Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

Let $ABC$ be a triangle with $\omega$ its circumcircle. Let $S$ be the midpoint of the arc $BC$ not containing $A$ . Let $X \in AB$ and $Y \in AC$ such that $XY \parallel BC$ . We take $P$ (resp. $Q$ ), the second intersection of $SX$ (resp. $SY$ ) with $\omega$ and $R = PQ \cap XY$ .

Show that $AR$ is tangent to $\omega$ .

Solutions — 2

Solution 1

Let's show that $PQ$ , $XY$ , and the tangent at $A$ to $\omega$ are the radical axes of three circles: the first is $\omega$ , the second is the circle passing through $A$ , $X$ , and $Y$ , and the third is the circle passing through $P$ , $Q$ , $Y$ , and $X$ . This quadrilateral is cyclic because, by angle chasing, as $\widehat{BCS} = \widehat{CBS}$ , we have $\widehat{PQY} = \widehat{PQB} + \widehat{QBS} = \widehat{PSB} + \widehat{BCS} = \widehat{PSB} + \widehat{CBS} = 180 - \widehat{PXY}$ . Since $\widehat{AYX} = \widehat{ACB}$ , the tangent at $A$ to $\omega$ is the same as the tangent to the circumcircle of $A$ , $X$ , and $Y$ , so the latter is the radical axis. Finally, the three radical axes intersect at a point, thus we obtain the desired result.

Solution 2

Let's show that $PQ$ , $XY$ , and the tangent at $A$ to $\omega$ are the radical axes of three circles, the first being $\omega$ , the second the circumcircle of $AXY$ , and the third the circumcircle of $PQYX$ . This quadrilateral is cyclic because, by angle chasing like $\widehat{BOS}=\widehat{CBS}$ , we have

$\widehat{PQY}=\widehat{PQB}+\widehat{QBS}=\widehat{PSB}+\widehat{BCS}=\widehat{PSB}+\widehat{CBS}=180-\widehat{PXY}$

Since $\widehat{AYX}=\widehat{ACB}$ , the tangent at $A$ to $\omega$ is the same as the tangent to the circumcircle of $AXY$ , so this is the radical axis. Finally, the three radical axes intersect at a point, so we obtain the desired result.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.