AoPS - Problem

Source: Nordic MO 2012 Q2

Tags: geometry, circumcircle, geometric transformation, reflection, geometry unsolved

djb86

21.04.2013 20:00

Given a triangle $ABC$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $BC$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $AB$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $AB$ (if there is no second point of intersection, choose $Q = B$). Prove that $AQ = AC$.

First of all, note that $P$ is the intersection of the bisector of $ \angle BAC$ with the circumcircle of $\triangle ABC$ , let $O_k$ be the center of circle $(k)$. We know that $(l)$ is tangent to $(k)$, so $(O_kP) \perp (l) $ and since $(l) \parallel (AB) $ it follows that $(O_kP) \parallel (AB)$ , and we have $O_kB=O_kQ$ so $(O_kP)$ is the segment bisector of $[BQ]$ and hence $PB=PQ$. We have : $ \angle ACP= \pi - \angle ABP = \angle QBP = \angle AQP $ and $ \angle PAC = \angle PAQ$ so: $\triangle APC = \triangle APQ$, and hence: $AC=AQ$.