Solution: 34. Let $A$ be the set of the $2^{n}$ sequences of $n$ terms equal to $\pm 1$. Since there are $k^{2}$ products $a b$ with $a, b \in B$, by the pigeonhole principle there exists $c \in A$ such that $a b=c$ holds for at most $k^{2} / 2^{n}$ pairs $(a, b) \in B \times B$. Then $c b \in B$ holds for at most $k^{2} / 2^{n}$ values $b \in B$, which means that $|B \cap c B| \leq k^{2} / 2^{n}$.


35. (ROM 1) Find all numbers $N=\overline{a_{1} a_{2} \ldots a_{n}}$ for which $9 \times \overline{a_{1} a_{2} \ldots a_{n}}=$ $\overline{a_{n} \ldots a_{2} a_{1}}$ such that at most one of the digits $a_{1}, a_{2}, \ldots, a_{n}$ is zero.

Solution: 35. The solutions are 0 and $N_{k}=10 \underbrace{99 \ldots 9}_{k} 89$, where $k=0,1,2, \ldots$.  Remark. If we omit the condition that at most one of the digits is zero, the solutions are numbers of the form $N_{k_{1}} N_{k_{2}} \ldots N_{k_{r}}$, where $k_{1}=k_{r}$, $k_{2}=k_{r-1}$ etc. The more general problem $k \cdot \overline{a_{1} a_{2} \ldots a_{n}}=\overline{a_{n} \ldots a_{2} a_{1}}$ has solutions only for $k=9$ and for $k=4$ (namely $0,2199 \ldots 978$ and combinations as above).


36. (ROM 2) Consider a sequence of numbers $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$. Define the operation  $$ S\left(\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)\right)=\left(a_{1} a_{2}, a_{2} a_{3}, \ldots, a_{2^{n}-1} a_{2^{n}}, a_{2^{n}} a_{1}\right) $$  Prove that whatever the sequence $\left(a_{1}, a_{2}, \ldots, a_{2^{n}}\right)$ is, with $a_{i} \in\{-1,1\}$ for $i=1,2, \ldots, 2^{n}$, after finitely many applications of the operation we get the sequence $(1,1, \ldots, 1)$.

Solution: 36. It can be shown by simple induction that $S^{m}\left(a_{1}, \ldots, a_{2^{n}}\right)=\left(b_{1}, \ldots, b_{2^{n}}\right)$, where  $$ \left.b_{k}=\prod_{i=0}^{m} a_{k+i}^{\binom{m}{i}} \text { (assuming that } a_{k+2^{n}}=a_{k}\right) $$  If we take $m=2^{n}$ all the binomial coefficients $\binom{m}{i}$ apart from $i=0$ and $i=m$ will be even, and thus $b_{k}=a_{k} a_{k+m}=1$ for all $k$.


37. (ROM 3) Let $A_{1}, A_{2}, \ldots, A_{n+1}$ be positive integers such that $\left(A_{i}, A_{n+1}\right)$ $=1$ for every $i=1,2, \ldots, n$. Show that the equation  $$ x_{1}^{A_{1}}+x_{2}^{A_{2}}+\cdots+x_{n}^{A_{n}}=x_{n+1}^{A_{n+1}} $$  has an infinite set of solutions $\left(x_{1}, x_{2}, \ldots, x_{n+1}\right)$ in positive integers.

Solution: 37. We look for a solution with $x_{1}^{A_{1}}=\cdots=x_{n}^{A_{n}}=n^{A_{1} A_{2} \cdots A_{n} x}$ and $x_{n+1}=$ $n^{y}$. In order for this to be a solution we must have $A_{1} A_{2} \cdots A_{n} x+1=$ $A_{n+1} y$. This equation has infinitely many solutions $(x, y)$ in $\mathbb{N}$, since $A_{1} A_{2} \cdots A_{n}$ and $A_{n+1}$ are coprime.


38. (ROM 4) Let $m_{j}>0$ for $j=1,2, \ldots, n$ and $a_{1} \leq \cdots \leq a_{n}<b_{1} \leq \cdots \leq$ $b_{n}<c_{1} \leq \cdots \leq c_{n}$ be real numbers. Prove:  $$ \left[\sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)\right]^{2}>3\left(\sum_{j=1}^{n} m_{j}\right)\left[\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right)\right] . $$

Solution: 38. The condition says that the quadratic equation $f(x)=0$ has distinct real solutions, where  $$ f(x)=3 x^{2} \sum_{j=1}^{n} m_{j}-2 x \sum_{j=1}^{n} m_{j}\left(a_{j}+b_{j}+c_{j}\right)+\sum_{j=1}^{n} m_{j}\left(a_{j} b_{j}+b_{j} c_{j}+c_{j} a_{j}\right) $$  It is easy to verify that the function $f$ is the derivative of  $$ F(x)=\sum_{j=1}^{n} m_{j}\left(x-a_{j}\right)\left(x-b_{j}\right)\left(x-c_{j}\right) $$  Since $F\left(a_{1}\right) \leq 0 \leq F\left(a_{n}\right), F\left(b_{1}\right) \leq 0 \leq F\left(b_{n}\right)$ and $F\left(c_{1}\right) \leq 0 \leq F\left(c_{n}\right)$, $F(x)$ has three distinct real roots, and hence by Rolle's theorem its derivative $f(x)$ has two distinct real roots.


39. (ROM 5) Consider 37 distinct points in space, all with integer coordinates. Prove that we may find among them three distinct points such that their barycenter has integers coordinates.

Solution: 39. By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their $x$-coordinates are congruent and their $y$-coordinates are congruent modulo 3 . Now among these 5 points either there exist three with $z$-coordinates congruent modulo 3 , or there exist three whose $z$ coordinates are congruent to $0,1,2$ modulo 3 . These three points are the desired ones. Remark. The minimum number $n$ such that among any $n$ integer points in space one can find three points whose barycenter is an integer point is $n=19$. Each proof of this result seems to consist in studying a great number of cases.


4. (BUL 4) We are given $n$ points in space. Some pairs of these points are connected by line segments so that the number of segments equals $\left[n^{2} / 4\right]$, and a connected triangle exists. Prove that any point from which the maximal number of segments starts is a vertex of a connected triangle.

Solution: 4. Consider any vertex $v_{n}$ from which the maximal number $d$ of segments start, and suppose it is not a vertex of a triangle. Let $\mathcal{A}=$ $\left\{v_{1}, v_{2}, \ldots, v_{d}\right\}$ be the set of points that are connected to $v_{n}$, and let $\mathcal{B}=\left\{v_{d+1}, v_{d+2}, \ldots, v_{n}\right\}$ be the set of the other points. Since $v_{n}$ is not a vertex of a triangle, there is no segment both of whose vertices lie in $\mathcal{A}$; i.e., each segment has an end in $\mathcal{B}$. Thus, if $d_{j}$ denotes the number of segments at $v_{j}$ and $m$ denotes the total number of segments, we have  $$ m \leq d_{d+1}+d_{d+2}+\cdots+d_{n} \leq d(n-d) \leq\left[\frac{n^{2}}{4}\right]=m $$  This means that each inequality must be equality, implying that each point in $\mathcal{B}$ is a vertex of $d$ segments, and each of these segments has the other end in $\mathcal{A}$. Then there is no triangle at all, which is a contradiction.


40. (SWE 1) The numbers $1,2,3, \ldots, 64$ are placed on a chessboard, one number in each square. Consider all squares on the chessboard of size $2 \times 2$. Prove that there are at least three such squares for which the sum of the 4 numbers contained exceeds 100.

Solution: 40. Let us divide the chessboard into 16 squares $Q_{1}, Q_{2}, \ldots, Q_{16}$ of size $2 \times 2$. Let $s_{k}$ be the sum of numbers in $Q_{k}$, and let us assume that $s_{1} \geq s_{2} \geq$ $\cdots \geq s_{16}$. Since $s_{4}+s_{5}+\cdots+s_{16} \geq 1+2+\cdots+52=1378$, we must have $s_{4} \geq 100$ and hence $s_{1}, s_{2}, s_{3} \geq 100$ as well.


41. (SWE 2) A wheel consists of a fixed circular disk and a mobile circular ring. On the disk the numbers $1,2,3, \ldots, N$ are marked, and on the ring $N$ integers $a_{1}, a_{2}, \ldots, a_{N}$ of sum 1 are marked (see the figure). The ring can be turned into $N$ different positions in which the numbers on the disk and on the ring match each other. Multiply every number on the ring with the corresponding number on the disk and form the sum of $N$ products. In this way a ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-127.jpg?height=349&width=362&top_left_y=824&top_left_x=919) sum is obtained for every position of the ring. Prove that the $N$ sums are different.

Solution: 41. The considered sums are congruent modulo $n$ to $S_{k}=\sum_{i=1}^{N}(i+k) a_{i}$, $k=0,1, \ldots, N-1$. Since $S_{k}=S_{0}+k\left(a_{1}+\cdots+a_{n}\right)=S_{0}+k$, all these sums give distinct residues modulo $n$ and therefore are distinct.


42. (SWE 3) The sequence $a_{n, k}, k=1,2,3, \ldots, 2^{n}, n=0,1,2, \ldots$, is defined by the following recurrence formula:  $$ \begin{aligned} a_{1} & =2, \quad a_{n, k}=2 a_{n-1, k}^{3}, \quad a_{n, k+2^{n-1}}=\frac{1}{2} a_{n-1, k}^{3} \\ \text { for } k & =1,2,3, \ldots, 2^{n-1}, n=0,1,2, \ldots . \end{aligned} $$  Prove that the numbers $a_{n, k}$ are all different.

Solution: 42. It can be proved by induction on $n$ that $\left\{a_{n, k} \mid 1 \leq k \leq 2^{n}\right\}=\left\{2^{m} \mid m=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}\left(s_{i}= \pm 1\right)\right\}$. Thus the result is an immediate consequence of the following lemma. Lemma. Each positive integer $s$ can be uniquely represented in the form  $$ s=3^{n}+3^{n-1} s_{1}+\cdots+3^{1} s_{n-1}+s_{n}, \quad \text { where } s_{i} \in\{-1,0,1\} $$  Proof. Both the existence and the uniqueness can be shown by simple induction on $s$. The statement is trivial for $s=1$, while for $s>1$ there exist $q \in \mathbb{N}, r \in\{-1,0,1\}$ such that $s=3 q+r$, and $q$ has a unique representation of the form (1).


43. (FIN 1) Evaluate  $$ S=\sum_{k=1}^{n} k(k+1) \cdots(k+p), $$  where $n$ and $p$ are positive integers.

Solution: 43. Since $\left.k(k+1) \cdots(k+p)=(p+1)!\binom{k+p}{p+1}=(p+1)!\left[\begin{array}{c}k+p+1 \\ p+2\end{array}\right)-\binom{k+p}{p+2}\right]$, it follows that $\sum_{k=1}^{n} k(k+1) \cdots(k+p)=(p+1)!\binom{n+p+1}{p+2}=\frac{n(n+1) \cdots(n+p+1)}{p+2}$.


44. (FIN 2) Let $E$ be a finite set of points in space such that $E$ is not contained in a plane and no three points of $E$ are collinear. Show that $E$ contains the vertices of a tetrahedron $T=A B C D$ such that $T \cap E=$ $\{A, B, C, D\}$ (including interior points of $T$ ) and such that the projection of $A$ onto the plane $B C D$ is inside a triangle that is similar to the triangle $B C D$ and whose sides have midpoints $B, C, D$.

Solution: 44. Let $d(X, \sigma)$ denote the distance from a point $X$ to a plane $\sigma$. Let us consider the pair $(A, \pi)$ where $A \in E$ and $\pi$ is a plane containing some three points $B, C, D \in E$ such that $d(A, \pi)$ is the smallest possible. We may suppose that $B, C, D$ are selected such that $\triangle B C D$ contains no other points of $E$. Let $A^{\prime}$ be the projection of $A$ on $\pi$, and let $l_{b}, l_{c}, l_{d}$ be lines through $B, C, D$ parallel to $C D, D B, B C$ respectively. If $A^{\prime}$ is in the half-plane determined by $l_{d}$ not containing $B C$, then $d(D, A B C) \leq d\left(A^{\prime}, A B C\right)<d(A, B C D)$, which is impossible. Similarly, $A^{\prime}$ lies in the half-planes determined by $l_{b}, l_{c}$ that contain $D$, and hence $A^{\prime}$ is inside the triangle bordered by $l_{b}, l_{c}, l_{d}$. The minimality property of $(A, \pi)$ and the way in which $B C D$ was selected guarantee that $E \cap T=\{A, B, C, D\}$.


Solution: 45. As in the previous problem, let us choose the pair $(A, \pi)$ such that $d(A, \pi)$ is minimal. If $\pi$ contains only three points of $E$, we are done. If not, there are four points in $E \cap P$, say $A_{1}, A_{2}, A_{3}, A_{4}$, such that the quadrilateral $Q=A_{1} A_{2} A_{3} A_{4}$ contains no other points of $E$. Suppose $Q$ is not convex, and that w.l.o.g. $A_{1}$ is inside the triangle $A_{2} A_{3} A_{4}$. If $A_{0}$ is the projection of $A$ on $P$, the point $A_{1}$ belongs to one of the triangles $A_{0} A_{2} A_{3}, A_{0} A_{3} A_{4}$, $A_{0} A_{4} A_{2}$, say $A_{0} A_{2} A_{3}$. Then $d\left(A_{1}, A A_{2} A_{3}\right) \leq d\left(A_{0}, A A_{2} A_{3}\right)<A A_{0}$, which is impossible. Hence $Q$ is convex. Also, by the minimality property of $(A, \pi)$ the pyramid $A A_{1} A_{2} A_{3} A_{4}$ contains no other points of $E$.


46. (FIN 3) Let $f$ be a strictly increasing function defined on the set of real numbers. For $x$ real and $t$ positive, set  $$ g(x, t)=\frac{f(x+t)-f(x)}{f(x)-f(x-t)} $$  Assume that the inequalities  $$ 2^{-1}<g(x, t)<2 $$  hold for all positive $t$ if $x=0$, and for all $t \leq|x|$ otherwise. Show that  $$ 14^{-1}<g(x, t)<14 $$  for all real $x$ and positive $t$.

Solution: 46. We need to consider only the case $t>|x|$. There is no loss of generality in assuming $x>0$. To obtain the estimate from below, set  $$ \begin{aligned} a_{1} & =f\left(-\frac{x+t}{2}\right)-f(-(x+t)), & a_{2}=f(0)-f\left(-\frac{x+t}{2}\right), \\ a_{3} & =f\left(\frac{x+t}{2}\right)-f(0), & a_{4}=f(x+t)-f\left(\frac{x+t}{2}\right) . \end{aligned} $$  Since $-(x+t)<x-t$ and $x<(x+t) / 2$, we have $f(x)-f(x-t) \leq$ $a_{1}+a_{2}+a_{3}$. Since $2^{-1}<a_{j+1} / a_{j}<2$, it follows that  $$ g(x, t)>\frac{a_{4}}{a_{1}+a_{2}+a_{3}}>\frac{a_{3} / 2}{4 a_{3}+2 a_{3}+a_{3}}=14^{-1} . $$  To obtain the estimate from above, set  $$ \begin{aligned} b_{1} & =f(0)-f\left(-\frac{x+t}{3}\right), & b_{2}=f\left(\frac{x+t}{3}\right)-f(0), \\ b_{3} & =f\left(\frac{2(x+t)}{3}\right)-f\left(\frac{x+t}{3}\right), & b_{4}=f(x+t)-f\left(\frac{2(x+t)}{3}\right) . \end{aligned} $$  If $t<2 x$, then $x-t<-(x+t) / 3$ and therefore $f(x)-f(x-t) \geq b_{1}$. If $t \geq 2 x$, then $(x+t) / 3 \leq x$ and therefore $f(x)-f(x-t) \geq b_{2}$. Since $2^{-1}<b_{j+1} / b_{j}<2$, we get  $$ g(x, t)<\frac{b_{2}+b_{3}+b_{4}}{\min \left\{b_{1}, b_{2}\right\}}<\frac{b_{2}+2 b_{2}+4 b_{2}}{b_{2} / 2}=14 $$


47. (USS 1) A square $A B C D$ is given. A line passing through $A$ intersects $C D$ at $Q$. Draw a line parallel to $A Q$ that intersects the boundary of the square at points $M$ and $N$ such that the area of the quadrilateral $A M N Q$ is maximal.

Solution: 47. $M$ lies on $A B$ and $N$ lies on $B C$. If $C Q \leq 2 C D / 3$, then $B M=C Q / 2$. If $C Q>2 C D / 3$, then $N$ coincides with C .


48. (USS 2) The intersection of a plane with a regular tetrahedron with edge $a$ is a quadrilateral with perimeter $P$. Prove that $2 a \leq P \leq 3 a$.

Solution: 48. Let a plane cut the edges $A B, B C, C D, D A$ at points $K, L, M, N$ respectively. Let $D^{\prime}, A^{\prime}, B^{\prime}$ be distinct points in the plane $A B C$ such that the triangles $B C D^{\prime}, C D^{\prime} A^{\prime}, D^{\prime} A^{\prime} B^{\prime}$ are equilateral, and $M^{\prime} \in\left[C D^{\prime}\right], N^{\prime} \in\left[D^{\prime} A^{\prime}\right]$, and $K^{\prime} \in\left[A^{\prime} B^{\prime}\right]$ such that $C M^{\prime}=C M$, $A^{\prime} N^{\prime}=A N$, and $A^{\prime} K^{\prime}=A K$. The perimeter $P$ of the quadrilateral $K L M N$ is equal to the length of the polygonal line $K L M^{\prime} N^{\prime} K^{\prime}$, which is not less than $K K^{\prime}$. It follows that $P \geq 2 a$. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-430.jpg?height=265&width=548&top_left_y=997&top_left_x=810)  Let us consider all quadrilaterals $K L M N$ that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane $\alpha$. The lengths of the segments $K L, L M, M N, N K$ are linear functions in $A K$, and so is $P$. Thus $P$ takes its maximum at an endpoint of the interval, i.e., when the plane $K L M N$ passes through one of the vertices $A, B, C, D$, and it is easy to see that in this case $P \leq 3 a$.


49. (USS 3) Find all pairs of integers $(p, q)$ for which all roots of the trinomials $x^{2}+p x+q$ and $x^{2}+q x+p$ are integers.

Solution: 49. If one of $p, q$, say $p$, is zero, then $-q$ is a perfect square. Conversely, $(p, q)=\left(0,-t^{2}\right)$ and $(p, q)=\left(-t^{2}, 0\right)$ satisfy the conditions for $t \in \mathbb{Z}$. We now assume that $p, q$ are nonzero. If the trinomial $x^{2}+p x+q$ has two integer roots $x_{1}, x_{2}$, then $|q|=\left|x_{1} x_{2}\right| \geq\left|x_{1}\right|+\left|x_{2}\right|-1 \geq|p|-1$. Similarly, if $x^{2}+q x+p$ has integer roots, then $|p| \geq|q|-1$ and $q^{2}-4 p$ is a square. Thus we have two cases to investigate: (i) $|p|=|q|$. Then $p^{2}-4 q=p^{2} \pm 4 p$ is a square, so $(p, q)=(4,4)$. (ii) $|p|=|q| \pm 1$. The solutions for $(p, q)$ are $(t,-1-t)$ for $t \in \mathbb{Z}$ and $(5,6)$, $(6,5)$.


Solution: 5. Let us denote by $I$ and $E$ the sets of interior boundary points and exterior boundary points. Let $A B C D$ be the square inscribed in the circle $k$ with sides parallel to the coordinate axes. Lines $A B, B C, C D, D A$ divide the plane into 9 regions: $\mathcal{R}, \mathcal{R}_{A}, \mathcal{R}_{B}$, $\mathcal{R}_{C}, \mathcal{R}_{D}, \mathcal{R}_{A B}, \mathcal{R}_{B C}, \mathcal{R}_{C D}, \mathcal{R}_{D A}$. There is a unique pair of lattice points $A_{I} \in \mathcal{R}, A_{E} \in$ $\mathcal{R}_{A}$ that are opposite vertices of a unit square. We similarly define $B_{I}, C_{I}, D_{I}, B_{E}, C_{E}, D_{E}$. Let us form a graph $G$ by connecting each point from $E$ lying in $\mathcal{R}_{A B}$ (respectively $\mathcal{R}_{B C}, \mathcal{R}_{C D}, \mathcal{R}_{D A}$ ) to its up- ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-420.jpg?height=389&width=505&top_left_y=271&top_left_x=825) per (respectively left, lower, right) neighbor point (which clearly belongs to $I$ ). It is easy to see that: (i) All vertices from $I$ other than $A_{I}, B_{I}, C_{I}, D_{I}$ have degree 1. (ii) $A_{E}$ is not in $E$ if and only if $A_{I} \in I$ and $\operatorname{deg} A_{I}=2$. (iii) No other lattice points inside $\mathcal{R}_{A}$ belong to $E$.  Thus if $m$ is the number of edges of the graph $G$ and $s$ is the number of points among $A_{E}, B_{E}, C_{E}$, and $D_{E}$ that are in $E$, using (i)-(iii) we easily obtain $|E|=m+s$ and $|I|=m-(4-s)=|E|+4$.


50. (USS 4) Determine all positive integers $n$ for which there exists a polynomial $P_{n}(x)$ of degree $n$ with integer coefficients that is equal to $n$ at $n$ different integer points and that equals zero at zero.

Solution: 50. Suppose that $P_{n}(x)=n$ for $x \in\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Then  $$ P_{n}(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)+n . $$  From $P_{n}(0)=0$ we obtain $n=\left|x_{1} x_{2} \cdots x_{n}\right| \geq 2^{n-2}$ (because at least $n-2$ factors are different from $\pm 1$ ) and therefore $n \geq 2^{n-2}$. It follows that $n \leq 4$. For each positive integer $n \leq 4$ there exists a polynomial $P_{n}$. Here is the list of such polynomials:  $$ \begin{array}{ll} n=1: \pm x, & n=2: 2 x^{2}, x^{2} \pm x,-x^{2} \pm 3 x \\ n=3: \pm\left(x^{3}-x\right)+3 x^{2}, & n=4:-x^{4}+5 x^{2} . \end{array} $$


51. (USS 5) Several segments, which we shall call white, are given, and the sum of their lengths is 1 . Several other segments, which we shall call black, are given, and the sum of their lengths is 1 . Prove that every such system of segments can be distributed on the segment that is 1.51 long in the following way: Segments of the same color are disjoint, and segments of different colors are either disjoint or one is inside the other. Prove that there exists a system that cannot be distributed in that way on the segment that is 1.49 long.

Solution: 51. We shall use the following algorithm:  Choose a segment of maximum length ("basic" segment) and put on it unused segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used. Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one-half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1.5 and can be distributed on a segment of length 1.51. On the other hand, if we are given two white segments of lengths 0.5 and two black segments of lengths 0.999 and 0.001 , we cannot distribute them on a segment of length less than 1.499.


52. (USA 1) Two perpendicular chords are drawn through a given interior point $P$ of a circle with radius $R$. Determine, with proof, the maximum and the minimum of the sum of the lengths of these two chords if the distance from $P$ to the center of the circle is $k R$.

Solution: 52. The maximum and minimum are $2 R \sqrt{4-2 k^{2}}$ and $2 R\left(1+\sqrt{1-k^{2}}\right)$ respectively.


53. (USA 2) Find all pairs of integers $a$ and $b$ for which  $$ 7 a+14 b=5 a^{2}+5 a b+5 b^{2} $$

Solution: 53. The discriminant of the given equation considered as a quadratic equation in $b$ is $196-75 a^{2}$. Thus $75 a^{2} \leq 196$ and hence $-1 \leq a \leq 1$. Now the integer solutions of the given equation are easily found: $(-1,3),(0,0),(1,2)$.


54. (USA 3) If $0 \leq a \leq b \leq c \leq d$, prove that  $$ a^{b} b^{c} c^{d} d^{a} \geq b^{a} c^{b} d^{c} a^{d} $$

Solution: 54. We shall use the following lemma.  Lemma. If a real function $f$ is convex on the interval $I$ and $x, y, z \in I$, $x \leq y \leq z$, then  $$ (y-z) f(x)+(z-x) f(y)+(x-y) f(z) \leq 0 . $$  Proof. The inequality is obvious for $x=y=z$. If $x<z$, then there exist $p, r$ such that $p+r=1$ and $y=p x+r z$. Then by Jensen's inequality $f(p x+r z) \leq p f(x)+r f(z)$, which is equivalent to the statement of the lemma. By applying the lemma to the convex function $-\ln x$ we obtain $x^{y} y^{z} z^{x} \geq$ $y^{x} z^{y} x^{z}$ for any $0<x \leq y \leq z$. Multiplying the inequalities $a^{b} b^{c} c^{a} \geq b^{a} c^{b} a^{c}$ and $a^{c} c^{d} d^{a} \geq c^{a} d^{c} a^{d}$ we get the desired inequality. Remark. Similarly, for $0<a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ it holds that $a_{1}^{a_{2}} a_{2}^{a_{3}} \cdots a_{n}^{a_{1}} \geq a_{2}^{a_{1}} a_{3}^{a_{2}} \cdots a_{1}^{a_{n}}$.


55. (USA 4) Through a point $O$ on the diagonal $B D$ of a parallelogram $A B C D$, segments $M N$ parallel to $A B$, and $P Q$ parallel to $A D$, are drawn, with $M$ on $A D$, and $Q$ on $A B$. Prove that diagonals $A O, B P, D N$ (extended if necessary) will be concurrent.

Solution: 55. The statement is true without the assumption that $O \in B D$. Let $B P \cap$ $D N=\{K\}$. If we denote $\overrightarrow{A B}=a, \overrightarrow{A D}=b$ and $\overrightarrow{A O}=\alpha a+\beta b$ for some $\alpha, \beta \in \mathbb{R}, 1 / \alpha+1 / \beta \neq 1$, by straightforward calculation we obtain that  $$ \overrightarrow{A K}=\frac{\alpha}{\alpha+\beta-\alpha \beta} a+\frac{\beta}{\alpha+\beta-\alpha \beta} b=\frac{1}{\alpha+\beta-\alpha \beta} \overrightarrow{A O} $$  Hence $A, K, O$ are collinear.


56. (USA 5) The four circumcircles of the four faces of a tetrahedron have equal radii. Prove that the four faces of the tetrahedron are congruent triangles.

Solution: 56. See the solution to (LL67-38).


Solution: 57. Suppose that there exists a sequence of 17 terms $a_{1}, a_{2}, \ldots, a_{17}$ satisfying the required conditions. Then the sum of terms in each row of the rectangular array below is positive, while the sum of terms in each column is negative, which is a contradiction.  $$ \begin{array}{cccc} a_{1} & a_{2} & \ldots & a_{11} \\ a_{2} & a_{3} & \ldots & a_{12} \\ \vdots & \vdots & & \vdots \\ a_{7} & a_{8} & \ldots & a_{17} \end{array} $$  On the other hand, there exist 16 -term sequences with the required property. An example is $5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5$ which can be obtained by solving the system of equations $\sum_{i=k}^{k+10} a_{i}=1$ $(k=1,2, \ldots, 6)$ and $\sum_{i=l}^{l+6} a_{i}=-1(l=1,2, \ldots, 10)$. Second solution. We shall prove a stronger statement: If 7 and 11 in the question are replaced by any positive integers $m, n$, then the maximum number of terms is $m+n-(m, n)-1$. Let $a_{1}, a_{2}, \ldots, a_{l}$ be a sequence of real numbers, and let us define $s_{0}=0$ and $s_{k}=a_{1}+\cdots+a_{k}(k=1, \ldots, l)$. The given conditions are equivalent to $s_{k}>s_{k+m}$ for $0 \leq k \leq l-m$ and $s_{k}<s_{k+n}$ for $0 \leq k \leq l-n$. Let $d=(m, n)$ and $m=m^{\prime} d, n=n^{\prime} d$. Suppose that there exists a sequence $\left(a_{k}\right)$ of length greater than or equal to $l=m+n-d$ satisfying the required conditions. Then the $m^{\prime}+n^{\prime}$ numbers $s_{0}, s_{d}, \ldots, s_{\left(m^{\prime}+n^{\prime}-1\right) d}$ satisfy $n^{\prime}$ inequalities $s_{k+m}<s_{k}$ and $m^{\prime}$ inequalities $s_{k}<s_{k+n}$. Moreover, each term $s_{k d}$ appears twice in these inequalities: once on the left-hand and once on the right-hand side. It follows that there exists a ring of inequalities $s_{i_{1}}<s_{i_{2}}<\cdots<s_{i_{k}}<s_{i_{1}}$, giving a contradiction. On the other hand, suppose that such a ring of inequalities can be made also for $l=m+n-d-1$, say $s_{i_{1}}<s_{i_{2}}<\cdots<s_{i_{k}}<s_{i_{1}}$. If there are $p$ inequalities of the form $a_{k+m}<a_{k}$ and $q$ inequalities of the form $a_{k+n}>a_{k}$ in the ring, then $q n=r m$, which implies $m^{\prime}\left|q, n^{\prime}\right| p$ and thus $k=p+q \geq m^{\prime}+n^{\prime}$. But since all $i_{1}, i_{2}, \ldots, i_{k}$ are congruent modulo $d$, we have $k \leq m^{\prime}+n^{\prime}-1$, a contradiction. Hence there exists a sequence of length $m+n-d-1$ with the required property.


58. (VIE 2) Prove that for every triangle the following inequality holds:  $$ \frac{a b+b c+c a}{4 S} \geq \cot \frac{\pi}{6} $$  where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.

Solution: 58. The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove:  $$ 2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2} \geq 4 S \sqrt{3} $$  First proof. Let us set $2 x=b+c-a, 2 y=c+a-b, 2 z=a+b-c$.  Then $x, y, z>0$ and the inequality (1) becomes  $$ y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2} \geq x y z(x+y+z) $$  which is equivalent to the obvious inequality $(x y-y z)^{2}+(y z-z x)^{2}+$ $(z x-x y)^{2} \geq 0$. Second proof. Using the known relations for a triangle  $$ \begin{aligned} a^{2}+b^{2}+c^{2} & =2 s^{2}-2 r^{2}-8 r R, \\ a b+b c+c a & =s^{2}+r^{2}+4 r R, \\ S & =r s, \end{aligned} $$  where $r$ and $R$ are the radii of the incircle and the circumcircle, $s$ the semiperimeter and $S$ the area, we can transform (1) into  $$ s \sqrt{3} \leq 4 R+r . $$  The last inequality is a consequence of the inequalities $2 r \leq R$ and $s^{2} \leq$ $4 R^{2}+4 R r+3 r^{2}$, where the last one follows from the equality $H I^{2}=$ $4 R^{2}+4 R r+3 r^{2}-s^{2}$ ( $H$ and $I$ being the orthocenter and the incenter of the triangle).


Solution: 59. Let us consider the set $R$ of pairs of coordinates of the points from $E$ reduced modulo 3 . If some element of $R$ occurs thrice, then the corresponding points are vertices of a triangle with integer barycenter. Also, no three elements from $E$ can have distinct $x$-coordinates and distinct $y$ coordinates. By an easy discussion we can conclude that the set $R$ contains at most four elements. Hence $|E| \leq 8$. An example of a set $E$ consisting of 8 points that satisfies the required condition is  $$ E=\{(0,0),(1,0),(0,1),(1,1),(3,6),(4,6),(3,7),(4,7)\} $$


6. (CZS 2) Let $x_{1}, x_{2}, \ldots, x_{n}(n \geq 1)$ be real numbers such that $0 \leq x_{j} \leq \pi$, $j=1,2, \ldots, n$. Prove that if $\sum_{j=1}^{n}\left(\cos x_{j}+1\right)$ is an odd integer, then $\sum_{j=1}^{n} \sin x_{j} \geq 1$.

Solution: 6. Let $\langle y\rangle$ denote the distance from $y \in \mathbb{R}$ to the closest even integer. We claim that  $$ \langle 1+\cos x\rangle \leq \sin x \quad \text { for all } x \in[0, \pi] $$  Indeed, if $\cos x \geq 0$, then $\langle 1+\cos x\rangle=1-\cos x \leq 1-\cos ^{2} x=\sin ^{2} x \leq$ $\sin x$; the proof is similar if $\cos x<0$. We note that $\langle x+y\rangle \leq\langle x\rangle+\langle y\rangle$ holds for all $x, y \in \mathbb{R}$. Therefore  $$ \sum_{j=1}^{n} \sin x_{j} \geq \sum_{j=1}^{n}\left\langle 1+\cos x_{j}\right\rangle \geq\left\langle\sum_{j=1}^{n}\left(1+\cos x_{j}\right)\right\rangle=1 $$


60. (VIE 4) Suppose $x_{0}, x_{1}, \ldots, x_{n}$ are integers and $x_{0}>x_{1}>\cdots>x_{n}$. Prove that at least one of the numbers $\left|F\left(x_{0}\right)\right|,\left|F\left(x_{1}\right)\right|,\left|F\left(x_{2}\right)\right|, \ldots$, $\left|F\left(x_{n}\right)\right|$, where  $$ F(x)=x^{n}+a_{1} x^{n-1}+\cdots+a_{n}, \quad a_{i} \in \mathbb{R}, \quad i=1, \ldots, n $$  is greater than $\frac{n!}{2^{n}}$.

Solution: 60. By Lagrange's interpolation formula we have  $$ F(x)=\sum_{j=0}^{n} F\left(x_{j}\right) \frac{\prod_{i \neq j}\left(x-x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} . $$  Since the leading coefficient in $F(x)$ is 1 , it follows that  $$ 1=\sum_{j=0}^{n} \frac{F\left(x_{j}\right)}{\prod_{i \neq j}\left(x_{i}-x_{j}\right)} $$  Since  $$ \left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|=\prod_{i=0}^{j-1}\left|x_{i}-x_{j}\right| \prod_{i=j+1}^{n}\left|x_{i}-x_{j}\right| \geq j!(n-j)! $$  we have  $$ 1 \leq \sum_{j=0}^{n} \frac{\left|F\left(x_{j}\right)\right|}{\left|\prod_{i \neq j}\left(x_{i}-x_{j}\right)\right|} \leq \frac{1}{n!} \sum_{j=0}^{n}\binom{n}{j}\left|F\left(x_{j}\right)\right| \leq \frac{2^{n}}{n!} \max \left|F\left(x_{j}\right)\right| . $$  Now the required inequality follows immediately.


7. (CZS 3) Prove the following assertion: If $c_{1}, c_{2}, \ldots, c_{n}(n \geq 2)$ are real numbers such that  $$ (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right)=\left(c_{1}+c_{2}+\cdots+c_{n}\right)^{2} $$  then either all these numbers are nonnegative or all these numbers are nonpositive.

Solution: 7. Let us suppose that $c_{1} \leq c_{2} \leq \cdots \leq c_{n}$ and that $c_{1}<0<c_{n}$. There exists $k, 1 \leq k<n$, such that $c_{k} \leq 0<c_{k+1}$. Then we have  $$ \begin{aligned} (n-1)\left(c_{1}^{2}+c_{2}^{2}+\cdots+c_{n}^{2}\right) \geq & k\left(c_{1}^{2}+\cdots+c_{k}^{2}\right)+(n-k)\left(c_{k+1}^{2}+\cdots+c_{n}^{2}\right) \\ \geq & \left(c_{1}+\cdots+c_{k}\right)^{2}+\left(c_{k+1}+\cdots+c_{n}\right)^{2} \\ = & \left(c_{1}+\cdots+c_{n}\right)^{2} \\ & -2\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right), \end{aligned} $$  from which we obtain $\left(c_{1}+\cdots+c_{k}\right)\left(c_{k+1}+\cdots+c_{n}\right) \geq 0$, a contradiction. Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have  $$ \begin{aligned} \left(c_{1}+\cdots+c_{n}\right)^{2} & =(n-1)\left(c_{1}^{2}+\cdots+c_{n-1}^{2}\right)+(n-1) c_{n}^{2} \\ & \geq\left(c_{1}+\cdots+c_{n-1}\right)^{2}+(n-1) c_{n}^{2} \end{aligned} $$  which is equivalent to $2\left(c_{1}+c_{2}+\cdots+c_{n}\right) c_{n} \geq n c_{n}^{2}$. Similarly, $2\left(c_{1}+c_{2}+\right.$ $\left.\cdots+c_{n}\right) c_{i} \geq n c_{i}^{2}$ for all $i=1, \ldots, n$. Hence all $c_{i}$ are of the same sign.


8. (CZS 4) A hexahedron $A B C D E$ is made of two regular congruent tetrahedra $A B C D$ and $A B C E$. Prove that there exists only one isometry $\mathcal{Z}$ that maps points $A, B, C, D, E$ onto $B, C, A, E, D$, respectively. Find all points $X$ on the surface of hexahedron whose distance from $\mathcal{Z}(X)$ is minimal.

Solution: 8. There is exactly one point satisfying the given condition on each face of the hexahedron. Namely, on the face $A B D$ it is the point that divides the median from $D$ in the ratio $32: 3$.


9. (CZS 5) Let $A B C D$ be a regular tetrahedron and $\mathcal{Z}$ an isometry mapping $A, B, C, D$ into $B, C, D, A$, respectively. Find the set $\mathcal{M}$ of all points $X$ of the face $A B C$ whose distance from $\mathcal{Z}(X)$ is equal to a given number $t$. Find necessary and sufficient conditions for the set $\mathcal{M}$ to be nonempty.

Solution: 9. A necessary and sufficient condition for $\mathcal{M}$ to be nonempty is that $1 / \sqrt{10} \leq t \leq 1$.


1. (BUL 1) The set $M=\{1,2, \ldots, 2 n\}$ is partitioned into $k$ nonintersecting subsets $M_{1}, M_{2}, \ldots, M_{k}$, where $n \geq k^{3}+k$. Prove that there exist even numbers $2 j_{1}, 2 j_{2}, \ldots, 2 j_{k+1}$ in $M$ that are in one and the same subset $M_{i}$ $(1 \leq i \leq k)$ such that the numbers $2 j_{1}-1,2 j_{2}-1, \ldots, 2 j_{k+1}-1$ are also in one and the same subset $M_{j}(1 \leq j \leq k)$.

Solution: 1. There exists an $M_{s}$ that contains at least $2 n / k=2\left(k^{2}+1\right)$ elements. It follows that $M_{s}$ contains either at least $k^{2}+1$ even numbers or at least $k^{2}+1$ odd numbers. In the former case, consider the predecessors of those $k^{2}+1$ numbers: among them, at least $\frac{k^{2}+1}{k+1}>k$, i.e., at least $k+1$, belong to the same subset, say $M_{t}$. Then we choose $s, t$. The latter case is similar. Second solution. For all $i, j \in\{1,2, \ldots, k\}$, consider the set $N_{i j}=\{r \mid$ $\left.2 r \in M_{i}, 2 r-1 \in M_{j}\right\}$. Then $\left\{N_{i j} \mid i, j\right\}$ is a partition of $\{1,2, \ldots, n\}$ into $k^{2}$ subsets. For $n \geq k^{3}+1$ one of these subsets contains at least $k+1$ elements, and the statement follows. Remark. The statement is not necessarily true when $n=k^{3}$.


10. (NET 1) ${ }^{\text {IMO6 }}$ An international society has its members in 6 different countries. The list of members contains 1978 names, numbered $1,2, \ldots$, 1978. Prove that there is at least one member whose number is the sum of the numbers of two, not necessarily distinct, of his compatriots.

Solution: 10. Assume the opposite. One of the countries, say $A$, contains at least 330 members $a_{1}, a_{2}, \ldots, a_{330}$ of the society $(6 \cdot 329=1974)$. Consider the differences $a_{330}-a_{i},=1,2, \ldots, 329$ : the members with these numbers are not in $A$, so at least 66 of them, $a_{330}-a_{i_{1}}, \ldots, a_{330}-a_{i_{66}}$, belong to the same country, say $B$. Then the differences $\left(a_{i_{66}}-a_{330}\right)-\left(a_{i_{j}}-a_{330}\right)=$ $a_{i_{66}}-a_{i_{j}}, j=1,2, \ldots, 65$, are neither in $A$ nor in $B$. Continuing this procedure, we find that 17 of these differences are in the same country, say $C$, then 6 among 16 differences of themselves in a country $D$, and 3 among 5 differences of themselves in $E$; finally, one among two differences of these 3 differences belong to country $F$, so that the difference of themselves cannot be in any country. This is a contradiction. Remark. The following stronger $([6!e]=1957)$ statement can be proved in the same way. Schurr's lemma. If $n$ is a natural number and $e$ the logarithm base, then for every partition of the set $\{1,2, \ldots,[e n!]\}$ into $n$ subsets one of these subsets contains some two elements and their difference.


11. (SWE 2) A function $f: I \rightarrow \mathbb{R}$, defined on an interval $I$, is called concave if $f(\theta x+(1-\theta) y) \geq \theta f(x)+(1-\theta) f(y)$ for all $x, y \in I$ and $0 \leq \theta \leq 1$. Assume that the functions $f_{1}, \ldots, f_{n}$, having all nonnegative values, are concave. Prove that the function $\left(f_{1} f_{2} \ldots f_{n}\right)^{1 / n}$ is concave.

Solution: 11. Set $F(x)=f_{1}(x) f_{2}(x) \cdots f_{n}(x)$ : we must prove concavity of $F^{1 / n}$. By the assumption,  $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \prod_{i=1}^{n}\left[\theta f_{i}(x)+(1-\theta) f(y)\right] \\ & =\sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k} \sum f_{i_{1}}(x) \ldots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \end{aligned} $$  where the second sum goes through all $\binom{n}{k} k$-subsets $\left\{i_{1}, \ldots, i_{k}\right\}$ of $\{1, \ldots, n\}$. The inequality between the arithmetic and geometric means now gives us  $$ \sum f_{i_{1}}(x) f_{i_{2}}(x) \cdots f_{i_{k}}(x) f_{i_{k+1}}(y) f_{i_{n}}(y) \geq\binom{ n}{k} F(x)^{k / n} F(y)^{(n-k) / n} $$  Inserting this in the above inequality and using the binomial formula, we finally obtain  $$ \begin{aligned} F(\theta x+(1-\theta) y) & \geq \sum_{k=0}^{n} \theta^{k}(1-\theta)^{n-k}\binom{n}{k} F(x)^{k / n} F(y)^{(n-k) / n} \\ & =\left(\theta F(x)^{1 / n}+(1-\theta) F(y)^{1 / n}\right)^{n}, \end{aligned} $$  which proves the assertion.


12. (USA 1) ${ }^{\mathrm{IMO} 4}$ In a triangle $A B C$ we have $A B=A C$. A circle is tangent internally to the circumcircle of $A B C$ and also to the sides $A B, A C$, at $P, Q$ respectively. Prove that the midpoint of $P Q$ is the center of the incircle of $A B C$.

Solution: 12. Let $O$ be the center of the smaller circle, $T$ its contact point with the circumcircle of $A B C$, and $J$ the midpoint of segment $B C$. The figure is symmetric with respect to the line through $A, O, J, T$. A homothety centered at $A$ taking $T$ into $J$ will take the smaller circle into the incircle of $A B C$, hence will take $O$ into the incenter $I$. On the other hand, $\angle A B T=\angle A C T=90^{\circ}$ implies that the quadrilaterals $A B T C$ and $A P O Q$ are similar. Hence the above homothety also maps $O$ to the midpoint of $P Q$. This finishes the proof. Remark. The assertion is true for a nonisosceles triangle $A B C$ as well, and this (more difficult) case is a matter of SL93-3.


13. (USA 6) ${ }^{\mathrm{IMO} 2}$ Given any point $P$ in the interior of a sphere with radius $R$, three mutually perpendicular segments $P A, P B, P C$ are drawn terminating on the sphere and having one common vertex in $P$. Consider the rectangular parallelepiped of which $P A, P B, P C$ are coterminal edges. Find the locus of the point $Q$ that is diagonally opposite $P$ in the parallelepiped when $P$ and the sphere are fixed.

Solution: 13. Lemma. If $M N P Q$ is a rectangle and $O$ any point in space, then $O M^{2}+$ $O P^{2}=O N^{2}+O Q^{2}$. Proof. Let $O_{1}$ be the projection of $O$ onto $M N P Q$, and $m, n, p, q$ denote the distances of $O_{1}$ from $M N, N P, P Q, Q M$, respectively. Then $O M^{2}=O O_{1}^{2}+q^{2}+m^{2}, O N^{2}=O O_{1}^{2}+m^{2}+n^{2}, O P^{2}=O O_{1}^{2}+n^{2}+p^{2}$, $O Q^{2}=O O_{1}^{2}+p^{2}+q^{2}$, and the lemma follows immediately. Now we return to the problem. Let $O$ be the center of the given sphere $S$, and $X$ the point opposite $P$ in the face of the parallelepiped through $P, A, B$. By the lemma, we have $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$ and $O P^{2}+$ $O X^{2}=O A^{2}+O B^{2}$. Hence $2 O P^{2}+O Q^{2}=O A^{2}+O B^{2}+O C^{2}=3 R^{2}$, i.e. $O Q=\sqrt{3 R^{2}-O P^{2}}>R$.  We claim that the locus of $Q$ is the whole sphere $\left(O, \sqrt{3 R^{2}-O P^{2}}\right)$. Choose any point $Q$ on this sphere. Since $O Q>R>O P$, the sphere with diameter $P Q$ intersects $S$ on a circle. Let $C$ be an arbitrary point on this circle, and $X$ the point opposite $C$ in the rectangle $P C Q X$. By the lemma, $O P^{2}+O Q^{2}=O C^{2}+O X^{2}$, hence $O X^{2}=2 R^{2}-O P^{2}>R^{2}$. The plane passing through $P$ and perpendicular to $P C$ intersects $S$ in a circle $\gamma$; both $P, X$ belong to this plane, $P$ being inside and $X$ outside the circle, so that the circle with diameter $P X$ intersects $\gamma$ at some point $B$. Finally, we choose $A$ to be the point opposite $B$ in the rectangle $P B X A$ : we deduce that $O A^{2}+O B^{2}=O P^{2}+O X^{2}$, and consequently $A \in S$. By the construction, there is a rectangular parallelepiped through $P, A, B, C, X, Q$.


14. (VIE 2) Prove that it is possible to place $2 n(2 n+1)$ parallelepipedic (rectangular) pieces of soap of dimensions $1 \times 2 \times(n+1)$ in a cubic box with edge $2 n+1$ if and only if $n$ is even or $n=1$. Remark. It is assumed that the edges of the pieces of soap are parallel to the edges of the box.

Solution: 14. We label the cells of the cube by $\left(a_{1}, a_{2}, a_{3}\right), a_{i} \in\{1,2, \ldots, 2 n+1\}$, in a natural way: for example, as Cartesian coordinates of centers of the cells $\left((1,1,1)\right.$ is one corner, etc.). Notice that there should be $(2 n+1)^{3}-$ $2 n(2 n+1) \cdot 2(n+1)=2 n+1$ void cells, i.e., those not covered by any piece of soap. $n=1$. In this case, six pieces of soap $1 \times 2 \times 2$ can be placed on the following positions: $[(1,1,1),(2,2,1)],[(3,1,1),(3,2,2)],[(2,3,1),(3,3,2)]$ and the symmetric ones with respect to the center of the box. (Here $[A, B]$ denotes the rectangle with opposite corners at $A, B$.) $n$ is even. Each of the $2 n+1$ planes $P_{k}=\left\{\left(a_{1}, a_{2}, k\right) \mid a_{i}=1, \ldots, 2 n+1\right\}$ can receive $2 n$ pieces of soap: In fact, $P_{k}$ can be partitioned into four $n \times(n+1)$ rectangles at the corners and the central cell, while an $n \times(n+1)$ rectangle can receive $n / 2$ pieces of soap. $n$ is odd, $n>1$. Let us color a cell $\left(a_{1}, a_{2}, a_{3}\right)$ blue, red, or yellow if exactly three, two or one $a_{i}$ respectively is equal to $n+1$. Thus there are 1 blue, $6 n$ red, and $12 n^{2}$ yellow cells. We notice that each piece of soap must contain at least one colored cell (because $2(n+1)>2 n+1)$. Also, every piece of soap contains an even number (actually, $1 \cdot 2,1(n+1)$, or $2(n+1)$ ) of cells in $P_{k}$. On the other hand, $2 n+1$ cells are void, i.e., one in each plane.  There are several cases for a piece of soap $S$ : (i) $S$ consists of 1 blue, $n+1$ red and $n$ yellow cells; (ii) $S$ consists of 2 red and $2 n$ yellow cells (and no blue cells); (iii) $S$ contains 1 red cell, $n+1$ yellow cells, and the are rest uncolored; (iv) $S$ contains 2 yellow cells and no blue or red ones.  From the descriptions of the last three cases, we can deduce that if $S$ contains $r$ red cells and no blue, then it contains exactly $2+(n-1) r$ red ones. $\quad(*)$ Now, let $B_{1}, \ldots, B_{k}$ be all boxes put in the cube, with a possible exception for the one covering the blue cell: thus $k=2 n(2 n+1)$ if the blue cell is void, or $k=2 n(2 n+1)-1$ otherwise. Let $r_{i}$ and $y_{i}$ respectively be the numbers of red and yellow cells inside $B_{i}$. By (*) we have $y_{1}+\cdots+y_{k}=2 k+(n-1)\left(r_{1}+\cdots+r_{k}\right)$. If the blue cell is void, then $r_{1}+\cdots+r_{k}=6 n$ and consequently $y_{1}+\cdots+y_{k}=$ $4 n(2 n+1)+6 n(n-1)=14 n^{2}-2 n$, which is impossible because there are only $12 n^{2}<14 n^{2}-2 n$ yellow cells. Otherwise, $r_{1}+\cdots+r_{k} \geq 5 n-2$ (because $n+1$ red cells are covered by the box containing the blue cell, and one can be void) and consequently $y_{1}+\cdots+y_{k} \geq 4 n(2 n+$ $1)-2+(n-1)(5 n-2)=13 n^{2}-3 n$; since there are $n$ more yellow cells in the box containing the blue one, this counts for $13 n^{2}-2 n>12 n^{2}$ ( $n \geq 3$ ), again impossible. Remark. The following solution of the case $n$ odd is simpler, but does not work for $n=3$. For $k=1,2,3$, let $m_{k}$ be the number of pieces whose long sides are perpendicular to the plane $\pi_{k}\left(a_{k}=n+1\right)$. Each of these $m_{k}$ pieces covers exactly 2 cells of $\pi_{k}$, while any other piece covers $n+1$, $2(n+1)$, or none. It follows that $4 n^{2}+4 n-2 m_{k}$ is divisible by $n+1$, and so is $2 m_{k}$. This further implies that $2 m_{1}+2 m_{2}+2 m_{3}=4 n(2 n+1)$ is a multiple of $n+1$, which is impossible for each odd $n$ except $n=1$ and $n=3$.


15. (YUG 1) Let $p$ be a prime and $A=\left\{a_{1}, \ldots, a_{p-1}\right\}$ an arbitrary subset of the set of natural numbers such that none of its elements is divisible by $p$. Let us define a mapping $f$ from $\mathcal{P}(A)$ (the set of all subsets of $A$ ) to the set $P=\{0,1, \ldots, p-1\}$ in the following way: (i) if $B=\left\{a_{i_{1}}, \ldots, a_{i_{k}}\right\} \subset A$ and $\sum_{j=1}^{k} a_{i_{j}} \equiv n(\bmod p)$, then $f(B)=n$, (ii) $f(\emptyset)=0, \emptyset$ being the empty set.  Prove that for each $n \in P$ there exists $B \subset A$ such that $f(B)=n$.

Solution: 15. Let $C_{n}=\left\{a_{1}, \ldots, a_{n}\right\}\left(C_{0}=\emptyset\right)$ and $P_{n}=\left\{f(B) \mid B \subseteq C_{n}\right\}$. We claim that $P_{n}$ contains at least $n+1$ distinct elements. First note that $P_{0}=\{0\}$ contains one element. Suppose that $P_{n+1}=P_{n}$ for some $n$. Since $P_{n+1}=$ $\left\{a_{n+1}+r \mid r \in P_{n}\right\}$, it follows that for each $r \in P_{n}$, also $r+b_{n} \in P_{n}$. Then obviously $0 \in P_{n}$ implies $k b_{n} \in P_{n}$ for all $k$; therefore $P_{n}=P$ has at least $p \geq n+1$ elements. Otherwise, if $P_{n+1} \supset P_{n}$ for all $n$, then $\left|P_{n+1}\right| \geq\left|P_{n}\right|+1$ and hence $\left|P_{n}\right| \geq n+1$, as claimed. Consequently, $\left|P_{p-1}\right| \geq p$. (All the operations here are performed modulo $p$.)


16. (YUG 2) Determine all the triples $(a, b, c)$ of positive real numbers such that the system  $$ \begin{array}{r} a x+b y-c z=0 \\ a \sqrt{1-x^{2}}+b \sqrt{1-y^{2}}-c \sqrt{1-z^{2}}=0 \end{array} $$  is compatible in the set of real numbers, and then find all its real solutions.

Solution: 16. Clearly $|x| \leq 1$. As $x$ runs over $[-1,1]$, the vector $u=\left(a x, a \sqrt{1-x^{2}}\right)$ runs over all vectors of length $a$ in the plane having a nonnegative vertical component. Putting $v=\left(b y, b \sqrt{1-y^{2}}\right), w=\left(c z, c \sqrt{1-z^{2}}\right)$, the system becomes $u+v=w$, with vectors $u, v, w$ of lengths $a, b, c$ respectively in the upper half-plane. Then $a, b, c$ are sides of a (possibly degenerate) triangle; i.e, $|a-b| \leq c \leq a+b$ is a necessary condition.  Conversely, if $a, b, c$ satisfy this condition, one constructs a triangle $O M N$ with $O M=a, O N=b, M N=c$. If the vectors $\overrightarrow{O M}, \overrightarrow{O N}$ have a positive nonnegative component, then so does their sum. For every such triangle, putting $u=\overrightarrow{O M}, v=\overrightarrow{O N}$, and $w=\overrightarrow{O M}+\overrightarrow{O N}$ gives a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: $\alpha=\angle M O N=\angle(u, v)$ and $\beta=\angle(u, w)$. Therefore, all solutions of the system are  $$ \begin{aligned} & x=\cos t, \quad y=\cos (t+\alpha), \quad z=y=\cos (t+\beta), \quad t \in[0, \pi-\alpha] \quad \text { or } \\ & x=\cos t, \quad y=\cos (t-\alpha), \quad z=y=\cos (t-\beta), \quad t \in[\alpha, \pi] . \end{aligned} $$


17. (FRA 3) Prove that for any positive integers $x, y, z$ with $x y-z^{2}=1$ one can find nonnegative integers $a, b, c, d$ such that $x=a^{2}+b^{2}, y=c^{2}+d^{2}$, $z=a c+b d$. Set $z=(2 q)$ ! to deduce that for any prime number $p=4 q+1, p$ can be represented as the sum of squares of two integers.

Solution: 17. Let $z_{0} \geq 1$ be a positive integer. Supposing that the statement is true for all triples $(x, y, z)$ with $z<z_{0}$, we shall prove that it is true for $z=z_{0}$ too.  If $z_{0}=1$, verification is trivial, while $x_{0}=y_{0}$ is obviously impossible. So let there be given a triple $\left(x_{0}, y_{0}, z_{0}\right)$ with $z_{0}>1$ and $x_{0}<y_{0}$, and define another triple $(x, y, z)$ by  $$ x=z_{0}, \quad y=x_{0}+y_{0}-2 z_{0}, \quad \text { and } \quad z=z_{0}-x_{0} $$  Then $x, y, z$ are positive integers. This is clear for $x, z$, while $y=x_{0}+y_{0}-$ $2 z_{0} \geq 2\left(\sqrt{x_{0} y_{0}}-z_{0}\right)>2\left(z_{0}-z_{0}\right)=0$. Moreover, $x y-z^{2}=x_{0}\left(x_{0}+y_{0}-\right.$ $\left.2 z_{0}\right)-\left(z_{0}-x_{0}\right)^{2}=x_{0} y_{0}-z_{0}^{2}=1$ and $z<z_{0}$, so that by the assumption, the statement holds for $x, y, z$. Thus for some nonnegative integers $a, b, c, d$ we have  $$ x=a^{2}+b^{2}, \quad y=c^{2}+d^{2}, \quad z=a c+b d $$  But then we obtain representations of this sort for $x_{0}, y_{0}, z_{0}$ too:  $$ x_{0}=a^{2}+b^{2}, \quad y_{0}=(a+c)^{2}+(b+d)^{2}, \quad z_{0}=a(a+c)+b(b+d) $$  For the second part of the problem, we note that for $z=(2 q)!$,  $$ \begin{aligned} z^{2} & =(2 q)!(2 q)(2 q-1) \cdots 1 \equiv(2 q)!\cdot(-(2 q+1))(-(2 q+2)) \cdots(-4 q) \\ & =(-1)^{2 q}(4 q)!\equiv-1(\bmod p) \end{aligned} $$  by Wilson's theorem. Hence $p \mid z^{2}+1=p y$ for some positive integer $y>0$. Now it follows from the first part that there exist integers $a, b$ such that $x=p=a^{2}+b^{2}$. Second solution. Another possibility is using arithmetic of Gaussian integers. Lemma. Suppose $m, n, p, q$ are elements of $\mathbb{Z}$ or any other unique factorization domain, with $m n=p q$. then there exist elements $a, b, c, d$ such that $m=a b, n=c d, p=a c, q=b d$. Proof is direct, for example using factorization of $a, b, c, d$ into primes. We now apply this lemma to the Gaussian integers in our case (because $\mathbb{Z}[i]$ has the unique factorization property), having in mind that $x y=$ $z^{2}+1=(z+i)(z-i)$. We obtain (1) $x=a b$, (2) $y=c d$, (3) $z+i=a c$, (4) $z-i=b d$ for some $a, b, c, d \in \mathbb{Z}[i]$. Let $a=a_{1}+a_{2} i$, etc. By (3) and (4), $\operatorname{gcd}\left(a_{1}, a_{2}\right)=$ $\cdots=\operatorname{gcd}\left(d_{1}, d_{2}\right)$. Then (1) and (2) give us $b=\bar{a}, c=\bar{d}$. The statement follows at once: $x=a b=a \bar{a}=a_{1}^{2}+a_{2}^{2}, y=d \bar{d}=d_{1}^{2}+d_{2}^{2}$ and $z+i=$ $\left(a_{1} d_{1}+a_{2} d_{2}\right)+\imath\left(a_{2} d_{1}-a_{1} d_{2}\right) \Rightarrow z=a_{1} d_{1}+a_{2} d_{2}$.


2. (BUL 4) Two identically oriented equilateral triangles, $A B C$ with center $S$ and $A^{\prime} B^{\prime} C$, are given in the plane. We also have $A^{\prime} \neq S$ and $B^{\prime} \neq S$. If $M$ is the midpoint of $A^{\prime} B$ and $N$ the midpoint of $A B^{\prime}$, prove that the triangles $S B^{\prime} M$ and $S A^{\prime} N$ are similar.

Solution: 2. Consider the transformation $\phi$ of the plane defined as the homothety $\mathcal{H}$ with center $B$ and coefficient 2 followed by the rotation $\mathcal{R}$ about the center $O$ through an angle of $60^{\circ}$. Being direct, this mapping must be a rotational homothety. We also see that $\mathcal{H}$ maps $S$ into the point symmetric to $S$ with respect to $O A$, and $\mathcal{R}$ takes it back to $S$. Hence $S$ is a fixed point, and is consequently also the center of $\phi$. Therefore $\phi$ is the rotational homothety about $S$ with the angle $60^{\circ}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-435.jpg?height=319&width=462&top_left_y=920&top_left_x=858) and coefficient 2. (In fact, this could also be seen from the fact that $\phi$ preserves angles of triangles and maps the segment $S R$ onto $S B$, where $R$ is the midpoint of $A B$.) Since $\phi(M)=B^{\prime}$, we conclude that $\angle M S B^{\prime}=60^{\circ}$ and $S B^{\prime} / S M=2$. Similarly, $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$, so triangles $M S B^{\prime}$ and $N S A^{\prime}$ are indeed similar. Second solution. Probably the simplest way here is using complex numbers. Put the origin at $O$ and complex numbers $a, a^{\prime}$ at points $A, A^{\prime}$, and denote the primitive sixth root of 1 by $\omega$. Then the numbers at $B, B^{\prime}$, $S$ and $N$ are $\omega a, \omega a^{\prime},(a+\omega a) / 3$, and $\left(a+\omega a^{\prime}\right) / 2$ respectively. Now it is easy to verify that $(n-s)=\omega\left(a^{\prime}-s\right) / 2$, i.e., that $\angle N S A^{\prime}=60^{\circ}$ and $S A^{\prime} / S N=2$.


3. (CUB 3) ${ }^{\mathrm{IMO} 1}$ Let $n>m \geq 1$ be natural numbers such that the groups of the last three digits in the decimal representation of $1978^{m}, 1978^{n}$ coincide. Find the ordered pair $(m, n)$ of such $m, n$ for which $m+n$ is minimal.

Solution: 3. What we need are $m, n$ for which $1978^{m}\left(1978^{n-m}-1\right)$ is divisible by $1000=8 \cdot 125$. Since $1978^{n-m}-1$ is odd, it follows that $1978^{m}$ is divisible by 8 , so $m \geq 3$. Also, $1978^{n-m}-1$ is divisible by 125 , i.e., $1978^{n-m} \equiv 1(\bmod 125)$. Note that $1978 \equiv-2(\bmod 5)$, and consequently also $-2^{n-m} \equiv 1$. Hence $4 \mid n-m=4 k, k \geq 1$. It remains to find the least $k$ such that $1978^{4 k} \equiv 1$ $(\bmod 125)$. Since $1978^{4} \equiv(-22)^{4}=484^{2} \equiv(-16)^{2}=256 \equiv 6$, we reduce it to $6^{k} \equiv 1$. Now $6^{k}=(1+5)^{k} \equiv 1+5 k+25\binom{k}{2}(\bmod 125)$, which reduces to $125 \mid 5 k(5 k-3)$. But $5 k-3$ is not divisible by 5 , and so $25 \mid k$. Therefore $100 \mid n-m$, and the desired values are $m=3, n=103$.


4. (CZS 2) Let $T_{1}$ be a triangle having $a, b, c$ as lengths of its sides and let $T_{2}$ be another triangle having $u, v, w$ as lengths of its sides. If $P, Q$ are the areas of the two triangles, prove that  $$ 16 P Q \leq a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) $$  When does equality hold?

Solution: 4. Let $\gamma, \varphi$ be the angles of $T_{1}$ and $T_{2}$ opposite to $c$ and $w$ respectively. By the cosine theorem, the inequality is transformed into  $$ \begin{aligned} & a^{2}\left(2 v^{2}-2 u v \cos \varphi\right)+b^{2}\left(2 u^{2}-2 u v \cos \varphi\right) \\ & \quad+2\left(a^{2}+b^{2}-2 a b \cos \gamma\right) u v \cos \varphi \geq 4 a b u v \sin \gamma \sin \varphi \end{aligned} $$  This is equivalent to $2\left(a^{2} v^{2}+b^{2} u^{2}\right)-4 a b u v(\cos \gamma \cos \varphi+\sin \gamma \sin \varphi) \geq 0$, i.e., to  $$ 2(a v-b u)^{2}+4 a b u v(1-\cos (\gamma-\varphi)) \geq 0 $$  which is clearly satisfied. Equality holds if and only if $\gamma=\varphi$ and $a / b=$ $u / v$, i.e., when the triangles are similar, $a$ corresponding to $u$ and $b$ to $v$.


5. (GDR 2) For every integer $d \geq 1$, let $M_{d}$ be the set of all positive integers that cannot be written as a sum of an arithmetic progression with difference $d$, having at least two terms and consisting of positive integers. Let $A=M_{1}, B=M_{2} \backslash\{2\}, C=M_{3}$. Prove that every $c \in C$ may be written in a unique way as $c=a b$ with $a \in A, b \in B$.

Solution: 5. We first explicitly describe the elements of the sets $M_{1}, M_{2}$. $x \notin M_{1}$ is equivalent to $x=a+(a+1)+\cdots+(a+n-1)=n(2 a+n-1) / 2$ for some natural numbers $n, a, n \geq 2$. Among $n$ and $2 a+n-1$, one is odd and the other even, and both are greater than 1 ; so $x$ has an odd factor $\geq 3$. On the other hand, for every $x$ with an odd divisor $p>3$ it is easy to see that there exist corresponding $a, n$. Therefore $M_{1}=\left\{2^{k} \mid k=0,1,2, \ldots\right\}$. $x \notin M_{2}$ is equivalent to $x=a+(a+2)+\cdots+(a+2(n-1))=n(a+n-1)$, where $n \geq 2$, i.e. to $x$ being composite. Therefore $M_{2}=\{1\} \cup\{p \mid$ $p=$ prime $\}$. $x \notin M_{3}$ is equivalent to $x=a+(a+3)+\cdots+(a+3(n-1))=$ $n(2 a+3(n-1)) / 2$. It remains to show that every $c \in M_{3}$ can be written as $c=2^{k} p$ with $p$ prime. Suppose the opposite, that $c=2^{k} p q$, where $p, q$ are odd and $q \geq p \geq 3$. Then there exist positive integers $a, n(n \geq 2)$ such that $c=n(2 a+3(n-1)) / 2$ and hence $c \notin M_{3}$. Indeed, if $k=0$, then $n=2$ and $2 a+3=p q$ work; otherwise, setting $n=p$ one obtains $a=2^{k} q-$ $3(p-1) / 2 \geq 2 q-3(p-1) / 2 \geq(p+3) / 2>1$.


6. (FRA 2) $)^{\mathrm{IMO} 5}$ Let $\varphi:\{1,2,3, \ldots\} \rightarrow\{1,2,3, \ldots\}$ be injective. Prove that for all $n$,  $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{1}{k} $$

Solution: 6. For fixed $n$ and the set $\{\varphi(1), \ldots, \varphi(n)\}$, there are finitely many possibilities for mapping $\varphi$ to $\{1, \ldots, n\}$. Suppose $\varphi$ is the one among these for which $\sum_{k=1}^{n} \varphi(k) / k^{2}$ is minimal. If $i<j$ and $\varphi(i)<\varphi(j)$ for some $i, j \in\{1, \ldots, n\}$, define $\psi$ as $\psi(i)=\varphi(j), \psi(j)=\varphi(i)$, and $\psi(k)=\varphi(k)$ for all other $k$. Then  $$ \begin{aligned} \sum \frac{\varphi(k)}{k^{2}}-\sum \frac{\psi(k)}{k^{2}} & =\left(\frac{\varphi(i)}{i^{2}}+\frac{\varphi(j)}{j^{2}}\right)-\left(\frac{\varphi(i)}{j^{2}}+\frac{\varphi(j)}{i^{2}}\right) \\ & =(i-j)(\varphi(j)-\varphi(i)) \frac{i+j}{i^{2} j^{2}}>0 \end{aligned} $$  which contradicts the assumption. This shows that $\varphi(1)<\cdots<\varphi(n)$, and consequently $\varphi(k) \geq k$ for all $k$. Hence  $$ \sum_{k=1}^{n} \frac{\varphi(k)}{k^{2}} \geq \sum_{k=1}^{n} \frac{k}{k^{2}}=\sum_{k=1}^{n} \frac{1}{k} $$


7. (FRA 5) We consider three distinct half-lines $O x, O y, O z$ in a plane. Prove the existence and uniqueness of three points $A \in O x, B \in O y$, $C \in O z$ such that the perimeters of the triangles $O A B, O B C, O C A$ are all equal to a given number $2 p>0$.

Solution: 7. Let $x=O A, y=O B, z=O C, \alpha=\angle B O C, \beta=\angle C O A, \gamma=\angle A O B$. The conditions yield the equation $x+y+\sqrt{x^{2}+y^{2}-2 x y \cos \gamma}=2 p$, which transforms to $(2 p-x-y)^{2}=x^{2}+y^{2}-2 x y \cos \gamma$, i.e. $(p-x)(p-y)=$ $x y(1-\cos \gamma)$. Thus  $$ \frac{p-x}{x} \cdot \frac{p-y}{y}=1-\cos \gamma $$  and analogously $\frac{p-y}{y} \cdot \frac{p-z}{z}=1-\cos \alpha, \frac{p-z}{z} \cdot \frac{p-x}{x}=1-\cos \beta$. Setting $u=\frac{p-x}{x}, v=\frac{p-y}{y}, w=\frac{p-z}{z}$, the above system becomes  $$ u v=1-\cos \gamma, \quad v w=1-\cos \alpha, \quad w u=1-\cos \beta $$  This system has a unique solution in positive real numbers $u, v, w$ : $u=\sqrt{\frac{(1-\cos \beta)(1-\cos \gamma)}{1-\cos \alpha}}$, etc. Finally, the values of $x, y, z$ are uniquely determined from $u, v, w$. Remark. It is not necessary that the three lines be in the same plane. Also, there could be any odd number of lines instead of three.


8. (GBR 4) Let $S$ be the set of all the odd positive integers that are not multiples of 5 and that are less than $30 \mathrm{~m}, \mathrm{~m}$ being an arbitrary positive integer. What is the smallest integer $k$ such that in any subset of $k$ integers from $S$ there must be two different integers, one of which divides the other?

Solution: 8. Take the subset $\left\{a_{i}\right\}=\{1,7,11,13,17,19,23,29, \ldots, 30 m-1\}$ of $S$ containing all the elements of $S$ that are not multiples of 3 . There are 8 m such elements. Every element in $S$ can be uniquely expressed as $3^{t} a_{i}$ for some $i$ and $t \geq 0$. In a subset of $S$ with $8 m+1$ elements, two of them will have the same $a_{i}$, hance one will divide the other. On the other hand, for each $i=1,2, \ldots, 8 m$ choose $t \geq 0$ such that $10 \mathrm{~m}<$ $b_{i}=3^{t} a_{i}<30 \mathrm{~m}$. Then there are $8 \mathrm{~m} b_{i}$ 's in the interval $(10 \mathrm{~m}, 30 \mathrm{~m})$, and the quotient of any two of them is less than 3 , so none of them can divide any other. Thus the answer is 8 m .


9. $\mathbf{( G B R} \mathbf{5})^{\mathrm{IMO} 3}$ Let $\{f(n)\}$ be a strictly increasing sequence of positive integers: $0<f(1)<f(2)<f(3)<\cdots$. Of the positive integers not belonging to the sequence, the $n$th in order of magnitude is $f(f(n))+1$. Determine $f(240)$.

Solution: 9. Since the $n$th missing number (gap) is $f(f(n))+1$ and $f(f(n))$ is a member of the sequence, there are exactly $n-1$ gaps less than $f(f(n))$. This leads to  $$ f(f(n))=f(n)+n-1 $$  Since 1 is not a gap, we have $f(1)=1$. The first gap is $f(f(1))+1=2$. Two consecutive integers cannot both be gaps (the predecessor of a gap is of the form $f(f(m))$ ). Now we deduce $f(2)=3$; a repeated application of the formula above gives $f(3)=3+1=4, f(4)=4+2=6, f(6)=9$, $f(9)=14, f(14)=22, f(22)=35, f(35)=56, f(56)=90, f(90)=145$, $f(145)=234, f(234)=378$. Also, $f(f(35))+1=91$ is a gap, so $f(57)=92$. Then by $(1), f(92)=148$, $f(148)=239, f(239)=386$. Finally, here $f(f(148))+1=387$ is a gap, so $f(240)=388$.  Second solution. As above, we arrive at formula (1). Then by simple induction it follows that $f\left(F_{n}+1\right)=F_{n+1}+1$, where $F_{k}$ is the Fibonacci sequence ( $F_{1}=F_{2}=1$ ). We now prove by induction (on $n$ ) that $f\left(F_{n}+x\right)=F_{n+1}+f(x)$ for all $x$ with $1 \leq x \leq F_{n-1}$. This is trivially true for $n=0,1$. Supposing that it holds for $n-1$, we shall prove it for $n$ : (i) If $x=f(y)$ for some $y$, then by the inductive assumption and (1)  $$ \begin{aligned} f\left(F_{n}+x\right) & =f\left(F_{n}+f(y)\right)=f\left(f\left(F_{n-1}+y\right)\right) \\ & =F_{n}+f(y)+F_{n-1}+y-1=F_{n+1}+f(x) \end{aligned} $$  (ii) If $x=f(f(y))+1$ is a gap, then $f\left(F_{n}+x-1\right)+1=F_{n+1}+f(x-1)+1$ is a gap also:  $$ \begin{aligned} F_{n+1}+f(x)+1 & =F_{n+1}+f(f(f(y)))+1 \\ & =f\left(F_{n}+f(f(y))\right)+1=f\left(f\left(F_{n-1}+f(y)\right)\right)+1 \end{aligned} $$  It follows that $f\left(F_{n}+x\right)=F_{n+1}+f(x-1)+2=F_{n+1}+f(x)$. Now, since we know that each positive integer $x$ is expressible as $x=$ $F_{k_{1}}+F_{k_{2}}+\cdots+F_{k_{r}}$, where $0<k_{r} \neq 2, k_{i} \geq k_{i+1}+2$, we obtain $f(x)=F_{k_{1}+1}+F_{k_{2}+1}+\cdots+F_{k_{r}+1}$. Particularly, $240=233+5+2$, so $f(240)=377+8+3=388$. Remark. It can be shown that $f(x)=[\alpha x]$, where $\alpha=(1+\sqrt{5}) / 2$.


1. (BEL 1) Prove that in the Euclidean plane every regular polygon having an even number of sides can be dissected into lozenges. (A lozenge is a quadrilateral whose four sides are all of equal length).

Solution: 1. We prove more generally, by induction on $n$, that any $2 n$-gon with equal edges and opposite edges parallel to each other can be dissected. For $n=2$ the only possible such $2 n$-gon is a single lozenge, so our theorem holds in this case. We will now show that it holds for general $n$. Assume by induction that it holds for $n-1$. Let $A_{1} A_{2} \ldots A_{2 n}$ be an arbitrary $2 n$-gon with equal edges and opposite edges parallel to each other. Then we can construct points $B_{i}$ for $i=3,4, \ldots, n$ such that $\overrightarrow{A_{i} B_{i}}=\overrightarrow{A_{2} A_{1}}=\overrightarrow{A_{n+1} A_{n+2}}$. We set $B_{2}=A_{2 n+1}=A_{1}$ and $B_{n+1}=A_{n+2}$. It follows that $A_{i} B_{i} B_{i+1} A_{i+1}$ for $i=2,3,4, \ldots, n$ are all lozenges. It also follows that $B_{i} B_{i+1}$ for $i=2,3,4, \ldots, n$ are equal to the edges of $A_{1} A_{2} \ldots A_{2 n}$ and parallel to $A_{i} A_{i+1}$ and hence to $A_{n+i} A_{n+i+1}$. Thus $B_{2} \ldots B_{n+1} A_{n+3} \ldots A_{2 n}$ is a $2(n-1)$-gon with equal edges and opposite sides parallel and hence, by the induction hypothesis, can be dissected into lozenges. We have thus provided a dissection for $A_{1} A_{2} \ldots A_{2 n}$. This completes the proof.


10. (FIN 3) Show that for any vectors $a, b$ in Euclidean space,  $$ |a \times b|^{3} \leq \frac{3 \sqrt{3}}{8}|a|^{2}|b|^{2}|a-b|^{2} $$  Remark. Here $\times$ denotes the vector product.

Solution: 10. In the cases $a=\overrightarrow{0}, b=\overrightarrow{0}$, and $a \| b$ the inequality is trivial. Otherwise, let us consider a triangle $A B C$ such that $\overrightarrow{C B}=a$ and $\overrightarrow{C A}=b$. From this point on we shall refer to $\alpha, \beta, \gamma$ as angles of $A B C$. Since $|a \times b|=$ $|a||b| \sin \gamma$, our inequality reduces to $|a||b| \sin ^{3} \gamma \leq 3 \sqrt{3}|c|^{2} / 8$, which is further reduced to  $$ \sin \alpha \sin \beta \sin \gamma \leq \frac{3 \sqrt{3}}{8} $$  using the sine law. The last inequality follows immediately from Jensen's inequality applied to the function $f(x)=\ln \sin x$, which is concave for $0<x<\pi$ because $f^{\prime}(x)=\cot x$ is strictly decreasing.


11. (GDR 1) Given real numbers $x_{1}, x_{2}, \ldots, x_{n}(n \geq 2)$, with $x_{i} \geq 1 / n$ $(i=1,2, \ldots, n)$ and with $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find whether the product $P=x_{1} x_{2} x_{3} \cdots x_{n}$ has a greatest and/or least value and if so, give these values.

Solution: 11. Let us define $y_{i}=x_{i}^{2}$. We thus have $y_{1}+y_{2}+\cdots+y_{n}=1, y_{i} \geq 1 / n^{2}$, and $P=\sqrt{y_{1} y_{2} \ldots y_{n}}$. The upper bound is obtained immediately from the AM-GM inequality: $P \leq 1 / n^{n / 2}$, where equality holds when $x_{i}=\sqrt{y_{i}}=1 / \sqrt{n}$. For the lower bound, let us assume w.l.o.g. that $y_{1} \geq y_{2} \geq \cdots \geq y_{n}$. We note that if $a \geq b \geq 1 / n^{2}$ and $s=a+b>2 / n^{2}$ is fixed, then $a b=\left(s^{2}-(a-b)^{2}\right) / 4$ is minimized when $|a-b|$ is maximized, i.e., when $b=1 / n^{2}$. Hence $y_{1} y_{2} \cdots y_{n}$ is minimal when $y_{2}=y_{3}=\cdots=y_{n}=1 / n^{2}$. Then $y_{1}=\left(n^{2}-n+1\right) / n^{2}$ and therefore $P_{\min }=\sqrt{n^{2}-n+1} / n^{n}$.


12. (GDR 3) Let $R$ be a set of exactly 6 elements. A set $F$ of subsets of $R$ is called an $S$-family over $R$ if and only if it satisfies the following three conditions: (i) For no two sets $X, Y$ in $F$ is $X \subseteq Y$; (ii) For any three sets $X, Y, Z$ in $F, X \cup Y \cup Z \neq R$, (iii) $\bigcup_{X \in F} X=R$.  We define $|F|$ to be the number of elements of $F$ (i.e., the number of subsets of $R$ belonging to $F$ ). Determine, if it exists, $h=\max |F|$, the maximum being taken over all S-families over $R$.

Solution: 12. The first criterion ensures that all sets in an $S$-family are distinct. Since the number of different families of subsets is finite, $h$ has to exist. In fact, we will show that $h=11$. First of all, if there exists $X \in F$ such that $|X| \geq 5$, then by (3) there exists $Y \in F$ such that $X \cup Y=R$. In this case $|F|$ is at most 2. Similarly, for $|X|=4$, for the remaining two elements either there exists a subset in $F$ that contains both, in which case we obtain the previous case, or there exist different $Y$ and $Z$ containing them, in which case $X \cup Y \cup Z=R$, which must not happen. Hence we can assume $|X| \leq 4$ for all $X \in F$. Assume $|X|=1$ for some $X$. In that case other sets must not contain that subset and hence must be contained in the remaining 5 -element subset. These elements must not be subsets of each other. From elementary combinatorics, the largest number of subsets of a 5 -element set of which none is subset of another is $\binom{5}{2}=10$. This occurs when we take all 2-element subsets. These subsets also satisfy (2). Hence $|F|_{\max }=11$ in this case. Otherwise, let us assume $|X|=3$ for some $X$. Let us define the following families of subsets: $G=\{Z=Y \backslash X \mid Y \in F\}$ and $H=\{Z=Y \cap X \mid Y \in$ $F\}$. Then no two sets in $G$ must complement each other in $R \backslash X$, and $G$ must cover this set. Hence $G$ contains exactly the sets of each of the remaining 3 elements. For each element of $G$ no two sets in $H$ of which one is a subset of another may be paired with it. There can be only 3 such subsets selected within a 3 -element set $X$. Hence the number of remaining sets is smaller than $3 \cdot 3=9$. Hence in this case $|F|_{\max }=10$.  In the remaining case all subsets have two elements. There are $\binom{6}{2}=15$ of them. But for every three that complement each other one must be discarded; hence the maximal number for $F$ in this case is $2 \cdot 15 / 3=10$. It follows that $h=11$.


Solution: 13. From elementary trigonometry we have $\sin 3 t=3 \sin t-4 \sin ^{3} t$. Hence, if we denote $y=\sin 20^{\circ}$, we have $\sqrt{3} / 2=\sin 60^{\circ}=3 y-4 y^{3}$. Obviously $0<y<1 / 2=\sin 30^{\circ}$. The function $f(x)=3 x-4 x^{3}$ is strictly increasing on $[0,1 / 2)$ because $f^{\prime}(x)=3-12 x^{2}>0$ for $0 \leq x<1 / 2$. Now the desired inequality $\frac{20}{60}=\frac{1}{3}<\sin 20^{\circ}<\frac{21}{60}=\frac{7}{20}$ follows from  $$ f\left(\frac{1}{3}\right)<\frac{\sqrt{3}}{2}<f\left(\frac{7}{20}\right) $$  which is directly verified.


14. (GRE 5) Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist.

Solution: 14. Let us assume that $a \in \mathbb{R} \backslash\{1\}$ is such that there exist $a$ and $x$ such that $x=\log _{a} x$, or equivalently $f(x):=\ln x / x=\ln a$. Then $a$ is a value of the function $f(x)$ for $x \in \mathbb{R}^{+} \backslash\{1\}$, and the converse also holds. First we observe that $f(x)$ tends to $-\infty$ as $x \rightarrow 0$ and $f(x)$ tends to 0 as $x \rightarrow 1$. Since $f(x)>0$ for $x>1$, the function $f(x)$ takes its maximum at a point $x$ for which $f^{\prime}(x)=(1-\ln x) / x^{2}=0$. Hence  $$ \max f(x)=f(e)=e^{1 / e} . $$  It follows that the set of values of $f(x)$ for $x \in \mathbb{R}^{+}$is the interval $\left(-\infty, e^{1 / e}\right)$, and consequently the desired set of bases $a$ of logarithms is $(0,1) \cup\left(1, e^{1 / e}\right]$.


15. (ISR 2) ${ }^{\mathrm{IMO} 5}$ The nonnegative real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, a$ satisfy the following relations:  $$ \sum_{i=1}^{5} i x_{i}=a, \quad \sum_{i=1}^{5} i^{3} x_{i}=a^{2}, \quad \sum_{i=1}^{5} i^{5} x_{i}=a^{3} $$  What are the possible values of $a$ ?

Solution: 15. We note that $\sum_{i=1}^{5} i\left(a-i^{2}\right)^{2} x_{i}=a^{2} \sum_{i=1}^{5} i x_{i}-2 a \sum_{i=1}^{5} i^{3} x_{i}+\sum_{i=1}^{5} i^{5} x_{i}=a^{2} \cdot a-2 a \cdot a^{2}+a^{3}=0$. Since the terms in the sum on the left are all nonnegative, it follows that all the terms have to be 0 . Thus, either $x_{i}=0$ for all $i$, in which case $a=0$, or $a=j^{2}$ for some $j$ and $x_{i}=0$ for $i \neq j$. In this case, $x_{j}=a / j=j$. Hence, the only possible values of $a$ are $\{0,1,4,9,16,25\}$.


16. (ISR 4) Let $K$ denote the set $\{a, b, c, d, e\} . F$ is a collection of 16 different subsets of $K$, and it is known that any three members of $F$ have at least one element in common. Show that all 16 members of $F$ have exactly one element in common.

Solution: 16. Obviously, no two elements of $F$ can be complements of each other. If one of the sets has one element, then the conclusion is trivial. If there exist two different 2-element sets, then they must contain a common element, which in turn must then be contained in all other sets. Thus we can assume that there exists at most one 2 -element subset of $K$ in $F$. Since there can be at most 6 subsets of more than 3 elements of a 5 -element set, it follows that at least 9 out of 10 possible 3 -element subsets of $K$ belong to $F$. Let us assume, without loss of generality, that all sets but $\{c, d, e\}$ belong to $F$. Then sets $\{a, b, c\},\{a, d, e\}$, and $\{b, c, d\}$ have no common element, which is a contradiction. Hence it follows that all sets have a common element.


17. (NET 1) Inside an equilateral triangle $A B C$ one constructs points $P$, $Q$ and $R$ such that  $$ \begin{aligned} & \angle Q A B=\angle P B A=15^{\circ}, \\ & \angle R B C=\angle Q C B=20^{\circ}, \\ & \angle P C A=\angle R A C=25^{\circ} . \end{aligned} $$  Determine the angles of triangle $P Q R$.

Solution: 17. Let $K, L$, and $M$ be intersections of $C Q$ and $B R, A R$ and $C P$, and $A Q$ and $B P$, respectively. Let $\angle X$ denote the angle of the hexagon $K Q M P L R$ at the vertex $X$, where $X$ is one of the six points. By an elementary calculation of angles we get $\angle K=140^{\circ}, \angle L=130^{\circ}, \angle M=150^{\circ}, \angle P=100^{\circ}, \angle Q=95^{\circ}, \angle R=105^{\circ}$. Since $\angle K B C=\angle K C B$, it follows that $K$ is on the symmetry line of $A B C$ through $A$. Analogous statements hold for $L$ and $M$. Let $K_{R}$ and $K_{Q}$ be points symmetric to $K$ with respect to $A R$ and $A Q$, respectively. Since $\angle A K_{Q} Q=\angle A K_{Q} K_{R}=70^{\circ}$ and $\angle A K_{R} R=\angle A K_{R} K_{Q}=70^{\circ}$, it follows that $K_{R}, R, Q$, and $K_{Q}$ are collinear. Hence $\angle Q R K=$ $2 \angle R-180^{\circ}$ and $\angle R Q K=2 \angle Q-$ $180^{\circ}$. We analogously get $\angle P R L=$ $2 \angle R-180^{\circ}, \angle R P L=2 \angle P-$ $180^{\circ}, \angle Q P M=2 \angle P-180^{\circ}$ and $\angle P Q M=2 \angle Q-180^{\circ}$. From these formulas we easily get $\angle R P Q=$ $60^{\circ}, \angle R Q P=75^{\circ}$, and $\angle Q R P=$ $45^{\circ}$.


18. (POL 1) Let $m$ positive integers $a_{1}, \ldots, a_{m}$ be given. Prove that there exist fewer than $2^{m}$ positive integers $b_{1}, \ldots, b_{n}$ such that all sums of distinct $b_{k}$ 's are distinct and all $a_{i}(i \leq m)$ occur among them.

Solution: 18. Let us write all $a_{i}$ in binary representation. For $S \subseteq\{1,2, \ldots, m\}$ let us define $b(S)$ as the number in whose binary representation ones appear in exactly the slots where ones appear in all $a_{i}$ where $i \subseteq S$ and don't appear in any other $a_{i}$. Some $b(S)$, including $b(\emptyset)$, will equal 0 , and hence there are fewer than $2^{m}$ different positive $b(S)$. We note that no two positive $b\left(S_{1}\right)$ and $b\left(S_{2}\right)\left(S_{1} \neq S_{2}\right)$ have ones in the same decimal places. Hence sums of distinct $b(S)$ 's are distinct. Moreover  $$ a_{i}=\sum_{i \in S} b(S) $$  and hence the positive $b(S)$ are indeed the numbers $b_{1}, \ldots, b_{n}$ whose existence we had to prove.


19. (ROM 1) Consider the sequences $\left(a_{n}\right),\left(b_{n}\right)$ defined by  $$ a_{1}=3, \quad b_{1}=100, \quad a_{n+1}=3^{a_{n}}, \quad b_{n+1}=100^{b_{n}} . $$  Find the smallest integer $m$ for which $b_{m}>a_{100}$.

Solution: 19. Let us define $i_{j}$ for two positive integers $i$ and $j$ in the following way: $i_{1}=i$ and $i_{j+1}=i^{i_{j}}$ for all positive integers $j$. Thus we must find the smallest $m$ such that $100_{m}>3_{100}$. Since $100_{1}=100>27=3_{2}$, we inductively have $100_{j}=10^{100_{j-1}}>3^{100_{j-1}}>3^{3_{j}}=3_{j+1}$ and hence $m \leq 99$. We now prove that $m=99$ by proving $100_{98}<3_{100}$. We note that $\left(100_{1}\right)^{2}=10^{4}<27^{4}=3^{12}<3^{27}=3_{3}$. We also note for $d>12$ (which trivially holds for all $d=100_{i}$ ) that if $c>d^{2}$, then we have  $$ 3^{c}>3^{d^{2}}>3^{12 d}=\left(3^{12}\right)^{d}>10000^{d}=\left(100^{d}\right)^{2} $$  Hence from $3_{3}>\left(100_{1}\right)^{2}$ it inductively follows that $3_{j}>\left(100_{j-2}\right)^{2}>$ $100_{j-2}$ and hence that $100_{99}>3_{100}>100_{98}$. Hence $m=99$.


2. (BEL 4) From a bag containing 5 pairs of socks, each pair a different color, a random sample of 4 single socks is drawn. Any complete pairs in the sample are discarded and replaced by a new pair draw from the bag. The process continues until the bag is empty or there are 4 socks of different colors held outside the bag. What is the probability of the latter alternative?

Solution: 2. The only way to arrive at the latter alternative is to draw four different socks in the first drawing or to draw only one pair in the first drawing and then draw two different socks in the last drawing. We will call these probabilities respectively $p_{1}, p_{2}, p_{3}$. We calculate them as follows:  $$ p_{1}=\frac{\binom{5}{4} 2^{4}}{\binom{10}{4}}=\frac{8}{21}, \quad p_{2}=\frac{5\binom{4}{2} 2^{2}}{\binom{10}{4}}=\frac{4}{7}, \quad p_{3}=\frac{4}{\binom{6}{2}}=\frac{4}{15} . $$  We finally calculate the desired probability: $P=p_{1}+p_{2} p_{3}=\frac{8}{15}$.


20. (SWE 2) Given the integer $n>1$ and the real number $a>0$ determine the maximum of $\sum_{i=1}^{n-1} x_{i} x_{i+1}$ taken over all nonnegative numbers $x_{i}$ with sum $a$.

Solution: 20. Let $x_{k}=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Then $x_{i} x_{i+1} \leq x_{i} x_{k}$ for $i=1,2, \ldots, k-1$ and $x_{i} x_{i+1} \leq x_{k} x_{i+1}$ for $i=k, \ldots, n-1$. Summing up these inequalities for $i=1,2, \ldots, n-1$ we obtain  $$ \sum_{i=1}^{n-1} \leq x_{k}\left(x_{1}+\cdots+x_{k-1}+x_{k+1}+\cdots+x_{n}\right)=x_{k}\left(a-x_{k}\right) \leq \frac{a^{2}}{4} $$  We note that the value $a^{2} / 4$ is attained for $x_{1}=x_{2}=a / 2$ and $x_{3}=\cdots=$ $x_{n}=0$. Hence $a^{2} / 4$ is the required maximum.


21. (USS 1) Let $N$ be the number of integral solutions of the equation  $$ x^{2}-y^{2}=z^{3}-t^{3} $$  satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$, and let $M$ be the number of integral solutions of the equation  $$ x^{2}-y^{2}=z^{3}-t^{3}+1 $$  satisfying the condition $0 \leq x, y, z, t \leq 10^{6}$. Prove that $N>M$.

Solution: 21. Let $f(n)$ be the number of different ways $n \in \mathbb{N}$ can be expressed as $x^{2}+y^{3}$ where $x, y \in\left\{0,1, \ldots, 10^{6}\right\}$. Clearly $f(n)=0$ for $n<0$ or $n>10^{12}+10^{18}$. The first equation can be written as $x^{2}+t^{3}=y^{2}+z^{3}=n$, whereas the second equation can be written as $x^{2}+t^{3}=n+1, y^{2}+z^{3}=n$. Hence we obtain the following formulas for $M$ and $N$ :  $$ M=\sum_{i=0}^{m} f(i)^{2}, \quad N=\sum_{i=0}^{m-1} f(i) f(i+1) . $$  Using the AM-GM inequality we get  $$ \begin{aligned} N & =\sum_{i=0}^{m-1} f(i) f(i+1) \\ & \leq \sum_{i=0}^{m-1} \frac{f(i)^{2}+f(i+1)^{2}}{2}=\frac{f(0)^{2}}{2}+\sum_{i=1}^{m-1} f(i)^{2}+\frac{f(m)^{2}}{2}<M . \end{aligned} $$  The last inequality is strong, since $f(0)=1>0$. This completes our proof.


22. (USS 3) ${ }^{\mathrm{IMO} 3}$ There are two circles in the plane. Let a point $A$ be one of the points of intersection of these circles. Two points begin moving simultaneously with constant speeds from the point $A$, each point along its own circle. The two points return to the point $A$ at the same time. Prove that there is a point $P$ in the plane such that at every moment of time the distances from the point $P$ to the moving points are equal.

Solution: 22. Let the centers of the two circles be denoted by $O$ and $O_{1}$ and their respective radii by $r$ and $r_{1}$, and let the positions of the points on the circles at time $t$ be denoted by $M(t)$ and $N(t)$. Let $Q$ be the point such that $O A O_{1} Q$ is a parallelogram. We will show that $Q$ is the point $P$ we are looking for, i.e., that $Q M(t)=$ $Q N(t)$ for all $t$. We note that $O Q=$ $O_{1} A=r_{1}, O_{1} Q=O A=r$ and ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-449.jpg?height=353&width=469&top_left_y=1504&top_left_x=850) $\angle Q O A=\angle Q O_{1} A=\phi$. Since the two points return to $A$ at the same time, it follows that $\angle M(t) O A=\angle N(t) O_{1} A=\omega t$. Therefore $\angle Q O M(t)=$ $\angle Q O_{1} N(t)=\phi+\omega t$, from which it follows that $\triangle Q O M(t) \cong \triangle Q O_{1} N(t)$. Hence $Q M(t)=Q N(t)$, as we claimed.


23. (USA 4) Find all natural numbers $n$ for which $2^{8}+2^{11}+2^{n}$ is a perfect square.

Solution: 23. It is easily verified that no solutions exist for $n \leq 8$. Let us now assume that $n>8$. We note that $2^{8}+2^{11}+2^{n}=2^{8} \cdot\left(9+2^{n-8}\right)$. Hence $9+2^{n-8}$ must also be a square, say $9+2^{n-8}=x^{2}, x \in \mathbb{N}$, i.e., $2^{n-8}=x^{2}-9=$ $(x-3)(x+3)$. Thus $x-3$ and $x+3$ are both powers of 2 , which is possible only for $x=5$ and $n=12$. Hence, $n=12$ is the only solution.


24. (USA 5) A circle O with center $O$ on base $B C$ of an isosceles triangle $A B C$ is tangent to the equal sides $A B, A C$. If point $P$ on $A B$ and point $Q$ on $A C$ are selected such that $P B \times C Q=(B C / 2)^{2}$, prove that line segment $P Q$ is tangent to circle O , and prove the converse.

Solution: 24. Clearly $O$ is the midpoint of $B C$. Let $M$ and $N$ be the points of tangency of the circle with $A B$ and $A C$, respectively, and let $\angle B A C=2 \varphi$. Then $\angle B O M=\angle C O N=\varphi$. Let us assume that $P Q$ touches the circle in $X$. If we set $\angle P O M=$ $\angle P O X=x$ and $\angle Q O N=\angle Q O X=y$, then $2 x+2 y=\angle M O N=$ $180^{\circ}-2 \varphi$, i.e., $y=90^{\circ}-\varphi-x$. It follows that $\angle O Q C=180^{\circ}-\angle Q O C-$ $\angle O C Q=180^{\circ}-(\varphi+y)-\left(90^{\circ}-\varphi\right)=90^{\circ}-y=x+\varphi=\angle B O P$. Hence the triangles $B O P$ and $C Q O$ are similar, and consequently $B P \cdot C Q=$ $B O \cdot C O=(B C / 2)^{2}$. Conversely, let $B P \cdot C Q=(B C / 2)^{2}$ and let $Q^{\prime}$ be the point on $(A C)$ such that $P Q^{\prime}$ is tangent to the circle. Then $B P \cdot C Q^{\prime}=(B C / 2)^{2}$, which implies $Q \equiv Q^{\prime}$.


25. (USA 6) ${ }^{\mathrm{IMO} 4}$ Given a point $P$ in a given plane $\pi$ and also a given point $Q$ not in $\pi$, show how to determine a point $R$ in $\pi$ such that $\frac{Q P+P R}{Q R}$ is a maximum.

Solution: 25. Let us first look for such a point $R$ on a line $l$ in $\pi$ going through $P$. Let $\angle Q P R=2 \theta$. Consider a point $Q^{\prime}$ on $l$ such that $Q^{\prime} P=Q P$. Then we have  $$ \frac{Q P+P R}{Q R}=\frac{R Q^{\prime}}{Q R}=\frac{\sin Q^{\prime} Q R}{\sin Q Q^{\prime} R} $$  Since $Q Q^{\prime} P$ is fixed, the maximal value of the expression occurs when $\angle Q Q^{\prime} R=90^{\circ}$. In this case $(Q P+P R) / Q R=1 / \sin \theta$. Looking at all possible lines $l$, we see that $\theta$ is minimized when $l$ equals the projection of $P Q$ onto $\pi$. Hence, the point $R$ is the intersection of the projection of $P Q$ onto $\pi$ and the plane through $Q$ perpendicular to $P Q$.


26. (YUG 4) Prove that the functional equations  $$ \begin{aligned} f(x+y) & =f(x)+f(y) \\ \text { and } \quad f(x+y+x y) & =f(x)+f(y)+f(x y) \quad(x, y \in \mathbb{R}) \end{aligned} $$  are equivalent.

Solution: 26. Let us assume that $f(x+y)=f(x)+f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+x y)=f(x+y)+f(x y)=$ $f(x)+f(y)+f(x y)$. Hence the equivalence is proved in the first direction. Now let us assume that $f(x+y+x y)=f(x)+f(y)+f(x y)$ for all reals. Plugging in $x=y=0$ we get $f(0)=0$. Plugging in $y=-1$ we get $f(x)=-f(-x)$. Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ and hence $f(2(u+v+u v)+1)=2 f(u+v+u v)+f(1)=2 f(u v)+$ $2 f(u)+2 f(v)+f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x=u$ and $y=2 v+1$ we get $f(2(u+v+u v)+1)=f(u+(2 v+$ 1) $+u(2 v+1))=f(u)+2 f(v)+f(1)+f(2 u v+u)$. Hence it follows that $2 f(u v)+2 f(u)+2 f(v)+f(1)=f(u)+2 f(v)+f(1)+f(2 u v+u)$, i.e.,  $$ f(2 u v+u)=2 f(u v)+f(u) $$  Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all reals. Now (1) reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is 0 .


3. (BUL 1) Find all polynomials $f(x)$ with real coefficients for which  $$ f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right) $$

Solution: 3. An obvious solution is $f(x)=0$. We now look for nonzero solutions. We note that plugging in $x=0$ we get $f(0)^{2}=f(0)$; hence $f(0)=0$ or $f(0)=1$. If $f(0)=0$, then $f$ is of the form $f(x)=x^{k} g(x)$, where $g(0) \neq 0$. Plugging this formula into $f(x) f\left(2 x^{2}\right)=f\left(2 x^{3}+x\right)$ we get $2^{k} x^{2 k} g(x) g\left(2 x^{2}\right)=\left(2 x^{2}+1\right)^{k} g\left(2 x^{3}+x\right)$. Plugging in $x=0$ gives us $g(0)=0$, which is a contradiction. Hence $f(0)=1$. For an arbitrary root $\alpha$ of the polynomial $f, 2 \alpha^{3}+\alpha$ must also be a root. Let $\alpha$ be a root of the largest modulus. If $|\alpha|>1$ then $\left|2 \alpha^{3}+\alpha\right|>$ $2|\alpha|^{3}-|\alpha|>|\alpha|$, which is impossible. It follows that $|\alpha| \leq 1$ and hence all roots of $f$ have modules less than or equal to 1 . But the product of all roots of $f$ is $|f(0)|=1$, which implies that all the roots have modulus 1. Consequently, for a root $\alpha$ it holds that $|\alpha|=\left|2 \alpha^{3}-\alpha\right|=1$. This is possible only if $\alpha= \pm \imath$. Since the coefficients of $f$ are real it follows that $f$ must be of the form $f(x)=\left(x^{2}+1\right)^{k}$ where $k \in \mathbb{N}_{0}$. These polynomials satisfy the original formula. Hence, the solutions for $f$ are $f(x)=0$ and $f(x)=\left(x^{2}+1\right)^{k}, k \in \mathbb{N}_{0}$.


4. (BUL 3) ${ }^{\mathrm{IMO} 2} \mathrm{~A}$ pentagonal prism $A_{1} A_{2} \ldots A_{5} B_{1} B_{2} \ldots B_{5}$ is given. The edges, the diagonals of the lateral walls and the internal diagonals of the prism are each colored either red or green in such a way that no triangle whose vertices are vertices of the prism has its three edges of the same color. Prove that all edges of the bases are of the same color.

Solution: 4. Let us prove first that the edges $A_{1} A_{2}, A_{2} A_{3}, \ldots, A_{5} A_{1}$ are of the same color. Assume the contrary, and let w.l.o.g. $A_{1} A_{2}$ be red and $A_{2} A_{3}$ be green. Three of the segments $A_{2} B_{l}(l=1,2,3,4,5)$, say $A_{2} B_{i}, A_{2} B_{j}, A_{2} B_{k}$, have to be of the same color, let it w.l.o.g. be red. Then $A_{1} B_{i}, A_{1} B_{j}, A_{1} B_{k}$ must be green. At least one of the sides of triangle $B_{i} B_{j} B_{k}$, say $B_{i} B_{j}$, must be an edge of the prism. Then looking at the triangles $A_{1} B_{i} B_{j}$ and $A_{2} B_{i} B_{j}$ we deduce that $B_{i} B_{j}$ can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$ have the same color. Similarly, all five edges of $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the $A$ edges are painted red and the $B$ edges green. Let us call segments of the form $A_{i} B_{j}$ diagonal ( $i$ and $j$ may be equal). We now count the diagonal segments by grouping the red segments based on their $A$ point, and the green segments based on their $B$ point. As above, the assumption that three of $A_{i} B_{j}$ for fixed $i$ are red leads to a contradiction. Hence at most two diagonal segments out of each $A_{i}$ may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25 , which is a contradiction. Hence all edges in the pentagons $A_{1} A_{2} A_{3} A_{4} A_{5}$ and $B_{1} B_{2} B_{3} B_{4} B_{5}$ have the same color.


5. (CZS 2) Let $n \geq 2$ be an integer. Find the maximal cardinality of a set $M$ of pairs $(j, k)$ of integers, $1 \leq j<k \leq n$, with the following property: If $(j, k) \in M$, then $(k, m) \notin M$ for any $m$.

Solution: 5. Let $A=\{x \mid(x, y) \in M\}$ and $B=\{y \mid(x, y) \in M$. Then $A$ and $B$ are disjoint and hence  $$ |M| \leq|A| \cdot|B| \leq \frac{(|A|+|B|)^{2}}{4} \leq\left[\frac{n^{2}}{4}\right] $$  These cardinalities can be achieved for $M=\{(a, b) \mid a=1,2, \ldots,[n / 2]$, $b=[n / 2]+1, \ldots, n\}$.


6. (CZS 4) Find the real values of $p$ for which the equation  $$ \sqrt{2 p+1-x^{2}}+\sqrt{3 x+p+4}=\sqrt{x^{2}+9 x+3 p+9} $$  in $x$ has exactly two real distinct roots $(\sqrt{t}$ means the positive square root of $t$ ).

Solution: 6. Setting $q=x^{2}+x-p$, the given equation becomes  $$ \sqrt{(x+1)^{2}-2 q}+\sqrt{(x+2)^{2}-q}=\sqrt{(2 x+3)^{2}-3 q} . $$  Taking squares of both sides we get $2 \sqrt{\left((x+1)^{2}-2 q\right)\left((x+2)^{2}-q\right)}=$ $2(x+1)(x+2)$. Taking squares again we get  $$ q\left(2 q-2(x+2)^{2}-(x+1)^{2}\right)=0 $$  If $2 q=2(x+2)^{2}+(x+1)^{2}$, at least one of the expressions under the three square roots in (1) is negative, and in that case the square root is not well-defined. Thus, we must have $q=0$. Now (1) is equivalent to $|x+1|+|x+2|=|2 x+3|$, which holds if and only if $x \notin(-2,-1)$. The number of real solutions $x$ of $q=x^{2}+x-p=0$ which are not in the interval $(-2,-1)$ is zero if $p<-1 / 4$, one if $p=-1 / 4$ or $0<p<2$, and two otherwise. Hence, the answer is $-1 / 4<p \leq 0$ or $p \geq 2$.


7. (FRG 1) ${ }^{\mathrm{IMO} 1}$ Given that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}$, where $p$ and $q$ are natural numbers having no common factor, prove that $p$ is divisible by 1979.

Solution: 7. We denote the sum mentioned above by $S$. We have the following equalities:  $$ \begin{aligned} S & =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319} \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right) \\ & =1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right) \\ & =\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \\ & =\sum_{i=660}^{989} \frac{1}{i}+\frac{1}{1979-i}=\sum_{i=660}^{989} \frac{1979}{i \cdot(1979-i)} \end{aligned} $$  Since no term in the sum contains a denominator divisible by 1979 (1979 is a prime number), it follows that when $S$ is represented as $p / q$ the numerator $p$ will have to be divisible by 1979.


8. (FRG 3) For all rational $x$ satisfying $0 \leq x<1, f$ is defined by  $$ f(x)= \begin{cases}f(2 x) / 4, & \text { for } 0 \leq x<1 / 2 \\ 3 / 4+f(2 x-1) / 4, & \text { for } 1 / 2 \leq x<1\end{cases} $$  Given that $x=0 . b_{1} b_{2} b_{3} \ldots$ is the binary representation of $x$, find $f(x)$.

Solution: 8. By the definition of $f$, it holds that $f\left(0 . b_{1} b_{2} \ldots\right)=3 b_{1} / 4+f\left(0 . b_{2} b_{3} \ldots\right) / 4$ $=0 . b_{1} b_{1}+f\left(0 . b_{2} b_{3} \ldots\right) / 4$. Continuing this argument we obtain  $$ f\left(0 . b_{1} b_{2} b_{3} \ldots\right)=0 . b_{1} b_{1} \ldots b_{n} b_{n}+\frac{1}{2^{2 n}} f\left(0 . b_{n+1} b_{n+2} \ldots\right) $$  The binary representation of every rational number is eventually periodic. Let us first determine $f(x)$ for a rational $x$ with the periodic representation $x=0 . \overline{b_{1} b_{2} \ldots b_{n}}$. Using (1) we obtain $f(x)=0 . b_{1} b_{1} \ldots b_{n} b_{n}+f(x) / 2^{2 n}$, and hence $f(x)=\frac{2^{n}}{2^{n}-1} 0 . b_{1} b_{1} \ldots b_{n} b_{n}=0 . \overline{b_{1} b_{1} \ldots b_{n} b_{n}}$. Now let $x=0 . a_{1} a_{2} \ldots a_{k} \overline{b_{1} b_{2} \ldots b_{n}}$ be an arbitrary rational number. Then it follows from (1) that $f(x)=0 . a_{1} a_{1} \ldots a_{k} a_{k}+\frac{1}{2^{2 n}} f\left(0 . \overline{b_{1} b_{2} \ldots b_{n}}\right)=0 . a_{1} a_{1} \ldots a_{k} a_{k} \overline{b_{1} b_{1} \ldots b_{n} b_{n}}$. Hence $f\left(0 . b_{1} b_{2} \ldots\right)=0 . b_{1} b_{1} b_{2} b_{2} \ldots$ for every rational number $0 . b_{1} b_{2} \ldots$.


9. (FRG 4) ${ }^{\mathrm{IMO6}}$ Let $S$ and $F$ be two opposite vertices of a regular octagon. A counter starts at $S$ and each second is moved to one of the two neighboring vertices of the octagon. The direction is determined by the toss of a coin. The process ends when the counter reaches $F$. We define $a_{n}$ to be the number of distinct paths of duration $n$ seconds that the counter may take to reach $F$ from $S$. Prove that for $n=1,2,3, \ldots$, $a_{2 n-1}=0, \quad a_{2 n}=\frac{1}{\sqrt{2}}\left(x^{n-1}-y^{n-1}\right), \quad$ where $x=2+\sqrt{2}, y=2-\sqrt{2}$.

Solution: 9. Let us number the vertices, starting from $S$ and moving clockwise. In that case $S=1$ and $F=5$. After an odd number of moves to a neighboring point we can be only on an even point, and hence it follows that $a_{2 n-1}=0$ for all $n \in \mathbb{N}$. Let us define respectively $z_{n}$ and $w_{n}$ as the number of paths from $S$ to $S$ in $2 n$ moves and the number of paths from $S$ to points 3 and 7 in $2 n$ moves. We easily derive the following recurrence relations:  $$ a_{2 n+2}=w_{n}, \quad w_{n+1}=2 w_{n}+2 z_{n}, \quad z_{n+1}=2 z_{n}+w_{n}, \quad n=0,1,2, \ldots $$  By subtracting the second equation from the third we get $z_{n+1}=w_{n+1}-$ $w_{n}$. By plugging this equation into the formula for $w_{n+2}$ we get $w_{n+2}-$ $4 w_{n+1}+2 w_{n}=0$. The roots of the characteristic equation $r^{2}-4 r+2=0$ are $x=2+\sqrt{2}$ and $y=2-\sqrt{2}$. From the conditions $w_{0}=0$ and $w_{1}=2$ we easily obtain $a_{2 n}=w_{n-1}=\left(x^{n-1}-y^{n-1}\right) / \sqrt{2}$.


1. (BEL) ${ }^{\mathrm{IMO} 4}$ (a) For which values of $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers? (b) For which values of $n>2$ is there a unique set having the stated property?

Solution: 1. Assume that the set $\{a-n+1, a-n+2, \ldots, a\}$ of $n$ consecutive numbers satisfies the condition $a \mid \operatorname{lcm}[a-n+1, \ldots, a-1]$. Let $a=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{r}^{\alpha_{r}}$ be the canonic representation of $a$, where $p_{1}<p_{2}<\cdots<p_{r}$ are primes and $\alpha_{1}, \cdots, \alpha_{r}>0$. Then for each $j=1,2, \ldots, r$, there exists $m, m=$ $1,2, \ldots, n-1$, such that $p_{j}^{\alpha_{j}} \mid a-m$, i.e., such that $p_{j}^{\alpha_{j}} \mid m$. Thus $p_{j}^{\alpha_{j}} \leq$ $n-1$. If $r=1$, then $a=p_{1}^{\alpha_{1}} \leq n-1$, which is impossible. Therefore $r \geq 2$. But then there must exist two distinct prime numbers less than $n$; hence $n \geq 4$. For $n=4$, we must have $p_{1}^{\alpha_{1}}, p_{2}^{\alpha_{2}} \leq 3$, which leads to $p_{1}=2, p_{2}=3$, $\alpha_{1}=\alpha_{2}=1$. Therefore $a=6$, and $\{3,4,5,6\}$ is a unique set satisfying the condition of the problem. For every $n \geq 5$ there exist at least two such sets. In fact, for $n=5$ we easily find two sets: $\{2,3,4,5,6\}$ and $\{8,9,10,11,12\}$. Suppose that $n \geq 6$. Let $r, s, t$ be natural numbers such that $2^{r} \leq n-1<2^{r+1}$, $3^{s} \leq n-1<3^{s+1}, 5^{t} \leq n-1<5^{t+1}$. Taking $a=2^{r} \cdot 3^{s}$ and $a=2^{r} \cdot 5^{t}$ we obtain two distinct sets with the required property. Thus the answers are (a) $n \geq 4$ and (b) $n=4$.


10. (FRA) Determine the smallest natural number $n$ having the following property: For every integer $p, p \geq n$, it is possible to subdivide (partition) a given square into $p$ squares (not necessarily equal).

Solution: 10. It is easy to see that partitioning into $p=2 k$ squares is possible for $k \geq 2$ (Fig. 1). Furthermore, whenever it is possible to partition the square into $p$ squares, there is a partition of the square into $p+3$ squares: namely, in the partition into $p$ squares, divide one of them into four new squares. ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-455.jpg?height=272&width=240&top_left_y=1678&top_left_x=393)  Fig. 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_8e985d6b9c83aa3e9d0eg-455.jpg?height=272&width=308&top_left_y=1678&top_left_x=919)  Fig. 2  This implies that both $p=2 k$ and $p=2 k+3$ are possible if $k \geq 2$, and therefore all $p \geq 6$ are possible.  On the other hand, partitioning the square into 5 squares is not possible. Assuming it is possible, one of its sides would be covered by exactly two squares, which cannot be of the same size (Fig. 2). The rest of the big square cannot be partitioned into three squares. Hence, the answer is $n=6$.


11. (NET) On a semicircle with unit radius four consecutive chords $A B, B C$, $C D, D E$ with lengths $a, b, c, d$, respectively, are given. Prove that  $$ a^{2}+b^{2}+c^{2}+d^{2}+a b c+b c d<4 $$

Solution: 11. Let us denote the center of the semicircle by $O$, and $\angle A O B=2 \alpha$, $\angle B O C=2 \beta, A C=m, C E=n$. We claim that $a^{2}+b^{2}+n^{2}+a b n=4$. Indeed, since $a=2 \sin \alpha, b=2 \sin \beta$, $n=2 \cos (\alpha+\beta)$, we have  $$ \begin{aligned} a^{2} & +b^{2}+n^{2}+a b n \\ & =4\left(\sin ^{2} \alpha+\sin ^{2} \beta+\cos ^{2}(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)\right) \\ & =4+4\left(-\frac{\cos 2 \alpha}{2}-\frac{\cos 2 \beta}{2}+\cos (\alpha+\beta) \cos (\alpha-\beta)\right) \\ & =4+4(\cos (\alpha+\beta) \cos (\alpha-\beta)-\cos (\alpha+\beta) \cos (\alpha-\beta))=4 \end{aligned} $$  Analogously, $c^{2}+d^{2}+m^{2}+c d m=4$. By adding both equalities and subtracting $m^{2}+n^{2}=4$ we obtain  $$ a^{2}+b^{2}+c^{2}+d^{2}+a b n+c d m=4 $$  Since $n>c$ and $m>b$, the desired inequality follows.


12. (NET) ${ }^{\mathrm{IMO} 3}$ Determine the maximum value of $m^{2}+n^{2}$ where $m$ and $n$ are integers satisfying  $$ m, n \in\{1,2, \ldots, 100\} \quad \text { and } \quad\left(n^{2}-m n-m^{2}\right)^{2}=1 $$

Solution: 12. We will solve the contest problem (in which $m, n \in\{1,2, \ldots, 1981\}$ ). For $m=1, n$ can be either 1 or 2 . If $m>1$, then $n(n-m)=m^{2} \pm 1>0$; hence $n-m>0$. Set $p=n-m$. Since $m^{2}-m p-p^{2}=m^{2}-p(m+p)=$ $-\left(n^{2}-n m-m^{2}\right)$, we see that $(m, n)$ is a solution of the equation if and only if $(p, m)$ is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence  $$ 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584, \ldots $$  So the required maximum is $987^{2}+1597^{2}$.


13. (ROM) Let $P$ be a polynomial of degree $n$ satisfying  $$ P(k)=\binom{n+1}{k}^{-1} \quad \text { for } k=0,1, \ldots, n $$  Determine $P(n+1)$.

Solution: 13. Lemma. For any polynomial $P$ of degree at most $n$,  $$ \sum_{i=0}^{n+1}(-1)^{i}\binom{n+1}{i} P(i)=0 $$  Proof. We shall use induction on $n$. For $n=0$ it is trivial. Assume that it is true for $n=k$ and suppose that $P(x)$ is a polynomial of degree $k+1$. Then $P(x)-P(x+1)$ clearly has degree at most $k$; hence (1) gives  $$ \begin{aligned} 0 & =\sum_{i=0}^{k+1}(-1)^{i}\binom{k+1}{i}(P(i)-P(i+1)) \\ & =\sum_{i=0}^{k+1}(-1)^{i}\binom{k+1}{i} P(i)+\sum_{i=1}^{k+2}(-1)^{i}\binom{k+1}{i-1} P(i) \\ & =\sum_{i=0}^{k+2}(-1)^{i}\binom{k+2}{i} P(i) \end{aligned} $$  This completes the proof of the lemma. Now we apply the lemma to obtain the value of $P(n+1)$. Since $P(i)=$ $\binom{n+1}{i}^{-1}$ for $i=0,1, \ldots, n$, we have  $$ 0=\sum_{i=0}^{n+1}(-1)^{i}\binom{n+1}{i} P(i)=(-1)^{n+1} P(n+1)+ \begin{cases}1, & 2 \mid n ; \\ 0, & 2 \nmid n .\end{cases} $$  It follows that $P(n+1)= \begin{cases}1, & 2 \mid n ; \\ 0, & 2 \nmid n .\end{cases}$


14. (ROM) Prove that a convex pentagon (a five-sided polygon) $A B C D E$ with equal sides and for which the interior angles satisfy the condition $\angle A \geq \angle B \geq \angle C \geq \angle D \geq \angle E$ is a regular pentagon.

Solution: 14. We need the following lemma.  Lemma. If a convex quadrilateral $P Q R S$ satisfies $P S=Q R$ and $\angle S P Q \geq$ $\angle R Q P$, then $\angle Q R S \geq \angle P S R$. Proof. If the lines $P S$ and $Q R$ are parallel, then this quadrilateral is a parallelogram, and the statement is trivial. Otherwise, let $X$ be the point of intersection of lines $P S$ and $Q R$. Assume that $\angle S P Q+\angle R Q P>180^{\circ}$. Then $\angle X P Q \leq \angle X Q P$ implies that $X P \geq X Q$, and consequently $X S \geq X R$. Hence, $\angle Q R S=$ $\angle X R S \geq \angle X S R=\angle P S R$. Similarly, if $\angle S P Q+\angle R Q P<180^{\circ}$, then $\angle X P Q \geq \angle X Q P$, from which it follows that $X P \leq X Q$, and thus $X S \leq X R$; hence $\angle Q R S=$ $180^{\circ}-\angle X R S \geq 180^{\circ}-\angle X S R=\angle P S R$. Now we apply the lemma to the quadrilateral $A B C D$. Since $\angle B \geq \angle C$ and $A B=C D$, it follows that $\angle C D A \geq \angle B A D$, which together with $\angle E D A=\angle E A D$ gives $\angle D \geq \angle A$. Thus $\angle A=\angle B=\angle C=\angle D$. Analogously, by applying the lemma to $B C D E$ we obtain $\angle E \geq \angle B$, and hence $\angle B=\angle C=\angle D=\angle E$.


15. (GBR) ${ }^{\mathrm{IMO} 1}$ Find the point $P$ inside the triangle $A B C$ for which  $$ \frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} $$  is minimal, where $P D, P E, P F$ are the perpendiculars from $P$ to $B C, C A$, $A B$ respectively.

Solution: 15. Set $B C=a, C A=b, A B=c$, and denote the area of $\triangle A B C$ by $P$, and $a / P D+b / P E+c / P F$ by $S$. Since $a \cdot P D+b \cdot P E+c \cdot P F=2 P$, by the Cauchy-Schwarz inequality we have  $$ 2 P S=(a \cdot P D+b \cdot P E+c \cdot P F)\left(\frac{a}{P D}+\frac{b}{P E}+\frac{c}{P F}\right) \geq(a+b+c)^{2} $$  with equality if and only if $P D=P E=P F$, i.e., $P$ is the incenter of $\triangle A B C$. In that case, $S$ attains its minimum:  $$ S_{\min }=\frac{(a+b+c)^{2}}{2 P} $$