Suppose that a sequence $a_{1}, a_{2}, \ldots$ of positive real numbers satisfies  $$ a_{k+1} \geqslant \frac{k a_{k}}{a_{k}^{2}+(k-1)} $$  for every positive integer $k$. Prove that $a_{1}+a_{2}+\cdots+a_{n} \geqslant n$ for every $n \geqslant 2$.

Solution: From the constraint (1), it can be seen that  $$ \frac{k}{a_{k+1}} \leqslant \frac{a_{k}^{2}+(k-1)}{a_{k}}=a_{k}+\frac{k-1}{a_{k}} $$  and so  $$ a_{k} \geqslant \frac{k}{a_{k+1}}-\frac{k-1}{a_{k}} . $$  Summing up the above inequality for $k=1, \ldots, m$, we obtain  $$ a_{1}+a_{2}+\cdots+a_{m} \geqslant\left(\frac{1}{a_{2}}-\frac{0}{a_{1}}\right)+\left(\frac{2}{a_{3}}-\frac{1}{a_{2}}\right)+\cdots+\left(\frac{m}{a_{m+1}}-\frac{m-1}{a_{m}}\right)=\frac{m}{a_{m+1}} $$  Now we prove the problem statement by induction on $n$. The case $n=2$ can be done by applying (1) to $k=1$ :  $$ a_{1}+a_{2} \geqslant a_{1}+\frac{1}{a_{1}} \geqslant 2 $$  For the induction step, assume that the statement is true for some $n \geqslant 2$. If $a_{n+1} \geqslant 1$, then the induction hypothesis yields  $$ \left(a_{1}+\cdots+a_{n}\right)+a_{n+1} \geqslant n+1 $$  Otherwise, if $a_{n+1}<1$ then apply (2) as  $$ \left(a_{1}+\cdots+a_{n}\right)+a_{n+1} \geqslant \frac{n}{a_{n+1}}+a_{n+1}=\frac{n-1}{a_{n+1}}+\left(\frac{1}{a_{n+1}}+a_{n+1}\right)>(n-1)+2 $$  That completes the solution. Comment 1. It can be seen easily that having equality in the statement requires $a_{1}=a_{2}=1$ in the base case $n=2$, and $a_{n+1}=1$ in (3). So the equality $a_{1}+\cdots+a_{n}=n$ is possible only in the trivial case $a_{1}=\cdots=a_{n}=1$.  Comment 2. After obtaining (2), there are many ways to complete the solution. We outline three such possibilities.  - With defining $s_{n}=a_{1}+\cdots+a_{n}$, the induction step can be replaced by  $$ s_{n+1}=s_{n}+a_{n+1} \geqslant s_{n}+\frac{n}{s_{n}} \geqslant n+1 $$  because the function $x \mapsto x+\frac{n}{x}$ increases on $[n, \infty)$.  - By applying the AM-GM inequality to the numbers $a_{1}+\cdots+a_{k}$ and $k a_{k+1}$, we can conclude  $$ a_{1}+\cdots+a_{k}+k a_{k+1} \geqslant 2 k $$  and sum it up for $k=1, \ldots, n-1$.  - We can derive the symmetric estimate  $$ \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{j}=\sum_{j=2}^{n}\left(a_{1}+\cdots+a_{j-1}\right) a_{j} \geqslant \sum_{j=2}^{n}(j-1)=\frac{n(n-1)}{2} $$  and combine it with the AM-QM inequality.


Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ with the property that  $$ f(x-f(y))=f(f(x))-f(y)-1 $$  holds for all $x, y \in \mathbb{Z}$. (Croatia)

Solution: It is immediately checked that both functions mentioned in the answer are as desired.  Now let $f$ denote any function satisfying (1) for all $x, y \in \mathbb{Z}$. Substituting $x=0$ and $y=f(0)$ into (1) we learn that the number $z=-f(f(0))$ satisfies $f(z)=-1$. So by plugging $y=z$ into (1) we deduce that  $$ f(x+1)=f(f(x)) $$  holds for all $x \in \mathbb{Z}$. Thereby (1) simplifies to  $$ f(x-f(y))=f(x+1)-f(y)-1 $$  We now work towards showing that $f$ is linear by contemplating the difference $f(x+1)-f(x)$ for any $x \in \mathbb{Z}$. By applying (3) with $y=x$ and (2) in this order, we obtain  $$ f(x+1)-f(x)=f(x-f(x))+1=f(f(x-1-f(x)))+1 $$  Since (3) shows $f(x-1-f(x))=f(x)-f(x)-1=-1$, this simplifies to  $$ f(x+1)=f(x)+A, $$  where $A=f(-1)+1$ is some absolute constant. Now a standard induction in both directions reveals that $f$ is indeed linear and that in fact we have $f(x)=A x+B$ for all $x \in \mathbb{Z}$, where $B=f(0)$. Substituting this into (2) we obtain that  $$ A x+(A+B)=A^{2} x+(A B+B) $$  holds for all $x \in \mathbb{Z}$; applying this to $x=0$ and $x=1$ we infer $A+B=A B+B$ and $A^{2}=A$. The second equation leads to $A=0$ or $A=1$. In case $A=1$, the first equation gives $B=1$, meaning that $f$ has to be the successor function. If $A=0$, then $f$ is constant and (1) shows that its constant value has to be -1 . Thereby the solution is complete.  Comment. After (2) and (3) have been obtained, there are several other ways to combine them so as to obtain linearity properties of $f$. For instance, using (2) thrice in a row and then (3) with $x=f(y)$ one may deduce that  $$ f(y+2)=f(f(y+1))=f(f(f(y)))=f(f(y)+1)=f(y)+f(0)+1 $$  holds for all $y \in \mathbb{Z}$. It follows that $f$ behaves linearly on the even numbers and on the odd numbers separately, and moreover that the slopes of these two linear functions coincide. From this point, one may complete the solution with some straightforward case analysis.  A different approach using the equations (2) and (3) will be presented in Solution 2. To show that it is also possible to start in a completely different way, we will also present a third solution that avoids these equations entirely.


Solution: We commence by deriving (2) and (3) as in the first solution. Now provided that $f$ is injective, (2) tells us that $f$ is the successor function. Thus we may assume from now on that $f$ is not injective, i.e., that there are two integers $a>b$ with $f(a)=f(b)$. A straightforward induction using (2) in the induction step reveals that we have $f(a+n)=f(b+n)$ for all nonnegative integers $n$. Consequently, the sequence $\gamma_{n}=f(b+n)$ is periodic and thus in particular bounded, which means that the numbers  $$ \varphi=\min _{n \geqslant 0} \gamma_{n} \quad \text { and } \quad \psi=\max _{n \geqslant 0} \gamma_{n} $$  exist. Let us pick any integer $y$ with $f(y)=\varphi$ and then an integer $x \geqslant a$ with $f(x-f(y))=\varphi$. Due to the definition of $\varphi$ and (3) we have  $$ \varphi \leqslant f(x+1)=f(x-f(y))+f(y)+1=2 \varphi+1 $$  whence $\varphi \geqslant-1$. The same reasoning applied to $\psi$ yields $\psi \leqslant-1$. Since $\varphi \leqslant \psi$ holds trivially, it follows that $\varphi=\psi=-1$, or in other words that we have $f(t)=-1$ for all integers $t \geqslant a$.  Finally, if any integer $y$ is given, we may find an integer $x$ which is so large that $x+1 \geqslant a$ and $x-f(y) \geqslant a$ hold. Due to (3) and the result from the previous paragraph we get  $$ f(y)=f(x+1)-f(x-f(y))-1=(-1)-(-1)-1=-1 . $$  Thereby the problem is solved.


Solution: Set $d=f(0)$. By plugging $x=f(y)$ into (1) we obtain  $$ f^{3}(y)=f(y)+d+1 $$  for all $y \in \mathbb{Z}$, where the left-hand side abbreviates $f(f(f(y)))$. When we replace $x$ in (1) by $f(x)$ we obtain $f(f(x)-f(y))=f^{3}(x)-f(y)-1$ and as a consequence of (4) this simplifies to  $$ f(f(x)-f(y))=f(x)-f(y)+d $$  Now we consider the set  $$ E=\{f(x)-d \mid x \in \mathbb{Z}\} $$  Given two integers $a$ and $b$ from $E$, we may pick some integers $x$ and $y$ with $f(x)=a+d$ and $f(y)=b+d$; now (5) tells us that $f(a-b)=(a-b)+d$, which means that $a-b$ itself exemplifies $a-b \in E$. Thus,  $$ E \text { is closed under taking differences. } $$  Also, the definitions of $d$ and $E$ yield $0 \in E$. If $E=\{0\}$, then $f$ is a constant function and (1) implies that the only value attained by $f$ is indeed -1 .  So let us henceforth suppose that $E$ contains some number besides zero. It is known that in this case (6) entails $E$ to be the set of all integer multiples of some positive integer $k$. Indeed, this holds for  $$ k=\min \{|x| \mid x \in E \text { and } x \neq 0\} $$  as one may verify by an argument based on division with remainder. Thus we have  $$ \{f(x) \mid x \in \mathbb{Z}\}=\{k \cdot t+d \mid t \in \mathbb{Z}\} $$  Due to (5) and (7) we get  $$ f(k \cdot t)=k \cdot t+d $$  for all $t \in \mathbb{Z}$, whence in particular $f(k)=k+d$. So by comparing the results of substituting $y=0$ and $y=k$ into (1) we learn that  $$ f(z+k)=f(z)+k $$  holds for all integers $z$. In plain English, this means that on any residue class modulo $k$ the function $f$ is linear with slope 1.  Now by (7) the set of all values attained by $f$ is such a residue class. Hence, there exists an absolute constant $c$ such that $f(f(x))=f(x)+c$ holds for all $x \in \mathbb{Z}$. Thereby (1) simplifies to  $$ f(x-f(y))=f(x)-f(y)+c-1 $$  On the other hand, considering (1) modulo $k$ we obtain $d \equiv-1(\bmod k)$ because of (7). So by (7) again, $f$ attains the value -1 .  Thus we may apply (9) to some integer $y$ with $f(y)=-1$, which gives $f(x+1)=f(x)+c$. So $f$ is a linear function with slope $c$. Hence, (8) leads to $c=1$, wherefore there is an absolute constant $d^{\prime}$ with $f(x)=x+d^{\prime}$ for all $x \in \mathbb{Z}$. Using this for $x=0$ we obtain $d^{\prime}=d$ and finally (4) discloses $d=1$, meaning that $f$ is indeed the successor function.


Let $n$ be a fixed positive integer. Find the maximum possible value of  $$ \sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s} $$  where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$.

Solution: Let $Z$ be the expression to be maximized. Since this expression is linear in every variable $x_{i}$ and $-1 \leqslant x_{i} \leqslant 1$, the maximum of $Z$ will be achieved when $x_{i}=-1$ or 1 . Therefore, it suffices to consider only the case when $x_{i} \in\{-1,1\}$ for all $i=1,2, \ldots, 2 n$.  For $i=1,2, \ldots, 2 n$, we introduce auxiliary variables  $$ y_{i}=\sum_{r=1}^{i} x_{r}-\sum_{r=i+1}^{2 n} x_{r} $$  Taking squares of both sides, we have  $$ \begin{aligned} y_{i}^{2} & =\sum_{r=1}^{2 n} x_{r}^{2}+\sum_{r<s \leqslant i} 2 x_{r} x_{s}+\sum_{i<r<s} 2 x_{r} x_{s}-\sum_{r \leqslant i<s} 2 x_{r} x_{s} \\ & =2 n+\sum_{r<s \leqslant i} 2 x_{r} x_{s}+\sum_{i<r<s} 2 x_{r} x_{s}-\sum_{r \leqslant i<s} 2 x_{r} x_{s}, \end{aligned} $$  where the last equality follows from the fact that $x_{r} \in\{-1,1\}$. Notice that for every $r<s$, the coefficient of $x_{r} x_{s}$ in (1) is 2 for each $i=1, \ldots, r-1, s, \ldots, 2 n$, and this coefficient is -2 for each $i=r, \ldots, s-1$. This implies that the coefficient of $x_{r} x_{s}$ in $\sum_{i=1}^{2 n} y_{i}^{2}$ is $2(2 n-s+r)-2(s-r)=$ $4(n-s+r)$. Therefore, summing (1) for $i=1,2, \ldots, 2 n$ yields  $$ \sum_{i=1}^{2 n} y_{i}^{2}=4 n^{2}+\sum_{1 \leqslant r<s \leqslant 2 n} 4(n-s+r) x_{r} x_{s}=4 n^{2}-4 Z $$  Hence, it suffices to find the minimum of the left-hand side. Since $x_{r} \in\{-1,1\}$, we see that $y_{i}$ is an even integer. In addition, $y_{i}-y_{i-1}=2 x_{i}= \pm 2$, and so $y_{i-1}$ and $y_{i}$ are consecutive even integers for every $i=2,3, \ldots, 2 n$. It follows that $y_{i-1}^{2}+y_{i}^{2} \geqslant 4$, which implies  $$ \sum_{i=1}^{2 n} y_{i}^{2}=\sum_{j=1}^{n}\left(y_{2 j-1}^{2}+y_{2 j}^{2}\right) \geqslant 4 n $$  Combining (2) and (3), we get  $$ 4 n \leqslant \sum_{i=1}^{2 n} y_{i}^{2}=4 n^{2}-4 Z $$  Hence, $Z \leqslant n(n-1)$. If we set $x_{i}=1$ for odd indices $i$ and $x_{i}=-1$ for even indices $i$, then we obtain equality in (3) (and thus in (4)). Therefore, the maximum possible value of $Z$ is $n(n-1)$, as desired.  Comment 1. $Z=n(n-1)$ can be achieved by several other examples. In particular, $x_{i}$ needs not be $\pm 1$. For instance, setting $x_{i}=(-1)^{i}$ for all $2 \leqslant i \leqslant 2 n$, we find that the coefficient of $x_{1}$ in $Z$ is 0 . Therefore, $x_{1}$ can be chosen arbitrarily in the interval $[-1,1]$.  Nevertheless, if $x_{i} \in\{-1,1\}$ for all $i=1,2, \ldots, 2 n$, then the equality $Z=n(n-1)$ holds only when $\left(y_{1}, y_{2}, \ldots, y_{2 n}\right)=(0, \pm 2,0, \pm 2, \ldots, 0, \pm 2)$ or $( \pm 2,0, \pm 2,0, \ldots, \pm 2,0)$. In each case, we can reconstruct $x_{i}$ accordingly. The sum $\sum_{i=1}^{2 n} x_{i}$ in the optimal cases needs not be 0 , but it must equal 0 or $\pm 2$.  Comment 2. Several variations in setting up the auxiliary variables are possible. For instance, one may let $x_{2 n+i}=-x_{i}$ and $y_{i}^{\prime}=x_{i}+x_{i+1}+\cdots+x_{i+n-1}$ for any $1 \leqslant i \leqslant 2 n$. Similarly to Solution 1 , we obtain $Y:=y_{1}^{\prime 2}+y_{2}^{\prime 2}+\cdots+y_{2 n}^{\prime 2}=2 n^{2}-2 Z$. Then, it suffices to show that $Y \geqslant 2 n$. If $n$ is odd, then each $y_{i}^{\prime}$ is odd, and so $y_{i}^{\prime 2} \geqslant 1$. If $n$ is even, then each $y_{i}^{\prime}$ is even. We can check that at least one of $y_{i}^{\prime}, y_{i+1}^{\prime}, y_{n+i}^{\prime}$, and $y_{n+i+1}^{\prime}$ is nonzero, so that $y_{i}^{\prime 2}+y_{i+1}^{\prime 2}+y_{n+i}^{\prime 2}+y_{n+i+1}^{\prime 2} \geqslant 4$; summing these up for $i=1,3, \ldots, n-1$ yields $Y \geqslant 2 n$.


Solution: We present a different method of obtaining the bound $Z \leqslant n(n-1)$. As in the previous solution, we reduce the problem to the case $x_{i} \in\{-1,1\}$. For brevity, we use the notation $[2 n]=\{1,2, \ldots, 2 n\}$.  Consider any $x_{1}, x_{2}, \ldots, x_{2 n} \in\{-1,1\}$. Let  $$ A=\left\{i \in[2 n]: x_{i}=1\right\} \quad \text { and } \quad B=\left\{i \in[2 n]: x_{i}=-1\right\} $$  For any subsets $X$ and $Y$ of [2n] we define  $$ e(X, Y)=\sum_{r<s, r \in X, s \in Y}(s-r-n) $$  One may observe that  $$ e(A, A)+e(A, B)+e(B, A)+e(B, B)=e([2 n],[2 n])=\sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n)=-\frac{(n-1) n(2 n-1)}{3} . $$  Therefore, we have  $$ Z=e(A, A)-e(A, B)-e(B, A)+e(B, B)=2(e(A, A)+e(B, B))+\frac{(n-1) n(2 n-1)}{3} . $$  Thus, we need to maximize $e(A, A)+e(B, B)$, where $A$ and $B$ form a partition of [2n]. Due to the symmetry, we may assume that $|A|=n-p$ and $|B|=n+p$, where $0 \leqslant p \leqslant n$. From now on, we fix the value of $p$ and find an upper bound for $Z$ in terms of $n$ and $p$.  Let $a_{1}<a_{2}<\cdots<a_{n-p}$ and $b_{1}<b_{2}<\cdots<b_{n+p}$ list all elements of $A$ and $B$, respectively. Then  $$ e(A, A)=\sum_{1 \leqslant i<j \leqslant n-p}\left(a_{j}-a_{i}-n\right)=\sum_{i=1}^{n-p}(2 i-1-n+p) a_{i}-\binom{n-p}{2} \cdot n $$  and similarly  $$ e(B, B)=\sum_{i=1}^{n+p}(2 i-1-n-p) b_{i}-\binom{n+p}{2} \cdot n $$  Thus, now it suffices to maximize the value of  $$ M=\sum_{i=1}^{n-p}(2 i-1-n+p) a_{i}+\sum_{i=1}^{n+p}(2 i-1-n-p) b_{i} $$  In order to get an upper bound, we will apply the rearrangement inequality to the sequence $a_{1}, a_{2}, \ldots, a_{n-p}, b_{1}, b_{2}, \ldots, b_{n+p}$ (which is a permutation of $1,2, \ldots, 2 n$ ), together with the sequence of coefficients of these numbers in (8). The coefficients of $a_{i}$ form the sequence  $$ n-p-1, n-p-3, \ldots, 1-n+p $$  and those of $b_{i}$ form the sequence  $$ n+p-1, n+p-3, \ldots, 1-n-p $$  Altogether, these coefficients are, in descending order:  - $n+p+1-2 i$, for $i=1,2, \ldots, p$; - $n-p+1-2 i$, counted twice, for $i=1,2, \ldots, n-p$; and - $-(n+p+1-2 i)$, for $i=p, p-1, \ldots, 1$.  Thus, the rearrangement inequality yields  $$ \begin{aligned} & M \leqslant \sum_{i=1}^{p}(n+p+1-2 i)(2 n+1-i) \\ & \quad+\sum_{i=1}^{n-p}(n-p+1-2 i)((2 n+2-p-2 i)+(2 n+1-p-2 i)) \\ & \quad-\sum_{i=1}^{p}(n+p+1-2 i) i \end{aligned} $$  Finally, combining the information from (5), (6), (7), and (9), we obtain  $$ \begin{aligned} Z \leqslant & \frac{(n-1) n(2 n-1)}{3}-2 n\left(\binom{n-p}{2}+\binom{n+p}{2}\right) \\ & +2 \sum_{i=1}^{p}(n+p+1-2 i)(2 n+1-2 i)+2 \sum_{i=1}^{n-p}(n-p+1-2 i)(4 n-2 p+3-4 i), \end{aligned} $$  which can be simplified to  $$ Z \leqslant n(n-1)-\frac{2}{3} p(p-1)(p+1) $$  Since $p$ is a nonnegative integer, this yields $Z \leqslant n(n-1)$.


Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the equation  $$ f(x+f(x+y))+f(x y)=x+f(x+y)+y f(x) $$  for all real numbers $x$ and $y$.

Solution: Clearly, each of the functions $x \mapsto x$ and $x \mapsto 2-x$ satisfies (1). It suffices now to show that they are the only solutions to the problem.  Suppose that $f$ is any function satisfying (1). Then setting $y=1$ in (1), we obtain  $$ f(x+f(x+1))=x+f(x+1) $$  in other words, $x+f(x+1)$ is a fixed point of $f$ for every $x \in \mathbb{R}$. We distinguish two cases regarding the value of $f(0)$. Case 1. $\quad f(0) \neq 0$. By letting $x=0$ in (1), we have  $$ f(f(y))+f(0)=f(y)+y f(0) $$  So, if $y_{0}$ is a fixed point of $f$, then substituting $y=y_{0}$ in the above equation we get $y_{0}=1$. Thus, it follows from (2) that $x+f(x+1)=1$ for all $x \in \mathbb{R}$. That is, $f(x)=2-x$ for all $x \in \mathbb{R}$. Case 2. $\quad f(0)=0$.  By letting $y=0$ and replacing $x$ by $x+1$ in (1), we obtain  $$ f(x+f(x+1)+1)=x+f(x+1)+1 $$  From (1), the substitution $x=1$ yields  $$ f(1+f(y+1))+f(y)=1+f(y+1)+y f(1) $$  By plugging $x=-1$ into (2), we see that $f(-1)=-1$. We then plug $y=-1$ into (4) and deduce that $f(1)=1$. Hence, (4) reduces to  $$ f(1+f(y+1))+f(y)=1+f(y+1)+y $$  Accordingly, if both $y_{0}$ and $y_{0}+1$ are fixed points of $f$, then so is $y_{0}+2$. Thus, it follows from (2) and (3) that $x+f(x+1)+2$ is a fixed point of $f$ for every $x \in \mathbb{R}$; i.e.,  $$ f(x+f(x+1)+2)=x+f(x+1)+2 $$  Replacing $x$ by $x-2$ simplifies the above equation to  $$ f(x+f(x-1))=x+f(x-1) $$  On the other hand, we set $y=-1$ in (1) and get  $$ f(x+f(x-1))=x+f(x-1)-f(x)-f(-x) $$  Therefore, $f(-x)=-f(x)$ for all $x \in \mathbb{R}$. Finally, we substitute $(x, y)$ by $(-1,-y)$ in (1) and use the fact that $f(-1)=-1$ to get  $$ f(-1+f(-y-1))+f(y)=-1+f(-y-1)+y $$  Since $f$ is an odd function, the above equation becomes  $$ -f(1+f(y+1))+f(y)=-1-f(y+1)+y $$  By adding this equation to (5), we conclude that $f(y)=y$ for all $y \in \mathbb{R}$.


Let $2 \mathbb{Z}+1$ denote the set of odd integers. Find all functions $f: \mathbb{Z} \rightarrow 2 \mathbb{Z}+1$ satisfying  $$ f(x+f(x)+y)+f(x-f(x)-y)=f(x+y)+f(x-y) $$  for every $x, y \in \mathbb{Z}$.

Solution: Throughout the solution, all functions are assumed to map integers to integers. For any function $g$ and any nonzero integer $t$, define  $$ \Delta_{t} g(x)=g(x+t)-g(x) $$  For any nonzero integers $a$ and $b$, notice that $\Delta_{a} \Delta_{b} g=\Delta_{b} \Delta_{a} g$. Moreover, if $\Delta_{a} g=0$ and $\Delta_{b} g=0$, then $\Delta_{a+b} g=0$ and $\Delta_{a t} g=0$ for all nonzero integers $t$. We say that $g$ is $t$-quasiperiodic if $\Delta_{t} g$ is a constant function (in other words, if $\Delta_{1} \Delta_{t} g=0$, or $\Delta_{1} g$ is $t$-periodic). In this case, we call $t$ a quasi-period of $g$. We say that $g$ is quasi-periodic if it is $t$-quasi-periodic for some nonzero integer $t$.  Notice that a quasi-period of $g$ is a period of $\Delta_{1} g$. So if $g$ is quasi-periodic, then its minimal positive quasi-period $t$ divides all its quasi-periods.  We now assume that $f$ satisfies (1). First, by setting $a=x+y$, the problem condition can be rewritten as  $$ \Delta_{f(x)} f(a)=\Delta_{f(x)} f(2 x-a-f(x)) \quad \text { for all } x, a \in \mathbb{Z} $$  Let $b$ be an arbitrary integer and let $k$ be an arbitrary positive integer. Applying (2) when $a$ is substituted by $b, b+f(x), \ldots, b+(k-1) f(x)$ and summing up all these equations, we get  $$ \Delta_{k f(x)} f(b)=\Delta_{k f(x)} f(2 x-b-k f(x)) $$  Notice that a similar argument works when $k$ is negative, so that  $$ \Delta_{M} f(b)=\Delta_{M} f(2 x-b-M) \quad \text { for any nonzero integer } M \text { such that } f(x) \mid M $$  We now prove two lemmas. Lemma 1. For any distinct integers $x$ and $y$, the function $\Delta_{\operatorname{lcm}(f(x), f(y))} f$ is $2(y-x)$-periodic. Proof. Denote $L=\operatorname{lcm}(f(x), f(y))$. Applying (3) twice, we obtain  $$ \Delta_{L} f(b)=\Delta_{L} f(2 x-b-L)=\Delta_{L} f(2 y-(b+2(y-x))-L)=\Delta_{L} f(b+2(y-x)) $$  Thus, the function $\Delta_{L} f$ is $2(y-x)$-periodic, as required. Lemma 2. Let $g$ be a function. If $t$ and $s$ are nonzero integers such that $\Delta_{t s} g=0$ and $\Delta_{t} \Delta_{t} g=0$, then $\Delta_{t} g=0$. Proof. Assume, without loss of generality, that $s$ is positive. Let $a$ be an arbitrary integer. Since $\Delta_{t} \Delta_{t} g=0$, we have  $$ \Delta_{t} g(a)=\Delta_{t} g(a+t)=\cdots=\Delta_{t} g(a+(s-1) t) $$  The sum of these $s$ equal numbers is $\Delta_{t s} g(a)=0$, so each of them is zero, as required.  We now return to the solution. Step 1. We prove that $f$ is quasi-periodic. Let $Q=\operatorname{lcm}(f(0), f(1))$. Applying Lemma 1, we get that the function $g=\Delta_{Q} f$ is 2-periodic. In other words, the values of $g$ are constant on even numbers and on odd numbers separately. Moreover, setting $M=Q$ and $x=b=0$ in (3), we get $g(0)=g(-Q)$. Since 0 and $-Q$ have different parities, the value of $g$ at even numbers is the same as that at odd numbers. Thus, $g$ is constant, which means that $Q$ is a quasi-period of $f$. Step 2. Denote the minimal positive quasi-period of $f$ by $T$. We prove that $T \mid f(x)$ for all integers $x$.  Since an odd number $Q$ is a quasi-period of $f$, the number $T$ is also odd. Now suppose, to the contrary, that there exist an odd prime $p$, a positive integer $\alpha$, and an integer $u$ such that $p^{\alpha} \mid T$ but $p^{\alpha} \nmid f(u)$. Setting $x=u$ and $y=0$ in (1), we have $2 f(u)=f(u+f(u))+f(u-f(u))$, so $p^{\alpha}$ does not divide the value of $f$ at one of the points $u+f(u)$ or $u-f(u)$. Denote this point by $v$.  Let $L=\operatorname{lcm}(f(u), f(v))$. Since $|u-v|=f(u)$, from Lemma 1 we get $\Delta_{2 f(u)} \Delta_{L} f=0$. Hence the function $\Delta_{L} f$ is $2 f(u)$-periodic as well as $T$-periodic, so it is $\operatorname{gcd}(T, 2 f(u))$-periodic, or $\Delta_{\operatorname{gcd}(T, 2 f(u))} \Delta_{L} f=0$. Similarly, observe that the function $\Delta_{\operatorname{gcd}(T, 2 f(u))} f$ is $L$-periodic as well as $T$-periodic, so we may conclude that $\Delta_{\operatorname{gcd}(T, L)} \Delta_{\operatorname{gcd}(T, 2 f(u))} f=0$. Since $p^{\alpha} \nmid L$, both $\operatorname{gcd}(T, 2 f(u))$ and $\operatorname{gcd}(T, L)$ divide $T / p$. We thus obtain $\Delta_{T / p} \Delta_{T / p} f=0$, which yields  $$ \Delta_{T / p} \Delta_{T / p} \Delta_{1} f=0 $$  Since $\Delta_{T} \Delta_{1} f=0$, we can apply Lemma 2 to the function $\Delta_{1} f$, obtaining $\Delta_{T / p} \Delta_{1} f=0$. However, this means that $f$ is $(T / p)$-quasi-periodic, contradicting the minimality of $T$. Our claim is proved.


Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences  $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014) \end{aligned} $$  are permutations of each other. (a) Prove that there exist distinct block-similar polynomials of degree $n+1$. (b) Prove that there do not exist distinct block-similar polynomials of degree $n$.

Solution: For convenience, we set $k=2015=2 \ell+1$. Part (a). Consider the following polynomials of degree $n+1$ :  $$ P(x)=\prod_{i=0}^{n}(x-i k) \quad \text { and } \quad Q(x)=\prod_{i=0}^{n}(x-i k-1) $$  Since $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\cdots=P(n k)$, these polynomials are block-similar (and distinct).  Part (b). For every polynomial $F(x)$ and every nonnegative integer $m$, define $\Sigma_{F}(m)=$ $\sum_{i=1}^{m} F(i)$; in particular, $\Sigma_{F}(0)=0$. It is well-known that for every nonnegative integer $d$ the sum $\sum_{i=1}^{m} i^{d}$ is a polynomial in $m$ of degree $d+1$. Thus $\Sigma_{F}$ may also be regarded as a real polynomial of degree $\operatorname{deg} F+1$ (with the exception that if $F=0$, then $\Sigma_{F}=0$ as well). This allows us to consider the values of $\Sigma_{F}$ at all real points (where the initial definition does not apply).  Assume for the sake of contradiction that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Then both polynomials $\Sigma_{P-Q}(x)$ and $\Sigma_{P^{2}-Q^{2}}(x)$ have roots at the points $0, k, 2 k, \ldots, n k$. This motivates the following lemma, where we use the special polynomial  $$ T(x)=\prod_{i=0}^{n}(x-i k) $$  Lemma. Assume that $F(x)$ is a nonzero polynomial such that $0, k, 2 k, \ldots, n k$ are among the roots of the polynomial $\Sigma_{F}(x)$. Then $\operatorname{deg} F \geqslant n$, and there exists a polynomial $G(x)$ such that $\operatorname{deg} G=\operatorname{deg} F-n$ and $F(x)=T(x) G(x)-T(x-1) G(x-1)$. Proof. If $\operatorname{deg} F<n$, then $\Sigma_{F}(x)$ has at least $n+1$ roots, while its degree is less than $n+1$. Therefore, $\Sigma_{F}(x)=0$ and hence $F(x)=0$, which is impossible. Thus $\operatorname{deg} F \geqslant n$.  The lemma condition yields that $\Sigma_{F}(x)=T(x) G(x)$ for some polynomial $G(x)$ such that $\operatorname{deg} G=\operatorname{deg} \Sigma_{F}-(n+1)=\operatorname{deg} F-n$.  Now, let us define $F_{1}(x)=T(x) G(x)-T(x-1) G(x-1)$. Then for every positive integer $n$ we have  $$ \Sigma_{F_{1}}(n)=\sum_{i=1}^{n}(T(x) G(x)-T(x-1) G(x-1))=T(n) G(n)-T(0) G(0)=T(n) G(n)=\Sigma_{F}(n) $$  so the polynomial $\Sigma_{F-F_{1}}(x)=\Sigma_{F}(x)-\Sigma_{F_{1}}(x)$ has infinitely many roots. This means that this polynomial is zero, which in turn yields $F(x)=F_{1}(x)$, as required.  First, we apply the lemma to the nonzero polynomial $R_{1}(x)=P(x)-Q(x)$. Since the degree of $R_{1}(x)$ is at most $n$, we conclude that it is exactly $n$. Moreover, $R_{1}(x)=\alpha \cdot(T(x)-T(x-1))$ for some nonzero constant $\alpha$.  Our next aim is to prove that the polynomial $S(x)=P(x)+Q(x)$ is constant. Assume the contrary. Then, notice that the polynomial $R_{2}(x)=P(x)^{2}-Q(x)^{2}=R_{1}(x) S(x)$ is also nonzero and satisfies the lemma condition. Since $n<\operatorname{deg} R_{1}+\operatorname{deg} S=\operatorname{deg} R_{2} \leqslant 2 n$, the lemma yields  $$ R_{2}(x)=T(x) G(x)-T(x-1) G(x-1) $$  with some polynomial $G(x)$ with $0<\operatorname{deg} G \leqslant n$. Since the polynomial $R_{1}(x)=\alpha(T(x)-T(x-1))$ divides the polynomial  $$ R_{2}(x)=T(x)(G(x)-G(x-1))+G(x-1)(T(x)-T(x-1)) $$  we get $R_{1}(x) \mid T(x)(G(x)-G(x-1))$. On the other hand,  $$ \operatorname{gcd}\left(T(x), R_{1}(x)\right)=\operatorname{gcd}(T(x), T(x)-T(x-1))=\operatorname{gcd}(T(x), T(x-1))=1 $$  since both $T(x)$ and $T(x-1)$ are the products of linear polynomials, and their roots are distinct. Thus $R_{1}(x) \mid G(x)-G(x-1)$. However, this is impossible since $G(x)-G(x-1)$ is a nonzero polynomial of degree less than $n=\operatorname{deg} R_{1}$.  Thus, our assumption is wrong, and $S(x)$ is a constant polynomial, say $S(x)=\beta$. Notice that the polynomials $(2 P(x)-\beta) / \alpha$ and $(2 Q(x)-\beta) / \alpha$ are also block-similar and distinct. So we may replace the initial polynomials by these ones, thus obtaining two block-similar polynomials $P(x)$ and $Q(x)$ with $P(x)=-Q(x)=T(x)-T(x-1)$. It remains to show that this is impossible.  For every $i=1,2 \ldots, n$, the values $T(i k-k+1)$ and $T(i k-1)$ have the same sign. This means that the values $P(i k-k+1)=T(i k-k+1)$ and $P(i k)=-T(i k-1)$ have opposite signs, so $P(x)$ has a root in each of the $n$ segments $[i k-k+1, i k]$. Since $\operatorname{deg} P=n$, it must have exactly one root in each of them.  Thus, the sequence $P(1), P(2), \ldots, P(k)$ should change sign exactly once. On the other hand, since $P(x)$ and $-P(x)$ are block-similar, this sequence must have as many positive terms as negative ones. Since $k=2 \ell+1$ is odd, this shows that the middle term of the sequence above must be zero, so $P(\ell+1)=0$, or $T(\ell+1)=T(\ell)$. However, this is not true since  $$ |T(\ell+1)|=|\ell+1| \cdot|\ell| \cdot \prod_{i=2}^{n}|\ell+1-i k|<|\ell| \cdot|\ell+1| \cdot \prod_{i=2}^{n}|\ell-i k|=|T(\ell)| $$  where the strict inequality holds because $n \geqslant 2$. We come to the final contradiction.  Comment 1. In the solution above, we used the fact that $k>1$ is odd. One can modify the arguments of the last part in order to work for every (not necessarily odd) sufficiently large value of $k$; namely, when $k$ is even, one may show that the sequence $P(1), P(2), \ldots, P(k)$ has different numbers of positive and negative terms.  On the other hand, the problem statement with $k$ replaced by 2 is false, since the polynomials $P(x)=T(x)-T(x-1)$ and $Q(x)=T(x-1)-T(x)$ are block-similar in this case, due to the fact that $P(2 i-1)=-P(2 i)=Q(2 i)=-Q(2 i-1)=T(2 i-1)$ for all $i=1,2, \ldots, n$. Thus, every complete solution should use the relation $k>2$.  One may easily see that the condition $n \geqslant 2$ is also substantial, since the polynomials $x$ and $k+1-x$ become block-similar if we set $n=1$.  It is easily seen from the solution that the result still holds if we assume that the polynomials have degree at most $n$.


Solution: We provide an alternative argument for part (b). Assume again that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Let $R(x)=P(x)-Q(x)$ and $S(x)=P(x)+Q(x)$. For brevity, we also denote the segment $[(i-1) k+1, i k]$ by $I_{i}$, and the set $\{(i-1) k+1,(i-1) k+2, \ldots, i k\}$ of all integer points in $I_{i}$ by $Z_{i}$. Step 1. We prove that $R(x)$ has exactly one root in each segment $I_{i}, i=1,2, \ldots, n$, and all these roots are simple.  Indeed, take any $i \in\{1,2, \ldots, n\}$ and choose some points $p^{-}, p^{+} \in Z_{i}$ so that  $$ P\left(p^{-}\right)=\min _{x \in Z_{i}} P(x) \quad \text { and } \quad P\left(p^{+}\right)=\max _{x \in Z_{i}} P(x) $$  Since the sequences of values of $P$ and $Q$ in $Z_{i}$ are permutations of each other, we have $R\left(p^{-}\right)=P\left(p^{-}\right)-Q\left(p^{-}\right) \leqslant 0$ and $R\left(p^{+}\right)=P\left(p^{+}\right)-Q\left(p^{+}\right) \geqslant 0$. Since $R(x)$ is continuous, there exists at least one root of $R(x)$ between $p^{-}$and $p^{+}$- thus in $I_{i}$.  So, $R(x)$ has at least one root in each of the $n$ disjoint segments $I_{i}$ with $i=1,2, \ldots, n$. Since $R(x)$ is nonzero and its degree does not exceed $n$, it should have exactly one root in each of these segments, and all these roots are simple, as required.


In Lineland there are $n \geqslant 1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2 n$ bulldozers are distinct. Every time when a right and a left bulldozer confront each other, the larger bulldozer pushes the smaller one off the road. On the other hand, the bulldozers are quite unprotected at their rears; so, if a bulldozer reaches the rear-end of another one, the first one pushes the second one off the road, regardless of their sizes.  Let $A$ and $B$ be two towns, with $B$ being to the right of $A$. We say that town $A$ can sweep town $B$ away if the right bulldozer of $A$ can move over to $B$ pushing off all bulldozers it meets. Similarly, $B$ can sweep $A$ away if the left bulldozer of $B$ can move to $A$ pushing off all bulldozers of all towns on its way.  Prove that there is exactly one town which cannot be swept away by any other one. (Estonia)

Solution: Let $T_{1}, T_{2}, \ldots, T_{n}$ be the towns enumerated from left to right. Observe first that, if town $T_{i}$ can sweep away town $T_{j}$, then $T_{i}$ also can sweep away every town located between $T_{i}$ and $T_{j}$.  We prove the problem statement by strong induction on $n$. The base case $n=1$ is trivial. For the induction step, we first observe that the left bulldozer in $T_{1}$ and the right bulldozer in $T_{n}$ are completely useless, so we may forget them forever. Among the other $2 n-2$ bulldozers, we choose the largest one. Without loss of generality, it is the right bulldozer of some town $T_{k}$ with $k<n$.  Surely, with this large bulldozer $T_{k}$ can sweep away all the towns to the right of it. Moreover, none of these towns can sweep $T_{k}$ away; so they also cannot sweep away any town to the left of $T_{k}$. Thus, if we remove the towns $T_{k+1}, T_{k+2}, \ldots, T_{n}$, none of the remaining towns would change its status of being (un)sweepable away by the others.  Applying the induction hypothesis to the remaining towns, we find a unique town among $T_{1}, T_{2}, \ldots, T_{k}$ which cannot be swept away. By the above reasons, it is also the unique such town in the initial situation. Thus the induction step is established.


Solution: We start with the same enumeration and the same observation as in   Clearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $j<k$, for otherwise $T_{i}$ would be able to sweep $T_{k}$ away as well.  Now we prove two claims, showing together that $T_{k}$ is the unique town which cannot be swept away, and thus establishing the problem statement. Claim 1. $T_{k}$ also cannot be swept away from the left. Proof. Let $T_{m}$ be some town to the left of $T_{k}$. By the choice of $T_{k}$, town $T_{m}$ can be swept away from the right by some town $T_{p}$ with $p>m$. As we have already observed, $p$ cannot be greater than $k$. On the other hand, $T_{m}$ cannot sweep $T_{p}$ away, so a fortiori it cannot sweep $T_{k}$ away.  Claim 2. Any town $T_{m}$ with $m \neq k$ can be swept away by some other town.  Proof. If $m<k$, then $T_{m}$ can be swept away from the right due to the choice of $T_{k}$. In the remaining case we have $m>k$.  Let $T_{p}$ be a town among $T_{k}, T_{k+1}, \ldots, T_{m-1}$ having the largest right bulldozer. We claim that $T_{p}$ can sweep $T_{m}$ away. If this is not the case, then $r_{p}<\ell_{q}$ for some $q$ with $p<q \leqslant m$. But this means that $\ell_{q}$ is greater than all the numbers $r_{i}$ with $k \leqslant i \leqslant m-1$, so $T_{q}$ can sweep $T_{k}$ away. This contradicts the choice of $T_{k}$.  Comment 1. One may employ the same ideas within the inductive approach. Here we sketch such a solution.  Assume that the problem statement holds for the collection of towns $T_{1}, T_{2}, \ldots, T_{n-1}$, so that there is a unique town $T_{i}$ among them which cannot be swept away by any other of them. Thus we need to prove that in the full collection $T_{1}, T_{2}, \ldots, T_{n}$, exactly one of the towns $T_{i}$ and $T_{n}$ cannot be swept away.  If $T_{n}$ cannot sweep $T_{i}$ away, then it remains to prove that $T_{n}$ can be swept away by some other town. This can be established as in the second paragraph of the proof of Claim 2.  If $T_{n}$ can sweep $T_{i}$ away, then it remains to show that $T_{n}$ cannot be swept away by any other town. Since $T_{n}$ can sweep $T_{i}$ away, it also can sweep all the towns $T_{i}, T_{i+1}, \ldots, T_{n-1}$ away, so $T_{n}$ cannot be swept away by any of those. On the other hand, none of the remaining towns $T_{1}, T_{2}, \ldots, T_{i-1}$ can sweep $T_{i}$ away, so that they cannot sweep $T_{n}$ away as well.  Comment 2. Here we sketch yet another inductive approach. Assume that $n>1$. Firstly, we find a town which can be swept away by each of its neighbors (each town has two neighbors, except for the bordering ones each of which has one); we call such town a loser. Such a town exists, because there are $n-1$ pairs of neighboring towns, and in each of them there is only one which can sweep the other away; so there exists a town which is a winner in none of these pairs.  Notice that a loser can be swept away, but it cannot sweep any other town away (due to its neighbors' protection). Now we remove a loser, and suggest its left bulldozer to its right neighbor (if it exists), and its right bulldozer to a left one (if it exists). Surely, a town accepts a suggestion if a suggested bulldozer is larger than the town's one of the same orientation.  Notice that suggested bulldozers are useless in attack (by the definition of a loser), but may serve for defensive purposes. Moreover, each suggested bulldozer's protection works for the same pairs of remaining towns as before the removal.  By the induction hypothesis, the new configuration contains exactly one town which cannot be swept away. The arguments above show that the initial one also satisfies this property.


Solution: We separately prove that $(i)$ there exists a town which cannot be swept away, and that (ii) there is at most one such town. We also make use of the two observations from the previous solutions. To prove ( $i$ ), assume contrariwise that every town can be swept away. Let $t_{1}$ be the leftmost town; next, for every $k=1,2, \ldots$ we inductively choose $t_{k+1}$ to be some town which can sweep $t_{k}$ away. Now we claim that for every $k=1,2, \ldots$, the town $t_{k+1}$ is to the right of $t_{k}$; this leads to the contradiction, since the number of towns is finite.  Induction on $k$. The base case $k=1$ is clear due to the choice of $t_{1}$. Assume now that for all $j$ with $1 \leqslant j<k$, the town $t_{j+1}$ is to the right of $t_{j}$. Suppose that $t_{k+1}$ is situated to the left of $t_{k}$; then it lies between $t_{j}$ and $t_{j+1}$ (possibly coinciding with $t_{j}$ ) for some $j<k$. Therefore, $t_{k+1}$ can be swept away by $t_{j+1}$, which shows that it cannot sweep $t_{j+1}$ away - so $t_{k+1}$ also cannot sweep $t_{k}$ away. This contradiction proves the induction step.  To prove (ii), we also argue indirectly and choose two towns $A$ and $B$ neither of which can be swept away, with $A$ being to the left of $B$. Consider the largest bulldozer $b$ between them (taking into consideration the right bulldozer of $A$ and the left bulldozer of $B$ ). Without loss of generality, $b$ is a left bulldozer; then it is situated in some town to the right of $A$, and this town may sweep $A$ away since nothing prevents it from doing that. A contradiction.  Comment 3. The Problem Selection Committee decided to reformulate this problem. The original formulation was as follows.  Let $n$ be a positive integer. There are $n$ cards in a deck, enumerated from bottom to top with numbers $1,2, \ldots, n$. For each $i=1,2, \ldots, n$, an even number $a_{i}$ is printed on the lower side and an odd number $b_{i}$ is printed on the upper side of the $i^{\text {th }}$ card. We say that the $i^{\text {th }}$ card opens the $j^{\text {th }}$ card, if $i<j$ and $b_{i}<a_{k}$ for every $k=i+1, i+2, \ldots, j$. Similarly, we say that the $i^{\text {th }}$ card closes the $j^{\text {th }}$ card, if $i>j$ and $a_{i}<b_{k}$ for every $k=i-1, i-2, \ldots, j$. Prove that the deck contains exactly one card which is neither opened nor closed by any other card.


Let $\mathcal{V}$ be a finite set of points in the plane. We say that $\mathcal{V}$ is balanced if for any two distinct points $A, B \in \mathcal{V}$, there exists a point $C \in \mathcal{V}$ such that $A C=B C$. We say that $\mathcal{V}$ is center-free if for any distinct points $A, B, C \in \mathcal{V}$, there does not exist a point $P \in \mathcal{V}$ such that $P A=P B=P C$. (a) Show that for all $n \geqslant 3$, there exists a balanced set consisting of $n$ points. (b) For which $n \geqslant 3$ does there exist a balanced, center-free set consisting of $n$ points? (Netherlands)

Solution: Part ( $\boldsymbol{a}$ ). Assume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \ldots, A_{n}$ in counter-clockwise order, and set $\mathcal{V}=\left\{A_{1}, \ldots, A_{n}\right\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, let $k \in\{1,2, \ldots, n\}$ be the solution of $2 k \equiv i+j(\bmod n)$. Then, since $k-i \equiv j-k(\bmod n)$, we have $A_{i} A_{k}=A_{j} A_{k}$, as required.  Now assume that $n$ is even. Consider a regular $(3 n-6)$-gon, and let $O$ be its circumcenter. Again, label its vertices as $A_{1}, \ldots, A_{3 n-6}$ in counter-clockwise order, and choose $\mathcal{V}=$ $\left\{O, A_{1}, A_{2}, \ldots, A_{n-1}\right\}$. We check that $\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, we always have $O A_{i}=O A_{j}$. We now consider the vertices $O$ and $A_{i}$. First note that the triangle $O A_{i} A_{n / 2-1+i}$ is equilateral for all $i \leqslant \frac{n}{2}$. Hence, if $i \leqslant \frac{n}{2}$, then we have $O A_{n / 2-1+i}=A_{i} A_{n / 2-1+i}$; otherwise, if $i>\frac{n}{2}$, then we have $O A_{i-n / 2+1}=A_{i} A_{i-n / 2+1}$. This completes the proof.  An example of such a construction when $n=10$ is shown in Figure 1. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-28.jpg?height=507&width=804&top_left_y=1574&top_left_x=272)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-28.jpg?height=600&width=617&top_left_y=1479&top_left_x=1185)  Figure 2  Comment (a). There are many ways to construct an example by placing equilateral triangles in a circle. Here we present one general method.  Let $O$ be the center of a circle and let $A_{1}, B_{1}, \ldots, A_{k}, B_{k}$ be distinct points on the circle such that the triangle $O A_{i} B_{i}$ is equilateral for each $i$. Then $\mathcal{V}=\left\{O, A_{1}, B_{1}, \ldots, A_{k}, B_{k}\right\}$ is balanced. To construct a set of even cardinality, put extra points $C, D, E$ on the circle such that triangles $O C D$ and $O D E$ are equilateral (see Figure 2). Then $\mathcal{V}=\left\{O, A_{1}, B_{1}, \ldots, A_{k}, B_{k}, C, D, E\right\}$ is balanced.  Part (b). We now show that there exists a balanced, center-free set containing $n$ points for all odd $n \geqslant 3$, and that one does not exist for any even $n \geqslant 3$.  If $n$ is odd, then let $\mathcal{V}$ be the set of vertices of a regular $n$-gon. We have shown in part ( $a$ ) that $\mathcal{V}$ is balanced. We claim that $\mathcal{V}$ is also center-free. Indeed, if $P$ is a point such that $P A=P B=P C$ for some three distinct vertices $A, B$ and $C$, then $P$ is the circumcenter of the $n$-gon, which is not contained in $\mathcal{V}$.  Now suppose that $\mathcal{V}$ is a balanced, center-free set of even cardinality $n$. We will derive a contradiction. For a pair of distinct points $A, B \in \mathcal{V}$, we say that a point $C \in \mathcal{V}$ is associated with the pair $\{A, B\}$ if $A C=B C$. Since there are $\frac{n(n-1)}{2}$ pairs of points, there exists a point $P \in \mathcal{V}$ which is associated with at least $\left\lceil\frac{n(n-1)}{2} / n\right\rceil=\frac{n}{2}$ pairs. Note that none of these $\frac{n}{2}$ pairs can contain $P$, so that the union of these $\frac{n}{2}$ pairs consists of at most $n-1$ points. Hence there exist two such pairs that share a point. Let these two pairs be $\{A, B\}$ and $\{A, C\}$. Then $P A=P B=P C$, which is a contradiction.  Comment (b). We can rephrase the argument in graph theoretic terms as follows. Let $\mathcal{V}$ be a balanced, center-free set consisting of $n$ points. For any pair of distinct vertices $A, B \in \mathcal{V}$ and for any $C \in \mathcal{V}$ such that $A C=B C$, draw directed edges $A \rightarrow C$ and $B \rightarrow C$. Then all pairs of vertices generate altogether at least $n(n-1)$ directed edges; since the set is center-free, these edges are distinct. So we must obtain a graph in which any two vertices are connected in both directions. Now, each vertex has exactly $n-1$ incoming edges, which means that $n-1$ is even. Hence $n$ is odd.


For a finite set $A$ of positive integers, we call a partition of $A$ into two disjoint nonempty subsets $A_{1}$ and $A_{2}$ good if the least common multiple of the elements in $A_{1}$ is equal to the greatest common divisor of the elements in $A_{2}$. Determine the minimum value of $n$ such that there exists a set of $n$ positive integers with exactly 2015 good partitions. (Ukraine)

Solution: Let $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, where $a_{1}<a_{2}<\cdots<a_{n}$. For a finite nonempty set $B$ of positive integers, denote by $\operatorname{lcm} B$ and $\operatorname{gcd} B$ the least common multiple and the greatest common divisor of the elements in $B$, respectively.  Consider any good partition $\left(A_{1}, A_{2}\right)$ of $A$. By definition, $\operatorname{lcm} A_{1}=d=\operatorname{gcd} A_{2}$ for some positive integer $d$. For any $a_{i} \in A_{1}$ and $a_{j} \in A_{2}$, we have $a_{i} \leqslant d \leqslant a_{j}$. Therefore, we have $A_{1}=\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ and $A_{2}=\left\{a_{k+1}, a_{k+2}, \ldots, a_{n}\right\}$ for some $k$ with $1 \leqslant k<n$. Hence, each good partition is determined by an element $a_{k}$, where $1 \leqslant k<n$. We call such $a_{k}$ partitioning.  It is convenient now to define $\ell_{k}=\operatorname{lcm}\left(a_{1}, a_{2}, \ldots, a_{k}\right)$ and $g_{k}=\operatorname{gcd}\left(a_{k+1}, a_{k+2}, \ldots, a_{n}\right)$ for $1 \leqslant k \leqslant n-1$. So $a_{k}$ is partitioning exactly when $\ell_{k}=g_{k}$.  We proceed by proving some properties of partitioning elements, using the following claim. Claim. If $a_{k-1}$ and $a_{k}$ are partitioning where $2 \leqslant k \leqslant n-1$, then $g_{k-1}=g_{k}=a_{k}$. Proof. Assume that $a_{k-1}$ and $a_{k}$ are partitioning. Since $\ell_{k-1}=g_{k-1}$, we have $\ell_{k-1} \mid a_{k}$. Therefore, $g_{k}=\ell_{k}=\operatorname{lcm}\left(\ell_{k-1}, a_{k}\right)=a_{k}$, and $g_{k-1}=\operatorname{gcd}\left(a_{k}, g_{k}\right)=a_{k}$, as desired. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-30.jpg?height=67&width=1692&top_left_y=1274&top_left_x=185) Proof. Suppose, to the contrary, that all three numbers $a_{k-1}, a_{k}$, and $a_{k+1}$ are partitioning. The claim yields that $a_{k+1}=g_{k}=a_{k}$, a contradiction. Property 2. The elements $a_{1}$ and $a_{2}$ cannot be simultaneously partitioning. Also, $a_{n-2}$ and $\overline{a_{n-1} \text { cannot be simultaneously partitioning }}$ Proof. Assume that $a_{1}$ and $a_{2}$ are partitioning. By the claim, it follows that $a_{2}=g_{1}=\ell_{1}=$ $\operatorname{lcm}\left(a_{1}\right)=a_{1}$, a contradiction.  Similarly, assume that $a_{n-2}$ and $a_{n-1}$ are partitioning. The claim yields that $a_{n-1}=g_{n-1}=$ $\operatorname{gcd}\left(a_{n}\right)=a_{n}$, a contradiction.  Now let $A$ be an $n$-element set with exactly 2015 good partitions. Clearly, we have $n \geqslant 5$. Using Property 2, we find that there is at most one partitioning element in each of $\left\{a_{1}, a_{2}\right\}$ and $\left\{a_{n-2}, a_{n-1}\right\}$. By Property 1 , there are at least $\left\lfloor\frac{n-5}{3}\right\rfloor$ non-partitioning elements in $\left\{a_{3}, a_{4}, \ldots, a_{n-3}\right\}$. Therefore, there are at most $(n-1)-2-\left\lfloor\frac{n-5}{3}\right\rfloor=\left\lceil\frac{2(n-2)}{3}\right\rceil$ partitioning elements in $A$. Thus, $\left\lceil\frac{2(n-2)}{3}\right\rceil \geqslant 2015$, which implies that $n \geqslant 3024$.  Finally, we show that there exists a set of 3024 positive integers with exactly 2015 partitioning elements. Indeed, in the set $A=\left\{2 \cdot 6^{i}, 3 \cdot 6^{i}, 6^{i+1} \mid 0 \leqslant i \leqslant 1007\right\}$, each element of the form $3 \cdot 6^{i}$ or $6^{i}$, except $6^{1008}$, is partitioning.  Therefore, the minimum possible value of $n$ is 3024 . Comment. Here we will work out the general case when 2015 is replaced by an arbitrary positive integer $m$. Note that the bound $\left\lceil\frac{2(n-2)}{3}\right\rceil \geqslant m$ obtained in the solution is, in fact, true for any positive integers $m$ and $n$. Using this bound, one can find that $n \geqslant\left\lceil\frac{3 m}{2}\right\rceil+1$.  To show that the bound is sharp, one constructs a set of $\left\lceil\frac{3 m}{2}\right\rceil+1$ elements with exactly $m$ good partitions. Indeed, the minimum is attained on the set $\left\{6^{i}, 2 \cdot 6^{i}, 3 \cdot 6^{i} \mid 0 \leqslant i \leqslant t-1\right\} \cup\left\{6^{t}\right\}$ for every even $m=2 t$, and $\left\{2 \cdot 6^{i}, 3 \cdot 6^{i}, 6^{i+1} \mid 0 \leqslant i \leqslant t-1\right\}$ for every odd $m=2 t-1$.


Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \leqslant n$. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of those the player has already chosen on any previous turn. (iii) The game is a draw if all numbers have been chosen; otherwise the player who cannot choose a number anymore loses the game.  The player $A$ takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies. (Finland)

Solution: For brevity, we denote by $[n]$ the set $\{1,2, \ldots, n\}$. Firstly, we show that $B$ wins whenever $n \neq 1,2,4,6$. For this purpose, we provide a strategy which guarantees that $B$ can always make a move after $A$ 's move, and also guarantees that the game does not end in a draw.  We begin with an important observation. Lemma. Suppose that $B$ 's first pick is $n$ and that $A$ has made the $k^{\text {th }}$ move where $k \geqslant 2$. Then $B$ can also make the $k^{\text {th }}$ move. Proof. Let $\mathcal{S}$ be the set of the first $k$ numbers chosen by $A$. Since $\mathcal{S}$ does not contain consecutive integers, we see that the set $[n] \backslash \mathcal{S}$ consists of $k$ "contiguous components" if $1 \in \mathcal{S}$, and $k+1$ components otherwise. Since $B$ has chosen only $k-1$ numbers, there is at least one component of $[n] \backslash \mathcal{S}$ consisting of numbers not yet picked by $B$. Hence, $B$ can choose a number from this component.  We will now describe a winning strategy for $B$, when $n \neq 1,2,4,6$. By symmetry, we may assume that $A$ 's first choice is a number not exceeding $\frac{n+1}{2}$. So $B$ can pick the number $n$ in $B$ 's first turn. We now consider two cases. Case 1. $n$ is odd and $n \geqslant 3$. The only way the game ends in a draw is that $A$ eventually picks all the odd numbers from the set $[n]$. However, $B$ has already chosen $n$, so this cannot happen. Thus $B$ can continue to apply the lemma until $A$ cannot make a move.  Case 2. $n$ is even and $n \geqslant 8$. Since $B$ has picked $n$, the game is a draw only if $A$ can eventually choose all the odd numbers from the set $[n-1]$. So $B$ picks a number from the set $\{1,3,5, \ldots, n-3\}$ not already chosen by $A$, on $B$ 's second move. This is possible since the set consists of $\frac{n-2}{2} \geqslant 3$ numbers and $A$ has chosen only 2 numbers. Hereafter $B$ can apply the lemma until $A$ cannot make a move.  Hence, in both cases $A$ loses. We are left with the cases $n=1,2,4,6$. The game is trivially a draw when $n=1,2$. When $n=4, A$ has to first pick 1 to avoid losing. Similarly, $B$ has to choose 4 as well. It then follows that the game ends in a draw.  When $n=6, B$ gets at least a draw by the lemma or by using a mirror strategy. On the other hand, $A$ may also get at least a draw in the following way. In the first turn, $A$ chooses 1. After $B$ 's response by a number $b, A$ finds a neighbor $c$ of $b$ which differs from 1 and 2 , and reserves $c$ for $A$ 's third move. Now, clearly $A$ can make the second move by choosing a number different from $1,2, c-1, c, c+1$. Therefore $A$ will not lose.  Comment 1. We present some explicit winning strategies for $B$. We start with the case $n$ is odd and $n \geqslant 3 . B$ starts by picking $n$ in the first turn. On the $k^{\text {th }}$ move for $k \geqslant 2, B$ chooses the number exactly 1 less than $A$ 's $k^{\text {th }}$ pick. The only special case is when $A$ 's $k^{\text {th }}$ choice is 1 . In this situation, $A^{\prime}$ 's first pick was a number $a>1$ and $B$ can respond by choosing $a-1$ on the $k^{\text {th }}$ move instead.  We now give an alternative winning strategy in the case $n$ is even and $n \geqslant 8$. We first present a winning strategy for the case when $A$ 's first pick is 1 . We consider two cases depending on $A$ 's second move.  Case 1. A's second pick is 3 . Then $B$ chooses $n-3$ on the second move. On the $k^{\text {th }}$ move, $B$ chooses the number exactly 1 less than $A^{\prime}$ 's $k^{\text {th }}$ pick except that $B$ chooses 2 if $A$ 's $k^{\text {th }}$ pick is $n-2$ or $n-1$.  Case 2. A's second pick is $a>3$. Then $B$ chooses $a-2$ on the second move. Afterwards on the $k^{\text {th }}$ move, $B$ picks the number exactly 1 less than $A^{\text {'s }} k^{\text {th }}$ pick.  One may easily see that this strategy guarantees $B$ 's victory, when $A$ 's first pick is 1 . The following claim shows how to extend the strategy to the general case. Claim. Assume that $B$ has an explicit strategy leading to a victory after $A$ picks 1 on the first move. Then $B$ also has an explicit strategy leading to a victory after any first moves of $A$. Proof. Let $S$ be an optimal strategy of $B$ after $A$ picks 1 on the first move. Assume that $A$ picks some number $a>1$ on this move; we show how $B$ can make use of $S$ in order to win in this case.  In parallel to the real play, $B$ starts an imaginary play. The positions in these plays differ by flipping the segment $[1, a]$; so, if a player chooses some number $x$ in the real play, then the same player chooses a number $x$ or $a+1-x$ in the imaginary play, depending on whether $x>a$ or $x \leqslant a$. Thus $A$ 's first pick in the imaginary play is 1.  Clearly, a number is chosen in the real play exactly if the corresponding number is chosen in the imaginary one. Next, if an unchosen number is neighboring to one chosen by $A$ in the imaginary play, then the corresponding number also has this property in the real play, so $A$ also cannot choose it. One can easily see that a similar statement with real and imaginary plays interchanged holds for $B$ instead of $A$.  Thus, when $A$ makes some move in the real play, $B$ may imagine the corresponding legal move in the imaginary one. Then $B$ chooses the response according to $S$ in the imaginary game and makes the corresponding legal move in the real one. Acting so, $B$ wins the imaginary game, thus $B$ will also win the real one.  Hence, $B$ has a winning strategy for all even $n$ greater or equal to 8 . Notice that the claim can also be used to simplify the argument when $n$ is odd. Comment 2. One may also employ symmetry when $n$ is odd. In particular, $B$ could use a mirror strategy. However, additional ideas are required to modify the strategy after $A$ picks $\frac{n+1}{2}$.


Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$.  Prove that there exist two positive integers $b$ and $N$ such that  $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant 1007^{2} $$  whenever $n>m \geqslant N$.

Solution: We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \neq n+a_{n}$ for $m \neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1 , that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \geqslant s$ at least once.  Suppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \leqslant b \leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required.  To see this, let any two integers $m$ and $n$ with $n>m \geqslant N$ be given. The sum $\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \ldots, x_{b}$ denote these numbers, and let $y_{1}, \ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain  $$ \sum_{i=m+1}^{n} a_{i}=\sum_{j=1}^{b}\left(y_{j}-x_{j}\right) $$  whence  $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum_{j=1}^{b}\left(y_{j}-n\right)-\sum_{j=1}^{b}\left(x_{j}-m\right) . $$  Now each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}-$ $m, \ldots, x_{b}-m$ are $b$ distinct members of the set $\{1,2, \ldots, 2015\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore  $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(x_{j}-m\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$  The same argument applied to $n$ and $y_{1}, \ldots, y_{b}$ yields  $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(y_{j}-n\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$  So altogether we get  $$ \begin{aligned} \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| & \leqslant \sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\ & \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2}, \end{aligned} $$  as desired.


Solution: Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have  $$ n+1 \leqslant s_{n} \leqslant n+2015 $$  for all $n \in \mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \ldots$ are distinct. We shall investigate the set  $$ M=\mathbb{Z}_{>0} \backslash\left\{s_{1}, s_{2}, \ldots\right\} $$  Claim. At most 2015 numbers belong to $M$. Proof. Otherwise let $m_{1}<m_{2}<\cdots<m_{2016}$ be any 2016 distinct elements from $M$. For $n=m_{2016}$ we have  $$ \left\{s_{1}, \ldots, s_{n}\right\} \cup\left\{m_{1}, \ldots, m_{2016}\right\} \subseteq\{1,2, \ldots, n+2015\} $$  where on the left-hand side we have a disjoint union containing altogether $n+2016$ elements. But the set on the right-hand side has only $n+2015$ elements. This contradiction proves our claim.  Now we work towards proving that the positive integers $b=|M|$ and $N=\max (M)$ are as required. Recall that we have just shown $b \leqslant 2015$.  Let us consider any integer $r \geqslant N$. As in the proof of the above claim, we see that  $$ B_{r}=M \cup\left\{s_{1}, \ldots, s_{r}\right\} $$  is a subset of $[1, r+2015] \cap \mathbb{Z}$ with precisely $b+r$ elements. Due to the definitions of $M$ and $N$, we also know $[1, r+1] \cap \mathbb{Z} \subseteq B_{r}$. It follows that there is a set $C_{r} \subseteq\{1,2, \ldots, 2014\}$ with $\left|C_{r}\right|=b-1$ and  $$ B_{r}=([1, r+1] \cap \mathbb{Z}) \cup\left\{r+1+x \mid x \in C_{r}\right\} $$  For any finite set of integers $J$ we denote the sum of its elements by $\sum J$. Now the equations (1) and (2) give rise to two ways of computing $\sum B_{r}$ and the comparison of both methods leads to  $$ \sum M+\sum_{i=1}^{r} s_{i}=\sum_{i=1}^{r} i+b(r+1)+\sum C_{r} $$  or in other words to  $$ \sum M+\sum_{i=1}^{r}\left(a_{i}-b\right)=b+\sum C_{r} $$  After this preparation, we consider any two integers $m$ and $n$ with $n>m \geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce  $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum C_{n}-\sum C_{m} $$  Since $C_{n}$ and $C_{m}$ are subsets of $\{1,2, \ldots, 2014\}$ with $\left|C_{n}\right|=\left|C_{m}\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\{1,2, \ldots, b-1\}$ and $C_{n}=\{2016-b, \ldots, 2014\}$, or the other way around. In these two cases we have  $$ \left|\sum C_{n}-\sum C_{m}\right|=(b-1)(2015-b) $$  so in the general case we find  $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant(b-1)(2015-b) \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2} $$  as desired.  Comment. The sets $C_{n}$ may be visualized by means of the following process: Start with an empty blackboard. For $n \geqslant 1$, the following happens during the $n^{\text {th }}$ step. The number $a_{n}$ gets written on the blackboard, then all numbers currently on the blackboard are decreased by 1 , and finally all zeros that have arisen get swept away.  It is not hard to see that the numbers present on the blackboard after $n$ steps are distinct and form the set $C_{n}$. Moreover, it is possible to complete a solution based on this idea.


Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.

Solution: Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\mathbb{Z}_{>0}$ denote the set of all positive integers.  Suppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n>N$ has exactly one odd representation.  Clearly, $S$ is infinite. We now claim the following properties of odd and even representations. $\underline{P r o p e r t y}$ 1. Any positive integer $n$ has at most one odd and at most one even representation. Proof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \in S$ such that $x>\max \{n, N\}$. Then, the number $n+x$ must be clean, and $x$ does not appear in any even representation of $n$. If $n$ has more than one even representation, then we obtain two distinct odd representations of $n+x$ by adding $x$ to the even representations of $n$, which is impossible. Therefore, $n$ can have at most one even representation.  Similarly, there exist two distinct elements $y, z \in S$ such that $y, z>\max \{n, N\}$. If $n$ has more than one odd representation, then we obtain two distinct odd representations of $n+y+z$ by adding $y$ and $z$ to the odd representations of $n$. This is again a contradiction.  Property 2. Fix $s \in S$. Suppose that a number $n>N$ has no even representation. Then $n+2 a s$ has an even representation containing $s$ for all integers $a \geqslant 1$. Proof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n+2 s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1).  Notice that the odd representation of $n+s$ does not contain $s$; otherwise, we have an even representation of $n$ without $s$. Then, adding $s$ to this odd representation of $n+s$, we get that $n+2 s$ has an even representation containing $s$, as desired.  Property 3. Every sufficiently large integer has an even representation. Proof. Fix any $s \in S$, and let $r$ be an arbitrary element in $\{1,2, \ldots, 2 s\}$. Then, Property 2 implies that the set $Z_{r}=\{r+2 a s: a \geqslant 0\}$ contains at most one number exceeding $N$ with no even representation. Therefore, $Z_{r}$ contains finitely many positive integers with no even representation, and so does $\mathbb{Z}_{>0}=\bigcup_{r=1}^{2 s} Z_{r}$.  In view of Properties 1 and 3 , we may assume that $N$ is chosen such that every $n>N$ has exactly one odd and exactly one even representation. In particular, each element $s>N$ of $S$ has an even representation. Property 4. For any $s, t \in S$ with $N<s<t$, the even representation of $t$ contains $s$. Proof. Suppose the contrary. Then, $s+t$ has at least two odd representations: one obtained by adding $s$ to the even representation of $t$ and one obtained by adding $t$ to the even representation of $s$. Since the latter does not contain $s$, these two odd representations of $s+t$ are distinct, a contradiction.  Let $s_{1}<s_{2}<\cdots$ be all the elements of $S$, and set $\sigma_{n}=\sum_{i=1}^{n} s_{i}$ for each nonnegative integer $n$. Fix an integer $k$ such that $s_{k}>N$. Then, Property 4 implies that for every $i>k$ the even representation of $s_{i}$ contains all the numbers $s_{k}, s_{k+1}, \ldots, s_{i-1}$. Therefore,  $$ s_{i}=s_{k}+s_{k+1}+\cdots+s_{i-1}+R_{i}=\sigma_{i-1}-\sigma_{k-1}+R_{i} $$  where $R_{i}$ is a sum of some of $s_{1}, \ldots, s_{k-1}$. In particular, $0 \leqslant R_{i} \leqslant s_{1}+\cdots+s_{k-1}=\sigma_{k-1}$.  Let $j_{0}$ be an integer satisfying $j_{0}>k$ and $\sigma_{j_{0}}>2 \sigma_{k-1}$. Then (1) shows that, for every $j>j_{0}$,  $$ s_{j+1} \geqslant \sigma_{j}-\sigma_{k-1}>\sigma_{j} / 2 . $$  Next, let $p>j_{0}$ be an index such that $R_{p}=\min _{i>j_{0}} R_{i}$. Then,  $$ s_{p+1}=s_{k}+s_{k+1}+\cdots+s_{p}+R_{p+1}=\left(s_{p}-R_{p}\right)+s_{p}+R_{p+1} \geqslant 2 s_{p} $$  Therefore, there is no element of $S$ larger than $s_{p}$ but smaller than $2 s_{p}$. It follows that the even representation $\tau$ of $2 s_{p}$ does not contain any element larger than $s_{p}$. On the other hand, inequality (2) yields $2 s_{p}>s_{1}+\cdots+s_{p-1}$, so $\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_{p}$. After removing $s_{p}$ from $\tau$, we have that $s_{p}$ has an odd representation not containing $s_{p}$, which contradicts Property 1 since $s_{p}$ itself also forms an odd representation of $s_{p}$.


Solution: We will also use Property 1 from  We first define some terminology and notations used in this solution. Let $\mathbb{Z}_{\geqslant 0}$ denote the set of all nonnegative integers. All sums mentioned are regarded as sums of distinct elements of $S$. Moreover, a sum is called even or odd depending on the parity of the number of terms in it. All closed or open intervals refer to sets of all integers inside them, e.g., $[a, b]=\{x \in \mathbb{Z}: a \leqslant x \leqslant b\}$.  Again, let $s_{1}<s_{2}<\cdots$ be all elements of $S$, and denote $\sigma_{n}=\sum_{i=1}^{n} s_{i}$ for each positive integer $n$. Let $O_{n}$ (respectively, $E_{n}$ ) be the set of numbers representable as an odd (respectively, even) sum of elements of $\left\{s_{1}, \ldots, s_{n}\right\}$. Set $E=\bigcup_{n=1}^{\infty} E_{n}$ and $O=\bigcup_{n=1}^{\infty} O_{n}$. We assume that $0 \in E_{n}$ since 0 is representable as a sum of 0 terms.  We now proceed to our proof. Assume, to the contrary, that there exist only finitely many positive integers that are not clean and denote the number of non-clean positive integers by $m-1$. Clearly, $S$ is infinite. By Property 1 from Solution 1 , every positive integer $n$ has at most one odd and at most one even representation. Step 1. We estimate $s_{n+1}$ and $\sigma_{n+1}$. Upper bounds: Property 1 yields $\left|O_{n}\right|=\left|E_{n}\right|=2^{n-1}$, so $\left|\left[1,2^{n-1}+m\right] \backslash O_{n}\right| \geqslant m$. Hence, there exists a clean integer $x_{n} \in\left[1,2^{n-1}+m\right] \backslash O_{n}$. The definition of $O_{n}$ then yields that the odd representation of $x_{n}$ contains a term larger than $s_{n}$. Therefore, $s_{n+1} \leqslant x_{n} \leqslant 2^{n-1}+m$ for every positive integer $n$. Moreover, since $s_{1}$ is the smallest clean number, we get $\sigma_{1}=s_{1} \leqslant m$. Then,  $$ \sigma_{n+1}=\sum_{i=2}^{n+1} s_{i}+s_{1} \leqslant \sum_{i=2}^{n+1}\left(2^{i-2}+m\right)+m=2^{n}-1+(n+1) m $$  for every positive integer $n$. Notice that this estimate also holds for $n=0$. Lower bounds: Since $O_{n+1} \subseteq\left[1, \sigma_{n+1}\right]$, we have $\sigma_{n+1} \geqslant\left|O_{n+1}\right|=2^{n}$ for all positive integers $n$. Then,  $$ s_{n+1}=\sigma_{n+1}-\sigma_{n} \geqslant 2^{n}-\left(2^{n-1}-1+n m\right)=2^{n-1}+1-n m $$  for every positive integer $n$. Combining the above inequalities, we have  $$ 2^{n-1}+1-n m \leqslant s_{n+1} \leqslant 2^{n-1}+m \quad \text { and } \quad 2^{n} \leqslant \sigma_{n+1} \leqslant 2^{n}-1+(n+1) m $$  for every positive integer $n$. Step 2. We prove Property 3 from Solution 1. For every integer $x$ and set of integers $Y$, define $x \pm Y=\{x \pm y: y \in Y\}$. In view of Property 1, we get  $$ E_{n+1}=E_{n} \sqcup\left(s_{n+1}+O_{n}\right) \quad \text { and } \quad O_{n+1}=O_{n} \sqcup\left(s_{n+1}+E_{n}\right), $$  where $\sqcup$ denotes the disjoint union operator. Notice also that $s_{n+2} \geqslant 2^{n}+1-(n+1) m>$ $2^{n-1}-1+n m \geqslant \sigma_{n}$ for every sufficiently large $n$. We now claim the following.  Claim 1. $\left(\sigma_{n}-s_{n+1}, s_{n+2}-s_{n+1}\right) \subseteq E_{n}$ for every sufficiently large $n$. Proof. For sufficiently large $n$, all elements of $\left(\sigma_{n}, s_{n+2}\right)$ are clean. Clearly, the elements of $\left(\sigma_{n}, s_{n+2}\right)$ can be in neither $O_{n}$ nor $O \backslash O_{n+1}$. So, $\left(\sigma_{n}, s_{n+2}\right) \subseteq O_{n+1} \backslash O_{n}=s_{n+1}+E_{n}$, which yields the claim.  Now, Claim 1 together with inequalities (3) implies that, for all sufficiently large $n$,  $$ E \supseteq E_{n} \supseteq\left(\sigma_{n}-s_{n+1}, s_{n+2}-s_{n+1}\right) \supseteq\left(2 n m, 2^{n-1}-(n+2) m\right) . $$  This easily yields that $\mathbb{Z}_{\geqslant 0} \backslash E$ is also finite. Since $\mathbb{Z}_{\geqslant 0} \backslash O$ is also finite, by Property 1 , there exists a positive integer $N$ such that every integer $n>N$ has exactly one even and one odd representation.  Step 3. We investigate the structures of $E_{n}$ and $O_{n}$. Suppose that $z \in E_{2 n}$. Since $z$ can be represented as an even sum using $\left\{s_{1}, s_{2}, \ldots, s_{2 n}\right\}$, so can its complement $\sigma_{2 n}-z$. Thus, we get $E_{2 n}=\sigma_{2 n}-E_{2 n}$. Similarly, we have  $$ E_{2 n}=\sigma_{2 n}-E_{2 n}, \quad O_{2 n}=\sigma_{2 n}-O_{2 n}, \quad E_{2 n+1}=\sigma_{2 n+1}-O_{2 n+1}, \quad O_{2 n+1}=\sigma_{2 n+1}-E_{2 n+1} $$  Claim 2. For every sufficiently large $n$, we have  $$ \left[0, \sigma_{n}\right] \supseteq O_{n} \supseteq\left(N, \sigma_{n}-N\right) \quad \text { and } \quad\left[0, \sigma_{n}\right] \supseteq E_{n} \supseteq\left(N, \sigma_{n}-N\right) $$  Proof. Clearly $O_{n}, E_{n} \subseteq\left[0, \sigma_{n}\right]$ for every positive integer $n$. We now prove $O_{n}, E_{n} \supseteq\left(N, \sigma_{n}-N\right)$. Taking $n$ sufficiently large, we may assume that $s_{n+1} \geqslant 2^{n-1}+1-n m>\frac{1}{2}\left(2^{n-1}-1+n m\right) \geqslant \sigma_{n} / 2$. Therefore, the odd representation of every element of ( $\left.N, \sigma_{n} / 2\right]$ cannot contain a term larger than $s_{n}$. Thus, $\left(N, \sigma_{n} / 2\right] \subseteq O_{n}$. Similarly, since $s_{n+1}+s_{1}>\sigma_{n} / 2$, we also have $\left(N, \sigma_{n} / 2\right] \subseteq E_{n}$. Equations (4) then yield that, for sufficiently large $n$, the interval $\left(N, \sigma_{n}-N\right)$ is a subset of both $O_{n}$ and $E_{n}$, as desired.  Step 4. We obtain a final contradiction. Notice that $0 \in \mathbb{Z}_{\geqslant 0} \backslash O$ and $1 \in \mathbb{Z}_{\geqslant 0} \backslash E$. Therefore, the sets $\mathbb{Z}_{\geqslant 0} \backslash O$ and $\mathbb{Z}_{\geqslant 0} \backslash E$ are nonempty. Denote $o=\max \left(\mathbb{Z}_{\geqslant 0} \backslash O\right)$ and $e=\max \left(\mathbb{Z}_{\geqslant 0} \backslash E\right)$. Observe also that $e, o \leqslant N$.  Taking $k$ sufficiently large, we may assume that $\sigma_{2 k}>2 N$ and that Claim 2 holds for all $n \geqslant 2 k$. Due to (4) and Claim 2, we have that $\sigma_{2 k}-e$ is the minimal number greater than $N$ which is not in $E_{2 k}$, i.e., $\sigma_{2 k}-e=s_{2 k+1}+s_{1}$. Similarly,  $$ \sigma_{2 k}-o=s_{2 k+1}, \quad \sigma_{2 k+1}-e=s_{2 k+2}, \quad \text { and } \quad \sigma_{2 k+1}-o=s_{2 k+2}+s_{1} $$  Therefore, we have  $$ \begin{aligned} s_{1} & =\left(s_{2 k+1}+s_{1}\right)-s_{2 k+1}=\left(\sigma_{2 k}-e\right)-\left(\sigma_{2 k}-o\right)=o-e \\ & =\left(\sigma_{2 k+1}-e\right)-\left(\sigma_{2 k+1}-o\right)=s_{2 k+2}-\left(s_{2 k+2}+s_{1}\right)=-s_{1} \end{aligned} $$  which is impossible since $s_{1}>0$.


In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least 3 , and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most 2015 unsociable groups, prove that it is possible to partition the company into 11 parts so that no two enemies are in the same part.

Solution: Let $G=(V, E)$ be a graph where $V$ is the set of people in the company and $E$ is the set of the enemy pairs - the edges of the graph. In this language, partitioning into 11 disjoint enemy-free subsets means properly coloring the vertices of this graph with 11 colors.  We will prove the following more general statement. Claim. Let $G$ be a graph with chromatic number $k \geqslant 3$. Then $G$ contains at least $2^{k-1}-k$ unsociable groups.  Recall that the chromatic number of $G$ is the least $k$ such that a proper coloring  $$ V=V_{1} \sqcup \cdots \sqcup V_{k} $$  exists. In view of $2^{11}-12>2015$, the claim implies the problem statement. Let $G$ be a graph with chromatic number $k$. We say that a proper coloring (1) of $G$ is leximinimal, if the $k$-tuple $\left(\left|V_{1}\right|,\left|V_{2}\right|, \ldots,\left|V_{k}\right|\right)$ is lexicographically minimal; in other words, the following conditions are satisfied: the number $n_{1}=\left|V_{1}\right|$ is minimal; the number $n_{2}=\left|V_{2}\right|$ is minimal, subject to the previously chosen value of $n_{1} ; \ldots$; the number $n_{k-1}=\left|V_{k-1}\right|$ is minimal, subject to the previously chosen values of $n_{1}, \ldots, n_{k-2}$.  The following lemma is the core of the proof. Lemma 1. Suppose that $G=(V, E)$ is a graph with odd chromatic number $k \geqslant 3$, and let (1) be one of its leximinimal colorings. Then $G$ contains an odd cycle which visits all color classes $V_{1}, V_{2}, \ldots, V_{k}$. Proof of Lemma 1. Let us call a cycle colorful if it visits all color classes. Due to the definition of the chromatic number, $V_{1}$ is nonempty. Choose an arbitrary vertex $v \in V_{1}$. We construct a colorful odd cycle that has only one vertex in $V_{1}$, and this vertex is $v$.  We draw a subgraph of $G$ as follows. Place $v$ in the center, and arrange the sets $V_{2}, V_{3}, \ldots, V_{k}$ in counterclockwise circular order around it. For convenience, let $V_{k+1}=V_{2}$. We will draw arrows to add direction to some edges of $G$, and mark the vertices these arrows point to. First we draw arrows from $v$ to all its neighbors in $V_{2}$, and mark all those neighbors. If some vertex $u \in V_{i}$ with $i \in\{2,3, \ldots, k\}$ is already marked, we draw arrows from $u$ to all its neighbors in $V_{i+1}$ which are not marked yet, and we mark all of them. We proceed doing this as long as it is possible. The process of marking is exemplified in Figure 1.  Notice that by the rules of our process, in the final state, marked vertices in $V_{i}$ cannot have unmarked neighbors in $V_{i+1}$. Moreover, $v$ is connected to all marked vertices by directed paths.  Now move each marked vertex to the next color class in circular order (see an example in Figure 3). In view of the arguments above, the obtained coloring $V_{1} \sqcup W_{2} \sqcup \cdots \sqcup W_{k}$ is proper. Notice that $v$ has a neighbor $w \in W_{2}$, because otherwise  $$ \left(V_{1} \backslash\{v\}\right) \sqcup\left(W_{2} \cup\{v\}\right) \sqcup W_{3} \sqcup \cdots \sqcup W_{k} $$  would be a proper coloring lexicographically smaller than (1). If $w$ was unmarked, i.e., $w$ was an element of $V_{2}$, then it would be marked at the beginning of the process and thus moved to $V_{3}$, which did not happen. Therefore, $w$ is marked and $w \in V_{k}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-40.jpg?height=780&width=1658&top_left_y=181&top_left_x=219)  Figure 1 Figure 2 Since $w$ is marked, there exists a directed path from $v$ to $w$. This path moves through the sets $V_{2}, \ldots, V_{k}$ in circular order, so the number of edges in it is divisible by $k-1$ and thus even. Closing this path by the edge $w \rightarrow v$, we get a colorful odd cycle, as required.  Proof of the claim. Let us choose a leximinimal coloring (1) of $G$. For every set $C \subseteq\{1,2, \ldots, k\}$ such that $|C|$ is odd and greater than 1 , we will provide an odd cycle visiting exactly those color classes whose indices are listed in the set $C$. This property ensures that we have different cycles for different choices of $C$, and it proves the claim because there are $2^{k-1}-k$ choices for the set $C$.  Let $V_{C}=\bigcup_{c \in C} V_{c}$, and let $G_{C}$ be the induced subgraph of $G$ on the vertex set $V_{C}$. We also have the induced coloring of $V_{C}$ with $|C|$ colors; this coloring is of course proper. Notice further that the induced coloring is leximinimal: if we had a lexicographically smaller coloring $\left(W_{c}\right)_{c \in C}$ of $G_{C}$, then these classes, together the original color classes $V_{i}$ for $i \notin C$, would provide a proper coloring which is lexicographically smaller than (1). Hence Lemma 1, applied to the subgraph $G_{C}$ and its leximinimal coloring $\left(V_{c}\right)_{c \in C}$, provides an odd cycle that visits exactly those color classes that are listed in the set $C$.


Solution: We provide a different proof of the claim from the previous solution. We say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number. Lemma 2. Suppose that $G=(V, E)$ is a critical graph with chromatic number $k \geqslant 3$. Then every vertex $v$ of $G$ is contained in at least $2^{k-2}-1$ unsociable groups. Proof. For every set $X \subseteq V$, denote by $n(X)$ the number of neighbors of $v$ in the set $X$. Since $G$ is critical, there exists a proper coloring of $G \backslash\{v\}$ with $k-1$ colors, so there exists a proper coloring $V=V_{1} \sqcup V_{2} \sqcup \cdots \sqcup V_{k}$ of $G$ such that $V_{1}=\{v\}$. Among such colorings, take one for which the sequence $\left(n\left(V_{2}\right), n\left(V_{3}\right), \ldots, n\left(V_{k}\right)\right)$ is lexicographically minimal. Clearly, $n\left(V_{i}\right)>0$ for every $i=2,3, \ldots, k$; otherwise $V_{2} \sqcup \ldots \sqcup V_{i-1} \sqcup\left(V_{i} \cup V_{1}\right) \sqcup V_{i+1} \sqcup \ldots V_{k}$ would be a proper coloring of $G$ with $k-1$ colors.  We claim that for every $C \subseteq\{2,3, \ldots, k\}$ with $|C| \geqslant 2$ being even, $G$ contains an unsociable group so that the set of its members' colors is precisely $C \cup\{1\}$. Since the number of such sets $C$ is $2^{k-2}-1$, this proves the lemma. Denote the elements of $C$ by $c_{1}, \ldots, c_{2 \ell}$ in increasing order. For brevity, let $U_{i}=V_{c_{i}}$. Denote by $N_{i}$ the set of neighbors of $v$ in $U_{i}$.  We show that for every $i=1, \ldots, 2 \ell-1$ and $x \in N_{i}$, the subgraph induced by $U_{i} \cup U_{i+1}$ contains a path that connects $x$ with another point in $N_{i+1}$. For the sake of contradiction, suppose that no such path exists. Let $S$ be the set of vertices that lie in the connected component of $x$ in the subgraph induced by $U_{i} \cup U_{i+1}$, and let $P=U_{i} \cap S$, and $Q=U_{i+1} \cap S$ (see Figure 3). Since $x$ is separated from $N_{i+1}$, the sets $Q$ and $N_{i+1}$ are disjoint. So, if we re-color $G$ by replacing $U_{i}$ and $U_{i+1}$ by $\left(U_{i} \cup Q\right) \backslash P$ and $\left(U_{i+1} \cup P\right) \backslash Q$, respectively, we obtain a proper coloring such that $n\left(U_{i}\right)=n\left(V_{c_{i}}\right)$ is decreased and only $n\left(U_{i+1}\right)=n\left(V_{c_{i+1}}\right)$ is increased. That contradicts the lexicographical minimality of $\left(n\left(V_{2}\right), n\left(V_{3}\right), \ldots, n\left(V_{k}\right)\right)$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-41.jpg?height=438&width=851&top_left_y=632&top_left_x=611)  Figure 3 Next, we build a path through $U_{1}, U_{2}, \ldots, U_{2 \ell}$ as follows. Let the starting point of the path be an arbitrary vertex $v_{1}$ in the set $N_{1}$. For $i \leqslant 2 \ell-1$, if the vertex $v_{i} \in N_{i}$ is already defined, connect $v_{i}$ to some vertex in $N_{i+1}$ in the subgraph induced by $U_{i} \cup U_{i+1}$, and add these edges to the path. Denote the new endpoint of the path by $v_{i+1}$; by the construction we have $v_{i+1} \in N_{i+1}$ again, so the process can be continued. At the end we have a path that starts at $v_{1} \in N_{1}$ and ends at some $v_{2 \ell} \in N_{2 \ell}$. Moreover, all edges in this path connect vertices in neighboring classes: if a vertex of the path lies in $U_{i}$, then the next vertex lies in $U_{i+1}$ or $U_{i-1}$. Notice that the path is not necessary simple, so take a minimal subpath of it. The minimal subpath is simple and connects the same endpoints $v_{1}$ and $v_{2 \ell}$. The property that every edge steps to a neighboring color class (i.e., from $U_{i}$ to $U_{i+1}$ or $U_{i-1}$ ) is preserved. So the resulting path also visits all of $U_{1}, \ldots, U_{2 \ell}$, and its length must be odd. Closing the path with the edges $v v_{1}$ and $v_{2 \ell} v$ we obtain the desired odd cycle (see Figure 4). ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-41.jpg?height=518&width=1489&top_left_y=1803&top_left_x=286)  Figure 4 Now we prove the claim by induction on $k \geqslant 3$. The base case $k=3$ holds by applying Lemma 2 to a critical subgraph. For the induction step, let $G_{0}$ be a critical $k$-chromatic subgraph of $G$, and let $v$ be an arbitrary vertex of $G_{0}$. By Lemma 2, $G_{0}$ has at least $2^{k-2}-1$ unsociable groups containing $v$. On the other hand, the graph $G_{0} \backslash\{v\}$ has chromatic number $k-1$, so it contains at least $2^{k-2}-(k-1)$ unsociable groups by the induction hypothesis. Altogether, this gives $2^{k-2}-1+2^{k-2}-(k-1)=2^{k-1}-k$ distinct unsociable groups in $G_{0}$ (and thus in $G$ ).  Comment 1. The claim we proved is sharp. The complete graph with $k$ vertices has chromatic number $k$ and contains exactly $2^{k-1}-k$ unsociable groups.  Comment 2. The proof of Lemma 2 works for odd values of $|C| \geqslant 3$ as well. Hence, the second solution shows the analogous statement that the number of even sized unsociable groups is at least $2^{k}-1-\binom{k}{2}$.


Let $A B C$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $A B G H$ is a parallelogram. Let $I$ be the point on the line $G H$ such that $A C$ bisects $H I$. Suppose that the line $A C$ intersects the circumcircle of the triangle $G C I$ at $C$ and $J$. Prove that $I J=A H$. (Australia)

Solution: Since $H G \| A B$ and $B G \| A H$, we have $B G \perp B C$ and $C H \perp G H$. Therefore, the quadrilateral $B G C H$ is cyclic. Since $H$ is the orthocenter of the triangle $A B C$, we have $\angle H A C=90^{\circ}-\angle A C B=\angle C B H$. Using that $B G C H$ and $C G J I$ are cyclic quadrilaterals, we get  $$ \angle C J I=\angle C G H=\angle C B H=\angle H A C . $$  Let $M$ be the intersection of $A C$ and $G H$, and let $D \neq A$ be the point on the line $A C$ such that $A H=H D$. Then $\angle M J I=\angle H A C=\angle M D H$.  Since $\angle M J I=\angle M D H, \angle I M J=\angle H M D$, and $I M=M H$, the triangles $I M J$ and $H M D$ are congruent, and thus $I J=H D=A H$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-44.jpg?height=863&width=901&top_left_y=1062&top_left_x=583)  Comment. Instead of introducing the point $D$, one can complete the solution by using the law of sines in the triangles $I J M$ and $A M H$, yielding  $$ \frac{I J}{I M}=\frac{\sin \angle I M J}{\sin \angle M J I}=\frac{\sin \angle A M H}{\sin \angle H A M}=\frac{A H}{M H}=\frac{A H}{I M} . $$


Solution: Obtain $\angle C G H=\angle H A C$ as in the previous solution. In the parallelogram $A B G H$ we have $\angle B A H=\angle H G B$. It follows that  $$ \angle H M C=\angle B A C=\angle B A H+\angle H A C=\angle H G B+\angle C G H=\angle C G B . $$  So the right triangles $C M H$ and $C G B$ are similar. Also, in the circumcircle of triangle $G C I$ we have similar triangles $M I J$ and $M C G$. Therefore,  $$ \frac{I J}{C G}=\frac{M I}{M C}=\frac{M H}{M C}=\frac{G B}{G C}=\frac{A H}{C G} $$  Hence $I J=A H$.


Let $A B C$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $B C$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $A B$ of $\Omega$ not containing $C$, and $G$ lies on the arc $A C$ of $\Omega$ not containing $B$. The circumcircles of the triangles $B D F$ and $C E G$ meet the sides $A B$ and $A C$ again at $K$ and $L$, respectively. Suppose that the lines $F K$ and $G L$ are distinct and intersect at $X$. Prove that the points $A, X$, and $O$ are collinear. (Greece)

Solution: It suffices to prove that the lines $F K$ and $G L$ are symmetric about $A O$. Now the segments $A F$ and $A G$, being chords of $\Omega$ with the same length, are clearly symmetric with respect to $A O$. Hence it is enough to show  $$ \angle K F A=\angle A G L . $$  Let us denote the circumcircles of $B D F$ and $C E G$ by $\omega_{B}$ and $\omega_{C}$, respectively. To prove (1), we start from  $$ \angle K F A=\angle D F G+\angle G F A-\angle D F K $$  In view of the circles $\omega_{B}, \Gamma$, and $\Omega$, this may be rewritten as  $$ \angle K F A=\angle C E G+\angle G B A-\angle D B K=\angle C E G-\angle C B G . $$  Due to the circles $\omega_{C}$ and $\Omega$, we obtain $\angle K F A=\angle C L G-\angle C A G=\angle A G L$. Thereby the problem is solved. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-45.jpg?height=729&width=1220&top_left_y=1389&top_left_x=418)  Figure 1


Solution: Again, we denote the circumcircle of $B D K F$ by $\omega_{B}$. In addition, we set $\alpha=$ $\angle B A C, \varphi=\angle A B F$, and $\psi=\angle E D A=\angle A E D$ (see Figure 2). Notice that $A F=A G$ entails $\varphi=\angle G C A$, so all three of $\alpha, \varphi$, and $\psi$ respect the "symmetry" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1).  This time, we start from  $$ 2 \angle K F A=2(\angle D F A-\angle D F K) . $$  Since the triangle $A F D$ is isosceles, we have  $$ \angle D F A=\angle A D F=\angle E D F-\psi=\angle B F D+\angle E B F-\psi . $$  Moreover, because of the circle $\omega_{B}$ we have $\angle D F K=\angle C B A$. Altogether, this yields  $$ 2 \angle K F A=\angle D F A+(\angle B F D+\angle E B F-\psi)-2 \angle C B A, $$  which simplifies to  $$ 2 \angle K F A=\angle B F A+\varphi-\psi-\angle C B A . $$  Now the quadrilateral $A F B C$ is cyclic, so this entails $2 \angle K F A=\alpha+\varphi-\psi$. Due to the "symmetry" between $B$ and $C$ alluded to above, this argument also shows that $2 \angle A G L=\alpha+\varphi-\psi$. This concludes the proof of (1). ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-46.jpg?height=547&width=926&top_left_y=726&top_left_x=571)  Figure 2  Comment 1. As the first solution shows, the assumption that $A$ be the center of $\Gamma$ may be weakened to the following one: The center of $\Gamma$ lies on the line $O A$. The second solution may be modified to yield the same result.  Comment 2. It might be interesting to remark that $\angle G D K=90^{\circ}$. To prove this, let $G^{\prime}$ denote the point on $\Gamma$ diametrically opposite to $G$. Because of $\angle K D F=\angle K B F=\angle A G F=\angle G^{\prime} D F$, the points $D, K$, and $G^{\prime}$ are collinear, which leads to the desired result. Notice that due to symmetry we also have $\angle L E F=90^{\circ}$.  Moreover, a standard argument shows that the triangles $A G L$ and $B G E$ are similar. By symmetry again, also the triangles $A F K$ and $C D F$ are similar.  There are several ways to derive a solution from these facts. For instance, one may argue that  $$ \begin{aligned} \angle K F A & =\angle B F A-\angle B F K=\angle B F A-\angle E D G^{\prime}=\left(180^{\circ}-\angle A G B\right)-\left(180^{\circ}-\angle G^{\prime} G E\right) \\ & =\angle A G E-\angle A G B=\angle B G E=\angle A G L . \end{aligned} $$  Comment 3. The original proposal did not contain the point $X$ in the assumption and asked instead to prove that the lines $F K, G L$, and $A O$ are concurrent. This differs from the version given above only insofar as it also requires to show that these lines cannot be parallel. The Problem Selection Committee removed this part from the problem intending to make it thus more suitable for the Olympiad.  For the sake of completeness, we would still like to sketch one possibility for proving $F K \nVdash A O$ here. As the points $K$ and $O$ lie in the angular region $\angle F A G$, it suffices to check $\angle K F A+\angle F A O<180^{\circ}$. Multiplying by 2 and making use of the formulae from the second solution, we see that this is equivalent to $(\alpha+\varphi-\psi)+\left(180^{\circ}-2 \varphi\right)<360^{\circ}$, which in turn is an easy consequence of $\alpha<180^{\circ}$.


Let $A B C$ be a triangle with $\angle C=90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $C B H$ so that $C H$ bisects $A D$. Let $P$ be the intersection point of the lines $B D$ and $C H$. Let $\omega$ be the semicircle with diameter $B D$ that meets the segment $C B$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $C Q$ and $A D$ meet on $\omega$. (Georgia)

Solution: Let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 1). Since $P H \| D K$, we have  $$ \frac{P D}{P B}=\frac{H K}{H B}=\frac{A H}{H B} $$  Let $L$ be the projection of $Q$ onto $D B$. Since $P Q$ is tangent to $\omega$ and $\angle D Q B=\angle B L Q=$ $90^{\circ}$, we have $\angle P Q D=\angle Q B P=\angle D Q L$. Therefore, $Q D$ and $Q B$ are respectively the internal and the external bisectors of $\angle P Q L$. By the angle bisector theorem, we obtain  $$ \frac{P D}{D L}=\frac{P Q}{Q L}=\frac{P B}{B L} $$  The relations (1) and (2) yield $\frac{A H}{H B}=\frac{P D}{P B}=\frac{D L}{L B}$. So, the spiral similarity $\tau$ centered at $B$ and sending $A$ to $D$ maps $H$ to $L$. Moreover, $\tau$ sends the semicircle with diameter $A B$ passing through $C$ to $\omega$. Due to $C H \perp A B$ and $Q L \perp D B$, it follows that $\tau(C)=Q$.  Hence, the triangles $A B D$ and $C B Q$ are similar, so $\angle A D B=\angle C Q B$. This means that the lines $A D$ and $C Q$ meet at some point $T$, and this point satisfies $\angle B D T=\angle B Q T$. Therefore, $T$ lies on $\omega$, as needed. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-47.jpg?height=438&width=801&top_left_y=1397&top_left_x=225)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-47.jpg?height=464&width=798&top_left_y=1367&top_left_x=1046)  Figure 2  Comment 1. Since $\angle B A D=\angle B C Q$, the point $T$ lies also on the circumcircle of the triangle $A B C$.


Solution: Let $\Gamma$ be the circumcircle of $A B C$, and let $A D$ meet $\omega$ at $T$. Then $\angle A T B=$ $\angle A C B=90^{\circ}$, so $T$ lies on $\Gamma$ as well. As in the previous solution, let $K$ be the projection of $D$ onto $A B$; then $A H=H K$ (see Figure 2).  Our goal now is to prove that the points $C, Q$, and $T$ are collinear. Let $C T$ meet $\omega$ again at $Q^{\prime}$. Then, it suffices to show that $P Q^{\prime}$ is tangent to $\omega$, or that $\angle P Q^{\prime} D=\angle Q^{\prime} B D$.  Since the quadrilateral $B D Q^{\prime} T$ is cyclic and the triangles $A H C$ and $K H C$ are congruent, we have $\angle Q^{\prime} B D=\angle Q^{\prime} T D=\angle C T A=\angle C B A=\angle A C H=\angle H C K$. Hence, the right triangles $C H K$ and $B Q^{\prime} D$ are similar. This implies that $\frac{H K}{C K}=\frac{Q^{\prime} D}{B D}$, and thus $H K \cdot B D=C K \cdot Q^{\prime} D$.  Notice that $P H \| D K$; therefore, we have $\frac{P D}{B D}=\frac{H K}{B K}$, and so $P D \cdot B K=H K \cdot B D$. Consequently, $P D \cdot B K=H K \cdot B D=C K \cdot Q^{\prime} D$, which yields $\frac{P D}{Q^{\prime} D}=\frac{C K}{B K}$.  Since $\angle C K A=\angle K A C=\angle B D Q^{\prime}$, the triangles $C K B$ and $P D Q^{\prime}$ are similar, so $\angle P Q^{\prime} D=$ $\angle C B A=\angle Q^{\prime} B D$, as required.  Comment 2. There exist several other ways to prove that $P Q^{\prime}$ is tangent to $\omega$. For instance, one may compute $\frac{P D}{P B}$ and $\frac{P Q^{\prime}}{P B}$ in terms of $A H$ and $H B$ to verify that $P Q^{\prime 2}=P D \cdot P B$, concluding that $P Q^{\prime}$ is tangent to $\omega$.  Another possible approach is the following. As in Solution 2, we introduce the points $T$ and $Q^{\prime}$ and mention that the triangles $A B C$ and $D B Q^{\prime}$ are similar (see Figure 3).  Let $M$ be the midpoint of $A D$, and let $L$ be the projection of $Q^{\prime}$ onto $A B$. Construct $E$ on the line $A B$ so that $E P$ is parallel to $A D$. Projecting from $P$, we get $(A, B ; H, E)=(A, D ; M, \infty)=-1$. Since $\frac{E A}{A B}=\frac{P D}{D B}$, the point $P$ is the image of $E$ under the similarity transform mapping $A B C$ to $D B Q^{\prime}$. Therefore, we have $(D, B ; L, P)=(A, B ; H, E)=-1$, which means that $Q^{\prime} D$ and $Q^{\prime} B$ are respectively the internal and the external bisectors of $\angle P Q^{\prime} L$. This implies that $P Q^{\prime}$ is tangent to $\omega$, as required. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-48.jpg?height=415&width=1026&top_left_y=820&top_left_x=521)  Figure 3


Solution: Introduce the points $T$ and $Q^{\prime}$ as in the previous solution. Note that $T$ lies on the circumcircle of $A B C$. Here we present yet another proof that $P Q^{\prime}$ is tangent to $\omega$.  Let $\Omega$ be the circle completing the semicircle $\omega$. Construct a point $F$ symmetric to $C$ with respect to $A B$. Let $S \neq T$ be the second intersection point of $F T$ and $\Omega$ (see Figure 4). ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-48.jpg?height=772&width=855&top_left_y=1553&top_left_x=606)  Figure 4 Since $A C=A F$, we have $\angle D K C=\angle H C K=\angle C B A=\angle C T A=\angle D T S=180^{\circ}-$ $\angle S K D$. Thus, the points $C, K$, and $S$ are collinear. Notice also that $\angle Q^{\prime} K D=\angle Q^{\prime} T D=$ $\angle H C K=\angle K F H=180^{\circ}-\angle D K F$. This implies that the points $F, K$, and $Q^{\prime}$ are collinear.  Applying PASCAL's theorem to the degenerate hexagon $K Q^{\prime} Q^{\prime} T S S$, we get that the tangents to $\Omega$ passing through $Q^{\prime}$ and $S$ intersect on $C F$. The relation $\angle Q^{\prime} T D=\angle D T S$ yields that $Q^{\prime}$ and $S$ are symmetric with respect to $B D$. Therefore, the two tangents also intersect on $B D$. Thus, the two tangents pass through $P$. Hence, $P Q^{\prime}$ is tangent to $\omega$, as needed.


Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triangle $A B C$. Determine all possible values of $B T / B M$. (Russia)

Solution: Let $S$ be the center of the parallelogram $B P T Q$, and let $B^{\prime} \neq B$ be the point on the ray $B M$ such that $B M=M B^{\prime}$ (see Figure 1). It follows that $A B C B^{\prime}$ is a parallelogram. Then, $\angle A B B^{\prime}=\angle P Q M$ and $\angle B B^{\prime} A=\angle B^{\prime} B C=\angle M P Q$, and so the triangles $A B B^{\prime}$ and $M Q P$ are similar. It follows that $A M$ and $M S$ are corresponding medians in these triangles. Hence,  $$ \angle S M P=\angle B^{\prime} A M=\angle B C A=\angle B T A . $$  Since $\angle A C T=\angle P B T$ and $\angle T A C=\angle T B C=\angle B T P$, the triangles $T C A$ and $P B T$ are similar. Again, as $T M$ and $P S$ are corresponding medians in these triangles, we have  $$ \angle M T A=\angle T P S=\angle B Q P=\angle B M P . $$  Now we deal separately with two cases. Case 1. $\quad S$ does not lie on $B M$. Since the configuration is symmetric between $A$ and $C$, we may assume that $S$ and $A$ lie on the same side with respect to the line $B M$.  Applying (1) and (2), we get  $$ \angle B M S=\angle B M P-\angle S M P=\angle M T A-\angle B T A=\angle M T B, $$  and so the triangles $B S M$ and $B M T$ are similar. We now have $B M^{2}=B S \cdot B T=B T^{2} / 2$, so $B T=\sqrt{2} B M$. Case 2. $\quad S$ lies on $B M$. It follows from (2) that $\angle B C A=\angle M T A=\angle B Q P=\angle B M P$ (see Figure 2). Thus, $P Q \| A C$ and $P M \| A T$. Hence, $B S / B M=B P / B A=B M / B T$, so $B T^{2}=2 B M^{2}$ and $B T=\sqrt{2} B M$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-49.jpg?height=960&width=730&top_left_y=1713&top_left_x=246)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-49.jpg?height=749&width=729&top_left_y=1938&top_left_x=1092)  Figure 2  Comment 1. Here is another way to show that the triangles $B S M$ and $B M T$ are similar. Denote by $\Omega$ the circumcircle of the triangle $A B C$. Let $R$ be the second point of intersection of $\omega$ and $\Omega$, and let $\tau$ be the spiral similarity centered at $R$ mapping $\omega$ to $\Omega$. Then, one may show that $\tau$ maps each point $X$ on $\omega$ to a point $Y$ on $\Omega$ such that $B, X$, and $Y$ are collinear (see Figure 3). If we let $K$ and $L$ be the second points of intersection of $B M$ with $\Omega$ and of $B T$ with $\omega$, respectively, then it follows that the triangle $M K T$ is the image of $S M L$ under $\tau$. We now obtain $\angle B S M=\angle T M B$, which implies the desired result. ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-50.jpg?height=758&width=757&top_left_y=752&top_left_x=204)  Figure 3 ![](https://cdn.mathpix.com/cropped/2024_11_18_62dbd531c72d0ff8164cg-50.jpg?height=954&width=846&top_left_y=551&top_left_x=1002)  Figure 4