Let $ABCD$ be a square and let $\Gamma$ be the circumcircle of $ABCD$. $M$ is a point of $\Gamma$ belonging to the arc $CD$ which doesn't contain $A$. $P$ and $R$ are respectively the intersection points of $(AM)$ with $[BD]$ and $[CD]$, $Q$ and $S$ are respectively the intersection points of $(BM)$ with $[AC]$ and $[DC]$. Prove that $(PS)$ and $(QR)$ are perpendicular.

Let $a,b,c$ be three positive real numbers such that $abc=1$. Show that: \[ \displaystyle \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+ \frac{c}{(c+1)(a+1)} \geq \frac{3}{4}. \] When is there equality?

Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$. Proposed by Mohsen Jamali, Iran

Click for solution The main ideas (what exactly do you call a method¿) for mine are: If $p \nmid a,b$ is any prime and we have any $k$ with $p|a^{k}+k$, then $p|a^{k}+k|b^{k}+k \implies p|b^{k}+k$, thus $a^{k}+k \equiv 0 \equiv b^{k}+k \mod p \iff a^{k}\equiv b^{k}\mod p$. If we additionally would have $k \equiv 1 \mod (p-1)$, then Fermats Little Theorem gives $a \equiv a^{k}\equiv b^{k}\equiv b \mod p$. But for $k \equiv 1 \mod (p-1)$ we have also $a^{k}+k \equiv a+k$, so to get $p|a^{k}+k$ we just need $p|a+k$. So lets look for some $k \equiv 1 \mod (p-1)$ and $k \equiv-a \mod p$, which can be found by the Chinese Remainder Theorem or directly as $k=(p-1)(a+1)+1$. Till now we didn't use any special property of $p$ except that it doesn't divide $a,b$. But we know that for any such prime we have $p|(a-b)$ from chossing a suitable $k$ with $p|a^{k}+k$, thus if $a \neq b$ we would get a contradiction by choosing a big prime.

In a $2\times n$ array we have positive reals s.t. the sum of the numbers in each of the $n$ columns is $1$. Show that we can select a number in each column s.t. the sum of the selected numbers in each row is at most $\frac{n+1}4$.

Click for solution grobber wrote: In a $2\times n$ array we have positive reals s.t. the sum of the numbers in each of the $n$ columns is $1$. Show that we can select a number in each column s.t. the sum of the selected numbers in each row is at most $\frac{n+1}4$. Here is my solution: We denote the numbers from the first row by $a_1$, $a_2$, ..., $a_n$ in increasing order: $a_1\leq a_2\leq ...\leq a_n$. Then, the corresponding numbers from the second row are obviously $1-a_1$, $1-a_2$, ..., $1-a_n$. Now, let $k$ be the largest index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$. We WLOG assume that $k \neq n$ (since otherwise, we can simply select all numbers from the first row and no numbers from the second row, and we are done). Thus, $k+1 \leq n$, so that $a_{k+1}$ is well-defined. Moreover, $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$ (since $k$ was chosen to be the largest index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$). Now, we are going to prove that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$. In fact, recall that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$ is greater or equal than the number $a_{k+1}$ (the smallest of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$). In other words, $\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1}$. On the other hand, recall again that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_1$, $a_2$, ..., $a_{k+1}$ is smaller or equal than the number $a_{k+1}$ (the greatest of the numbers $a_1$, $a_2$, ..., $a_{k+1}$). In other words, $\frac{a_1+a_2+...+a_{k+1}}{k+1}\leq a_{k+1}$. Thus, $\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1} \geq \frac{a_1+a_2+...+a_{k+1}}{k+1}$, and thus $a_{k+1}+a_{k+2}+...+a_n\geq\left(n-k\right)\cdot\frac{a_1+a_2+...+a_{k+1}}{k+1} \geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1}$ (since $n-k\geq 0$ and $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$). In other words, $a_{k+1}+a_{k+2}+...+a_n\geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1} = \frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$. Hence, $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)$ $=\left(n-k\right)-\left(a_{k+1}+a_{k+2}+...+a_n\right)\leq\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$. Thus, in order to show that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$, it will be enough to prove that $\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}$. This, however, is straightforward: $\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}+\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$ $\Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\left(1+\frac{n-k}{k+1}\right)\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\cdot\frac{n+1}{k+1}$ $\Longleftrightarrow\ \ \ \ \ n-k\leq\left(\frac{n+1}{2}\right)^2\cdot\frac{1}{k+1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \left(n-k\right)\left(k+1\right)\leq\left(\frac{n+1}{2}\right)^2$. But this is clear from AM-GM: $\left(n-k\right)\left(k+1\right)\leq\left(\frac{\left(n-k\right)+\left(k+1\right)}{2}\right)^2 = \left(\frac{n+1}{2}\right)^2$. So we have proved the inequality $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$. Together with $a_1+a_2+...+a_k\leq\frac{n+1}{4}$, this shows that if we choose the numbers $a_1$, $a_2$, ..., $a_k$ from the first row and the numbers $1-a_{k+1}$, $1-a_{k+2}$, ..., $1-a_n$ from the second row, then the sum of the chosen numbers in each row is $\leq\frac{n+1}{4}$. And the problem is solved. Unrelated to my solution above, here is a question on seshadri's solution: In the case $k\geq m$, how do we make sure that $k\geq n/2$ (we need this to speak of $a_{2k-n+1}$)? Maybe $m\leq n/2$ should be replaced by $m\geq n/2$ ? In this case, seshadri's solution seems to work (one minor mistake, though: the proof of $2k\leq n+m$ does not work in the special case when $n=m$, because we get a $\leq$ sign instead of the $<$ sign in $k-1/2\leq \sum a_i = \sum_{i=1}^m a_i+\sum_{i=m+1}^n a_i < m+(n-m)/2=(n+m)/2$; but fortunately there is an easy reason why $2k\leq n+m$ holds in the case when $n=m$). Darij

Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle. Proposed by Dimitris Kontogiannis, Greece

Click for solution Let $C'$ be a symmetrical point to $C$ with respect to $I$, $A'$ lie on line $CA$ such that $CA' = 2AB$ and $B'$ lie on line $CB$ such that $CB' = 2AB$. $CA + CB = 3AB$ imply $CD = AB$, $BB' = DA$ and $AA' = EB$. We see, that $C'B' = C'A' = 2IE$ and $C'K = C'L = CE = AB$. Triangle $C'B'B$ is congruent to $ADK$ and $C'A'A$ to $BLE$. Thus $C'B = AK$ and $C'A = LB$. So triangle $C'BK$ is congurent to $ABK$ and $C'AL$ to $ABL$. Thus $\angle KC'B = \angle KAB$ and $\angle LC'A = \angle LBA$. It means that $AKBC'$ and $ALBC'$ are cyclic quadrilaterals thus $ALKB$ is cyclic quadrilateral.

Let $M=\{1,2,\ldots,3 \cdot n\}$. Partition $M$ into three sets $A,B,C$ which $card$ $A$ $=$ $card$ $B$ $=$ $card$ $C$ $=$ $n .$ Prove that there exists $a$ in $A,b$ in $B, c$ in $C$ such that or $a=b+c,$ or $b=c+a,$ or $c=a+b$ Edited by orl.

Click for solution I think he means more or less the following, which is the solution we gave when we used the problem in our training : We will say that $(a,b,c) \in A \times B \times C$ is good if one of $a,b,c$ is the sum of the two others. Assume that there is no good triples. Wlog, we may assume that $1 \in A$ and that the least number which is not in $A$, say $k$, belongs to $B$. Let $x \in C$. We prove that $x-1 \in A$ : Clearly, since there is no good triple, $x-1 \notin B$. Suppose that $x-1 \in C$, and from the minimality of $k$, it follows that $x-1 > k$. Since none of the triples $(x-k,k,x)$ and $(k-1,x-k,x-1)$ is good, it follows that $x-k \notin A \cup B$. Similarly, using the triples $(x-k-1,k,x-1)$ and $(1,x-k-1,x-k)$ we deduce that $x-k-1 \notin A \cup B$. Thus both of them belong to $C$. By induction, it is now easy to prove that all the positive integers of the form $x-ik$ or $x-ik-1$ belong to $C$. But, for some $i$, one of the numbers $x-ik$ or $x-ik-1$ is positive and not greater than $k$, so that it belongs to $A \cup B$. A contradiction. It follows that, for all $x \in C$, we have $x-1 \in A$. Let $C = \{c_1, \cdots, c_n \}$. Thus each of the numbers $c_1-1, \cdots , c_n -1$ belongs to $A$. But these numbers are pairwise distincts and all are greater than $1$ (since they all the $c_i$'s are greater than $k$). Thus $|A| \geq n+1$, a contradiction. And we are done. Pierre.