Solution: The solution consists of two parts. In Part 1 we show that it suffices to prove that  $$ \frac{X B}{X D}=\frac{A B}{C D} $$  and  $$ \frac{X A}{X C}=\frac{D A}{B C} $$  In Part 2 we establish these equalities.  Part 1. Using the sine law and applying (1) we obtain  $$ \frac{\sin \angle A X B}{\sin \angle X A B}=\frac{A B}{X B}=\frac{C D}{X D}=\frac{\sin \angle C X D}{\sin \angle X C D} $$  so $\sin \angle A X B=\sin \angle C X D$ by the problem conditions. Similarly, (2) yields $\sin \angle D X A=$ $\sin \angle B X C$. If at least one of the pairs $(\angle A X B, \angle C X D)$ and $(\angle B X C, \angle D X A)$ consists of supplementary angles, then we are done. Otherwise, $\angle A X B=\angle C X D$ and $\angle D X A=\angle B X C$. In this case $X=A C \cap B D$, and the problem conditions yield that $A B C D$ is a parallelogram and hence a rhombus. In this last case the claim also holds.  Part 2. To prove the desired equality (1), invert $A B C D$ at centre $X$ with unit radius; the images of points are denoted by primes.  We have  $$ \angle A^{\prime} B^{\prime} C^{\prime}=\angle X B^{\prime} A^{\prime}+\angle X B^{\prime} C^{\prime}=\angle X A B+\angle X C B=\angle X C D+\angle X C B=\angle B C D . $$  Similarly, the corresponding angles of quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are equal.  Moreover, we have  $$ A^{\prime} B^{\prime} \cdot C^{\prime} D^{\prime}=\frac{A B}{X A \cdot X B} \cdot \frac{C D}{X C \cdot X D}=\frac{B C}{X B \cdot X C} \cdot \frac{D A}{X D \cdot D A}=B^{\prime} C^{\prime} \cdot D^{\prime} A^{\prime} $$  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-52.jpg?height=586&width=1084&top_left_y=455&top_left_x=495)  Now we need the following Lemma.  Lemma. Assume that the corresponding angles of convex quadrilaterals $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ are equal, and that $X Y \cdot Z T=Y Z \cdot T X$ and $X^{\prime} Y^{\prime} \cdot Z^{\prime} T^{\prime}=Y^{\prime} Z^{\prime} \cdot T^{\prime} X^{\prime}$. Then the two quadrilaterals are similar.  Proof. Take the quadrilateral $X Y Z_{1} T_{1}$ similar to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ and sharing the side $X Y$ with $X Y Z T$, such that $Z_{1}$ and $T_{1}$ lie on the rays $Y Z$ and $X T$, respectively, and $Z_{1} T_{1} \| Z T$. We need to prove that $Z_{1}=Z$ and $T_{1}=T$. Assume the contrary. Without loss of generality, $T X>X T_{1}$. Let segments $X Z$ and $Z_{1} T_{1}$ intersect at $U$. We have  $$ \frac{T_{1} X}{T_{1} Z_{1}}<\frac{T_{1} X}{T_{1} U}=\frac{T X}{Z T}=\frac{X Y}{Y Z}<\frac{X Y}{Y Z_{1}} $$  thus $T_{1} X \cdot Y Z_{1}<T_{1} Z_{1} \cdot X Y$. A contradiction.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-52.jpg?height=343&width=440&top_left_y=1662&top_left_x=814)  It follows from the Lemma that the quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are similar, hence  $$ \frac{B C}{A B}=\frac{A^{\prime} B^{\prime}}{D^{\prime} A^{\prime}}=\frac{A B}{X A \cdot X B} \cdot \frac{X D \cdot X A}{D A}=\frac{A B}{A D} \cdot \frac{X D}{X B} $$  and therefore  $$ \frac{X B}{X D}=\frac{A B^{2}}{B C \cdot A D}=\frac{A B^{2}}{A B \cdot C D}=\frac{A B}{C D} $$  We obtain (1), as desired; (2) is proved similarly.  Comment. Part 1 is an easy one, while part 2 seems to be crucial. On the other hand, after the proof of the similarity $D^{\prime} A^{\prime} B^{\prime} C^{\prime} \sim A B C D$ one may finish the solution in different ways, e.g., as follows. The similarity taking $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ to $A B C D$ maps $X$ to the point $X^{\prime}$ isogonally conjugate of $X$ with respect to $A B C D$ (i.e. to the point $X^{\prime}$ inside $A B C D$ such that $\angle B A X=\angle D A X^{\prime}$, $\left.\angle C B X=\angle A B X^{\prime}, \angle D C X=\angle B C X^{\prime}, \angle A D X=\angle C D X^{\prime}\right)$. It is known that the required equality $\angle A X B+\angle C X D=180^{\circ}$ is one of known conditions on a point $X$ inside $A B C D$ equivalent to the existence of its isogonal conjugate.  This page is intentionally left blank


Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$.  (Russia)

Solution: As usual, we denote the directed angle between the lines $a$ and $b$ by $\Varangle(a, b)$. We frequently use the fact that $a_{1} \perp a_{2}$ and $b_{1} \perp b_{2}$ yield $\Varangle\left(a_{1}, b_{1}\right)=\Varangle\left(a_{2}, b_{2}\right)$.  Let the lines $\ell_{B}$ and $\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-54.jpg?height=1020&width=1384&top_left_y=895&top_left_x=336)  Claim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.  Proof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have  $$ \Varangle\left(O_{A} P, O_{C} P\right)=\Varangle\left(O_{C} O, O_{A} O\right)=\Varangle(C P, A P)=\Varangle(C B, A B)=\Varangle\left(O_{A} L_{B}, O_{C} L_{B}\right) . $$  Denote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\omega_{B}$. Define the circles $\omega_{A}$ and $\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.  Proof. From cyclic quadruples of points in the circles $\omega_{B}$ and $\omega_{C}$, we have  $$ \begin{aligned} \Varangle\left(L_{C} L_{A}, L_{C} P\right) & =\Varangle\left(L_{C} O_{B}, L_{C} P\right)=\Varangle\left(O_{A} O_{B}, O_{A} P\right) \\ & =\Varangle\left(O_{A} O_{C}, O_{A} P\right)=\Varangle\left(L_{B} O_{C}, L_{B} P\right)=\Varangle\left(L_{B} L_{A}, L_{B} P\right) . \end{aligned} $$  Claim 3. The points $P, L_{C}$, and $C$ are collinear.  Proof. We have $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=\Varangle\left(P L_{C}, L_{C} O_{B}\right)=\Varangle\left(P O_{A}, O_{A} O_{B}\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \Varangle\left(P O_{A}, O_{A} O_{B}\right)=\Varangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \Varangle(P A, A O)=\pi / 2-\Varangle(C A, C P)=\Varangle\left(C P, L_{C} L_{A}\right)$. We obtain $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=$ $\Varangle\left(C P, L_{C} L_{A}\right)$, which shows that $P \in C L_{C}$.  Similarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that  $$ \Varangle\left(O P, P L_{A}\right)=\Varangle(P A, A O)=\Varangle\left(P L_{C}, L_{C} L_{A}\right) $$  which means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.  Comment 1. The proof of Claim 2 may be replaced by the following remark: since $P$ belongs to the circles $\omega_{A}$ and $\omega_{C}, P$ is the Miquel point of the four lines $\ell_{A}, \ell_{B}, \ell_{C}$, and $O_{A} O_{B} O_{C}$.  Comment 2. Claims 2 and 3 can be proved in several different ways and, in particular, in the reverse order.  Claim 3 implies that the triangles $A B C$ and $L_{A} L_{B} L_{C}$ are perspective with perspector $P$. Claim 2 can be derived from this observation using spiral similarity. Consider the centre $Q$ of the spiral similarity that maps $A B C$ to $L_{A} L_{B} L_{C}$. From known spiral similarity properties, the points $L_{A}, L_{B}, P, Q$ are concyclic, and so are $L_{A}, L_{C}, P, Q$.  Comment 3. The final conclusion can also be proved it terms of spiral similarity: the spiral similarity with centre $Q$ located on the circle $A B C$ maps the circle $A B C$ to the circle $P L_{A} L_{B} L_{C}$. Thus these circles are orthogonal.  Comment 4. Notice that the homothety with centre $O$ and ratio 2 takes $O_{A}$ to $A^{\prime}$ that is the common point of tangents to $\Omega$ at $A$ and $P$. Similarly, let this homothety take $O_{B}$ to $B^{\prime}$ and $O_{C}$ to $C^{\prime}$. Let the tangents to $\Omega$ at $B$ and $C$ meet at $A^{\prime \prime}$, and define the points $B^{\prime \prime}$ and $C^{\prime \prime}$ similarly. Now, replacing labels $O$ with $I, \Omega$ with $\omega$, and swapping labels $A \leftrightarrow A^{\prime \prime}, B \leftrightarrow B^{\prime \prime}, C \leftrightarrow C^{\prime \prime}$ we obtain the following  Reformulation. Let $\omega$ be the incircle, and let $I$ be the incentre of a triangle $A B C$. Let $P$ be a point of $\omega$ (other than the points of contact of $\omega$ with the sides of $A B C$ ). The tangent to $\omega$ at $P$ meets the lines $A B, B C$, and $C A$ at $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. Line $\ell_{A}$ parallel to the internal angle bisector of $\angle B A C$ passes through $A^{\prime}$; define lines $\ell_{B}$ and $\ell_{C}$ similarly. Prove that the line $I P$ is tangent to the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$.  Though this formulation is equivalent to the original one, it seems more challenging, since the point of contact is now "hidden".


Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the numbers of divisors of $s n$ and of $s k$ are equal.  (Ukraine)

Solution: As usual, the number of divisors of a positive integer $n$ is denoted by $d(n)$. If $n=\prod_{i} p_{i}^{\alpha_{i}}$ is the prime factorisation of $n$, then $d(n)=\prod_{i}\left(\alpha_{i}+1\right)$.  We start by showing that one cannot find any suitable number $s$ if $k \mid n$ or $n \mid k$ (and $k \neq n$ ). Suppose that $n \mid k$, and choose any positive integer $s$. Then the set of divisors of $s n$ is a proper subset of that of $s k$, hence $d(s n)<d(s k)$. Therefore, the pair $(n, k)$ does not satisfy the problem requirements. The case $k \mid n$ is similar.  Now assume that $n \nmid k$ and $k \nmid n$. Let $p_{1}, \ldots, p_{t}$ be all primes dividing $n k$, and consider the prime factorisations  $$ n=\prod_{i=1}^{t} p_{i}^{\alpha_{i}} \quad \text { and } \quad k=\prod_{i=1}^{t} p_{i}^{\beta_{i}} $$  It is reasonable to search for the number $s$ having the form  $$ s=\prod_{i=1}^{t} p_{i}^{\gamma_{i}} $$  The (nonnegative integer) exponents $\gamma_{i}$ should be chosen so as to satisfy  $$ \frac{d(s n)}{d(s k)}=\prod_{i=1}^{t} \frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=1 $$  First of all, if $\alpha_{i}=\beta_{i}$ for some $i$, then, regardless of the value of $\gamma_{i}$, the corresponding factor in (1) equals 1 and does not affect the product. So we may assume that there is no such index $i$. For the other factors in (1), the following lemma is useful.  Lemma. Let $\alpha>\beta$ be nonnegative integers. Then, for every integer $M \geqslant \beta+1$, there exists a nonnegative integer $\gamma$ such that  $$ \frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M}=\frac{M+1}{M} . $$  Proof.  $$ \frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M} \Longleftrightarrow \frac{\alpha-\beta}{\beta+\gamma+1}=\frac{1}{M} \Longleftrightarrow \gamma=M(\alpha-\beta)-(\beta+1) \geqslant 0 . $$  Now we can finish the solution. Without loss of generality, there exists an index $u$ such that $\alpha_{i}>\beta_{i}$ for $i=1,2, \ldots, u$, and $\alpha_{i}<\beta_{i}$ for $i=u+1, \ldots, t$. The conditions $n \nmid k$ and $k \nmid n$ mean that $1 \leqslant u \leqslant t-1$.  Choose an integer $X$ greater than all the $\alpha_{i}$ and $\beta_{i}$. By the lemma, we can define the numbers $\gamma_{i}$ so as to satisfy  $$ \begin{array}{ll} \frac{\alpha_{i}+\gamma_{i}+1}{\beta_{i}+\gamma_{i}+1}=\frac{u X+i}{u X+i-1} & \text { for } i=1,2, \ldots, u, \text { and } \\ \frac{\beta_{u+i}+\gamma_{u+i}+1}{\alpha_{u+i}+\gamma_{u+i}+1}=\frac{(t-u) X+i}{(t-u) X+i-1} & \text { for } i=1,2, \ldots, t-u . \end{array} $$  Then we will have  $$ \frac{d(s n)}{d(s k)}=\prod_{i=1}^{u} \frac{u X+i}{u X+i-1} \cdot \prod_{i=1}^{t-u} \frac{(t-u) X+i-1}{(t-u) X+i}=\frac{u(X+1)}{u X} \cdot \frac{(t-u) X}{(t-u)(X+1)}=1, $$  as required.  Comment. The lemma can be used in various ways, in order to provide a suitable value of $s$. In particular, one may apply induction on the number $t$ of prime factors, using identities like  $$ \frac{n}{n-1}=\frac{n^{2}}{n^{2}-1} \cdot \frac{n+1}{n} $$


Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied:  (i) Each number in the table is congruent to 1 modulo $n$;  (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$.  Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$.  (Indonesia)

Solution: Let $A_{i, j}$ be the entry in the $i^{\text {th }}$ row and the $j^{\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that  $$ \sum_{i=1}^{n} R_{i} \equiv(n-1)+P \quad\left(\bmod n^{4}\right) $$  Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion.  By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence  $R_{i}=\prod_{j=1}^{n}\left(1+a_{i, j}\right)=1+\sum_{j=1}^{n} a_{i, j}+\sum_{1 \leqslant j_{1}<j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}}+\cdots \equiv 1+\sum_{j=1}^{n} a_{i, j} \equiv 1-n+\sum_{j=1}^{n} A_{i, j} \quad\left(\bmod n^{2}\right)$  for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\left(\bmod n^{2}\right)$, and so $n^{2} \mid r_{i}$.  Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain  $$ P=\prod_{i=1}^{n} R_{i}=\prod_{i=1}^{n}\left(1+r_{i}\right) \equiv 1+\sum_{i=1}^{n} r_{i} \quad\left(\bmod n^{4}\right) $$  whence  $$ \sum_{i=1}^{n} R_{i}=n+\sum_{i=1}^{n} r_{i} \equiv n+(P-1) \quad\left(\bmod n^{4}\right) $$  as desired.  Comment. The original version of the problem statement contained also the condition  (iii) The product of all the numbers in the table is congruent to 1 modulo $n^{4}$.  This condition appears to be superfluous, so it was omitted.


Solution: We present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$.  By condition (i), all the $a_{i, j}$ are divisible by $n$. Therefore, we have  $$ \begin{aligned} P=\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv 1+\sum_{(i, j)} a_{i, j} & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right), \end{aligned} $$  where the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution.  Similarly,  $$ \sum_{i=1}^{n} R_{i}=\sum_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv n+\sum_{i} \sum_{j} a_{i, j}+\sum_{i} \sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}}+\sum_{i} \sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \quad\left(\bmod n^{4}\right) $$  Therefore,  $$ \begin{aligned} P+(n-1)-\sum_{i} R_{i} \equiv \sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right) \\ i_{1} \neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2} \neq i_{3} \neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\ & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2}=i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right) . \end{aligned} $$  We show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\Sigma_{1}, \Sigma_{2}$, and $\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have  $$ \sum_{j} a_{i, j} \equiv 0 \quad\left(\bmod n^{2}\right) \quad \text { for all indices } i $$  For every two indices $i_{1}<i_{2}$ we have  $$ \sum_{j_{1}} \sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \equiv 0 \quad\left(\bmod n^{4}\right) $$  since each of the two factors is divisible by $n^{2}$. Summing over all pairs $\left(i_{1}, i_{2}\right)$ we obtain $n^{4} \mid \Sigma_{1}$.  Similarly, for every three indices $i_{1}<i_{2}<i_{3}$ we have  $$ \sum_{j_{1}} \sum_{j_{2}} \sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \cdot\left(\sum_{j_{3}} a_{i_{3}, j_{3}}\right) $$  which is divisible even by $n^{6}$. Hence $n^{4} \mid \Sigma_{2}$.  Finally, for every indices $i_{1} \neq i_{2}=i_{3}$ and $j_{2}<j_{3}$ we have  $$ a_{i_{2}, j_{2}} \cdot a_{i_{2}, j_{3}} \cdot \sum_{j_{1}} a_{i_{1}, j_{1}} \equiv 0 \quad\left(\bmod n^{4}\right) $$  since the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4 -tuples of indices $\left(i_{1}, i_{2}, j_{2}, j_{3}\right)$ we get $n^{4} \mid \Sigma_{3}$.


Define the sequence $a_{0}, a_{1}, a_{2}, \ldots$ by $a_{n}=2^{n}+2^{\lfloor n / 2\rfloor}$. Prove that there are infinitely many terms of the sequence which can be expressed as a sum of (two or more) distinct terms of the sequence, as well as infinitely many of those which cannot be expressed in such a way.  (Serbia)

Solution: Call a nonnegative integer representable if it equals the sum of several (possibly 0 or 1) distinct terms of the sequence. We say that two nonnegative integers $b$ and $c$ are equivalent (written as $b \sim c$ ) if they are either both representable or both non-representable.  One can easily compute  $$ S_{n-1}:=a_{0}+\cdots+a_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3 . $$  Indeed, we have $S_{n}-S_{n-1}=2^{n}+2^{\lfloor n / 2\rfloor}=a_{n}$ so we can use the induction. In particular, $S_{2 k-1}=2^{2 k}+2^{k+1}-3$.  Note that, if $n \geqslant 3$, then $2^{[n / 2]} \geqslant 2^{2}>3$, so  $$ S_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3>2^{n}+2^{\lfloor n / 2\rfloor}=a_{n} . $$  Also notice that $S_{n-1}-a_{n}=2^{[n / 2]}-3<a_{n}$.  The main tool of the solution is the following claim.  Claim 1. Assume that $b$ is a positive integer such that $S_{n-1}-a_{n}<b<a_{n}$ for some $n \geqslant 3$. Then $b \sim S_{n-1}-b$.  Proof. As seen above, we have $S_{n-1}>a_{n}$. Denote $c=S_{n-1}-b$; then $S_{n-1}-a_{n}<c<a_{n}$, so the roles of $b$ and $c$ are symmetrical.  Assume that $b$ is representable. The representation cannot contain $a_{i}$ with $i \geqslant n$, since $b<a_{n}$. So $b$ is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. The converse is obtained by swapping $b$ and $c$.  We also need the following version of this claim.  Claim 2. For any $n \geqslant 3$, the number $a_{n}$ can be represented as a sum of two or more distinct terms of the sequence if and only if $S_{n-1}-a_{n}=2^{[n / 2]}-3$ is representable.  Proof. Denote $c=S_{n-1}-a_{n}<a_{n}$. If $a_{n}$ satisfies the required condition, then it is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. Conversely, if $c$ is representable, then its representation consists only of the numbers from $\left\{a_{0}, \ldots, a_{n-1}\right\}$, so $a_{n}$ is the sum of the complement.  By Claim 2, in order to prove the problem statement, it suffices to find infinitely many representable numbers of the form $2^{t}-3$, as well as infinitely many non-representable ones.  Claim 3. For every $t \geqslant 3$, we have $2^{t}-3 \sim 2^{4 t-6}-3$, and $2^{4 t-6}-3>2^{t}-3$.  Proof. The inequality follows from $t \geqslant 3$. In order to prove the equivalence, we apply Claim 1 twice in the following manner.  First, since $S_{2 t-3}-a_{2 t-2}=2^{t-1}-3<2^{t}-3<2^{2 t-2}+2^{t-1}=a_{2 t-2}$, by Claim 1 we have $2^{t}-3 \sim S_{2 t-3}-\left(2^{t}-3\right)=2^{2 t-2}$.  Second, since $S_{4 t-7}-a_{4 t-6}=2^{2 t-3}-3<2^{2 t-2}<2^{4 t-6}+2^{2 t-3}=a_{4 t-6}$, by Claim 1 we have $2^{2 t-2} \sim S_{4 t-7}-2^{2 t-2}=2^{4 t-6}-3$.  Therefore, $2^{t}-3 \sim 2^{2 t-2} \sim 2^{4 t-6}-3$, as required.  Now it is easy to find the required numbers. Indeed, the number $2^{3}-3=5=a_{0}+a_{1}$ is representable, so Claim 3 provides an infinite sequence of representable numbers  $$ 2^{3}-3 \sim 2^{6}-3 \sim 2^{18}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots \text {. } $$  On the other hand, the number $2^{7}-3=125$ is non-representable (since by Claim 1 we have $125 \sim S_{6}-125=24 \sim S_{4}-24=17 \sim S_{3}-17=4$ which is clearly non-representable). So Claim 3 provides an infinite sequence of non-representable numbers  $$ 2^{7}-3 \sim 2^{22}-3 \sim 2^{82}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots . $$


Solution: We keep the notion of representability and the notation $S_{n}$ from the previous solution. We say that an index $n$ is good if $a_{n}$ writes as a sum of smaller terms from the sequence $a_{0}, a_{1}, \ldots$. Otherwise we say it is bad. We must prove that there are infinitely many good indices, as well as infinitely many bad ones.  Lemma 1. If $m \geqslant 0$ is an integer, then $4^{m}$ is representable if and only if either of $2 m+1$ and $2 m+2$ is good.  Proof. The case $m=0$ is obvious, so we may assume that $m \geqslant 1$. Let $n=2 m+1$ or $2 m+2$. Then $n \geqslant 3$. We notice that  $$ S_{n-1}<a_{n-2}+a_{n} $$  The inequality writes as $2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3<2^{n}+2^{\lfloor n / 2\rfloor}+2^{n-2}+2^{\lfloor n / 2\rfloor-1}$, i.e. as $2^{\lceil n / 2\rceil}<$ $2^{n-2}+2^{\lfloor n / 2\rfloor-1}+3$. If $n \geqslant 4$, then $n / 2 \leqslant n-2$, so $\lceil n / 2\rceil \leqslant n-2$ and $2^{\lceil n / 2\rceil} \leqslant 2^{n-2}$. For $n=3$ the inequality verifies separately.  If $n$ is good, then $a_{n}$ writes as $a_{n}=a_{i_{1}}+\cdots+a_{i_{r}}$, where $r \geqslant 2$ and $i_{1}<\cdots<i_{r}<n$. Then $i_{r}=n-1$ and $i_{r-1}=n-2$, for if $n-1$ or $n-2$ is missing from the sequence $i_{1}, \ldots, i_{r}$, then $a_{i_{1}}+\cdots+a_{i_{r}} \leqslant a_{0}+\cdots+a_{n-3}+a_{n-1}=S_{n-1}-a_{n-2}<a_{n}$. Thus, if $n$ is good, then both $a_{n}-a_{n-1}$ and $a_{n}-a_{n-1}-a_{n-2}$ are representable.  We now consider the cases $n=2 m+1$ and $n=2 m+2$ separately.  If $n=2 m+1$, then $a_{n}-a_{n-1}=a_{2 m+1}-a_{2 m}=\left(2^{2 m+1}+2^{m}\right)-\left(2^{2 m}+2^{m}\right)=2^{2 m}$. So we proved that, if $2 m+1$ is good, then $2^{2 m}$ is representable. Conversely, if $2^{2 m}$ is representable, then $2^{2 m}<a_{2 m}$, so $2^{2 m}$ is a sum of some distinct terms $a_{i}$ with $i<2 m$. It follows that $a_{2 m+1}=a_{2 m}+2^{2 m}$ writes as $a_{2 m}$ plus a sum of some distinct terms $a_{i}$ with $i<2 m$. Hence $2 m+1$ is good.  If $n=2 m+2$, then $a_{n}-a_{n-1}-a_{n-2}=a_{2 m+2}-a_{2 m+1}-a_{2 m}=\left(2^{2 m+2}+2^{m+1}\right)-\left(2^{2 m+1}+\right.$ $\left.2^{m}\right)-\left(2^{2 m}+2^{m}\right)=2^{2 m}$. So we proved that, if $2 m+2$ is good, then $2^{2 m}$ is representable. Conversely, if $2^{2 m}$ is representable, then, as seen in the previous case, it writes as a sum of some distinct terms $a_{i}$ with $i<2 m$. Hence $a_{2 m+2}=a_{2 m+1}+a_{2 m}+2^{2 m}$ writes as $a_{2 m+1}+a_{2 m}$ plus a sum of some distinct terms $a_{i}$ with $i<2 m$. Thus $2 m+2$ is good.  Lemma 2. If $k \geqslant 2$, then $2^{4 k-2}$ is representable if and only if $2^{k+1}$ is representable.  In particular, if $s \geqslant 2$, then $4^{s}$ is representable if and only if $4^{4 s-3}$ is representable. Also, $4^{4 s-3}>4^{s}$.  Proof. We have $2^{4 k-2}<a_{4 k-2}$, so in a representation of $2^{4 k-2}$ we can have only terms $a_{i}$ with $i \leqslant 4 k-3$. Notice that  $$ a_{0}+\cdots+a_{4 k-3}=2^{4 k-2}+2^{2 k}-3<2^{4 k-2}+2^{2 k}+2^{k}=2^{4 k-2}+a_{2 k} . $$  Hence, any representation of $2^{4 k-2}$ must contain all terms from $a_{2 k}$ to $a_{4 k-3}$. (If any of these terms is missing, then the sum of the remaining ones is $\leqslant\left(a_{0}+\cdots+a_{4 k-3}\right)-a_{2 k}<2^{4 k-2}$. Hence, if $2^{4 k-2}$ is representable, then $2^{4 k-2}-\sum_{i=2 k}^{4 k-3} a_{i}$ is representable. But  $2^{4 k-2}-\sum_{i=2 k}^{4 k-3} a_{i}=2^{4 k-2}-\left(S_{4 k-3}-S_{2 k-1}\right)=2^{4 k-2}-\left(2^{4 k-2}+2^{2 k}-3\right)+\left(2^{2 k}+2^{k+1}-3\right)=2^{k+1}$.  So, if $2^{4 k-2}$ is representable, then $2^{k+1}$ is representable. Conversely, if $2^{k+1}$ is representable, then $2^{k+1}<2^{2 k}+2^{k}=a_{2 k}$, so $2^{k+1}$ writes as a sum of some distinct terms $a_{i}$ with $i<2 k$. It follows that $2^{4 k-2}=\sum_{i=2 k}^{4 k-3} a_{i}+2^{k+1}$ writes as $a_{4 k-3}+a_{4 k-4}+\cdots+a_{2 k}$ plus the sum of some distinct terms $a_{i}$ with $i<2 k$. Hence $2^{4 k-2}$ is representable.  For the second statement, if $s \geqslant 2$, then we just take $k=2 s-1$ and we notice that $2^{k+1}=4^{s}$ and $2^{4 k-2}=4^{4 s-3}$. Also, $s \geqslant 2$ implies that $4 s-3>s$.  Now $4^{2}=a_{2}+a_{3}$ is representable, whereas $4^{6}=4096$ is not. Indeed, note that $4^{6}=2^{12}<a_{12}$, so the only available terms for a representation are $a_{0}, \ldots, a_{11}$, i.e., $2,3,6,10,20,36,72$, 136, 272, 528, 1056, 2080. Their sum is $S_{11}=4221$, which exceeds 4096 by 125 . Then any representation of 4096 must contain all the terms from $a_{0}, \ldots, a_{11}$ that are greater that 125 , i.e., $136,272,528,1056,2080$. Their sum is 4072 . Since $4096-4072=24$ and 24 is clearly not representable, 4096 is non-representable as well.  Starting with these values of $m$, by using Lemma 2, we can obtain infinitely many representable powers of 4, as well as infinitely many non-representable ones. By Lemma 1, this solves our problem.  This page is intentionally left blank


Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of positive integers such that  $$ \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}} $$  is an integer for all $n \geqslant k$, where $k$ is some positive integer. Prove that there exists a positive integer $m$ such that $a_{n}=a_{n+1}$ for all $n \geqslant m$.  (Mongolia)

Solution: The argument hinges on the following two facts: Let $a, b, c$ be positive integers such that $N=b / c+(c-b) / a$ is an integer.  (1) If $\operatorname{gcd}(a, c)=1$, then $c$ divides $b$; and  (2) If $\operatorname{gcd}(a, b, c)=1$, then $\operatorname{gcd}(a, b)=1$.  To prove (1), write $a b=c(a N+b-c)$. Since $\operatorname{gcd}(a, c)=1$, it follows that $c$ divides $b$. To prove (2), write $c^{2}-b c=a(c N-b)$ to infer that $a$ divides $c^{2}-b c$. Letting $d=\operatorname{gcd}(a, b)$, it follows that $d$ divides $c^{2}$, and since the two are relatively prime by hypothesis, $d=1$.  Now, let $s_{n}=a_{1} / a_{2}+a_{2} / a_{3}+\cdots+a_{n-1} / a_{n}+a_{n} / a_{1}$, let $\delta_{n}=\operatorname{gcd}\left(a_{1}, a_{n}, a_{n+1}\right)$ and write  $$ s_{n+1}-s_{n}=\frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}-a_{n}}{a_{1}}=\frac{a_{n} / \delta_{n}}{a_{n+1} / \delta_{n}}+\frac{a_{n+1} / \delta_{n}-a_{n} / \delta_{n}}{a_{1} / \delta_{n}} . $$  Let $n \geqslant k$. Since $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}, a_{n+1} / \delta_{n}\right)=1$, it follows by (2) that $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=1$. Let $d_{n}=\operatorname{gcd}\left(a_{1}, a_{n}\right)$. Then $d_{n}=\delta_{n} \cdot \operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=\delta_{n}$, so $d_{n}$ divides $a_{n+1}$, and therefore $d_{n}$ divides $d_{n+1}$.  Consequently, from some rank on, the $d_{n}$ form a nondecreasing sequence of integers not exceeding $a_{1}$, so $d_{n}=d$ for all $n \geqslant \ell$, where $\ell$ is some positive integer.  Finally, since $\operatorname{gcd}\left(a_{1} / d, a_{n+1} / d\right)=1$, it follows by (1) that $a_{n+1} / d$ divides $a_{n} / d$, so $a_{n} \geqslant a_{n+1}$ for all $n \geqslant \ell$. The conclusion follows.


Solution: We use the same notation $s_{n}$. This time, we explore the exponents of primes in the prime factorizations of the $a_{n}$ for $n \geqslant k$.  To start, for every $n \geqslant k$, we know that the number  $$ s_{n+1}-s_{n}=\frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}}{a_{1}}-\frac{a_{n}}{a_{1}} $$  is integer. Multiplying it by $a_{1}$ we obtain that $a_{1} a_{n} / a_{n+1}$ is integer as well, so that $a_{n+1} \mid a_{1} a_{n}$. This means that $a_{n} \mid a_{1}^{n-k} a_{k}$, so all prime divisors of $a_{n}$ are among those of $a_{1} a_{k}$. There are finitely many such primes; therefore, it suffices to prove that the exponent of each of them in the prime factorization of $a_{n}$ is eventually constant.  Choose any prime $p \mid a_{1} a_{k}$. Recall that $v_{p}(q)$ is the standard notation for the exponent of $p$ in the prime factorization of a nonzero rational number $q$. Say that an index $n \geqslant k$ is large if $v_{p}\left(a_{n}\right) \geqslant v_{p}\left(a_{1}\right)$. We separate two cases.  Case 1: There exists a large index $n$.  If $v_{p}\left(a_{n+1}\right)<v_{p}\left(a_{1}\right)$, then $v_{p}\left(a_{n} / a_{n+1}\right)$ and $v_{p}\left(a_{n} / a_{1}\right)$ are nonnegative, while $v_{p}\left(a_{n+1} / a_{1}\right)<0$; hence $(*)$ cannot be an integer. This contradiction shows that index $n+1$ is also large.  On the other hand, if $v_{p}\left(a_{n+1}\right)>v_{p}\left(a_{n}\right)$, then $v_{p}\left(a_{n} / a_{n+1}\right)<0$, while $v_{p}\left(\left(a_{n+1}-a_{n}\right) / a_{1}\right) \geqslant 0$, so $(*)$ is not integer again. Thus, $v_{p}\left(a_{1}\right) \leqslant v_{p}\left(a_{n+1}\right) \leqslant v_{p}\left(a_{n}\right)$.  The above arguments can now be applied successively to indices $n+1, n+2, \ldots$, showing that all the indices greater than $n$ are large, and the sequence $v_{p}\left(a_{n}\right), v_{p}\left(a_{n+1}\right), v_{p}\left(a_{n+2}\right), \ldots$ is nonincreasing — hence eventually constant.  Case 2: There is no large index.  We have $v_{p}\left(a_{1}\right)>v_{p}\left(a_{n}\right)$ for all $n \geqslant k$. If we had $v_{p}\left(a_{n+1}\right)<v_{p}\left(a_{n}\right)$ for some $n \geqslant k$, then $v_{p}\left(a_{n+1} / a_{1}\right)<v_{p}\left(a_{n} / a_{1}\right)<0<v_{p}\left(a_{n} / a_{n+1}\right)$ which would also yield that $(*)$ is not integer. Therefore, in this case the sequence $v_{p}\left(a_{k}\right), v_{p}\left(a_{k+1}\right), v_{p}\left(a_{k+2}\right), \ldots$ is nondecreasing and bounded by $v_{p}\left(a_{1}\right)$ from above; hence it is also eventually constant.  Comment. Given any positive odd integer $m$, consider the $m$-tuple $\left(2,2^{2}, \ldots, 2^{m-1}, 2^{m}\right)$. Appending an infinite string of 1 's to this $m$-tuple yields an eventually constant sequence of integers satisfying the condition in the statement, and shows that the rank from which the sequence stabilises may be arbitrarily large.  There are more sophisticated examples. The solution to part (b) of 10532, Amer. Math. Monthly, Vol. 105 No. 8 (Oct. 1998), 775-777 (available at https://www.jstor.org/stable/2589009), shows that, for every integer $m \geqslant 5$, there exists an $m$-tuple $\left(a_{1}, a_{2}, \ldots, a_{m}\right)$ of pairwise distinct positive integers such that $\operatorname{gcd}\left(a_{1}, a_{2}\right)=\operatorname{gcd}\left(a_{2}, a_{3}\right)=\cdots=\operatorname{gcd}\left(a_{m-1}, a_{m}\right)=\operatorname{gcd}\left(a_{m}, a_{1}\right)=1$, and the sum $a_{1} / a_{2}+a_{2} / a_{3}+\cdots+a_{m-1} / a_{m}+a_{m} / a_{1}$ is an integer. Letting $a_{m+k}=a_{1}, k=1,2, \ldots$, extends such an $m$-tuple to an eventually constant sequence of positive integers satisfying the condition in the statement of the problem at hand.  Here is the example given by the proposers of 10532. Let $b_{1}=2$, let $b_{k+1}=1+b_{1} \cdots b_{k}=$ $1+b_{k}\left(b_{k}-1\right), k \geqslant 1$, and set $B_{m}=b_{1} \cdots b_{m-4}=b_{m-3}-1$. The $m$-tuple $\left(a_{1}, a_{2}, \ldots, a_{m}\right)$ defined below satisfies the required conditions:  $$ \begin{aligned} a_{1}=1, \quad a_{2} & =\left(8 B_{m}+1\right) B_{m}+8, \quad a_{3}=8 B_{m}+1, \quad a_{k}=b_{m-k} \quad \text { for } 4 \leqslant k \leqslant m-1, \\ & a_{m}=\frac{a_{2}}{2} \cdot a_{3} \cdot \frac{B_{m}}{2}=\left(\frac{1}{2}\left(8 B_{m}+1\right) B_{m}+4\right) \cdot\left(8 B_{m}+1\right) \cdot \frac{B_{m}}{2} . \end{aligned} $$  It is readily checked that $a_{1}<a_{m-1}<a_{m-2}<\cdots<a_{3}<a_{2}<a_{m}$. For further details we refer to the solution mentioned above. Acquaintance with this example (or more elaborated examples derived from) offers no advantage in tackling the problem.


Four positive integers $x, y, z$, and $t$ satisfy the relations  $$ x y-z t=x+y=z+t $$  Is it possible that both $x y$ and $z t$ are perfect squares?  (Russia)

Solution: Arguing indirectly, assume that $x y=a^{2}$ and $z t=c^{2}$ with $a, c>0$.  Suppose that the number $x+y=z+t$ is odd. Then $x$ and $y$ have opposite parity, as well as $z$ and $t$. This means that both $x y$ and $z t$ are even, as well as $x y-z t=x+y$; a contradiction. Thus, $x+y$ is even, so the number $s=\frac{x+y}{2}=\frac{z+t}{2}$ is a positive integer.  Next, we set $b=\frac{|x-y|}{2}, d=\frac{|z-t|}{2}$. Now the problem conditions yield  $$ s^{2}=a^{2}+b^{2}=c^{2}+d^{2} $$  and  $$ 2 s=a^{2}-c^{2}=d^{2}-b^{2} $$  (the last equality in (2) follows from (1)). We readily get from (2) that $a, d>0$.  In the sequel we will use only the relations (1) and (2), along with the fact that $a, d, s$ are positive integers, while $b$ and $c$ are nonnegative integers, at most one of which may be zero. Since both relations are symmetric with respect to the simultaneous swappings $a \leftrightarrow d$ and $b \leftrightarrow c$, we assume, without loss of generality, that $b \geqslant c$ (and hence $b>0$ ). Therefore, $d^{2}=2 s+b^{2}>c^{2}$, whence  $$ d^{2}>\frac{c^{2}+d^{2}}{2}=\frac{s^{2}}{2} $$  On the other hand, since $d^{2}-b^{2}$ is even by (2), the numbers $b$ and $d$ have the same parity, so $0<b \leqslant d-2$. Therefore,  $$ 2 s=d^{2}-b^{2} \geqslant d^{2}-(d-2)^{2}=4(d-1), \quad \text { i.e., } \quad d \leqslant \frac{s}{2}+1 $$  Combining (3) and (4) we obtain  $$ 2 s^{2}<4 d^{2} \leqslant 4\left(\frac{s}{2}+1\right)^{2}, \quad \text { or } \quad(s-2)^{2}<8 $$  which yields $s \leqslant 4$.  Finally, an easy check shows that each number of the form $s^{2}$ with $1 \leqslant s \leqslant 4$ has a unique representation as a sum of two squares, namely $s^{2}=s^{2}+0^{2}$. Thus, (1) along with $a, d>0$ imply $b=c=0$, which is impossible.


Solution: We start with a complete description of all 4-tuples $(x, y, z, t)$ of positive integers satisfying (*). As in the solution above, we notice that the numbers  $$ s=\frac{x+y}{2}=\frac{z+t}{2}, \quad p=\frac{x-y}{2}, \quad \text { and } \quad q=\frac{z-t}{2} $$  are integers (we may, and will, assume that $p, q \geqslant 0$ ). We have  $$ 2 s=x y-z t=(s+p)(s-p)-(s+q)(s-q)=q^{2}-p^{2} $$  so $p$ and $q$ have the same parity, and $q>p$.  Set now $k=\frac{q-p}{2}, \ell=\frac{q+p}{2}$. Then we have $s=\frac{q^{2}-p^{2}}{2}=2 k \ell$ and hence  $$ \begin{array}{rlrl} x & =s+p=2 k \ell-k+\ell, & y & =s-p=2 k \ell+k-\ell, \\ z & =s+q=2 k \ell+k+\ell, & t=s-q=2 k \ell-k-\ell . \end{array} $$  Recall here that $\ell \geqslant k>0$ and, moreover, $(k, \ell) \neq(1,1)$, since otherwise $t=0$.  Assume now that both $x y$ and $z t$ are squares. Then $x y z t$ is also a square. On the other hand, we have  $$ \begin{aligned} x y z t=(2 k \ell-k+\ell) & (2 k \ell+k-\ell)(2 k \ell+k+\ell)(2 k \ell-k-\ell) \\ & =\left(4 k^{2} \ell^{2}-(k-\ell)^{2}\right)\left(4 k^{2} \ell^{2}-(k+\ell)^{2}\right)=\left(4 k^{2} \ell^{2}-k^{2}-\ell^{2}\right)^{2}-4 k^{2} \ell^{2} \end{aligned} $$  Denote $D=4 k^{2} \ell^{2}-k^{2}-\ell^{2}>0$. From (6) we get $D^{2}>x y z t$. On the other hand,  $$ \begin{array}{r} (D-1)^{2}=D^{2}-2\left(4 k^{2} \ell^{2}-k^{2}-\ell^{2}\right)+1=\left(D^{2}-4 k^{2} \ell^{2}\right)-\left(2 k^{2}-1\right)\left(2 \ell^{2}-1\right)+2 \\ =x y z t-\left(2 k^{2}-1\right)\left(2 \ell^{2}-1\right)+2<x y z t \end{array} $$  since $\ell \geqslant 2$ and $k \geqslant 1$. Thus $(D-1)^{2}<x y z t<D^{2}$, and $x y z t$ cannot be a perfect square; a contradiction.  Comment. The first part of Solution 2 shows that all 4-tuples of positive integers $x \geqslant y, z \geqslant t$ satisfying (*) have the form (5), where $\ell \geqslant k>0$ and $\ell \geqslant 2$. The converse is also true: every pair of positive integers $\ell \geqslant k>0$, except for the pair $k=\ell=1$, generates via (5) a 4-tuple of positive integers satisfying $(*)$.


Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$.  (Mexico)

Solution: For every positive integer $m$, define $S_{m}=\{n: m \mid f(n)\}$.  Lemma. If the set $S_{m}$ is infinite, then $S_{m}=\{d, 2 d, 3 d, \ldots\}=d \cdot \mathbb{Z}_{>0}$ for some positive integer $d$. Proof. Let $d=\min S_{m}$; the definition of $S_{m}$ yields $m \mid f(d)$.  Whenever $n \in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \mid f(n-d)$ and therefore $n-d \in S_{m}$. Let $r \leqslant d$ be the least positive integer with $n \equiv r(\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \ldots, r \in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \mid n$.  Starting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$.  The solution for the problem will be split into two cases.  Case 1: The function $f$ is bounded.  Call a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$.  Let $p_{1}, \ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \cdot \mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number  $$ n=N d_{1} d_{2} \cdots d_{k}+1 $$  Due to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \in S_{p_{i}}$, and therefore $d_{i} \mid n$. But $n \equiv 1\left(\bmod d_{i}\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \cdot \mathbb{Z}_{>0}=\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$.  Case 2: $f$ is unbounded.  We prove that $f(1)$ divides all $f(n)$.  Let $a=f(1)$. Since $1 \in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set.  Call a positive integer $p$ a peak if $f(p)>\max (f(1), \ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}<p_{2}<\ldots$ be the sequence of all peaks, and let $h_{k}=f\left(p_{k}\right)$. Notice that for any peak $p_{i}$ and for any $k<p_{i}$, we have $f\left(p_{i}\right) \mid f(k)+f\left(p_{i}-k\right)<$ $2 f\left(p_{i}\right)$, hence  $$ f(k)+f\left(p_{i}-k\right)=f\left(p_{i}\right)=h_{i} . $$  By the pigeonhole principle, among the numbers $h_{1}, h_{2}, \ldots$ there are infinitely many that are congruent modulo $a$. Let $k_{0}<k_{1}<k_{2}<\ldots$ be an infinite sequence of positive integers such that $h_{k_{0}} \equiv h_{k_{1}} \equiv \ldots(\bmod a)$. Notice that  $$ f\left(p_{k_{i}}-p_{k_{0}}\right)=f\left(p_{k_{i}}\right)-f\left(p_{k_{0}}\right)=h_{k_{i}}-h_{k_{0}} \equiv 0 \quad(\bmod a), $$  so $p_{k_{i}}-p_{k_{0}} \in S_{a}$ for all $i=1,2, \ldots$ This provides infinitely many elements in $S_{a}$.  Hence, $S_{a}$ is an infinite set, and therefore $f(1)=a$ divides $f(n)$ for every $n$.  Comment. As an extension of the solution above, it can be proven that if $f$ is not bounded then $f(n)=a n$ with $a=f(1)$.  Take an arbitrary positive integer $n$; we will show that $f(n+1)=f(n)+a$. Then it follows by induction that $f(n)=a n$.  Take a peak $p$ such that $p>n+2$ and $h=f(p)>f(n)+2 a$. By (1) we have $f(p-1)=$ $f(p)-f(1)=h-a$ and $f(n+1)=f(p)-f(p-n-1)=h-f(p-n-1)$. From $h-a=f(p-1) \mid$ $f(n)+f(p-n-1)<f(n)+h<2(h-a)$ we get $f(n)+f(p-n-1)=h-a$. Then  $$ f(n+1)-f(n)=(h-f(p-n-1))-(h-a-f(p-n-1))=a . $$  On the other hand, there exists a wide family of bounded functions satisfying the required properties. Here we present a few examples:  $$ f(n)=c ; \quad f(n)=\left\{\begin{array}{ll} 2 c & \text { if } n \text { is even } \\ c & \text { if } n \text { is odd; } \end{array} \quad f(n)= \begin{cases}2018 c & \text { if } n \leqslant 2018 \\ c & \text { if } n>2018\end{cases}\right. $$


Solution: Let $d_{n}=\operatorname{gcd}(f(n), f(1))$. From $d_{n+1} \mid f(1)$ and $d_{n+1}|f(n+1)| f(n)+f(1)$, we can see that $d_{n+1} \mid f(n)$; then $d_{n+1} \mid \operatorname{gcd}(f(n), f(1))=d_{n}$. So the sequence $d_{1}, d_{2}, \ldots$ is nonincreasing in the sense that every element is a divisor of the previous elements. Let $d=\min \left(d_{1}, d_{2}, \ldots\right)=\operatorname{gcd}\left(d_{1} \cdot d_{2}, \ldots\right)=\operatorname{gcd}(f(1), f(2), \ldots)$; we have to prove $d \geqslant 2$.  For the sake of contradiction, suppose that the statement is wrong, so $d=1$; that means there is some index $n_{0}$ such that $d_{n}=1$ for every $n \geqslant n_{0}$, i.e., $f(n)$ is coprime with $f(1)$.  Claim 1. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right) \leqslant 2^{k}$.  Proof. By the condition, $f(2 n) \mid 2 f(n)$; a trivial induction yields $f\left(2^{k}\right) \mid 2^{k} f(1)$. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right)$ is coprime with $f(1)$, so $f\left(2^{k}\right)$ is a divisor of $2^{k}$.  Claim 2. There is a constant $C$ such that $f(n)<n+C$ for every $n$.  Proof. Take the first power of 2 which is greater than or equal to $n_{0}$ : let $K=2^{k} \geqslant n_{0}$. By Claim 1, we have $f(K) \leqslant K$. Notice that $f(n+K) \mid f(n)+f(K)$ implies $f(n+K) \leqslant$ $f(n)+f(K) \leqslant f(n)+K$. If $n=t K+r$ for some $t \geqslant 0$ and $1 \leqslant r \leqslant K$, then we conclude  $f(n) \leqslant K+f(n-K) \leqslant 2 K+f(n-2 K) \leqslant \ldots \leqslant t K+f(r)<n+\max (f(1), f(2), \ldots, f(K))$,  so the claim is true with $C=\max (f(1), \ldots, f(K))$.  Claim 3. If $a, b \in \mathbb{Z}_{>0}$ are coprime then $\operatorname{gcd}(f(a), f(b)) \mid f(1)$. In particular, if $a, b \geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime.  Proof. Let $d=\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \mid f(1)$.  Without loss of generality, suppose $1<a<b$. Then $d \mid f(a)$ and $d|f(b)| f(a)+f(b-a)$, so $d \mid f(b-a)$. Therefore $d$ divides $\operatorname{gcd}(f(a), f(b-a))$ which is a divisor of $f(1)$ by the induction hypothesis.  Let $p_{1}<p_{2}<\ldots$ be the sequence of all prime numbers; for every $k$, let $q_{k}$ be the lowest power of $p_{k}$ with $q_{k} \geqslant n_{0}$. (Notice that there are only finitely many positive integers with $q_{k} \neq p_{k}$.)  Take a positive integer $N$, and consider the numbers  $$ f(1), f\left(q_{1}\right), f\left(q_{2}\right), \ldots, f\left(q_{N}\right) $$  Here we have $N+1$ numbers, each being greater than 1 , and they are pairwise coprime by Claim 3. Therefore, they have at least $N+1$ different prime divisors in total, and their greatest prime divisor is at least $p_{N+1}$. Hence, $\max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right) \geqslant p_{N+1}$.  Choose $N$ such that $\max \left(q_{1}, \ldots, q_{N}\right)=p_{N}$ (this is achieved if $N$ is sufficiently large), and $p_{N+1}-p_{N}>C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction  $$ p_{N+1} \leqslant \max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right)<\max \left(1+C, q_{1}+C, \ldots, q_{N}+C\right)=p_{N}+C<p_{N+1} $$  which proves the statement.


Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5 n$. Suppose that the sequence  $$ \frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}} $$  forms an arithmetic progression. Prove that the terms of the sequence are equal.

Solution: Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\Delta=\frac{c}{d}$, where $d>0$ and $c, d$ are coprime.  We will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \leqslant i \leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\left(b_{i}\right)<v_{p}(d)$. $\left(v_{p}(x)\right.$ stands for the exponent of $p$ in the prime factorisation of $x$.)  Claim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$.  Proof. Let $\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $\left.p^{\alpha}\right)$ such that $i \not \equiv j(\bmod p)$. Then the least common denominator of $\frac{a_{i}}{b_{i}}$ and $\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\alpha}$. But this is impossible because in their difference, $(i-j) \Delta=\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\alpha}$ divides the denominator $d$.  Claim 2. $d$ has no prime divisors greater than 5 .  Proof. Suppose that $p \geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \ldots, n$, at most $\left\lceil\frac{n}{p}\right\rceil<\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\frac{p-1}{p} n-1$ of $b_{1}, \ldots, b_{n}$. Since these denominators are distinct,  $$ 5 n \geqslant \max \left\{b_{i}: p \mid b_{i}\right\} \geqslant\left(\frac{p-1}{p} n-1\right) p=(p-1)(n-1)-1 \geqslant 6(n-1)-1>5 n, $$  a contradiction.  Claim 3. For every $0 \leqslant k \leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \ldots, b_{k+30}$, at least $\varphi(30)=8$ are divisible by $d$.  Proof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \cdot(3-1) \cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \mid b_{i}$ for every uncovered index $i$.  Claim 4. $|\Delta|<\frac{20}{n-2}$ and $d>\frac{n-2}{20}$.  Proof. From the sequence (1), remove all fractions with $b_{n}<\frac{n}{2}$, There remain at least $\frac{n}{2}$ fractions, and they cannot exceed $\frac{5 n}{n / 2}=10$. So we have at least $\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\frac{10}{n / 2-1}=\frac{20}{n-2}$.  The second inequality follows from $\frac{1}{d} \leqslant \frac{|c|}{d}=|\Delta|$.  Now we have everything to get the final contradiction. By Claim 3, we have $d \mid b_{i}$ for at least $\left\lfloor\frac{n}{30}\right\rfloor \cdot 8$ indices $i$. By Claim 4 , we have $d \geqslant \frac{n-2}{20}$. Therefore,  $$ 5 n \geqslant \max \left\{b_{i}: d \mid b_{i}\right\} \geqslant\left(\left\lfloor\frac{n}{30}\right\rfloor \cdot 8\right) \cdot d>\left(\frac{n}{30}-1\right) \cdot 8 \cdot \frac{n-2}{20}>5 n . $$  Comment 1. It is possible that all terms in (1) are equal, for example with $a_{i}=2 i-1$ and $b_{i}=4 i-2$ we have $\frac{a_{i}}{b_{i}}=\frac{1}{2}$.  Comment 2. The bound $5 n$ in the statement is far from sharp; the solution above can be modified to work for $9 n$. For large $n$, the bound $5 n$ can be replaced by $n^{\frac{3}{2}-\varepsilon}$.  The activities of the Problem Selection Committee were supported by  DEDEMAN  LA FANTANA vine la tine


Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions $f: \mathbb{Q}_{>0} \rightarrow \mathbb{Q}_{>0}$ satisfying  $$ f\left(x^{2} f(y)^{2}\right)=f(x)^{2} f(y) $$  for all $x, y \in \mathbb{Q}_{>0}$.  (Switzerland)  Answer: $f(x)=1$ for all $x \in \mathbb{Q}_{>0}$.

Solution: Take any $a, b \in \mathbb{Q}_{>0}$. By substituting $x=f(a), y=b$ and $x=f(b), y=a$ into $(*)$ we get  $$ f(f(a))^{2} f(b)=f\left(f(a)^{2} f(b)^{2}\right)=f(f(b))^{2} f(a) $$  which yields  $$ \frac{f(f(a))^{2}}{f(a)}=\frac{f(f(b))^{2}}{f(b)} \quad \text { for all } a, b \in \mathbb{Q}_{>0} $$  In other words, this shows that there exists a constant $C \in \mathbb{Q}_{>0}$ such that $f(f(a))^{2}=C f(a)$, or  $$ \left(\frac{f(f(a))}{C}\right)^{2}=\frac{f(a)}{C} \quad \text { for all } a \in \mathbb{Q}_{>0} \text {. } $$  Denote by $f^{n}(x)=\underbrace{f(f(\ldots(f}_{n}(x)) \ldots))$ the $n^{\text {th }}$ iteration of $f$. Equality (1) yields  $$ \frac{f(a)}{C}=\left(\frac{f^{2}(a)}{C}\right)^{2}=\left(\frac{f^{3}(a)}{C}\right)^{4}=\cdots=\left(\frac{f^{n+1}(a)}{C}\right)^{2^{n}} $$  for all positive integer $n$. So, $f(a) / C$ is the $2^{n}$-th power of a rational number for all positive integer $n$. This is impossible unless $f(a) / C=1$, since otherwise the exponent of some prime in the prime decomposition of $f(a) / C$ is not divisible by sufficiently large powers of 2 . Therefore, $f(a)=C$ for all $a \in \mathbb{Q}_{>0}$.  Finally, after substituting $f \equiv C$ into (*) we get $C=C^{3}$, whence $C=1$. So $f(x) \equiv 1$ is the unique function satisfying $(*)$.  Comment 1. There are several variations of the solution above. For instance, one may start with finding $f(1)=1$. To do this, let $d=f(1)$. By substituting $x=y=1$ and $x=d^{2}, y=1$ into (*) we get $f\left(d^{2}\right)=d^{3}$ and $f\left(d^{6}\right)=f\left(d^{2}\right)^{2} \cdot d=d^{7}$. By substituting now $x=1, y=d^{2}$ we obtain $f\left(d^{6}\right)=d^{2} \cdot d^{3}=d^{5}$. Therefore, $d^{7}=f\left(d^{6}\right)=d^{5}$, whence $d=1$.  After that, the rest of the solution simplifies a bit, since we already know that $C=\frac{f(f(1))^{2}}{f(1)}=1$. Hence equation (1) becomes merely $f(f(a))^{2}=f(a)$, which yields $f(a)=1$ in a similar manner.  Comment 2. There exist nonconstant functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$satisfying $(*)$ for all real $x, y>0-$ e.g., $f(x)=\sqrt{x}$.


Find all positive integers $n \geqslant 3$ for which there exist real numbers $a_{1}, a_{2}, \ldots, a_{n}$, $a_{n+1}=a_{1}, a_{n+2}=a_{2}$ such that  $$ a_{i} a_{i+1}+1=a_{i+2} $$  for all $i=1,2, \ldots, n$.  (Slovakia)  Answer: $n$ can be any multiple of 3 .

Solution: For the sake of convenience, extend the sequence $a_{1}, \ldots, a_{n+2}$ to an infinite periodic sequence with period $n$. ( $n$ is not necessarily the shortest period.)  If $n$ is divisible by 3 , then $\left(a_{1}, a_{2}, \ldots\right)=(-1,-1,2,-1,-1,2, \ldots)$ is an obvious solution.  We will show that in every periodic sequence satisfying the recurrence, each positive term is followed by two negative values, and after them the next number is positive again. From this, it follows that $n$ is divisible by 3 .  If the sequence contains two consecutive positive numbers $a_{i}, a_{i+1}$, then $a_{i+2}=a_{i} a_{i+1}+1>1$, so the next value is positive as well; by induction, all numbers are positive and greater than 1 . But then $a_{i+2}=a_{i} a_{i+1}+1 \geqslant 1 \cdot a_{i+1}+1>a_{i+1}$ for every index $i$, which is impossible: our sequence is periodic, so it cannot increase everywhere.  If the number 0 occurs in the sequence, $a_{i}=0$ for some index $i$, then it follows that $a_{i+1}=a_{i-1} a_{i}+1$ and $a_{i+2}=a_{i} a_{i+1}+1$ are two consecutive positive elements in the sequences and we get the same contradiction again.  Notice that after any two consecutive negative numbers the next one must be positive: if $a_{i}<0$ and $a_{i+1}<0$, then $a_{i+2}=a_{1} a_{i+1}+1>1>0$. Hence, the positive and negative numbers follow each other in such a way that each positive term is followed by one or two negative values and then comes the next positive term.  Consider the case when the positive and negative values alternate. So, if $a_{i}$ is a negative value then $a_{i+1}$ is positive, $a_{i+2}$ is negative and $a_{i+3}$ is positive again.  Notice that $a_{i} a_{i+1}+1=a_{i+2}<0<a_{i+3}=a_{i+1} a_{i+2}+1$; by $a_{i+1}>0$ we conclude $a_{i}<a_{i+2}$. Hence, the negative values form an infinite increasing subsequence, $a_{i}<a_{i+2}<a_{i+4}<\ldots$, which is not possible, because the sequence is periodic.  The only case left is when there are consecutive negative numbers in the sequence. Suppose that $a_{i}$ and $a_{i+1}$ are negative; then $a_{i+2}=a_{i} a_{i+1}+1>1$. The number $a_{i+3}$ must be negative. We show that $a_{i+4}$ also must be negative.  Notice that $a_{i+3}$ is negative and $a_{i+4}=a_{i+2} a_{i+3}+1<1<a_{i} a_{i+1}+1=a_{i+2}$, so  $$ a_{i+5}-a_{i+4}=\left(a_{i+3} a_{i+4}+1\right)-\left(a_{i+2} a_{i+3}+1\right)=a_{i+3}\left(a_{i+4}-a_{i+2}\right)>0, $$  therefore $a_{i+5}>a_{i+4}$. Since at most one of $a_{i+4}$ and $a_{i+5}$ can be positive, that means that $a_{i+4}$ must be negative.  Now $a_{i+3}$ and $a_{i+4}$ are negative and $a_{i+5}$ is positive; so after two negative and a positive terms, the next three terms repeat the same pattern. That completes the solution.


Solution: We prove that the shortest period of the sequence must be 3 . Then it follows that $n$ must be divisible by 3 .  Notice that the equation $x^{2}+1=x$ has no real root, so the numbers $a_{1}, \ldots, a_{n}$ cannot be all equal, hence the shortest period of the sequence cannot be 1 .  By applying the recurrence relation for $i$ and $i+1$,  $$ \begin{gathered} \left(a_{i+2}-1\right) a_{i+2}=a_{i} a_{i+1} a_{i+2}=a_{i}\left(a_{i+3}-1\right), \quad \text { so } \\ a_{i+2}^{2}-a_{i} a_{i+3}=a_{i+2}-a_{i} . \end{gathered} $$  By summing over $i=1,2, \ldots, n$, we get  $$ \sum_{i=1}^{n}\left(a_{i}-a_{i+3}\right)^{2}=0 $$  That proves that $a_{i}=a_{i+3}$ for every index $i$, so the sequence $a_{1}, a_{2}, \ldots$ is indeed periodic with period 3. The shortest period cannot be 1 , so it must be 3 ; therefore, $n$ is divisible by 3 .  Comment. By solving the system of equations $a b+1=c, \quad b c+1=a, \quad c a+1=b$, it can be seen that the pattern $(-1,-1,2)$ is repeated in all sequences satisfying the problem conditions.


Given any set $S$ of positive integers, show that at least one of the following two assertions holds:  (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$;  (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.  ## (Luxembourg)

Solution: Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below.  Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$.  Proof. If $x$ is a member of $F_{q}$, then  $$ \sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y} $$  so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then  $$ \sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y} $$  so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$.  Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=\lfloor r x\rfloor$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd.  Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible.  Comment. The solution above can be adapted to show that the problem statement still holds, if the condition $r<1$ in (2) is replaced with $r<\delta$, for an arbitrary positive $\delta$. This yields that, if $S$ does not satisfy (1), then there exist infinitely many positive rational numbers $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.


Solution: A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$.  We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so  $$ s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}} $$  If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2).  Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2).  We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so  $$ \sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}} $$  Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).


Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of real numbers such that $a_{0}=0, a_{1}=1$, and for every $n \geqslant 2$ there exists $1 \leqslant k \leqslant n$ satisfying  $$ a_{n}=\frac{a_{n-1}+\cdots+a_{n-k}}{k} $$  Find the maximal possible value of $a_{2018}-a_{2017}$.  (Belgium)  Answer: The maximal value is $\frac{2016}{2017^{2}}$.

Solution: The claimed maximal value is achieved at  $$ \begin{gathered} a_{1}=a_{2}=\cdots=a_{2016}=1, \quad a_{2017}=\frac{a_{2016}+\cdots+a_{0}}{2017}=1-\frac{1}{2017}, \\ a_{2018}=\frac{a_{2017}+\cdots+a_{1}}{2017}=1-\frac{1}{2017^{2}} . \end{gathered} $$  Now we need to show that this value is optimal. For brevity, we use the notation  $$ S(n, k)=a_{n-1}+a_{n-2}+\cdots+a_{n-k} \quad \text { for nonnegative integers } k \leqslant n \text {. } $$  In particular, $S(n, 0)=0$ and $S(n, 1)=a_{n-1}$. In these terms, for every integer $n \geqslant 2$ there exists a positive integer $k \leqslant n$ such that $a_{n}=S(n, k) / k$.  For every integer $n \geqslant 1$ we define  $$ M_{n}=\max _{1 \leqslant k \leqslant n} \frac{S(n, k)}{k}, \quad m_{n}=\min _{1 \leqslant k \leqslant n} \frac{S(n, k)}{k}, \quad \text { and } \quad \Delta_{n}=M_{n}-m_{n} \geqslant 0 . $$  By definition, $a_{n} \in\left[m_{n}, M_{n}\right]$ for all $n \geqslant 2$; on the other hand, $a_{n-1}=S(n, 1) / 1 \in\left[m_{n}, M_{n}\right]$. Therefore,  $$ a_{2018}-a_{2017} \leqslant M_{2018}-m_{2018}=\Delta_{2018}, $$  and we are interested in an upper bound for $\Delta_{2018}$.  Also by definition, for any $0<k \leqslant n$ we have $k m_{n} \leqslant S(n, k) \leqslant k M_{n}$; notice that these inequalities are also valid for $k=0$.  Claim 1. For every $n>2$, we have $\Delta_{n} \leqslant \frac{n-1}{n} \Delta_{n-1}$.  Proof. Choose positive integers $k, \ell \leqslant n$ such that $M_{n}=S(n, k) / k$ and $m_{n}=S(n, \ell) / \ell$. We have $S(n, k)=a_{n-1}+S(n-1, k-1)$, so  $$ k\left(M_{n}-a_{n-1}\right)=S(n, k)-k a_{n-1}=S(n-1, k-1)-(k-1) a_{n-1} \leqslant(k-1)\left(M_{n-1}-a_{n-1}\right), $$  since $S(n-1, k-1) \leqslant(k-1) M_{n-1}$. Similarly, we get  $$ \ell\left(a_{n-1}-m_{n}\right)=(\ell-1) a_{n-1}-S(n-1, \ell-1) \leqslant(\ell-1)\left(a_{n-1}-m_{n-1}\right) . $$  Since $m_{n-1} \leqslant a_{n-1} \leqslant M_{n-1}$ and $k, \ell \leqslant n$, the obtained inequalities yield  $$ \begin{array}{ll} M_{n}-a_{n-1} \leqslant \frac{k-1}{k}\left(M_{n-1}-a_{n-1}\right) \leqslant \frac{n-1}{n}\left(M_{n-1}-a_{n-1}\right) \quad \text { and } \\ a_{n-1}-m_{n} \leqslant \frac{\ell-1}{\ell}\left(a_{n-1}-m_{n-1}\right) \leqslant \frac{n-1}{n}\left(a_{n-1}-m_{n-1}\right) . \end{array} $$  Therefore,  $$ \Delta_{n}=\left(M_{n}-a_{n-1}\right)+\left(a_{n-1}-m_{n}\right) \leqslant \frac{n-1}{n}\left(\left(M_{n-1}-a_{n-1}\right)+\left(a_{n-1}-m_{n-1}\right)\right)=\frac{n-1}{n} \Delta_{n-1} . $$  Back to the problem, if $a_{n}=1$ for all $n \leqslant 2017$, then $a_{2018} \leqslant 1$ and hence $a_{2018}-a_{2017} \leqslant 0$. Otherwise, let $2 \leqslant q \leqslant 2017$ be the minimal index with $a_{q}<1$. We have $S(q, i)=i$ for all $i=1,2, \ldots, q-1$, while $S(q, q)=q-1$. Therefore, $a_{q}<1$ yields $a_{q}=S(q, q) / q=1-\frac{1}{q}$.  Now we have $S(q+1, i)=i-\frac{1}{q}$ for $i=1,2, \ldots, q$, and $S(q+1, q+1)=q-\frac{1}{q}$. This gives us  $$ m_{q+1}=\frac{S(q+1,1)}{1}=\frac{S(q+1, q+1)}{q+1}=\frac{q-1}{q} \quad \text { and } \quad M_{q+1}=\frac{S(q+1, q)}{q}=\frac{q^{2}-1}{q^{2}} $$  so $\Delta_{q+1}=M_{q+1}-m_{q+1}=(q-1) / q^{2}$. Denoting $N=2017 \geqslant q$ and using Claim 1 for $n=q+2, q+3, \ldots, N+1$ we finally obtain  $$ \Delta_{N+1} \leqslant \frac{q-1}{q^{2}} \cdot \frac{q+1}{q+2} \cdot \frac{q+2}{q+3} \cdots \frac{N}{N+1}=\frac{1}{N+1}\left(1-\frac{1}{q^{2}}\right) \leqslant \frac{1}{N+1}\left(1-\frac{1}{N^{2}}\right)=\frac{N-1}{N^{2}} $$  as required.  Comment 1. One may check that the maximal value of $a_{2018}-a_{2017}$ is attained at the unique sequence, which is presented in the solution above.  Comment 2. An easier question would be to determine the maximal value of $\left|a_{2018}-a_{2017}\right|$. In this version, the answer $\frac{1}{2018}$ is achieved at  $$ a_{1}=a_{2}=\cdots=a_{2017}=1, \quad a_{2018}=\frac{a_{2017}+\cdots+a_{0}}{2018}=1-\frac{1}{2018} . $$  To prove that this value is optimal, it suffices to notice that $\Delta_{2}=\frac{1}{2}$ and to apply Claim 1 obtaining  $$ \left|a_{2018}-a_{2017}\right| \leqslant \Delta_{2018} \leqslant \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{2017}{2018}=\frac{1}{2018} . $$


Solution: We present a different proof of the estimate $a_{2018}-a_{2017} \leqslant \frac{2016}{2017^{2}}$. We keep the same notations of $S(n, k), m_{n}$ and $M_{n}$ from the previous solution.  Notice that $S(n, n)=S(n, n-1)$, as $a_{0}=0$. Also notice that for $0 \leqslant k \leqslant \ell \leqslant n$ we have $S(n, \ell)=S(n, k)+S(n-k, \ell-k)$.  Claim 2. For every positive integer $n$, we have $m_{n} \leqslant m_{n+1}$ and $M_{n+1} \leqslant M_{n}$, so the segment $\left[m_{n+1}, M_{n+1}\right]$ is contained in $\left[m_{n}, M_{n}\right]$.  Proof. Choose a positive integer $k \leqslant n+1$ such that $m_{n+1}=S(n+1, k) / k$. Then we have  $$ k m_{n+1}=S(n+1, k)=a_{n}+S(n, k-1) \geqslant m_{n}+(k-1) m_{n}=k m_{n}, $$  which establishes the first inequality in the Claim. The proof of the second inequality is similar.  Claim 3. For every positive integers $k \geqslant n$, we have $m_{n} \leqslant a_{k} \leqslant M_{n}$.  Proof. By Claim 2, we have $\left[m_{k}, M_{k}\right] \subseteq\left[m_{k-1}, M_{k-1}\right] \subseteq \cdots \subseteq\left[m_{n}, M_{n}\right]$. Since $a_{k} \in\left[m_{k}, M_{k}\right]$, the claim follows.  Claim 4. For every integer $n \geqslant 2$, we have $M_{n}=S(n, n-1) /(n-1)$ and $m_{n}=S(n, n) / n$.  Proof. We use induction on $n$. The base case $n=2$ is routine. To perform the induction step, we need to prove the inequalities  $$ \frac{S(n, n)}{n} \leqslant \frac{S(n, k)}{k} \quad \text { and } \quad \frac{S(n, k)}{k} \leqslant \frac{S(n, n-1)}{n-1} $$  for every positive integer $k \leqslant n$. Clearly, these inequalities hold for $k=n$ and $k=n-1$, as $S(n, n)=S(n, n-1)>0$. In the sequel, we assume that $k<n-1$.  Now the first inequality in (1) rewrites as $n S(n, k) \geqslant k S(n, n)=k(S(n, k)+S(n-k, n-k))$, or, cancelling the terms occurring on both parts, as  $$ (n-k) S(n, k) \geqslant k S(n-k, n-k) \Longleftrightarrow S(n, k) \geqslant k \cdot \frac{S(n-k, n-k)}{n-k} $$  By the induction hypothesis, we have $S(n-k, n-k) /(n-k)=m_{n-k}$. By Claim 3, we get $a_{n-i} \geqslant m_{n-k}$ for all $i=1,2, \ldots, k$. Summing these $k$ inequalities we obtain  $$ S(n, k) \geqslant k m_{n-k}=k \cdot \frac{S(n-k, n-k)}{n-k} $$  as required.  The second inequality in (1) is proved similarly. Indeed, this inequality is equivalent to  $$ \begin{aligned} (n-1) S(n, k) \leqslant k S(n, n-1) & \Longleftrightarrow(n-k-1) S(n, k) \leqslant k S(n-k, n-k-1) \\ & \Longleftrightarrow S(n, k) \leqslant k \cdot \frac{S(n-k, n-k-1)}{n-k-1}=k M_{n-k} ; \end{aligned} $$  the last inequality follows again from Claim 3, as each term in $S(n, k)$ is at most $M_{n-k}$.  Now we can prove the required estimate for $a_{2018}-a_{2017}$. Set $N=2017$. By Claim 4 ,  $$ \begin{aligned} a_{N+1}-a_{N} \leqslant M_{N+1}-a_{N}=\frac{S(N+1, N)}{N}-a_{N} & =\frac{a_{N}+S(N, N-1)}{N}-a_{N} \\ & =\frac{S(N, N-1)}{N}-\frac{N-1}{N} \cdot a_{N} . \end{aligned} $$  On the other hand, the same Claim yields  $$ a_{N} \geqslant m_{N}=\frac{S(N, N)}{N}=\frac{S(N, N-1)}{N} . $$  Noticing that each term in $S(N, N-1)$ is at most 1 , so $S(N, N-1) \leqslant N-1$, we finally obtain  $$ a_{N+1}-a_{N} \leqslant \frac{S(N, N-1)}{N}-\frac{N-1}{N} \cdot \frac{S(N, N-1)}{N}=\frac{S(N, N-1)}{N^{2}} \leqslant \frac{N-1}{N^{2}} . $$  Comment 1. Claim 1 in Solution 1 can be deduced from Claims 2 and 4 in Solution 2.  By Claim 4 we have $M_{n}=\frac{S(n, n-1)}{n-1}$ and $m_{n}=\frac{S(n, n)}{n}=\frac{S(n, n-1)}{n}$. It follows that $\Delta_{n}=M_{n}-m_{n}=$ $\frac{S(n, n-1)}{(n-1) n}$ and so $M_{n}=n \Delta_{n}$ and $m_{n}=(n-1) \Delta_{n}$  Similarly, $M_{n-1}=(n-1) \Delta_{n-1}$ and $m_{n-1}=(n-2) \Delta_{n-1}$. Then the inequalities $m_{n-1} \leqslant m_{n}$ and $M_{n} \leqslant M_{n-1}$ from Claim 2 write as $(n-2) \Delta_{n-1} \leqslant(n-1) \Delta_{n}$ and $n \Delta_{n} \leqslant(n-1) \Delta_{n-1}$. Hence we have the double inequality  $$ \frac{n-2}{n-1} \Delta_{n-1} \leqslant \Delta_{n} \leqslant \frac{n-1}{n} \Delta_{n-1} $$  Comment 2. Both solutions above discuss the properties of an arbitrary sequence satisfying the problem conditions. Instead, one may investigate only an optimal sequence which maximises the value of $a_{2018}-a_{2017}$. Here we present an observation which allows to simplify such investigation - for instance, the proofs of Claim 1 in Solution 1 and Claim 4 in Solution 2.  The sequence $\left(a_{n}\right)$ is uniquely determined by choosing, for every $n \geqslant 2$, a positive integer $k(n) \leqslant n$ such that $a_{n}=S(n, k(n)) / k(n)$. Take an arbitrary $2 \leqslant n_{0} \leqslant 2018$, and assume that all such integers $k(n)$, for $n \neq n_{0}$, are fixed. Then, for every $n$, the value of $a_{n}$ is a linear function in $a_{n_{0}}$ (whose possible values constitute some discrete subset of $\left[m_{n_{0}}, M_{n_{0}}\right]$ containing both endpoints). Hence, $a_{2018}-a_{2017}$ is also a linear function in $a_{n_{0}}$, so it attains its maximal value at one of the endpoints of the segment $\left[m_{n_{0}}, M_{n_{0}}\right]$.  This shows that, while dealing with an optimal sequence, we may assume $a_{n} \in\left\{m_{n}, M_{n}\right\}$ for all $2 \leqslant n \leqslant 2018$. Now one can easily see that, if $a_{n}=m_{n}$, then $m_{n+1}=m_{n}$ and $M_{n+1} \leqslant \frac{m_{n}+n M_{n}}{n+1} ;$ similar estimates hold in the case $a_{n}=M_{n}$. This already establishes Claim 1, and simplifies the inductive proof of Claim 4, both applied to an optimal sequence.


Determine all functions $f:(0, \infty) \rightarrow \mathbb{R}$ satisfying  $$ \left(x+\frac{1}{x}\right) f(y)=f(x y)+f\left(\frac{y}{x}\right) $$  for all $x, y>0$.  (South Korea)  Answer: $f(x)=C_{1} x+\frac{C_{2}}{x}$ with arbitrary constants $C_{1}$ and $C_{2}$.

Solution: Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\left(t^{2}\right)$, $f(a t)$ and $f\left(a^{2} t^{2}\right)$, the relation (1) provides a system of linear equations:  $$ \begin{array}{llll} x=y=t: & \left(t+\frac{1}{t}\right) f(t) & =f\left(t^{2}\right)+f(1) \\ x=\frac{t}{a}, y=a t: & \left(\frac{t}{a}+\frac{a}{t}\right) f(a t) & =f\left(t^{2}\right)+f\left(a^{2}\right) \\ x=a^{2} t, y=t: & \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t) & =f\left(a^{2} t^{2}\right)+f\left(\frac{1}{a^{2}}\right) \\ x=y=a t: & \left(a t+\frac{1}{a t}\right) f(a t) & =f\left(a^{2} t^{2}\right)+f(1) \end{array} $$  In order to eliminate $f\left(t^{2}\right)$, take the difference of (2a) and (2b); from (2c) and (2d) eliminate $f\left(a^{2} t^{2}\right)$; then by taking a linear combination, eliminate $f(a t)$ as well:  $$ \begin{gathered} \left(t+\frac{1}{t}\right) f(t)-\left(\frac{t}{a}+\frac{a}{t}\right) f(a t)=f(1)-f\left(a^{2}\right) \text { and } \\ \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t)-\left(a t+\frac{1}{a t}\right) f(a t)=f\left(1 / a^{2}\right)-f(1), \text { so } \\ \left(\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)\right) f(t) \\ =\left(a t+\frac{1}{a t}\right)\left(f(1)-f\left(a^{2}\right)\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(f\left(1 / a^{2}\right)-f(1)\right) . \end{gathered} $$  Notice that on the left-hand side, the coefficient of $f(t)$ is nonzero and does not depend on $t$ :  $$ \left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)=a+\frac{1}{a}-\left(a^{3}+\frac{1}{a^{3}}\right)<0 . $$  After dividing by this fixed number, we get  $$ f(t)=C_{1} t+\frac{C_{2}}{t} $$  where the numbers $C_{1}$ and $C_{2}$ are expressed in terms of $a, f(1), f\left(a^{2}\right)$ and $f\left(1 / a^{2}\right)$, and they do not depend on $t$.  The functions of the form (3) satisfy the equation:  $$ \left(x+\frac{1}{x}\right) f(y)=\left(x+\frac{1}{x}\right)\left(C_{1} y+\frac{C_{2}}{y}\right)=\left(C_{1} x y+\frac{C_{2}}{x y}\right)+\left(C_{1} \frac{y}{x}+C_{2} \frac{x}{y}\right)=f(x y)+f\left(\frac{y}{x}\right) . $$


Solution: We start with an observation. If we substitute $x=a \neq 1$ and $y=a^{n}$ in (1), we obtain  $$ f\left(a^{n+1}\right)-\left(a+\frac{1}{a}\right) f\left(a^{n}\right)+f\left(a^{n-1}\right)=0 . $$  For the sequence $z_{n}=a^{n}$, this is a homogeneous linear recurrence of the second order, and its characteristic polynomial is $t^{2}-\left(a+\frac{1}{a}\right) t+1=(t-a)\left(t-\frac{1}{a}\right)$ with two distinct nonzero roots, namely $a$ and $1 / a$. As is well-known, the general solution is $z_{n}=C_{1} a^{n}+C_{2}(1 / a)^{n}$ where the index $n$ can be as well positive as negative. Of course, the numbers $C_{1}$ and $C_{2}$ may depend of the choice of $a$, so in fact we have two functions, $C_{1}$ and $C_{2}$, such that  $$ f\left(a^{n}\right)=C_{1}(a) \cdot a^{n}+\frac{C_{2}(a)}{a^{n}} \quad \text { for every } a \neq 1 \text { and every integer } n \text {. } $$  The relation (4) can be easily extended to rational values of $n$, so we may conjecture that $C_{1}$ and $C_{2}$ are constants, and whence $f(t)=C_{1} t+\frac{C_{2}}{t}$. As it was seen in the previous solution, such functions indeed satisfy (1).  The equation (1) is linear in $f$; so if some functions $f_{1}$ and $f_{2}$ satisfy (1) and $c_{1}, c_{2}$ are real numbers, then $c_{1} f_{1}(x)+c_{2} f_{2}(x)$ is also a solution of (1). In order to make our formulas simpler, define  $$ f_{0}(x)=f(x)-f(1) \cdot x \text {. } $$  This function is another one satisfying (1) and the extra constraint $f_{0}(1)=0$. Repeating the same argument on linear recurrences, we can write $f_{0}(a)=K(a) a^{n}+\frac{L(a)}{a^{n}}$ with some functions $K$ and $L$. By substituting $n=0$, we can see that $K(a)+L(a)=f_{0}(1)=0$ for every $a$. Hence,  $$ f_{0}\left(a^{n}\right)=K(a)\left(a^{n}-\frac{1}{a^{n}}\right) $$  Now take two numbers $a>b>1$ arbitrarily and substitute $x=(a / b)^{n}$ and $y=(a b)^{n}$ in (1):  $$ \begin{aligned} \left(\frac{a^{n}}{b^{n}}+\frac{b^{n}}{a^{n}}\right) f_{0}\left((a b)^{n}\right) & =f_{0}\left(a^{2 n}\right)+f_{0}\left(b^{2 n}\right), \quad \text { so } \\ \left(\frac{a^{n}}{b^{n}}+\frac{b^{n}}{a^{n}}\right) K(a b)\left((a b)^{n}-\frac{1}{(a b)^{n}}\right) & =K(a)\left(a^{2 n}-\frac{1}{a^{2 n}}\right)+K(b)\left(b^{2 n}-\frac{1}{b^{2 n}}\right), \quad \text { or equivalently } \\ K(a b)\left(a^{2 n}-\frac{1}{a^{2 n}}+b^{2 n}-\frac{1}{b^{2 n}}\right) & =K(a)\left(a^{2 n}-\frac{1}{a^{2 n}}\right)+K(b)\left(b^{2 n}-\frac{1}{b^{2 n}}\right) . \end{aligned} $$  By dividing (5) by $a^{2 n}$ and then taking limit with $n \rightarrow+\infty$ we get $K(a b)=K(a)$. Then (5) reduces to $K(a)=K(b)$. Hence, $K(a)=K(b)$ for all $a>b>1$.  Fix $a>1$. For every $x>0$ there is some $b$ and an integer $n$ such that $1<b<a$ and $x=b^{n}$. Then  $$ f_{0}(x)=f_{0}\left(b^{n}\right)=K(b)\left(b^{n}-\frac{1}{b^{n}}\right)=K(a)\left(x-\frac{1}{x}\right) . $$  Hence, we have $f(x)=f_{0}(x)+f(1) x=C_{1} x+\frac{C_{2}}{x}$ with $C_{1}=K(a)+f(1)$ and $C_{2}=-K(a)$.  Comment. After establishing (5), there are several variants of finishing the solution. For example, instead of taking a limit, we can obtain a system of linear equations for $K(a), K(b)$ and $K(a b)$ by substituting two positive integers $n$ in (5), say $n=1$ and $n=2$. This approach leads to a similar ending as in the first solution.  Optionally, we define another function $f_{1}(x)=f_{0}(x)-C\left(x-\frac{1}{x}\right)$ and prescribe $K(c)=0$ for another fixed $c$. Then we can choose $a b=c$ and decrease the number of terms in (5).


Find the maximal value of  $$ S=\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}}+\sqrt[3]{\frac{d}{a+7}} $$  where $a, b, c, d$ are nonnegative real numbers which satisfy $a+b+c+d=100$.  (Taiwan)  Answer: $\frac{8}{\sqrt[3]{7}}$, reached when $(a, b, c, d)$ is a cyclic permutation of $(1,49,1,49)$.

Solution: Since the value $8 / \sqrt[3]{7}$ is reached, it suffices to prove that $S \leqslant 8 / \sqrt[3]{7}$.  Assume that $x, y, z, t$ is a permutation of the variables, with $x \leqslant y \leqslant z \leqslant t$. Then, by the rearrangement inequality,  $$ S \leqslant\left(\sqrt[3]{\frac{x}{t+7}}+\sqrt[3]{\frac{t}{x+7}}\right)+\left(\sqrt[3]{\frac{y}{z+7}}+\sqrt[3]{\frac{z}{y+7}}\right) $$  Claim. The first bracket above does not exceed $\sqrt[3]{\frac{x+t+14}{7}}$.  Proof. Since  $$ X^{3}+Y^{3}+3 X Y Z-Z^{3}=\frac{1}{2}(X+Y-Z)\left((X-Y)^{2}+(X+Z)^{2}+(Y+Z)^{2}\right) $$  the inequality $X+Y \leqslant Z$ is equivalent (when $X, Y, Z \geqslant 0$ ) to $X^{3}+Y^{3}+3 X Y Z \leqslant Z^{3}$. Therefore, the claim is equivalent to  $$ \frac{x}{t+7}+\frac{t}{x+7}+3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}} \leqslant \frac{x+t+14}{7} $$  Notice that  $$ \begin{array}{r} 3 \sqrt[3]{\frac{x t(x+t+14)}{7(x+7)(t+7)}}=3 \sqrt[3]{\frac{t(x+7)}{7(t+7)} \cdot \frac{x(t+7)}{7(x+7)} \cdot \frac{7(x+t+14)}{(t+7)(x+7)}} \\ \leqslant \frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \end{array} $$  by the AM-GM inequality, so it suffices to prove  $$ \frac{x}{t+7}+\frac{t}{x+7}+\frac{t(x+7)}{7(t+7)}+\frac{x(t+7)}{7(x+7)}+\frac{7(x+t+14)}{(t+7)(x+7)} \leqslant \frac{x+t+14}{7} . $$  A straightforward check verifies that the last inequality is in fact an equality.  The claim leads now to  $$ S \leqslant \sqrt[3]{\frac{x+t+14}{7}}+\sqrt[3]{\frac{y+z+14}{7}} \leqslant 2 \sqrt[3]{\frac{x+y+z+t+28}{14}}=\frac{8}{\sqrt[3]{7}} $$  the last inequality being due to the AM-CM inequality (or to the fact that $\sqrt[3]{ }$ is concave on $[0, \infty))$.


Solution: We present a different proof for the estimate $S \leqslant 8 / \sqrt[3]{7}$.  Start by using Hölder's inequality:  $$ S^{3}=\left(\sum_{\mathrm{cyc}} \frac{\sqrt[6]{a} \cdot \sqrt[6]{a}}{\sqrt[3]{b+7}}\right)^{3} \leqslant \sum_{\mathrm{cyc}}(\sqrt[6]{a})^{3} \cdot \sum_{\mathrm{cyc}}(\sqrt[6]{a})^{3} \cdot \sum_{\mathrm{cyc}}\left(\frac{1}{\sqrt[3]{b+7}}\right)^{3}=\left(\sum_{\mathrm{cyc}} \sqrt{a}\right)^{2} \sum_{\mathrm{cyc}} \frac{1}{b+7} $$  Notice that  $$ \frac{(x-1)^{2}(x-7)^{2}}{x^{2}+7} \geqslant 0 \Longleftrightarrow x^{2}-16 x+71 \geqslant \frac{448}{x^{2}+7} $$  yields  $$ \sum \frac{1}{b+7} \leqslant \frac{1}{448} \sum(b-16 \sqrt{b}+71)=\frac{1}{448}\left(384-16 \sum \sqrt{b}\right)=\frac{48-2 \sum \sqrt{b}}{56} $$  Finally,  $$ S^{3} \leqslant \frac{1}{56}\left(\sum \sqrt{a}\right)^{2}\left(48-2 \sum \sqrt{a}\right) \leqslant \frac{1}{56}\left(\frac{\sum \sqrt{a}+\sum \sqrt{a}+\left(48-2 \sum \sqrt{a}\right)}{3}\right)^{3}=\frac{512}{7} $$  by the AM-GM inequality. The conclusion follows.  Comment. All the above works if we replace 7 and 100 with $k>0$ and $2\left(k^{2}+1\right)$, respectively; in this case, the answer becomes  $$ 2 \sqrt[3]{\frac{(k+1)^{2}}{k}} $$  Even further, a linear substitution allows to extend the solutions to a version with 7 and 100 being replaced with arbitrary positive real numbers $p$ and $q$ satisfying $q \geqslant 4 p$.


Queenie and Horst play a game on a $20 \times 20$ chessboard. In the beginning the board is empty. In every turn, Horst places a black knight on an empty square in such a way that his new knight does not attack any previous knights. Then Queenie places a white queen on an empty square. The game gets finished when somebody cannot move.  Find the maximal positive $K$ such that, regardless of the strategy of Queenie, Horst can put at least $K$ knights on the board.  (Armenia)  Answer: $K=20^{2} / 4=100$. In case of a $4 N \times 4 M$ board, the answer is $K=4 N M$.

Solution: We show two strategies, one for Horst to place at least 100 knights, and another strategy for Queenie that prevents Horst from putting more than 100 knights on the board.  A strategy for Horst: Put knights only on black squares, until all black squares get occupied.  Colour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is $20^{2} / 2=200$. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first 100 steps.  A strategy for Queenie: Group the squares into cycles of length 4, and after each step of Horst, occupy the opposite square in the same cycle.  Consider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a $4 \times 4$ board the squares can be grouped into 4 cycles of length 4 , as shown in Figure 1. Divide the board into parts of size $4 \times 4$, and perform the same grouping in every part; this way we arrange the 400 squares of the board into 100 cycles (Figure 2).  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-27.jpg?height=360&width=351&top_left_y=1579&top_left_x=310)  Figure 1  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-27.jpg?height=362&width=420&top_left_y=1578&top_left_x=772)  Figure 2  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-27.jpg?height=357&width=463&top_left_y=1578&top_left_x=1296)  Figure 3  The strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square $A$, which is part of some cycle $A-B-C-D-A$, let Queenie put her queen on the opposite square $C$ in that cycle (Figure 3). From this point, Horst cannot put any knight on $A$ or $C$ because those squares are already occupied, neither on $B$ or $D$ because those squares are attacked by the knight standing on $A$. Hence, Horst can put at most one knight on each cycle, that is at most 100 knights in total.  Comment 1. Queenie's strategy can be prescribed by a simple rule: divide the board into $4 \times 4$ parts; whenever Horst puts a knight in a part $P$, Queenie reflects that square about the centre of $P$ and puts her queen on the reflected square.  Comment 2. The result remains the same if Queenie moves first. In the first turn, she may put her first queen arbitrarily. Later, if she has to put her next queen on a square that already contains a queen, she may move arbitrarily again.


An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:  $$ \begin{array}{ccccc}  & & & \\ & 2 & 6 & \\ & 5 & 7 & \\ 8 & 3 & 10 & 9 . \end{array} $$  Is it possible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once?  (Iran)  Answer: No, it is not possible.

Solution: Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-29.jpg?height=391&width=465&top_left_y=1415&top_left_x=430)  Figure 1  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-29.jpg?height=394&width=463&top_left_y=1411&top_left_x=1162)  Figure 2  Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently,  $$ \begin{array}{r} b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\ \geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8}, \end{array} $$  which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.  Comment. The above estimate may be slightly improved by noticing that $b_{\ell}^{\prime} \neq b_{n}$. This implies $n(n+1) / 2=b_{n}>b_{\ell}^{\prime} \geqslant\lceil(n-2) / 2\rceil(2 n+\lceil(n-2) / 2\rceil+1) / 2$, so $n \leqslant 7$ if $n$ is odd, and $n \leqslant 12$ if $n$ is even. It seems that the largest anti-Pascal pyramid whose entries are a permutation of the integers from 1 to $1+2+\cdots+n$ has 5 rows.


Let $k$ be a positive integer. The organising committee of a tennis tournament is to schedule the matches for $2 k$ players so that every two players play once, each day exactly one match is played, and each player arrives to the tournament site the day of his first match, and departs the day of his last match. For every day a player is present on the tournament, the committee has to pay 1 coin to the hotel. The organisers want to design the schedule so as to minimise the total cost of all players' stays. Determine this minimum cost.  (Russia)  Answer: The required minimum is $k\left(4 k^{2}+k-1\right) / 2$.

Solution: Enumerate the days of the tournament $1,2, \ldots,\left(\begin{array}{c}2 k \\ 2\end{array}\right)$. Let $b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{2 k}$ be the days the players arrive to the tournament, arranged in nondecreasing order; similarly, let $e_{1} \geqslant \cdots \geqslant e_{2 k}$ be the days they depart arranged in nonincreasing order (it may happen that a player arrives on day $b_{i}$ and departs on day $e_{j}$, where $i \neq j$ ). If a player arrives on day $b$ and departs on day $e$, then his stay cost is $e-b+1$. Therefore, the total stay cost is  $$ \Sigma=\sum_{i=1}^{2 k} e_{i}-\sum_{i=1}^{2 k} b_{i}+n=\sum_{i=1}^{2 k}\left(e_{i}-b_{i}+1\right) $$  Bounding the total cost from below. To this end, estimate $e_{i+1}-b_{i+1}+1$. Before day $b_{i+1}$, only $i$ players were present, so at most $\left(\begin{array}{l}i \\ 2\end{array}\right)$ matches could be played. Therefore, $b_{i+1} \leqslant\left(\begin{array}{l}i \\ 2\end{array}\right)+1$. Similarly, at most $\left(\begin{array}{l}i \\ 2\end{array}\right)$ matches could be played after day $e_{i+1}$, so $e_{i} \geqslant\left(\begin{array}{c}2 k \\ 2\end{array}\right)-\left(\begin{array}{l}i \\ 2\end{array}\right)$. Thus,  $$ e_{i+1}-b_{i+1}+1 \geqslant\left(\begin{array}{c} 2 k \\ 2 \end{array}\right)-2\left(\begin{array}{l} i \\ 2 \end{array}\right)=k(2 k-1)-i(i-1) $$  This lower bound can be improved for $i>k$ : List the $i$ players who arrived first, and the $i$ players who departed last; at least $2 i-2 k$ players appear in both lists. The matches between these players were counted twice, though the players in each pair have played only once. Therefore, if $i>k$, then  $$ e_{i+1}-b_{i+1}+1 \geqslant\left(\begin{array}{c} 2 k \\ 2 \end{array}\right)-2\left(\begin{array}{c} i \\ 2 \end{array}\right)+\left(\begin{array}{c} 2 i-2 k \\ 2 \end{array}\right)=(2 k-i)^{2} $$  An optimal tournament, We now describe a schedule in which the lower bounds above are all achieved simultaneously. Split players into two groups $X$ and $Y$, each of cardinality $k$. Next, partition the schedule into three parts. During the first part, the players from $X$ arrive one by one, and each newly arrived player immediately plays with everyone already present. During the third part (after all players from $X$ have already departed) the players from $Y$ depart one by one, each playing with everyone still present just before departing.  In the middle part, everyone from $X$ should play with everyone from $Y$. Let $S_{1}, S_{2}, \ldots, S_{k}$ be the players in $X$, and let $T_{1}, T_{2}, \ldots, T_{k}$ be the players in $Y$. Let $T_{1}, T_{2}, \ldots, T_{k}$ arrive in this order; after $T_{j}$ arrives, he immediately plays with all the $S_{i}, i>j$. Afterwards, players $S_{k}$, $S_{k-1}, \ldots, S_{1}$ depart in this order; each $S_{i}$ plays with all the $T_{j}, i \leqslant j$, just before his departure, and $S_{k}$ departs the day $T_{k}$ arrives. For $0 \leqslant s \leqslant k-1$, the number of matches played between $T_{k-s}$ 's arrival and $S_{k-s}$ 's departure is  $$ \sum_{j=k-s}^{k-1}(k-j)+1+\sum_{j=k-s}^{k-1}(k-j+1)=\frac{1}{2} s(s+1)+1+\frac{1}{2} s(s+3)=(s+1)^{2} $$  Thus, if $i>k$, then the number of matches that have been played between $T_{i-k+1}$ 's arrival, which is $b_{i+1}$, and $S_{i-k+1}$ 's departure, which is $e_{i+1}$, is $(2 k-i)^{2}$; that is, $e_{i+1}-b_{i+1}+1=(2 k-i)^{2}$, showing the second lower bound achieved for all $i>k$.  If $i \leqslant k$, then the matches between the $i$ players present before $b_{i+1}$ all fall in the first part of the schedule, so there are $\left(\begin{array}{l}i \\ 2\end{array}\right)$ such, and $b_{i+1}=\left(\begin{array}{l}i \\ 2\end{array}\right)+1$. Similarly, after $e_{i+1}$, there are $i$ players left, all $\left(\begin{array}{l}i \\ 2\end{array}\right)$ matches now fall in the third part of the schedule, and $e_{i+1}=\left(\begin{array}{c}2 k \\ 2\end{array}\right)-\left(\begin{array}{c}i \\ 2\end{array}\right)$. The first lower bound is therefore also achieved for all $i \leqslant k$.  Consequently, all lower bounds are achieved simultaneously, and the schedule is indeed optimal.  Evaluation. Finally, evaluate the total cost for the optimal schedule:  $$ \begin{aligned} \Sigma & =\sum_{i=0}^{k}(k(2 k-1)-i(i-1))+\sum_{i=k+1}^{2 k-1}(2 k-i)^{2}=(k+1) k(2 k-1)-\sum_{i=0}^{k} i(i-1)+\sum_{j=1}^{k-1} j^{2} \\ & =k(k+1)(2 k-1)-k^{2}+\frac{1}{2} k(k+1)=\frac{1}{2} k\left(4 k^{2}+k-1\right) . \end{aligned} $$


Solution: Consider any tournament schedule. Label players $P_{1}, P_{2}, \ldots, P_{2 k}$ in order of their arrival, and label them again $Q_{2 k}, Q_{2 k-1}, \ldots, Q_{1}$ in order of their departure, to define a permutation $a_{1}, a_{2}, \ldots, a_{2 k}$ of $1,2, \ldots, 2 k$ by $P_{i}=Q_{a_{i}}$.  We first describe an optimal tournament for any given permutation $a_{1}, a_{2}, \ldots, a_{2 k}$ of the indices $1,2, \ldots, 2 k$. Next, we find an optimal permutation and an optimal tournament.  Optimisation for a fixed $a_{1}, \ldots, a_{2 k}$. We say that the cost of the match between $P_{i}$ and $P_{j}$ is the number of players present at the tournament when this match is played. Clearly, the Committee pays for each day the cost of the match of that day. Hence, we are to minimise the total cost of all matches.  Notice that $Q_{2 k}$ 's departure does not precede $P_{2 k}$ 's arrival. Hence, the number of players at the tournament monotonically increases (non-strictly) until it reaches $2 k$, and then monotonically decreases (non-strictly). So, the best time to schedule the match between $P_{i}$ and $P_{j}$ is either when $P_{\max (i, j)}$ arrives, or when $Q_{\max \left(a_{i}, a_{j}\right)}$ departs, in which case the cost is $\min \left(\max (i, j), \max \left(a_{i}, a_{j}\right)\right)$.  Conversely, assuming that $i>j$, if this match is scheduled between the arrivals of $P_{i}$ and $P_{i+1}$, then its cost will be exactly $i=\max (i, j)$. Similarly, one can make it cost $\max \left(a_{i}, a_{j}\right)$. Obviously, these conditions can all be simultaneously satisfied, so the minimal cost for a fixed sequence $a_{1}, a_{2}, \ldots, a_{2 k}$ is  $$ \Sigma\left(a_{1}, \ldots, a_{2 k}\right)=\sum_{1 \leqslant i<j \leqslant 2 k} \min \left(\max (i, j), \max \left(a_{i}, a_{j}\right)\right) $$  Optimising the sequence $\left(a_{i}\right)$. Optimisation hinges on the lemma below.  Lemma. If $a \leqslant b$ and $c \leqslant d$, then  $$ \begin{aligned} \min (\max (a, x), \max (c, y))+\min & (\max (b, x), \max (d, y)) \\ & \geqslant \min (\max (a, x), \max (d, y))+\min (\max (b, x), \max (c, y)) \end{aligned} $$  Proof. Write $a^{\prime}=\max (a, x) \leqslant \max (b, x)=b^{\prime}$ and $c^{\prime}=\max (c, y) \leqslant \max (d, y)=d^{\prime}$ and check that $\min \left(a^{\prime}, c^{\prime}\right)+\min \left(b^{\prime}, d^{\prime}\right) \geqslant \min \left(a^{\prime}, d^{\prime}\right)+\min \left(b^{\prime}, c^{\prime}\right)$.  Consider a permutation $a_{1}, a_{2}, \ldots, a_{2 k}$ such that $a_{i}<a_{j}$ for some $i<j$. Swapping $a_{i}$ and $a_{j}$ does not change the $(i, j)$ th summand in (1), and for $\ell \notin\{i, j\}$ the sum of the $(i, \ell)$ th and the $(j, \ell)$ th summands does not increase by the Lemma. Hence the optimal value does not increase, but the number of disorders in the permutation increases. This process stops when $a_{i}=2 k+1-i$ for all $i$, so the required minimum is  $$ \begin{aligned} S(2 k, 2 k-1, \ldots, 1) & =\sum_{1 \leqslant i<j \leqslant 2 k} \min (\max (i, j), \max (2 k+1-i, 2 k+1-j)) \\ & =\sum_{1 \leqslant i<j \leqslant 2 k} \min (j, 2 k+1-i) . \end{aligned} $$  The latter sum is fairly tractable and yields the stated result; we omit the details.  Comment. If the number of players is odd, say, $2 k-1$, the required minimum is $k(k-1)(4 k-1) / 2$. In this case, $|X|=k,|Y|=k-1$, the argument goes along the same lines, but some additional technicalities are to be taken care of.  This page is intentionally left blank


A circle $\omega$ of radius 1 is given. A collection $T$ of triangles is called good, if the following conditions hold:  (i) each triangle from $T$ is inscribed in $\omega$;  (ii) no two triangles from $T$ have a common interior point.  Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.  (South Africa)  Answer: $t \in(0,4]$.

Solution: First, we show how to construct a good collection of $n$ triangles, each of perimeter greater than 4 . This will show that all $t \leqslant 4$ satisfy the required conditions.  Construct inductively an $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ inscribed in $\omega$ such that $B C$ is a diameter, and $B A_{1} A_{2}, B A_{2} A_{3}, \ldots, B A_{n-1} A_{n}, B A_{n} C$ is a good collection of $n$ triangles. For $n=1$, take any triangle $B A_{1} C$ inscribed in $\omega$ such that $B C$ is a diameter; its perimeter is greater than $2 B C=4$. To perform the inductive step, assume that the $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ is already constructed. Since $A_{n} B+A_{n} C+B C>4$, one can choose a point $A_{n+1}$ on the small arc $\widehat{C A_{n}}$, close enough to $C$, so that $A_{n} B+A_{n} A_{n+1}+B A_{n+1}$ is still greater than 4 . Thus each of these new triangles $B A_{n} A_{n+1}$ and $B A_{n+1} C$ has perimeter greater than 4, which completes the induction step.  ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-42.jpg?height=366&width=691&top_left_y=1353&top_left_x=682)  We proceed by showing that no $t>4$ satisfies the conditions of the problem. To this end, we assume that there exists a good collection $T$ of $n$ triangles, each of perimeter greater than $t$, and then bound $n$ from above.  Take $\varepsilon>0$ such that $t=4+2 \varepsilon$.  Claim. There exists a positive constant $\sigma=\sigma(\varepsilon)$ such that any triangle $\Delta$ with perimeter $2 s \geqslant 4+2 \varepsilon$, inscribed in $\omega$, has area $S(\Delta)$ at least $\sigma$.  Proof. Let $a, b, c$ be the side lengths of $\Delta$. Since $\Delta$ is inscribed in $\omega$, each side has length at most 2. Therefore, $s-a \geqslant(2+\varepsilon)-2=\varepsilon$. Similarly, $s-b \geqslant \varepsilon$ and $s-c \geqslant \varepsilon$. By Heron's formula, $S(\Delta)=\sqrt{s(s-a)(s-b)(s-c)} \geqslant \sqrt{(2+\varepsilon) \varepsilon^{3}}$. Thus we can set $\sigma(\varepsilon)=\sqrt{(2+\varepsilon) \varepsilon^{3}}$.  Now we see that the total area $S$ of all triangles from $T$ is at least $n \sigma(\varepsilon)$. On the other hand, $S$ does not exceed the area of the disk bounded by $\omega$. Thus $n \sigma(\varepsilon) \leqslant \pi$, which means that $n$ is bounded from above.  Comment 1. One may prove the Claim using the formula $S=\frac{a b c}{4 R}$ instead of Heron's formula.  Comment 2. In the statement of the problem condition $(i)$ could be replaced by a weaker one: each triangle from $T$ lies within $\omega$. This does not affect the solution above, but reduces the number of ways to prove the Claim.  This page is intentionally left blank