Problem 1
Let $O$ be the center of the circumcircle of an acute triangle $ABC$. The line $AC$ intersects the circumcircle of the triangle $ABO$ a second time at $S$. Prove that the line $OS$ is perpendicular to the line $BC$.
Solution
Let us put $\angle ACB=\gamma$. Since $\angle AOB$ is central, then $\angle AOB=2\gamma$. This implies that $\angle ABO=90^{\circ}-\gamma$. Since $ABOS$ is cyclic, then $\angle CSO = \angle ABO = 90^{\circ}-\gamma$. Therefore, we have that $\angle OSC + \angle SCB = 90^{\circ}$ and $SO \perp BC$.
Problem 2
Let $ABC$ be an acute triangle with $BC > AC$. The perpendicular bisector of the segment $AB$ intersects the line $BC$ at $X$ and the line $AC$ at $Y$. Let $P$ be the projection of $X$ on $AC$ and let $Q$ be the projection of $Y$ on $BC$. Prove that the line $PQ$ intersects the segment $AB$ at its midpoint.
Solution
Let $M$ be the midpoint of $AB$. It is sufficient to prove that $M$, $P$ and $Q$ are collinear. Let us put $\angle ABC=\beta$. Since $AXB$ is isosceles, then $\angle XAB = \angle XBA =\beta$. The quadrilateral $XPAM$ is cyclic and therefore $\angle XPM = \angle XAM =\beta$. The quadrilateral $YQMB$ is cyclic and therefore $\angle QYM = \angle QBM =\beta$. The quadrilateral $YQPX$ is cyclic and therefore $\angle XPQ = 180^{\circ}-\beta$ and $M$, $P$ and $Q$ are collinear.
Problem 3
Anaelle has $2n$ stones labelled $1$, $2$, $3$, … , $2n$ as well as a red box and a blue box. She wants to
put each of the $2n$ stones into one of the two boxes such that the stones $k$ and $2k$ are indifferent
boxes for all $k = 1,2,…,n$. How many possibilities does Anaelle have to do so?
Solution
Answer: $2^n$.
Let us prove it by induction. For $n=1$, the statement is true, since there are only two possibilities to put the numbers $1$ and $2$ into two boxes, such that the stones $1$ and $2k$ are in different boxes. Let us assume that the statement is true for $n=k$, i.e. there are $2^k$ possibilities for Anaelle to do the distribution. Let us now place two more numbers: $2k+1$ and $2k+2$. Notice that the box for the number $2k+2$ is fixed (it is a different box from where $k+1$ is). Since the number $2k+1$ can be placed in either box, then the number of possibilities is $2 \cdot 2^k=2^{k+1}$ and the statement is proven.
Problem 4
Prove that for every integer $n \geq 3$ there exist positive integers $a_1 < a_2 < … < a_n$, such that
$$ a_k | \left(a_1 + a_2 + … + a_n \right) $$
holds for every $k = 1,2,…,n$.
Solution
Let us put $a_1=1$, $a_2=2$ and $a_{n+1}=a_1+a_2+…+a_n$ for $n \geq 2$. The sequence is obviously increasing and satisfies the conditions of the problem.
Problem 5
Find all positive integers $n \geq 2$, such that, for every divisor $d > 1$ of $n$, we have that
$d^2 + n$ divides $n^2 + d $
Solution
Answer: $n$ is prime.
If $n$ is prime, then $d=n$ and $n^2+n$ divides $n^2+n$. This implies that all primes work. If $n$ is composite, then there exists a divisor $d$, such that $n=dk$ and $d,k>1$. We have that therefore $d^2+dk$ divides $d^2k^2+d$. This implies that $d+k | dk^2+1$. If we now switch $d$ and $k$, then we will have that also $d+k | d^2k+1$. By adding the last two expressions we have that $d+k | dk(d+k)+2$. Therefore, $d+k|2$ and we obtained a contradiction.
