, 2010

Geometry Difficulty 6.6 National olympiad Prove it Ukraine

Acute-angled triangle $ABC$ is given. On the perpendicular bisectors to sides $AB$ and $BC$ respectively, points $P$ and $Q$ are chosen. Let $M$ and $N$ be the projections of $P$ and $Q$ onto $AC$ (Fig.09). It turns out, that $2MN = AC$ . Prove, that circumcircle of triangle $PBQ$ passes through the circumcenter of triangle $ABC$ .

Fig.09

Solution

Let $O$ be the circumcenter of $ABC$ . Let our perpendicular bisectors meet sides $AB$ and $BC$ at points $K$ and $L$ respectively. Then $KL$ is a midline of triangle $ABC$ . Therefore, $KL \parallel AC$ and $2KL = AC$ . Thus, $MKLN$ is parallelogram, because $KL \parallel MN$ and $KL = MN$ . Due to the fact, that $PM \perp AC$ and $QN \perp AC$ , we have $PN \parallel QN$ . Moreover, $PM \parallel QN$ and $KM \parallel LN$ , thus, $\angle PMK = \angle QNL$ (Fig.10).

$\angle AKP = \angle AMP = 90^\circ$ , thus $P, K, M$ and $A$ lie on the same circle with diameter $PA$ . By analogy, we obtain that $Q, N, C$ and $L$ lie on the same circle with diameter $CQ$ , points $O, K, B$ and $L$ belong to the circle with diameter $BO$ . From these observations, it follows $\angle PAK = \angle PMK = \angle QNL = \angle QCL$ .

$P$ and $Q$ belong to perpendicular bisectors of the segments $AB$ and $BC$ respectively, thus $APB$ and $BQC$ are isosceles triangles. From this, we obtain $\angle PBA = \angle PAB = \angle QCB = \angle QBC$ . Hence, $\angle PBQ = \angle KBL = 180^\circ - \angle KOL = 180^\circ - \angle POQ$ , in other words, points $P, B, Q, O$ are cyclic, which provides desired result.

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