} x_{5} $$  which contradicts the assumption $x_{1}>x_{2}$. Second solution.  $$ \begin{aligned} 0 & \geq L_{1}=\left(x_{1}^{2}-x_{3} x_{5}\right)\left(x_{2}^{2}-x_{3} x_{5}\right)=x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\left(x_{1}^{2}+x_{2}^{2}\right) x_{3} x_{5} \\ & \geq x_{1}^{2} x_{2}^{2}+x_{3}^{2} x_{5}^{2}-\frac{1}{2}\left(x_{1}^{2} x_{3}^{2}+x_{1}^{2} x_{5}^{2}+x_{2}^{2} x_{3}^{2}+x_{2}^{2} x_{5}^{2}\right) \end{aligned} $$  and analogously for $L_{2}, \ldots, L_{5}$. Therefore $L_{1}+L_{2}+L_{3}+L_{4}+L_{5} \geq 0$, with the only case of equality $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}$.


1. (BUL 6) Let a tetrahedron $A B C D$ be inscribed in a sphere $S$. Find the locus of points $P$ inside the sphere $S$ for which the equality  $$ \frac{A P}{P A_{1}}+\frac{B P}{P B_{1}}+\frac{C P}{P C_{1}}+\frac{D P}{P D_{1}}=4 $$  holds, where $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are the intersection points of $S$ with the lines $A P, B P, C P$, and $D P$, respectively.

Solution: 1. The condition of the point $P$ can be written in the form $\frac{A P^{2}}{A P \cdot P A_{1}}+\frac{B P^{2}}{B P \cdot P B_{1}}+$ $\frac{C P^{2}}{C P \cdot P C_{1}}+\frac{D P^{2}}{D P \cdot P D_{1}}=4$. All the four denominators are equal to $R^{2}-O P^{2}$, i.e., to the power of $P$ with respect to $S$. Thus the condition becomes  $$ A P^{2}+B P^{2}+C P^{2}+D P^{2}=4\left(R^{2}-O P^{2}\right) $$  Let $M$ and $N$ be the midpoints of segments $A B$ and $C D$ respectively, and $G$ the midpoint of $M N$, or the centroid of $A B C D$. By Stewart's formula, an arbitrary point $P$ satisfies  $$ \begin{aligned} A P^{2}+B P^{2}+C P^{2}+D P^{2} & =2 M P^{2}+2 N P^{2}+\frac{1}{2} A B^{2}+\frac{1}{2} C D^{2} \\ & =4 G P^{2}+M N^{2}+\frac{1}{2}\left(A B^{2}+C D^{2}\right) \end{aligned} $$  Particularly, for $P \equiv O$ we get $4 R^{2}=4 O G^{2}+M N^{2}+\frac{1}{2}\left(A B^{2}+C D^{2}\right)$, and the above equality becomes  $$ A P^{2}+B P^{2}+C P^{2}+D P^{2}=4 G P^{2}+4 R^{2}-4 O G^{2} $$  Therefore (1) is equivalent to $O G^{2}=O P^{2}+G P^{2} \Leftrightarrow \angle O P G=90^{\circ}$. Hence the locus of points $P$ is the sphere with diameter $O G$. Now the converse is easy.


10. (SWE 3) ${ }^{\mathrm{IMO} 6}$ Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive numbers and $q$ a given real number, $0<q<1$. Find $n$ real numbers $b_{1}, b_{2}, \ldots, b_{n}$ that satisfy: (1) $a_{k}<b_{k}$ for all $k=1,2, \ldots, n$; (2) $q<\frac{b_{k+1}}{b_{k}}<\frac{1}{q}$ for all $k=1,2, \ldots, n-1$; (3) $b_{1}+b_{2}+\cdots+b_{n}<\frac{1+q}{1-q}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$.

Solution: 10. Let $b_{k}=a_{1} q^{k-1}+\cdots+a_{k-1} q+a_{k}+a_{k+1} q+\cdots+a_{n} q^{n-k}, \quad k=1,2, \ldots, n$. We show that these numbers satisfy the required conditions. Obviously $b_{k} \geq a_{k}$. Further,  $$ b_{k+1}-q b_{k}=-\left[\left(q^{2}-1\right) a_{k+1}+\cdots+q^{n-k-1}\left(q^{2}-1\right) a_{n}\right]>0 $$  we analogously obtain $q b_{k+1}-b_{k}<0$. Finally,  $$ \begin{aligned} b_{1}+b_{2}+\cdots+b_{n}= & a_{1}\left(q^{n-1}+\cdots+q+1\right)+\ldots \\ & +a_{k}\left(q^{n-k}+\cdots+q+1+q+\cdots+q^{k-1}\right)+\ldots \\ \leq & \left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1+2 q+2 q^{2}+\cdots+2 q^{n-1}\right) \\ < & \frac{1+q}{1-q}\left(a_{1}+\cdots+a_{n}\right) \end{aligned} $$


11. (SWE 4) ${ }^{\mathrm{IMO} 3}$ Determine the minimum of $a^{2}+b^{2}$ if $a$ and $b$ are real numbers for which the equation  $$ x^{4}+a x^{3}+b x^{2}+a x+1=0 $$  has at least one real solution.

Solution: 11. Putting $x+\frac{1}{x}=t$ we also get $x^{2}+\frac{1}{x^{2}}=t^{2}-2$, and the given equation reduces to $t^{2}+a t+b-2=0$. Since $x=\frac{t \pm \sqrt{t^{2}-4}}{2}, x$ will be real if and only if $|t| \geq 2, t \in \mathbb{R}$. Thus we need the minimum value of $a^{2}+b^{2}$ under the condition $a t+b=-\left(t^{2}-2\right),|t| \geq 2$. However, by the Cauchy-Schwarz inequality we have  $$ \left(a^{2}+b^{2}\right)\left(t^{2}+1\right) \geq(a t+b)^{2}=\left(t^{2}-2\right)^{2} $$  It follows that $a^{2}+b^{2} \geq h(t)=\frac{\left(t^{2}-2\right)^{2}}{t^{2}+1}$. Since $h(t)=\left(t^{2}+1\right)+\frac{9}{t^{2}+1}-6$ is increasing for $t \geq 2$, we conclude that $a^{2}+b^{2} \geq h(2)=\frac{4}{5}$. The cases of equality are easy to examine: These are $a= \pm \frac{4}{5}$ and $b=-\frac{2}{5}$. Second solution. In fact, there was no need for considering $x=t+1 / t$. By the Cauchy-Schwarz inequality we have $\left(a^{2}+2 b^{2}+a^{2}\right)\left(x^{6}+x^{4} / 2+x^{2}\right) \geq$ $\left(a x^{3}+b x^{2}+a x\right)^{2}=\left(x^{4}+1\right)^{2}$. Hence  $$ a^{2}+b^{2} \geq \frac{\left(x^{4}+1\right)^{2}}{2 x^{6}+x^{4}+2 x^{2}} \geq \frac{4}{5} $$  with equality for $x=1$.


12. (SWE 6) Consider the two square matrices  $$ A=\left[\begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & -1 & 1 \\ 1 & 1 & -1 & 1 & -1 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 \\ 1 & 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 \end{array}\right] $$  with entries 1 and -1 . The following operations will be called elementary: (1) Changing signs of all numbers in one row; (2) Changing signs of all numbers in one column; (3) Interchanging two rows (two rows exchange their positions); (4) Interchanging two columns.  Prove that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.

Solution: 12. Observe that the absolute values of the determinants of the given matrices are invariant under all the admitted operations. The statement follows from $\operatorname{det} A=16 \neq \operatorname{det} B=0$.


13. (YUG 4) Find the sphere of maximal radius that can be placed inside every tetrahedron that has all altitudes of length greater than or equal to 1.

Solution: 13. Let $S_{1}, S_{2}, S_{3}, S_{4}$ denote the areas of the faces of the tetrahedron, $V$ its volume, $h_{1}, h_{2}, h_{3}, h_{4}$ its altitudes, and $r$ the radius of its inscribed sphere. Since  $$ 3 V=S_{1} h_{1}=S_{2} h_{2}=S_{3} h_{3}=S_{4} h_{4}=\left(S_{1}+S_{2}+S_{3}+S_{4}\right) r, $$  it follows that  $$ \frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}}+\frac{1}{h_{4}}=\frac{1}{r} . $$  In our case, $h_{1}, h_{2}, h_{3}, h_{4} \geq 1$, hence $r \geq 1 / 4$. On the other hand, it is clear that a sphere of radius greater than $1 / 4$ cannot be inscribed in a tetrahedron all of whose altitudes have length equal to 1 . Thus the answer is $1 / 4$.


14. (YUG 5) ${ }^{\mathrm{IMO} 4} \mathrm{~A}$ soldier has to investigate whether there are mines in an area that has the form of an equilateral triangle. The radius of his detector is equal to one-half of an altitude of the triangle. The soldier starts from one vertex of the triangle. Determine the shortest path that the soldier has to traverse in order to check the whole region.

Solution: 14. Suppose that the soldier starts at the vertex $A$ of the equilateral triangle $A B C$ of side length $a$. Let $\varphi, \psi$ be the arcs of circles with centers $B$ and $C$ and radii $a \sqrt{3} / 4$ respectively, that lie inside the triangle. In order to check the vertices $B, C$, he must visit some points $D \in \varphi$ and $E \in \psi$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-401.jpg?height=324&width=394&top_left_y=1496&top_left_x=887)  Thus his path cannot be shorter than the path $A D E$ (or $A E D$ ) itself. The length of the path $A D E$ is $A D+D E \geq A D+D C-a \sqrt{3} / 4$. Let $F$ be the reflection of $C$ across the line $M N$, where $M, N$ are the midpoints of $A B$ and $B C$. Then $D C \geq D F$ and hence $A D+D C \geq A D+D F \geq A F$. Consequently $A D+D E \geq A F-a \frac{\sqrt{3}}{4}=a\left(\frac{\sqrt{7}}{2}-\frac{\sqrt{3}}{4}\right)$, with equality if and only if $D$ is the midpoint of $\operatorname{arc} \varphi$ and $E=(C D) \cap \psi$.  Moreover, it is easy to verify that, in following the path $A D E$, the soldier will check the whole region. Therefore this path (as well as the one symmetric to it) is shortest possible path that the soldier can take in order to check the entire field.


15. (CUB 1) Prove that for all $n \in \mathbb{N}$ the following is true:  $$ 2^{n} \prod_{k=1}^{n} \sin \frac{k \pi}{2 n+1}=\sqrt{2 n+1} $$

Solution: 15. If $z=\cos \theta+i \sin \theta$, then $z-z^{-1}=2 i \sin \theta$. Now put $z=\cos \frac{\pi}{2 n+1}+$ $i \sin \frac{\pi}{2 n+1}$. Using de Moivre's formula we transform the required equality into  $$ A=\prod_{k=1}^{n}\left(z^{k}-z^{-k}\right)=i^{n} \sqrt{2 n+1} $$  On the other hand, the complex numbers $z^{2 k}(k=-n,-n+1, \ldots, n)$ are the roots of $x^{2 n+1}-1$, and hence  $$ \prod_{k=1}^{n}\left(x-z^{2 k}\right)\left(x-z^{-2 k}\right)=\frac{x^{2 n+1}-1}{x-1}=x^{2 n}+\cdots+x+1 $$  Now we go back to proving (1). We have  $$ (-1)^{n} z^{n(n+1) / 2} A=\prod_{k=1}^{n}\left(1-z^{2 k}\right) \quad \text { and } \quad z^{-n(n+1) / 2} A=\prod_{k=1}^{n}\left(1-z^{-2 k}\right) $$  Multiplying these two equalities, we obtain $(-1)^{n} A^{2}=\prod_{k=1}^{n}\left(1-z^{2 k}\right)(1-$ $\left.z^{-2 k}\right)=2 n+1$, by (2). Therefore $A= \pm i^{-n} \sqrt{2 n+1}$. This actually implies that the required product is $\pm \sqrt{2 n+1}$, but it must be positive, since all the sines are, and the result follows.


16. (CUB 2) Given $a, \theta \in \mathbb{R}, m \in \mathbb{N}$, and $P(x)=x^{2 m}-2|a|^{m} x^{m} \cos \theta+a^{2 m}$, factorize $P(x)$ as a product of $m$ real quadratic polynomials.

Solution: 16. First, we have $P(x)=Q(x) R(x)$ for $Q(x)=x^{m}-|a|^{m} e^{i \theta}$ and $R(x)=$ $x^{m}-|a|^{m} e^{-i \theta}$, where $e^{i \varphi}$ means of course $\cos \varphi+i \sin \varphi$. It remains to factor both $Q$ and $R$. Suppose that $Q(x)=\left(x-q_{1}\right) \cdots\left(x-q_{m}\right)$ and $R(x)=\left(x-r_{1}\right) \cdots\left(x-r_{m}\right)$. Considering $Q(x)$, we see that $\left|q_{k}^{m}\right|=|a|^{m}$ and also $\left|q_{k}\right|=|a|$ for $k=$ $1, \ldots, m$. Thus we may put $q_{k}=|a| e^{i \beta_{k}}$ and obtain by de Moivre's formula $q_{k}^{m}=|a|^{m} e^{i m \beta_{k}}$. It follows that $m \beta_{k}=\theta+2 j \pi$ for some $j \in \mathbb{Z}$, and we have exactly $m$ possibilities for $\beta_{k}$ modulo $2 \pi$ : $\beta_{k}=\frac{\theta+2(k-1) \pi}{m}$ for $k=1,2, \ldots, m$. Thus $q_{k}=|a| e^{i \beta_{k}}$; analogously we obtain for $R(x)$ that $r_{k}=|a| e^{-i \beta_{k}}$. Consequently, $x^{m}-|a|^{m} e^{i \theta}=\prod_{k=1}^{m}\left(x-|a| e^{i \beta_{k}}\right) \quad$ and $\quad x^{m}-|a|^{m} e^{-i \theta}=\prod_{k=1}^{m}\left(x-|a| e^{-i \beta_{k}}\right)$. Finally, grouping the $k$ th factors of both polynomials, we get  $$ \begin{aligned} P(x) & =\prod_{k=1}^{m}\left(x-|a| e^{i \beta_{k}}\right)\left(x-|a| e^{-i \beta_{k}}\right)=\prod_{k=1}^{m}\left(x^{2}-2|a| x \cos \beta_{k}+a^{2}\right) \\ & =\prod_{k=1}^{m}\left(x^{2}-2|a| x \cos \frac{\theta+2(k-1) \pi}{m}+a^{2}\right) \end{aligned} $$


17. (POL 1) ${ }^{\mathrm{IMO} 5}$ Let $\mathcal{F}$ be a nonempty set of functions $f: \mathbb{R} \rightarrow \mathbb{R}$ of the form $f(x)=a x+b$, where $a$ and $b$ are real numbers and $a \neq 0$. Suppose that $\mathcal{F}$ satisfies the following conditions: (1) If $f, g \in \mathcal{F}$, then $g \circ f \in \mathcal{F}$, where $(g \circ f)(x)=g[f(x)]$. (2) If $f \in \mathcal{F}$ and $f(x)=a x+b$, then the inverse $f^{-1}$ of $f$ belongs to $\mathcal{F}$ $\left(f^{-1}(x)=(x-b) / a\right)$. (3) None of the functions $f(x)=x+c$, for $c \neq 0$, belong to $\mathcal{F}$.  Prove that there exists $x_{0} \in \mathbb{R}$ such that $f\left(x_{0}\right)=x_{0}$ for all $f \in \mathcal{F}$.

Solution: 17. Let $f_{1}(x)=a x+b$ and $f_{2}(x)=c x+d$ be two functions from $\mathcal{F}$. We define $g(x)=f_{1} \circ f_{2}(x)=a c x+(a d+b)$ and $h(x)=f_{2} \circ f_{1}(x)=a c x+(b c+d)$.  By the condition for $\mathcal{F}$, both $g(x)$ and $h(x)$ belong to $\mathcal{F}$. Moreover, there exists $h^{-1}(x)=\frac{x-(b c+d)}{a c}$, and  $$ h^{-1} \circ g(x)=\frac{a c x+(a d+b)-(b c+d)}{a c}=x+\frac{(a d+b)-(b c+d)}{a c} $$  belongs to $\mathcal{F}$. Now it follows that we must have $a d+b=b c+d$ for every $f_{1}, f_{2} \in \mathcal{F}$, which is equivalent to $\frac{b}{a-1}=\frac{d}{c-1}=k$. But these formulas exactly describe the fixed points of $f_{1}$ and $f_{2}: f_{1}(x)=a x+b=x \Rightarrow x=$ $\frac{b}{a-1}$. Hence all the functions in $\mathcal{F}$ fix the point $k$.


2. (CZS 1) Given a circle $K$, find the locus of vertices $A$ of parallelograms $A B C D$ with diagonals $A C \leq B D$, such that $B D$ is inside $K$.

Solution: 2. Let $D^{\prime}$ be the reflection of $D$ across $A$. Since $B C A D^{\prime}$ is then a parallelogram, the condition $B D \geq A C$ is equivalent to $B D \geq B D^{\prime}$, which is in turn equivalent to $\angle B A D \geq \angle B A D^{\prime}$, i.e. to $\angle B A D \geq 90^{\circ}$. Thus the needed locus is actually the locus of points $A$ for which there exist points $B, D$ inside $K$ with $\angle B A D=90^{\circ}$. Such points $B, D$ exist if and only if the two tangents from $A$ to $K$, say $A P$ and $A Q$, determine an obtuse angle. Then if $P, Q \in K$, we have $\angle P A O=\angle Q A O=\varphi>45^{\circ}$; hence $O A=\frac{O P}{\sin \varphi}<O P \sqrt{2}$. Therefore the locus of $A$ is the interior of the circle $K^{\prime}$ with center $O$ and radius $\sqrt{2}$ times the radius of $K$.


3. (CZS 6) ${ }^{\mathrm{IMO}}$ Prove that the sum of an odd number of unit vectors passing through the same point $O$ and lying in the same half-plane whose border passes through $O$ has length greater than or equal to 1 .

Solution: 3 . We use induction on odd numbers $n$. For $n=1$ there is nothing to prove. Suppose that the result holds for $n-2$ vectors, and let us be given vectors $v_{1}, v_{2}, \ldots, v_{n}$ arranged clockwise. Set $v^{\prime}=v_{2}+v_{3}+\cdots+v_{n-1}, u=v_{1}+v_{n}$, and $v=v_{1}+v_{2}+\cdots+v_{n}=v^{\prime}+u$. By the inductive hypothesis we have $\left|v^{\prime}\right| \geq 1$. Now if the angles between $v^{\prime}$ and the vectors $v_{1}, v_{n}$ are $\alpha$ and $\beta$ respectively, then the angle between $u$ and $v^{\prime}$ is $|\alpha-\beta| / 2 \leq 90^{\circ}$. Hence $\left|v^{\prime}+u\right| \geq\left|v^{\prime}\right| \geq 1$. Second solution. Again by induction, it can be easily shown that all possible values of the sum $v=v_{1}+v_{2}+\cdots+v_{n}$, for $n$ vectors $v_{1}, \ldots, v_{n}$ in the upper half-plane (with $y \geq 0$ ), are those for which $|v| \leq n$ and $|v-k e| \geq 1$ for every integer $k$ for which $n-k$ is odd, where $e$ is the unit vector on the $x$ axis.


4. (GBR 1) Let $P$ be a set of 7 different prime numbers and $C$ a set of 28 different composite numbers each of which is a product of two (not necessarily different) numbers from $P$. The set $C$ is divided into 7 disjoint four-element subsets such that each of the numbers in one set has a common prime divisor with at least two other numbers in that set. How many such partitions of $C$ are there?

Solution: 4. Each of the subsets must be of the form $\left\{a^{2}, a b, a c, a d\right\}$ or $\left\{a^{2}, a b, a c, b c\right\}$. It is now easy to count up the partitions. The result is 26460 .


5. (FRA 2) A circle of radius 1 is located in a right-angled trihedron and touches all its faces. Find the locus of centers of such circles.

Solution: 5. Let $O$ be the vertex of the trihedron, $Z$ the center of a circle $k$ inscribed in the trihedron, and $A, B, C$ points in which the plane of the circle meets the edges of the trihedron. We claim that the distance $O Z$ is constant. Set $O A=x, O B=y, O C=z, B C=a, C A=b, A B=c$, and let $S$ and $r=1$ be the area and inradius of $\triangle A B C$. Since $Z$ is the incenter of $A B C$, we have $(a+b+c) \overrightarrow{O Z}=a \overrightarrow{O A}+b \overrightarrow{O B}+c \overrightarrow{O C}$. Hence  $$ (a+b+c)^{2} O Z^{2}=(a \overrightarrow{O A}+b \overrightarrow{O B}+c \overrightarrow{O C})^{2}=a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2} $$  But since $y^{2}+z^{2}=a^{2}, z^{2}+x^{2}=b^{2}$ and $x^{2}+y^{2}=c^{2}$, we obtain $x^{2}=\frac{-a^{2}+b^{2}+c^{2}}{2}, y^{2}=\frac{a^{2}-b^{2}+c^{2}}{2}, z^{2}=\frac{a^{2}+b^{2}-c^{2}}{2}$. Substituting these values in (1) yields  $$ \begin{aligned} (a+b+c)^{2} O Z^{2} & =\frac{2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}}{2} \\ & =8 S^{2}=2(a+b+c)^{2} r^{2} \end{aligned} $$  Hence $O Z=r \sqrt{2}=\sqrt{2}$, and $Z$ belongs to a sphere $\sigma$ with center $O$ and radius $\sqrt{2}$. Moreover, the distances of $Z$ from the faces of the trihedron do not exceed 1 ; hence $Z$ belongs to a part of $\sigma$ that lies inside the unit cube with three faces lying on the faces of the trihedron. It is easy to see that this part of $\sigma$ is exactly the required locus.


6. (POL 2) ${ }^{\mathrm{IMO} 2}$ Does there exist a finite set $M$ of points in space, not all in the same plane, such that for each two points $A, B \in M$ there exist two other points $C, D \in M$ such that lines $A B$ and $C D$ are parallel?

Solution: 6. Yes. Take for $\mathcal{M}$ the set of vertices of a cube $A B C D E F G H$ and two points $I, J$ symmetric to the center $O$ of the cube with respect to the laterals $A B C D$ and $E F G H$. Remark. We prove a stronger result: Given an arbitrary finite set of points $\mathcal{S}$, then there is a finite set $\mathcal{M} \supset \mathcal{S}$ with the described property. Choose a point $A \in \mathcal{S}$ and any point $O$ such that $A O \| B C$ for some two points $B, C \in \mathcal{S}$. Now let $X^{\prime}$ be the point symmetric to $X$ with respect to $\underline{O}$, and $\mathcal{S}^{\prime}=\left\{X, X^{\prime} \mid X \in S\right\}$. Finally, take $\mathcal{M}=\left\{X, \bar{X} \mid X \in S^{\prime}\right\}$, where $\bar{X}$ denotes the point symmetric to $X$ with respect to $A$. This $\mathcal{M}$ has the desired property: If $X, Y \in \mathcal{M}$ and $Y \neq \bar{X}$, then $X Y \| \overline{X Y}$; otherwise, $X \bar{X}$, i.e., $X A$ is parallel to $X^{\prime} A^{\prime}$ if $X \neq A^{\prime}$, or to $B C$ otherwise.


7. (POL 3) Given a tetrahedron $A B C D$, let $x=A B \cdot C D, y=A C \cdot B D$, and $z=A D \cdot B C$. Prove that there exists a triangle with edges $x, y, z$.

Solution: 7. The result follows immediately from Ptolemy's inequality.


8. (ROM 1) Prove that there are exactly $\binom{k}{[k / 2]}$ arrays $a_{1}, a_{2}, \ldots, a_{k+1}$ of nonnegative integers such that $a_{1}=0$ and $\left|a_{i}-a_{i+1}\right|=1$ for $i=1,2, \ldots, k$.

Solution: 8. Let $f_{n}$ be the required total number, and let $f_{n}(k)$ denote the number of sequences $a_{1}, \ldots, a_{n}$ of nonnegative integers such that $a_{1}=0, a_{n}=k$, and $\left|a_{i}-a_{i+1}\right|=1$ for $i=1, \ldots, n-1$. In particular, $f_{1}(0)=1$ and $f_{n}(k)=0$ if $k<0$ or $k \geq n$. Since $a_{n-1}$ is either $k-1$ or $k+1$, we have  $$ f_{n}(k)=f_{n-1}(k+1)+f_{n-1}(k-1) \quad \text { for } k \geq 1 $$  By successive application of (1) we obtain  $$ f_{n}(k)=\sum_{i=0}^{r}\left[\binom{r}{i}-\binom{r}{i-k-1}\right] f_{n-r}(k+r-2 i) . $$  This can be verified by direct induction. Substituting $r=n-1$ in (2), we get at most one nonzero summand, namely the one for which $i=\frac{k+n-1}{2}$. Therefore $f_{n}(n-1-2 j)=\binom{n-1}{j}-\binom{n-1}{n-1}$. Adding up these equalities for $j=0,1, \ldots,\left[\frac{n-1}{2}\right]$ we obtain $f_{n}=\binom{n-1}{\left[\frac{n-1}{2}\right]}$, as required.


9. (ROM 2) Let $O x, O y, O z$ be three rays, and $G$ a point inside the trihedron $O x y z$. Consider all planes passing through $G$ and cutting $O x, O y, O z$ at points $A, B, C$, respectively. How is the plane to be placed in order to yield a tetrahedron $O A B C$ with minimal perimeter?

Solution: 9. Let $a, b, c$ be vectors going along $O x, O y, O z$, respectively, such that $\overrightarrow{O G}=$ $\xrightarrow{a+b}+c$. Now let $A \in O x, B \in O y, C \in O z$ and let $\overrightarrow{O A}=\alpha a, \overrightarrow{O B}=\beta b$, $\overrightarrow{O C}=\gamma c$, where $\alpha, \beta, \gamma>0$. Point $G$ belongs to a plane $A B C$ with $A \in O x, B \in O y, C \in O z$ if and only if there exist positive real numbers $\lambda, \mu, \nu$ with sum 1 such that $\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}+\nu \overrightarrow{O C}=\overrightarrow{O G}$, which is equivalent to $\lambda \alpha=\mu \beta=\nu \gamma=1$. Such $\lambda, \mu, \nu$ exist if and only if  $$ \alpha, \beta, \gamma>0 \quad \text { and } \quad \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=1 $$  Since the volume of $O A B C$ is proportional to the product $\alpha \beta \gamma$, it is minimized when $\frac{1}{\alpha} \cdot \frac{1}{\beta} \cdot \frac{1}{\gamma}$ is maximized, which occurs when $\alpha=\beta=\gamma=3$ and $G$ is the centroid of $\triangle A B C$.


1. I 1 (USA 4) ${ }^{\mathrm{IMO}}$ Alice, Betty, and Carol took the same series of examinations. There was one grade of $A$, one grade of $B$, and one grade of $C$ for each examination, where $A, B, C$ are different positive integers. The final test scores were  | Alice | Betty | Carol | | :---: | :---: | :---: | | 20 | 10 | 9 |  If Betty placed first in the arithmetic examination, who placed second in the spelling examination?

Solution: 1. Denote by $n$ the number of exams. We have $n(A+B+C)=20+10+9=39$, and since $A, B, C$ are distinct, their sum is at least 6 ; therefore $n=3$ and $A+B+C=13$. Assume w.l.o.g. that $A>B>C$. Since Betty gained $A$ points in arithmetic, but fewer than 13 points in total, she had $C$ points in both remaining exams (in spelling as well). Furthermore, Carol also gained fewer than 13 points, but with at least $B$ points on two examinations (on which Betty scored $C$ ), including spelling. If she had $A$ in spelling, then she would have at least $A+B+C=13$ points in total, a contradiction. Hence, Carol scored $B$ and placed second in spelling. Remark. Moreover, it follows that Alice, Betty, and Carol scored $B+A+A$, $A+C+C$, and $C+B+B$ respectively, and that $A=8, B=4, C=1$.


10. II 4 (FIN 3) ${ }^{\mathrm{IMO} 2}$ Let $\triangle A B C$ be a triangle. Prove that there exists a point $D$ on the side $A B$ such that $C D$ is the geometric mean of $A D$ and $B D$ if and only if $\sqrt{\sin A \sin B} \leq \sin \frac{C}{2}$.

Solution: 10. If we set $\angle A C D=\gamma_{1}$ and $\angle B C D=\gamma_{2}$ for a point $D$ on the segment $A B$, then by the sine theorem,  $$ f(D)=\frac{C D^{2}}{A D \cdot B D}=\frac{C D}{A D} \cdot \frac{C D}{B D}=\frac{\sin \alpha \sin \beta}{\sin \gamma_{1} \sin \gamma_{2}} . $$  The denominator of the last fraction is  $$ \begin{aligned} \sin \gamma_{1} \sin \gamma_{2} & =\frac{1}{2}\left(\cos \left(\gamma_{1}-\gamma_{2}\right)-\cos \left(\gamma_{1}+\gamma_{2}\right)\right) \\ & =\frac{1}{2}\left(\cos \left(\gamma_{1}-\gamma_{2}\right)-\cos \gamma\right) \leq \frac{1-\cos \gamma}{2}=\sin ^{2} \frac{\gamma}{2} \end{aligned} $$  from which we deduce that the set of values of $f(D)$ is the interval $\left[\frac{\sin \alpha \sin \beta}{\sin ^{2} \frac{1}{2}},+\infty\right.$ ). Hence $f(D)=1$ (equivalently, $\left.C D^{2}=A D \cdot B D\right)$ is possible if and only if $\sin \alpha \sin \beta \leq \sin ^{2} \frac{\gamma}{2}$, i.e.,  $$ \sqrt{\sin \alpha \sin \beta} \leq \sin \frac{\gamma}{2} $$  Second solution. Let $E$ be the second point of intersection of the line $C D$ with the circumcircle $k$ of $A B C$. Since $A D \cdot B D=C D \cdot E D$ (power of $D$ with respect to $k$ ), $C D^{2}=A D \cdot B D$ ie equivalent to $E D \geq C D$. Clearly the ratio $\frac{E D}{C D}(D \in A B)$ takes a minimal value when $E$ is the midpoint of the arc $A B$ not containing $C$. (This follows from $E D: C D=E^{\prime} D: C^{\prime} D$ when $C^{\prime}$ and $E^{\prime}$ are respectively projections from $C$ and $E$ onto $A B$.) On the other hand, it is directly shown that in this case  $$ \frac{E D}{C D}=\frac{\sin ^{2} \frac{\gamma}{2}}{\sin \alpha \sin \beta} $$  and the assertion follows.


11. II 5 (BUL 1) ${ }^{\mathrm{IMO} 4}$ Consider a partition of an $8 \times 8$ chessboard into $p$ rectangles whose interiors are disjoint such that each of them has an equal number of white and black cells. Assume that $a_{1}<a_{2}<\cdots<a_{p}$, where $a_{i}$ denotes the number of white cells in the $i$ th rectangle. Find the maximal $p$ for which such a partition is possible and for that $p$ determine all possible corresponding sequences $a_{1}, a_{2}, \ldots, a_{p}$.

Solution: 11. First, we notice that $a_{1}+a_{2}+\cdots+a_{p}=32$. The numbers $a_{i}$ are distinct, and consequently $a_{i} \geq i$ and $a_{1}+\cdots+a_{p} \geq p(p+1) / 2$. Therefore $p \leq 7$. The number 32 can be represented as a sum of 7 mutually distinct positive integers in the following ways: (1) $32=1+2+3+4+5+6+11$; (2) $32=1+2+3+4+5+7+10$; (3) $32=1+2+3+4+5+8+9$; (4) $32=1+2+3+4+6+7+9$; (5) $32=1+2+3+5+6+7+8$.  The case (1) is eliminated because there is no rectangle with 22 cells on an $8 \times 8$ chessboard. In the other cases the partitions are realized as below. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=239&width=250&top_left_y=1281&top_left_x=234)  Case (2) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=239&width=243&top_left_y=1281&top_left_x=518)  Case (3) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=236&width=241&top_left_y=1282&top_left_x=808)  Case (4) ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-408.jpg?height=238&width=243&top_left_y=1279&top_left_x=1098)  Case (5)


12. II 6 (USS 1) In a certain language words are formed using an alphabet of three letters. Some words of two or more letters are not allowed, and any two such distinct words are of different lengths. Prove that one can form a word of arbitrary length that does not contain any nonallowed word.

Solution: 12. We say that a word is good if it doesn't contain any nonallowed word. Let $a_{n}$ be the number of good words of length $n$. If we prolong any good word of length $n$ by adding one letter to its end (there are $3 a_{n}$ words that can be so obtained), we get either (i) a good word of length $n+1$, or (ii) an $(n+1)$-letter word of the form $X Y$, where $X$ is a good word and $Y$ a nonallowed word. The number of words of type (ii) with word $Y$ of length $k$ is exactly $a_{n+1-k}$; hence the total number of words of kind (ii) doesn't exceed $a_{n-1}+$ $\cdots+a_{1}+a_{0}$ (where $a_{0}=1$ ). Hence  $$ a_{n+1} \geq 3 a_{n}-\left(a_{n-1}+\cdots+a_{1}+a_{0}\right), \quad a_{0}=1, a_{1}=3 $$  We prove by induction that $a_{n+1}>2 a_{n}$ for all $n$. For $n=1$ the claim is trivial. If it holds for $i \leq n$, then $a_{i} \leq 2^{i-n} a_{n}$; thus we obtain from (1)  $$ a_{n+1}>a_{n}\left(3-\frac{1}{2}-\frac{1}{2^{2}}-\cdots-\frac{1}{2^{n}}\right)>2 a_{n} $$  Therefore $a_{n} \geq 2^{n}$ for all $n$ (moreover, one can show from (1) that $a_{n} \geq$ $(n+2) 2^{n-1}$ ); hence there exist good words of length $n$. Remark. If there are two nonallowed words (instead of one) of each length greater than 1, the statement of the problem need not remain true.


2. I 2 (POL 1) Prove that the squares with sides $1 / 1,1 / 2,1 / 3, \ldots$ may be put into the square with side $3 / 2$ in such a way that no two of them have any interior point in common.

Solution: 2. We denote by $q_{i}$ the square with side $\frac{1}{i}$. Let us divide the big square into rectangles $r_{i}$ by parallel lines, where the size of $r_{i}$ is $\frac{3}{2} \times \frac{1}{2^{i}}$ for $i=2,3, \ldots$ and $\frac{3}{2} \times 1$ for $i=1$ (this can be done because $1+\sum_{i=2}^{\infty} \frac{1}{2^{i}}=\frac{3}{2}$ ). In rectangle $r_{1}$, one can put the squares $q_{1}, q_{2}, q_{3}$, as is done on the figure. Also, since $\frac{1}{2^{i}}+\cdots+\frac{1}{2^{i+1}-1}<2^{i} \cdot \frac{1}{2^{i}}=1<\frac{3}{2}$, in each $r_{i}, i \geq 2$, one can put $q_{2^{i}}, \ldots, q_{2^{i+1}-1}$. This completes the proof. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-404.jpg?height=252&width=254&top_left_y=1103&top_left_x=664)  Remark. It can be shown that the squares $q_{1}, q_{2}$ cannot fit in any square of side less than $\frac{3}{2}$.


3. I 3 (SWE 3) ${ }^{\mathrm{IMO}}$ Let $P(x)$ be a polynomial with integer coefficients. If $n(P)$ is the number of (distinct) integers $k$ such that $P^{2}(k)=1$, prove that  $$ n(P)-\operatorname{deg}(P) \leq 2 $$  where $\operatorname{deg}(P)$ denotes the degree of the polynomial $P$.

Solution: 3. For $\operatorname{deg}(P) \leq 2$ the statement is obvious, since $n(P) \leq \operatorname{deg}\left(P^{2}\right)=$ $2 \operatorname{deg}(P) \leq \operatorname{deg}(P)+2$ 。 Suppose now that $\operatorname{deg}(P) \geq 3$ and $n(P)>\operatorname{deg}(P)+2$. Then there is at least one integer $b$ for which $P(b)=-1$, and at least one $x$ with $P(x)=1$. We may assume w.l.o.g. that $b=0$ (if necessary, we consider the polynomial $P(x+b)$ instead). If $k_{1}, \ldots, k_{m}$ are all integers for which $P\left(k_{i}\right)=1$, then $P(x)=Q(x)\left(x-k_{1}\right) \cdots\left(x-k_{m}\right)+1$ for some polynomial $Q(x)$ with integer coefficients. Setting $x=0$ we obtain $(-1)^{m} Q(0) k_{1} \cdots k_{m}=1-P(0)=2$. It follows that $k_{1} \cdots k_{m} \mid 2$, and hence $m$ is at most 3 . The same holds for the polynomial $-P(x)$, and thus $P(x)=-1$ also has at most 3 integer solutions. This counts for 6 solutions of $P^{2}(x)=1$ in total, implying the statement for $\operatorname{deg}(P) \geq 4$. It remains to verify the statement for $n=3$. If $\operatorname{deg}(P)=3$ and $n(P)=6$, then it follows from the above consideration that $P(x)$ is either $-\left(x^{2}-\right.$ $1)(x-2)+1$ or $\left(x^{2}-1\right)(x+2)+1$. It is directly checked that $n(P)$ equals only 4 in both cases.


4. I 4 (USS 4) The sum of the squares of five real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ equals 1. Prove that the least of the numbers $\left(a_{i}-a_{j}\right)^{2}$, where $i, j=$ $1,2,3,4,5$ and $i \neq j$, does not exceed $1 / 10$.

Solution: 4. Assume w.l.o.g. that $a_{1} \leq a_{2} \leq a_{3} \leq a_{4} \leq a_{5}$. If $m$ is the least value of $\left|a_{i}-a_{j}\right|, i \neq j$, then $a_{i+1}-a_{i} \geq m$ for $i=1,2, \ldots, 5$, and consequently $a_{i}-a_{j} \geq(i-j) m$ for any $i, j \in\{1, \ldots, 5\}, i>j$. Then it follows that  $$ \sum_{i>j}\left(a_{i}-a_{j}\right)^{2} \geq m^{2} \sum_{i>j}(i-j)^{2}=50 m^{2} $$  On the other hand, by the condition of the problem,  $$ \sum_{i>j}\left(a_{i}-a_{j}\right)^{2}=5 \sum_{i=1}^{5} a_{i}^{2}-\left(a_{1}+\cdots+a_{5}\right)^{2} \leq 5 $$  Therefore $50 m^{2} \leq 5$; i.e., $m^{2} \leq \frac{1}{10}$.


5. I 5 (GBR 3) Let $A_{r}, B_{r}, C_{r}$ be points on the circumference of a given circle $S$. From the triangle $A_{r} B_{r} C_{r}$, called $\triangle_{r}$, the triangle $\triangle_{r+1}$ is obtained by constructing the points $A_{r+1}, B_{r+1}, C_{r+1}$ on $S$ such that $A_{r+1} A_{r}$ is parallel to $B_{r} C_{r}, B_{r+1} B_{r}$ is parallel to $C_{r} A_{r}$, and $C_{r+1} C_{r}$ is parallel to $A_{r} B_{r}$. Each angle of $\triangle_{1}$ is an integer number of degrees and those integers are not multiples of 45 . Prove that at least two of the triangles $\triangle_{1}, \triangle_{2}, \ldots, \triangle_{15}$ are congruent.

Solution: 5. All the angles are assumed to be oriented and measured modulo $180^{\circ}$. Denote by $\alpha_{i}, \beta_{i}, \gamma_{i}$ the angles of triangle $\triangle_{i}$, at $A_{i}, B_{i}, C_{i}$ respectively. Let us determine the angles of $\triangle_{i+1}$. If $D_{i}$ is the intersection of lines $B_{i} B_{i+1}$ and $C_{i} C_{i+1}$, we have $\angle B_{i+1} A_{i+1} C_{i+1}=\angle D_{i} B_{i} C_{i+1}=\angle B_{i} D_{i} C_{i+1}+$ $\angle D_{i} C_{i+1} B_{i}=\angle B_{i} D_{i} C_{i}-\angle B_{i} C_{i+1} C_{i}=-2 \angle B_{i} A_{i} C_{i}$. We conclude that  $$ \alpha_{i+1}=-2 \alpha_{i}, \quad \text { and analogously } \quad \beta_{i+1}=-2 \beta_{i}, \quad \gamma_{i+1}=-2 \gamma_{i} $$  Therefore $\alpha_{r+t}=(-2)^{t} \alpha_{r}$. However, since $(-2)^{12} \equiv 1(\bmod 45)$ and consequently $(-2)^{14} \equiv(-2)^{2}(\bmod 180)$, it follows that $\alpha_{15}=\alpha_{3}$, since all values are modulo $180^{\circ}$. Analogously, $\beta_{15}=\beta_{3}$ and $\gamma_{15}=\gamma_{3}$, and moreover, $\triangle_{3}$ and $\triangle_{15}$ are inscribed in the same circle; hence $\triangle_{3} \cong \triangle_{15}$.


6. I 6 (ROM 4) ${ }^{\mathrm{IMO} 3}$ Does there exist a natural number $n$ for which the number  $$ \sum_{k=0}^{n}\binom{2 n+1}{2 k+1} 2^{3 k} $$  is divisible by 5 ?

Solution: 6. We set  $$ \begin{aligned} & x=\sum_{k=0}^{n}\binom{2 n+1}{2 k+1} 2^{3 k}=\frac{1}{\sqrt{8}} \sum_{k=0}^{n}\binom{2 n+1}{2 k+1} \sqrt{8}^{2 k+1}, \\ & y=\sum_{k=0}^{n}\binom{2 n+1}{2 k} 2^{3 k}=\sum_{k=0}^{n}\binom{2 n+1}{2 k} \sqrt{8}^{2 k} \end{aligned} $$  Both $x$ and $y$ are positive integers. Also, from the binomial formula we obtain  $$ y+x \sqrt{8}=\sum_{i=0}^{2 n+1}\binom{2 n+1}{i} \sqrt{8}^{i}=(1+\sqrt{8})^{2 n+1} $$  and similarly  $$ y-x \sqrt{8}=(1-\sqrt{8})^{2 n+1} . $$  Multiplying these equalities, we get $y^{2}-8 x^{2}=(1+\sqrt{8})^{2 n+1}(1-\sqrt{8})^{2 n+1}=$ $-7^{2 n+1}$. Reducing modulo 5 gives us  $$ 3 x^{2}-y^{2} \equiv 2^{2 n+1} \equiv 2 \cdot(-1)^{n} $$  Now we see that if $x$ is divisible by 5 , then $y^{2} \equiv \pm 2(\bmod 5)$, which is impossible. Therefore $x$ is never divisible by 5 . Second solution. Another standard way is considering recurrent formulas. If we set  $$ x_{m}=\sum_{k}\binom{m}{2 k+1} 8^{k}, \quad y_{m}=\sum_{k}\binom{m}{2 k} 8^{k} $$  then since $\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$, it follows that $x_{m+1}=x_{m}+y_{m}$ and $y_{m+1}=8 x_{m}+y_{m}$; therefore $x_{m+1}=2 x_{m}+7 x_{m-1}$. We need to show that none of $x_{2 n+1}$ are divisible by 5 . Considering the sequence $\left\{x_{m}\right\}$ modulo 5 , we get that $x_{m}=0,1,2,1,1,4,0,3,1,3,3,2,0,4,3,4,4,1, \ldots$ Zeros occur in the initial position of blocks of length 6 , where each subsequent block is obtained by multiplying the previous one by 3 (modulo 5 ). Consequently, $x_{m}$ is divisible by 5 if and only if $m$ is a multiple of 6 , which cannot happen if $m=2 n+1$.


7. II 1 (POL 2) Let $a_{i}, b_{i}$ be coprime positive integers for $i=1,2, \ldots, k$, and $m$ the least common multiple of $b_{1}, \ldots, b_{k}$. Prove that the greatest common divisor of $a_{1} \frac{m}{b_{1}}, \ldots, a_{k} \frac{m}{b_{k}}$ equals the greatest common divisor of $a_{1}, \ldots, a_{k}$.

Solution: 7. Consider an arbitrary prime number $p$. If $p \mid m$, then there exists $b_{i}$ that is divisible by the same power of $p$ as $m$. Then $p$ divides neither $a_{i} \frac{m}{b_{i}}$ nor $a_{i}$, because $\left(a_{i}, b_{i}\right)=1$. If otherwise $p \nmid m$, then $\frac{m}{b_{i}}$ is not divisible by $p$ for any $i$, hence $p$ divides $a_{i}$ and $a_{i} \frac{m}{b_{i}}$ to the same power. Therefore $\left(a_{1}, \ldots, a_{k}\right)$ and $\left(a_{1} \frac{m}{b_{1}}, \ldots, a_{k} \frac{m}{b_{k}}\right)$ have the same factorization; hence they are equal. Second solution. For $k=2$ we easily verify the formula $\left(m \frac{a_{1}}{b_{1}}, m \frac{a_{2}}{b_{2}}\right)=$ $\frac{m}{b_{1} b_{2}}\left(a_{1} b_{2}, a_{2} b_{1}\right)=\frac{1}{b_{1} b_{2}}\left[b_{1}, b_{2}\right]\left(a_{1}, a_{2}\right)\left(b_{1}, b_{2}\right)=\left(a_{1}, a_{2}\right)$, since $\left[b_{1}, b_{2}\right]$. $\left(b_{1}, b_{2}\right)=b_{1} b_{2}$. We proceed by induction:  $$ \begin{aligned} \left(a_{1} \frac{m}{b_{1}}, \ldots, a_{k} \frac{m}{b_{k}}, a_{k+1} \frac{m}{b_{k+1}}\right) & =\left(\frac{m}{\left[b_{1}, \ldots, b_{k}\right]}\left(a_{1}, \ldots, a_{k}\right), a_{k+1} \frac{m}{b_{k+1}}\right) \\ & =\left(a_{1}, \ldots, a_{k}, a_{k+1}\right) . \end{aligned} $$


8. II 2 (NET 3) ${ }^{\mathrm{IMO} 5}$ If $a, b, c, d$ are arbitrary positive real numbers, find all possible values of  $$ S=\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d} . $$

Solution: 8. It is clear that  $$ \begin{gathered} \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}<S \\ \text { and } \quad S<\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{c+d}+\frac{d}{c+d} \end{gathered} $$  or equivalently, $1<S<2$. On the other hand, all values from $(1,2)$ are attained. Since $S=1$ for $(a, b, c, d)=(0,0,1,1)$ and $S=2$ for $(a, b, c, d)=(0,1,0,1)$, due to continuity all the values from $(1,2)$ are obtained, for example, for $(a, b, c, d)=(x(1-x), x, 1-x, 1)$, where $x$ goes through $(0,1)$. Second solution. Set  $$ S_{1}=\frac{a}{a+b+d}+\frac{c}{b+c+d} \quad \text { and } \quad S_{2}=\frac{b}{a+b+c}+\frac{d}{a+c+d} $$  We may assume without loss of generality that $a+b+c+d=1$. Putting $a+c=x$ and $b+d=y$ (then $x+y=1$ ), we obtain that the set of values of  $$ S_{1}=\frac{a}{1-c}+\frac{c}{1-a}=\frac{2 a c+x-x^{2}}{a c+1-x} $$  is $\left(x, \frac{2 x}{2-x}\right)$. Having the analogous result for $S_{2}$ in mind, we conclude that the values that $S=S_{1}+S_{2}$ can take are $\left(x+y, \frac{2 x}{2-x}+\frac{2 y}{2-y}\right]$. Since $x+y=1$ and  $$ \frac{2 x}{2-x}+\frac{2 y}{2-y}=\frac{4-4 x y}{2+x y} \leq 2 $$  with equality for $x y=0$, the desired set of values for $S$ is $(1,2)$.


Solution: 9. There exist real numbers $a, b, c$ with $\tan a=x, \tan b=y, \tan c=z$. Then using the additive formula for tangents we obtain  $$ \tan (a+b+c)=\frac{x+y+z-x y z}{1-x y-x z-y z} $$  We are given that $x y z=x+y+z$. In this case $x y+y z+z x=1$ is impossible; otherwise, $x, y, z$ would be the zeros of a cubic polynomial $t^{3}-\lambda t^{2}+t-\lambda=\left(t^{2}+1\right)(t-\lambda)$ (where $\left.\lambda=x y z\right)$, which has only one real root. It follows that  $$ x+y+z=x y z \Longleftrightarrow \tan (a+b+c)=0 . $$  Hence $a+b+c=k \pi$ for some $k \in \mathbb{Z}$. We note that $\frac{3 x-x^{3}}{1-3 x^{2}}$ actually expresses $\tan 3 a$. Since $3 a+3 b+3 c=3 k \pi$, the result follows from (1) for the numbers $\frac{3 x-x^{3}}{1-3 x^{2}}, \frac{3 y-y^{3}}{1-3 y^{2}}, \frac{3 z-z^{3}}{1-3 z^{2}}$.


1. (FRA) There are six ports on a lake. Is it possible to organize a series of routes satisfying the following conditions: (i) Every route includes exactly three ports; (ii) No two routes contain the same three ports; (iii) The series offers exactly two routes to each tourist who desires to visit two different arbitrary ports?

Solution: 1. First, we observe that there cannot exist three routes of the form $(A, B, C)$, $(A, B, D),(A, C, D)$, for if $E, F$ are the remaining two ports, there can be only one route covering $A, E$, namely, $(A, E, F)$. Thus if $(A, B, C)$, $(A, B, D)$ are two routes, the one covering $A, C$ must be w.l.o.g. $(A, C, E)$. The other roots are uniquely determined: These are $(A, D, F),(A, E, F)$, $(B, D, E),(B, E, F),(B, C, F),(C, D, E),(C, D, F)$.


10. (GBR) ${ }^{\mathrm{IMO} 6}$ The function $f(x, y)$ is a homogeneous polynomial of the $n$th degree in $x$ and $y$. If $f(1,0)=1$ and for all $a, b, c$,  $$ f(a+b, c)+f(b+c, a)+f(c+a, b)=0 $$  prove that $f(x, y)=(x-2 y)(x+y)^{n-1}$.

Solution: 10. We shall prove that for all $x, y$ with $x+y=1$ it holds that $f(x, y)=x-2 y$. In this case $f(x, y)=f(x, 1-x)$ can be regarded as a polynomial in $z=x-2 y=3 x-2$, say $f(x, 1-x)=F(z)$. Putting in the given relation $a=b=x / 2, c=1-x$, we obtain $f(x, 1-x)+2 f(1-x / 2, x / 2)=0$; hence $F(z)+2 F(-z / 2)=0$. Now $F(1)=1$, and we get that for all $k$, $F\left((-2)^{k}\right)=(-2)^{k}$. Thus $F(z)=z$ for infinitely many values of $z$; hence $F(z) \equiv z$. Consequently $f(x, y)=x-2 y$ if $x+y=1$. For general $x, y$ with $x+y \neq 0$, since $f$ is homogeneous, we have $f(x, y)=$ $(x+y)^{n} f\left(\frac{x}{x+y}, \frac{y}{x+y}\right)=(x+y)^{n}\left(\frac{x}{x+y}-2 \frac{y}{x+y}\right)=(x+y)^{n-1}(x-2 y)$. The same is true for $x+y=0$, because $f$ is a polynomial.


11. $(\mathbf{G B R})^{\mathrm{IMO} 2}$ Let $a_{1}, a_{2}, a_{3}, \ldots$ be any infinite increasing sequence of positive integers. (For every integer $i>0, a_{i+1}>a_{i}$.) Prove that there are infinitely many $m$ for which positive integers $x, y, h, k$ can be found such that $0<h<k<m$ and $a_{m}=x a_{h}+y a_{k}$.

Solution: 11. Let $\left(a_{k_{i}}\right)$ be the subsequence of $\left(a_{k}\right)$ consisting of all $a_{k}$ 's that give remainder $r$ upon division by $a_{1}$. For every $i>1, a_{k_{i}} \equiv a_{k_{1}}\left(\bmod a_{1}\right)$; hence $a_{k_{i}}=a_{k_{1}}+y a_{1}$ for some integer $y>0$. It follows that for every $r=0,1, \ldots, a_{1}-1$ there is exactly one member of the corresponding $\left(a_{k_{i}}\right)_{i \geq 1}$ that cannot be represented as $x a_{l}+y a_{m}$, and hence at most $a_{1}+1$ members of $\left(a_{k}\right)$ in total are not representable in the given form.


12. (GRE) Consider on the first quadrant of the trigonometric circle the $\operatorname{arcs} A M_{1}=x_{1}, A M_{2}=x_{2}, A M_{3}=x_{3}, \ldots, A M_{\nu}=x_{\nu}$, such that $x_{1}<$ $x_{2}<x_{3}<\cdots<x_{\nu}$. Prove that  $$ \sum_{i=0}^{\nu-1} \sin 2 x_{i}-\sum_{i=0}^{\nu-1} \sin \left(x_{i}-x_{i+1}\right)<\frac{\pi}{2}+\sum_{i=0}^{\nu-1} \sin \left(x_{i}+x_{i+1}\right) $$

Solution: 12. Since $\sin 2 x_{i}=2 \sin x_{i} \cos x_{i}$ and $\sin \left(x_{i}+x_{i+1}\right)+\sin \left(x_{i}-x_{i+1}\right)=$ $2 \sin x_{i} \cos x_{i+1}$, the inequality from the problem is equivalent to  $$ \begin{gathered} \left(\cos x_{1}-\cos x_{2}\right) \sin x_{1}+\left(\cos x_{2}-\cos x_{3}\right) \sin x_{2}+\cdots \\ \cdots+\left(\cos x_{\nu-1}-\cos x_{\nu}\right) \sin x_{\nu-1}<\frac{\pi}{4} \end{gathered} $$  Consider the unit circle with center at $O(0,0)$ and points $M_{i}\left(\cos x_{i}, \sin x_{i}\right)$ on it. Also, choose the points $N_{i}\left(\cos x_{i}, 0\right)$ and $M_{i}^{\prime}\left(\cos x_{i+1}, \sin x_{i}\right)$. It is clear that $\left(\cos x_{i}-\cos x_{i+1}\right) \sin x_{i}$ is equal to the area of the rectangle $M_{i} N_{i} N_{i+1} M_{i}^{\prime}$. Since all these rectangles are disjoint and lie inside the quarter circle in the first quadrant whose area is $\frac{\pi}{4}$, inequality (1) follows.


13. (ROM) Let $A_{0}, A_{1}, \ldots, A_{n}$ be points in a plane such that (i) $A_{0} A_{1} \leq \frac{1}{2} A_{1} A_{2} \leq \cdots \leq \frac{1}{2^{n-1}} A_{n-1} A_{n}$ and (ii) $0<\measuredangle A_{0} A_{1} A_{2}<\measuredangle A_{1} A_{2} A_{3}<\cdots<\measuredangle A_{n-2} A_{n-1} A_{n}<180^{\circ}$, where all these angles have the same orientation. Prove that the segments $A_{k} A_{k+1}, A_{m} A_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \leq k \leq$ $m-2<n-2$.

Solution: 13. Suppose that $A_{k} A_{k+1} \cap A_{m} A_{m+1} \neq \emptyset$ for some $k, m>k+1$. Without loss of generality we may suppose that $k=0, m=n-1$ and that no two segments $A_{k} A_{k+1}$ and $A_{m} A_{m+1}$ intersect for $0 \leq k<m-1<n-1$ except for $k=0, m=n-1$. Also, shortening $A_{0} A_{1}$, we may suppose that $A_{0} \in A_{n-1} A_{n}$. Finally, we may reduce the problem to the case that $A_{0} \ldots A_{n-1}$ is convex: Otherwise, the segment $A_{n-1} A_{n}$ can be prolonged so that it intersects some $A_{k} A_{k+1}, 0<k<n-2$. If $n=3$, then $A_{1} A_{2} \geq 2 A_{0} A_{1}$ implies $A_{0} A_{2}>A_{0} A_{1}$, hence $\angle A_{0} A_{1} A_{2}>$ $\angle A_{1} A_{2} A_{3}$, a contradiction. Let $n=4$. From $A_{3} A_{2}>A_{1} A_{2}$ we conclude that $\angle A_{3} A_{1} A_{2}>\angle A_{1} A_{3} A_{2}$. Using the inequality $\angle A_{0} A_{3} A_{2}>\angle A_{0} A_{1} A_{2}$ we obtain that $\angle A_{0} A_{3} A_{1}>$ $\angle A_{0} A_{1} A_{3}$ implying $A_{0} A_{1}>A_{0} A_{3}$. Now we have $A_{2} A_{3}<A_{3} A_{0}+A_{0} A_{1}+$ $A_{1} A_{2}<2 A_{0} A_{1}+A_{1} A_{2} \leq 2 A_{1} A_{2} \leq A_{2} A_{3}$, which is not possible. Now suppose $n \geq 5$. If $\alpha_{i}$ is the exterior angle at $A_{i}$, then $\alpha_{1}>\cdots>\alpha_{n-1}$; hence $\alpha_{n-1}<\frac{360^{\circ}}{n-1} \leq 90^{\circ}$. Consequently $\angle A_{n-2} A_{n-1} A_{0} \geq 90^{\circ}$ and $A_{0} A_{n-2}>A_{n-1} A_{n-2}$. On the other hand, $A_{0} A_{n-2}<A_{0} A_{1}+A_{1} A_{2}+\cdots+$ $A_{n-3} A_{n-2}<\left(\frac{1}{2^{n-2}}+\frac{1}{2^{n-3}}+\cdots+\frac{1}{2}\right) A_{n-1} A_{n-2}<A_{n-1} A_{n-2}$, which contradicts the previous relation.


Solution: 14. We shall prove that for every $n \in \mathbb{N}, \sqrt{2 n+25} \leq x_{n} \leq \sqrt{2 n+25}+0.1$. Note that for $n=1000$ this gives us exactly the desired inequalities.  First, notice that the recurrent relation is equivalent to  $$ 2 x_{k}\left(x_{k+1}-x_{k}\right)=2 $$  Since $x_{0}<x_{1}<\cdots<x_{k}<\cdots$, from (1) we get $x_{k+1}^{2}-x_{k}^{2}=\left(x_{k+1}+\right.$ $\left.x_{k}\right)\left(x_{k+1}-x_{k}\right)>2$. Adding these up we obtain $x_{n}^{2} \geq x_{0}^{2}+2 n$, which proves the first inequality. On the other hand, $x_{k+1}=x_{k}+\frac{1}{x_{k}} \leq x_{k}+0.2$ (for $x_{k} \geq 5$ ), and one also deduces from (1) that $x_{k+1}^{2}-x_{k}^{2}-0.2\left(x_{k+1}-x_{k}\right)=\left(x_{k+1}+x_{k}-\right.$ $0.2)\left(x_{k+1}-x_{k}\right) \leq 2$. Again, adding these inequalities up, $(k=0, \ldots, n-1)$ yields  $$ x_{n}^{2} \leq 2 n+x_{0}^{2}+0.2\left(x_{n}-x_{0}\right)=2 n+24+0.2 x_{n} $$  Solving the corresponding quadratic equation, we obtain $x_{n}<0.1+$ $\sqrt{2 n+24.01}<0.1++\sqrt{2 n+25}$.


15. (USS) ${ }^{\mathrm{IMO} 5}$ Is it possible to plot 1975 points on a circle with radius 1 so that the distance between any two of them is a rational number (distances have to be measured by chords)?

Solution: 15. Assume that the center of the circle is at the origin $O(0,0)$, and that the points $A_{1}, A_{2}, \ldots, A_{1975}$ are arranged on the upper half-circle so that $\angle A_{i} O A_{1}=\alpha_{i}\left(\alpha_{1}=0\right)$. The distance $A_{i} A_{j}$ equals $2 \sin \frac{\alpha_{j}-\alpha_{i}}{2}=$ $2 \sin \frac{\alpha_{j}}{2} \cos \frac{\alpha_{i}}{2}-\cos \frac{\alpha_{j}}{2} \sin \frac{\alpha_{i}}{2}$, and it will be rational if all $\sin \frac{\alpha_{k}}{2}, \cos \frac{\alpha_{k}}{2}$ are rational. Finally, observe that there exist infinitely many angles $\alpha$ such that both $\sin \alpha, \cos \alpha$ are rational, and that such $\alpha$ can be arbitrarily small. For example, take $\alpha$ so that $\sin \alpha=\frac{2 t}{t^{2}+1}$ and $\cos \alpha=\frac{t^{2}-1}{t^{2}+1}$ for any $t \in \mathbb{Q}$.


2. (CZS) ${ }^{\mathrm{IMO} 1}$ Let $x_{1} \geq x_{2} \geq \cdots \geq x_{n}$ and $y_{1} \geq y_{2} \geq \cdots \geq y_{n}$ be two $n$-tuples of numbers. Prove that  $$ \sum_{i=1}^{n}\left(x_{i}-y_{i}\right)^{2} \leq \sum_{i=1}^{n}\left(x_{i}-z_{i}\right)^{2} $$  is true when $z_{1}, z_{2}, \ldots, z_{n}$ denote $y_{1}, y_{2}, \ldots, y_{n}$ taken in another order.

Solution: 2. Since there are finitely many arrangements of the $z_{i}$ 's, assume that $z_{1}, \ldots, z_{n}$ is the one for which $\sum_{i=1}^{n}\left(x_{i}-z_{i}\right)^{2}$ is minimal. We claim that in this case $i<j \Rightarrow z_{i} \geq z_{j}$, from which the claim of the problem directly follows. Indeed, otherwise we would have  $$ \begin{aligned} \left(x_{i}-z_{j}\right)^{2}+\left(x_{j}-z_{i}\right)^{2}= & \left(x_{i}-z_{i}\right)^{2}+\left(x_{j}-z_{j}\right)^{2} \\ & +2\left(x_{i} z_{i}+x_{j} z_{j}-x_{i} z_{j}-x_{j} z_{i}\right) \\ = & \left(x_{i}-z_{i}\right)^{2}+\left(x_{j}-z_{j}\right)^{2}+2\left(x_{i}-x_{j}\right)\left(z_{i}-z_{j}\right) \\ \leq & \left(x_{i}-z_{i}\right)^{2}+\left(x_{j}-z_{j}\right)^{2} \end{aligned} $$  contradicting the assumption.


3. (USA) Find the integer represented by $\left[\sum_{n=1}^{10^{9}} n^{-2 / 3}\right]$. Here $[x]$ denotes the greatest integer less than or equal to $x$ (e.g. $[\sqrt{2}]=1$ ).

Solution: 3. From $\left((k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}\right)\left((k+1)^{1 / 3}-k^{1 / 3}\right)=1$ and $3 k^{2 / 3}<(k+1)^{2 / 3}+(k+1)^{1 / 3} k^{1 / 3}+k^{2 / 3}<3(k+1)^{2 / 3}$ we obtain  $$ 3\left((k+1)^{1 / 3}-k^{1 / 3}\right)<k^{-2 / 3}<3\left(k^{1 / 3}-(k-1)^{1 / 3}\right) . $$  Summing from 1 to $n$ we get  $$ 1+3\left((n+1)^{1 / 3}-2^{1 / 3}\right)<\sum_{k=1}^{n} k^{-2 / 3}<1+3\left(n^{1 / 3}-1\right) $$  In particular, for $n=10^{9}$ this inequality gives  $$ 2997<1+3\left(\left(10^{9}+1\right)^{1 / 3}-2^{1 / 3}\right)<\sum_{k=1}^{10^{9}} k^{-2 / 3}<2998 $$  Therefore $\left[\sum_{k=1}^{10^{9}} k^{-2 / 3}\right]=2997$.


4. (SWE) Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of real numbers such that $0 \leq a_{n} \leq 1$ and $a_{n}-2 a_{n+1}+a_{n+2} \geq 0$ for $n=1,2,3, \ldots$. Prove that  $$ 0 \leq(n+1)\left(a_{n}-a_{n+1}\right) \leq 2 \quad \text { for } n=1,2,3, \ldots $$

Solution: 4. Put $\Delta a_{n}=a_{n}-a_{n+1}$. By the imposed condition, $\Delta a_{n}>\Delta a_{n+1}$. Suppose that for some $n, \Delta a_{n}<0$ : Then for each $k \geq n, \Delta a_{k}<\Delta a_{n}$; hence $a_{n}-a_{n+m}=\Delta a_{n}+\cdots+\Delta a_{n+m-1}<m \Delta a_{n}$. Thus for sufficiently large $m$ it holds that $a_{n}-a_{n+m}<-1$, which is impossible. This proves the first part of the inequality. Next one observes that $n \geq \sum_{k=1}^{n} a_{k}=n a_{n+1}+\sum_{k=1}^{n} k \Delta a_{k} \geq(1+2+\cdots+n) \Delta a_{n}=\frac{n(n+1)}{2} \Delta a_{n}$. Hence $(n+1) \Delta a_{n} \leq 2$.


5. (SWE) Let $M$ be the set of all positive integers that do not contain the digit 9 (base 10). If $x_{1}, \ldots, x_{n}$ are arbitrary but distinct elements in $M$, prove that  $$ \sum_{j=1}^{n} \frac{1}{x_{j}}<80 $$

Solution: 5. There are exactly $8 \cdot 9^{k-1} k$-digit numbers in $M$ (the first digit can be chosen in 8 ways, while any other position admits 9 possibilities). The least of them is $10^{k}$, and hence  $$ \begin{aligned} \sum_{x_{j}<10^{k}} \frac{1}{x_{j}} & =\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{x_{j}}<\sum_{i=1}^{k} \sum_{10^{i-1} \leq x_{j}<10^{i}} \frac{1}{10^{i-1}} \\ & =\sum_{i=1}^{k} \frac{8 \cdot 9^{i-1}}{10^{i-1}}=80\left(1-\frac{9^{k}}{10^{k}}\right)<80 . \end{aligned} $$


6. (USS) ${ }^{\mathrm{IMO} 04}$ Let $A$ be the sum of the digits of the number $16^{16}$ and $B$ the sum of the digits of the number $A$. Find the sum of the digits of the number $B$ without calculating $16^{16}$.

Solution: 6. Let us denote by $C$ the sum of digits of $B$. We know that $16^{16} \equiv A \equiv$ $B \equiv C(\bmod 9)$. Since $16^{16}=2^{64}=2^{6 \cdot 10+4} \equiv 2^{4} \equiv 7$, we get $C \equiv 7(\bmod$ 9). Moreover, $16^{16}<100^{16}=10^{32}$, hence $A$ cannot exceed $9 \cdot 32=288$; consequently, $B$ cannot exceed 19 and $C$ is at most 10 . Therefore $C=7$.


7. (GDR) Prove that from $x+y=1(x, y \in \mathbb{R})$ it follows that  $$ x^{m+1} \sum_{j=0}^{n}\binom{m+j}{j} y^{j}+y^{n+1} \sum_{i=0}^{m}\binom{n+i}{i} x^{i}=1 \quad(m, n=0,1,2, \ldots) . $$

Solution: 7. We use induction on $m$. Denote by $S_{m}$ the left-hand side of the equality to be proved. First $S_{0}=(1-y)\left(1+y+\cdots+y^{n}\right)+y^{n+1}=1$, since $x=1-y$. Furthermore,  $$ \begin{aligned} & S_{m+1}-S_{m} \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1}+x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} x y^{j}-\binom{m+j}{j} y^{j}\right) \\ = & \binom{m+n+1}{m+1} x^{m+1} y^{n+1} \\ & +x^{m+1} \sum_{j=0}^{n}\left(\binom{m+1+j}{j} y^{j}-\binom{m+j}{j} y^{j}-\binom{m+1+j}{j} y^{j+1}\right) \\ = & x^{m+1}\left[\binom{m+n+1}{n} y^{n+1}+\sum_{j=0}^{n}\left(\binom{m+j}{j-1} y^{j}-\binom{m+j+1}{j} y^{j+1}\right)\right] \\ = & 0 \end{aligned} $$  i.e., $S_{m+1}=S_{m}=1$ for every $m$.  Second solution. Let us be given an unfair coin that, when tossed, shows heads with probability $x$ and tails with probability $y$. Note that $x^{m+1}\binom{m+j}{j} y^{j}$ is the probability that until the moment when the $(m+1)$ th head appears, exactly $j$ tails $(j<n+1)$ have appeared. Similarly, $y^{n+1}\binom{n+i}{i} x^{i}$ is the probability that exactly $i$ heads will appear before the $(n+1)$ th tail occurs. Therefore, the above sum is the probability that either $m+1$ heads will appear before $n+1$ tails, or vice versa, and this probability is clearly 1.


8. (NET) ${ }^{\mathrm{IMO} 3}$ On the sides of an arbitrary triangle $A B C$, triangles $B P C$, $C Q A$, and $A R B$ are externally erected such that $\measuredangle P B C=\measuredangle C A Q=45^{\circ}$, $\measuredangle B C P=\measuredangle Q C A=30^{\circ}$, $\measuredangle A B R=\measuredangle B A R=15^{\circ}$. Prove that $\measuredangle Q R P=90^{\circ}$ and $Q R=R P$.

Solution: 8. Let $K$ and $L$ be the feet of perpendiculars from $P$ and $Q$ to $B C$ and $A C$ respectively.  Let $M, N$ be points on $A B$ (ordered $A-N-M-B$ ) such that $R M N$ is a right isosceles triangle with $\angle R=90^{\circ}$. By sine theorem we have $\frac{B M}{B A}=\frac{B M}{B R} \cdot \frac{B R}{B A}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$. Since $\frac{B K}{B C}=\frac{\sin 45^{\circ} \sin 30^{\circ}}{\cos 15^{\circ}}=\frac{\sin 15^{\circ}}{\sin 45^{\circ}}$, we deduce that $M K \| A C$ and $M K=A L$. Similarly, $N L \| B C$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-412.jpg?height=308&width=396&top_left_y=250&top_left_x=886) and $N L=B K$. It follows that the vectors $\overrightarrow{R N}, \overrightarrow{N L}, \overrightarrow{L Q}$ are the images of $\overrightarrow{R M}, \overrightarrow{K P}, \overrightarrow{M K}$ respectively under a rotation of $90^{\circ}$, and consequently the same holds for their sums $\overrightarrow{R Q}$ and $\overrightarrow{R P}$. Therefore, $Q R=R P$ and $\angle Q R P=90^{\circ}$. Second solution. Let $A B S$ be the equilateral triangle constructed in the exterior of $\triangle A B C$. Obviously, the triangles $B P C, B R S, A R S, A Q C$ are similar. Let $f$ be the rotational homothety centered at $B$ that maps $P$ onto $C$, and let $g$ be the rotational homothety about $A$ that maps $C$ onto $Q$. The composition $h=g \circ f$ is also a rotational homothety; its angle is $\angle P B C+\angle C A Q=90^{\circ}$, and the coefficient is $\frac{B C}{B P} \cdot \frac{A Q}{A C}=1$. Moreover, $R$ is a fixed point of $h$ because $f(R)=S$ and $g(S)=R$. Hence $R$ is the center of $h$, and the statement follows from $h(P)=Q$. Remark. There are two more possible approaches: One includes using complex numbers and the other one is mere calculating of $R P, R Q, P Q$ by the cosine theorem. Second remark. The problem allows a generalization: Given that $\angle C B P=$ $\angle C A Q=\alpha, \angle B C P=\angle A C Q=\beta$, and $\angle R A B=\angle R B A=90^{\circ}-\alpha-\beta$, show that $R P=R Q$ and $\angle P R Q=2 \alpha$.


9. (NET) Let $f(x)$ be a continuous function defined on the closed interval $0 \leq x \leq 1$. Let $G(f)$ denote the graph of $f(x): G(f)=\left\{(x, y) \in \mathbb{R}^{2} \mid 0 \leq\right.$ $x \leq 1, y=f(x)\}$. Let $G_{a}(f)$ denote the graph of the translated function $f(x-a)$ (translated over a distance $a$ ), defined by $G_{a}(f)=\{(x, y) \in$ $\left.\mathbb{R}^{2} \mid a \leq x \leq a+1, y=f(x-a)\right\}$. Is it possible to find for every $a$, $0<a<1$, a continuous function $f(x)$, defined on $0 \leq x \leq 1$, such that $f(0)=f(1)=0$ and $G(f)$ and $G_{a}(f)$ are disjoint point sets?

Solution: 9. Suppose $n$ is the natural number with $n a \leq 1<(n+1) a$. If a function $f$ with the desired properties exists, then $f_{a}(a)=0$ and let w.l.o.g. $f(a)>0$, or equivalently, let the graph of $f_{a}$ lie below the graph of $f$. In this case also $f(2 a)>f(a)$, since otherwise, the graphs of $f$ and $f_{a}$ would intersect between $a$ and $2 a$. Continuing in this way we are led to $0=f(0)<$ $f(a)<f(2 a)<\cdots<f(n a)$. Thus if $n a=1$, i.e., $a=1 / n$, such an $f$ does not exist. On the other hand, if $a \neq 1 / n$, then we similarly obtain $f(1)>f(1-a)>f(1-2 a)>\cdots>f(1-n a)$. Choosing values of $f$ at $i a, 1-i a, i=1, \ldots, n$, so that they satisfy $f(1-n a)<\cdots<f(1-a)<$ $0<f(a)<\cdots<f(n a)$, we can extend $f$ to other values of $[0,1]$ by linear interpolation. A function obtained this way has the desired property.


1. (BUL 1) Let $A B C$ be a triangle with bisectors $A A_{1}, B B_{1}, C C_{1}\left(A_{1} \in\right.$ $B C$, etc.) and $M$ their common point. Consider the triangles $M B_{1} A$, $M C_{1} A, M C_{1} B, M A_{1} B, M A_{1} C, M B_{1} C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $A B=$ $B C=C A$.

Solution: 1. Let $r$ denote the common inradius. Some two of the four triangles with the inradii $\rho$ have cross angles at $M$ : Suppose these are $\triangle A M B_{1}$ and $\triangle B M A_{1}$. We shall show that $\triangle A M B_{1} \cong \triangle B M A_{1}$. Indeed, the altitudes of these two triangles are both equal to $r$, the inradius of $\triangle A B C$, and their interior angles at $M$ are equal to some angle $\varphi$. If $P$ is the point of tangency of the incircle of $\triangle A_{1} M B$ with $M B$, then $\frac{r}{\rho}=\frac{A_{1} M+B M+A_{1} B}{A_{1} B}$, which also implies $\frac{r-2 \rho}{\rho}=\frac{A_{1} M+B M-A_{1} B}{A_{1} B}=\frac{2 M P}{A_{1} B}=\frac{2 r \cot (\varphi / 2)}{A_{1} B}$. Since similarly $\frac{r-2 \rho}{\rho}=\frac{2 r \cot (\varphi / 2)}{B_{1} A}$, we obtain $A_{1} B=B_{1} A$ and consequently $\triangle A M B_{1} \cong \triangle B M A_{1}$. Thus $\angle B A C=\angle A B C$ and $C C_{1} \perp A B$. There are two alternatives for the other two incircles: (i) If the inradii of $A M C_{1}$ and $A M B_{1}$ are equal to $r$, it is easy to obtain that $\triangle A M C_{1} \cong \triangle A M B_{1}$. Hence $\angle A B_{1} M=\angle A C_{1} M=90^{\circ}$, and $\triangle A B C$ is equilateral. (ii) The inradii of $A M B_{1}$ and $C M B_{1}$ are equal to $r$. Put $x=\angle M A C_{1}=$ $\angle M B C_{1}$. In this case $\varphi=2 x$ and $\angle B_{1} M C=90^{\circ}-x$. Now we have $\frac{A B_{1}}{C B_{1}}=\frac{S_{A M B_{1}}}{S_{C M B_{1}}}=\frac{A M+M B_{1}+A B_{1}}{C M+M B_{1}+C B_{1}}=\frac{A M+M B_{1}-A B_{1}}{C M+M B_{1}-C B_{1}}=\frac{\mathrm{cot} x}{\cot \left(45^{\circ}-x / 2\right)}$. On the other hand, $\frac{A B_{1}}{C B_{1}}=\frac{A B}{B C}=2 \cos 2 x$. Thus we have an equation for $x: \tan \left(45^{\circ}-x / 2\right)=2 \cos 2 x \tan x$, or equivalently  $$ 2 \tan \left(45^{\circ}-\frac{x}{2}\right) \sin \left(45^{\circ}-\frac{x}{2}\right) \cos \left(45^{\circ}-\frac{x}{2}\right)=2 \cos 2 x \sin x . $$  Hence $\sin 3 x-\sin x=2 \sin ^{2}\left(45^{\circ}-\frac{x}{2}\right)=1-\sin x$, implying $\sin 3 x=1$, i.e., $x=30^{\circ}$. Therefore $\triangle A B C$ is equilateral.


10. (USA 6) ${ }^{\mathrm{IMO}}$ Find the largest number obtainable as the product of positive integers whose sum is 1976.

Solution: 10. Let $a_{1}<a_{2}<\cdots<a_{n}$ be positive integers whose sum is 1976 . Let $M$ denote the maximal value of $a_{1} a_{2} \cdots a_{n}$. We make the following observations: (1) $a_{1}=1$ does not yield the maximum, since replacing $1, a_{2}$ by $1+a_{2}$ increases the product. (2) $a_{j}-a_{i} \geq 2$ does not yield the maximal value, since replacing $a_{i}, a_{j}$ by $a_{i}+1, a_{j}-1$ increases the product. (3) $a_{i} \geq 5$ does not yield the maximal value, since $2\left(a_{i}-2\right)=2 a_{i}-4>a_{i}$. Since $4=2^{2}$, we may assume that all $a_{i}$ are either 2 or 3 , and $M=2^{k} 3^{l}$, where $2 k+3 l=1976$. (4) $k \geq 3$ does not yield the maximal value, since $2 \cdot 2 \cdot 2<3 \cdot 3$.  Hence $k \leq 2$ and $2 k \equiv 1976(\bmod 3)$ gives us $k=1, l=658$ and $M=2 \cdot 3^{658}$.


11. (VIE 1) Prove that there exist infinitely many positive integers $n$ such that the decimal representation of $5^{n}$ contains a block of 1976 consecutive zeros.

Solution: 11. We shall show by induction that $5^{2^{k}}-1=2^{k+2} q_{k}$ for each $k=0,1, \ldots$, where $q_{k} \in \mathbb{N}$. Indeed, the statement is true for $k=0$, and if it holds for some $k$ then $5^{2^{k+1}}-1=\left(5^{2^{k}}+1\right)\left(5^{2^{k}}-1\right)=2^{k+3} d_{k+1}$ where $d_{k+1}=$ $\left(5^{2^{k}}+1\right) d_{k} / 2$ is an integer by the inductive hypothesis. Let us now choose $n=2^{k}+k+2$. We have $5^{n}=10^{k+2} q_{k}+5^{k+2}$. It follows from $5^{4}<10^{3}$ that $5^{k+2}$ has at most $[3(k+2) / 4]+2$ nonzero digits, while $10^{k+2} q_{k}$ ends in $k+2$ zeros. Hence the decimal representation of $5^{n}$ contains at least $[(k+2) / 4]-2$ consecutive zeros. Now it suffices to take $k>4 \cdot 1978$.


12. (VIE 2) The polynomial $1976\left(x+x^{2}+\cdots+x^{n}\right)$ is decomposed into a sum of polynomials of the form $a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}$, where $a_{1}, a_{2}, \cdots, a_{n}$ are distinct positive integers not greater than $n$. Find all values of $n$ for which such a decomposition is possible.

Solution: 12. Suppose the decomposition into $k$ polynomials is possible. The sum of coefficients of each polynomial $a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}$ equals $1+\cdots+$ $n=n(n+1) / 2$ while the sum of coefficients of $1976\left(x+x^{2}+\cdots+x^{n}\right)$ is $1976 n$. Hence we must have $1976 n=k n(n+1) / 2$, which reduces to $(n+1) \mid 3952=2^{4} \cdot 13 \cdot 19$. In other words, $n$ is of the form $n=2^{\alpha} 13^{\beta} 19^{\gamma}-1$, with $0 \leq \alpha \leq 4,0 \leq \beta \leq 1,0 \leq \gamma \leq 1$. We can immediately eliminate the values $n=0$ and $n=3951$ that correspond to $\alpha=\beta=\gamma=0$ and $\alpha=4, \beta=\gamma=1$. We claim that all other values $n$ are permitted. There are two cases. $\alpha \leq 3$. In this case $k=3952 /(n+1)$ is even. The simple choice of the polynomials $P=x+2 x^{2}+\cdots+n x^{n}$ and $P^{\prime}=n x+(n-1) x^{2}+\cdots+x^{n}$ suffices, since $k\left(P+P^{\prime}\right) / 2=1976\left(x+x^{2}+\cdots+x^{n}\right)$. $\alpha=4$. Then $k$ is odd. Consider $(k-3) / 2$ pairs $\left(P, P^{\prime}\right)$ of the former case and  $$ \begin{aligned} P_{1}= & {\left[n x+(n-1) x^{3}+\cdots+\frac{n+1}{2} x^{n}\right] } \\ & +\left[\frac{n-1}{2} x^{2}+\frac{n-3}{2} x^{4}+\cdots+x^{n-1}\right] \\ P_{2}= & {\left[\frac{n+1}{2} x+\frac{n-1}{2} x^{3}+\cdots+x^{n}\right] } \\ & +\left[n x^{2}+(n-1) x^{4}+\cdots+\frac{n+3}{2} x^{n-1}\right] \end{aligned} $$  Then $P+P_{1}+P_{2}=3(n+1)\left(x+x^{2}+\cdots+x^{n}\right) / 2$ and therefore $(k-3)\left(P+P^{\prime}\right) / 2+\left(P+P_{1}+P_{2}\right)=1976\left(x+x^{2}+\cdots+x^{n}\right)$. It follows that the desired decomposition is possible if and only if $1<n<$ 3951 and $n+1 \mid 2 \cdot 1976$.


2. (BUL 3) Let $a_{0}, a_{1}, \ldots, a_{n}, a_{n+1}$ be a sequence of real numbers satisfying the following conditions:  $$ \begin{aligned} a_{0} & =a_{n+1}=0 \\ \left|a_{k-1}-2 a_{k}+a_{k+1}\right| & \leq 1 \quad(k=1,2, \ldots, n) \end{aligned} $$  Prove that $\left|a_{k}\right| \leq \frac{k(n+1-k)}{2}(k=0,1, \ldots, n+1)$.

Solution: 2. Let us put $b_{i}=i(n+1-i) / 2$, and let $c_{i}=a_{i}-b_{i}, i=0,1, \ldots, n+1$. It is easy to verify that $b_{0}=b_{n+1}=0$ and $b_{i-1}-2 b_{i}+b_{i+1}=-1$. Subtracting this inequality from $a_{i-1}-2 a_{i}+a_{i+1} \geq-1$, we obtain $c_{i-1}-2 c_{i}+c_{i+1} \geq 0$, i.e., $2 c_{i} \leq c_{i-1}+c_{i+1}$. We also have $c_{0}=c_{n+1}=0$.  Suppose that there exists $i \in\{1, \ldots, n\}$ for which $c_{i}>0$, and let $c_{k}$ be the maximal such $c_{i}$. Assuming w.l.o.g. that $c_{k-1}<c_{k}$, we obtain $c_{k-1}+c_{k+1}<2 c_{k}$, which is a contradiction. Hence $c_{i} \leq 0$ for all $i$; i.e., $a_{i} \leq b_{i}$. Similarly, considering the sequence $c_{i}^{\prime}=a_{i}+b_{i}$ one can show that $c_{i}^{\prime} \geq 0$, i.e., $a_{i} \geq-b_{i}$ for all $i$. This completes the proof.


3. (CZS 3) ${ }^{\mathrm{IMO} 1}$ In a convex quadrangle with area $32 \mathrm{~cm}^{2}$, the sum of the lengths of two nonadjacent edges and of the length of one diagonal is equal to 16 cm . (a) What is the length of the other diagonal? (b) What are the lengths of the edges of the quadrangle if the perimeter is a minimum? (c) Is it possible to choose the edges in such a way that the perimeter is a maximum?

Solution: 3. (a) Let $A B C D$ be a quadrangle with $16=d=A B+C D+A C$, and let $S$ be its area. Then $S \leq(A C \cdot A B+A C \cdot C D) / 2=A C(d-A C) / 2 \leq$ $d^{2} / 8=32$, where equality occurs if and only if $A B \perp A C \perp C D$ and $A C=A B+C D=8$. In this case $B D=8 \sqrt{2}$. (b) Let $A^{\prime}$ be the point with $\overrightarrow{D A^{\prime}}=\overrightarrow{A C}$. The triangular inequality implies $A D+B C \geq A A^{\prime}=8 \sqrt{5}$. Thus the perimeter attains its minimum for $A B=C D=4$. (c) Let us assume w.l.o.g. that $C D \leq A B$. Then $C$ lies inside $\triangle B D A^{\prime}$ and hence $B C+A D=B C+C A^{\prime}<B D+D A^{\prime}$. The maximal value $B D+D A^{\prime}$ of $B C+A D$ is attained when $C$ approaches $D$, making a degenerate quadrangle.


4. (GBR 1a) ${ }^{\mathrm{IMO} 6}$ For all positive integral $n$, $u_{n+1}=u_{n}\left(u_{n-1}^{2}-2\right)-u_{1}$, $u_{0}=2$, and $u_{1}=5 / 2$. Prove that  $$ 3 \log _{2}\left[u_{n}\right]=2^{n}-(-1)^{n} $$  where $[x]$ is the integral part of $x$.

Solution: 4. The first few values are easily verified to be $2^{r_{n}}+2^{-r_{n}}$, where $r_{0}=0$, $r_{1}=r_{2}=1, r_{3}=3, r_{4}=5, r_{5}=11, \ldots$. Let us put $u_{n}=2^{r_{n}}+2^{-r_{n}}$ (we will show that $r_{n}$ exists and is integer for each $n$ ). A simple calculation gives us $u_{n}\left(u_{n-1}^{2}-2\right)=2^{r_{n}+2 r_{n-1}}+2^{-r_{n}-2 r_{n-1}}+2^{r_{n}-2 r_{n-1}}+2^{-r_{n}+2 r_{n-1}}$. If an array $q_{n}$, with $q_{0}=0$ and $q_{1}=1$, is set so as to satisfy the linear recurrence $q_{n+1}=q_{n}+2 q_{n-1}$, then it also satisfies $q_{n}-2 q_{n-1}=-\left(q_{n-1}-\right.$ $\left.2 q_{n-2}\right)=\cdots=(-1)^{n-1}\left(q_{1}-2 q_{0}\right)=(-1)^{n-1}$. Assuming inductively up to $n r_{i}=q_{i}$, the expression for $u_{n}\left(u_{n-1}^{2}-2\right)=u_{n+1}+u_{1}$ reduces to $2^{q_{n+1}}+2^{-q_{n+1}}+u_{1}$. Therefore, $r_{n+1}=q_{n+1}$. The solution to this linear recurrence with $r_{0}=0, r_{1}=1$ is $r_{n}=q_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and since $\left[u_{n}\right]=2^{r_{n}}$ for $n \geq 0$, the result follows. Remark. One could simply guess that $u_{n}=2^{r_{n}}+2^{-r_{n}}$ for $r_{n}=\frac{2^{n}-(-1)^{n}}{3}$, and then prove this result by induction.


5. (NET 3) ${ }^{\mathrm{IMO} 5}$ Let a set of $p$ equations be given,  $$ \begin{gathered} a_{11} x_{1}+\cdots+a_{1 q} x_{q}=0 \\ a_{21} x_{1}+\cdots+a_{2 q} x_{q}=0 \\ \vdots \\ a_{p 1} x_{1}+\cdots+a_{p q} x_{q}=0, \end{gathered} $$  with coefficients $a_{i j}$ satisfying $a_{i j}=-1,0$, or +1 for all $i=1, \ldots, p$ and $j=1, \ldots, q$. Prove that if $q=2 p$, there exists a solution $x_{1}, \ldots, x_{q}$ of this system such that all $x_{j}(j=1, \ldots, q)$ are integers satisfying $\left|x_{j}\right| \leq q$ and $x_{j} \neq 0$ for at least one value of $j$.

Solution: 5. If one substitutes an integer $q$-tuple $\left(x_{1}, \ldots, x_{q}\right)$ satisfying $\left|x_{i}\right| \leq p$ for all $i$ in an equation of the given system, the absolute value of the right-hand member never exceeds $p q$. So for the right-hand member of the system there are $(2 p q+1)^{p}$ possibilities There are $(2 p+1)^{q}$ possible $q$-tuples $\left(x_{1}, \ldots, x_{q}\right)$. Since $(2 p+1)^{q} \geq(2 p q+1)^{p}$, there are at least two $q$-tuples $\left(y_{1}, \ldots, y_{q}\right)$ and $\left(z_{1}, \ldots, z_{q}\right)$ giving the same right-hand members in the given system. The difference $\left(x_{1}, \ldots, x_{q}\right)=\left(y_{1}-z_{1}, \ldots, y_{q}-z_{q}\right)$ thus satisfies all the requirements of the problem.


6. (NET 4) ${ }^{\mathrm{IMO} 3} \mathrm{~A}$ rectangular box can be filled completely with unit cubes. If one places cubes with volume 2 in the box such that their edges are parallel to the edges of the box, one can fill exactly $40 \%$ of the box. Determine all possible (interior) sizes of the box.

Solution: 6. Suppose $a_{1} \leq a_{2} \leq a_{3}$ are the dimensions of the box. If we set $b_{i}=$ $\left[a_{i} / \sqrt[3]{2}\right]$, the condition of the problem is equivalent to $\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}}=5$. We list some values of $a, b=[a / \sqrt[3]{2}]$ and $a / b$ :  | $a$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | $b$ | 1 | 2 | 3 | 3 | 4 | 5 | 6 | 7 | 7 | | $a / b$ | 2 | 1.5 | 1.33 | 1.67 | 1.5 | 1.4 | 1.33 | 1.29 | 1.43 |  We note that if $a>2$, then $a / b \leq 5 / 3$, and if $a>5$, then $a / b \leq 3 / 2$. If $a_{1}>2$, then $\frac{a_{1}}{b_{1}} \cdot \frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}}<(5 / 3)^{3}<5$, a contradiction. Hence $a_{1}=2$. If also $a_{2}=2$, then $a_{3} / b_{3}=5 / 4 \leq \sqrt[3]{2}$, which is impossible. Also, if $a_{2} \geq 6$, then $\frac{a_{2}}{b_{2}} \cdot \frac{a_{3}}{b_{3}} \leq(1.5)^{2}<2.5$, again a contradiction. We thus have the following cases: (i) $a_{1}=2, a_{2}=3$, then $a_{3} / b_{3}=5 / 3$, which holds only if $a_{3}=5$; (ii) $a_{1}=2, a_{2}=4$, then $a_{3} / b_{3}=15 / 8$, which is impossible; (iii) $a_{1}=2, a_{2}=5$, then $a_{3} / b_{3}=3 / 2$, which holds only if $a_{3}=6$.  The only possible sizes of the box are therefore $(2,3,5)$ and $(2,5,6)$.


7. (POL 1b) Let $I=(0,1]$ be the unit interval of the real line. For a given number $a \in(0,1)$ we define a map $T: I \rightarrow I$ by the formula  $$ T(x, y)= \begin{cases}x+(1-a) & \text { if } 0<x \leq a \\ x-a & \text { if } a<x \leq 1\end{cases} $$  Show that for every interval $J \subset I$ there exists an integer $n>0$ such that $T^{n}(J) \cap J \neq \emptyset$.

Solution: 7. The map $T$ transforms the interval $(0, a]$ onto $(1-a, 1]$ and the interval $(a, 1]$ onto $(0,1-a]$. Clearly $T$ preserves the measure. Since the measure of the interval $[0,1]$ is finite, there exist two positive integers $k, l>k$ such that $T^{k}(J)$ and $T^{l}(J)$ are not disjoint. But the map $T$ is bijective; hence $T^{l-k}(J)$ and $J$ are not disjoint.


8. (SWE 3) Let $P$ be a polynomial with real coefficients such that $P(x)>0$ if $x>0$. Prove that there exist polynomials $Q$ and $R$ with nonnegative coefficients such that $P(x)=\frac{Q(x)}{R(x)}$ if $x>0$.

Solution: 8. Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result only for a quadratic polynomial $P(x)=x^{2}-2 a x+b^{2}$, with $a>0$ and $b^{2}>a^{2}$. Using the identity  $$ \left(x^{2}+b^{2}\right)^{2 n}-(2 a x)^{2 n}=\left(x^{2}-2 a x+b^{2}\right) \sum_{k=0}^{2 n-1}\left(x^{2}+b^{2}\right)^{k}(2 a x)^{2 n-k-1} $$  we have solved the problem if we can choose $n$ such that $b^{2 n}\binom{2 n}{n}>2^{2 n} a^{2 n}$. However, it is is easy to show that $2 n\binom{2 n}{n}<2^{2 n}$; hence it is enough to take $n$ such that $(b / a)^{2 n}>2 n$. Since $\lim _{n \rightarrow \infty}(2 n)^{1 /(2 n)}=1<b / a$, such an $n$ always exists.


9. (FIN 2) ${ }^{\mathrm{IMO} 2}$ Let $P_{1}(x)=x^{2}-2, P_{j}(x)=P_{1}\left(P_{j-1}(x)\right), j=2,3, \ldots$. Show that for arbitrary $n$ the roots of the equation $P_{n}(x)=x$ are real and different from one another.

Solution: 9. The equation $P_{n}(x)=x$ is of degree $2^{n}$, and has at most $2^{n}$ distinct roots. If $x>2$, then by simple induction $P_{n}(x)>x$ for all $n$. Similarly, if $x<-1$, then $P_{1}(x)>2$, which implies $P_{n}(x)>2$ for all $n$. It follows that all real roots of the equation $P_{n}(x)=x$ lie in the interval $[-2,2]$, and thus have the form $x=2 \cos t$. Now we observe that $P_{1}(2 \cos t)=4 \cos ^{2} t-2=2 \cos 2 t$, and in general $P_{n}(2 \cos t)=2 \cos 2^{n} t$. Our equation becomes  $$ \cos 2^{n} t=\cos t $$  which indeed has $2^{n}$ different solutions $t=\frac{2 \pi m}{2^{n}-1}\left(m=0,1, \ldots, 2^{n-1}-1\right)$ and $t=\frac{2 \pi m}{2^{n}+1}\left(m=1,2, \ldots, 2^{n-1}\right)$.


1. (BUL 1) A pentagon $A B C D E$ inscribed in a circle for which $B C<C D$ and $A B<D E$ is the base of a pyramid with vertex $S$. If $A S$ is the longest edge starting from $S$, prove that $B S>C S$.

Solution: 1. Let $P$ be the projection of $S$ onto the plane $A B C D E$. Obviously $B S>C S$ is equivalent to $B P>C P$. The conditions of the problem imply that $P A>P B$ and $P A>P E$. The locus of such points $P$ is the region of the plane that is determined by the perpendicular bisectors of segments $A B$ and $A E$ and that contains the point diametrically opposite $A$. But since $A B<D E$, the whole of this region lies on one side of the perpendicular bisector of $B C$. The result follows immediately. Remark. The assumption $B C<C D$ is redundant.


Solution: 10. Integers $a, b, q, r$ satisfy  $$ a^{2}+b^{2}=(a+b) q+r, \quad 0 \leq r<a+b, \quad q^{2}+r=1977 . $$  From $q^{2} \leq 1977$ it follows that $q \leq 44$, and consequently $a^{2}+b^{2}<$ $45(a+b)$. Having in mind the inequality $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$, we get $(a+b)^{2}<90(a+b)$, i.e., $a+b<90$ and consequently $r<90$. Now from $q^{2}=1977-r>1977-90=1887$ it follows that $q>43$; hence $q=44$ and $r=41$. It remains to find positive integers $a$ and $b$ satisfying $a^{2}+b^{2}=44(a+b)+41$, or equivalently  $$ (a-22)^{2}+(b-22)^{2}=1009 $$  The only solutions to this Diophantine equation are $(|a-22|,|b-22|) \in$ $\{(15,28),(28,15)\}$, which yield $(a, b) \in\{(7,50),(37,50),(50,7),(50,37)\}$.


11. (FRG 2) Let $n$ and $z$ be integers greater than 1 and $(n, z)=1$. Prove: (a) At least one of the numbers $z_{i}=1+z+z^{2}+\cdots+z^{i}, i=0,1, \ldots, n-1$, is divisible by $n$. (b) If $(z-1, n)=1$, then at least one of the numbers $z_{i}, i=0,1, \ldots, n-2$, is divisible by $n$.

Solution: 11. (a) Suppose to the contrary that none of the numbers $z_{0}, z_{1}, \ldots, z_{n-1}$ is divisible by $n$. Then two of these numbers, say $z_{k}$ and $z_{l}(0 \leq k<l \leq$ $n-1$ ), are congruent modulo $n$, and thus $n \mid z_{l}-z_{k}=z^{k+1} z_{l-k-1}$. But since $(n, z)=1$, this implies $n \mid z_{l-k-1}$, which is a contradiction. (b) Again suppose the contrary, that none of $z_{0}, z_{1}, \ldots, z_{n-2}$ is divisible by $n$. Since $(z-1, n)=1$, this is equivalent to $n \nmid(z-1) z_{j}$, i.e., $z^{k} \not \equiv 1$ $(\bmod n)$ for all $k=1,2, \ldots, n-1$. But since $(z, n)=1$, we also have that $z^{k} \not \equiv 0(\bmod n)$. It follows that there exist $k, l, 1 \leq k<l \leq n-1$ such that $z^{k} \equiv z^{l}$, i.e., $z^{l-k} \equiv 1(\bmod n)$, which is a contradiction.


12. (FRG 3) Let $z$ be an integer $>1$ and let $M$ be the set of all numbers of the form $z_{k}=1+z+\cdots+z^{k}, k=0,1, \ldots$. Determine the set $T$ of divisors of at least one of the numbers $z_{k}$ from $M$.

Solution: 12. According to part (a) of the previous problem we can conclude that $T=$ $\{n \in \mathbb{N} \mid(n, z)=1\}$.


Solution: 13. The figure $\Phi$ contains two points $A$ and $B$ having maximum distance. Let $h$ be the semicircle with diameter $A B$ that lies in $\Phi$, and let $k$ be the circle containing $h$. Consider any point $M$ inside $k$. The line passing through $M$ that is orthogonal to $A M$ meets $h$ in some point $P$ (because $\angle A M B>90^{\circ}$ ). Let $h^{\prime}$ and $\overline{h^{\prime}}$ be the two semicircles with diameter $A P$, where $M \in h^{\prime}$. Since $\overline{h^{\prime}}$ contains a point $C$ such that $B C>A B$, it cannot be contained in $\Phi$, implying that $h^{\prime} \subset \Phi$. Hence $M$ belongs to $\Phi$. Since $\Phi$ contains no points outside the circle $k$, it must coincide with the disk determined by $k$. On the other hand, any disk has the required property.


Solution: 14. We prove by induction on $n$ that independently of the word $w_{0}$, the given algorithm generates all words of length $n$. This is clear for $n=1$. Suppose now the statement is true for $n-1$, and that we are given a word $w_{0}=$ $c_{1} c_{2} \ldots c_{n}$ of length $n$. Obviously, the words $w_{0}, w_{1}, \ldots, w_{2^{n-1}-1}$ all have the $n$th digit $c_{n}$, and by the inductive hypothesis these are all words whose $n$th digit is $c_{n}$. Similarly, by the inductive hypothesis $w_{2^{n-1}}, \ldots, w_{2^{n-1}}$ are all words whose $n$th digit is $1-c_{n}$, and the induction is complete.


15. (GDR 1) Let $n$ be an integer greater than 1 . In the Cartesian coordinate system we consider all squares with integer vertices $(x, y)$ such that $1 \leq$ $x, y \leq n$. Denote by $p_{k}(k=0,1,2, \ldots)$ the number of pairs of points that are vertices of exactly $k$ such squares. Prove that $\sum_{k}(k-1) p_{k}=0$.

Solution: 15. Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore $p_{k}=0$ for $k>3$, and we have to prove that  $$ p_{0}=p_{2}+2 p_{3} $$  Let us calculate the number $q(n)$ of considered squares. Each of these squares is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are $(n-s)^{2}$ squares of side $s$ with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly $s$ of the considered squares. It follows that $q(n)=\sum_{s=1}^{n-1}(n-s)^{2} s=n^{2}\left(n^{2}-1\right) / 12$. Computing the number of edges and diagonals of the considered squares in two ways, we obtain that  $$ p_{1}+2 p_{2}+3 p_{3}=6 q(n) $$  On the other hand, the total number of segments with endpoints in the considered integer points is given by  $$ p_{0}+p_{1}+p_{2}+p_{3}=\binom{n^{2}}{2}=\frac{n^{2}\left(n^{2}-1\right)}{2}=6 q(n) . $$  Now (1) follows immediately from (2) and (3).


Solution: 16. For $i=k$ and $j=l$ the system is reduced to $1 \leq i, j \leq n$, and has exactly $n^{2}$ solutions. Let us assume that $i \neq k$ or $j \neq l$. The points $A(i, j), B(k, l)$, $C(-j+k+l, i-k+l), D(i-j+l, i+j-k)$ are vertices of a negatively oriented square with integer vertices lying inside the square $[1, n] \times[1, n]$, and each of these squares corresponds to exactly 4 solutions to the system. By the previous problem there are exactly $q(n)=n^{2}\left(n^{2}-1\right) / 12$ such squares. Hence the number of solutions is equal to $n^{2}+4 q(n)=n^{2}\left(n^{2}+2\right) / 3$.


17. (GDR 3) A ball $K$ of radius $r$ is touched from the outside by mutually equal balls of radius $R$. Two of these balls are tangent to each other. Moreover, for two balls $K_{1}$ and $K_{2}$ tangent to $K$ and tangent to each other there exist two other balls tangent to $K_{1}, K_{2}$ and also to $K$. How many balls are tangent to $K$ ? For a given $r$ determine $R$.

Solution: 17. Centers of the balls that are tangent to $K$ are vertices of a regular polyhedron with triangular faces, with edge length $2 R$ and radius of circumscribed sphere $r+R$. Therefore the number $n$ of these balls is 4,6 , or 20 . It is straightforward to obtain that: (i) If $n=4$, then $r+R=2 R(\sqrt{6} / 4)$, whence $R=r(2+\sqrt{6})$. (ii) If $n=6$, then $r+R=2 R(\sqrt{2} / 2)$, whence $R=r(1+\sqrt{2})$. (iii) If $n=20$, then $r+R=2 R \sqrt{5+\sqrt{5}} / 8$, whence $R=r[\sqrt{5-2 \sqrt{5}}+$  $$ (3-\sqrt{5}) / 2] $$


18. (GDR 4) Given an isosceles triangle $A B C$ with a right angle at $C$, construct the center $M$ and radius $r$ of a circle cutting on segments $A B, B C, C A$ the segments $D E, F G$, and $H K$, respectively, such that $\angle D M E+\angle F M G+\angle H M K=180^{\circ}$ and $D E: F G: H K=A B: B C:$ $C A$.

Solution: 18. Let $U$ be the midpoint of the segment $A B$. The point $M$ belongs to $C U$ and $C M=(\sqrt{5}-1) C U / 2, r=C U \sqrt{\sqrt{5}-2}$.


19. (GBR 1) Given any integer $m>1$ prove that there exist infinitely many positive integers $n$ such that the last $m$ digits of $5^{n}$ are a sequence $a_{m}, a_{m-1}, \ldots, a_{1}=5\left(0 \leq a_{j}<10\right)$ in which each digit except the last is of opposite parity to its successor (i.e., if $a_{i}$ is even, then $a_{i-1}$ is odd, and if $a_{i}$ is odd, then $a_{i-1}$ is even).

Solution: 19. We shall prove the statement by induction on $m$. For $m=2$ it is trivial, since each power of 5 greater than 5 ends in 25 . Suppose that the statement is true for some $m \geq 2$, and that the last $m$ digits of $5^{n}$ alternate in parity. It can be shown by induction that the maximum power of 2 that divides $5^{2^{m-2}}-1$ is $2^{m}$, and consequently the difference $5^{n+2^{m-2}}-5^{n}$ is divisible by $10^{m}$ but not by $2 \cdot 10^{m}$. It follows that the last $m$ digits of the numbers $5^{n+2^{m-2}}$ and $5^{n}$ coincide, but the digits at the position $m+1$ have opposite parity. Hence the last $m+1$ digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.


Solution: 2. We shall prove by induction on $n$ that $f(x)>f(n)$ whenever $x>n$. The case $n=0$ is trivial. Suppose that $n \geq 1$ and that $x>k$ implies $f(x)>f(k)$ for all $k<n$. It follows that $f(x) \geq n$ holds for all $x \geq n$. Let $f(m)=\min _{x \geq n} f(x)$. If we suppose that $m>n$, then $m-1 \geq n$ and consequently $f(m-1) \geq n$. But in this case the inequality $f(m)>$ $f(f(m-1))$ contradicts the minimality property of $m$. The inductive proof is thus completed. It follows that $f$ is strictly increasing, so $f(n+1)>f(f(n))$ implies that $n+1>f(n)$. But since $f(n) \geq n$ we must have $f(n)=n$.


Solution: 20. There exist $u, v$ such that $a \cos x+b \sin x=r \cos (x-u)$ and $A \cos 2 x+$ $B \sin 2 x=R \cos 2(x-v)$, where $r=\sqrt{a^{2}+b^{2}}$ and $R=\sqrt{A^{2}+B^{2}}$. Then $1-f(x)=r \cos (x-u)+R \cos 2(x-v) \leq 1$ holds for all $x \in \mathbb{R}$. There exists $x \in \mathbb{R}$ such that $\cos (x-u) \geq 0$ and $\cos 2(x-v)=1$ (indeed, either $x=v$ or $x=v+\pi$ works). It follows that $R \leq 1$. Similarly, there exists $x \in \mathbb{R}$ such that $\cos (x-u)=1 / \sqrt{2}$ and $\cos 2(x-v) \geq 0$ (either $x=u-\pi / 4$ or $x=u+\pi / 4$ works). It follows that $r \leq \sqrt{2}$. Remark. The proposition of this problem contained as an addendum the following, more difficult, inequality:  $$ \sqrt{a^{2}+b^{2}}+\sqrt{A^{2}+B^{2}} \leq 2 . $$  The proof follows from the existence of $x \in \mathbb{R}$ such that $\cos (x-u) \geq 1 / 2$ and $\cos 2(x-v) \geq 1 / 2$.


Solution: 21. Let us consider the vectors $v_{1}=\left(x_{1}, x_{2}, x_{3}\right), v_{2}=\left(y_{1}, y_{2}, y_{3}\right), v_{3}=(1,1,1)$ in space. The given equalities express the condition that these three vectors are mutually perpendicular. Also, $\frac{x_{1}^{2}}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}, \frac{y_{1}^{2}}{y_{1}^{2}+y_{2}^{2}+y_{3}^{2}}$, and $1 / 3$ are the squares of the projections of the vector $(1,0,0)$ onto the directions of $v_{1}, v_{2}, v_{3}$, respectively. The result follows from the fact that the sum of squares of projections of a unit vector on three mutually perpendicular directions is 1.


Solution: 22. Since the quadrilateral $O A_{1} B B_{1}$ is cyclic, $\angle O A_{1} B_{1}=\angle O B C$. By using the analogous equalities we obtain $\angle O A_{4} B_{4}=\angle O B_{3} C_{3}=\angle O C_{2} D_{2}=$ $\angle O D_{1} A_{1}=\angle O A B$, and similarly $\angle O B_{4} A_{4}=\angle O B A$. Hence $\triangle O A_{4} B_{4} \sim$ $\triangle O A B$. Analogously, we have for the other three pairs of triangles $\triangle O B_{4} C_{4} \sim \triangle O B C, \triangle O C_{4} D_{4} \sim \triangle O C D, \triangle O D_{4} A_{4} \sim \triangle O D A$, and consequently $A B C D \sim A_{4} B_{4} C_{4} D_{4}$.


Solution: 23. Every polynomial $q\left(x_{1}, \ldots, x_{n}\right)$ with integer coefficients can be expressed in the form $q=r_{1}+x_{1} r_{2}$, where $r_{1}, r_{2}$ are polynomials in $x_{1}, \ldots, x_{n}$ with integer coefficients in which the variable $x_{1}$ occurs only with even exponents. Thus if $q_{1}=r_{1}-x_{1} r_{2}$, the polynomial $q q_{1}=r_{1}^{2}-x_{1}^{2} r_{2}^{2}$ contains $x_{1}$ only with even exponents. We can continue inductively constructing polynomials $q_{j}, j=2,3, \ldots, n$, such that $q q_{1} q_{2} \cdots q_{j}$ contains each of variables $x_{1}, x_{2}, \ldots, x_{j}$ only with even exponents. Thus the polynomial $q q_{1} \cdots q_{n}$ is a polynomial in $x_{1}^{2}, \ldots, x_{n}^{2}$. The polynomials $f$ and $g$ exist for every $n \in \mathbb{N}$. In fact, it suffices to construct $q_{1}, \ldots, q_{n}$ for the polynomial $q=x_{1}+\cdots+x_{n}$ and take $f=$ $q_{1} q_{2} \cdots q_{n}$.


24. (HUN 2) Determine all real functions $f(x)$ that are defined and continuous on the interval $(-1,1)$ and that satisfy the functional equation  $$ f(x+y)=\frac{f(x)+f(y)}{1-f(x) f(y)} \quad(x, y, x+y \in(-1,1)) . $$

Solution: 24. Setting $x=y=0$ gives us $f(0)=0$. Let us put $g(x)=\arctan f(x)$. The given functional equation becomes $\tan g(x+y)=\tan (g(x)+g(y))$; hence  $$ g(x+y)=g(x)+g(y)+k(x, y) \pi $$  where $k(x, y)$ is an integer function. But $k(x, y)$ is continuous and $k(0,0)=$ 0 , therefore $k(x, y)=0$. Thus we obtain the classical Cauchy's functional equation $g(x+y)=g(x)+g(y)$ on the interval $(-1,1)$, all of whose continuous solutions are of the form $g(x)=a x$ for some real $a$. Moreover, $g(x) \in(-\pi, \pi)$ implies $|a| \leq \pi / 2$. Therefore $f(x)=\tan a x$ for some $|a| \leq \pi / 2$, and this is indeed a solution to the given equation.


Solution: 25. Let  $$ f_{n}(z)=z^{n}+a \sum_{k=1}^{n}\binom{n}{k}(a-k b)^{k-1}(z+k b)^{n-k} . $$  We shall prove by induction on $n$ that $f_{n}(z)=(z+a)^{n}$. This is trivial for $n=1$. Suppose that the statement is true for some positive integer $n-1$. Then  $$ \begin{aligned} f_{n}^{\prime}(z) & =n z^{n-1}+a \sum_{k=1}^{n-1}\binom{n}{k}(n-k)(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n z^{n-1}+n a \sum_{k=1}^{n-1}\binom{n-1}{k}(a-k b)^{k-1}(z+k b)^{n-k-1} \\ & =n f_{n-1}(z)=n(z+a)^{n-1} \end{aligned} $$  It remains to prove that $f_{n}(-a)=0$. For $z=-a$ we have by the lemma of (SL81-13),  $$ \begin{aligned} f_{n}(-a) & =(-a)^{n}+a \sum_{k=1}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1} \\ & =a \sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}(a-k b)^{n-1}=0 . \end{aligned} $$


26. (NET 1) Let $p$ be a prime number greater than 5 . Let $V$ be the collection of all positive integers $n$ that can be written in the form $n=k p+1$ or $n=k p-1(k=1,2, \ldots)$. A number $n \in V$ is called indecomposable in $V$ if it is impossible to find $k, l \in V$ such that $n=k l$. Prove that there exists a number $N \in V$ that can be factorized into indecomposable factors in $V$ in more than one way.

Solution: 26. The result is an immediate consequence (for $G=\{-1,1\}$ ) of the following generalization. (1) Let $G$ be a proper subgroup of $\mathbb{Z}_{n}^{*}$ (the multiplicative group of residue classes modulo $n$ coprime to $n$ ), and let $V$ be the union of elements of $G$. A number $m \in V$ is called indecomposable in $V$ if there do not exist numbers $p, q \in V, p, q \notin\{-1,1\}$, such that $p q=m$. There exists a number $r \in V$ that can be expressed as a product of elements indecomposable in $V$ in more than one way. First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in $V$ that do not divide $n$. Proof. There is at least one such prime: In fact, any number other than $\pm 1$ not in $V$ must have a prime factor not in $V$, since $V$ is closed under multiplication. If there were a finite number of such primes, say $p_{1}, p_{2}, \ldots, p_{k}$, then one of the numbers $p_{1} p_{2} \cdots p_{k}+n, p_{1}^{2} p_{2} \cdots p_{k}+n$ is not in $V$ and is coprime to $n$ and $p_{1}, \ldots, p_{k}$, which is a contradiction. [This lemma is actually a direct consequence of Dirichlet's theorem.] Let us consider two such primes $p, q$ that are congruent modulo $n$. Let $p^{k}$ be the least power of $p$ that is in $V$. Then $p^{k}, q^{k}, p^{k-1} q, p q^{k-1}$ belong to $V$ and are indecomposable in $V$. It follows that  $$ r=p^{k} \cdot q^{k}=p^{k-1} q \cdot p q^{k-1} $$  has the desired property. Second proof. Let $p$ be any prime not in $V$ that does not divide $n$, and let $p^{k}$ be the least power of $p$ that is in $V$. Obviously $p^{k}$ is indecomposable in $V$. Then the number  $$ r=p^{k} \cdot\left(p^{k-1}+n\right)(p+n)=p\left(p^{k-1}+n\right) \cdot p^{k-1}(p+n) $$  has at least two different factorizations into indecomposable factors.


Solution: 27. The result is a consequence of the generalization from the previous problem for $G=\{1\}$. Remark. There is an explicit example: $r=(n-1)^{2} \cdot(2 n-1)^{2}=[(n-$ 1) $(2 n-1)]^{2}$.


Solution: 28. The recurrent relations give us that  $$ x_{i+1}=\left[\frac{x_{i}+\left[n / x_{i}\right]}{2}\right]=\left[\frac{x_{i}+n / x_{i}}{2}\right] \geq[\sqrt{n}] $$  On the other hand, if $x_{i}>[\sqrt{n}]$ for some $i$, then we have $x_{i+1}<x_{i}$. This follows from the fact that $x_{i+1}<x_{i}$ is equivalent to $x_{i}>\left(x_{i}+n / x_{i}\right) / 2$, i.e., to $x_{i}^{2}>n$. Therefore $x_{i}=[\sqrt{n}]$ holds for at least one $i \leq n-[\sqrt{n}]+1$. Remark. If $n+1$ is a perfect square, then $x_{i}=[\sqrt{n}]$ implies $x_{i+1}=$ $[\sqrt{n}]+1$. Otherwise, $x_{i}=[\sqrt{n}]$ implies $x_{i+1}=[\sqrt{n}]$.


Solution: 29. Let us denote the midpoints of segments $L M, A N, B L, M N, B K, C M$, $N K, C L, D N, K L, D M, A K$ by $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}, P_{8}, P_{9}, P_{10}$, $P_{11}, P_{12}$, respectively.  We shall prove that the dodecagon $P_{1} P_{2} P_{3} \ldots P_{11} P_{12}$ is regular. From $B L=B A$ and $\angle A B L=30^{\circ}$ it follows that $\angle B A L=75^{\circ}$. Similarly $\angle D A M=75^{\circ}$, and therefore $\angle L A M=60^{\circ}$, which together with $A L=A M$ implies that the triangle $A L M$ is equilateral. Now, from the triangles $O L M$ and $A L N$, we get ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-426.jpg?height=383&width=433&top_left_y=233&top_left_x=863) $O P_{1}=L M / 2, O P_{2}=A L / 2$ and $O P_{2} \| A L$. Hence $O P_{1}=O P_{2}$, $\angle P_{1} O P_{2}=\angle P_{1} A L=30^{\circ}$ and $\angle P_{2} O M=\angle L A D=15^{\circ}$. The desired result follows from symmetry.


3. (BUL 3) In a company of $n$ persons, each person has no more than $d$ acquaintances, and in that company there exists a group of $k$ persons, $k \geq d$, who are not acquainted with each other. Prove that the number of acquainted pairs is not greater than $\left[n^{2} / 4\right]$.

Solution: 3. Let $v_{1}, v_{2}, \ldots, v_{k}$ be $k$ persons who are not acquainted with each other. Let us denote by $m$ the number of acquainted couples and by $d_{j}$ the number of acquaintances of person $v_{j}$. Then $m \leq d_{k+1}+d_{k+2}+\cdots+d_{n} \leq d(n-k) \leq k(n-k) \leq\left(\frac{k+(n-k)}{2}\right)^{2}=\frac{n^{2}}{4}$.


30. (NET 5) A triangle $A B C$ with $\angle A=30^{\circ}$ and $\angle C=54^{\circ}$ is given. On $B C$ a point $D$ is chosen such that $\angle C A D=12^{\circ}$. On $A B$ a point $E$ is chosen such that $\angle A C E=6^{\circ}$. Let $S$ be the point of intersection of $A D$ and $C E$. Prove that $B S=B C$.

Solution: 30. Suppose $\angle S B A=x$. By the trigonometric form of Ceva's theorem we have  $$ \frac{\sin \left(96^{\circ}-x\right)}{\sin x} \frac{\sin 18^{\circ}}{\sin 12^{\circ}} \frac{\sin 6^{\circ}}{\sin 48^{\circ}}=1 $$  We claim that $x=12^{\circ}$ is a solution of this equation. To prove this, it is enough to show that $\sin 84^{\circ} \sin 6^{\circ} \sin 18^{\circ}=\sin 48^{\circ} \sin 12^{\circ} \sin 12^{\circ}$, which is equivalent to $\sin 18^{\circ}=2 \sin 48^{\circ} \sin 12^{\circ}=\cos 36^{\circ}-\cos 60^{\circ}$. The last equality can be checked directly. Since the equation is equivalent to $\left(\sin 96^{\circ} \cot x-\cos 96^{\circ}\right) \sin 6^{\circ} \sin 18^{\circ}=$ $\sin 48^{\circ} \sin 12^{\circ}$, the solution $x \in[0, \pi)$ is unique. Hence $x=12^{\circ}$. Second solution. We know that if $a, b, c, a^{\prime}, b^{\prime}, c^{\prime}$ are points on the unit circle in the complex plane, the lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent if and only if  $$ \left(a-b^{\prime}\right)\left(b-c^{\prime}\right)\left(c-a^{\prime}\right)=\left(a-c^{\prime}\right)\left(b-a^{\prime}\right)\left(c-b^{\prime}\right) $$  We shall prove that $x=12^{\circ}$. We may suppose that $A B C$ is the triangle in the complex plane with vertices $a=1, b=\epsilon^{9}, c=\epsilon^{14}$, where $\epsilon=$ $\cos \frac{\pi}{15}+i \sin \frac{\pi}{15}$. If $a^{\prime}=\epsilon^{12}, b^{\prime}=\epsilon^{28}, c^{\prime}=\epsilon$, our task is the same as proving that lines $a a^{\prime}, b b^{\prime}, c c^{\prime}$ are concurrent, or by (1) that  $$ \left(1-\epsilon^{28}\right)\left(\epsilon^{9}-\epsilon\right)\left(\epsilon^{14}-\epsilon^{12}\right)-(1-\epsilon)\left(\epsilon^{9}-\epsilon^{12}\right)\left(\epsilon^{14}-\epsilon^{28}\right)=0 . $$  The last equality holds, since the left-hand side is divisible by the minimum polynomial of $\epsilon: z^{8}+z^{7}-z^{5}-z^{4}-z^{3}+z+1$.


31. (POL 1) Let $f$ be a function defined on the set of pairs of nonzero rational numbers whose values are positive real numbers. Suppose that $f$ satisfies the following conditions: (1) $f(a b, c)=f(a, c) f(b, c), f(c, a b)=f(c, a) f(c, b)$; (2) $f(a, 1-a)=1$.  Prove that $f(a, a)=f(a,-a)=1, f(a, b) f(b, a)=1$.

Solution: 31. We obtain from (1) that $f(1, c)=f(1, c) f(1, c)$; hence $f(1, c)=1$ and consequently $f(-1, c) f(-1, c)=f(1, c)=1$, i.e. $f(-1, c)=1$. Analogously, $f(c, 1)=f(c,-1)=1$. Clearly $f(1,1)=f(-1,1)=f(1,-1)=1$. Now let us assume that $a \neq 1$. Observe that $f\left(x^{-1}, y\right)=f\left(x, y^{-1}\right)=f(x, y)^{-1}$. Thus by (1) and (2) we get  $$ \begin{aligned} 1 & =f(a, 1-a) f(1 / a, 1-1 / a) \\ & =f(a, 1-a) f\left(a, \frac{1}{1-1 / a}\right)=f\left(a, \frac{1-a}{1-1 / a}\right)=f(a,-a) \end{aligned} $$  We now have $f(a, a)=f(a,-1) f(a,-a)=1 \cdot 1=1$ and $1=f(a b, a b)=$ $f(a, a b) f(b, a b)=f(a, a) f(a, b) f(b, a) f(b, b)=f(a, b) f(b, a)$.


32. (POL 2) In a room there are nine men. Among every three of them there are two mutually acquainted. Prove that some four of them are mutually acquainted.

Solution: 32. It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.


33. (POL 3) A circle $K$ centered at $(0,0)$ is given. Prove that for every vector $\left(a_{1}, a_{2}\right)$ there is a positive integer $n$ such that the circle $K$ translated by the vector $n\left(a_{1}, a_{2}\right)$ contains a lattice point (i.e., a point both of whose coordinates are integers).

Solution: 33. Let $r$ be the radius of $K$ and $s>\sqrt{2} / r$ an integer. Consider the points $A_{k}\left(k a_{1}-\left[k a_{1}\right], k a_{2}-\left[k a_{2}\right]\right)$, where $k=0,1,2, \ldots, s^{2}$. Since all these points are in the unit square, two of them, say $A_{p}, A_{q}, q>p$, are in a small square with side $1 / s$, and consequently $A_{p} A_{q} \leq \sqrt{2} / s<r$. Therefore, for $n=q-p, m_{1}=\left[q a_{1}\right]-\left[p a_{1}\right]$ and $m_{2}=\left[q a_{2}\right]-\left[p a_{2}\right]$ the distance between the points $n\left(a_{1}, a_{2}\right)$ and $\left(m_{1}, m_{2}\right)$ is less then $r$, i.e., the point $\left(m_{1}, m_{2}\right)$ is in the circle $K+n\left(a_{1}, a_{2}\right)$.