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Geometry Difficulty 4.9 AIME Prove it Taiwan

Let the orthocenter of triangle $ABC$ be point $H$ . Let line $BH$ and line $AC$ intersect at point $E$ , and let line $CH$ and line $AB$ intersect at point $F$ . Let point $X$ be an arbitrary point on line $BC$ . Let the circumcircle of triangle $BEX$ intersect line $AB$ again at point $Y$ , and let the circumcircle of triangle $CFX$ intersect line $AC$ again at point $Z$ . Prove that the circumcircle of triangle $AYZ$ is tangent to line $AH$ .

Solution

By Miquel's theorem (applied to $X$ , $F$ , $E$ on sides of $HBC$ ), we know that the circumcircles of $HEF$ , $BEX$ , $CFX$ intersect at a point $M$ . Now again, by Miquel's theorem (applied to $Y$ , $E$ , $A$ on sides of $HAB$ where $A$ is viewed as a point on $HA$ ), we know that the circumcircles of $HEA$ , $BEY$ and the circle passing through $A$ , $Y$ that is tangent to $HA$ intersect at a point. Since $AHEF$ and $BEXY$ are concyclic, we know that the intersection has to be $M$ . Therefore the circumcircle of $AMY$ is tangent to $AH$ . Similarly, the circumcircle of $AMZ$ is tangent to $AH$ . As a consequence, $AMYZ$ are on a circle that tangents $AH$ , as desired.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.