# $1^{\text {sT }}$ European Mathematical Cup <br> $24^{\text {th }}$ November 2012-1 $1^{\text {st }}$ December 2012 <br> Junior Category 

## Problems and solutions

Problem 1. Let $A B C$ be a triangle and $Q$ a point on the internal angle bisector of $\angle B A C$. Circle $\omega_{1}$ is circumscribed to triangle $B A Q$ and intersects the segment $A C$ in point $P \neq C$. Circle $\omega_{2}$ is circumscribed to the triangle $C Q P$. Radius of the cirlce $\omega_{1}$ is larger than the radius of $\omega_{2}$. Circle centered at $Q$ with radius $Q A$ intersects the circle $\omega_{1}$ in points $A$ and $A_{1}$. Circle centered at $Q$ with radius $Q C$ intersects $\omega_{1}$ in points $C_{1}$ and $C_{2}$. Prove $\angle A_{1} B C_{1}=\angle C_{2} P A$.

(Matija Bucić)

Solution. From the conditions in the problem we have $\left|Q C_{1}\right|=\left|Q C_{2}\right|$ and $|Q A|=\left|Q A_{1}\right|$. Also as $Q$ lies on the internal angle bisector of $\angle C A B$ we have $\angle P A Q=\angle Q A B \Longrightarrow|Q P|=|Q B|$.

Now noting from this that pairs of points $A$ and $A_{1}, C_{1}$ and $C_{2}, B$ and $P$ are symmetric in line $Q S_{1}$, where $S_{1}$ is the center of $\omega_{1}$. We can directly conclude $\angle A_{1} B C_{1}=\angle A P C_{2}$ as these is the image of the angle in symmetry.

This way we have avoided checking many cases but there are many ways to prove this problem.

Problem 2. Let $S$ be the set of positive integers. For any $a$ and $b$ in the set we have $G C D(a, b)>1$. For any $a, b$ and $c$ in the set we have $G C D(a, b, c)=1$. Is it possible that $S$ has 2012 elements?

$G C D(x, y)$ and $G C D(x, y, z)$ stand for the greatest common divisor of the numbers $x$ and $y$ and numbers $x, y$ and $z$ respectively.

(Ognjen Stipetić)

Solution. There is such a set.

We will construct it in the following way: Let $a_{1}, a_{2}, \ldots a_{2012}$ equal to 1 in the begining. Then we take $\frac{2012 \cdot 2011}{2}$ different prime numbers, and assign a different prime to every pair $a_{i}, a_{j}$ (where $i \neq j$ ) and multiply them with this assigned number. (I.e. for the set of 4 elements we can take $2,3,5,7,11,13$, so $S$ would be $\{2 \cdot 3 \cdot 5,2 \cdot 7 \cdot 11,3 \cdot 7 \cdot 13,5 \cdot 11 \cdot 13\}$. The construction works as we have multiplied any pair of numbers with some prime so the condition $\operatorname{gcd}(a, b)>1$ is satisfied for all $a, b$. As well as each prime divides exactly 2 primes so no three numbers $a, b, c$ can have $\operatorname{gcd}(a, b, c)>1$.

Problem 3. Do there exist positive real numbers $x, y$ and $z$ such that

$$
\begin{gathered}
x^{4}+y^{4}+z^{4}=13 \\
x^{3} y^{3} z+y^{3} z^{3} x+z^{3} x^{3} y=6 \sqrt{3} \\
x^{3} y z+y^{3} z x+z^{3} x y=5 \sqrt{3} ?
\end{gathered}
$$

(Matko Ljulj)

Solution. Let's assume that such $x, y, z$ exist. Let $a=x^{2}, b=y^{2}, c=z^{2}$. As well, let $A=a+b+c, B=a b+b c+c a$, $C=a b c$. The upper system can be rewritten as:

$$
\begin{gathered}
a^{2}+b^{2}+c^{2}=13 \Longrightarrow(a+b+c)^{2}-2(a b+b c+c a)=13 \Longrightarrow A^{2}-2 B=13 \\
x y z\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}\right)=6 \sqrt{3} \Longrightarrow \sqrt{C} B=6 \sqrt{3} \\
x y z\left(x^{2}+y^{2}+z^{2}\right)=5 \sqrt{3} \Longrightarrow \sqrt{C} A=5 \sqrt{3} .
\end{gathered}
$$

We can note that $a, b$ and $c$ are positive reals (They are not negaitve from the definition; and as $\sqrt{C} B=6 \sqrt{3}$ they are not 0 ).

When we cancel out $\sqrt{C}$ from the second and third equation we get $5 B=6 A$. When we express $B$ in terms of $A$ and put int the first equation we get a quadratic equation

$$
A^{2}-\frac{12}{5} A-13=0
$$

with solutions 5 and $-\frac{13}{5}$. As $a, b$ and $c$ are positive reals, and the sum must be positive so their sum is poistive real number as well. So $A=5 \Longrightarrow B=6 \Longrightarrow C=3$.

By $A M$ - $G M$ inequality we get

$$
\begin{gathered}
\frac{a b+b c+c a}{3} \geqslant \sqrt[3]{a b \cdot b c \cdot c a} \\
\Longleftrightarrow \frac{B}{3} \geqslant \sqrt[3]{C^{2}} \\
\Longleftrightarrow \frac{6}{3} \geqslant \sqrt[3]{9} /^{3} \\
\Longleftrightarrow 8 \geqslant 9
\end{gathered}
$$

so we reached a contradiction, thus such $x, y, z$ don't exist.

Problem 4. Let $k$ be a positive integer. At the European Chess Cup every pair of players played a game in which somebody won (there were no draws). For any $k$ players there was a player against whom they all lost, and the number of players was the least possible for such $k$. Is it possible that at the Closing Ceremony all the participants were seated at the round table in such a way that every participant was seated next to both a person he won against and a person he lost against.

(Matija Bucić)

Solution. The answer is yes.

In this problem we could use graph theory terminology but as this problem was intended for younger students we shall avoid mentioning any specific graph theory terms.

Let's take the largest number of participants whom we can seat around the table as desired. If we have seated all the participants we are done. Otherwise there is a person not seated at the table. As well there is at least one person seated at the table so let's name it $a$.

WLOG we can assume that for each person seated at the table to his right there is a person he won against and to his left a person he lost against.

Denote by $W$ the set of people who won against person $a$, and are not seated at the table. Similarly, let $L$ denote the set of all people who lost against $a$ and are not seated at the table.

Let's consider any person $p$ from $W$. If person $p$ lost against the left neighbour of $a$, then we could seat $p$ in between $a$ and his (former) left neighbour, which is a contradiction with the assumption that we have seated the maximal possible number of people. So $p$ won against the left neighbour of $a$. Using similar deduction we conclude that $p$ won against the next left neighbour as well etc. So $p$ must have won against everybody seated at the table.

In the same way if we consider any person $q$ from $L$ and consider the right neighbour of $a$, we can conclude that $q$ lost against every person seated at the table.

If some person $r$ from $W$ lost against some person $s$ in $L$, then instead of seating $a$ we can seat $s$ and $r$ respectively by which we would reach a contradiction to the number of people seated being maximal.

So we conclude that all the people in $W$ won against all people not in $W$ and all the people in $L$ lost against all people not in $L$.

As there is a someone who is not seated either $W$ or $L$ is non-empty. If $W$ is non-empty, we can consider the set $W$ as an independent chess cup. It is a cup with smaller number of participants but still satisfying problem conditions which would be the contradiction with the fact that our starting cup is the smallest such cup.

As well if $L$ is non-empty, the smaller cup made by people seated at the table and people in $W$ also satisfies the problem conditions and gives us a contradiction.

So the only possibility is that both $W$ and $L$ are empty so indeed it is possible to seat everyone at such table.

## $1^{\text {ST }}$ EURopeAn Mathematical CUP $24^{\text {th }}$ November 2012-1 ${ }^{\text {st }}$ December 2012 <br> Senior Category

## Problems and solutions

Problem 1. Find all positive integers $a, b, n$ and prime numbers $p$ that satisfy

$$
a^{2013}+b^{2013}=p^{n}
$$

(Matija Bucić)

First solution. Let's denote $d=D(a, b), x=\frac{a}{d}, y=\frac{b}{d}$. With this we get

$$
d^{2013}\left(a^{2013}+b^{2013}\right)=p^{n}
$$

So $d$ must be a power of $p$, so let $d=p^{k}, k \in \mathbb{N}_{0}$. We can divide the equality by $p^{2013 k}$. Now let's denote $m=$ $n-2013 k, A=x^{671}, B=y^{671}$. So we get

$$
A^{3}+B^{3}=p^{m}
$$

and after factorisation

$$
(A+B)\left(A^{2}-A B+B^{2}\right)=p^{m}
$$

(From the definition, $A$ and $B$ are coprime.)

Let's observe the case when some factor is 1: $A+B=1$ is impossible as both $A$ and $B$ are positive integers. And $A^{2}-A B+B^{2}=1 \Leftrightarrow(A-B)^{2}+A B=1 \Leftrightarrow A=B=1$, so we get a solution $a=b=2^{k}, n=2013 k+1, p=2, \forall k \in \mathbb{N}_{0}$. If both factors are larger than 1 we have

$$
\begin{gathered}
p \mid A+B \\
p \mid A^{2}-A B+B^{2}=(A+B)^{2}-3 A B \\
\Longrightarrow p \mid 3 A B
\end{gathered}
$$

If $p \mid A B$, in accordance with $p \mid A+B$ we get $p \mid A$ and $p \mid B$, which is in contradiction with $A$ and $B$ being coprime. So, $p \mid 3 \Longrightarrow p=3$.

Now we are left with 2 cases:

- First case: $A^{2}-A B+B^{2}=3 \Leftrightarrow(A-B)^{2}+A B=3$ - so the only possible solutions are $A=2, B=1 \mathrm{i} A=1, B=2$, but this turns out not to be a solution as $2=x^{671}$ does not have a solution in positive integers.
- Second case: $3^{2} \mid A^{2}-A B+B^{2}$ - then we have:

$$
\begin{gathered}
3\left|A+B \Longrightarrow 3^{2}\right|(A+B)^{2} \\
3^{2} \mid A^{2}-A B+B^{2}=(A+B)^{2}-3 A B \\
\Longrightarrow 3^{2} \mid 3 A B \\
\Longrightarrow 3 \mid A B
\end{gathered}
$$

And as we have already commented the case $p \nmid A B \Longrightarrow$ doesn't have any solutions.

So all the solutions are given by

$$
a=b=2^{k}, n=2013 k+1, p=2, \forall k \in \mathbb{N}_{0}
$$

Second solution. As in the first solution, we take the highest common factor of $a$ and $b$ (which must be of the form $\left.p^{k}\right)$. Factorising the given equality we get

$$
(x+y)\left(x^{2012}-x^{2011} y+x^{2010} y^{2}-\cdots-x y^{2011}+y^{2012}\right)=p^{m}
$$

(We're using the same notation as in the first solution.) Denote the right hand side factor by $A$. As $x$ and $y$ are natural numbers, we have $x+y>1 \Longrightarrow p \mid x+y$. So $p \nmid x$ and $p \nmid y$ (as $x$ and $y$ are coprime). Now by applying LTE (Lifting the Exponent Lemma):

$$
\nu_{p}\left(x^{2013}+y^{2013}\right)=\nu_{p}(x+y)+\nu_{p}(2013)
$$

Now we know $\nu_{p}(2013)=0$ fo all primes $p$ except $3,11,61$, and in the remaining cases $\nu_{p}(2013)=1$. Note $A=1$ and $(x, y)=(1,1)$ and $A>61$ for $(x, y) \neq(1,1)$. This inequality holds because for $(x, y) \neq(1,1)$ (WLOG $x \geqslant y)$, we can write $A$ as

$$
x^{2011}(x-y)+x^{2009} y^{2}(x-y)+\cdots+x y^{2010}(x-y)+y^{2012}
$$

which is greater than 61 in cases $x>y$ and $y \neq 1$.

- If $\nu_{p}(2013)=1 \Longrightarrow \nu_{p}(A)=1 \Longrightarrow A \in\{3,11,61\}$ which is clearly impossible.
- If $\nu_{p}(2013)=0 \Longrightarrow \nu_{p}(A)=0 \Longrightarrow A=1 \Longrightarrow(x, y)=(1,1)$, so we get a solution

$$
a=b=2^{k}, n=2013 k+1, p=2, \forall k \in \mathbb{N}_{0}
$$

Problem 2. Let $A B C$ be an acute triangle with orthocenter $H$. Segments $A H$ and $C H$ intersect segments $B C$ and $A B$ in points $A_{1}$ and $C_{1}$ respectively. The segments $B H$ and $A_{1} C_{1}$ meet at point $D$. Let $P$ be the midpoint of the segment $B H$. Let $D^{\prime}$ be the reflection of the point $D$ in $A C$. Prove that quadrilateral $A P C D^{\prime}$ is cyclic.

(Matko Ljulj)

First solution. We shall prove that $D$ is the orthocenter of triangle $A P C$. From that the problem statement follows as

$$
\begin{gathered}
\angle A D^{\prime} C=\angle A D C=180^{\circ}-\angle D A C-\angle D C A=\left(90^{\circ}-\angle D A C\right)+\left(90^{\circ}-\angle D C A\right)= \\
=\angle P C A+\angle P A C=180^{\circ}-\angle A P C
\end{gathered}
$$

We can note that quadrilateral $B A_{1} H C_{1}$ is cyclic. Lines $B A_{1}$ and $C_{1} H$ intersect in $C$, lines $B C_{1}$ and $A_{1} H$ intersect in $A$, lines $B H$ and $C_{1} A_{1}$ intersect in $D$, and point $P$ is the circumcenter of $B A_{1} H C_{1}$. So by the corollary of the Brocard's theorem point $D$ is indeed the orthocenter of triangle $A P C$ as desired.

Second solution. Denote by $B_{1}$ the orthogonal projection of $B$ on $A C$. By cyclic quadrilaterals $B_{1} C_{1} P A_{1}$ (Euler's circle), $H A_{1} C B_{1}, A C_{1} A_{1} C$ and $C_{1} H B_{1} A$ we get the following equations:

$$
\begin{gathered}
\angle A_{1} P B_{1}=\angle D C_{1} B_{1} \\
\angle A_{1} B_{1} P=\angle A_{1} C C_{1}=\angle A_{1} A C_{1}=\angle D B_{1} C_{1}
\end{gathered}
$$

From these equalities we get that triangles $B_{1} P A_{1}$ and $B_{1} C_{1} D$ are similar, which implies

$$
\frac{\left|B_{1} D\right|}{\left|B_{1} A_{1}\right|}=\frac{\left|B_{1} C_{1}\right|}{\left|B_{1} P\right|} \Longrightarrow\left|B_{1} A_{1}\right| \cdot\left|B_{1} C_{1}\right|=\left|B_{1} D\right| \cdot\left|B_{1} P\right|
$$

Analogously, using cyclic quadrilateral $A B A_{1} B_{1}$ and $C_{1} B C B_{1}$ we get the following angle equations:

$$
\begin{gathered}
\angle B_{1} A C_{1}=180^{\circ}-\angle B_{1} A_{1} B=\angle B_{1} A_{1} C \\
\angle A B_{1} C_{1}=180^{\circ}-\angle C_{1} B_{1} C=\angle C B A=180^{\circ}-\angle A_{1} B_{1} A=\angle A_{1} B_{1} C
\end{gathered}
$$

From these equalities we get that triangles $B_{1} A C_{1}$ and $B_{1} A_{C}$ are similar so

$$
\frac{\left|B_{1} C_{1}\right|}{\left|B_{1} C\right|}=\frac{\left|A B_{1}\right|}{\left|A_{1} B_{1}\right|} \Longrightarrow\left|B_{1} A_{1}\right| \cdot\left|B_{1} C_{1}\right|=\left|B_{1} A\right| \cdot\left|B_{1} C\right|
$$

Thus we get $\left|B_{1} D^{\prime}\right| \cdot\left|B_{1} P\right|=\left|B_{1} D\right| \cdot\left|B_{1} P\right|=\left|B_{1} A_{1}\right| \cdot\left|B_{1} C_{1}\right|=\left|B_{1} A\right| \cdot\left|B_{1} C\right|$ so by the reverse of the power of the point theorem the quadrilateral $A P C D^{\prime}$ is cyclic as desired.

Problem 3. Prove that the following inequality holds for all positive real numbers $a, b, c, d, e$ and $f$ :

$$
\sqrt[3]{\frac{a b c}{a+b+d}}+\sqrt[3]{\frac{d e f}{c+e+f}}<\sqrt[3]{(a+b+d)(c+e+f)}
$$

(Dimitar Trenevski)

Solution. The inequality is equivalent to

$$
\sqrt[3]{\frac{a b c}{(a+b+d)^{2}(c+e+f)}}+\sqrt[3]{\frac{d e f}{(a+b+d)(c+e+f)^{2}}}<1
$$

By AM-GM inequality we have

$$
\begin{aligned}
& \sqrt[3]{\frac{a b c}{(a+b+d)^{2}(c+e+f)}} \leqslant \frac{1}{3}\left(\frac{a}{a+b+d}+\frac{b}{a+b+d}+\frac{c}{c+e+f}\right) \\
& \sqrt[3]{\frac{d e f}{(a+b+d)(c+e+f)^{2}}} \leqslant \frac{1}{3}\left(\frac{d}{a+b+d}+\frac{e}{c+e+f}+\frac{f}{c+e+f}\right)
\end{aligned}
$$

Adding the inequalities we get

$$
\sqrt[3]{\frac{a b c}{(a+b+d)^{2}(c+e+f)}}+\sqrt[3]{\frac{d e f}{(a+b+d)(c+e+f)^{2}}} \leqslant \frac{1}{3}\left(\frac{a+b+d}{a+b+d}+\frac{c+e+f}{c+e+f}\right)=\frac{2}{3}<1
$$

as desired.

Problem 4. Olja writes down $n$ positive integers $a_{1}, a_{2}, \ldots, a_{n}$ smaller than $p_{n}$ where $p_{n}$ denotes the $n$-th prime number. Oleg can choose two (not necessarily different) numbers $x$ and $y$ and replace one of them with their product $x y$. If there are two equal numbers Oleg wins. Can Oleg guarantee a win?

(Matko Ljulj)

Solution. For $n=1$, Oleg won't be able to write 2 equal numbers on the board as there will be only one number written on the board. We shall now consider the case $n>2$.

Let's note that as all the numbers are strictly smaller than $p_{n}$ we have all their prime factors are from the set $\left\{p_{1}, p_{2}, \ldots, p_{n-1}\right\}$, so there are at most $n-1$ of them in total. We will represent each number $a_{1}, a_{2}, \ldots, a_{n}$ by the ordered $(n-1)$-tuple of non-negative integers in the following way if $a_{i}=p_{1}^{\alpha_{i, 1}} \cdot p_{2}^{\alpha_{i, 2}} \cdot \ldots \cdot p_{n-1}^{\alpha_{i,(n-1)}}$, then we assign $v_{i}=\left(\alpha_{i, 1}, \alpha_{i, 2}, \ldots, \alpha_{i,(n-1)}\right)$, for all $i \in\{1,2, \ldots, n\}$.

Let's consider the following system of equations:

$$
\begin{gathered}
\alpha_{1,1} x_{1}+\alpha_{2,1} x_{2}+\cdots+\alpha_{n, 1} x_{n}=0 \\
\alpha_{1,2} x_{1}+\alpha_{2,2} x_{2}+\cdots+\alpha_{n, 2} x_{n}=0 \\
\cdots \\
\alpha_{1,(n-1)} x_{1}+\alpha_{2,(n-1)} x_{2}+\cdots+\alpha_{n,(n-1)} x_{n}=0
\end{gathered}
$$

There is a trivial solution $x_{1}=x_{2}=\cdots=x_{n}=0$. But as this system has less equalities than variables we can deduce that it has infinitely many solutions in the set of rational numbers (as all the coefficients are rational). Let $\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ be a not trivial solution (so the solution in which not all of $y_{i}$ equal 0 ). Then we can rewrite the initial system using $a_{1}, a_{2}, \ldots, a_{n}$ :

$$
\begin{gathered}
\prod_{i=1}^{n} a_{i}^{y_{i}}=\prod_{i=1}^{n} p_{1}^{\alpha_{i, 1} y_{i}} \cdot p_{2}^{\alpha_{i, 2} y_{i}} \cdot \ldots \cdot p_{n-1}^{\alpha_{i,(n-1)} y_{i}}=\prod_{j=1}^{n-1} p_{j}^{\alpha_{1, j} y_{1}+\alpha_{2, j} y_{2}+\cdots+\alpha_{n, j} y_{n}}=\prod_{j=1}^{n-1} p_{j}^{0}=1 \\
\Longrightarrow \prod_{i=1}^{n} a_{i}^{y_{i}}=1
\end{gathered}
$$

Considering the numbers $y_{1}, y_{2}, \ldots, y_{n}$ as rational numbers in which the respective nominator and denominator are coprime, Denote by $L$ the lowest common multiplier of their denominators. Taking the $L$-th power of the upper equality we get integer exponents in the upper equation (which don't have a common factor). Furthermore, WLOG we can assume that $a_{1}, a_{2}, \ldots, a_{k}$ are those elements $a_{i}$ whose exponents are negative and numbers $a_{k+1}, a_{k+2}, \ldots, a_{k+l}$ are
those elements with postivie exponent (for some $k, l \in \mathbb{N}, k+l \leqslant n$ ). Then, when we shift all $a_{i}$-s with negative exponent to the opposite side of the equation and when those with zero exponent get ruled out we get that the following equality

$$
\prod_{i=1}^{k} a_{i}^{r_{i}}=\prod_{i=k+1}^{l} a_{i}^{r_{i}}
$$

holds for some positive integers $r_{1}, r_{2}, \ldots, r_{k+l}$ for which $D\left(r_{1}, r_{2}, \ldots, r_{k+l}\right)=1$ and for some numbers $a_{1}, a_{2}, \ldots, a_{k+l}$. (We can note that there is at least one number $a_{i}$ on both sides of the equality otherwise we have only ones on the board.)

We shall prove that there is a sequence of transformations by which using this relation we will get two equal numbers among $a_{1}, a_{2}, \ldots a_{n}$.

Lemma 1. Let $(a, b) \in \mathbb{N}^{2}$ and $\left(x_{1}, x_{2}\right) \in \mathbb{N}^{2}$ be such that $G C D\left(x_{1}, x_{2}\right)=1$. Then there exists a sequence of transformations which replaces the numbers $(a, b)$ with $\left(a^{\prime}, b^{\prime}\right)$, where one of these numbers $a^{\prime}, b^{\prime}$ is equal to $a^{x_{1}} b^{x_{2}}$.

Proof. We'll prove this by induction on $x_{1}+x_{2}$, for all $(a, b) \in \mathbb{N}^{2}$. As the basis consider $x_{1}+x_{2}=2 \Longrightarrow x_{1}=x_{2}=1$. The number $a b$ we can get by applying transformation $(a, b) \rightarrow(a, a b)$.

Let's assume that the claim holds for all $\left(x_{1}, x_{2}\right)$ such that $x_{1}+x_{2}<n$, and for all $(a, b)$. Let's take some numbers $\left(x_{1}, x_{2}\right)$ such that $x_{1}+x_{2}=n$ and some arbitrary numbers $(a, b)$. If $x_{1}=x_{2}$ is satisfied, since $x_{1}$ and $x_{2}$ are coprime, we could conclude that both numbers are equal to 1 , but we have already proved this case in basis. Let's assume $x_{1} \neq x_{2}$. WLOG $x_{1}>x_{2}$. Then we apply the transformation $(a, b) \rightarrow(a, a b)$, and then apply the induction hypothesis on numbers $(a, a b)$ and $\left(x_{1}-x_{2}, x_{2}\right):$

$$
(a, b) \rightarrow(a, a b) \rightarrow\left(\gamma, a^{x_{1}-x_{2}}(a b)^{x_{2}}\right)=\left(\gamma, a^{x_{1}} b^{x_{2}}\right)
$$

where $\gamma$ is some positive integer, what we wanted to prove.

Lemma 2. Let $k \in \mathbb{N},\left(b_{1}, b_{2}, \ldots b_{k}\right) \in \mathbb{N}^{k}$ and $\left(x_{1}, x_{2}, \ldots x_{k}\right) \in \mathbb{N}^{k}$. Then there exists sequence of transformations which instead of numbers $\left(b_{1}, b_{2}, \ldots b_{k}\right)$ writes down numbers $\left(b_{1}^{\prime}, b_{2}^{\prime}, \ldots b_{k}^{\prime}\right)$ such that one of those numbers is equal to

$$
\left(b_{1}^{x_{1}} b_{2}^{x_{2}} \cdots b_{k}^{x_{k}}\right)^{\frac{1}{d}}
$$

where denotes greatest common divisor of numbers $x_{1}, x_{2}, \ldots x_{k}$.

Proof. Intuitively, this lemma is just Lemma 1 repeated $(k-1)$ times.

We'll prove this by induction on $k$, for all $b_{1}, b_{2}, \ldots b_{k}$ and $x_{1}, x_{2}, \ldots x_{k}$. In the basis, for $k=1$, it holds $d=x_{1}$, so it we don't have to do any transformation to reach desired situation.

Let's assume that the claim holds for some $k \in \mathbb{N}$. Let's take arbitrary $\left(b_{1}, b_{2}, \ldots b_{k}, b_{k+1}\right)$ and $\left(x_{1}, x_{2}, \ldots x_{k}, x_{k+1}\right)$. Then we apply Lemma 1 on numbers $\left(b_{k}, b_{k+1}\right)$ and $\left(x_{k}^{\prime}, x_{k+1}^{\prime}\right)$, where $x_{k}^{\prime}=\frac{x_{k}}{d_{1}}, x_{k+1}^{\prime}=\frac{x_{k+1}}{d_{1}}, d_{1}=G C D\left(x_{k}, x_{k+1}\right)$, and then we apply the induction hypothesis on numbers $\left(b_{1}, b_{2}, \ldots b_{k}^{x_{k}^{\prime}} b_{k+1}^{x_{k+1}^{\prime}}\right)$ and $\left(x_{1}, x_{2}, \ldots x_{k-1}, d_{1}\right)$ :

$$
\left(b_{1}, b_{2}, \ldots b_{k}, b_{k+1}\right) \rightarrow\left(b_{1}, b_{2}, \ldots b_{k-1}, \gamma_{k}, b_{k}^{x_{k}^{\prime}} b_{k+1}^{x_{k+1}^{\prime}}\right) \rightarrow\left(\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k},\left(b_{1}^{x_{1}} b_{2}^{x_{2}} \cdots b_{k-1}^{x_{k-1}}\left(b_{k}^{x_{k}^{\prime}} b_{k+1}^{x_{k+1}^{\prime}}\right)^{d_{1}}\right)^{\frac{1}{d_{2}}}\right)
$$

where $\gamma_{1}, \gamma_{2}, \ldots, \gamma_{k}$ are some positive integers and $d_{2}=G C D\left(x_{1}, x_{2}, \ldots x_{k-1}, d\right)=G C D\left(x_{1}, x_{2}, \ldots x_{k-1}, x_{k}, x_{k+1}\right)=d$. Notice that last number in upper relation is the one we wanted to get.

Lemma 3. Let $(a, b) \in \mathbb{N}^{2}$ and $\left(x_{1}, x_{2}\right) \in \mathbb{N}^{2}$ such that $G C D\left(x_{1}, x_{2}\right)=1$. Then there exists sequence of transformations which instead of numbers $(a, b)$ writes down numbers $\left(a^{\prime}, b^{\prime}\right)$ for which it is satisfied $a^{\prime} / b^{\prime}=a^{x_{1}} / b^{x_{2}}$.

Proof. We'll prove this by induction on $x_{1}+x_{2}$, for all $(a, b) \in \mathbb{N}^{2}$. In the basis is $x_{1}+x_{2}=2 \Longrightarrow x_{1}=x_{2}=1$, so we don't have to do any transformation to reach desired situation.

Ler's assume that the claim hold for all $\left(x_{1}, x_{2}\right)$ such that $x_{1}+x_{2}<n$, and for all $(a, b)$. Let's take some numbers $\left(x_{1}, x_{2}\right)$ such that $x_{1}+x_{2}=n$ and arbitrary numbers $(a, b)$.

- If one of the numbers $x_{1}$ and $x_{2}$ is even (WLOG $x_{1}$ is even): we apply tranformation $(a, b) \rightarrow\left(a^{2}, b\right)$, and then we apply induction hypothesis on numbers $\left(a^{2}, b\right)$ and $\left(\frac{x_{1}}{2}, x_{2}\right)$.
- Both numbers $x_{1}$ and $x_{2}$ are odd, and they are equal: then they are both equal to 1 , which we have already solved in the basis.
- Numbers $x_{1}$ and $x_{2}$ are odd and distinct (WLOG $x_{1}>x_{2}$ ): we make following transformations $(a, b) \rightarrow(a, a b) \rightarrow$ $\left(a^{2}, a b\right)$, and then we apply induction hypothesis on numbers $\left(a^{2}, a b\right)$ and $\left(\frac{x_{1}+x_{2}}{2}, x_{2}\right)$ :

$$
(a, b) \rightarrow(a, a b) \rightarrow\left(a^{2}, a b\right) \rightarrow\left(c \cdot\left(a^{2}\right)^{\frac{x_{1}+x_{2}}{2}}, c \cdot(a b)^{x_{2}}\right)=\left(\left(a^{x_{2}} c\right) \cdot a^{x_{1}},\left(a^{x_{2}} c\right) \cdot b^{x_{2}}\right)
$$

where $c$ is some positive integer, what we wanted to prove.

In the equality (1), let $d_{1}=G C D\left(r_{1}, r_{2}, \ldots, r_{k}\right), d_{2}=G C D\left(r_{k+1}, r_{k+2}, \ldots, r_{k+l}\right), z_{i}=\frac{r_{i}}{d_{1}}, \forall i \in\{1,2, \ldots, k\}, z_{i}=\frac{r_{i}}{d_{2}}$, $\forall i \in\{k+1, k+2, \ldots, k+l\}$. As well let $A$ be the left hand side of the equality (1), and let $B$ be the right hand side. Let $A^{\prime}=A^{\frac{1}{d_{1}}}$ and $B^{\prime}=B^{\frac{1}{d_{2}}}$. We want to do such transformations that we get $x$ i $y$ which will have same ratio as $A$ and $B$. If we apply Lemma 2 on the numbers $\left(a_{1}, a_{2}, \ldots, a_{k}\right)$ and $\left(z_{1}, z_{2}, \ldots, z_{k}\right)$; we get (among other numbers we get) the number $A^{\prime}$. As well applying the same lemma on the numbers $\left(a_{k+1}, a_{k+2}, \ldots, a_{k+l}\right)$ and $\left(z_{k+1}, z_{k+2}, \ldots, z_{k+l}\right)$, we will get the number $B^{\prime}$ on the board.

Numbers $d_{1}$ and $d_{2}$ are coprime (otherwise there would be some prime $p$ which would divide $d_{1}$ and $d_{2}$ which would imply it divides $r_{1}, r_{2}, \ldots, r_{k+l}$ as well which is in contradiction to the assumption they do not have a common factor). So we can apply Lemma 3 on the numbers $\left(A^{\prime}, B^{\prime}\right)$ and $\left(d_{1}, d_{2}\right)$. Now we get two numbers with the same ratio as $A$ i $B$. But as by (1) we have $A=B$, we get 2 equal numbers on the board.

Thus Oleg can guarantee a win for any $n>1$.

Comment: We can get to the relation (1) by concluding that the set $\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}$ is linearly dependant subset of $(n-1)$-dimensional space $\mathbb{Q}^{n-1}$.