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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Romania

Let $A$ be a point outside the circle $\mathcal{C}$ . The tangents from $A$ touch the circle at $B$ and $C$ . Let $P$ be an arbitrary point on $AC$ produced, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\mathcal{C}$ . Prove that $\angle PEQ = 2\angle APB$ .

Cuban Olympiad, 2003

Solution

Let $\{D\} = (BP \cap \mathcal{C})$ and $\{M\} = DE \cap AC$ .

Then $\angle BDE = \angle BCE = 180^\circ - \angle ACB - \angle PCE = 180^\circ - \angle ABC - \angle CBE = \angle APE$ , hence $\Delta MPD \sim \Delta MEP$ (AA), which leads to $\angle MPD \equiv \angle MEP$ and $MP^2 = MD \cdot ME$ .

On the other hand, from the power of the point $M$ with respect to the circle $\mathcal{C}$ it follows that $MC^2 = MD \cdot ME$ , i.e. $MP^2 = MC^2$ , which means that $M$ is the midpoint of $[PC]$ .

It follows that $\angle MEP \equiv \angle MPD \equiv \angle MQP$ , i.e., the quadrilateral $MQEP$ is cyclic. We obtain that $\angle MEQ \equiv \angle MPQ \equiv \angle MEP$ , and the conclusion.

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