Junior Balkan Mathematical Olympiad 2021 - Problem 3 | Solution - 2 July 2021



Problem 3. Let $ABC$ be an acute scalene triangle with circumcenter $O$ . Let $D$ be the foot of the altitude from $A$ to the side $BC$ . The lines $BC$ and $AO$ intersect at $E$ . Let $s$ be the line through $E$ perpendicular to $AO$ . The line $s$ intersects $AB$ and $AC$ at $K$ and $L$ , respectively. Denote by $\omega$ the circumcircle of triangle $AKL$ . Line $AD$ intersects $\omega$ again at $X$ . Prove that $\omega$ and the circumcircles of triangles $ABC$ and $DEX$ have a common point.

Solution Let $AO \cap (ABC) = Y$ . Since $\angle AKE = \angle ACB$ , we have $KBLC$ is cyclic. So by PoP, we have $AE\cdot EY = BE \cdot EC = KE \cdot EL$ , implying that $ALYK$ is cyclic. Now note that $\angle AXK = \angle ALK = \angle ABC$ , giving $BDXK$ is cyclic. Similarly $\angle AYK = \angle ALK = \angle ABC$ , giving $BEYK$ is cyclic. Thus by PoP, we have $AE\cdot AY = AB \cdot AK = AD \cdot AX$ , implying that $DEYX$ is cyclic, so the circumcircles of $ABC , DEX$ and $AKL$ have a common point, which is $Y$ . :