Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that  $$ \frac{f(p)^{2}+f(q)^{2}}{f\left(r^{2}\right)+f\left(s^{2}\right)}=\frac{p^{2}+q^{2}}{r^{2}+s^{2}} $$  for all $p, q, r, s>0$ with $p q=r s$.

Solution: Let $f$ satisfy the given condition. Setting $p=q=r=s=1$ yields $f(1)^{2}=f(1)$ and hence $f(1)=1$. Now take any $x>0$ and set $p=x, q=1, r=s=\sqrt{x}$ to obtain  $$ \frac{f(x)^{2}+1}{2 f(x)}=\frac{x^{2}+1}{2 x} . $$  This recasts into  $$ \begin{gathered} x f(x)^{2}+x=x^{2} f(x)+f(x), \\ (x f(x)-1)(f(x)-x)=0 . \end{gathered} $$  And thus,  $$ \text { for every } x>0, \text { either } f(x)=x \text { or } f(x)=\frac{1}{x} \text {. } $$  Obviously, if  $$ f(x)=x \quad \text { for all } x>0 \quad \text { or } \quad f(x)=\frac{1}{x} \quad \text { for all } x>0 $$  then the condition of the problem is satisfied. We show that actually these two functions are the only solutions.  So let us assume that there exists a function $f$ satisfying the requirement, other than those in (2). Then $f(a) \neq a$ and $f(b) \neq 1 / b$ for some $a, b>0$. By (1), these values must be $f(a)=1 / a, f(b)=b$. Applying now the equation with $p=a, q=b, r=s=\sqrt{a b}$ we obtain $\left(a^{-2}+b^{2}\right) / 2 f(a b)=\left(a^{2}+b^{2}\right) / 2 a b ;$ equivalently,  $$ f(a b)=\frac{a b\left(a^{-2}+b^{2}\right)}{a^{2}+b^{2}} . $$  We know however (see (1)) that $f(a b)$ must be either $a b$ or $1 / a b$. If $f(a b)=a b$ then by (3) $a^{-2}+b^{2}=a^{2}+b^{2}$, so that $a=1$. But, as $f(1)=1$, this contradicts the relation $f(a) \neq a$. Likewise, if $f(a b)=1 / a b$ then (3) gives $a^{2} b^{2}\left(a^{-2}+b^{2}\right)=a^{2}+b^{2}$, whence $b=1$, in contradiction to $f(b) \neq 1 / b$. Thus indeed the functions listed in (2) are the only two solutions.  Comment. The equation has as many as four variables with only one constraint $p q=r s$, leaving three degrees of freedom and providing a lot of information. Various substitutions force various useful properties of the function searched. We sketch one more method to reach conclusion (1); certainly there are many others.  Noticing that $f(1)=1$ and setting, first, $p=q=1, r=\sqrt{x}, s=1 / \sqrt{x}$, and then $p=x, q=1 / x$, $r=s=1$, we obtain two relations, holding for every $x>0$,  $$ f(x)+f\left(\frac{1}{x}\right)=x+\frac{1}{x} \quad \text { and } \quad f(x)^{2}+f\left(\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}} . $$  Squaring the first and subtracting the second gives $2 f(x) f(1 / x)=2$. Subtracting this from the second relation of (4) leads to  $$ \left(f(x)-f\left(\frac{1}{x}\right)\right)^{2}=\left(x-\frac{1}{x}\right)^{2} \quad \text { or } \quad f(x)-f\left(\frac{1}{x}\right)= \pm\left(x-\frac{1}{x}\right) . $$  The last two alternatives combined with the first equation of (4) imply the two alternatives of (1).


(a) Prove the inequality  $$ \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geq 1 $$  for real numbers $x, y, z \neq 1$ satisfying the condition $x y z=1$.  (b) Show that there are infinitely many triples of rational numbers $x, y, z$ for which this inequality turns into equality.

Solution: (a) We start with the substitution  $$ \frac{x}{x-1}=a, \quad \frac{y}{y-1}=b, \quad \frac{z}{z-1}=c, \quad \text { i.e., } \quad x=\frac{a}{a-1}, \quad y=\frac{b}{b-1}, \quad z=\frac{c}{c-1} \text {. } $$  The inequality to be proved reads $a^{2}+b^{2}+c^{2} \geq 1$. The new variables are subject to the constraints $a, b, c \neq 1$ and the following one coming from the condition $x y z=1$,  $$ (a-1)(b-1)(c-1)=a b c . $$  This is successively equivalent to  $$ \begin{aligned} a+b+c-1 & =a b+b c+c a, \\ 2(a+b+c-1) & =(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right), \\ a^{2}+b^{2}+c^{2}-2 & =(a+b+c)^{2}-2(a+b+c), \\ a^{2}+b^{2}+c^{2}-1 & =(a+b+c-1)^{2} . \end{aligned} $$  Thus indeed $a^{2}+b^{2}+c^{2} \geq 1$, as desired.  (b) From the equation $a^{2}+b^{2}+c^{2}-1=(a+b+c-1)^{2}$ we see that the proposed inequality becomes an equality if and only if both sums $a^{2}+b^{2}+c^{2}$ and $a+b+c$ have value 1 . The first of them is equal to $(a+b+c)^{2}-2(a b+b c+c a)$. So the instances of equality are described by the system of two equations  $$ a+b+c=1, \quad a b+b c+c a=0 $$  plus the constraint $a, b, c \neq 1$. Elimination of $c$ leads to $a^{2}+a b+b^{2}=a+b$, which we regard as a quadratic equation in $b$,  $$ b^{2}+(a-1) b+a(a-1)=0, $$  with discriminant  $$ \Delta=(a-1)^{2}-4 a(a-1)=(1-a)(1+3 a) . $$  We are looking for rational triples $(a, b, c)$; it will suffice to have $a$ rational such that $1-a$ and $1+3 a$ are both squares of rational numbers (then $\Delta$ will be so too). Set $a=k / m$. We want $m-k$ and $m+3 k$ to be squares of integers. This is achieved for instance by taking $m=k^{2}-k+1$ (clearly nonzero); then $m-k=(k-1)^{2}, m+3 k=(k+1)^{2}$. Note that distinct integers $k$ yield distinct values of $a=k / m$.  And thus, if $k$ is any integer and $m=k^{2}-k+1, a=k / m$ then $\Delta=\left(k^{2}-1\right)^{2} / m^{2}$ and the quadratic equation has rational roots $b=\left(m-k \pm k^{2} \mp 1\right) /(2 m)$. Choose e.g. the larger root,  $$ b=\frac{m-k+k^{2}-1}{2 m}=\frac{m+(m-2)}{2 m}=\frac{m-1}{m} . $$  Computing $c$ from $a+b+c=1$ then gives $c=(1-k) / m$. The condition $a, b, c \neq 1$ eliminates only $k=0$ and $k=1$. Thus, as $k$ varies over integers greater than 1 , we obtain an infinite family of rational triples $(a, b, c)$ - and coming back to the original variables $(x=a /(a-1)$ etc. $)$-an infinite family of rational triples $(x, y, z)$ with the needed property. (A short calculation shows that the resulting triples are $x=-k /(k-1)^{2}, y=k-k^{2}, z=(k-1) / k^{2}$; but the proof was complete without listing them.)  Comment 1. There are many possible variations in handling the equation system $a^{2}+b^{2}+c^{2}=1$, $a+b+c=1(a, b, c \neq 1)$ which of course describes a circle in the $(a, b, c)$-space (with three points excluded), and finding infinitely many rational points on it.  Also the initial substitution $x=a /(a-1)$ (etc.) can be successfully replaced by other similar substitutions, e.g. $x=1-1 / \alpha$ (etc.); or $x=x^{\prime}-1$ (etc.); or $1-y z=u$ (etc.) - eventually reducing the inequality to $(\cdots)^{2} \geq 0$, the expression in the parentheses depending on the actual substitution.  Depending on the method chosen, one arrives at various sequences of rational triples $(x, y, z)$ as needed; let us produce just one more such example: $x=(2 r-2) /(r+1)^{2}, y=(2 r+2) /(r-1)^{2}$, $z=\left(r^{2}-1\right) / 4$ where $r$ can be any rational number different from 1 or -1 .  Solution 2 (an outline). (a) Without changing variables, just setting $z=1 / x y$ and clearing fractions, the proposed inequality takes the form  $$ (x y-1)^{2}\left(x^{2}(y-1)^{2}+y^{2}(x-1)^{2}\right)+(x-1)^{2}(y-1)^{2} \geq(x-1)^{2}(y-1)^{2}(x y-1)^{2} . $$  With the notation $p=x+y, q=x y$ this becomes, after lengthy routine manipulation and a lot of cancellation  $$ q^{4}-6 q^{3}+2 p q^{2}+9 q^{2}-6 p q+p^{2} \geq 0 \text {. } $$  It is not hard to notice that the expression on the left is just $\left(q^{2}-3 q+p\right)^{2}$, hence nonnegative.  (Without introducing $p$ and $q$, one is of course led with some more work to the same expression, just written in terms of $x$ and $y$; but then it is not that easy to see that it is a square.)  (b) To have equality, one needs $q^{2}-3 q+p=0$. Note that $x$ and $y$ are the roots of the quadratic trinomial (in a formal variable $t$ ): $t^{2}-p t+q$. When $q^{2}-3 q+p=0$, the discriminant equals  $$ \delta=p^{2}-4 q=\left(3 q-q^{2}\right)^{2}-4 q=q(q-1)^{2}(q-4) . $$  Now it suffices to have both $q$ and $q-4$ squares of rational numbers (then $p=3 q-q^{2}$ and $\sqrt{\delta}$ are also rational, and so are the roots of the trinomial). On setting $q=(n / m)^{2}=4+(l / m)^{2}$ the requirement becomes $4 m^{2}+l^{2}=n^{2}$ (with $l, m, n$ being integers). This is just the Pythagorean equation, known to have infinitely many integer solutions.  Comment 2. Part (a) alone might also be considered as a possible contest problem (in the category of easy problems).


Let $S \subseteq \mathbb{R}$ be a set of real numbers. We say that a pair $(f, g)$ of functions from $S$ into $S$ is a Spanish Couple on $S$, if they satisfy the following conditions:  (i) Both functions are strictly increasing, i.e. $f(x)<f(y)$ and $g(x)<g(y)$ for all $x, y \in S$ with $x<y$;  (ii) The inequality $f(g(g(x)))<g(f(x))$ holds for all $x \in S$.  Decide whether there exists a Spanish Couple  (a) on the set $S=\mathbb{N}$ of positive integers;  (b) on the set $S=\{a-1 / b: a, b \in \mathbb{N}\}$.

Solution: We show that the answer is NO for part (a), and YES for part (b).  (a) Throughout the solution, we will use the notation $g_{k}(x)=\overbrace{g(g(\ldots g}^{k}(x) \ldots))$, including $g_{0}(x)=x$ as well.  Suppose that there exists a Spanish Couple $(f, g)$ on the set $\mathbb{N}$. From property (i) we have $f(x) \geq x$ and $g(x) \geq x$ for all $x \in \mathbb{N}$.  We claim that $g_{k}(x) \leq f(x)$ for all $k \geq 0$ and all positive integers $x$. The proof is done by induction on $k$. We already have the base case $k=0$ since $x \leq f(x)$. For the induction step from $k$ to $k+1$, apply the induction hypothesis on $g_{2}(x)$ instead of $x$, then apply (ii):  $$ g\left(g_{k+1}(x)\right)=g_{k}\left(g_{2}(x)\right) \leq f\left(g_{2}(x)\right)<g(f(x)) $$  Since $g$ is increasing, it follows that $g_{k+1}(x)<f(x)$. The claim is proven.  If $g(x)=x$ for all $x \in \mathbb{N}$ then $f(g(g(x)))=f(x)=g(f(x))$, and we have a contradiction with (ii). Therefore one can choose an $x_{0} \in S$ for which $x_{0}<g\left(x_{0}\right)$. Now consider the sequence $x_{0}, x_{1}, \ldots$ where $x_{k}=g_{k}\left(x_{0}\right)$. The sequence is increasing. Indeed, we have $x_{0}<g\left(x_{0}\right)=x_{1}$, and $x_{k}<x_{k+1}$ implies $x_{k+1}=g\left(x_{k}\right)<g\left(x_{k+1}\right)=x_{k+2}$.  Hence, we obtain a strictly increasing sequence $x_{0}<x_{1}<\ldots$ of positive integers which on the other hand has an upper bound, namely $f\left(x_{0}\right)$. This cannot happen in the set $\mathbb{N}$ of positive integers, thus no Spanish Couple exists on $\mathbb{N}$.  (b) We present a Spanish Couple on the set $S=\{a-1 / b: a, b \in \mathbb{N}\}$.  Let  $$ \begin{aligned} & f(a-1 / b)=a+1-1 / b, \\ & g(a-1 / b)=a-1 /\left(b+3^{a}\right) . \end{aligned} $$  These functions are clearly increasing. Condition (ii) holds, since  $$ f(g(g(a-1 / b)))=(a+1)-1 /\left(b+2 \cdot 3^{a}\right)<(a+1)-1 /\left(b+3^{a+1}\right)=g(f(a-1 / b)) . $$  Comment. Another example of a Spanish couple is $f(a-1 / b)=3 a-1 / b, g(a-1 / b)=a-1 /(a+b)$. More generally, postulating $f(a-1 / b)=h(a)-1 / b, \quad g(a-1 / b)=a-1 / G(a, b)$ with $h$ increasing and $G$ increasing in both variables, we get that $f \circ g \circ g<g \circ f$ holds if $G(a, G(a, b))<G(h(a), b)$. A search just among linear functions $h(a)=C a, G(a, b)=A a+B b$ results in finding that any integers $A>0, C>2$ and $B=1$ produce a Spanish couple (in the example above, $A=1, C=3$ ). The proposer's example results from taking $h(a)=a+1, G(a, b)=3^{a}+b$.


For an integer $m$, denote by $t(m)$ the unique number in $\{1,2,3\}$ such that $m+t(m)$ is a multiple of 3. A function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfies $f(-1)=0, f(0)=1, f(1)=-1$ and  $$ f\left(2^{n}+m\right)=f\left(2^{n}-t(m)\right)-f(m) \quad \text { for all integers } m, n \geq 0 \text { with } 2^{n}>m \text {. } $$  Prove that $f(3 p) \geq 0$ holds for all integers $p \geq 0$.

Solution: The given conditions determine $f$ uniquely on the positive integers. The signs of $f(1), f(2), \ldots$ seem to change quite erratically. However values of the form $f\left(2^{n}-t(m)\right)$ are sufficient to compute directly any functional value. Indeed, let $n>0$ have base 2 representation $n=2^{a_{0}}+2^{a_{1}}+\cdots+2^{a_{k}}, a_{0}>a_{1}>\cdots>a_{k} \geq 0$, and let $n_{j}=2^{a_{j}}+2^{a_{j-1}}+\cdots+2^{a_{k}}, j=0, \ldots, k$. Repeated applications of the recurrence show that $f(n)$ is an alternating sum of the quantities $f\left(2^{a_{j}}-t\left(n_{j+1}\right)\right)$ plus $(-1)^{k+1}$. (The exact formula is not needed for our proof.)  So we focus attention on the values $f\left(2^{n}-1\right), f\left(2^{n}-2\right)$ and $f\left(2^{n}-3\right)$. Six cases arise; more specifically,  $t\left(2^{2 k}-3\right)=2, t\left(2^{2 k}-2\right)=1, t\left(2^{2 k}-1\right)=3, t\left(2^{2 k+1}-3\right)=1, t\left(2^{2 k+1}-2\right)=3, t\left(2^{2 k+1}-1\right)=2$.  Claim. For all integers $k \geq 0$ the following equalities hold:  $$ \begin{array}{lll} f\left(2^{2 k+1}-3\right)=0, & f\left(2^{2 k+1}-2\right)=3^{k}, & f\left(2^{2 k+1}-1\right)=-3^{k}, \\ f\left(2^{2 k+2}-3\right)=-3^{k}, & f\left(2^{2 k+2}-2\right)=-3^{k}, & f\left(2^{2 k+2}-1\right)=2 \cdot 3^{k} . \end{array} $$  Proof. By induction on $k$. The base $k=0$ comes down to checking that $f(2)=-1$ and $f(3)=2$; the given values $f(-1)=0, f(0)=1, f(1)=-1$ are also needed. Suppose the claim holds for $k-1$. For $f\left(2^{2 k+1}-t(m)\right)$, the recurrence formula and the induction hypothesis yield  $$ \begin{aligned} & f\left(2^{2 k+1}-3\right)=f\left(2^{2 k}+\left(2^{2 k}-3\right)\right)=f\left(2^{2 k}-2\right)-f\left(2^{2 k}-3\right)=-3^{k-1}+3^{k-1}=0, \\ & f\left(2^{2 k+1}-2\right)=f\left(2^{2 k}+\left(2^{2 k}-2\right)\right)=f\left(2^{2 k}-1\right)-f\left(2^{2 k}-2\right)=2 \cdot 3^{k-1}+3^{k-1}=3^{k}, \\ & f\left(2^{2 k+1}-1\right)=f\left(2^{2 k}+\left(2^{2 k}-1\right)\right)=f\left(2^{2 k}-3\right)-f\left(2^{2 k}-1\right)=-3^{k-1}-2 \cdot 3^{k-1}=-3^{k} . \end{aligned} $$  For $f\left(2^{2 k+2}-t(m)\right)$ we use the three equalities just established:  $$ \begin{aligned} & f\left(2^{2 k+2}-3\right)=f\left(2^{2 k+1}+\left(2^{2 k+1}-3\right)\right)=f\left(2^{2 k+1}-1\right)-f\left(2^{2 k+1}-3\right)=-3^{k}-0=-3^{k}, \\ & f\left(2^{2 k+2}-2\right)=f\left(2^{2 k+1}+\left(2^{2 k+1}-2\right)\right)=f\left(2^{2 k+1}-3\right)-f\left(2^{2 k}-2\right)=0-3^{k}=-3^{k}, \\ & f\left(2^{2 k+2}-1\right)=f\left(2^{2 k+1}+\left(2^{2 k+1}-1\right)\right)=f\left(2^{2 k+1}-2\right)-f\left(2^{2 k+1}-1\right)=3^{k}+3^{k}=2 \cdot 3^{k} . \end{aligned} $$  The claim follows.  A closer look at the six cases shows that $f\left(2^{n}-t(m)\right) \geq 3^{(n-1) / 2}$ if $2^{n}-t(m)$ is divisible by 3 , and $f\left(2^{n}-t(m)\right) \leq 0$ otherwise. On the other hand, note that $2^{n}-t(m)$ is divisible by 3 if and only if $2^{n}+m$ is. Therefore, for all nonnegative integers $m$ and $n$,  (i) $f\left(2^{n}-t(m)\right) \geq 3^{(n-1) / 2}$ if $2^{n}+m$ is divisible by 3 ;  (ii) $f\left(2^{n}-t(m)\right) \leq 0$ if $2^{n}+m$ is not divisible by 3 .  One more (direct) consequence of the claim is that $\left|f\left(2^{n}-t(m)\right)\right| \leq \frac{2}{3} \cdot 3^{n / 2}$ for all $m, n \geq 0$.  The last inequality enables us to find an upper bound for $|f(m)|$ for $m$ less than a given power of 2 . We prove by induction on $n$ that $|f(m)| \leq 3^{n / 2}$ holds true for all integers $m, n \geq 0$ with $2^{n}>m$.  The base $n=0$ is clear as $f(0)=1$. For the inductive step from $n$ to $n+1$, let $m$ and $n$ satisfy $2^{n+1}>m$. If $m<2^{n}$, we are done by the inductive hypothesis. If $m \geq 2^{n}$ then $m=2^{n}+k$ where $2^{n}>k \geq 0$. Now, by $\left|f\left(2^{n}-t(k)\right)\right| \leq \frac{2}{3} \cdot 3^{n / 2}$ and the inductive assumption,  $$ |f(m)|=\left|f\left(2^{n}-t(k)\right)-f(k)\right| \leq\left|f\left(2^{n}-t(k)\right)\right|+|f(k)| \leq \frac{2}{3} \cdot 3^{n / 2}+3^{n / 2}<3^{(n+1) / 2} . $$  The induction is complete.  We proceed to prove that $f(3 p) \geq 0$ for all integers $p \geq 0$. Since $3 p$ is not a power of 2 , its binary expansion contains at least two summands. Hence one can write $3 p=2^{a}+2^{b}+c$ where $a>b$ and $2^{b}>c \geq 0$. Applying the recurrence formula twice yields  $$ f(3 p)=f\left(2^{a}+2^{b}+c\right)=f\left(2^{a}-t\left(2^{b}+c\right)\right)-f\left(2^{b}-t(c)\right)+f(c) . $$  Since $2^{a}+2^{b}+c$ is divisible by 3 , we have $f\left(2^{a}-t\left(2^{b}+c\right)\right) \geq 3^{(a-1) / 2}$ by (i). Since $2^{b}+c$ is not divisible by 3 , we have $f\left(2^{b}-t(c)\right) \leq 0$ by (ii). Finally $|f(c)| \leq 3^{b / 2}$ as $2^{b}>c \geq 0$, so that $f(c) \geq-3^{b / 2}$. Therefore $f(3 p) \geq 3^{(a-1) / 2}-3^{b / 2}$ which is nonnegative because $a>b$.


Let $a, b, c, d$ be positive real numbers such that  $$ a b c d=1 \quad \text { and } \quad a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$  Prove that  $$ a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d} $$

Solution: We show that if $a b c d=1$, the sum $a+b+c+d$ cannot exceed a certain weighted mean of the expressions $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$.  By applying the AM-GM inequality to the numbers $\frac{a}{b}, \frac{a}{b}, \frac{b}{c}$ and $\frac{a}{d}$, we obtain  $$ a=\sqrt[4]{\frac{a^{4}}{a b c d}}=\sqrt[4]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{a}{d}} \leq \frac{1}{4}\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}+\frac{a}{d}\right) $$  Analogously,  $$ b \leq \frac{1}{4}\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{d}+\frac{b}{a}\right), \quad c \leq \frac{1}{4}\left(\frac{c}{d}+\frac{c}{d}+\frac{d}{a}+\frac{c}{b}\right) \quad \text { and } \quad d \leq \frac{1}{4}\left(\frac{d}{a}+\frac{d}{a}+\frac{a}{b}+\frac{d}{c}\right) . $$  Summing up these estimates yields  $$ a+b+c+d \leq \frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)+\frac{1}{4}\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right) . $$  In particular, if $a+b+c+d>\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$ then $a+b+c+d<\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}$.  Comment. The estimate in the above solution was obtained by applying the AM-GM inequality to each column of the $4 \times 4$ array  $$ \begin{array}{llll} a / b & b / c & c / d & d / a \\ a / b & b / c & c / d & d / a \\ b / c & c / d & d / a & a / b \\ a / d & b / a & c / b & d / c \end{array} $$  and adding up the resulting inequalities. The same table yields a stronger bound: If $a, b, c, d>0$ and $a b c d=1$ then  $$ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^{3}\left(\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\right) \geq(a+b+c+d)^{4} $$  It suffices to apply Hölder's inequality to the sequences in the four rows, with weights $1 / 4$ :  $$ \begin{gathered} \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^{1 / 4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^{1 / 4}\left(\frac{b}{c}+\frac{c}{d}+\frac{d}{a}+\frac{a}{b}\right)^{1 / 4}\left(\frac{a}{d}+\frac{b}{a}+\frac{c}{b}+\frac{d}{c}\right)^{1 / 4} \\ \geq\left(\frac{a a b a}{b b c d}\right)^{1 / 4}+\left(\frac{b b c b}{c c d a}\right)^{1 / 4}+\left(\frac{c c d c}{d d a b}\right)^{1 / 4}+\left(\frac{d d a d}{a a b c}\right)^{1 / 4}=a+b+c+d \end{gathered} $$


Let $f: \mathbb{R} \rightarrow \mathbb{N}$ be a function which satisfies  $$ f\left(x+\frac{1}{f(y)}\right)=f\left(y+\frac{1}{f(x)}\right) \quad \text { for all } x, y \in \mathbb{R} . $$  Prove that there is a positive integer which is not a value of $f$.

Solution: Suppose that the statement is false and $f(\mathbb{R})=\mathbb{N}$. We prove several properties of the function $f$ in order to reach a contradiction.  To start with, observe that one can assume $f(0)=1$. Indeed, let $a \in \mathbb{R}$ be such that $f(a)=1$, and consider the function $g(x)=f(x+a)$. By substituting $x+a$ and $y+a$ for $x$ and $y$ in (1), we have  $$ g\left(x+\frac{1}{g(y)}\right)=f\left(x+a+\frac{1}{f(y+a)}\right)=f\left(y+a+\frac{1}{f(x+a)}\right)=g\left(y+\frac{1}{g(x)}\right) . $$  So $g$ satisfies the functional equation (1), with the additional property $g(0)=1$. Also, $g$ and $f$ have the same set of values: $g(\mathbb{R})=f(\mathbb{R})=\mathbb{N}$. Henceforth we assume $f(0)=1$.  Claim 1. For an arbitrary fixed $c \in \mathbb{R}$ we have $\left\{f\left(c+\frac{1}{n}\right): n \in \mathbb{N}\right\}=\mathbb{N}$.  Proof. Equation (1) and $f(\mathbb{R})=\mathbb{N}$ imply  $f(\mathbb{R})=\left\{f\left(x+\frac{1}{f(c)}\right): x \in \mathbb{R}\right\}=\left\{f\left(c+\frac{1}{f(x)}\right): x \in \mathbb{R}\right\} \subset\left\{f\left(c+\frac{1}{n}\right): n \in \mathbb{N}\right\} \subset f(\mathbb{R})$.  The claim follows.  We will use Claim 1 in the special cases $c=0$ and $c=1 / 3$ :  $$ \left\{f\left(\frac{1}{n}\right): n \in \mathbb{N}\right\}=\left\{f\left(\frac{1}{3}+\frac{1}{n}\right): n \in \mathbb{N}\right\}=\mathbb{N} $$  Claim 2. If $f(u)=f(v)$ for some $u, v \in \mathbb{R}$ then $f(u+q)=f(v+q)$ for all nonnegative rational $q$. Furthermore, if $f(q)=1$ for some nonnegative rational $q$ then $f(k q)=1$ for all $k \in \mathbb{N}$.  Proof. For all $x \in \mathbb{R}$ we have by (1)  $$ f\left(u+\frac{1}{f(x)}\right)=f\left(x+\frac{1}{f(u)}\right)=f\left(x+\frac{1}{f(v)}\right)=f\left(v+\frac{1}{f(x)}\right) $$  Since $f(x)$ attains all positive integer values, this yields $f(u+1 / n)=f(v+1 / n)$ for all $n \in \mathbb{N}$. Let $q=k / n$ be a positive rational number. Then $k$ repetitions of the last step yield  $$ f(u+q)=f\left(u+\frac{k}{n}\right)=f\left(v+\frac{k}{n}\right)=f(v+q) $$  Now let $f(q)=1$ for some nonnegative rational $q$, and let $k \in \mathbb{N}$. As $f(0)=1$, the previous conclusion yields successively $f(q)=f(2 q), f(2 q)=f(3 q), \ldots, f((k-1) q)=f(k q)$, as needed.  Claim 3. The equality $f(q)=f(q+1)$ holds for all nonnegative rational $q$.  Proof. Let $m$ be a positive integer such that $f(1 / m)=1$. Such an $m$ exists by (2). Applying the second statement of Claim 2 with $q=1 / m$ and $k=m$ yields $f(1)=1$.  Given that $f(0)=f(1)=1$, the first statement of Claim 2 implies $f(q)=f(q+1)$ for all nonnegative rational $q$.  Claim 4. The equality $f\left(\frac{1}{n}\right)=n$ holds for every $n \in \mathbb{N}$.  Proof. For a nonnegative rational $q$ we set $x=q, y=0$ in (1) and use Claim 3 to obtain  $$ f\left(\frac{1}{f(q)}\right)=f\left(q+\frac{1}{f(0)}\right)=f(q+1)=f(q) . $$  By (2), for each $n \in \mathbb{N}$ there exists a $k \in \mathbb{N}$ such that $f(1 / k)=n$. Applying the last equation with $q=1 / k$, we have  $$ n=f\left(\frac{1}{k}\right)=f\left(\frac{1}{f(1 / k)}\right)=f\left(\frac{1}{n}\right) $$  Now we are ready to obtain a contradiction. Let $n \in \mathbb{N}$ be such that $f(1 / 3+1 / n)=1$. Such an $n$ exists by (2). Let $1 / 3+1 / n=s / t$, where $s, t \in \mathbb{N}$ are coprime. Observe that $t>1$ as $1 / 3+1 / n$ is not an integer. Choose $k, l \in \mathbb{N}$ so that that $k s-l t=1$.  Because $f(0)=f(s / t)=1$, Claim 2 implies $f(k s / t)=1$. Now $f(k s / t)=f(1 / t+l)$; on the other hand $f(1 / t+l)=f(1 / t)$ by $l$ successive applications of Claim 3 . Finally, $f(1 / t)=t$ by Claim 4, leading to the impossible $t=1$. The solution is complete.


Prove that for any four positive real numbers $a, b, c, d$ the inequality  $$ \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geq 0 $$  holds. Determine all cases of equality.

Solution: Denote the four terms by  $$ A=\frac{(a-b)(a-c)}{a+b+c}, \quad B=\frac{(b-c)(b-d)}{b+c+d}, \quad C=\frac{(c-d)(c-a)}{c+d+a}, \quad D=\frac{(d-a)(d-b)}{d+a+b} . $$  The expression $2 A$ splits into two summands as follows,  $$ 2 A=A^{\prime}+A^{\prime \prime} \quad \text { where } \quad A^{\prime}=\frac{(a-c)^{2}}{a+b+c}, \quad A^{\prime \prime}=\frac{(a-c)(a-2 b+c)}{a+b+c} ; $$  this is easily verified. We analogously represent $2 B=B^{\prime}+B^{\prime \prime}, 2 C=C^{\prime}+C^{\prime \prime}, 2 B=D^{\prime}+D^{\prime \prime}$ and examine each of the sums $A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}$ and $A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}$ separately.  Write $s=a+b+c+d$; the denominators become $s-d, s-a, s-b, s-c$. By the CauchySchwarz inequality,  $$ \begin{aligned} & \left(\frac{|a-c|}{\sqrt{s-d}} \cdot \sqrt{s-d}+\frac{|b-d|}{\sqrt{s-a}} \cdot \sqrt{s-a}+\frac{|c-a|}{\sqrt{s-b}} \cdot \sqrt{s-b}+\frac{|d-b|}{\sqrt{s-c}} \cdot \sqrt{s-c}\right)^{2} \\ & \quad \leq\left(\frac{(a-c)^{2}}{s-d}+\frac{(b-d)^{2}}{s-a}+\frac{(c-a)^{2}}{s-b}+\frac{(d-b)^{2}}{s-c}\right)(4 s-s)=3 s\left(A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}\right) . \end{aligned} $$  Hence  $$ A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime} \geq \frac{(2|a-c|+2|b-d|)^{2}}{3 s} \geq \frac{16 \cdot|a-c| \cdot|b-d|}{3 s} . $$  Next we estimate the absolute value of the other sum. We couple $A^{\prime \prime}$ with $C^{\prime \prime}$ to obtain  $$ \begin{aligned} A^{\prime \prime}+C^{\prime \prime} & =\frac{(a-c)(a+c-2 b)}{s-d}+\frac{(c-a)(c+a-2 d)}{s-b} \\ & =\frac{(a-c)(a+c-2 b)(s-b)+(c-a)(c+a-2 d)(s-d)}{(s-d)(s-b)} \\ & =\frac{(a-c)(-2 b(s-b)-b(a+c)+2 d(s-d)+d(a+c))}{s(a+c)+b d} \\ & =\frac{3(a-c)(d-b)(a+c)}{M}, \quad \text { with } \quad M=s(a+c)+b d . \end{aligned} $$  Hence by cyclic shift  $$ B^{\prime \prime}+D^{\prime \prime}=\frac{3(b-d)(a-c)(b+d)}{N}, \quad \text { with } \quad N=s(b+d)+c a $$  Thus  $$ A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}=3(a-c)(b-d)\left(\frac{b+d}{N}-\frac{a+c}{M}\right)=\frac{3(a-c)(b-d) W}{M N} $$  where  $$ W=(b+d) M-(a+c) N=b d(b+d)-a c(a+c) . $$  Note that  $$ M N>(a c(a+c)+b d(b+d)) s \geq|W| \cdot s . $$  Now (2) and (4) yield  $$ \left|A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}\right| \leq \frac{3 \cdot|a-c| \cdot|b-d|}{s} $$  Combined with (1) this results in  $$ \begin{aligned} 2(A+B & +C+D)=\left(A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}\right)+\left(A^{\prime \prime}+B^{\prime \prime}+C^{\prime \prime}+D^{\prime \prime}\right) \\ & \geq \frac{16 \cdot|a-c| \cdot|b-d|}{3 s}-\frac{3 \cdot|a-c| \cdot|b-d|}{s}=\frac{7 \cdot|a-c| \cdot|b-d|}{3(a+b+c+d)} \geq 0 \end{aligned} $$  This is the required inequality. From the last line we see that equality can be achieved only if either $a=c$ or $b=d$. Since we also need equality in (1), this implies that actually $a=c$ and $b=d$ must hold simultaneously, which is obviously also a sufficient condition.


Solution: We keep the notations $A, B, C, D, s$, and also $M, N, W$ from the preceding solution; the definitions of $M, N, W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from  $$ 2 A=\frac{(a-c)^{2}+3(a+c)(a-c)}{a+b+c}-2 a+2 c $$  we get  $$ \begin{aligned} 2(A & +C)=(a-c)^{2}\left(\frac{1}{s-d}+\frac{1}{s-b}\right)+3(a+c)(a-c)\left(\frac{1}{s-d}-\frac{1}{s-b}\right) \\ & =(a-c)^{2} \frac{2 s-b-d}{M}+3(a+c)(a-c) \cdot \frac{d-b}{M}=\frac{p(a-c)^{2}-3(a+c)(a-c)(b-d)}{M} \end{aligned} $$  where $p=2 s-b-d=s+a+c$. Similarly, writing $q=s+b+d$ we have  $$ 2(B+D)=\frac{q(b-d)^{2}-3(b+d)(b-d)(c-a)}{N} ; $$  specific grouping of terms in the numerators has its aim. Note that $p q>2 s^{2}$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,  $$ 2(A+B+C+D)=\frac{p(a-c)^{2}}{M}+\frac{3(a-c)(b-d) W}{M N}+\frac{q(b-d)^{2}}{N} $$  with $W$ defined by (3).  Substitution $x=(a-c) / M, y=(b-d) / N$ brings the required inequality to the form  $$ 2(A+B+C+D)=M p x^{2}+3 W x y+N q y^{2} \geq 0 . $$  It will be enough to verify that the discriminant $\Delta=9 W^{2}-4 M N p q$ of the quadratic trinomial $M p t^{2}+3 W t+N q$ is negative; on setting $t=x / y$ one then gets (6). The first inequality in (4) together with $p q>2 s^{2}$ imply $4 M N p q>8 s^{3}(a c(a+c)+b d(b+d))$. Since  $$ (a+c) s^{3}>(a+c)^{4} \geq 4 a c(a+c)^{2} \quad \text { and likewise } \quad(b+d) s^{3}>4 b d(b+d)^{2} $$  the estimate continues as follows,  $$ 4 M N p q>8\left(4(a c)^{2}(a+c)^{2}+4(b d)^{2}(b+d)^{2}\right)>32(b d(b+d)-a c(a+c))^{2}=32 W^{2} \geq 9 W^{2} $$  Thus indeed $\Delta<0$. The desired inequality (6) hence results. It becomes an equality if and only if $x=y=0$; equivalently, if and only if $a=c$ and simultaneously $b=d$.  Comment. The two solutions presented above do not differ significantly; large portions overlap. The properties of the number $W$ turn out to be crucial in both approaches. The Cauchy-Schwarz inequality, applied in the first solution, is avoided in the second, which requires no knowledge beyond quadratic trinomials.  The estimates in the proof of $\Delta<0$ in the second solution seem to be very wasteful. However, they come close to sharp when the terms in one of the pairs $(a, c),(b, d)$ are equal and much bigger than those in the other pair.  In attempts to prove the inequality by just considering the six cases of arrangement of the numbers $a, b, c, d$ on the real line, one soon discovers that the cases which create real trouble are precisely those in which $a$ and $c$ are both greater or both smaller than $b$ and $d$.


Solution: $$ \begin{gathered} (a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\ =((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\ =\left(a^{2}+a d-b^{2}-b d\right)\left(a^{2}+a d-c^{2}-c d\right)(b+c+d)= \\ =\left(a^{4}+2 a^{3} d-a^{2} b^{2}-a^{2} b d-a^{2} c^{2}-a^{2} c d+a^{2} d^{2}-a b^{2} d-a b d^{2}-a c^{2} d-a c d^{2}+b^{2} c^{2}+b^{2} c d+b c^{2} d+b c d^{2}\right)(b+c+d)= \\ =a^{4} b+a^{4} c+a^{4} d+\left(b^{3} c^{2}+a^{2} d^{3}\right)-a^{2} c^{3}+\left(2 a^{3} d^{2}-b^{3} a^{2}+c^{3} b^{2}\right)+ \\ +\left(b^{3} c d-c^{3} d a-d^{3} a b\right)+\left(2 a^{3} b d+c^{3} d b-d^{3} a c\right)+\left(2 a^{3} c d-b^{3} d a+d^{3} b c\right) \\ +\left(-a^{2} b^{2} c+3 b^{2} c^{2} d-2 a c^{2} d^{2}\right)+\left(-2 a^{2} b^{2} d+2 b c^{2} d^{2}\right)+\left(-a^{2} b c^{2}-2 a^{2} c^{2} d-2 a b^{2} d^{2}+2 b^{2} c d^{2}\right)+ \\ +\left(-2 a^{2} b c d-a b^{2} c d-a b c^{2} d-2 a b c d^{2}\right) \\ \text { Introducing the notation } S_{x y z w}=\sum_{c y c} a^{x} b^{y} c^{z} d^{w}, \text { one can write } \\ \sum_{c y c}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\ =S_{4100}+S_{4010}+S_{4001}+2 S_{3200}-S_{3020}+2 S_{3002}-S_{3110}+2 S_{3101}+2 S_{3011}-3 S_{2120}-6 S_{2111}= \\ +\left(S_{4100}+S_{4001}+\frac{1}{2} S_{3110}+\frac{1}{2} S_{3011}-3 S_{2120}\right)+ \\ +\left(S_{4010}-S_{3020}-\frac{3}{2} S_{3110}+\frac{3}{2} S_{3011}+\frac{9}{16} S_{2210}+\frac{9}{16} S_{2201}-\frac{9}{8} S_{2111}\right)+ \\ +\frac{9}{16}\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\right)+\frac{23}{16}\left(S_{3200}-2 S_{3101}+S_{3002}\right)+\frac{39}{8}\left(S_{3101}-S_{2111}\right), \end{gathered} $$  where the expressions  $$ \begin{gathered} S_{4100}+S_{4001}+\frac{1}{2} S_{3110}+\frac{1}{2} S_{3011}-3 S_{2120}=\sum_{c y c}\left(a^{4} b+b c^{4}+\frac{1}{2} a^{3} b c+\frac{1}{2} a b c^{3}-3 a^{2} b c^{2}\right), \\ S_{4010}-S_{3020}-\frac{3}{2} S_{3110}+\frac{3}{2} S_{3011}+\frac{9}{16} S_{2210}+\frac{9}{16} S_{2201}-\frac{9}{8} S_{2111}=\sum_{c y c} a^{2} c\left(a-c-\frac{3}{4} b+\frac{3}{4} d\right)^{2}, \\ S_{3200}-S_{2210}-S_{2201}+S_{3002}=\sum_{c y c} b^{2}\left(a^{3}-a^{2} c-a c^{2}+c^{3}\right)=\sum_{c y c} b^{2}(a+c)(a-c)^{2}, \\ S_{3200}-2 S_{3101}+S_{3002}=\sum_{c y c} a^{3}(b-d)^{2} \quad \text { and } \quad S_{3101}-S_{2111}=\frac{1}{3} \sum_{c y c} b d\left(2 a^{3}+c^{3}-3 a^{2} c\right) \end{gathered} $$  are all nonnegative.


In the plane we consider rectangles whose sides are parallel to the coordinate axes and have positive length. Such a rectangle will be called a box. Two boxes intersect if they have a common point in their interior or on their boundary.  Find the largest $n$ for which there exist $n$ boxes $B_{1}, \ldots, B_{n}$ such that $B_{i}$ and $B_{j}$ intersect if and only if $i \not \equiv j \pm 1(\bmod n)$.

Solution: The maximum number of such boxes is 6 . One example is shown in the figure.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-22.jpg?height=437&width=457&top_left_y=1135&top_left_x=777)  Now we show that 6 is the maximum. Suppose that boxes $B_{1}, \ldots, B_{n}$ satisfy the condition. Let the closed intervals $I_{k}$ and $J_{k}$ be the projections of $B_{k}$ onto the $x$ - and $y$-axis, for $1 \leq k \leq n$.  If $B_{i}$ and $B_{j}$ intersect, with a common point $(x, y)$, then $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$. So the intersections $I_{i} \cap I_{j}$ and $J_{i} \cap J_{j}$ are nonempty. Conversely, if $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$ for some real numbers $x, y$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Putting it around, $B_{i}$ and $B_{j}$ are disjoint if and only if their projections on at least one coordinate axis are disjoint.  For brevity we call two boxes or intervals adjacent if their indices differ by 1 modulo $n$, and nonadjacent otherwise.  The adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect for each $k=1, \ldots, n$. Hence $\left(I_{k}, I_{k+1}\right)$ or $\left(J_{k}, J_{k+1}\right)$ is a pair of disjoint intervals, $1 \leq k \leq n$. So there are at least $n$ pairs of disjoint intervals among $\left(I_{1}, I_{2}\right), \ldots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right) ;\left(J_{1}, J_{2}\right), \ldots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$.  Next, every two nonadjacent boxes intersect, hence their projections on both axes intersect, too. Then the claim below shows that at most 3 pairs among $\left(I_{1}, I_{2}\right), \ldots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$ are disjoint, and the same holds for $\left(J_{1}, J_{2}\right) \ldots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$. Consequently $n \leq 3+3=6$, as stated. Thus we are left with the claim and its justification.  Claim. Let $\Delta_{1}, \Delta_{2}, \ldots, \Delta_{n}$ be intervals on a straight line such that every two nonadjacent intervals intersect. Then $\Delta_{k}$ and $\Delta_{k+1}$ are disjoint for at most three values of $k=1, \ldots, n$.  Proof. Denote $\Delta_{k}=\left[a_{k}, b_{k}\right], 1 \leq k \leq n$. Let $\alpha=\max \left(a_{1}, \ldots, a_{n}\right)$ be the rightmost among the left endpoints of $\Delta_{1}, \ldots, \Delta_{n}$, and let $\beta=\min \left(b_{1}, \ldots, b_{n}\right)$ be the leftmost among their right endpoints. Assume that $\alpha=a_{2}$ without loss of generality.  If $\alpha \leq \beta$ then $a_{i} \leq \alpha \leq \beta \leq b_{i}$ for all $i$. Every $\Delta_{i}$ contains $\alpha$, and thus no disjoint pair $\left(\Delta_{i}, \Delta_{i+1}\right)$ exists.  If $\beta<\alpha$ then $\beta=b_{i}$ for some $i$ such that $a_{i}<b_{i}=\beta<\alpha=a_{2}<b_{2}$, hence $\Delta_{2}$ and $\Delta_{i}$ are disjoint. Now $\Delta_{2}$ intersects all remaining intervals except possibly $\Delta_{1}$ and $\Delta_{3}$, so $\Delta_{2}$ and $\Delta_{i}$ can be disjoint only if $i=1$ or $i=3$. Suppose by symmetry that $i=3$; then $\beta=b_{3}$. Since each of the intervals $\Delta_{4}, \ldots, \Delta_{n}$ intersects $\Delta_{2}$, we have $a_{i} \leq \alpha \leq b_{i}$ for $i=4, \ldots, n$. Therefore $\alpha \in \Delta_{4} \cap \ldots \cap \Delta_{n}$, in particular $\Delta_{4} \cap \ldots \cap \Delta_{n} \neq \emptyset$. Similarly, $\Delta_{5}, \ldots, \Delta_{n}, \Delta_{1}$ all intersect $\Delta_{3}$, so that $\Delta_{5} \cap \ldots \cap \Delta_{n} \cap \Delta_{1} \neq \emptyset$ as $\beta \in \Delta_{5} \cap \ldots \cap \Delta_{n} \cap \Delta_{1}$. This leaves $\left(\Delta_{1}, \Delta_{2}\right),\left(\Delta_{2}, \Delta_{3}\right)$ and $\left(\Delta_{3}, \Delta_{4}\right)$ as the only candidates for disjoint interval pairs, as desired.  Comment. The problem is a two-dimensional version of the original proposal which is included below. The extreme shortage of easy and appropriate submissions forced the Problem Selection Committee to shortlist a simplified variant. The same one-dimensional Claim is used in both versions.  Original proposal. We consider parallelepipeds in three-dimensional space, with edges parallel to the coordinate axes and of positive length. Such a parallelepiped will be called a box. Two boxes intersect if they have a common point in their interior or on their boundary.  Find the largest $n$ for which there exist $n$ boxes $B_{1}, \ldots, B_{n}$ such that $B_{i}$ and $B_{j}$ intersect if and only if $i \not \equiv j \pm 1(\bmod n)$.  The maximum number of such boxes is 9 . Suppose that boxes $B_{1}, \ldots, B_{n}$ satisfy the condition. Let the closed intervals $I_{k}, J_{k}$ and $K_{k}$ be the projections of box $B_{k}$ onto the $x-, y-$ and $z$-axis, respectively, for $1 \leq k \leq n$. As before, $B_{i}$ and $B_{j}$ are disjoint if and only if their projections on at least one coordinate axis are disjoint.  We call again two boxes or intervals adjacent if their indices differ by 1 modulo $n$, and nonadjacent otherwise.  The adjacent boxes $B_{i}$ and $B_{i+1}$ do not intersect for each $i=1, \ldots, n$. Hence at least one of the pairs $\left(I_{i}, I_{i+1}\right),\left(J_{i}, J_{i+1}\right)$ and $\left(K_{i}, K_{i+1}\right)$ is a pair of disjoint intervals. So there are at least $n$ pairs of disjoint intervals among $\left(I_{i}, I_{i+1}\right),\left(J_{i}, J_{i+1}\right),\left(K_{i}, K_{i+1}\right), 1 \leq i \leq n$.  Next, every two nonadjacent boxes intersect, hence their projections on the three axes intersect, too. Referring to the Claim in the solution of the two-dimensional version, we cocnclude that at most 3 pairs among $\left(I_{1}, I_{2}\right) \ldots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$ are disjoint; the same holds for $\left(J_{1}, J_{2}\right), \ldots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$ and $\left(K_{1}, K_{2}\right), \ldots,\left(K_{n-1}, K_{n}\right),\left(K_{n}, K_{1}\right)$. Consequently $n \leq 3+3+3=9$, as stated.  For $n=9$, the desired system of boxes exists. Consider the intervals in the following table:  | $i$ | $I_{i}$ | $J_{i}$ | $K_{i}$ | | :---: | :---: | :---: | :---: | | 1 | $[1,4]$ | $[1,6]$ | $[3,6]$ | | 2 | $[5,6]$ | $[1,6]$ | $[1,6]$ | | 3 | $[1,2]$ | $[1,6]$ | $[1,6]$ | | 4 | $[3,6]$ | $[1,4]$ | $[1,6]$ | | 5 | $[1,6]$ | $[5,6]$ | $[1,6]$ | | 6 | $[1,6]$ | $[1,2]$ | $[1,6]$ | | 7 | $[1,6]$ | $[3,6]$ | $[1,4]$ | | 8 | $[1,6]$ | $[1,6]$ | $[5,6]$ | | 9 | $[1,6]$ | $[1,6]$ | $[1,2]$ |  We have $I_{1} \cap I_{2}=I_{2} \cap I_{3}=I_{3} \cap I_{4}=\emptyset, J_{4} \cap J_{5}=J_{5} \cap J_{6}=J_{6} \cap J_{7}=\emptyset$, and finally $K_{7} \cap K_{8}=K_{8} \cap K_{9}=K_{9} \cap K_{1}=\emptyset$. The intervals in each column intersect in all other cases. It follows that the boxes $B_{i}=I_{i} \times J_{i} \times K_{i}, i=1, \ldots, 9$, have the stated property.


For every positive integer $n$ determine the number of permutations $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of the set $\{1,2, \ldots, n\}$ with the following property:  $$ 2\left(a_{1}+a_{2}+\cdots+a_{k}\right) \quad \text { is divisible by } k \text { for } k=1,2, \ldots, n \text {. } $$

Solution: For each $n$ let $F_{n}$ be the number of permutations of $\{1,2, \ldots, n\}$ with the required property; call them nice. For $n=1,2,3$ every permutation is nice, so $F_{1}=1, F_{2}=2, F_{3}=6$.  Take an $n>3$ and consider any nice permutation $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of $\{1,2, \ldots, n\}$. Then $n-1$ must be a divisor of the number  $$ \begin{aligned} & 2\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)=2\left((1+2+\cdots+n)-a_{n}\right) \\ & \quad=n(n+1)-2 a_{n}=(n+2)(n-1)+\left(2-2 a_{n}\right) . \end{aligned} $$  So $2 a_{n}-2$ must be divisible by $n-1$, hence equal to 0 or $n-1$ or $2 n-2$. This means that  $$ a_{n}=1 \quad \text { or } \quad a_{n}=\frac{n+1}{2} \quad \text { or } \quad a_{n}=n \text {. } $$  Suppose that $a_{n}=(n+1) / 2$. Since the permutation is nice, taking $k=n-2$ we get that $n-2$ has to be a divisor of  $$ \begin{aligned} 2\left(a_{1}+a_{2}+\cdots+a_{n-2}\right) & =2\left((1+2+\cdots+n)-a_{n}-a_{n-1}\right) \\ & =n(n+1)-(n+1)-2 a_{n-1}=(n+2)(n-2)+\left(3-2 a_{n-1}\right) . \end{aligned} $$  So $2 a_{n-1}-3$ should be divisible by $n-2$, hence equal to 0 or $n-2$ or $2 n-4$. Obviously 0 and $2 n-4$ are excluded because $2 a_{n-1}-3$ is odd. The remaining possibility $\left(2 a_{n-1}-3=n-2\right)$ leads to $a_{n-1}=(n+1) / 2=a_{n}$, which also cannot hold. This eliminates $(n+1) / 2$ as a possible value of $a_{n}$. Consequently $a_{n}=1$ or $a_{n}=n$.  If $a_{n}=n$ then $\left(a_{1}, a_{2}, \ldots, a_{n-1}\right)$ is a nice permutation of $\{1,2, \ldots, n-1\}$. There are $F_{n-1}$ such permutations. Attaching $n$ to any one of them at the end creates a nice permutation of $\{1,2, \ldots, n\}$.  If $a_{n}=1$ then $\left(a_{1}-1, a_{2}-1, \ldots, a_{n-1}-1\right)$ is a permutation of $\{1,2, \ldots, n-1\}$. It is also nice because the number  $$ 2\left(\left(a_{1}-1\right)+\cdots+\left(a_{k}-1\right)\right)=2\left(a_{1}+\cdots+a_{k}\right)-2 k $$  is divisible by $k$, for any $k \leq n-1$. And again, any one of the $F_{n-1}$ nice permutations $\left(b_{1}, b_{2}, \ldots, b_{n-1}\right)$ of $\{1,2, \ldots, n-1\}$ gives rise to a nice permutation of $\{1,2, \ldots, n\}$ whose last term is 1 , namely $\left(b_{1}+1, b_{2}+1, \ldots, b_{n-1}+1,1\right)$.  The bijective correspondences established in both cases show that there are $F_{n-1}$ nice permutations of $\{1,2, \ldots, n\}$ with the last term 1 and also $F_{n-1}$ nice permutations of $\{1,2, \ldots, n\}$ with the last term $n$. Hence follows the recurrence $F_{n}=2 F_{n-1}$. With the base value $F_{3}=6$ this gives the outcome formula $F_{n}=3 \cdot 2^{n-2}$ for $n \geq 3$.


In the coordinate plane consider the set $S$ of all points with integer coordinates. For a positive integer $k$, two distinct points $A, B \in S$ will be called $k$-friends if there is a point $C \in S$ such that the area of the triangle $A B C$ is equal to $k$. A set $T \subset S$ will be called a $k$-clique if every two points in $T$ are $k$-friends. Find the least positive integer $k$ for which there exists a $k$-clique with more than 200 elements.

Solution: To begin, let us describe those points $B \in S$ which are $k$-friends of the point $(0,0)$. By definition, $B=(u, v)$ satisfies this condition if and only if there is a point $C=(x, y) \in S$ such that $\frac{1}{2}|u y-v x|=k$. (This is a well-known formula expressing the area of triangle $A B C$ when $A$ is the origin.)  To say that there exist integers $x, y$ for which $|u y-v x|=2 k$, is equivalent to saying that the greatest common divisor of $u$ and $v$ is also a divisor of $2 k$. Summing up, a point $B=(u, v) \in S$ is a $k$-friend of $(0,0)$ if and only if $\operatorname{gcd}(u, v)$ divides $2 k$.  Translation by a vector with integer coordinates does not affect $k$-friendship; if two points are $k$-friends, so are their translates. It follows that two points $A, B \in S, A=(s, t), B=(u, v)$, are $k$-friends if and only if the point $(u-s, v-t)$ is a $k$-friend of $(0,0)$; i.e., if $\operatorname{gcd}(u-s, v-t) \mid 2 k$.  Let $n$ be a positive integer which does not divide $2 k$. We claim that a $k$-clique cannot have more than $n^{2}$ elements.  Indeed, all points $(x, y) \in S$ can be divided into $n^{2}$ classes determined by the remainders that $x$ and $y$ leave in division by $n$. If a set $T$ has more than $n^{2}$ elements, some two points $A, B \in T, A=(s, t), B=(u, v)$, necessarily fall into the same class. This means that $n \mid u-s$ and $n \mid v-t$. Hence $n \mid d$ where $d=\operatorname{gcd}(u-s, v-t)$. And since $n$ does not divide $2 k$, also $d$ does not divide $2 k$. Thus $A$ and $B$ are not $k$-friends and the set $T$ is not a $k$-clique.  Now let $M(k)$ be the least positive integer which does not divide $2 k$. Write $M(k)=m$ for the moment and consider the set $T$ of all points $(x, y)$ with $0 \leq x, y<m$. There are $m^{2}$ of them. If $A=(s, t), B=(u, v)$ are two distinct points in $T$ then both differences $|u-s|,|v-t|$ are integers less than $m$ and at least one of them is positive. By the definition of $m$, every positive integer less than $m$ divides $2 k$. Therefore $u-s$ (if nonzero) divides $2 k$, and the same is true of $v-t$. So $2 k$ is divisible by $\operatorname{gcd}(u-s, v-t)$, meaning that $A, B$ are $k$-friends. Thus $T$ is a $k$-clique.  It follows that the maximum size of a $k$-clique is $M(k)^{2}$, with $M(k)$ defined as above. We are looking for the minimum $k$ such that $M(k)^{2}>200$.  By the definition of $M(k), 2 k$ is divisible by the numbers $1,2, \ldots, M(k)-1$, but not by $M(k)$ itself. If $M(k)^{2}>200$ then $M(k) \geq 15$. Trying to hit $M(k)=15$ we get a contradiction immediately ( $2 k$ would have to be divisible by 3 and 5 , but not by 15 ).  So let us try $M(k)=16$. Then $2 k$ is divisible by the numbers $1,2, \ldots, 15$, hence also by their least common multiple $L$, but not by 16 . And since $L$ is not a multiple of 16 , we infer that $k=L / 2$ is the least $k$ with $M(k)=16$.  Finally, observe that if $M(k) \geq 17$ then $2 k$ must be divisible by the least common multiple of $1,2, \ldots, 16$, which is equal to $2 L$. Then $2 k \geq 2 L$, yielding $k>L / 2$.  In conclusion, the least $k$ with the required property is equal to $L / 2=180180$.


Let $n$ and $k$ be fixed positive integers of the same parity, $k \geq n$. We are given $2 n$ lamps numbered 1 through $2 n$; each of them can be on or off. At the beginning all lamps are off. We consider sequences of $k$ steps. At each step one of the lamps is switched (from off to on or from on to off).  Let $N$ be the number of $k$-step sequences ending in the state: lamps $1, \ldots, n$ on, lamps $n+1, \ldots, 2 n$ off.  Let $M$ be the number of $k$-step sequences leading to the same state and not touching lamps $n+1, \ldots, 2 n$ at all.  Find the ratio $N / M$.

Solution: A sequence of $k$ switches ending in the state as described in the problem statement (lamps $1, \ldots, n$ on, lamps $n+1, \ldots, 2 n$ off) will be called an admissible process. If, moreover, the process does not touch the lamps $n+1, \ldots, 2 n$, it will be called restricted. So there are $N$ admissible processes, among which $M$ are restricted.  In every admissible process, restricted or not, each one of the lamps $1, \ldots, n$ goes from off to on, so it is switched an odd number of times; and each one of the lamps $n+1, \ldots, 2 n$ goes from off to off, so it is switched an even number of times.  Notice that $M>0$; i.e., restricted admissible processes do exist (it suffices to switch each one of the lamps $1, \ldots, n$ just once and then choose one of them and switch it $k-n$ times, which by hypothesis is an even number).  Consider any restricted admissible process $\mathbf{p}$. Take any lamp $\ell, 1 \leq \ell \leq n$, and suppose that it was switched $k_{\ell}$ times. As noticed, $k_{\ell}$ must be odd. Select arbitrarily an even number of these $k_{\ell}$ switches and replace each of them by the switch of lamp $n+\ell$. This can be done in $2^{k_{\ell}-1}$ ways (because a $k_{\ell}$-element set has $2^{k_{\ell}-1}$ subsets of even cardinality). Notice that $k_{1}+\cdots+k_{n}=k$.  These actions are independent, in the sense that the action involving lamp $\ell$ does not affect the action involving any other lamp. So there are $2^{k_{1}-1} \cdot 2^{k_{2}-1} \cdots 2^{k_{n}-1}=2^{k-n}$ ways of combining these actions. In any of these combinations, each one of the lamps $n+1, \ldots, 2 n$ gets switched an even number of times and each one of the lamps $1, \ldots, n$ remains switched an odd number of times, so the final state is the same as that resulting from the original process $\mathbf{p}$.  This shows that every restricted admissible process $\mathbf{p}$ can be modified in $2^{k-n}$ ways, giving rise to $2^{k-n}$ distinct admissible processes (with all lamps allowed).  Now we show that every admissible process $\mathbf{q}$ can be achieved in that way. Indeed, it is enough to replace every switch of a lamp with a label $\ell>n$ that occurs in $\mathbf{q}$ by the switch of the corresponding lamp $\ell-n$; in the resulting process $\mathbf{p}$ the lamps $n+1, \ldots, 2 n$ are not involved.  Switches of each lamp with a label $\ell>n$ had occurred in $\mathbf{q}$ an even number of times. So the performed replacements have affected each lamp with a label $\ell \leq n$ also an even number of times; hence in the overall effect the final state of each lamp has remained the same. This means that the resulting process $\mathbf{p}$ is admissible - and clearly restricted, as the lamps $n+1, \ldots, 2 n$ are not involved in it any more.  If we now take process $\mathbf{p}$ and reverse all these replacements, then we obtain process $\mathbf{q}$. These reversed replacements are nothing else than the modifications described in the foregoing paragraphs.  Thus there is a one-to $-\left(2^{k-n}\right)$ correspondence between the $M$ restricted admissible processes and the total of $N$ admissible processes. Therefore $N / M=2^{k-n}$.


Let $S=\left\{x_{1}, x_{2}, \ldots, x_{k+\ell}\right\}$ be a $(k+\ell)$-element set of real numbers contained in the interval $[0,1] ; k$ and $\ell$ are positive integers. A $k$-element subset $A \subset S$ is called nice if  $$ \left|\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{\ell} \sum_{x_{j} \in S \backslash A} x_{j}\right| \leq \frac{k+\ell}{2 k \ell} . $$  Prove that the number of nice subsets is at least $\frac{2}{k+\ell}\left(\begin{array}{c}k+\ell \\ k\end{array}\right)$.

Solution: For a $k$-element subset $A \subset S$, let $f(A)=\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{\ell} \sum_{x_{j} \in S \backslash A} x_{j}$. Denote $\frac{k+\ell}{2 k \ell}=d$. By definition a subset $A$ is nice if $|f(A)| \leq d$.  To each permutation $\left(y_{1}, y_{2}, \ldots, y_{k+\ell}\right)$ of the set $S=\left\{x_{1}, x_{2}, \ldots, x_{k+\ell}\right\}$ we assign $k+\ell$ subsets of $S$ with $k$ elements each, namely $A_{i}=\left\{y_{i}, y_{i+1}, \ldots, y_{i+k-1}\right\}, i=1,2, \ldots, k+\ell$. Indices are taken modulo $k+\ell$ here and henceforth. In other words, if $y_{1}, y_{2}, \ldots, y_{k+\ell}$ are arranged around a circle in this order, the sets in question are all possible blocks of $k$ consecutive elements.  Claim. At least two nice sets are assigned to every permutation of $S$.  Proof. Adjacent sets $A_{i}$ and $A_{i+1}$ differ only by the elements $y_{i}$ and $y_{i+k}, i=1, \ldots, k+\ell$. By the definition of $f$, and because $y_{i}, y_{i+k} \in[0,1]$,  $$ \left|f\left(A_{i+1}\right)-f\left(A_{i}\right)\right|=\left|\left(\frac{1}{k}+\frac{1}{\ell}\right)\left(y_{i+k}-y_{i}\right)\right| \leq \frac{1}{k}+\frac{1}{\ell}=2 d \text {. } $$  Each element $y_{i} \in S$ belongs to exactly $k$ of the sets $A_{1}, \ldots, A_{k+\ell}$. Hence in $k$ of the expressions $f\left(A_{1}\right), \ldots, f\left(A_{k+\ell}\right)$ the coefficient of $y_{i}$ is $1 / k$; in the remaining $\ell$ expressions, its coefficient is $-1 / \ell$. So the contribution of $y_{i}$ to the sum of all $f\left(A_{i}\right)$ equals $k \cdot 1 / k-\ell \cdot 1 / \ell=0$. Since this holds for all $i$, it follows that $f\left(A_{1}\right)+\cdots+f\left(A_{k+\ell}\right)=0$.  If $f\left(A_{p}\right)=\min f\left(A_{i}\right), f\left(A_{q}\right)=\max f\left(A_{i}\right)$, we obtain in particular $f\left(A_{p}\right) \leq 0, f\left(A_{q}\right) \geq 0$. Let $p<q$ (the case $p>q$ is analogous; and the claim is true for $p=q$ as $f\left(A_{i}\right)=0$ for all $i$ ).  We are ready to prove that at least two of the sets $A_{1}, \ldots, A_{k+\ell}$ are nice. The interval $[-d, d]$ has length $2 d$, and we saw that adjacent numbers in the circular arrangement $f\left(A_{1}\right), \ldots, f\left(A_{k+\ell}\right)$ differ by at most $2 d$. Suppose that $f\left(A_{p}\right)<-d$ and $f\left(A_{q}\right)>d$. Then one of the numbers $f\left(A_{p+1}\right), \ldots, f\left(A_{q-1}\right)$ lies in $[-d, d]$, and also one of the numbers $f\left(A_{q+1}\right), \ldots, f\left(A_{p-1}\right)$ lies there. Consequently, one of the sets $A_{p+1}, \ldots, A_{q-1}$ is nice, as well as one of the sets $A_{q+1}, \ldots, A_{p-1}$. If $-d \leq f\left(A_{p}\right)$ and $f\left(A_{q}\right) \leq d$ then $A_{p}$ and $A_{q}$ are nice.  Let now $f\left(A_{p}\right)<-d$ and $f\left(A_{q}\right) \leq d$. Then $f\left(A_{p}\right)+f\left(A_{q}\right)<0$, and since $\sum f\left(A_{i}\right)=0$, there is an $r \neq q$ such that $f\left(A_{r}\right)>0$. We have $0<f\left(A_{r}\right) \leq f\left(A_{q}\right) \leq d$, so the sets $f\left(A_{r}\right)$ and $f\left(A_{q}\right)$ are nice. The only case remaining, $-d \leq f\left(A_{p}\right)$ and $d<f\left(A_{q}\right)$, is analogous.  Apply the claim to each of the $(k+\ell)$ ! permutations of $S=\left\{x_{1}, x_{2}, \ldots, x_{k+\ell}\right\}$. This gives at least $2(k+\ell)$ ! nice sets, counted with repetitions: each nice set is counted as many times as there are permutations to which it is assigned.  On the other hand, each $k$-element set $A \subset S$ is assigned to exactly $(k+\ell) k ! \ell$ ! permutations. Indeed, such a permutation $\left(y_{1}, y_{2}, \ldots, y_{k+\ell}\right)$ is determined by three independent choices: an in$\operatorname{dex} i \in\{1,2, \ldots, k+\ell\}$ such that $A=\left\{y_{i}, y_{i+1}, \ldots, y_{i+k-1}\right\}$, a permutation $\left(y_{i}, y_{i+1}, \ldots, y_{i+k-1}\right)$ of the set $A$, and a permutation $\left(y_{i+k}, y_{i+k+1}, \ldots, y_{i-1}\right)$ of the set $S \backslash A$.  In summary, there are at least $\frac{2(k+\ell) !}{(k+\ell) k ! \ell !}=\frac{2}{k+\ell}\left(\begin{array}{c}k+\ell \\ k\end{array}\right)$ nice sets.


For $n \geq 2$, let $S_{1}, S_{2}, \ldots, S_{2^{n}}$ be $2^{n}$ subsets of $A=\left\{1,2,3, \ldots, 2^{n+1}\right\}$ that satisfy the following property: There do not exist indices $a$ and $b$ with $a<b$ and elements $x, y, z \in A$ with $x<y<z$ such that $y, z \in S_{a}$ and $x, z \in S_{b}$. Prove that at least one of the sets $S_{1}, S_{2}, \ldots, S_{2^{n}}$ contains no more than $4 n$ elements.

Solution: We prove that there exists a set $S_{a}$ with at most $3 n+1$ elements.  Given a $k \in\{1, \ldots, n\}$, we say that an element $z \in A$ is $k$-good to a set $S_{a}$ if $z \in S_{a}$ and $S_{a}$ contains two other elements $x$ and $y$ with $x<y<z$ such that $z-y<2^{k}$ and $z-x \geq 2^{k}$. Also, $z \in A$ will be called $\operatorname{good}$ to $S_{a}$ if $z$ is $k$-good to $S_{a}$ for some $k=1, \ldots, n$.  We claim that each $z \in A$ can be $k$-good to at most one set $S_{a}$. Indeed, suppose on the contrary that $z$ is $k$-good simultaneously to $S_{a}$ and $S_{b}$, with $a<b$. Then there exist $y_{a} \in S_{a}$, $y_{a}<z$, and $x_{b} \in S_{b}, x_{b}<z$, such that $z-y_{a}<2^{k}$ and $z-x_{b} \geq 2^{k}$. On the other hand, since $z \in S_{a} \cap S_{b}$, by the condition of the problem there is no element of $S_{a}$ strictly between $x_{b}$ and $z$. Hence $y_{a} \leq x_{b}$, implying $z-y_{a} \geq z-x_{b}$. However this contradicts $z-y_{a}<2^{k}$ and $z-x_{b} \geq 2^{k}$. The claim follows.  As a consequence, a fixed $z \in A$ can be good to at most $n$ of the given sets (no more than one of them for each $k=1, \ldots, n$ ).  Furthermore, let $u_{1}<u_{2}<\cdots<u_{m}<\cdots<u_{p}$ be all elements of a fixed set $S_{a}$ that are not good to $S_{a}$. We prove that $u_{m}-u_{1}>2\left(u_{m-1}-u_{1}\right)$ for all $m \geq 3$.  Indeed, assume that $u_{m}-u_{1} \leq 2\left(u_{m-1}-u_{1}\right)$ holds for some $m \geq 3$. This inequality can be written as $2\left(u_{m}-u_{m-1}\right) \leq u_{m}-u_{1}$. Take the unique $k$ such that $2^{k} \leq u_{m}-u_{1}<2^{k+1}$. Then $2\left(u_{m}-u_{m-1}\right) \leq u_{m}-u_{1}<2^{k+1}$ yields $u_{m}-u_{m-1}<2^{k}$. However the elements $z=u_{m}, x=u_{1}$, $y=u_{m-1}$ of $S_{a}$ then satisfy $z-y<2^{k}$ and $z-x \geq 2^{k}$, so that $z=u_{m}$ is $k$-good to $S_{a}$.  Thus each term of the sequence $u_{2}-u_{1}, u_{3}-u_{1}, \ldots, u_{p}-u_{1}$ is more than twice the previous one. Hence $u_{p}-u_{1}>2^{p-1}\left(u_{2}-u_{1}\right) \geq 2^{p-1}$. But $u_{p} \in\left\{1,2,3, \ldots, 2^{n+1}\right\}$, so that $u_{p} \leq 2^{n+1}$. This yields $p-1 \leq n$, i. e. $p \leq n+1$.  In other words, each set $S_{a}$ contains at most $n+1$ elements that are not good to it.  To summarize the conclusions, mark with red all elements in the sets $S_{a}$ that are good to the respective set, and with blue the ones that are not good. Then the total number of red elements, counting multiplicities, is at most $n \cdot 2^{n+1}$ (each $z \in A$ can be marked red in at most $n$ sets). The total number of blue elements is at most $(n+1) 2^{n}$ (each set $S_{a}$ contains at most $n+1$ blue elements). Therefore the sum of cardinalities of $S_{1}, S_{2}, \ldots, S_{2^{n}}$ does not exceed $(3 n+1) 2^{n}$. By averaging, the smallest set has at most $3 n+1$ elements.


Solution: We show that one of the sets $S_{a}$ has at most $2 n+1$ elements. In the sequel $|\cdot|$ denotes the cardinality of a (finite) set.  Claim. For $n \geq 2$, suppose that $k$ subsets $S_{1}, \ldots, S_{k}$ of $\left\{1,2, \ldots, 2^{n}\right\}$ (not necessarily different) satisfy the condition of the problem. Then  $$ \sum_{i=1}^{k}\left(\left|S_{i}\right|-n\right) \leq(2 n-1) 2^{n-2} $$  Proof. Observe that if the sets $S_{i}(1 \leq i \leq k)$ satisfy the condition then so do their arbitrary subsets $T_{i}(1 \leq i \leq k)$. The condition also holds for the sets $t+S_{i}=\left\{t+x \mid x \in S_{i}\right\}$ where $t$ is arbitrary.  Note also that a set may occur more than once among $S_{1}, \ldots, S_{k}$ only if its cardinality is less than 3, in which case its contribution to the sum $\sum_{i=1}^{k}\left(\left|S_{i}\right|-n\right)$ is nonpositive (as $n \geq 2$ ).  The proof is by induction on $n$. In the base case $n=2$ we have subsets $S_{i}$ of $\{1,2,3,4\}$. Only the ones of cardinality 3 and 4 need to be considered by the remark above; each one of them occurs at most once among $S_{1}, \ldots, S_{k}$. If $S_{i}=\{1,2,3,4\}$ for some $i$ then no $S_{j}$ is a 3 -element subset in view of the condition, hence $\sum_{i=1}^{k}\left(\left|S_{i}\right|-2\right) \leq 2$. By the condition again, it is impossible that $S_{i}=\{1,3,4\}$ and $S_{j}=\{2,3,4\}$ for some $i, j$. So if $\left|S_{i}\right| \leq 3$ for all $i$ then at most 3 summands $\left|S_{i}\right|-2$ are positive, corresponding to 3 -element subsets. This implies $\sum_{i=1}^{k}\left(\left|S_{i}\right|-2\right) \leq 3$, therefore the conclusion is true for $n=2$.  Suppose that the claim holds for some $n \geq 2$, and let the sets $S_{1}, \ldots, S_{k} \subseteq\left\{1,2, \ldots, 2^{n+1}\right\}$ satisfy the given property. Denote $U_{i}=S_{i} \cap\left\{1,2, \ldots, 2^{n}\right\}, V_{i}=S_{i} \cap\left\{2^{n}+1, \ldots, 2^{n+1}\right\}$. Let  $$ I=\left\{i|1 \leq i \leq k,| U_{i} \mid \neq 0\right\}, \quad J=\{1, \ldots, k\} \backslash I $$  The sets $S_{j}$ with $j \in J$ are all contained in $\left\{2^{n}+1, \ldots, 2^{n+1}\right\}$, so the induction hypothesis applies to their translates $-2^{n}+S_{j}$ which have the same cardinalities. Consequently, this gives $\sum_{j \in J}\left(\left|S_{j}\right|-n\right) \leq(2 n-1) 2^{n-2}$, so that  $$ \sum_{j \in J}\left(\left|S_{j}\right|-(n+1)\right) \leq \sum_{j \in J}\left(\left|S_{j}\right|-n\right) \leq(2 n-1) 2^{n-2} $$  For $i \in I$, denote by $v_{i}$ the least element of $V_{i}$. Observe that if $V_{a}$ and $V_{b}$ intersect, with $a<b$, $a, b \in I$, then $v_{a}$ is their unique common element. Indeed, let $z \in V_{a} \cap V_{b} \subseteq S_{a} \cap S_{b}$ and let $m$ be the least element of $S_{b}$. Since $b \in I$, we have $m \leq 2^{n}$. By the condition, there is no element of $S_{a}$ strictly between $m \leq 2^{n}$ and $z>2^{n}$, which implies $z=v_{a}$.  It follows that if the element $v_{i}$ is removed from each $V_{i}$, a family of pairwise disjoint sets $W_{i}=V_{i} \backslash\left\{v_{i}\right\}$ is obtained, $i \in I$ (we assume $W_{i}=\emptyset$ if $V_{i}=\emptyset$ ). As $W_{i} \subseteq\left\{2^{n}+1, \ldots, 2^{n+1}\right\}$ for all $i$, we infer that $\sum_{i \in I}\left|W_{i}\right| \leq 2^{n}$. Therefore $\sum_{i \in I}\left(\left|V_{i}\right|-1\right) \leq \sum_{i \in I}\left|W_{i}\right| \leq 2^{n}$.  On the other hand, the induction hypothesis applies directly to the sets $U_{i}, i \in I$, so that $\sum_{i \in \mathcal{I}}\left(\left|U_{i}\right|-n\right) \leq(2 n-1) 2^{n-2}$. In summary,  $$ \sum_{i \in I}\left(\left|S_{i}\right|-(n+1)\right)=\sum_{i \in I}\left(\left|U_{i}\right|-n\right)+\sum_{i \in I}\left(\left|V_{i}\right|-1\right) \leq(2 n-1) 2^{n-2}+2^{n} $$  The estimates (1) and (2) are sufficient to complete the inductive step:  $$ \begin{aligned} \sum_{i=1}^{k}\left(\left|S_{i}\right|-(n+1)\right) & =\sum_{i \in I}\left(\left|S_{i}\right|-(n+1)\right)+\sum_{j \in J}\left(\left|S_{j}\right|-(n+1)\right) \\ & \leq(2 n-1) 2^{n-2}+2^{n}+(2 n-1) 2^{n-2}=(2 n+1) 2^{n-1} \end{aligned} $$  Returning to the problem, consider $k=2^{n}$ subsets $S_{1}, S_{2}, \ldots, S_{2^{n}}$ of $\left\{1,2,3, \ldots, 2^{n+1}\right\}$. If they satisfy the given condition, the claim implies $\sum_{i=1}^{2^{n}}\left(\left|S_{i}\right|-(n+1)\right) \leq(2 n+1) 2^{n-1}$. By averaging again, we see that the smallest set has at most $2 n+1$ elements.  Comment. It can happen that each set $S_{i}$ has cardinality at least $n+1$. Here is an example by the proposer.  For $i=1, \ldots, 2^{n}$, let $S_{i}=\left\{i+2^{k} \mid 0 \leq k \leq n\right\}$. Then $\left|S_{i}\right|=n+1$ for all $i$. Suppose that there exist $a<b$ and $x<y<z$ such that $y, z \in S_{a}$ and $x, z \in S_{b}$. Hence $z=a+2^{k}=b+2^{l}$ for some $k>l$. Since $y \in S_{a}$ and $y<z$, we have $y \leq a+2^{k-1}$. So the element $x \in S_{b}$ satisfies  $$ x<y \leq a+2^{k-1}=z-2^{k-1} \leq z-2^{l}=b . $$  However the least element of $S_{b}$ is $b+1$, a contradiction.


In an acute-angled triangle $A B C$, point $H$ is the orthocentre and $A_{0}, B_{0}, C_{0}$ are the midpoints of the sides $B C, C A, A B$, respectively. Consider three circles passing through $H$ : $\omega_{a}$ around $A_{0}, \omega_{b}$ around $B_{0}$ and $\omega_{c}$ around $C_{0}$. The circle $\omega_{a}$ intersects the line $B C$ at $A_{1}$ and $A_{2} ; \omega_{b}$ intersects $C A$ at $B_{1}$ and $B_{2} ; \omega_{c}$ intersects $A B$ at $C_{1}$ and $C_{2}$. Show that the points $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle.

Solution: The perpendicular bisectors of the segments $A_{1} A_{2}, B_{1} B_{2}, C_{1} C_{2}$ are also the perpendicular bisectors of $B C, C A, A B$. So they meet at $O$, the circumcentre of $A B C$. Thus $O$ is the only point that can possibly be the centre of the desired circle.  From the right triangle $O A_{0} A_{1}$ we get  $$ O A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2} $$  Let $K$ be the midpoint of $A H$ and let $L$ be the midpoint of $C H$. Since $A_{0}$ and $B_{0}$ are the midpoints of $B C$ and $C A$, we see that $A_{0} L \| B H$ and $B_{0} L \| A H$. Thus the segments $A_{0} L$ and $B_{0} L$ are perpendicular to $A C$ and $B C$, hence parallel to $O B_{0}$ and $O A_{0}$, respectively. Consequently $O A_{0} L B_{0}$ is a parallelogram, so that $O A_{0}$ and $B_{0} L$ are equal and parallel. Also, the midline $B_{0} L$ of triangle $A H C$ is equal and parallel to $A K$ and $K H$.  It follows that $A K A_{0} O$ and $H A_{0} O K$ are parallelograms. The first one gives $A_{0} K=O A=R$, where $R$ is the circumradius of $A B C$. From the second one we obtain  $$ 2\left(O A_{0}^{2}+A_{0} H^{2}\right)=O H^{2}+A_{0} K^{2}=O H^{2}+R^{2} . $$  (In a parallelogram, the sum of squares of the diagonals equals the sum of squares of the sides).  From (1) and (2) we get $O A_{1}^{2}=\left(O H^{2}+R^{2}\right) / 2$. By symmetry, the same holds for the distances $O A_{2}, O B_{1}, O B_{2}, O C_{1}$ and $O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ all lie on a circle with centre at $O$ and radius $\left(O H^{2}+R^{2}\right) / 2$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-30.jpg?height=614&width=820&top_left_y=2103&top_left_x=595)


Solution: We are going to show again that the circumcentre $O$ is equidistant from the six points in question.  Let $A^{\prime}$ be the second intersection point of $\omega_{b}$ and $\omega_{c}$. The line $B_{0} C_{0}$, which is the line of centers of circles $\omega_{b}$ and $\omega_{c}$, is a midline in triangle $A B C$, parallel to $B C$ and perpendicular to the altitude $A H$. The points $A^{\prime}$ and $H$ are symmetric with respect to the line of centers. Therefore $A^{\prime}$ lies on the line $A H$.  From the two circles $\omega_{b}$ and $\omega_{c}$ we obtain $A C_{1} \cdot A C_{2}=A A^{\prime} \cdot A H=A B_{1} \cdot A B_{2}$. So the quadrilateral $B_{1} B_{2} C_{1} C_{2}$ is cyclic. The perpendicular bisectors of the sides $B_{1} B_{2}$ and $C_{1} C_{2}$ meet at $O$. Hence $O$ is the circumcentre of $B_{1} B_{2} C_{1} C_{2}$ and so $O B_{1}=O B_{2}=O C_{1}=O C_{2}$.  Analogous arguments yield $O A_{1}=O A_{2}=O B_{1}=O B_{2}$ and $O A_{1}=O A_{2}=O C_{1}=O C_{2}$. Thus $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$ lie on a circle centred at $O$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-31.jpg?height=648&width=854&top_left_y=847&top_left_x=635)  Comment. The problem can be solved without much difficulty in many ways by calculation, using trigonometry, coordinate geometry or complex numbers. As an example we present a short proof using vectors.


Solution: Let again $O$ and $R$ be the circumcentre and circumradius. Consider the vectors  $$ \overrightarrow{O A}=\mathbf{a}, \quad \overrightarrow{O B}=\mathbf{b}, \quad \overrightarrow{O C}=\mathbf{c}, \quad \text { where } \quad \mathbf{a}^{2}=\mathbf{b}^{2}=\mathbf{c}^{2}=R^{2} $$  It is well known that $\overrightarrow{O H}=\mathbf{a}+\mathbf{b}+\mathbf{c}$. Accordingly,  $$ \overrightarrow{A_{0} H}=\overrightarrow{O H}-\overrightarrow{O A_{0}}=(\mathbf{a}+\mathbf{b}+\mathbf{c})-\frac{\mathbf{b}+\mathbf{c}}{2}=\frac{2 \mathbf{a}+\mathbf{b}+\mathbf{c}}{2} $$  and  $$ \begin{gathered} O A_{1}^{2}=O A_{0}^{2}+A_{0} A_{1}^{2}=O A_{0}^{2}+A_{0} H^{2}=\left(\frac{\mathbf{b}+\mathbf{c}}{2}\right)^{2}+\left(\frac{2 \mathbf{a}+\mathbf{b}+\mathbf{c}}{2}\right)^{2} \\ =\frac{1}{4}\left(\mathbf{b}^{2}+2 \mathbf{b} \mathbf{c}+\mathbf{c}^{2}\right)+\frac{1}{4}\left(4 \mathbf{a}^{2}+4 \mathbf{a} \mathbf{b}+4 \mathbf{a} \mathbf{c}+\mathbf{b}^{2}+2 \mathbf{b} \mathbf{c}+\mathbf{c}^{2}\right)=2 R^{2}+(\mathbf{a b}+\mathbf{a c}+\mathbf{b c}) \end{gathered} $$  here $\mathbf{a b}$, bc, etc. denote dot products of vectors. We get the same for the distances $O A_{2}, O B_{1}$, $O B_{2}, O C_{1}$ and $O C_{2}$.


Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$, assume that there exist points $E$ on line $B C$ outside segment $B C$, and $F$ inside segment $A D$, such that $\angle D A E=\angle C B F$. Denote by $I$ the point of intersection of $C D$ and $E F$, and by $J$ the point of intersection of $A B$ and $E F$. Let $K$ be the midpoint of segment $E F$; assume it does not lie on line $A B$.  Prove that $I$ belongs to the circumcircle of $A B K$ if and only if $K$ belongs to the circumcircle of $C D J$.

Solution: Assume that the disposition of points is as in the diagram.  Since $\angle E B F=180^{\circ}-\angle C B F=180^{\circ}-\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \cdot J B=F J \cdot J E$. In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $I J \cdot J K=F J \cdot J E$. Expressing $I J=I F+F J, J E=F E-F J$, and $J K=\frac{1}{2} F E-F J$, we find that $I$ belongs to the circumcircle of $A B K$ if and only if  $$ F J=\frac{I F \cdot F E}{2 I F+F E} . $$  Since $A E B F$ is cyclic and $A B, C D$ are parallel, $\angle F E C=\angle F A B=180^{\circ}-\angle C D F$. Then $C D F E$ is also cyclic, yielding $I D \cdot I C=I F \cdot I E$. It follows that $K$ belongs to the circumcircle of $C D J$ if and only if $I J \cdot I K=I F \cdot I E$. Expressing $I J=I F+F J, I K=I F+\frac{1}{2} F E$, and $I E=I F+F E$, we find that $K$ is on the circumcircle of $C D J$ if and only if  $$ F J=\frac{I F \cdot F E}{2 I F+F E} $$  The conclusion follows.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-32.jpg?height=705&width=914&top_left_y=1469&top_left_x=548)  Comment. While the figure shows $B$ inside segment $C E$, it is possible that $C$ is inside segment $B E$. Consequently, $I$ would be inside segment $E F$ and $J$ outside segment $E F$. The position of point $K$ on line $E F$ with respect to points $I, J$ may also vary.  Some case may require that an angle $\varphi$ be replaced by $180^{\circ}-\varphi$, and in computing distances, a sum may need to become a difference. All these cases can be covered by the proposed solution if it is clearly stated that signed distances and angles are used.


Let $A B C D$ be a convex quadrilateral and let $P$ and $Q$ be points in $A B C D$ such that $P Q D A$ and $Q P B C$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $P Q$ such that $\angle P A E=\angle Q D E$ and $\angle P B E=\angle Q C E$. Show that the quadrilateral $A B C D$ is cyclic.

Solution: Let $F$ be the point on the line $A D$ such that $E F \| P A$. By hypothesis, the quadrilateral $P Q D A$ is cyclic. So if $F$ lies between $A$ and $D$ then $\angle E F D=\angle P A D=180^{\circ}-\angle E Q D$; the points $F$ and $Q$ are on distinct sides of the line $D E$ and we infer that $E F D Q$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\angle E F D=\angle E Q D$; but now the points $F$ and $Q$ lie on the same side of $D E$, so that $E D F Q$ is a cyclic quadrilateral.  In either case we obtain the equality $\angle E F Q=\angle E D Q=\angle P A E$ which implies that $F Q \| A E$. So the triangles $E F Q$ and $P A E$ are either homothetic or parallel-congruent. More specifically, triangle $E F Q$ is the image of $P A E$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.  Let now $G$ be the point on the line $B C$ such that $E G \| P B$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $E G Q$ is the image of $P B E$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.  If $P E \neq Q E$, so that $f$ is a homothety with a centre $X$, then the lines $A F, P E, B G$-i.e. the lines $A D, P Q, B C$ - are concurrent at $X$. And since $P Q D A$ and $Q P B C$ are cyclic quadrilaterals, the equalities $X A \cdot X D=X P \cdot X Q=X B \cdot X C$ hold, showing that the quadrilateral $A B C D$ is cyclic.  Finally, if $P E=Q E$, so that $f$ is a translation, then $A D\|P Q\| B C$. Thus $P Q D A$ and $Q P B C$ are isosceles trapezoids. Then also $A B C D$ is an isosceles trapezoid, hence a cyclic quadrilateral.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-33.jpg?height=868&width=1256&top_left_y=1602&top_left_x=434)


Solution: Here is another way to reach the conclusion that the lines $A D, B C$ and $P Q$ are either concurrent or parallel. From the cyclic quadrilateral $P Q D A$ we get  $$ \angle P A D=180^{\circ}-\angle P Q D=\angle Q D E+\angle Q E D=\angle P A E+\angle Q E D . $$  Hence $\angle Q E D=\angle P A D-\angle P A E=\angle E A D$. This in view of the tangent-chord theorem means that the circumcircle of triangle $E A D$ is tangent to the line $P Q$ at $E$. Analogously, the circumcircle of triangle $E B C$ is tangent to $P Q$ at $E$.  Suppose that the line $A D$ intersects $P Q$ at $X$. Since $X E$ is tangent to the circle $(E A D)$, $X E^{2}=X A \cdot X D$. Also, $X A \cdot X D=X P \cdot X Q$ because $P, Q, D, A$ lie on a circle. Therefore $X E^{2}=X P \cdot X Q$.  It is not hard to see that this equation determines the position of the point $X$ on the line $P Q$ uniquely. Thus, if $B C$ also cuts $P Q$, say at $Y$, then the analogous equation for $Y$ yields $X=Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $A D\|P Q\| B C$, the concluding argument is the same as in the first solution.  It remains to eliminate the possibility that e.g. $A D$ meets $P Q$ at $X$ while $B C \| P Q$. Indeed, $Q P B C$ would then be an isosceles trapezoid and the angle equality $\angle P B E=\angle Q C E$ would force that $E$ is the midpoint of $P Q$. So the length of $X E$, which is the geometric mean of the lengths of $X P$ and $X Q$, should also be their arithmetic mean-impossible, as $X P \neq X Q$. The proof is now complete.  Comment. After reaching the conclusion that the circles $(E D A)$ and $(E B C)$ are tangent to $P Q$ one may continue as follows. Denote the circles $(P Q D A),(E D A),(E B C),(Q P B C)$ by $\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}$ respectively. Let $\ell_{i j}$ be the radical axis of the pair $\left(\omega_{i}, \omega_{j}\right)$ for $i<j$. As is well-known, the lines $\ell_{12}, \ell_{13}, \ell_{23}$ concur, possibly at infinity (let this be the meaning of the word concur in this comment). So do the lines $\ell_{12}, \ell_{14}, \ell_{24}$. Note however that $\ell_{23}$ and $\ell_{14}$ both coincide with the line $P Q$. Hence the pair $\ell_{12}, P Q$ is in both triples; thus the four lines $\ell_{12}, \ell_{13}, \ell_{24}$ and $P Q$ are concurrent.  Similarly, $\ell_{13}, \ell_{14}, \ell_{34}$ concur, $\ell_{23}, \ell_{24}, \ell_{34}$ concur, and since $\ell_{14}=\ell_{23}=P Q$, the four lines $\ell_{13}, \ell_{24}, \ell_{34}$ and $P Q$ are concurrent. The lines $\ell_{13}$ and $\ell_{24}$ are present in both quadruples, therefore all the lines $\ell_{i j}$ are concurrent. Hence the result.


In an acute triangle $A B C$ segments $B E$ and $C F$ are altitudes. Two circles passing through the points $A$ and $F$ are tangent to the line $B C$ at the points $P$ and $Q$ so that $B$ lies between $C$ and $Q$. Prove that the lines $P E$ and $Q F$ intersect on the circumcircle of triangle $A E F$.

Solution: To approach the desired result we need some information about the slopes of the lines $P E$ and $Q F$; this information is provided by formulas (1) and (2) which we derive below.  The tangents $B P$ and $B Q$ to the two circles passing through $A$ and $F$ are equal, as $B P^{2}=B A \cdot B F=B Q^{2}$. Consider the altitude $A D$ of triangle $A B C$ and its orthocentre $H$. From the cyclic quadrilaterals $C D F A$ and $C D H E$ we get $B A \cdot B F=B C \cdot B D=B E \cdot B H$. Thus $B P^{2}=B E \cdot B H$, or $B P / B H=B E / B P$, implying that the triangles $B P H$ and $B E P$ are similar. Hence  $$ \angle B P E=\angle B H P \text {. } $$  The point $P$ lies between $D$ and $C$; this follows from the equality $B P^{2}=B C \cdot B D$. In view of this equality, and because $B P=B Q$,  $$ D P \cdot D Q=(B P-B D) \cdot(B P+B D)=B P^{2}-B D^{2}=B D \cdot(B C-B D)=B D \cdot D C . $$  Also $A D \cdot D H=B D \cdot D C$, as is seen from the similar triangles $B D H$ and $A D C$. Combining these equalities we obtain $A D \cdot D H=D P \cdot D Q$. Therefore $D H / D P=D Q / D A$, showing that the triangles $H D P$ and $Q D A$ are similar. Hence $\angle H P D=\angle Q A D$, which can be rewritten as $\angle B P H=\angle B A D+\angle B A Q$. And since $B Q$ is tangent to the circumcircle of triangle $F A Q$,  $$ \angle B Q F=\angle B A Q=\angle B P H-\angle B A D . $$  From (1) and (2) we deduce  $$ \begin{aligned} & \angle B P E+\angle B Q F=(\angle B H P+\angle B P H)-\angle B A D=\left(180^{\circ}-\angle P B H\right)-\angle B A D \\ & =\left(90^{\circ}+\angle B C A\right)-\left(90^{\circ}-\angle A B C\right)=\angle B C A+\angle A B C=180^{\circ}-\angle C A B . \end{aligned} $$  Thus $\angle B P E+\angle B Q F<180^{\circ}$, which means that the rays $P E$ and $Q F$ meet. Let $S$ be the point of intersection. Then $\angle P S Q=180^{\circ}-(\angle B P E+\angle B Q F)=\angle C A B=\angle E A F$.  If $S$ lies between $P$ and $E$ then $\angle P S Q=180^{\circ}-\angle E S F$; and if $E$ lies between $P$ and $S$ then $\angle P S Q=\angle E S F$. In either case the equality $\angle P S Q=\angle E A F$ which we have obtained means that $S$ lies on the circumcircle of triangle $A E F$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-35.jpg?height=748&width=1305&top_left_y=1970&top_left_x=410)


Solution: Let $H$ be the orthocentre of triangle $A B C$ and let $\omega$ be the circle with diameter $A H$, passing through $E$ and $F$. Introduce the points of intersection of $\omega$ with the following lines emanating from $P: P A \cap \omega=\{A, U\}, P H \cap \omega=\{H, V\}, P E \cap \omega=\{E, S\}$. The altitudes of triangle $A H P$ are contained in the lines $A V, H U, B C$, meeting at its orthocentre $Q^{\prime}$.  By Pascal's theorem applied to the (tied) hexagon $A E S F H V$, the points $A E \cap F H=C$, $E S \cap H V=P$ and $S F \cap V A$ are collinear, so $F S$ passes through $Q^{\prime}$.  Denote by $\omega_{1}$ and $\omega_{2}$ the circles with diameters $B C$ and $P Q^{\prime}$, respectively. Let $D$ be the foot of the altitude from $A$ in triangle $A B C$. Suppose that $A D$ meets the circles $\omega_{1}$ and $\omega_{2}$ at the respective points $K$ and $L$.  Since $H$ is the orthocentre of $A B C$, the triangles $B D H$ and $A D C$ are similar, and so $D A \cdot D H=D B \cdot D C=D K^{2}$; the last equality holds because $B K C$ is a right triangle. Since $H$ is the orthocentre also in triangle $A Q^{\prime} P$, we analogously have $D L^{2}=D A \cdot D H$. Therefore $D K=D L$ and $K=L$.  Also, $B D \cdot B C=B A \cdot B F$, from the similar triangles $A B D, C B F$. In the right triangle $B K C$ we have $B K^{2}=B D \cdot B C$. Hence, and because $B A \cdot B F=B P^{2}=B Q^{2}$ (by the definition of $P$ and $Q$ in the problem statement), we obtain $B K=B P=B Q$. It follows that $B$ is the centre of $\omega_{2}$ and hence $Q^{\prime}=Q$. So the lines $P E$ and $Q F$ meet at the point $S$ lying on the circumcircle of triangle $A E F$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-36.jpg?height=743&width=1328&top_left_y=1202&top_left_x=341)  Comment 1. If $T$ is the point defined by $P F \cap \omega=\{F, T\}$, Pascal's theorem for the hexagon $A F T E H V$ will analogously lead to the conclusion that the line $E T$ goes through $Q^{\prime}$. In other words, the lines $P F$ and $Q E$ also concur on $\omega$.  Comment 2. As is known from algebraic geometry, the points of the circle $\omega$ form a commutative groups with the operation defined as follows. Choose any point $0 \in \omega$ (to be the neutral element of the group) and a line $\ell$ exterior to the circle. For $X, Y \in \omega$, draw the line from the point $X Y \cap \ell$ through 0 to its second intersection with $\omega$ and define this point to be $X+Y$.  In our solution we have chosen $H$ to be the neutral element in this group and line $B C$ to be $\ell$. The fact that the lines $A V, H U, E T, F S$ are concurrent can be deduced from the identities $A+A=0$, $F=E+A, V=U+A=S+E=T+F$.  Comment 3. The problem was submitted in the following equivalent formulation:  Let $B E$ and $C F$ be altitudes of an acute triangle $A B C$. We choose $P$ on the side $B C$ and $Q$ on the extension of $C B$ beyond $B$ such that $B Q^{2}=B P^{2}=B F \cdot A B$. If $Q F$ and $P E$ intersect at $S$, prove that $E S A F$ is cyclic.


Let $k$ and $n$ be integers with $0 \leq k \leq n-2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersection points of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$.  A point $X \in I$ is colored red if the open line segment $O X$ intersects at most $k$ lines in $L$. Prove that $I$ contains at least $\frac{1}{2}(k+1)(k+2)$ red points.

Solution: There are at least $\frac{1}{2}(k+1)(k+2)$ points in the intersection set $I$ in view of the condition $n \geq k+2$.  For each point $P \in I$, define its order as the number of lines that intersect the open line segment $O P$. By definition, $P$ is red if its order is at most $k$. Note that there is always at least one point $X \in I$ of order 0 . Indeed, the lines in $L$ divide the plane into regions, bounded or not, and $O$ belongs to one of them. Clearly any corner of this region is a point of $I$ with order 0 .  Claim. Suppose that two points $P, Q \in I$ lie on the same line of $L$, and no other line of $L$ intersects the open line segment $P Q$. Then the orders of $P$ and $Q$ differ by at most 1 .  Proof. Let $P$ and $Q$ have orders $p$ and $q$, respectively, with $p \geq q$. Consider triangle $O P Q$. Now $p$ equals the number of lines in $L$ that intersect the interior of side $O P$. None of these lines intersects the interior of side $P Q$, and at most one can pass through $Q$. All remaining lines must intersect the interior of side $O Q$, implying that $q \geq p-1$. The conclusion follows.  We prove the main result by induction on $k$. The base $k=0$ is clear since there is a point of order 0 which is red. Assuming the statement true for $k-1$, we pass on to the inductive step. Select a point $P \in I$ of order 0 , and consider one of the lines $\ell \in L$ that pass through $P$. There are $n-1$ intersection points on $\ell$, one of which is $P$. Out of the remaining $n-2$ points, the $k$ closest to $P$ have orders not exceeding $k$ by the Claim. It follows that there are at least $k+1$ red points on $\ell$.  Let us now consider the situation with $\ell$ removed (together with all intersection points it contains). By hypothesis of induction, there are at least $\frac{1}{2} k(k+1)$ points of order not exceeding $k-1$ in the resulting configuration. Restoring $\ell$ back produces at most one new intersection point on each line segment joining any of these points to $O$, so their order is at most $k$ in the original configuration. The total number of points with order not exceeding $k$ is therefore at least $(k+1)+\frac{1}{2} k(k+1)=\frac{1}{2}(k+1)(k+2)$. This completes the proof.  Comment. The steps of the proof can be performed in reverse order to obtain a configuration of $n$ lines such that equality holds simultaneously for all $0 \leq k \leq n-2$. Such a set of lines is illustrated in the Figure.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-37.jpg?height=400&width=1240&top_left_y=2059&top_left_x=448)


There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that  $$ \angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ} $$  if and only if the diagonals $A C$ and $B D$ are perpendicular.

Solution: For a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $A K P N$ and $D N P M$ give  $$ \angle P A B+\angle P D C=\angle P N K+\angle P N M=\angle K N M . $$  Analogously, $\angle P B C+\angle P A D=\angle L K N$ and $\angle P C D+\angle P B A=\angle M L K$. Hence the equalities (1) imply $\angle K N M=\angle L K N=\angle M L K=90^{\circ}$, so that $K L M N$ is a rectangle. The converse also holds true, provided that $K, L, M, N$ are interior to sides $A B, B C, C D, D A$.  (i) Suppose that there exists a point $P$ in $A B C D$ such that $K L M N$ is a rectangle. We show that $A C$ and $B D$ are parallel to the respective sides of $K L M N$.  Let $O_{A}$ and $O_{C}$ be the circumcentres of the cyclic quadrilaterals $A K P N$ and $C M P L$. Line $O_{A} O_{C}$ is the common perpendicular bisector of $L M$ and $K N$, therefore $O_{A} O_{C}$ is parallel to $K L$ and $M N$. On the other hand, $O_{A} O_{C}$ is the midline in the triangle $A C P$ that is parallel to $A C$. Therefore the diagonal $A C$ is parallel to the sides $K L$ and $M N$ of the rectangle. Likewise, $B D$ is parallel to $K N$ and $L M$. Hence $A C$ and $B D$ are perpendicular.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-38.jpg?height=751&width=1100&top_left_y=1395&top_left_x=455)  (ii) Suppose that $A C$ and $B D$ are perpendicular and meet at $R$. If $A B C D$ is a rhombus, $P$ can be chosen to be its centre. So assume that $A B C D$ is not a rhombus, and let $B R<D R$ without loss of generality.  Denote by $U_{A}$ and $U_{C}$ the circumcentres of the triangles $A B D$ and $C D B$, respectively. Let $A V_{A}$ and $C V_{C}$ be the diameters through $A$ and $C$ of the two circumcircles. Since $A R$ is an altitude in triangle $A D B$, lines $A C$ and $A V_{A}$ are isogonal conjugates, i. e. $\angle D A V_{A}=\angle B A C$. Now $B R<D R$ implies that ray $A U_{A}$ lies in $\angle D A C$. Similarly, ray $C U_{C}$ lies in $\angle D C A$. Both diameters $A V_{A}$ and $C V_{C}$ intersect $B D$ as the angles at $B$ and $D$ of both triangles are acute. Also $U_{A} U_{C}$ is parallel to $A C$ as it is the perpendicular bisector of $B D$. Hence $V_{A} V_{C}$ is parallel to $A C$, too. We infer that $A V_{A}$ and $C V_{C}$ intersect at a point $P$ inside triangle $A C D$, hence inside $A B C D$.  Construct points $K, L, M, N, O_{A}$ and $O_{C}$ in the same way as in the introduction. It follows from the previous paragraph that $K, L, M, N$ are interior to the respective sides. Now $O_{A} O_{C}$ is a midline in triangle $A C P$ again. Therefore lines $A C, O_{A} O_{C}$ and $U_{A} U_{C}$ are parallel.  The cyclic quadrilateral $A K P N$ yields $\angle N K P=\angle N A P$. Since $\angle N A P=\angle D A U_{A}=$ $\angle B A C$, as specified above, we obtain $\angle N K P=\angle B A C$. Because $P K$ is perpendicular to $A B$, it follows that $N K$ is perpendicular to $A C$, hence parallel to $B D$. Likewise, $L M$ is parallel to $B D$.  Consider the two homotheties with centres $A$ and $C$ which transform triangles $A B D$ and $C D B$ into triangles $A K N$ and $C M L$, respectively. The images of points $U_{A}$ and $U_{C}$ are $O_{A}$ and $O_{C}$, respectively. Since $U_{A} U_{C}$ and $O_{A} O_{C}$ are parallel to $A C$, the two ratios of homothety are the same, equal to $\lambda=A N / A D=A K / A B=A O_{A} / A U_{A}=C O_{C} / C U_{C}=C M / C D=C L / C B$. It is now straightforward that $D N / D A=D M / D C=B K / B A=B L / B C=1-\lambda$. Hence $K L$ and $M N$ are parallel to $A C$, implying that $K L M N$ is a rectangle and completing the proof.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-39.jpg?height=759&width=1122&top_left_y=954&top_left_x=501)


Solution: For a point $P$ distinct from $A, B, C, D$, let circles $(A P D)$ and $(B P C)$ intersect again at $Q(Q=P$ if the circles are tangent). Next, let circles $(A Q B)$ and $(C Q D)$ intersect again at $R$. We show that if $P$ lies in $A B C D$ and satisfies (1) then $A C$ and $B D$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\measuredangle(U V, X Y)$ denote the angle of counterclockwise rotation that makes line $U V$ parallel to line $X Y$. Recall that four noncollinear points $U, V, X, Y$ are concyclic if and only if $\measuredangle(U X, V X)=\measuredangle(U Y, V Y)$.  The definitions of points $P, Q$ and $R$ imply  $$ \begin{aligned} \measuredangle(A R, B R) & =\measuredangle(A Q, B Q)=\measuredangle(A Q, P Q)+\measuredangle(P Q, B Q)=\measuredangle(A D, P D)+\measuredangle(P C, B C), \\ \measuredangle(C R, D R) & =\measuredangle(C Q, D Q)=\measuredangle(C Q, P Q)+\measuredangle(P Q, D Q)=\measuredangle(C B, P B)+\measuredangle(P A, D A), \\ \measuredangle(B R, C R) & =\measuredangle(B R, R Q)+\measuredangle(R Q, C R)=\measuredangle(B A, A Q)+\measuredangle(D Q, C D) \\ & =\measuredangle(B A, A P)+\measuredangle(A P, A Q)+\measuredangle(D Q, D P)+\measuredangle(D P, C D) \\ & =\measuredangle(B A, A P)+\measuredangle(D P, C D) . \end{aligned} $$  Observe that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles $(A R B)$ and $(C R D)$, and then define $P$ as the second intersection of circles $(A Q D)$ and $(B Q C)$. The equalities above will still hold true.  Assume in addition that $P$ is interior to $A B C D$. Then  $$ \begin{gathered} \measuredangle(A D, P D)=\angle P D A, \measuredangle(P C, B C)=\angle P C B, \measuredangle(C B, P B)=\angle P B C, \measuredangle(P A, D A)=\angle P A D, \\ \measuredangle(B A, A P)=\angle P A B, \measuredangle(D P, C D)=\angle P D C . \end{gathered} $$  (i) Suppose that $P$ lies in $A B C D$ and satisfies (1). Then $\measuredangle(A R, B R)=\angle P D A+\angle P C B=90^{\circ}$ and similarly $\measuredangle(B R, C R)=\measuredangle(C R, D R)=90^{\circ}$. It follows that $R$ is the common point of lines $A C$ and $B D$, and that these lines are perpendicular.  (ii) Suppose that $A C$ and $B D$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$ ) lies in $A B C D$. This is enough to finish the solution, because then the angle equalities above will imply (1).  One can assume that $Q$, the second common point of circles $(A B R)$ and $(C D R)$, lies in $\angle A R D$. Then in fact $Q$ lies in triangle $A D R$ as angles $A Q R$ and $D Q R$ are obtuse. Hence $\angle A Q D$ is obtuse, too, so that $B$ and $C$ are outside circle $(A D Q)(\angle A B D$ and $\angle A C D$ are acute).  Now $\angle C A B+\angle C D B=\angle B Q R+\angle C Q R=\angle C Q B$ implies $\angle C A B<\angle C Q B$ and $\angle C D B<$ $\angle C Q B$. Hence $A$ and $D$ are outside circle $(B C Q)$. In conclusion, the second common point $P$ of circles $(A D Q)$ and $(B C Q)$ lies on their arcs $A D Q$ and $B C Q$.  We can assume that $P$ lies in $\angle C Q D$. Since  $$ \begin{gathered} \angle Q P C+\angle Q P D=\left(180^{\circ}-\angle Q B C\right)+\left(180^{\circ}-\angle Q A D\right)= \\ =360^{\circ}-(\angle R B C+\angle Q B R)-(\angle R A D-\angle Q A R)=360^{\circ}-\angle R B C-\angle R A D>180^{\circ}, \end{gathered} $$  point $P$ lies in triangle $C D Q$, and hence in $A B C D$. The proof is complete.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-40.jpg?height=757&width=1008&top_left_y=1489&top_left_x=501)


Let $A B C D$ be a convex quadrilateral with $A B \neq B C$. Denote by $\omega_{1}$ and $\omega_{2}$ the incircles of triangles $A B C$ and $A D C$. Suppose that there exists a circle $\omega$ inscribed in angle $A B C$, tangent to the extensions of line segments $A D$ and $C D$. Prove that the common external tangents of $\omega_{1}$ and $\omega_{2}$ intersect on $\omega$.

Solution: The proof below is based on two known facts.  Lemma 1. Given a convex quadrilateral $A B C D$, suppose that there exists a circle which is inscribed in angle $A B C$ and tangent to the extensions of line segments $A D$ and $C D$. Then $A B+A D=C B+C D$.  Proof. The circle in question is tangent to each of the lines $A B, B C, C D, D A$, and the respective points of tangency $K, L, M, N$ are located as with circle $\omega$ in the figure. Then  $$ A B+A D=(B K-A K)+(A N-D N), \quad C B+C D=(B L-C L)+(C M-D M) . $$  Also $B K=B L, D N=D M, A K=A N, C L=C M$ by equalities of tangents. It follows that $A B+A D=C B+C D$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_e32cef0f3c5b05a0a6b1g-41.jpg?height=1068&width=1079&top_left_y=1156&top_left_x=523)  For brevity, in the sequel we write "excircle $A C$ " for the excircle of a triangle with side $A C$ which is tangent to line segment $A C$ and the extensions of the other two sides.  Lemma 2. The incircle of triangle $A B C$ is tangent to its side $A C$ at $P$. Let $P P^{\prime}$ be the diameter of the incircle through $P$, and let line $B P^{\prime}$ intersect $A C$ at $Q$. Then $Q$ is the point of tangency of side $A C$ and excircle $A C$.  Proof. Let the tangent at $P^{\prime}$ to the incircle $\omega_{1}$ meet $B A$ and $B C$ at $A^{\prime}$ and $C^{\prime}$. Now $\omega_{1}$ is the excircle $A^{\prime} C^{\prime}$ of triangle $A^{\prime} B C^{\prime}$, and it touches side $A^{\prime} C^{\prime}$ at $P^{\prime}$. Since $A^{\prime} C^{\prime} \| A C$, the homothety with centre $B$ and ratio $B Q / B P^{\prime}$ takes $\omega_{1}$ to the excircle $A C$ of triangle $A B C$. Because this homothety takes $P^{\prime}$ to $Q$, the lemma follows.  Recall also that if the incircle of a triangle touches its side $A C$ at $P$, then the tangency point $Q$ of the same side and excircle $A C$ is the unique point on line segment $A C$ such that $A P=C Q$.  We pass on to the main proof. Let $\omega_{1}$ and $\omega_{2}$ touch $A C$ at $P$ and $Q$, respectively; then $A P=(A C+A B-B C) / 2, C Q=(C A+C D-A D) / 2$. Since $A B-B C=C D-A D$ by Lemma 1, we obtain $A P=C Q$. It follows that in triangle $A B C$ side $A C$ and excircle $A C$ are tangent at $Q$. Likewise, in triangle $A D C$ side $A C$ and excircle $A C$ are tangent at $P$. Note that $P \neq Q$ as $A B \neq B C$.  Let $P P^{\prime}$ and $Q Q^{\prime}$ be the diameters perpendicular to $A C$ of $\omega_{1}$ and $\omega_{2}$, respectively. Then Lemma 2 shows that points $B, P^{\prime}$ and $Q$ are collinear, and so are points $D, Q^{\prime}$ and $P$.  Consider the diameter of $\omega$ perpendicular to $A C$ and denote by $T$ its endpoint that is closer to $A C$. The homothety with centre $B$ and ratio $B T / B P^{\prime}$ takes $\omega_{1}$ to $\omega$. Hence $B, P^{\prime}$ and $T$ are collinear. Similarly, $D, Q^{\prime}$ and $T$ are collinear since the homothety with centre $D$ and ratio $-D T / D Q^{\prime}$ takes $\omega_{2}$ to $\omega$.  We infer that points $T, P^{\prime}$ and $Q$ are collinear, as well as $T, Q^{\prime}$ and $P$. Since $P P^{\prime} \| Q Q^{\prime}$, line segments $P P^{\prime}$ and $Q Q^{\prime}$ are then homothetic with centre $T$. The same holds true for circles $\omega_{1}$ and $\omega_{2}$ because they have $P P^{\prime}$ and $Q Q^{\prime}$ as diameters. Moreover, it is immediate that $T$ lies on the same side of line $P P^{\prime}$ as $Q$ and $Q^{\prime}$, hence the ratio of homothety is positive. In particular $\omega_{1}$ and $\omega_{2}$ are not congruent.  In summary, $T$ is the centre of a homothety with positive ratio that takes circle $\omega_{1}$ to circle $\omega_{2}$. This completes the solution, since the only point with the mentioned property is the intersection of the the common external tangents of $\omega_{1}$ and $\omega_{2}$.


Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a, b, c$ are integers (not necessarily positive) satisfying the equations  $$ a^{n}+p b=b^{n}+p c=c^{n}+p a, $$  then $a=b=c$.

Solution: If two of $a, b, c$ are equal, it is immediate that all the three are equal. So we may assume that $a \neq b \neq c \neq a$. Subtracting the equations we get $a^{n}-b^{n}=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield  $$ \frac{a^{n}-b^{n}}{a-b} \cdot \frac{b^{n}-c^{n}}{b-c} \cdot \frac{c^{n}-a^{n}}{c-a}=-p^{3} . $$  If $n$ is odd then the differences $a^{n}-b^{n}$ and $a-b$ have the same sign and the product on the left is positive, while $-p^{3}$ is negative. So $n$ must be even.  Let $d$ be the greatest common divisor of the three differences $a-b, b-c, c-a$, so that $a-b=d u, b-c=d v, \quad c-a=d w ; \operatorname{gcd}(u, v, w)=1, u+v+w=0$.  From $a^{n}-b^{n}=-p(b-c)$ we see that $(a-b) \mid p(b-c)$, i.e., $u \mid p v$; and cyclically $v|p w, w| p u$. As $\operatorname{gcd}(u, v, w)=1$ and $u+v+w=0$, at most one of $u, v, w$ can be divisible by $p$. Supposing that the prime $p$ does not divide any one of them, we get $u|v, v| w, w \mid u$, whence $|u|=|v|=|w|=1$; but this quarrels with $u+v+w=0$.  Thus $p$ must divide exactly one of these numbers. Let e.g. $p \mid u$ and write $u=p u_{1}$. Now we obtain, similarly as before, $u_{1}|v, v| w, w \mid u_{1}$ so that $\left|u_{1}\right|=|v|=|w|=1$. The equation $p u_{1}+v+w=0$ forces that the prime $p$ must be even; i.e. $p=2$. Hence $v+w=-2 u_{1}= \pm 2$, implying $v=w(= \pm 1)$ and $u=-2 v$. Consequently $a-b=-2(b-c)$.  Knowing that $n$ is even, say $n=2 k$, we rewrite the equation $a^{n}-b^{n}=-p(b-c)$ with $p=2$ in the form  $$ \left(a^{k}+b^{k}\right)\left(a^{k}-b^{k}\right)=-2(b-c)=a-b . $$  The second factor on the left is divisible by $a-b$, so the first factor $\left(a^{k}+b^{k}\right)$ must be $\pm 1$. Then exactly one of $a$ and $b$ must be odd; yet $a-b=-2(b-c)$ is even. Contradiction ends the proof.


Solution: The beginning is as in the first solution. Assuming that $a, b, c$ are not all equal, hence are all distinct, we derive equation (1) with the conclusion that $n$ is even. Write $n=2 k$.  Suppose that $p$ is odd. Then the integer  $$ \frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2} b+\cdots+b^{n-1} $$  which is a factor in (1), must be odd as well. This sum of $n=2 k$ summands is odd only if $a$ and $b$ have different parities. The same conclusion holding for $b, c$ and for $c$, $a$, we get that $a, b, c, a$ alternate in their parities, which is clearly impossible.  Thus $p=2$. The original system shows that $a, b, c$ must be of the same parity. So we may divide (1) by $p^{3}$, i.e. $2^{3}$, to obtain the following product of six integer factors:  $$ \frac{a^{k}+b^{k}}{2} \cdot \frac{a^{k}-b^{k}}{a-b} \cdot \frac{b^{k}+c^{k}}{2} \cdot \frac{b^{k}-c^{k}}{b-c} \cdot \frac{c^{k}+a^{k}}{2} \cdot \frac{c^{k}-a^{k}}{c-a}=-1 . $$  Each one of the factors must be equal to $\pm 1$. In particular, $a^{k}+b^{k}= \pm 2$. If $k$ is even, this becomes $a^{k}+b^{k}=2$ and yields $|a|=|b|=1$, whence $a^{k}-b^{k}=0$, contradicting (2).  Let now $k$ be odd. Then the sum $a^{k}+b^{k}$, with value $\pm 2$, has $a+b$ as a factor. Since $a$ and $b$ are of the same parity, this means that $a+b= \pm 2$; and cyclically, $b+c= \pm 2, c+a= \pm 2$. In some two of these equations the signs must coincide, hence some two of $a, b, c$ are equal. This is the desired contradiction.  Comment. Having arrived at the equation (1) one is tempted to write down all possible decompositions of $-p^{3}$ (cube of a prime) into a product of three integers. This leads to cumbersome examination of many cases, some of which are unpleasant to handle. One may do that just for $p=2$, having earlier in some way eliminated odd primes from consideration.  However, the second solution shows that the condition of $p$ being a prime is far too strong. What is actually being used in that solution, is that $p$ is either a positive odd integer or $p=2$.