Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Brazil

Let $A$ be one of the intersection points of two circles with centers $X$ and $Y$ . The tangent lines to these circles passing through $A$ meet the circles again at $B$ and $C$ . Let $P$ be the point in the plane such that $PXAY$ is a parallelogram. Prove that $P$ is the circumcenter of the triangle $ABC$ .

Solution

The tangent lines passing through $A$ are perpendicular to $AX$ and $AY$ , so $AX \perp AC$ and $AY \perp AB$ . Since $AXPY$ is a parallelogram, $PX$ is parallel to $AY$ , so $PX \perp AB$ . Since $AB$ is a chord of a circle with center $X$ , $PX$ is the perpendicular bisector of $AB$ . Analogously, $PY$ is the perpendicular bisector of $AC$ . So $P$ is the intersection of the perpendicular bisectors of $AB$ and $AC$ .

and, therefore, circumcenter of the triangle $ABC$ .

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