Solutions to the Polish Mathematical Olympiad, 2021

Problem 1

Positive integers $a$, $b$, $n$ satisfy the equality

$$\frac{a}{b}=\frac{a^2+n^2}{b^2+n^2}$$

Prove that the number $\sqrt{ab}$ is an integer.

Solution

Let us cross-multiply and factor:

 

$$a(b^2+n^2)=(a^2+n^2)$$

$$a{b}^2+a{n}^2=a^2+b{n}^2$$

$$a{b}^2+a{n}^2–b{a}^2–b{n}^2=0$$

$$ab(b-a)–n^2(b-a)=0$$

$$(b-a)(ab-n^2)=0$$

If $b-a=0$, then $a=b$ and

$$\sqrt{ab}=\sqrt{b^2}=b$$

which is an integer.

If $ab-n^2=0$, then $ab=n^2$ and

$$\sqrt{ab}=\sqrt{n^2}=n$$

which is an integer.

Problem 2

In the right triangle $ABC$, the point $M$ is the midpoint of the hypotenuse $AB$. The points $P$ and $Q$ lie on the segments $AM$ and $MB$, respectively, such that $PQ=CQ$. Prove that $AP\le 2MQ$.

Solution

Let us start with the triangle inequality for the triangle $CMQ$:

$$CM\le MQ+CQ$$

Notice that since $AM=BM=CM$ and $PQ=CQ$, then we have

$$CM\le MQ+CQ$$

$$AM\le MQ+PQ$$

$$AM\le MQ+MQ+PM$$

$$AP\le 2MQ$$

which is what needed to be proven.

Problem 3

$16$ players participated in the badminton tournament. Each player has played at most one match with any other player, no match ended in a draw. After the tournament, it turned out that each player won a different number of matches. Prove that each player lost a different number of matches.

Solution

Notice that since the number of victories is at least $0$ and at most $15$, then there are only $16$ different number of victories. Therefore each number of victories from $0$ to $15$ appear exactly once. The total number of matches in a tournament is equal to the total number of all victories because each match has exactly one winner. Therefore, total the number of matches played is equal

$$0+1+2+\dots +15=120$$

On the other hand, the greatest possible number of matches played in the entire tournament is equal to the maximum number of different pairs that can be formed from among $16$ players, i.e.

$$\frac{16\cdot 15}{2}=120$$

Hence the conclusion that all possible matches were played, i.e. each player played exactly $15$ matches. This implies that each player who won $k$ matches also lost $15-k$ matches. Since the numbers of matches won are all different, so are the numbers of lost matches as well.

Problem 4

On the side $AB$ of the scalene triangle $ABC$, points $M$ and $N$ are chosen, such that $AN=AC$ and $BM=BC$. The line parallel to $BC$ passing through the point $M$, and the line parallel to $AC$ through the point $N$ intersect at the point $S$. Prove that $\mathrm{\angle }CSM=\mathrm{\angle }CSN$.

Solution (according to Carlos Rodriguez)

Let $SM$ intersect the line $AC$ at the point $P$ and $SN$ intersect the line $BC$ at the point $Q$. Since $SQ\parallel AC$, then $\mathrm{\angle }PMC=\mathrm{\angle }BCM=\mathrm{\angle }BMC$ and $MC$ is the external angle bisector of the triangle $SMN$. Since $SP\parallel BC$, then $\mathrm{\angle }QNC=\mathrm{\angle }ACN=\mathrm{\angle }ANC$ and $NC$ is the external angle bisector of the triangle $SMN$. Therefore the point $C$ is the excenter of the triangle $SMN$. Therefore, $CS$ is the angle bisector of the angle $MSN$ and $\mathrm{\angle }CSM=\mathrm{\angle }CSN$.

Problem 5

Given two natural numbers $a$ and $b$, which have the same digits in their decimal representation, i.e. each of the digits from $0$ to $9$ occurs the same number of times in the representation of $a$ and in the representation of $b$. Show that if $a+b={10}^{1000}$, then the numbers $a$ and $b$ are divisible by $10$.

Solution

First note that each of the numbers $a$, $b$ is exactly $1000$ digits long. Indeed, since the sum $a+b$ is the smallest $1001$-digit number, $a$ and $b$ cannot have more than $1000$ digits. If they both had less than $1000$ digits, then $a<{10}^{999}$ and $b<{10}^{999}$, from where we have

$$a+b<2\cdot {10}^{999}<10\cdot {10}^{999}={10}^{1000}$$

Contradiction.

Let $a=\overline{a_{1000}\dots a_2{a}_1}$ and $b=\overline{b_{1000}\dots b_2{b}_1}$, then we have that

$$a_1+a_2+\dots +a_{1000}=b_1+b_2+\dots +b_{1000}=S$$

Since the last digit of the sum $a+b$ is $0$, then $a_1+b_1=0$ or $a_1+b_1=10$. In the first case, we have $a_1=b_1=0$ and we are done. In the second case we have

$a_2+b_2=9$, $a_3+b_3=9$, … , $a_{1000}+b_{1000}=9$.

So we obtain

$$2S=a_1+a_2+\dots +a_{1000}+b_1+b_2+\dots +b_{1000}=10+9+9+\dots +9$$

where $9$ is repeared $999$ times. The left side of the last equality is even and the right side is odd. Contradiction.

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