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:
$$ a \left( b^2+n^2 \right) = \left( a^2+n^2 \right) \ $$
$$ a b^2 + a n^2 = a^2 + b n^2 \ $$
$$ a b^2 + a n^2 – b a^2 – b n^2 = 0 \\ $$
$$ ab(b-a) – n^2(b-a) = 0 \\ $$
$$ (b-a)\left(ab- n^2 \right) = 0 $$
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 $AP \leq 2MQ$.
Solution
Let us start with the triangle inequality for the triangle $CMQ$:
$$ CM \leq MQ+CQ $$
Notice that since $AM=BM=CM$ and $PQ=CQ$, then we have
$$ CM \leq MQ+CQ $$
$$ AM \leq MQ+PQ $$
$$ AM \leq MQ+MQ+PM $$
$$ AP \leq 2MQ $$
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
$$ 0 + 1 + 2 + … + 15 = 120$$
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 $\angle CSM = \angle CSN$.
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 $\angle PMC = \angle BCM = \angle BMC$ and $MC$ is the external angle bisector of the triangle $SMN$. Since $SP \parallel BC$, then $\angle QNC = \angle ACN = \angle ANC$ 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 $\angle CSM = \angle CSN$.
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 $a=\overline{a_{1000} … a_2 a_1}$ and $b=\overline{b_{1000} … b_2 b_1}$, then we have that
$$ a_1+a_2+…+a_{1000} = b_1+b_2+…+b_{1000} = S $$
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
$$ 2S = a_1+a_2+…+a_{1000} + b_1+b_2+…+b_{1000} = 10 + 9 + 9 + … + 9 $$
where $9$ is repeared $999$ times. The left side of the last equality is even and the right side is odd. Contradiction.
