CZE   Find the largest possible integer $k$, such that the following statement is true: Let 2009 arbitrary non-degenerated triangles be given. In every triangle the three sides are colored, such that one is blue, one is red and one is white. Now, for every color separately, let us sort the lengths of the sides. We obtain  $$ \begin{aligned} & b_{1} \leq b_{2} \\ & \leq \ldots \leq b_{2009} \quad \text { the lengths of the blue sides, } \\ r_{1} & \leq r_{2} \leq \ldots \leq r_{2009} \quad \text { the lengths of the red sides, } \\ \text { and } \quad w_{1} & \leq w_{2} \leq \ldots \leq w_{2009} \quad \text { the lengths of the white sides. } \end{aligned} $$  Then there exist $k$ indices $j$ such that we can form a non-degenerated triangle with side lengths $b_{j}, r_{j}, w_{j}$ 。

Solution: We will prove that the largest possible number $k$ of indices satisfying the given condition is one. Firstly we prove that $b_{2009}, r_{2009}, w_{2009}$ are always lengths of the sides of a triangle. Without loss of generality we may assume that $w_{2009} \geq r_{2009} \geq b_{2009}$. We show that the inequality $b_{2009}+r_{2009}>w_{2009}$ holds. Evidently, there exists a triangle with side lengths $w, b, r$ for the white, blue and red side, respectively, such that $w_{2009}=w$. By the conditions of the problem we have $b+r>w, b_{2009} \geq b$ and $r_{2009} \geq r$. From these inequalities it follows  $$ b_{2009}+r_{2009} \geq b+r>w=w_{2009} $$  Secondly we will describe a sequence of triangles for which $w_{j}, b_{j}, r_{j}$ with $j<2009$ are not the lengths of the sides of a triangle. Let us define the sequence $\Delta_{j}, j=1,2, \ldots, 2009$, of triangles, where $\Delta_{j}$ has a blue side of length $2 j$, a red side of length $j$ for all $j \leq 2008$ and 4018 for $j=2009$, and a white side of length $j+1$ for all $j \leq 2007$, 4018 for $j=2008$ and 1 for $j=2009$. Since  $$ \begin{array}{rrrl} (j+1)+j>2 j & \geq j+1>j, & \text { if } & j \leq 2007 \\ 2 j+j>4018>2 j \quad>j, & \text { if } & j=2008 \\ 4018+1>2 j & =4018>1, & \text { if } & j=2009 \end{array} $$  such a sequence of triangles exists. Moreover, $w_{j}=j, r_{j}=j$ and $b_{j}=2 j$ for $1 \leq j \leq 2008$. Then  $$ w_{j}+r_{j}=j+j=2 j=b_{j} $$  i.e., $b_{j}, r_{j}$ and $w_{j}$ are not the lengths of the sides of a triangle for $1 \leq j \leq 2008$.


EST   Let $a, b, c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that  $$ \frac{1}{(2 a+b+c)^{2}}+\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \leq \frac{3}{16} $$

Solution: For positive real numbers $x, y, z$, from the arithmetic-geometric-mean inequality,  $$ 2 x+y+z=(x+y)+(x+z) \geq 2 \sqrt{(x+y)(x+z)} $$  we obtain  $$ \frac{1}{(2 x+y+z)^{2}} \leq \frac{1}{4(x+y)(x+z)} $$  Applying this to the left-hand side terms of the inequality to prove, we get  $$ \begin{aligned} \frac{1}{(2 a+b+c)^{2}} & +\frac{1}{(2 b+c+a)^{2}}+\frac{1}{(2 c+a+b)^{2}} \\ & \leq \frac{1}{4(a+b)(a+c)}+\frac{1}{4(b+c)(b+a)}+\frac{1}{4(c+a)(c+b)} \\ & =\frac{(b+c)+(c+a)+(a+b)}{4(a+b)(b+c)(c+a)}=\frac{a+b+c}{2(a+b)(b+c)(c+a)} . \end{aligned} $$  A second application of the inequality of the arithmetic-geometric mean yields  $$ a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b \geq 6 a b c $$  or, equivalently,  $$ 9(a+b)(b+c)(c+a) \geq 8(a+b+c)(a b+b c+c a) $$  The supposition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$ can be written as  $$ a b+b c+c a=a b c(a+b+c) . $$  Applying the arithmetic-geometric-mean inequality $x^{2} y^{2}+x^{2} z^{2} \geq 2 x^{2} y z$ thrice, we get  $$ a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq a^{2} b c+a b^{2} c+a b c^{2} $$  which is equivalent to  $$ (a b+b c+c a)^{2} \geq 3 a b c(a+b+c) $$  Combining (1), (2), (3), and (4), we will finish the proof:  $$ \begin{aligned} \frac{a+b+c}{2(a+b)(b+c)(c+a)} & =\frac{(a+b+c)(a b+b c+c a)}{2(a+b)(b+c)(c+a)} \cdot \frac{a b+b c+c a}{a b c(a+b+c)} \cdot \frac{a b c(a+b+c)}{(a b+b c+c a)^{2}} \\ & \leq \frac{9}{2 \cdot 8} \cdot 1 \cdot \frac{1}{3}=\frac{3}{16} \end{aligned} $$


Solution: Equivalently, we prove the homogenized inequality  $$ \frac{(a+b+c)^{2}}{(2 a+b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+2 b+c)^{2}}+\frac{(a+b+c)^{2}}{(a+b+2 c)^{2}} \leq \frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $$  for all positive real numbers $a, b, c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a, b, c>0$, fulfilling this condition, the inequality  $$ \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}} \leq \frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) . $$  Applying Jensen's inequality to the function $f(x)=\frac{x}{(1+x)^{2}}$, which is concave for $0 \leq x \leq 2$ and increasing for $0 \leq x \leq 1$, we obtain  $$ \alpha \frac{a}{(1+a)^{2}}+\beta \frac{b}{(1+b)^{2}}+\gamma \frac{c}{(1+c)^{2}} \leq(\alpha+\beta+\gamma) \frac{A}{(1+A)^{2}}, \quad \text { where } \quad A=\frac{\alpha a+\beta b+\gamma c}{\alpha+\beta+\gamma} $$  Choosing $\alpha=\frac{1}{a}, \beta=\frac{1}{b}$, and $\gamma=\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality  $$ A=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \leq \frac{a+b+c}{3}=\frac{1}{3}<1 $$  Finally we prove (5):  $$ \begin{aligned} \frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}} & \leq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \frac{A}{(1+A)^{2}} \\ & \leq\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \frac{\frac{1}{3}}{\left(1+\frac{1}{3}\right)^{2}}=\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) . \end{aligned} $$


FRA   Determine all functions $f$ from the set of positive integers into the set of positive integers such that for all $x$ and $y$ there exists a non degenerated triangle with sides of lengths  $$ x, \quad f(y) \quad \text { and } \quad f(y+f(x)-1) $$

Solution: The identity function $f(x)=x$ is the only solution of the problem. If $f(x)=x$ for all positive integers $x$, the given three lengths are $x, y=f(y)$ and $z=$ $f(y+f(x)-1)=x+y-1$. Because of $x \geq 1, y \geq 1$ we have $z \geq \max \{x, y\}>|x-y|$ and $z<x+y$. From this it follows that a triangle with these side lengths exists and does not degenerate. We prove in several steps that there is no other solution.  Step 1. We show $f(1)=1$. If we had $f(1)=1+m>1$ we would conclude $f(y)=f(y+m)$ for all $y$ considering the triangle with the side lengths $1, f(y)$ and $f(y+m)$. Thus, $f$ would be $m$-periodic and, consequently, bounded. Let $B$ be a bound, $f(x) \leq B$. If we choose $x>2 B$ we obtain the contradiction $x>2 B \geq f(y)+f(y+f(x)-1)$.  Step 2. For all positive integers $z$, we have $f(f(z))=z$. Setting $x=z$ and $y=1$ this follows immediately from Step 1.  Step 3. For all integers $z \geq 1$, we have $f(z) \leq z$. Let us show, that the contrary leads to a contradiction. Assume $w+1=f(z)>z$ for some $z$. From Step 1 we know that $w \geq z \geq 2$. Let $M=\max \{f(1), f(2), \ldots, f(w)\}$ be the largest value of $f$ for the first $w$ integers. First we show, that no positive integer $t$ exists with  $$ f(t)>\frac{z-1}{w} \cdot t+M $$  otherwise we decompose the smallest value $t$ as $t=w r+s$ where $r$ is an integer and $1 \leq s \leq w$. Because of the definition of $M$, we have $t>w$. Setting $x=z$ and $y=t-w$ we get from the triangle inequality  $$ z+f(t-w)>f((t-w)+f(z)-1)=f(t-w+w)=f(t) $$  Hence,  $$ f(t-w) \geq f(t)-(z-1)>\frac{z-1}{w}(t-w)+M $$  a contradiction to the minimality of $t$. Therefore the inequality (1) fails for all $t \geq 1$, we have proven  $$ f(t) \leq \frac{z-1}{w} \cdot t+M $$  instead.  Now, using (2), we finish the proof of Step 3. Because of $z \leq w$ we have $\frac{z-1}{w}<1$ and we can choose an integer $t$ sufficiently large to fulfill the condition  $$ \left(\frac{z-1}{w}\right)^{2} t+\left(\frac{z-1}{w}+1\right) M<t . $$  Applying (2) twice we get  $$ f(f(t)) \leq \frac{z-1}{w} f(t)+M \leq \frac{z-1}{w}\left(\frac{z-1}{w} t+M\right)+M<t $$  in contradiction to Step 2, which proves Step 3.  Final step. Thus, following Step 2 and Step 3, we obtain  $$ z=f(f(z)) \leq f(z) \leq z $$  and $f(z)=z$ for all positive integers $z$ is proven.


BLR   Let $a, b, c$ be positive real numbers such that $a b+b c+c a \leq 3 a b c$. Prove that  $$ \sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3 \leq \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}) . $$

Solution: Starting with the terms of the right-hand side, the quadratic-arithmetic-mean inequality yields  $$ \begin{aligned} \sqrt{2} \sqrt{a+b} & =2 \sqrt{\frac{a b}{a+b}} \sqrt{\frac{1}{2}\left(2+\frac{a^{2}+b^{2}}{a b}\right)} \\ & \geq 2 \sqrt{\frac{a b}{a+b}} \cdot \frac{1}{2}\left(\sqrt{2}+\sqrt{\frac{a^{2}+b^{2}}{a b}}\right)=\sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{a^{2}+b^{2}}{a+b}} \end{aligned} $$  and, analogously,  $$ \sqrt{2} \sqrt{b+c} \geq \sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}, \quad \sqrt{2} \sqrt{c+a} \geq \sqrt{\frac{2 c a}{c+a}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}} . $$  Applying the inequality between the arithmetic mean and the squared harmonic mean will finish the proof:  $$ \sqrt{\frac{2 a b}{a+b}}+\sqrt{\frac{2 b c}{b+c}}+\sqrt{\frac{2 c a}{c+a}} \geq 3 \cdot \sqrt{\frac{3}{\sqrt{\frac{a+b}{2 a b}}^{2}+\sqrt{\frac{b+c}{2 b c}}^{2}+\sqrt{\frac{c+a}{2 c a}}}}{ }^{2}=3 \cdot \sqrt{\frac{3 a b c}{a b+b c+c a}} \geq 3 . $$


BLR   Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that  $$ f(x-f(y))>y f(x)+x $$

Solution: Assume that  $$ f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y $$  Let $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,  $$ f(y) \leq y+a \quad \text { for all real } y $$  Setting $x=f(y)$ in (1) yields in view of (2)  $$ a=f(0) \leq y f(f(y))+f(y) \leq y f(f(y))+y+a . $$  This implies $0 \leq y(f(f(y))+1)$ and thus  $$ f(f(y)) \geq-1 \quad \text { for all } y>0 $$  From (2) and (3) we obtain $-1 \leq f(f(y)) \leq f(y)+a$ for all $y>0$, so  $$ f(y) \geq-a-1 \quad \text { for all } y>0 $$  Now we show that  $$ f(x) \leq 0 \quad \text { for all real } x $$  Assume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that  $$ y<x-a \quad \text { and } \quad y<\frac{-a-x-1}{f(x)} $$  Then in view of (2)  $$ x-f(y) \geq x-(y+a)>0 $$  and with (1) and (4) we obtain  $$ y f(x)+x \geq f(x-f(y)) \geq-a-1, $$  whence  $$ y \geq \frac{-a-x-1}{f(x)} $$  contrary to our choice of $y$. Thereby, we have established (5). Setting $x=0$ in (5) leads to $a=f(0) \leq 0$ and (2) then yields  $$ f(x) \leq x \quad \text { for all real } x $$  Now choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (11), (5) and (6) we obtain  $$ f(-1)=f(x-f(y)) \leq y f(x)+x=y f(f(y)-1)+f(y)-1 \leq y(f(y)-1)-1 \leq-y-1, $$  i.e. $y \leq-f(-1)-1$, a contradiction to the choice of $y$.


Solution: Assume that  $$ f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y $$  Let $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,  $$ f(y) \leq y+a \quad \text { for all real } y $$  Now we show that  $$ f(z) \geq 0 \quad \text { for all } z \geq 1 $$  Let $z \geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives  $$ f(w)-z f(w+b) \leq w+b \quad \text { for all real } w $$  Applying (10) to $w, w+b, \ldots, w+(n-1) b$, where $n=1,2, \ldots$, leads to  $$ \begin{gathered} f(w)-z^{n} f(w+n b)=(f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\ +\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\ \leq(w+b)+z(w+2 b)+\cdots+z^{n-1}(w+n b) \end{gathered} $$  From (8) we obtain  $$ f(w+n b) \leq w+n b+a $$  and, thus, we have for all positive integers $n$  $$ f(w) \leq\left(1+z+\cdots+z^{n-1}+z^{n}\right) w+\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+z^{n} a . $$  With $w=0$ we get  $$ a \leq\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+a z^{n} . $$  In view of the assumption $b<0$ we find some $n$ such that  $$ a>(n b+a) z^{n} $$  because the right hand side tends to $-\infty$ as $n \rightarrow \infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality  $$ f(1)>0 \text {. } $$  Indeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to  $$ f(-1) \leq-(n+1)+a $$  which is false if $n$ is sufficiently large. To complete the proof we set $t=\min \{-a,-2 / f(1)\}$. Setting $x=1$ and $y=t$ in (7) gives  $$ f(1-f(t)) \leq t f(1)+1 \leq-2+1=-1 . $$  On the other hand, by (8) and the choice of $t$ we have $f(t) \leq t+a \leq 0$ and hence $1-f(t) \geq 1$. The inequality (9) yields  $$ f(1-f(t)) \geq 0 $$  which contradicts (15).


USA   Suppose that $s_{1}, s_{2}, s_{3}, \ldots$ is a strictly increasing sequence of positive integers such that the subsequences  $$ s_{s_{1}}, s_{s_{2}}, s_{s_{3}}, \ldots \quad \text { and } \quad s_{s_{1}+1}, s_{s_{2}+1}, s_{s_{3}+1}, \ldots $$  are both arithmetic progressions. Prove that $s_{1}, s_{2}, s_{3}, \ldots$ is itself an arithmetic progression.

Solution: Let $D$ be the common difference of the progression $s_{s_{1}}, s_{s_{2}}, \ldots$. Let for $n=$ 1, 2, ...  $$ d_{n}=s_{n+1}-s_{n} $$  We have to prove that $d_{n}$ is constant. First we show that the numbers $d_{n}$ are bounded. Indeed, by supposition $d_{n} \geq 1$ for all $n$. Thus, we have for all $n$  $$ d_{n}=s_{n+1}-s_{n} \leq d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n+1}-1}=s_{s_{n+1}}-s_{s_{n}}=D . $$  The boundedness implies that there exist  $$ m=\min \left\{d_{n}: n=1,2, \ldots\right\} \quad \text { and } \quad M=\max \left\{d_{n}: n=1,2, \ldots\right\} $$  It suffices to show that $m=M$. Assume that $m<M$. Choose $n$ such that $d_{n}=m$. Considering a telescoping sum of $m=d_{n}=s_{n+1}-s_{n}$ items not greater than $M$ leads to  $$ D=s_{s_{n+1}}-s_{s_{n}}=s_{s_{n}+m}-s_{s_{n}}=d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n}+m-1} \leq m M $$  and equality holds if and only if all items of the sum are equal to $M$. Now choose $n$ such that $d_{n}=M$. In the same way, considering a telescoping sum of $M$ items not less than $m$ we obtain  $$ D=s_{s_{n+1}}-s_{s_{n}}=s_{s_{n}+M}-s_{s_{n}}=d_{s_{n}}+d_{s_{n}+1}+\cdots+d_{s_{n}+M-1} \geq M m $$  and equality holds if and only if all items of the sum are equal to $m$. The inequalities (1) and (2) imply that $D=M m$ and that  $$ \begin{aligned} d_{s_{n}}=d_{s_{n}+1}=\cdots=d_{s_{n+1}-1}=M & \text { if } d_{n}=m, \\ d_{s_{n}}=d_{s_{n}+1}=\cdots=d_{s_{n+1}-1}=m & \text { if } d_{n}=M . \end{aligned} $$  Hence, $d_{n}=m$ implies $d_{s_{n}}=M$. Note that $s_{n} \geq s_{1}+(n-1) \geq n$ for all $n$ and moreover $s_{n}>n$ if $d_{n}=n$, because in the case $s_{n}=n$ we would have $m=d_{n}=d_{s_{n}}=M$ in contradiction to the assumption $m<M$. In the same way $d_{n}=M$ implies $d_{s_{n}}=m$ and $s_{n}>n$. Consequently, there is a strictly increasing sequence $n_{1}, n_{2}, \ldots$ such that  $$ d_{s_{n_{1}}}=M, \quad d_{s_{n_{2}}}=m, \quad d_{s_{n_{3}}}=M, \quad d_{s_{n_{4}}}=m, \quad \ldots $$  The sequence $d_{s_{1}}, d_{s_{2}}, \ldots$ is the sequence of pairwise differences of $s_{s_{1}+1}, s_{s_{2}+1}, \ldots$ and $s_{s_{1}}, s_{s_{2}}, \ldots$, hence also an arithmetic progression. Thus $m=M$.


Solution: Let the integers $D$ and $E$ be the common differences of the progressions $s_{s_{1}}, s_{s_{2}}, \ldots$ and $s_{s_{1}+1}, s_{s_{2}+1}, \ldots$, respectively. Let briefly $A=s_{s_{1}}-D$ and $B=s_{s_{1}+1}-E$. Then, for all positive integers $n$,  $$ s_{s_{n}}=A+n D, \quad s_{s_{n}+1}=B+n E $$  Since the sequence $s_{1}, s_{2}, \ldots$ is strictly increasing, we have for all positive integers $n$  $$ s_{s_{n}}<s_{s_{n}+1} \leq s_{s_{n+1}} $$  which implies  $$ A+n D<B+n E \leq A+(n+1) D $$  and thereby  $$ 0<B-A+n(E-D) \leq D $$  which implies $D-E=0$ and thus  $$ 0 \leq B-A \leq D $$  Let $m=\min \left\{s_{n+1}-s_{n}: n=1,2, \ldots\right\}$. Then  $$ B-A=\left(s_{s_{1}+1}-E\right)-\left(s_{s_{1}}-D\right)=s_{s_{1}+1}-s_{s_{1}} \geq m $$  and  $$ D=A+\left(s_{1}+1\right) D-\left(A+s_{1} D\right)=s_{s_{s_{1}+1}}-s_{s_{s_{1}}}=s_{B+D}-s_{A+D} \geq m(B-A) $$  From (3) we consider two cases. Case 1. $B-A=D$. Then, for each positive integer $n, s_{s_{n}+1}=B+n D=A+(n+1) D=s_{s_{n+1}}$, hence $s_{n+1}=s_{n}+1$ and $s_{1}, s_{2}, \ldots$ is an arithmetic progression with common difference 1 .  Case 2. $B-A<D$. Choose some positive integer $N$ such that $s_{N+1}-s_{N}=m$. Then  $$ \begin{aligned} m(A-B+D-1) & =m((A+(N+1) D)-(B+N D+1)) \\ & \leq s_{A+(N+1) D}-s_{B+N D+1}=s_{s_{s_{N+1}}}-s_{s_{s_{N}+1}+1} \\ & =\left(A+s_{N+1} D\right)-\left(B+\left(s_{N}+1\right) D\right)=\left(s_{N+1}-s_{N}\right) D+A-B-D \\ & =m D+A-B-D, \end{aligned} $$  i.e.,  $$ (B-A-m)+(D-m(B-A)) \leq 0 $$  The inequalities (4)-(6) imply that  $$ B-A=m \quad \text { and } \quad D=m(B-A) $$  Assume that there is some positive integer $n$ such that $s_{n+1}>s_{n}+m$. Then $\left.m(m+1) \leq m\left(s_{n+1}-s_{n}\right) \leq s_{s_{n+1}}-s_{s_{n}}=(A+(n+1) D)-(A+n D)\right)=D=m(B-A)=m^{2}$, a contradiction. Hence $s_{1}, s_{2}, \ldots$ is an arithmetic progression with common difference $m$.


JPN   Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity  $$ f(x f(x+y))=f(y f(x))+x^{2} $$

Solution: It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions. Let $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \neq 0$. For any real $t$, substituting $(x, y)=\left(0, \frac{t}{f(0)}\right)$ into the given functional equation, we obtain  $$ f(0)=f(t) $$  contradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get  $$ f(t f(t))=f(0)+t^{2}=t^{2} $$  and  $$ f(t f(0))=f(-t f(t))+t^{2} $$  respectively. Therefore, we conclude that  $$ f(t f(t))=t^{2}, \quad f(-t f(t))=-t^{2}, \quad \text { for every real } t $$  Consequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$. We next show that for any real number $s$,  $$ f(-s)=-f(s) $$  This is clear if $f(s)=0$. Suppose now $f(s)<0$, then we can find a number $t$ for which $f(s)=-t^{2}$. As $t \neq 0$ implies $f(t) \neq 0$, we can also find number $a$ such that $a f(t)=s$. Substituting $(x, y)=(t, a)$ into the given equation, we get  $$ f(t f(t+a))=f(a f(t))+t^{2}=f(s)+t^{2}=0 $$  and therefore, $t f(t+a)=0$, which implies $t+a=0$, and hence $s=-t f(t)$. Consequently, $f(-s)=f(t f(t))=t^{2}=-\left(-t^{2}\right)=-f(s)$ holds in this case. Finally, suppose $f(s)>0$ holds. Then there exists a real number $t \neq 0$ for which $f(s)=t^{2}$. Choose a number $a$ such that $t f(a)=s$. Substituting $(x, y)=(t, a-t)$ into the given equation, we get $f(s)=f(t f(a))=f((a-t) f(t))+t^{2}=f((a-t) f(t))+f(s)$. So we have $f((a-t) f(t))=0$, from which we conclude that $(a-t) f(t)=0$. Since $f(t) \neq 0$, we get $a=t$ so that $s=t f(t)$ and thus we see $f(-s)=f(-t f(t))=-t^{2}=-f(s)$ holds in this case also. This observation finishes the proof of (3). By substituting $(x, y)=(s, t),(x, y)=(t,-s-t)$ and $(x, y)=(-s-t, s)$ into the given equation, we obtain  $$ \begin{array}{r} f(s f(s+t)))=f(t f(s))+s^{2} \\ f(t f(-s))=f((-s-t) f(t))+t^{2} \end{array} $$  and  $$ f((-s-t) f(-t))=f(s f(-s-t))+(s+t)^{2} $$  respectively. Using the fact that $f(-x)=-f(x)$ holds for all $x$ to rewrite the second and the third equation, and rearranging the terms, we obtain  $$ \begin{aligned} f(t f(s))-f(s f(s+t)) & =-s^{2} \\ f(t f(s))-f((s+t) f(t)) & =-t^{2} \\ f((s+t) f(t))+f(s f(s+t)) & =(s+t)^{2} \end{aligned} $$  Adding up these three equations now yields $2 f(t f(s))=2 t s$, and therefore, we conclude that $f(t f(s))=t s$ holds for every pair of real numbers $s, t$. By fixing $s$ so that $f(s)=1$, we obtain $f(x)=s x$. In view of the given equation, we see that $s= \pm 1$. It is easy to check that both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation, so these are the desired solutions.


Solution: As in  Now we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r \neq s$. Then, by (2)  $$ r^{2}=f(r f(r))=f(r f(s))=f((s-r) f(r))+r^{2} $$  where the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r) f(r)$ satisfies $f(h)=0$ which implies $h^{2}=f(h f(h))=f(0)=0$, i.e., $h=0$. Then, by $s \neq r$ we have $f(r)=0$ which implies $r=0$, and finally $f(s)=f(r)=f(0)=0$. Analogously, it follows that $s=0$ which gives the contradiction $r=s$. To prove $|f(1)|=1$ we apply (2) with $t=1$ and also with $t=f(1)$ and obtain $f(f(1))=1$ and $(f(1))^{2}=f(f(1) \cdot f(f(1)))=f(f(1))=1$. Now we choose $\eta \in\{-1,1\}$ with $f(1)=\eta$. Using that $f$ is odd and the given equation with $x=1, y=z$ (second equality) and with $x=-1, y=z+2$ (fourth equality) we obtain  $$ \begin{aligned} & f(z)+2 \eta=\eta(f(z \eta)+2)=\eta(f(f(z+1))+1)=\eta(-f(-f(z+1))+1) \\ & =-\eta f((z+2) f(-1))=-\eta f((z+2)(-\eta))=\eta f((z+2) \eta)=f(z+2) \end{aligned} $$  Hence,  $$ f(z+2 \eta)=\eta f(\eta z+2)=\eta(f(\eta z)+2 \eta)=f(z)+2 $$  Using this argument twice we obtain  $$ f(z+4 \eta)=f(z+2 \eta)+2=f(z)+4 $$  Substituting $z=2 f(x)$ we have  $$ f(2 f(x))+4=f(2 f(x)+4 \eta)=f(2 f(x+2)) $$  where the last equality follows from (4). Applying the given functional equation we proceed to  $$ f(2 f(x+2))=f(x f(2))+4=f(2 \eta x)+4 $$  where the last equality follows again from (4) with $z=0$, i.e., $f(2)=2 \eta$. Finally, $f(2 f(x))=$ $f(2 \eta x)$ and by injectivity of $f$ we get $2 f(x)=2 \eta x$ and hence the two solutions.


NZL   Consider 2009 cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. (a) Does the game necessarily end? (b) Does there exist a winning strategy for the starting player?

Solution: (a) We interpret a card showing black as the digit 0 and a card showing gold as the digit 1. Thus each position of the 2009 cards, read from left to right, corresponds bijectively to a nonnegative integer written in binary notation of 2009 digits, where leading zeros are allowed. Each move decreases this integer, so the game must end. (b) We show that there is no winning strategy for the starting player. We label the cards from right to left by $1, \ldots, 2009$ and consider the set $S$ of cards with labels $50 i, i=1,2, \ldots, 40$. Let $g_{n}$ be the number of cards from $S$ showing gold after $n$ moves. Obviously, $g_{0}=40$. Moreover, $\left|g_{n}-g_{n+1}\right|=1$ as long as the play goes on. Thus, after an odd number of moves, the nonstarting player finds a card from $S$ showing gold and hence can make a move. Consequently, this player always wins.


ROU   For any integer $n \geq 2$, let $N(n)$ be the maximal number of triples $\left(a_{i}, b_{i}, c_{i}\right), i=1, \ldots, N(n)$, consisting of nonnegative integers $a_{i}, b_{i}$ and $c_{i}$ such that the following two conditions are satisfied: (1) $a_{i}+b_{i}+c_{i}=n$ for all $i=1, \ldots, N(n)$, (2) If $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$.  Determine $N(n)$ for all $n \geq 2$.  Comment. The original problem was formulated for $m$-tuples instead for triples. The numbers $N(m, n)$ are then defined similarly to $N(n)$ in the case $m=3$. The numbers $N(3, n)$ and $N(n, n)$ should be determined. The case $m=3$ is the same as in the present problem. The upper bound for $N(n, n)$ can be proved by a simple generalization. The construction of a set of triples attaining the bound can be easily done by induction from $n$ to $n+2$.

Solution: Let $n \geq 2$ be an integer and let $\left\{T_{1}, \ldots, T_{N}\right\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have  $$ \sum_{i=1}^{N} a_{i} \geq \sum_{i=1}^{N}(i-1)=\frac{N(N-1)}{2} $$  Analogously,  $$ \sum_{i=1}^{N} b_{i} \geq \frac{N(N-1)}{2} \quad \text { and } \quad \sum_{i=1}^{N} c_{i} \geq \frac{N(N-1)}{2} $$  Summing these three inequalities and applying (1) yields  $$ 3 \frac{N(N-1)}{2} \leq \sum_{i=1}^{N} a_{i}+\sum_{i=1}^{N} b_{i}+\sum_{i=1}^{N} c_{i}=\sum_{i=1}^{N}\left(a_{i}+b_{i}+c_{i}\right)=n N, $$  hence $3 \frac{N-1}{2} \leq n$ and, consequently,  $$ N \leq\left\lfloor\frac{2 n}{3}\right\rfloor+1 $$  By constructing examples, we show that this upper bound can be attained, so $N(n)=\left\lfloor\frac{2 n}{3}\right\rfloor+1$.  We distinguish the cases $n=3 k-1, n=3 k$ and $n=3 k+1$ for $k \geq 1$ and present the extremal examples in form of a table.  | $n=3 k-1$ |  |  | | :---: | :---: | :---: | | $\left\lfloor\frac{2 n}{3}\right\rfloor+1=2 k$ |  |  | | $a_{i}$ | $b_{i}$ | $c_{i}$ | | 0 | $k+1$ | $2 k-2$ | | 1 | $k+2$ | $2 k-4$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $k-1$ | $2 k$ | 0 | | $k$ | 0 | $2 k-1$ | | $k+1$ | 1 | $2 k-3$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $2 k-1$ | $k-1$ | 1 |   | $n=3 k$ |  |  | | :---: | :---: | :---: | | $\rfloor+1=2 k+1$ |  |  | | $a_{i}$ | $b_{i}$ | $c_{i}$ | | 0 | $k$ | $2 k$ | | 1 | $k+1$ | $2 k-2$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $k$ | $2 k$ | 0 | | $k+1$ | 0 | $2 k-1$ | | $k+2$ | 1 | $2 k-3$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $2 k$ | $k-1$ | 1 |   | $n=3 k+1$ |  |  | | :---: | :---: | :---: | | $\left\lfloor\left\lfloor\frac{2 n}{3}\right\rfloor+1=2 k+1\right.$ |  |  | | $a_{i}$ | $b_{i}$ | $c_{i}$ | | 0 | $k$ | $2 k+1$ | | 1 | $k+1$ | $2 k-1$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $k$ | $2 k$ | 1 | | $k+1$ | 0 | $2 k$ | | $k+2$ | 1 | $2 k-2$ | | $\vdots$ | $\vdots$ | $\vdots$ | | $2 k$ | $k-1$ | 2 |  It can be easily seen that the conditions (1) and (2) are satisfied and that we indeed have $\left\lfloor\frac{2 n}{3}\right\rfloor+1$ triples in each case.  Comment. A cute combinatorial model is given by an equilateral triangle, partitioned into $n^{2}$ congruent equilateral triangles by $n-1$ equidistant parallels to each of its three sides. Two chess-like bishops placed at any two vertices of the small triangles are said to menace one another if they lie on a same parallel. The problem is to determine the largest number of bishops that can be placed so that none menaces another. A bishop may be assigned three coordinates $a, b, c$, namely the numbers of sides of small triangles they are off each of the sides of the big triangle. It is readily seen that the sum of these coordinates is always $n$, therefore fulfilling the requirements.


RUS   Let $n$ be a positive integer. Given a sequence $\varepsilon_{1}, \ldots, \varepsilon_{n-1}$ with $\varepsilon_{i}=0$ or $\varepsilon_{i}=1$ for each $i=1, \ldots, n-1$, the sequences $a_{0}, \ldots, a_{n}$ and $b_{0}, \ldots, b_{n}$ are constructed by the following rules:  $$ \begin{gathered} a_{0}=b_{0}=1, \quad a_{1}=b_{1}=7, \\ a_{i+1}=\left\{\begin{array}{ll} 2 a_{i-1}+3 a_{i}, & \text { if } \varepsilon_{i}=0, \\ 3 a_{i-1}+a_{i}, & \text { if } \varepsilon_{i}=1, \end{array} \text { for each } i=1, \ldots, n-1,\right. \\ b_{i+1}=\left\{\begin{array}{ll} 2 b_{i-1}+3 b_{i}, & \text { if } \varepsilon_{n-i}=0, \\ 3 b_{i-1}+b_{i}, & \text { if } \varepsilon_{n-i}=1, \end{array} \text { for each } i=1, \ldots, n-1 .\right. \end{gathered} $$  Prove that $a_{n}=b_{n}$.

Solution: For a binary word $w=\sigma_{1} \ldots \sigma_{n}$ of length $n$ and a letter $\sigma \in\{0,1\}$ let $w \sigma=$ $\sigma_{1} \ldots \sigma_{n} \sigma$ and $\sigma w=\sigma \sigma_{1} \ldots \sigma_{n}$. Moreover let $\bar{w}=\sigma_{n} \ldots \sigma_{1}$ and let $\emptyset$ be the empty word (of length 0 and with $\bar{\emptyset}=\emptyset$ ). Let $(u, v)$ be a pair of two real numbers. For binary words $w$ we define recursively the numbers $(u, v)^{w}$ as follows:  $$ \begin{gathered} (u, v)^{\emptyset}=v, \quad(u, v)^{0}=2 u+3 v, \quad(u, v)^{1}=3 u+v, \\ (u, v)^{w \sigma \varepsilon}= \begin{cases}3(u, v)^{w}+3(u, v)^{w \sigma}, & \text { if } \varepsilon=0, \\ 3(u, v)^{w}+(u, v)^{w \sigma}, & \text { if } \varepsilon=1\end{cases} \end{gathered} $$  It easily follows by induction on the length of $w$ that for all real numbers $u_{1}, v_{1}, u_{2}, v_{2}, \lambda_{1}$ and $\lambda_{2}$  $$ \left(\lambda_{1} u_{1}+\lambda_{2} u_{2}, \lambda_{1} v_{1}+\lambda_{2} v_{2}\right)^{w}=\lambda_{1}\left(u_{1}, v_{1}\right)^{w}+\lambda_{2}\left(u_{2}, v_{2}\right)^{w} $$  and that for $\varepsilon \in\{0,1\}$  $$ (u, v)^{\varepsilon w}=\left(v,(u, v)^{\varepsilon}\right)^{w} . $$  Obviously, for $n \geq 1$ and $w=\varepsilon_{1} \ldots \varepsilon_{n-1}$, we have $a_{n}=(1,7)^{w}$ and $b_{n}=(1,7)^{\bar{w}}$. Thus it is sufficient to prove that  $$ (1,7)^{w}=(1,7)^{\bar{w}} $$  for each binary word $w$. We proceed by induction on the length of $w$. The assertion is obvious if $w$ has length 0 or 1 . Now let $w \sigma \varepsilon$ be a binary word of length $n \geq 2$ and suppose that the assertion is true for all binary words of length at most $n-1$. Note that $(2,1)^{\sigma}=7=(1,7)^{\emptyset}$ for $\sigma \in\{0,1\},(1,7)^{0}=23$, and $(1,7)^{1}=10$. First let $\varepsilon=0$. Then in view of the induction hypothesis and the equalities (1) and (2), we obtain  $$ \begin{aligned} &(1,7)^{w \sigma 0}=2(1,7)^{w}+3(1,7)^{w \sigma}=2(1,7)^{\bar{w}}+3(1,7)^{\sigma \bar{w}}=2(2,1)^{\sigma \bar{w}}+3(1,7)^{\sigma \bar{w}} \\ &=(7,23)^{\sigma \bar{w}}=(1,7)^{0 \sigma \bar{w}} \end{aligned} $$  Now let $\varepsilon=1$. Analogously, we obtain  $$ \begin{aligned} &(1,7)^{w \sigma 1}=3(1,7)^{w}+(1,7)^{w \sigma}=3(1,7)^{\bar{w}}+(1,7)^{\sigma \bar{w}}=3(2,1)^{\sigma \bar{w}}+(1,7)^{\sigma \bar{w}} \\ &=(7,10)^{\sigma \bar{w}}=(1,7)^{1 \sigma \bar{w}} \end{aligned} $$  Thus the induction step is complete, (3) and hence also $a_{n}=b_{n}$ are proved.  Comment. The original solution uses the relation  $$ (1,7)^{\alpha \beta w}=\left((1,7)^{w},(1,7)^{\beta w}\right)^{\alpha}, \quad \alpha, \beta \in\{0,1\} $$  which can be proved by induction on the length of $w$. Then (3) also follows by induction on the length of $w$ :  $$ (1,7)^{\alpha \beta w}=\left((1,7)^{w},(1,7)^{\beta w}\right)^{\alpha}=\left((1,7)^{\bar{w}},(1,7)^{\bar{w} \beta}\right)^{\alpha}=(1,7)^{\bar{w} \beta \alpha} . $$  Here $w$ may be the empty word.


NLD   For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a partition.

Solution: For a $k \times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{i j}$ in row $i$ and column $j$ has vertices $(i-1, j-1),(i-$ $1, j),(i, j-1),(i, j)$, where $i, j \in\{1, \ldots, k\}$. Without loss of generality assume that the cells $C_{i i}$, $i=1, \ldots, k$, form a separate rectangle. Then we may consider the boards $B_{k}=\bigcup_{1 \leq i<j \leq k} C_{i j}$ below that diagonal and the congruent board $B_{k}^{\prime}=\bigcup_{1 \leq j<i \leq k} C_{i j}$ above that diagonal separately because no rectangle can simultaneously cover cells from $B_{k}$ and $B_{k}^{\prime}$. We will show that for $k=2^{m}$ the smallest total perimeter of a rectangular partition of $B_{k}$ is $m 2^{m+1}$. Then the overall answer to the problem is $2 \cdot m 2^{m+1}+4 \cdot 2^{m}=(m+1) 2^{m+2}$. First we inductively construct for $m \geq 1$ a partition of $B_{2^{m}}$ with total perimeter $m 2^{m+1}$. If $m=0$, the board $B_{2^{m}}$ is empty and the total perimeter is 0 . For $m \geq 0$, the board $B_{2^{m+1}}$ consists of a $2^{m} \times 2^{m}$ square in the lower right corner with vertices $\left(2^{m}, 2^{m}\right),\left(2^{m}, 2^{m+1}\right),\left(2^{m+1}, 2^{m}\right)$, $\left(2^{m+1}, 2^{m+1}\right)$ to which two boards congruent to $B_{2^{m}}$ are glued along the left and the upper margin. The square together with the inductive partitions of these two boards yield a partition with total perimeter $4 \cdot 2^{m}+2 \cdot m 2^{m+1}=(m+1) 2^{m+2}$ and the induction step is complete. Let  $$ D_{k}=2 k \log _{2} k $$  Note that $D_{k}=m 2^{m+1}$ if $k=2^{m}$. Now we show by induction on $k$ that the total perimeter of a rectangular partition of $B_{k}$ is at least $D_{k}$. The case $k=1$ is trivial (see $m=0$ from above). Let the assertion be true for all positive integers less than $k$. We investigate a fixed rectangular partition of $B_{k}$ that attains the minimal total perimeter. Let $R$ be the rectangle that covers the cell $C_{1 k}$ in the lower right corner. Let $(i, j)$ be the upper left corner of $R$. First we show that $i=j$. Assume that $i<j$. Then the line from $(i, j)$ to $(i+1, j)$ or from $(i, j)$ to $(i, j-1)$ must belong to the boundary of some rectangle in the partition. Without loss of generality assume that this is the case for the line from $(i, j)$ to $(i+1, j)$. Case 1. No line from $(i, l)$ to $(i+1, l)$ where $j<l<k$ belongs to the boundary of some rectangle of the partition. Then there is some rectangle $R^{\prime}$ of the partition that has with $R$ the common side from $(i, j)$ to $(i, k)$. If we join these two rectangles to one rectangle we get a partition with smaller total perimeter, a contradiction. Case 2. There is some $l$ such that $j<l<k$ and the line from $(i, l)$ to $(i+1, l)$ belongs to the boundary of some rectangle of the partition. Then we replace the upper side of $R$ by the line $(i+1, j)$ to $(i+1, k)$ and for the rectangles whose lower side belongs to the line from $(i, j)$ to $(i, k)$ we shift the lower side upwards so that the new lower side belongs to the line from $(i+1, j)$ to $(i+1, k)$. In such a way we obtain a rectangular partition of $B_{k}$ with smaller total perimeter, a contradiction. Now the fact that the upper left corner of $R$ has the coordinates $(i, i)$ is established. Consequently, the partition consists of $R$, of rectangles of a partition of a board congruent to $B_{i}$ and of rectangles of a partition of a board congruent to $B_{k-i}$. By the induction hypothesis, its total perimeter is at least  $$ 2(k-i)+2 i+D_{i}+D_{k-i} \geq 2 k+2 i \log _{2} i+2(k-i) \log _{2}(k-i) . $$  Since the function $f(x)=2 x \log _{2} x$ is convex for $x>0$, JEnSEN's inequality immediately shows that the minimum of the right hand sight of (1) is attained for $i=k / 2$. Hence the total perimeter of the optimal partition of $B_{k}$ is at least $2 k+2 k / 2 \log _{2} k / 2+2(k / 2) \log _{2}(k / 2)=D_{k}$.


Solution: We start as in   $$ p=2\left(\sum_{i=1}^{M} r_{i}+\sum_{i=1}^{M} c_{i}\right) . $$  No rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{i i}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i, 1 \leq i \leq M$, if $S$ contains all $r_{i}$ rectangles that cover some cell from row $i$, but none of the $c_{i}$ rectangles that cover some cell from column $i$. Altogether there are $2^{n-r_{i}-c_{i}}$ subsets of type $i$. Now we show that no subset $S$ can be simultaneously of type $i$ and of type $j$ if $i \neq j$. Assume the contrary and let without loss of generality $i<j$. The cell $C_{i j}$ must be covered by some rectangle $R$. The subset $S$ is of type $i$, hence $R$ is contained in $S$. $S$ is of type $j$, thus $R$ does not belong to $S$, a contradiction. Since there are $2^{n}$ subsets of rectangles of the partition, we infer  $$ 2^{n} \geq \sum_{i=1}^{M} 2^{n-r_{i}-c_{i}}=2^{n} \sum_{i=1}^{M} 2^{-\left(r_{i}+c_{i}\right)} $$  By applying Jensen's inequality to the convex function $f(x)=2^{-x}$ we derive  $$ \frac{1}{M} \sum_{i=1}^{M} 2^{-\left(r_{i}+c_{i}\right)} \geq 2^{-\frac{1}{M} \sum_{i=1}^{M}\left(r_{i}+c_{i}\right)}=2^{-\frac{p}{2 M}} $$  From (2) and (3) we obtain  $$ 1 \geq M 2^{-\frac{p}{2 M}} $$  and equivalently  $$ p \geq m 2^{m+1} $$


NLD   Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighboring buckets, empties them into the river, and puts them back. Then the next round begins. The Stepmother's goal is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Solution: No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_{0}, B_{1}, B_{2}, B_{3}$, and $B_{4}$, where $B_{k}$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}(k=0,1,2,3,4)$ and all indices are taken modulo 5 . Cinderella enforces that the following three conditions are satisfied at the beginning of every round: (1) Two adjacent buckets (say $B_{1}$ and $B_{2}$ ) are empty. (2) The two buckets standing next to these adjacent buckets (here $B_{0}$ and $B_{3}$ ) have total contents at most 1. (3) The remaining bucket (here $B_{4}$ ) has contents at most 1 .  These conditions clearly hold at the beginning of the first round, when all buckets are empty. Assume that Cinderella manages to maintain them until the beginning of the $r$-th round $(r \geq 1)$. Denote by $x_{k}(k=0,1,2,3,4)$ the contents of bucket $B_{k}$ at the beginning of this round and by $y_{k}$ the corresponding contents after the Stepmother has distributed her liter of water in this round. By the conditions, we can assume $x_{1}=x_{2}=0, x_{0}+x_{3} \leq 1$ and $x_{4} \leq 1$. Then, since the Stepmother adds one liter, we conclude $y_{0}+y_{1}+y_{2}+y_{3} \leq 2$. This inequality implies $y_{0}+y_{2} \leq 1$ or $y_{1}+y_{3} \leq 1$. For reasons of symmetry, we only consider the second case. Then Cinderella empties buckets $B_{0}$ and $B_{4}$. At the beginning of the next round $B_{0}$ and $B_{4}$ are empty (condition (1) is fulfilled), due to $y_{1}+y_{3} \leq 1$ condition (2) is fulfilled and finally since $x_{2}=0$ we also must have $y_{2} \leq 1$ (condition (3) is fulfilled).  Therefore, Cinderella can indeed manage to maintain the three conditions (1)-(3) also at the beginning of the $(r+1)$-th round. By induction, she thus manages to maintain them at the beginning of every round. In particular she manages to keep the contents of every single bucket at most 1 liter. Therefore, the buckets of 2-liter capacity will never overflow.


Solution: We prove that Cinderella can maintain the following two conditions and hence she can prevent the buckets from overflow: (1') Every two non-adjacent buckets contain a total of at most 1. (2') The total contents of all five buckets is at most $\frac{3}{2}$. We use the same notations as in the first solution. The two conditions again clearly hold at the beginning. Assume that Cinderella maintained these two conditions until the beginning of the $r$-th round. A pair of non-neighboring buckets $\left(B_{i}, B_{i+2}\right), i=0,1,2,3,4$ is called critical if $y_{i}+y_{i+2}>1$. By condition $\left(2^{\prime}\right)$, after the Stepmother has distributed her water we have $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \leq \frac{5}{2}$. Therefore,  $$ \left(y_{0}+y_{2}\right)+\left(y_{1}+y_{3}\right)+\left(y_{2}+y_{4}\right)+\left(y_{3}+y_{0}\right)+\left(y_{4}+y_{1}\right)=2\left(y_{0}+y_{1}+y_{2}+y_{3}+y_{4}\right) \leq 5 $$  and hence there is a pair of non-neighboring buckets which is not critical, say $\left(B_{0}, B_{2}\right)$. Now, if both of the pairs $\left(B_{3}, B_{0}\right)$ and $\left(B_{2}, B_{4}\right)$ are critical, we must have $y_{1}<\frac{1}{2}$ and Cinderella can empty the buckets $B_{3}$ and $B_{4}$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_{1}+\left(y_{0}+y_{2}\right) \leq \frac{3}{2}$. Therefore, conditions $\left(1^{\prime}\right)$ and $\left(2^{\prime}\right)$ are fulfilled.  Now suppose that without loss of generality the pair $\left(B_{3}, B_{0}\right)$ is not critical. If in this case $y_{0} \leq \frac{1}{2}$, then one of the inequalities $y_{0}+y_{1}+y_{2} \leq \frac{3}{2}$ and $y_{0}+y_{3}+y_{4} \leq \frac{3}{2}$ must hold. But then Cinderella can empty $B_{3}$ and $B_{4}$ or $B_{1}$ and $B_{2}$, respectively and clearly fulfill the conditions. Finally consider the case $y_{0}>\frac{1}{2}$. By $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \leq \frac{5}{2}$, at least one of the pairs $\left(B_{1}, B_{3}\right)$ and $\left(B_{2}, B_{4}\right)$ is not critical. Without loss of generality let this be the pair $\left(B_{1}, B_{3}\right)$. Since the pair $\left(B_{3}, B_{0}\right)$ is not critical and $y_{0}>\frac{1}{2}$, we must have $y_{3} \leq \frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_{1}$ and $B_{2}$ or $B_{4}$ and $B_{0}$.  Comments on GREEDY approaches. A natural approach for Cinderella would be a GREEDY strategy as for example: Always remove as much water as possible from the system. It is straightforward to prove that GREEDY can avoid buckets of capacity $\frac{5}{2}$ from overflowing: If before the Stepmothers move one has $x_{0}+x_{1}+x_{2}+x_{3}+x_{4} \leq \frac{3}{2}$ then after her move the inequality $Y=y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \leq \frac{5}{2}$ holds. If now Cinderella removes the two adjacent buckets with maximum total contents she removes at least $\frac{2 Y}{5}$ and thus the remaining buckets contain at most $\frac{3}{5} \cdot Y \leq \frac{3}{2}$. But GREEDY is in general not strong enough to settle this problem as can be seen in the following example:  - In an initial phase, the Stepmother brings all the buckets (after her move) to contents of at least $\frac{1}{2}-2 \epsilon$, where $\epsilon$ is an arbitrary small positive number. This can be done by always splitting the 1 liter she has to distribute so that all buckets have the same contents. After her $r$-th move the total contents of each of the buckets is then $c_{r}$ with $c_{1}=1$ and $c_{r+1}=1+\frac{3}{5} \cdot c_{r}$ and hence $c_{r}=\frac{5}{2}-\frac{3}{2} \cdot\left(\frac{3}{5}\right)^{r-1}$. So the contents of each single bucket indeed approaches $\frac{1}{2}$ (from below). In particular, any two adjacent buckets have total contents strictly less than 1 which enables the Stepmother to always refill the buckets that Cinderella just emptied and then distribute the remaining water evenly over all buckets. - After that phase GREEDY faces a situation like this ( $\left.\frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon\right)$ and leaves a situation of the form $\left(x_{0}, x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(\frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon, \frac{1}{2}-2 \epsilon, 0,0\right)$. - Then the Stepmother can add the amounts $\left(0, \frac{1}{4}+\epsilon, \epsilon, \frac{3}{4}-2 \epsilon, 0\right)$ to achieve a situation like this: $\left(y_{0}, y_{1}, y_{2}, y_{3}, y_{4}\right)=\left(\frac{1}{2}-2 \epsilon, \frac{3}{4}-\epsilon, \frac{1}{2}-\epsilon, \frac{3}{4}-2 \epsilon, 0\right)$. - Now $B_{1}$ and $B_{2}$ are the adjacent buckets with the maximum total contents and thus GREEDY empties them to yield $\left(x_{0}, x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(\frac{1}{2}-2 \epsilon, 0,0, \frac{3}{4}-2 \epsilon, 0\right)$. - Then the Stepmother adds $\left(\frac{5}{8}, 0,0, \frac{3}{8}, 0\right)$, which yields $\left(\frac{9}{8}-2 \epsilon, 0,0, \frac{9}{8}-2 \epsilon, 0\right)$. - Now GREEDY can only empty one of the two nonempty buckets and in the next step the Stepmother adds her liter to the other bucket and brings it to $\frac{17}{8}-2 \epsilon$, i.e. an overflow.  A harder variant. Five identical empty buckets of capacity $b$ stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighboring buckets, empties them into the river, and puts them back. Then the next round begins. The Stepmother's goal is to make one of these buckets overflow. Cinderella's goal is to prevent this. Determine all bucket capacities $b$ for which the Stepmother can enforce a bucket to overflow.  Solution to the harder variant. The answer is $b<2$. The previous proof shows that for all $b \geq 2$ the Stepmother cannot enforce overflowing. Now if $b<2$, let $R$ be a positive integer such that $b<2-2^{1-R}$. In the first $R$ rounds the Stepmother now ensures that at least one of the (nonadjacent) buckets $B_{1}$ and $B_{3}$ have contents of at least $1-2^{1-r}$ at the beginning of round $r(r=1,2, \ldots, R)$. This is trivial for $r=1$ and if it holds at the beginning of round $r$, she can fill the bucket which contains at least $1-2^{1-r}$ liters with another $2^{-r}$ liters and put the rest of her water $-1-2^{-r}$ liters - in the other bucket. As Cinderella now can remove the water of at most one of the two buckets, the other bucket carries its contents into the next round. At the beginning of the $R$-th round there are $1-2^{1-R}$ liters in $B_{1}$ or $B_{3}$. The Stepmother puts the entire liter into that bucket and produces an overflow since $b<2-2^{1-R}$.


BGR   On a $999 \times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A nonintersecting route of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called cyclic, if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over. How many squares does the longest possible cyclic, non-intersecting route of a limp rook visit?

Solution: The answer is $998^{2}-4=4 \cdot\left(499^{2}-1\right)$ squares. First we show that this number is an upper bound for the number of cells a limp rook can visit. To do this we color the cells with four colors $A, B, C$ and $D$ in the following way: for $(i, j) \equiv(0,0) \bmod 2$ use $A$, for $(i, j) \equiv(0,1) \bmod 2$ use $B$, for $(i, j) \equiv(1,0) \bmod 2$ use $C$ and for $(i, j) \equiv(1,1) \bmod 2$ use $D$. From an $A$-cell the rook has to move to a $B$-cell or a $C$-cell. In the first case, the order of the colors of the cells visited is given by $A, B, D, C, A, B, D, C, A, \ldots$, in the second case it is $A, C, D, B, A, C, D, B, A, \ldots$ Since the route is closed it must contain the same number of cells of each color. There are only $499^{2} A$-cells. In the following we will show that the rook cannot visit all the $A$-cells on its route and hence the maximum possible number of cells in a route is $4 \cdot\left(499^{2}-1\right)$. Assume that the route passes through every single $A$-cell. Color the $A$-cells in black and white in a chessboard manner, i.e. color any two $A$-cells at distance 2 in different color. Since the number of $A$-cells is odd the rook cannot always alternate between visiting black and white $A$-cells along its route. Hence there are two $A$-cells of the same color which are four rook-steps apart that are visited directly one after the other. Let these two $A$-cells have row and column numbers $(a, b)$ and $(a+2, b+2)$ respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-38.jpg?height=449&width=524&top_left_y=1883&top_left_x=766)  There is up to reflection only one way the rook can take from $(a, b)$ to $(a+2, b+2)$. Let this way be $(a, b) \rightarrow(a, b+1) \rightarrow(a+1, b+1) \rightarrow(a+1, b+2) \rightarrow(a+2, b+2)$. Also let without loss of generality the color of the cell $(a, b+1)$ be $B$ (otherwise change the roles of columns and rows). Now consider the $A$-cell $(a, b+2)$. The only way the rook can pass through it is via $(a-1, b+2) \rightarrow$ $(a, b+2) \rightarrow(a, b+3)$ in this order, since according to our assumption after every $A$-cell the rook passes through a $B$-cell. Hence, to connect these two parts of the path, there must be a path connecting the cell $(a, b+3)$ and $(a, b)$ and also a path connecting $(a+2, b+2)$ and $(a-1, b+2)$.  But these four cells are opposite vertices of a convex quadrilateral and the paths are outside of that quadrilateral and hence they must intersect. This is due to the following fact: The path from $(a, b)$ to $(a, b+3)$ together with the line segment joining these two cells form a closed loop that has one of the cells $(a-1, b+2)$ and $(a+2, b+2)$ in its inside and the other one on the outside. Thus the path between these two points must cross the previous path. But an intersection is only possible if a cell is visited twice. This is a contradiction. Hence the number of cells visited is at most $4 \cdot\left(499^{2}-1\right)$. The following picture indicates a recursive construction for all $n \times n$-chessboards with $n \equiv 3$ $\bmod 4$ which clearly yields a path that misses exactly one $A$-cell (marked with a dot, the center cell of the $15 \times 15$-chessboard) and hence, in the case of $n=999$ crosses exactly $4 \cdot\left(499^{2}-1\right)$ cells. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-39.jpg?height=686&width=715&top_left_y=979&top_left_x=676)


RUS   Variant 1. A grasshopper jumps along the real axis. He starts at point 0 and makes 2009 jumps to the right with lengths $1,2, \ldots, 2009$ in an arbitrary order. Let $M$ be a set of 2008 positive integers less than 1005 - 2009. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.  Variant 2. Let $n$ be a nonnegative integer. A grasshopper jumps along the real axis. He starts at point 0 and makes $n+1$ jumps to the right with pairwise different positive integral lengths $a_{1}, a_{2}, \ldots, a_{n+1}$ in an arbitrary order. Let $M$ be a set of $n$ positive integers in the interval $(0, s)$, where $s=a_{1}+a_{2}+\cdots+a_{n+1}$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.


AUT   For any integer $n \geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$. (1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit 0 . (2) If $1 \leq r \leq 9$ we split the decimal representation of $n$ into a maximal right part $R$ that solely consists of digits not less than $r$ and into a left part $L$ that either is empty or ends with a digit strictly smaller than $r$. Then the decimal representation of $h(n)$ consists of the decimal representation of $L$, followed by two copies of the decimal representation of $R-1$. For instance, for the number $n=17,151,345,543$, we will have $L=17,151, R=345,543$ and $h(n)=17,151,345,542,345,542$. Prove that, starting with an arbitrary integer $n \geq 2$, iterated application of $h$ produces the integer 1 after finitely many steps.

Solution: We identify integers $n \geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. We recursively define ten functions $f_{0}, \ldots, f_{9}$ that map some strings into integers for $k=$ $9,8, \ldots, 1,0$. The function $f_{9}$ is only defined on strings $x$ (including the empty string $\varepsilon$ ) that entirely consist of nines. If $x$ consists of $m$ nines, then $f_{9}(x)=m+1, m=0,1, \ldots$ For $k \leq 8$, the domain of $f_{k}(x)$ is the set of all strings consisting only of digits that are $\geq k$. We write $x$ in the form $x_{0} k x_{1} k x_{2} k \ldots x_{m-1} k x_{m}$ where the strings $x_{s}$ only consist of digits $\geq k+1$. Note that some of these strings might equal the empty string $\varepsilon$ and that $m=0$ is possible, i.e. the digit $k$ does not appear in $x$. Then we define  $$ f_{k}(x)=\sum_{s=0}^{m} 4^{f_{k+1}\left(x_{s}\right)} $$  We will use the following obvious fact: Fact 1. If $x$ does not contain digits smaller than $k$, then $f_{i}(x)=4^{f_{i+1}(x)}$ for all $i=0, \ldots, k-1$. In particular, $f_{i}(\varepsilon)=4^{9-i}$ for all $i=0,1, \ldots, 9$. Moreover, by induction on $k=9,8, \ldots, 0$ it follows easily: Fact 2. If the nonempty string $x$ does not contain digits smaller than $k$, then $f_{i}(x)>f_{i}(\varepsilon)$ for all $i=0, \ldots, k$. We will show the essential fact: Fact 3. $f_{0}(n)>f_{0}(h(n))$. Then the empty string will necessarily be reached after a finite number of applications of $h$. But starting from a string without leading zeros, $\varepsilon$ can only be reached via the strings $1 \rightarrow 00 \rightarrow 0 \rightarrow \varepsilon$. Hence also the number 1 will appear after a finite number of applications of $h$. Proof of Fact 3. If the last digit $r$ of $n$ is 0 , then we write $n=x_{0} 0 \ldots 0 x_{m-1} 0 \varepsilon$ where the $x_{i}$ do not contain the digit 0 . Then $h(n)=x_{0} 0 \ldots 0 x_{m-1}$ and $f_{0}(n)-f_{0}(h(n))=f_{0}(\varepsilon)>0$. So let the last digit $r$ of $n$ be at least 1 . Let $L=y k$ and $R=z r$ be the corresponding left and right parts where $y$ is some string, $k \leq r-1$ and the string $z$ consists only of digits not less than $r$. Then $n=y k z r$ and $h(n)=y k z(r-1) z(r-1)$. Let $d(y)$ be the smallest digit of $y$. We consider two cases which do not exclude each other.  Case 1. $d(y) \geq k$. Then  $$ f_{k}(n)-f_{k}(h(n))=f_{k}(z r)-f_{k}(z(r-1) z(r-1)) . $$  In view of Fact 1 this difference is positive if and only if  $$ f_{r-1}(z r)-f_{r-1}(z(r-1) z(r-1))>0 $$  We have, using Fact 2,  $$ f_{r-1}(z r)=4^{f_{r}(z r)}=4^{f_{r}(z)+4^{f_{r+1}(z)}} \geq 4 \cdot 4^{f_{r}(z)}>4^{f_{r}(z)}+4^{f_{r}(z)}+4^{f_{r}(z)}=f_{r-1}(z(r-1) z(r-1)) . $$  Here we use the additional definition $f_{10}(\varepsilon)=0$ if $r=9$. Consequently, $f_{k}(n)-f_{k}(h(n))>0$ and according to Fact $1, f_{0}(n)-f_{0}(h(n))>0$. Case 2. $d(y) \leq k$. We prove by induction on $d(y)=k, k-1, \ldots, 0$ that $f_{i}(n)-f_{i}(h(n))>0$ for all $i=0, \ldots, d(y)$. By Fact 1, it suffices to do so for $i=d(y)$. The initialization $d(y)=k$ was already treated in Case 1. Let $t=d(y)<k$. Write $y$ in the form utv where $v$ does not contain digits $\leq t$. Then, in view of the induction hypothesis,  $$ f_{t}(n)-f_{t}(h(n))=f_{t}(v k z r)-f_{t}(v k z(r-1) z(r-1))=4^{f_{t+1}(v k z r)}-4^{f_{t+1}(v k z(r-1) z(r-1))}>0 $$  Thus the inequality $f_{d(y)}(n)-f_{d(y)}(h(n))>0$ is established and from Fact 1 it follows that $f_{0}(n)-f_{0}(h(n))>0$.


Solution: We identify integers $n \geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from 0 to 9. Moreover, let us define that the empty string, $\varepsilon$, is being mapped to the empty string. In the following all functions map the set of strings into the set of strings. For two functions $f$ and $g$ let $g \circ f$ be defined by $(g \circ f)(x)=g(f(x))$ for all strings $x$ and let, for non-negative integers $n$, $f^{n}$ denote the $n$-fold application of $f$. For any string $x$ let $s(x)$ be the smallest digit of $x$, and for the empty string let $s(\varepsilon)=\infty$. We define nine functions $g_{1}, \ldots, g_{9}$ as follows: Let $k \in\{1, \ldots, 9\}$ and let $x$ be a string. If $x=\varepsilon$ then $g_{k}(x)=\varepsilon$. Otherwise, write $x$ in the form $x=y z r$ where $y$ is either the empty string or ends with a digit smaller than $k, s(z) \geq k$ and $r$ is the rightmost digit of $x$. Then $g_{k}(x)=z r$.  Lemma 1. We have $g_{k} \circ h=g_{k} \circ h \circ g_{k}$ for all $k=1, \ldots, 9$. Proof of Lemma 1. Let $x=y z r$ be as in the definition of $g_{k}$. If $y=\varepsilon$, then $g_{k}(x)=x$, whence  $$ g_{k}(h(x))=g_{k}\left(h\left(g_{k}(x)\right) .\right. $$  So let $y \neq \varepsilon$. Case 1. $z$ contains a digit smaller than $r$. Let $z=u a v$ where $a<r$ and $s(v) \geq r$. Then  $$ h(x)= \begin{cases}\text { yuav } & \text { if } r=0 \\ \operatorname{yuav}(r-1) v(r-1) & \text { if } r>0\end{cases} $$  and  $$ h\left(g_{k}(x)\right)=h(z r)=h(\text { uavr })= \begin{cases}\operatorname{uav} & \text { if } r=0 \\ \operatorname{uav}(r-1) v(r-1) & \text { if } r>0\end{cases} $$  Since $y$ ends with a digit smaller than $k$, (1) is obviously true. Case 2. $z$ does not contain a digit smaller than $r$. Let $y=u v$ where $u$ is either the empty string or ends with a digit smaller than $r$ and $s(v) \geq r$. We have  $$ h(x)= \begin{cases}u v z & \text { if } r=0 \\ u v z(r-1) v z(r-1) & \text { if } r>0\end{cases} $$  and  $$ h\left(g_{k}(x)\right)=h(z r)= \begin{cases}z & \text { if } r=0 \\ z(r-1) z(r-1) & \text { if } r>0\end{cases} $$  Recall that $y$ and hence $v$ ends with a digit smaller than $k$, but all digits of $v$ are at least $r$. Now if $r>k$, then $v=\varepsilon$, whence the terminal digit of $u$ is smaller than $k$, which entails  $$ g_{k}(h(x))=z(r-1) z(r-1)=g_{k}\left(h\left(g_{k}(x)\right)\right) . $$  If $r \leq k$, then  $$ g_{k}(h(x))=z(r-1)=g_{k}\left(h\left(g_{k}(x)\right)\right), $$  so that in both cases (1) is true. Thus Lemma 1 is proved. Lemma 2. Let $k \in\{1, \ldots, 9\}$, let $x$ be a non-empty string and let $n$ be a positive integer. If $h^{n}(x)=\varepsilon$ then $\left(g_{k} \circ h\right)^{n}(x)=\varepsilon$. Proof of Lemma 2. We proceed by induction on $n$. If $n=1$ we have  $$ \varepsilon=h(x)=g_{k}(h(x))=\left(g_{k} \circ h\right)(x) . $$  Now consider the step from $n-1$ to $n$ where $n \geq 2$. Let $h^{n}(x)=\varepsilon$ and let $y=h(x)$. Then $h^{n-1}(y)=\varepsilon$ and by the induction hypothesis $\left(g_{k} \circ h\right)^{n-1}(y)=\varepsilon$. In view of Lemma 1 ,  $$ \begin{aligned} & \varepsilon=\left(g_{k} \circ h\right)^{n-2}\left(\left(g_{k} \circ h\right)(y)\right)=\left(g_{k} \circ h\right)^{n-2}\left(g_{k}(h(y))\right. \\ &=\left(g_{k} \circ h\right)^{n-2}\left(g_{k}\left(h\left(g_{k}(y)\right)\right)=\left(g_{k} \circ h\right)^{n-2}\left(g_{k}\left(h\left(g_{k}(h(x))\right)\right)=\left(g_{k} \circ h\right)^{n}(x)\right.\right. \end{aligned} $$  Thus the induction step is complete and Lemma 2 is proved. We say that the non-empty string $x$ terminates if $h^{n}(x)=\varepsilon$ for some non-negative integer $n$. Lemma 3. Let $x=y z r$ where $s(y) \geq k, s(z) \geq k, y$ ends with the digit $k$ and $z$ is possibly empty. If $y$ and $z r$ terminate then also $x$ terminates. Proof of Lemma 3. Suppose that $y$ and $z r$ terminate. We proceed by induction on $k$. Let $k=0$. Obviously, $h(y w)=y h(w)$ for any non-empty string $w$. Let $h^{n}(z r)=\epsilon$. It follows easily by induction on $m$ that $h^{m}(y z r)=y h^{m}(z r)$ for $m=1, \ldots, n$. Consequently, $h^{n}(y z r)=y$. Since $y$ terminates, also $x=y z r$ terminates. Now let the assertion be true for all nonnegative integers less than $k$ and let us prove it for $k$ where $k \geq 1$. It turns out that it is sufficient to prove that $y g_{k}(h(z r))$ terminates. Indeed: Case 1. $r=0$. Then $h(y z r)=y z=y g_{k}(h(z r))$. Case 2. $0<r \leq k$. We have $h(z r)=z(r-1) z(r-1)$ and $g_{k}(h(z r))=z(r-1)$. Then $h(y z r)=y z(r-1) y z(r-$ $1)=y g_{k}(h(z r)) y g_{k}(h(z r))$ and we may apply the induction hypothesis to see that if $\left.y g_{k} h(z r)\right)$ terminates, then $h(y z r)$ terminates.  Case 3. $r>k$. Then $h(y z r)=y h(z r)=y g_{k}(h(z r))$. Note that $y g_{k}(h(z r))$ has the form $y z^{\prime} r^{\prime}$ where $s\left(z^{\prime}\right) \geq k$. By the same arguments it is sufficient to prove that $y g_{k}\left(h\left(z^{\prime} r^{\prime}\right)\right)=y\left(g_{k} \circ h\right)^{2}(z r)$ terminates and, by induction, that $y\left(g_{k} \circ h\right)^{m}(z r)$ terminates for some positive integer $m$. In view of Lemma 2 there is some $m$ such that $\left(g_{k} \circ\right.$ $h)^{m}(z r)=\epsilon$, so $x=y z r$ terminates if $y$ terminates. Thus Lemma 3 is proved. Now assume that there is some string $x$ that does not terminate. We choose $x$ minimal. If $x \geq 10$, we can write $x$ in the form $x=y z r$ of Lemma 3 and by this lemma $x$ terminates since $y$ and $z r$ are smaller than $x$. If $x \leq 9$, then $h(x)=(x-1)(x-1)$ and $h(x)$ terminates again by Lemma 3 and the minimal choice of $x$.


Solution: We commence by introducing some terminology. Instead of integers, we will consider the set $S$ of all strings consisting of the digits $0,1, \ldots, 9$, including the empty string $\epsilon$. If $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is a nonempty string, we let $\rho(a)=a_{n}$ denote the terminal digit of $a$ and $\lambda(a)$ be the string with the last digit removed. We also define $\lambda(\epsilon)=\epsilon$ and denote the set of non-negative integers by $\mathbb{N}_{0}$. Now let $k \in\{0,1,2, \ldots, 9\}$ denote any digit. We define a function $f_{k}: S \longrightarrow S$ on the set of strings: First, if the terminal digit of $n$ belongs to $\{0,1, \ldots, k\}$, then $f_{k}(n)$ is obtained from $n$ by deleting this terminal digit, i.e $f_{k}(n)=\lambda(n)$. Secondly, if the terminal digit of $n$ belongs to $\{k+1, \ldots, 9\}$, then $f_{k}(n)$ is obtained from $n$ by the process described in the problem. We also define $f_{k}(\epsilon)=\epsilon$. Note that up to the definition for integers $n \leq 1$, the function $f_{0}$ coincides with the function $h$ in the problem, through interpreting integers as digit strings. The argument will be roughly as follows. We begin by introducing a straightforward generalization of our claim about $f_{0}$. Then it will be easy to see that $f_{9}$ has all these stronger properties, which means that is suffices to show for $k \in\{0,1, \ldots, 8\}$ that $f_{k}$ possesses these properties provided that $f_{k+1}$ does. We continue to use $k$ to denote any digit. The operation $f_{k}$ is said to be separating, if the followings holds: Whenever $a$ is an initial segment of $b$, there is some $N \in \mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. The following two notions only apply to the case where $f_{k}$ is indeed separating, otherwise they remain undefined. For every $a \in S$ we denote the least $N \in \mathbb{N}_{0}$ for which $f_{k}^{N}(a)=\epsilon$ occurs by $g_{k}(a)$ (because $\epsilon$ is an initial segment of $a$, such an $N$ exists if $f_{k}$ is separating). If for every two strings $a$ and $b$ such that $a$ is a terminal segment of $b$ one has $g_{k}(a) \leq g_{k}(b)$, we say that $f_{k}$ is coherent. In case that $f_{k}$ is separating and coherent we call the digit $k$ seductive. As $f_{9}(a)=\lambda(a)$ for all $a$, it is obvious that 9 is seductive. Hence in order to show that 0 is seductive, which clearly implies the statement of the problem, it suffices to take any $k \in\{0,1, \ldots, 8\}$ such that $k+1$ is seductive and to prove that $k$ has to be seductive as well. Note that in doing so, we have the function $g_{k+1}$ at our disposal. We have to establish two things and we begin with  Step 1. $f_{k}$ is separating.  Before embarking on the proof of this, we record a useful observation which is easily proved by induction on $M$.  Claim 1. For any strings $A, B$ and any positive integer $M$ such that $f_{k}^{M-1}(B) \neq \epsilon$, we have  $$ f_{k}^{M}(A k B)=A k f_{k}^{M}(B) $$  Now we call a pair $(a, b)$ of strings wicked provided that $a$ is an initial segment of $b$, but there is no $N \in \mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. We need to show that there are none, so assume that there were such pairs. Choose a wicked pair $(a, b)$ for which $g_{k+1}(b)$ attains its minimal possible value. Obviously $b \neq \epsilon$ for any wicked pair $(a, b)$. Let $z$ denote the terminal digit of $b$. Observe that $a \neq b$, which means that $a$ is also an initial segment of $\lambda(b)$. To facilitate the construction of the eventual contradiction, we prove Claim 2. There cannot be an $N \in \mathbb{N}_{0}$ such that  $$ f_{k}^{N}(b)=\lambda(b) $$  Proof of Claim 2. For suppose that such an $N$ existed. Because $g_{k+1}(\lambda(b))<g_{k+1}(b)$ in view of the coherency of $f_{k+1}$, the pair $(a, \lambda(b))$ is not wicked. But then there is some $N^{\prime}$ for which $f_{k}^{N^{\prime}}(\lambda(b))=a$ which entails $f_{k}^{N+N^{\prime}}(b)=a$, contradiction. Hence Claim 2 is proved.  It follows that $z \leq k$ is impossible, for otherwise $N=1$ violated Claim 2. Also $z>k+1$ is impossible: Set $B=f_{k}(b)$. Then also $f_{k+1}(b)=B$, but $g_{k+1}(B)<g_{k+1}(b)$ and $a$ is an initial segment of $B$. Thus the pair $(a, B)$ is not wicked. Hence there is some $N \in \mathbb{N}_{0}$ with $a=f_{k}^{N}(B)$, which, however, entails $a=f_{k}^{N+1}(b)$. We are left with the case $z=k+1$. Let $L$ denote the left part and $R=R^{*}(k+1)$ the right part of $b$. Then we have symbolically  $$ f_{k}(b)=L R^{*} k R^{*} k, f_{k}^{2}(b)=L R^{*} k R^{*} \quad \text { and } \quad f_{k+1}(b)=L R^{*} . $$  Using that $R^{*}$ is a terminal segment of $L R^{*}$ and the coherency of $f_{k+1}$, we infer  $$ g_{k+1}\left(R^{*}\right) \leq g_{k+1}\left(L R^{*}\right)<g_{k+1}(b) $$  Hence the pair $\left(\epsilon, R^{*}\right)$ is not wicked, so there is some minimal $M \in \mathbb{N}_{0}$ with $f_{k}^{M}\left(R^{*}\right)=\epsilon$ and by Claim 1 it follows that $f_{k}^{2+M}(b)=L R^{*} k$. Finally, we infer that $\lambda(b)=L R^{*}=f_{k}\left(L R^{*} k\right)=$ $f_{k}^{3+M}(b)$, which yields a contradiction to Claim 2. This final contradiction establishes that $f_{k}$ is indeed separating.  Step 2. $f_{k}$ is coherent.  To prepare the proof of this, we introduce some further pieces of terminology. A nonempty string $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is called a hypostasis, if $a_{n}<a_{i}$ for all $i=1, \ldots, n-1$. Reading an arbitrary string $a$ backwards, we easily find a, possibly empty, sequence $\left(A_{1}, A_{2}, \ldots, A_{m}\right)$ of hypostases such that $\rho\left(A_{1}\right) \leq \rho\left(A_{2}\right) \leq \cdots \leq \rho\left(A_{m}\right)$ and, symbolically, $a=A_{1} A_{2} \ldots A_{m}$. The latter sequence is referred to as the decomposition of $a$. So, for instance, $(20,0,9)$ is the decomposition of 2009 and the string 50 is a hypostasis. Next we explain when we say about two strings $a$ and $b$ that $a$ is injectible into $b$. The definition is by induction on the length of $b$. Let $\left(B_{1}, B_{2}, \ldots, B_{n}\right)$ be the decomposition of $b$ into hypostases. Then $a$ is injectible into $b$ if for the decomposition $\left(A_{1}, A_{2}, \ldots, A_{m}\right)$ of $a$ there is a strictly increasing function $H:\{1,2, \ldots, m\} \longrightarrow\{1,2, \ldots, n\}$ satisfying  $$ \rho\left(A_{i}\right)=\rho\left(B_{H(i)}\right) \text { for all } i=1, \ldots, m \text {; } $$  $\lambda\left(A_{i}\right)$ is injectible into $\lambda\left(B_{H(i)}\right)$ for all $i=1, \ldots, m$. If one can choose $H$ with $H(m)=n$, then we say that $a$ is strongly injectible into $b$. Obviously, if $a$ is a terminal segment of $b$, then $a$ is strongly injectible into $b$. Claim 3. If $a$ and $b$ are two nonempty strings such that $a$ is strongly injectible into $b$, then $\lambda(a)$ is injectible into $\lambda(b)$.  Proof of Claim 3. Let $\left(B_{1}, B_{2}, \ldots, B_{n}\right)$ be the decomposition of $b$ and let $\left(A_{1}, A_{2}, \ldots, A_{m}\right)$ be the decomposition of $a$. Take a function $H$ exemplifying that $a$ is strongly injectible into $b$. Let $\left(C_{1}, C_{2}, \ldots, C_{r}\right)$ be the decomposition of $\lambda\left(A_{m}\right)$ and let $\left(D_{1}, D_{2}, \ldots, D_{s}\right)$ be the decomposition of $\lambda\left(B_{n}\right)$. Choose a strictly increasing $H^{\prime}:\{1,2, \ldots, r\} \longrightarrow\{1,2, \ldots s\}$ witnessing that $\lambda\left(A_{m}\right)$ is injectible into $\lambda\left(B_{n}\right)$. Clearly, $\left(A_{1}, A_{2}, \ldots, A_{m-1}, C_{1}, C_{2}, \ldots, C_{r}\right)$ is the decomposition of $\lambda(a)$ and $\left(B_{1}, B_{2}, \ldots, B_{n-1}, D_{1}, D_{2}, \ldots, D_{s}\right)$ is the decomposition of $\lambda(b)$. Then the function $H^{\prime \prime}:\{1,2, \ldots, m+r-1\} \longrightarrow\{1,2, \ldots, n+s-1\}$ given by $H^{\prime \prime}(i)=H(i)$ for $i=1,2, \ldots, m-1$ and $H^{\prime \prime}(m-1+i)=n-1+H^{\prime}(i)$ for $i=1,2, \ldots, r$ exemplifies that $\lambda(a)$ is injectible into $\lambda(b)$, which finishes the proof of the claim.  A pair $(a, b)$ of strings is called aggressive if $a$ is injectible into $b$ and nevertheless $g_{k}(a)>g_{k}(b)$. Observe that if $f_{k}$ was incoherent, which we shall assume from now on, then such pairs existed. Now among all aggressive pairs we choose one, say $(a, b)$, for which $g_{k}(b)$ attains its least possible value. Obviously $f_{k}(a)$ cannot be injectible into $f_{k}(b)$, for otherwise the pair $\left(f_{k}(a), f_{k}(b)\right)$ was aggressive and contradicted our choice of $(a, b)$. Let $\left(A_{1}, A_{2}, \ldots, A_{m}\right)$ and $\left(B_{1}, B_{2}, \ldots, B_{n}\right)$ be the decompositions of $a$ and $b$ and take a function $H:\{1,2, \ldots, m\} \longrightarrow\{1,2, \ldots, n\}$ exemplifying that $a$ is indeed injectible into $b$. If we had $H(m)<n$, then $a$ was also injectible into the number $b^{\prime}$ whose decomposition is $\left(B_{1}, B_{2}, \ldots, B_{n-1}\right)$ and by separativity of $f_{k}$ we obtained $g_{k}\left(b^{\prime}\right)<g_{k}(b)$, whence the pair ( $a, b^{\prime}$ ) was also aggressive, contrary to the minimality condition imposed on $b$. Therefore $a$ is strongly injectible into $b$. In particular, $a$ and $b$ have a common terminal digit, say $z$. If we had $z \leq k$, then $f_{k}(a)=\lambda(a)$ and $f_{k}(b)=\lambda(b)$, so that by Claim $3, f_{k}(a)$ was injectible into $f_{k}(b)$, which is a contradiction. Hence, $z \geq k+1$. Now let $r$ be the minimal element of $\{1,2, \ldots, m\}$ for which $\rho\left(A_{r}\right)=z$. Then the maximal right part of $a$ consisting of digits $\geq z$ is equal to $R_{a}$, the string whose decomposition is $\left(A_{r}, A_{r+1}, \ldots, A_{m}\right)$. Then $R_{a}-1$ is a hypostasis and $\left(A_{1}, \ldots, A_{r-1}, R_{a}-1, R_{a}-1\right)$ is the decomposition of $f_{k}(a)$. Defining $s$ and $R_{b}$ in a similar fashion with respect to $b$, we see that $\left(B_{1}, \ldots, B_{s-1}, R_{b}-1, R_{b}-1\right)$ is the decomposition of $f_{k}(b)$. The definition of injectibility then easily entails that $R_{a}$ is strongly injectible into $R_{b}$. It follows from Claim 3 that $\lambda\left(R_{a}\right)=$ $\lambda\left(R_{a}-1\right)$ is injectible into $\lambda\left(R_{b}\right)=\lambda\left(R_{b}-1\right)$, whence the function $H^{\prime}:\{1,2, \ldots, r+1\} \longrightarrow$ $\{1,2, \ldots, s+1\}$, given by $H^{\prime}(i)=H(i)$ for $i=1,2, \ldots, r-1, H^{\prime}(r)=s$ and $H^{\prime}(r+1)=s+1$ exemplifies that $f_{k}(a)$ is injectible into $f_{k}(b)$, which yields a contradiction as before. This shows that aggressive pairs cannot exist, whence $f_{k}$ is indeed coherent, which finishes the proof of the seductivity of $k$, whereby the problem is finally solved.


BEL   Let $A B C$ be a triangle with $A B=A C$. The angle bisectors of $A$ and $B$ meet the sides $B C$ and $A C$ in $D$ and $E$, respectively. Let $K$ be the incenter of triangle $A D C$. Suppose that $\angle B E K=45^{\circ}$. Find all possible values of $\angle B A C$.

Solution: Answer: $\angle B A C=60^{\circ}$ or $\angle B A C=90^{\circ}$ are possible values and the only possible values.  Let $I$ be the incenter of triangle $A B C$, then $K$ lies on the line $C I$. Let $F$ be the point, where the incircle of triangle $A B C$ touches the side $A C$; then the segments $I F$ and $I D$ have the same length and are perpendicular to $A C$ and $B C$, respectively. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-49.jpg?height=549&width=615&top_left_y=1025&top_left_x=252)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-49.jpg?height=510&width=901&top_left_y=1067&top_left_x=906)  Figure 2  Let $P, Q$ and $R$ be the points where the incircle of triangle $A D C$ touches the sides $A D, D C$ and $C A$, respectively. Since $K$ and $I$ lie on the angle bisector of $\angle A C D$, the segments $I D$ and $I F$ are symmetric with respect to the line $I C$. Hence there is a point $S$ on $I F$ where the incircle of triangle $A D C$ touches the segment $I F$. Then segments $K P, K Q, K R$ and $K S$ all have the same length and are perpendicular to $A D, D C, C A$ and $I F$, respectively. So - regardless of the value of $\angle B E K$ - the quadrilateral $K R F S$ is a square and $\angle S F K=\angle K F C=45^{\circ}$. Consider the case $\angle B A C=60^{\circ}$ (see Figure 1). Then triangle $A B C$ is equilateral. Furthermore we have $F=E$, hence $\angle B E K=\angle I F K=\angle S E K=45^{\circ}$. So $60^{\circ}$ is a possible value for $\angle B A C$. Now consider the case $\angle B A C=90^{\circ}$ (see Figure 2). Then $\angle C B A=\angle A C B=45^{\circ}$. Furthermore, $\angle K I E=\frac{1}{2} \angle C B A+\frac{1}{2} \angle A C B=45^{\circ}, \angle A E B=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$ and $\angle E I A=\angle B I D=180^{\circ}-90^{\circ}-22.5^{\circ}=67.5^{\circ}$. Hence triangle $I E A$ is isosceles and a reflection of the bisector of $\angle I A E$ takes $I$ to $E$ and $K$ to itself. So triangle $I K E$ is symmetric with respect to this axis, i.e. $\angle K I E=\angle I E K=\angle B E K=45^{\circ}$. So $90^{\circ}$ is a possible value for $\angle B A C$, too. If, on the other hand, $\angle B E K=45^{\circ}$ then $\angle B E K=\angle I E K=\angle I F K=45^{\circ}$. Then  - either $F=E$, which makes the angle bisector $B I$ be an altitude, i.e., which makes triangle $A B C$ isosceles with base $A C$ and hence equilateral and so $\angle B A C=60^{\circ}$, - or $E$ lies between $F$ and $C$, which makes the points $K, E, F$ and $I$ concyclic, so $45^{\circ}=$ $\angle K F C=\angle K F E=\angle K I E=\angle C B I+\angle I C B=2 \cdot \angle I C B=90^{\circ}-\frac{1}{2} \angle B A C$, and so $\angle B A C=90^{\circ}$, - or $F$ lies between $E$ and $C$, then again, $K, E, F$ and $I$ are concyclic, so $45^{\circ}=\angle K F C=$ $180^{\circ}-\angle K F E=\angle K I E$, which yields the same result $\angle B A C=90^{\circ}$. (However, for $\angle B A C=90^{\circ} E$ lies, in fact, between $F$ and $C$, see Figure 2. So this case does not occur.) This proves $90^{\circ}$ and $60^{\circ}$ to be the only possible values for $\angle B A C$.


Solution: Denote angles at $A, B$ and $C$ as usual by $\alpha, \beta$ and $\gamma$. Since triangle $A B C$ is isosceles, we have $\beta=\gamma=90^{\circ}-\frac{\alpha}{2}<90^{\circ}$, so $\angle E C K=45^{\circ}-\frac{\alpha}{4}=\angle K C D$. Since $K$ is the incenter of triangle $A D C$, we have $\angle C D K=\angle K D A=45^{\circ}$; furthermore $\angle D I C=45^{\circ}+\frac{\alpha}{4}$. Now, if $\angle B E K=45^{\circ}$, easy calculations within triangles $B C E$ and $K C E$ yield  $$ \begin{aligned} & \angle K E C=180^{\circ}-\frac{\beta}{2}-45^{\circ}-\beta=135^{\circ}-\frac{3}{2} \beta=\frac{3}{2}\left(90^{\circ}-\beta\right)=\frac{3}{4} \alpha, \\ & \angle I K E=\frac{3}{4} \alpha+45^{\circ}-\frac{\alpha}{4}=45^{\circ}+\frac{\alpha}{2} . \end{aligned} $$  So in triangles $I C E, I K E, I D K$ and $I D C$ we have (see Figure 3)  $$ \begin{array}{ll} \frac{I C}{I E}=\frac{\sin \angle I E C}{\sin \angle E C I}=\frac{\sin \left(45^{\circ}+\frac{3}{4} \alpha\right)}{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}, & \frac{I E}{I K}=\frac{\sin \angle E K I}{\sin \angle I E K}=\frac{\sin \left(45^{\circ}+\frac{\alpha}{2}\right)}{\sin 45^{\circ}} \\ \frac{I K}{I D}=\frac{\sin \angle K D I}{\sin \angle I K D}=\frac{\sin 45^{\circ}}{\sin \left(90^{\circ}-\frac{\alpha}{4}\right)}, & \frac{I D}{I C}=\frac{\sin \angle I C D}{\sin \angle C D I}=\frac{\sin \left(45^{\circ}-\frac{\alpha}{4}\right)}{\sin 90^{\circ}} \end{array} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-50.jpg?height=581&width=972&top_left_y=1337&top_left_x=548)  Figure 3 Multiplication of these four equations yields  $$ 1=\frac{\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)}{\sin \left(90^{\circ}-\frac{\alpha}{4}\right)} $$  But, since  $$ \begin{aligned} \sin \left(90^{\circ}-\frac{\alpha}{4}\right) & =\cos \frac{\alpha}{4}=\cos \left(\left(45^{\circ}+\frac{3}{4} \alpha\right)-\left(45^{\circ}+\frac{\alpha}{2}\right)\right) \\ & =\cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)+\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right) \end{aligned} $$  this is equivalent to  $$ \sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)=\cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)+\sin \left(45^{\circ}+\frac{3}{4} \alpha\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right) $$  and finally  $$ \cos \left(45^{\circ}+\frac{3}{4} \alpha\right) \cos \left(45^{\circ}+\frac{\alpha}{2}\right)=0 $$  But this means $\cos \left(45^{\circ}+\frac{3}{4} \alpha\right)=0$, hence $45^{\circ}+\frac{3}{4} \alpha=90^{\circ}$, i.e. $\alpha=60^{\circ}$ or $\cos \left(45^{\circ}+\frac{\alpha}{2}\right)=0$, hence $45^{\circ}+\frac{\alpha}{2}=90^{\circ}$, i.e. $\alpha=90^{\circ}$. So these values are the only two possible values for $\alpha$. On the other hand, both $\alpha=90^{\circ}$ and $\alpha=60^{\circ}$ yield $\angle B E K=45^{\circ}$, this was shown in Solution 1.


RUS   Let $A B C$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $C A$ and $A B$, respectively. The circle $k$ passes through the midpoints of the segments $B P$, $C Q$, and $P Q$. Prove that if the line $P Q$ is tangent to circle $k$ then $O P=O Q$.

Solution: Let $K, L, M, B^{\prime}, C^{\prime}$ be the midpoints of $B P, C Q, P Q, C A$, and $A B$, respectively (see Figure 1). Since $C A \| L M$, we have $\angle L M P=\angle Q P A$. Since $k$ touches the segment $P Q$ at $M$, we find $\angle L M P=\angle L K M$. Thus $\angle Q P A=\angle L K M$. Similarly it follows from $A B \| M K$ that $\angle P Q A=\angle K L M$. Therefore, triangles $A P Q$ and $M K L$ are similar, hence  $$ \frac{A P}{A Q}=\frac{M K}{M L}=\frac{\frac{Q B}{2}}{\frac{P C}{2}}=\frac{Q B}{P C} $$  Now (1) is equivalent to $A P \cdot P C=A Q \cdot Q B$ which means that the power of points $P$ and $Q$ with respect to the circumcircle of $\triangle A B C$ are equal, hence $O P=O Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-52.jpg?height=769&width=803&top_left_y=1152&top_left_x=632)  Figure 1  Comment. The last argument can also be established by the following calculation:  $$ \begin{aligned} O P^{2}-O Q^{2} & =O B^{\prime 2}+B^{\prime} P^{2}-O C^{2}-C^{\prime} Q^{2} \\ & =\left(O A^{2}-A B^{2}\right)+B^{\prime} P^{2}-\left(O A^{2}-A C^{\prime 2}\right)-C^{\prime} Q^{2} \\ & =\left(A C^{\prime 2}-C^{\prime} Q^{2}\right)-\left(A B^{\prime 2}-B^{\prime} P^{2}\right) \\ & =\left(A C^{\prime}-C^{\prime} Q\right)\left(A C^{\prime}+C^{\prime} Q\right)-\left(A B^{\prime}-B^{\prime} P\right)\left(A B^{\prime}+B^{\prime} P\right) \\ & =A Q \cdot Q B-A P \cdot P C . \end{aligned} $$  With (1), we conclude $O P^{2}-O Q^{2}=0$, as desired.


Solution: Again, denote by $K, L, M$ the midpoints of segments $B P, C Q$, and $P Q$, respectively. Let $O, S, T$ be the circumcenters of triangles $A B C, K L M$, and $A P Q$, respectively (see Figure 2). Note that $M K$ and $L M$ are the midlines in triangles $B P Q$ and $C P Q$, respectively, so $\overrightarrow{M K}=\frac{1}{2} \overrightarrow{Q B}$ and $\overrightarrow{M L}=\frac{1}{2} \overrightarrow{P C}$. Denote by $\operatorname{pr}_{l}(\vec{v})$ the projection of vector $\vec{v}$ onto line $l$. Then $\operatorname{pr}_{A B}(\overrightarrow{O T})=\operatorname{pr}_{A B}(\overrightarrow{O A}-\overrightarrow{T A})=\frac{1}{2} \overrightarrow{B A}-\frac{1}{2} \overrightarrow{Q A}=\frac{1}{2} \overrightarrow{B Q}=\overrightarrow{K M}$ and $\operatorname{pr}_{A B}(\overrightarrow{S M})=\operatorname{pr}_{M K}(\overrightarrow{S M})=$ $\frac{1}{2} \overrightarrow{K M}=\frac{1}{2} \operatorname{pr}_{A B}(\overrightarrow{O T})$. Analogously we get $\operatorname{pr}_{C A}(\overrightarrow{S M})=\frac{1}{2} \operatorname{pr}_{C A}(\overrightarrow{O T})$. Since $A B$ and $C A$ are not parallel, this implies that $\overrightarrow{S M}=\frac{1}{2} \overrightarrow{O T}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-53.jpg?height=760&width=795&top_left_y=608&top_left_x=636)  Figure 2 Now, since the circle $k$ touches $P Q$ at $M$, we get $S M \perp P Q$, hence $O T \perp P Q$. Since $T$ is equidistant from $P$ and $Q$, the line $O T$ is a perpendicular bisector of segment $P Q$, and hence $O$ is equidistant from $P$ and $Q$ which finishes the proof.


IRN   Let $A B C$ be a triangle. The incircle of $A B C$ touches the sides $A B$ and $A C$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $B Y$ and $C Z$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $B C Y R$ and $B C S Z$ are parallelograms. Prove that $G R=G S$.

Solution: Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $A B C$. Let $k$ and $k_{a}$ touch the side $B C$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $A B$ and $A C$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude  $$ \begin{gathered} Z P=Z B+B P=X B+B T=B X+C X=Z S \text { and } \\ C Q=C T=B X=B Z=C S . \end{gathered} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-54.jpg?height=972&width=812&top_left_y=1276&top_left_x=628)  So for each of the points $Z, C$, their distances to $S$ equal the length of a tangent segment from this point to $k_{a}$. It is well-known, that all points with this property lie on the line $Z C$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $B Y$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $Z C$ and $B Y$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $G R=G S$ follows immediately.


Solution: Denote $x=A Z=A Y, y=B Z=B X, z=C X=C Y, p=Z G, q=G C$. Several lengthy calculations (Menelaos' theorem in triangle $A Z C$, law of Cosines in triangles $A B C$ and $A Z C$ and Stewart's theorem in triangle $Z C S$ ) give four equations for $p, q, \cos \alpha$ and $G S$ in terms of $x, y$, and $z$ that can be resolved for $G S$. The result is symmetric in $y$ and $z$, so $G R=G S$. More in detail this means: The line $B Y$ intersects the sides of triangle $A Z C$, so Menelaos' theorem yields $\frac{p}{q} \cdot \frac{z}{x} \cdot \frac{x+y}{y}=1$, hence  $$ \frac{p}{q}=\frac{x y}{y z+z x} . $$  Since we only want to show that the term for $G S$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N=x y+y z+z x$. So (1) implies  $$ \frac{p}{p+q}=\frac{x y}{x y+y z+z x}=\frac{x y}{N} \quad \text { and } \quad \frac{q}{p+q}=\frac{y z+z x}{x y+y z+z x}=\frac{y z+z x}{N} . $$  Now the law of Cosines in triangle $A B C$ yields  $$ \cos \alpha=\frac{(x+y)^{2}+(x+z)^{2}-(y+z)^{2}}{2(x+y)(x+z)}=\frac{2 x^{2}+2 x y+2 x z-2 y z}{2(x+y)(x+z)}=1-\frac{2 y z}{(x+y)(x+z)} . $$  We use this result to apply the law of Cosines in triangle $A Z C$ :  $$ \begin{aligned} (p+q)^{2} & =x^{2}+(x+z)^{2}-2 x(x+z) \cos \alpha \\ & =x^{2}+(x+z)^{2}-2 x(x+z) \cdot\left(1-\frac{2 y z}{(x+y)(x+z)}\right) \\ & =z^{2}+\frac{4 x y z}{x+y} . \end{aligned} $$  Now in triangle $Z C S$ the segment $G S$ is a cevian, so with StEWART's theorem we have $p y^{2}+q(y+z)^{2}=(p+q)\left(G S^{2}+p q\right)$, hence  $$ G S^{2}=\frac{p}{p+q} \cdot y^{2}+\frac{q}{p+q} \cdot(y+z)^{2}-\frac{p}{p+q} \cdot \frac{q}{p+q} \cdot(p+q)^{2} . $$  Replacing the $p$ 's and $q$ 's herein by (2) and (3) yields  $$ \begin{aligned} G S^{2} & =\frac{x y}{N} y^{2}+\frac{y z+z x}{N}(y+z)^{2}-\frac{x y}{N} \cdot \frac{y z+z x}{N} \cdot\left(z^{2}+\frac{4 x y z}{x+y}\right) \\ & =\frac{x y^{3}}{N}+\underbrace{\frac{y z(y+z)^{2}}{N}}_{M_{1}}+\frac{z x(y+z)^{2}}{N}-\frac{x y z^{3}(x+y)}{N^{2}}-\underbrace{\frac{4 x^{2} y^{2} z^{2}}{N^{2}}}_{M_{2}} \\ & =\frac{x y^{3}+z x(y+z)^{2}}{N}-\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\ & =\underbrace{\frac{x\left(y^{3}+y^{2} z+y z^{2}+z^{3}\right)}{N}}_{M_{3}}+\frac{x y z^{2} N}{N^{2}}-\frac{x y z^{3}(x+y)}{N^{2}}+M_{1}-M_{2} \\ & =\frac{x^{2} y^{2} z^{2}+x y^{2} z^{3}+x^{2} y z^{3}-x^{2} y z^{3}-x y^{2} z^{3}}{N^{2}}+M_{1}-M_{2}+M_{3} \\ & =\frac{x^{2} y^{2} z^{2}}{N^{2}}+M_{1}-M_{2}+M_{3}, \end{aligned} $$  a term that is symmetric in $y$ and $z$, indeed.  Comment. $G$ is known as GERGONnE's point of $\triangle A B C$.


UNK   Given a cyclic quadrilateral $A B C D$, let the diagonals $A C$ and $B D$ meet at $E$ and the lines $A D$ and $B C$ meet at $F$. The midpoints of $A B$ and $C D$ are $G$ and $H$, respectively. Show that $E F$ is tangent at $E$ to the circle through the points $E, G$, and $H$.

Solution: It suffices to show that $\angle H E F=\angle H G E$ (see Figure 1), since in circle $E G H$ the angle over the chord $E H$ at $G$ equals the angle between the tangent at $E$ and $E H$.  First, $\angle B A D=180^{\circ}-\angle D C B=\angle F C D$. Since triangles $F A B$ and $F C D$ have also a common interior angle at $F$, they are similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-56.jpg?height=592&width=1069&top_left_y=932&top_left_x=499)  Figure 1 Denote by $\mathcal{T}$ the transformation consisting of a reflection at the bisector of $\angle D F C$ followed by a dilation with center $F$ and factor of $\frac{F A}{F C}$. Then $\mathcal{T}$ maps $F$ to $F, C$ to $A, D$ to $B$, and $H$ to $G$. To see this, note that $\triangle F C A \sim \triangle F D B$, so $\frac{F A}{F C}=\frac{F B}{F D}$. Moreover, as $\angle A D B=\angle A C B$, the image of the line $D E$ under $\mathcal{T}$ is parallel to $A C$ (and passes through $B$ ) and similarly the image of $C E$ is parallel to $D B$ and passes through $A$. Hence $E$ is mapped to the point $X$ which is the fourth vertex of the parallelogram $B E A X$. Thus, in particular $\angle H E F=\angle F X G$. As $G$ is the midpoint of the diagonal $A B$ of the parallelogram $B E A X$, it is also the midpoint of $E X$. In particular, $E, G, X$ are collinear, and $E X=2 \cdot E G$. Denote by $Y$ the fourth vertex of the parallelogram $D E C Y$. By an analogous reasoning as before, it follows that $\mathcal{T}$ maps $Y$ to $E$, thus $E, H, Y$ are collinear with $E Y=2 \cdot E H$. Therefore, by the intercept theorem, $H G \| X Y$. From the construction of $\mathcal{T}$ it is clear that the lines $F X$ and $F E$ are symmetric with respect to the bisector of $\angle D F C$, as are $F Y$ and $F E$. Thus, $F, X, Y$ are collinear, which together with $H G \| X Y$ implies $\angle F X E=\angle H G E$. This completes the proof.


Solution: We use the following Lemma (Gauß). Let $A B C D$ be a quadrilateral. Let $A B$ and $C D$ intersect at $P$, and $B C$ and $D A$ intersect at $Q$. Then the midpoints $K, L, M$ of $A C, B D$, and $P Q$, respectively, are collinear.  Proof: Let us consider the points $Z$ that fulfill the equation  $$ (A B Z)+(C D Z)=(B C Z)+(D A Z) $$  where $(R S T)$ denotes the oriented area of the triangle $R S T$ (see Figure 2). ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-57.jpg?height=809&width=692&top_left_y=681&top_left_x=688)  Figure 2 As (1) is linear in $Z$, it can either characterize a line, or be contradictory, or be trivially fulfilled for all $Z$ in the plane. If (1) was fulfilled for all $Z$, then it would hold for $Z=A, Z=B$, which gives $(C D A)=(B C A),(C D B)=(D A B)$, respectively, i.e. the diagonals of $A B C D$ would bisect each other, thus $A B C D$ would be a parallelogram. This contradicts the hypothesis that $A D$ and $B C$ intersect. Since $E, F, G$ fulfill (1), it is the equation of a line which completes the proof of the lemma. Now consider the parallelograms $E A X B$ and $E C Y D$ (see Figure 1). Then $G, H$ are the midpoints of $E X, E Y$, respectively. Let $M$ be the midpoint of $E F$. By applying the Lemma to the (re-entrant) quadrilateral $A D B C$, it is evident that $G, H$, and $M$ are collinear. A dilation by a factor of 2 with center $E$ shows that $X, Y, F$ are collinear. Since $A X \| D E$ and $B X \| C E$, we have pairwise equal interior angles in the quadrilaterals $F D E C$ and $F B X A$. Since we have also $\angle E B A=\angle D C A=\angle C D Y$, the quadrilaterals are similar. Thus, $\angle F X A=\angle C E F$. Clearly the parallelograms $E C Y D$ and $E B X A$ are similar, too, thus $\angle E X A=\angle C E Y$. Consequently, $\angle F X E=\angle F X A-\angle E X A=\angle C E F-\angle C E Y=\angle Y E F$. By the converse of the tangent-chord angle theorem $E F$ is tangent to the circle $X E Y$. A dilation by a factor of $\frac{1}{2}$ completes the proof.


Solution: As in   $$ \left(\frac{\boldsymbol{e}-\boldsymbol{f}}{2}\right)^{2}=\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{a}+\boldsymbol{b}}{2}\right)\left(\frac{\boldsymbol{e}+\boldsymbol{f}}{2}-\frac{\boldsymbol{c}+\boldsymbol{d}}{2}\right) $$  or equivalently  $$ 4 \boldsymbol{e f}-(\boldsymbol{e}+\boldsymbol{f})(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}+\boldsymbol{d})+(\boldsymbol{a}+\boldsymbol{b})(\boldsymbol{c}+\boldsymbol{d})=0 $$  With $R$ as the circumradius of $A B C D$, we obtain for the powers $\mathcal{P}(E)$ and $\mathcal{P}(F)$ of $E$ and $F$, respectively, with respect to the circumcircle  $$ \begin{aligned} & \mathcal{P}(E)=(\boldsymbol{e}-\boldsymbol{a})(\boldsymbol{e}-\boldsymbol{c})=(\boldsymbol{e}-\boldsymbol{b})(\boldsymbol{e}-\boldsymbol{d})=\boldsymbol{e}^{2}-R^{2} \\ & \mathcal{P}(F)=(\boldsymbol{f}-\boldsymbol{a})(\boldsymbol{f}-\boldsymbol{d})=(\boldsymbol{f}-\boldsymbol{b})(\boldsymbol{f}-\boldsymbol{c})=\boldsymbol{f}^{2}-R^{2} \end{aligned} $$  hence  $$ \begin{aligned} & (\boldsymbol{e}-\boldsymbol{a})(\boldsymbol{e}-\boldsymbol{c})=\boldsymbol{e}^{2}-R^{2} \\ & (\boldsymbol{e}-\boldsymbol{b})(\boldsymbol{e}-\boldsymbol{d})=\boldsymbol{e}^{2}-R^{2} \\ & (\boldsymbol{f}-\boldsymbol{a})(\boldsymbol{f}-\boldsymbol{d})=\boldsymbol{f}^{2}-R^{2} \\ & (\boldsymbol{f}-\boldsymbol{b})(\boldsymbol{f}-\boldsymbol{c})=\boldsymbol{f}^{2}-R^{2} \end{aligned} $$  Since $F$ lies on the polar to $E$ with respect to the circumcircle, we have  $$ 4 \boldsymbol{e} \boldsymbol{f}=4 R^{2} $$  Adding up (3) to (7) yields (2), as desired.


POL   Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \subset R$ we have  $$ \frac{|R|}{|P|} \leq \sqrt{2} $$  where $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively.

Solution: We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $\left|R_{1}\right| \leq \sqrt{2}|P|$ and $\left|R_{3}\right| \leq \sqrt{2}|P|$ holds (see Figure 1). First we will construct a parallelogram $R_{1} \supseteq P$ with the property that the midpoints of the sides of $R_{1}$ are points of the boundary of $P$. Choose two points $A$ and $B$ of $P$ such that the triangle $O A B$ has maximal area. Let $a$ be the line through $A$ parallel to $O B$ and $b$ the line through $B$ parallel to $O A$. Let $A^{\prime}, B^{\prime}, a^{\prime}$ and $b^{\prime}$ be the points or lines, that are symmetric to $A, B, a$ and $b$, respectively, with respect to $O$. Now let $R_{1}$ be the parallelogram defined by $a, b, a^{\prime}$ and $b^{\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-59.jpg?height=652&width=1104&top_left_y=1319&top_left_x=476)  Figure 1 Obviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A, B, A^{\prime}$ and $B^{\prime}$ are midpoints of the sides of $R_{1}$. We note that $P \subseteq R_{1}$. Otherwise, there would be a point $Z \in P$ but $Z \notin R_{1}$, i.e., one of the lines $a, b, a^{\prime}$ or $b^{\prime}$ were between $O$ and $Z$. If it is $a$, we have $|O Z B|>|O A B|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b, a^{\prime}$ or $b^{\prime}$ almost identical arguments lead to a similar contradiction. Let $R_{2}$ be the parallelogram $A B A^{\prime} B^{\prime}$. Since $A$ and $B$ are points of $P$, segment $A B \subset P$ and so $R_{2} \subset R_{1}$. Since $A, B, A^{\prime}$ and $B^{\prime}$ are midpoints of the sides of $R_{1}$, an easy argument yields  $$ \left|R_{1}\right|=2 \cdot\left|R_{2}\right| $$  Let $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $A B$ and $B A^{\prime}$. Obviously $R_{2} \subset R_{3}$ and every side of $R_{3}$ contains at least one point of the boundary of $P$. Denote by $C$ the intersection point of $a$ and $b$, by $X$ the intersection point of $A B$ and $O C$, and by $X^{\prime}$ the intersection point of $X C$ and the boundary of $R_{3}$. In a similar way denote by $D$ the intersection point of $b$ and $a^{\prime}$, by $Y$ the intersection point of $A^{\prime} B$ and $O D$, and by $Y^{\prime}$ the intersection point of $Y D$ and the boundary of $R_{3}$. Note that $O C=2 \cdot O X$ and $O D=2 \cdot O Y$, so there exist real numbers $x$ and $y$ with $1 \leq x, y \leq 2$ and $O X^{\prime}=x \cdot O X$ and $O Y^{\prime}=y \cdot O Y$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields  $$ \left|R_{3}\right|=x y \cdot\left|R_{2}\right| $$  The side of $R_{3}$ containing $X^{\prime}$ contains at least one point $X^{*}$ of $P$; due to the convexity of $P$ we have $A X^{*} B \subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $A B$ we have $\left|A X^{*} B\right|=\left|A X^{\prime} B\right|$, so $\left|O A X^{\prime} B\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A$. In a similar way we conclude that $\left|O B^{\prime} Y^{\prime} A^{\prime}\right|$ does not exceed the area of $P$ confined to the sector defined by the rays $O B$ and $O A^{\prime}$. Putting things together we have $\left|O A X^{\prime} B\right|=x \cdot|O A B|,\left|O B D A^{\prime}\right|=y \cdot\left|O B A^{\prime}\right|$. Since $|O A B|=\left|O B A^{\prime}\right|$, we conclude that $|P| \geq 2 \cdot\left|A X^{\prime} B Y^{\prime} A^{\prime}\right|=2 \cdot\left(x \cdot|O A B|+y \cdot\left|O B A^{\prime}\right|\right)=4 \cdot \frac{x+y}{2} \cdot|O A B|=\frac{x+y}{2} \cdot R_{2}$; this is in short  $$ \frac{x+y}{2} \cdot\left|R_{2}\right| \leq|P| $$  Since all numbers concerned are positive, we can combine (1)-(3). Using the arithmetic-geometric-mean inequality we obtain  $$ \left|R_{1}\right| \cdot\left|R_{3}\right|=2 \cdot\left|R_{2}\right| \cdot x y \cdot\left|R_{2}\right| \leq 2 \cdot\left|R_{2}\right|^{2}\left(\frac{x+y}{2}\right)^{2} \leq 2 \cdot|P|^{2} $$  This implies immediately the desired result $\left|R_{1}\right| \leq \sqrt{2} \cdot|P|$ or $\left|R_{3}\right| \leq \sqrt{2} \cdot|P|$.


Solution: We construct the parallelograms $R_{1}, R_{2}$ and $R_{3}$ in the same way as in  ![](https://cdn.mathpix.com/cropped/2024_11_18_b191c9407e8cb3d672cbg-60.jpg?height=569&width=580&top_left_y=1720&top_left_x=738)  Figure 2 Recall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of generality we may assume that $R_{1}$ is a square (see Figure 2). Then $R_{2}$, whose vertices are the midpoints of the sides of $R_{1}$, is a square too, and $R_{3}$, whose sides are parallel to the diagonals of $R_{1}$, is a rectangle. Let $a>0, b \geq 0$ and $c \geq 0$ be the distances introduced in Figure 2. Then $\left|R_{1}\right|=2 a^{2}$ and $\left|R_{3}\right|=(a+2 b)(a+2 c)$. Points $A, A^{\prime}, B$ and $B^{\prime}$ are in the convex polygon $P$. Hence the square $A B A^{\prime} B^{\prime}$ is a subset of $P$. Moreover, each of the sides of the rectangle $R_{3}$ contains a point of $P$, otherwise $R_{3}$ would not be minimal. It follows that  $$ |P| \geq a^{2}+2 \cdot \frac{a b}{2}+2 \cdot \frac{a c}{2}=a(a+b+c) $$  Now assume that both $\frac{\left|R_{1}\right|}{|P|}>\sqrt{2}$ and $\frac{\left|R_{3}\right|}{|P|}>\sqrt{2}$, then  $$ 2 a^{2}=\left|R_{1}\right|>\sqrt{2} \cdot|P| \geq \sqrt{2} \cdot a(a+b+c) $$  and  $$ (a+2 b)(a+2 c)=\left|R_{3}\right|>\sqrt{2} \cdot|P| \geq \sqrt{2} \cdot a(a+b+c) $$  All numbers concerned are positive, so after multiplying these inequalities we get  $$ 2 a^{2}(a+2 b)(a+2 c)>2 a^{2}(a+b+c)^{2} $$  But the arithmetic-geometric-mean inequality implies the contradictory result  $$ 2 a^{2}(a+2 b)(a+2 c) \leq 2 a^{2}\left(\frac{(a+2 b)+(a+2 c)}{2}\right)^{2}=2 a^{2}(a+b+c)^{2} $$  Hence $\frac{\left|R_{1}\right|}{|P|} \leq \sqrt{2}$ or $\frac{\left|R_{3}\right|}{|P|} \leq \sqrt{2}$, as desired.