BMO Shortlist, 2022

Geometry Difficulty 8.9 Shortlist Prove it Balkan Mathematical Olympiad

Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Let $E$ be the midpoint of the minor arc $BC$ of $\omega$ , and $M$ the midpoint of $BC$ . Let $V$ be the other point of intersection of $AM$ with $\omega$ , $F$ the point of intersection of $AE$ with $BC$ , $X$ the other point of intersection of the circumcircle of $FEM$ with $\omega$ , $X'$ the reflection of $V$ with respect to $M$ , $A'$ the foot of the perpendicular from $A$ to $BC$ and $S$ the other point of intersection of $XA'$ with $\omega$ . If $Z \in \omega$ with $Z \neq X$ is such that $AX = AZ$ , then prove that $S$ , $X'$ and $Z$ are collinear.

Solution

Claim 1. $AX$ is the $A$ -symmedian of $\triangle ABC$ .
Proof of Claim 1. Let $Y \in \omega$ such that $AY$ is the $A$ -symmedian of triangle $ABC$ . We want to prove that $Y = X$ .

We have that $\angle BAY = \angle CAM$ and $\angle BYA = \angle BCA = \angle MCA$ , therefore the triangles $ABY$ and $AMC$ are similar. It follows that $(AY)(AM) = (AB)(AC)$ .
Since $AE$ is the bisector of $\angle BAC$ , then $\angle BAF = \angle CAE$ . We also have $\angle ABF = \angle ABC = \angle AEC$ , therefore the triangles $BAF$ and $EAC$ are similar. It follows that $(AE)(AF) = (AB)(AC)$ .
We get $(AY)(AM) = (AE)(AF)$ and since also $\angle YAF = \angle EAM$ , then the triangles $YAF$ and $EAM$ are similar. So $\angle AFY = \angle AME$ and $\angle YFE = \angle EMV$ . But as $E$ is the midpoint of the arc $YV$ , it follows that $\angle EMV = \angle YME$ . So $\angle YFE = \angle YME$ from which it follows that the quadrilateral $YFME$ is cyclic. But since $Y \in \omega$ , we finally get that $Y = X$ . $\square$
From Claim 1 we conclude that the triangles $XBC$ and $VCB$ are equal. Thus $MX = MV = MX'$ . So the triangle $X'XV$ is a right-angled triangle and $X'X$ is perpendicular to $XV$ and therefore also to $BC$ . Thus $X'$ is the reflection of $X$ on $BC$ .
Claim 2. The quadrilateral $ASA'M$ is cyclic.
Proof of Claim 2. We have $\angle ASA' = \angle ASX = \angle ABX$ . But from Claim 1 we also have $\angle ABX = \angle AMC$ . So $\angle ASA' = \angle AMC$ and the result follows. $\square$
Claim 3. The quadrilateral $XSX'M$ is cyclic.
Proof of Claim 3. From Claim 2 we have $\angle XSM = \angle A'SM = \angle A'AM$ . Since $XX'$ is parallel to $AA'$ we have $\angle A'AM = \angle XX'M$ . So $\angle XSM = \angle XX'M$ and the result follows. $\square$

Now from Claim 3 we have
$\angle XSX' = \angle XMV = \angle XX'M + \angle X'XM = 2\angle XX'M = 2\angle AA'M.$
So to conclude the proof it is enough to also show that $\angle XSZ = 2\angle AA'M$ . From Claim 1 we have $\angle ACX = \angle MAB$ and therefore
$\angle AZX = \angle ACX = \angle AMB = 90^\circ - \angle A'AM.$
Since the triangle $XAZ$ is isosceles, we deduce that
$\angle XSZ = \angle XAZ = 180^\circ - 2\angle AZX = 2\angle A'AM$
thus completing the proof.

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