China Western Mathematical Olympiad

By the angle bisector theorem, we have $$\frac{PF}{FC}\cdot \frac{CB}{BA}\cdot \frac{AD}{DP}=\frac{PB}{BC}\cdot \frac{BC}{BA}\cdot \frac{AB}{PB}=1.$$ By the converse of Ceva theorem, the lines $AF$, $BP$ and $CD$ are concurrent. Suppose there exists $\angle D^{\prime }B{F}^{\prime }$ satisfying the conditions: (a) $\angle D^{\prime }B{F}^{\prime }=90^{\circ }$, (b) the lines $A{F}^{\prime }$, $BP$ and $C{D}^{\prime }$ are concurrent. We may assume that $F^{\prime }$ lies on $PF$. Then, $D^{\prime }$ is on $AD$. So $$\frac{P{F}^{\prime }}{F^{\prime }C}<\frac{PF}{FC},\frac{A{D}^{\prime }}{P{D}^{\prime }}<\frac{AD}{PD}.$$ Thus $$\frac{P{F}^{\prime }}{F^{\prime }C}\cdot \frac{CB}{BA}\cdot \frac{A{D}^{\prime }}{D^{\prime }P}<\frac{PB}{BC}\cdot \frac{BC}{BA}\cdot \frac{AB}{PB}=1,$$ which leads to a contradiction. This completes the proof.

(2) We may assume that the circle $O$ has radius 1. Let

$\angle PCB=\alpha$. By (1), $\angle PBE=\angle EBC=30^{\circ }$. Therefore, $E$ is the midpoint of $\overline{PC}$. Since $\angle MPE=\angle PBE=30^{\circ }$, $\angle CPE=\angle CBE=30^{\circ }$ and $PD=PE$, we obtain $\angle PDE=\angle PED=15^{\circ }$, $PE=2\cdot 1\cdot \sin 30^{\circ }$ and $DE=2\cos 15^{\circ }$.



Since $$BE=2\sin \angle ECB=2\sin (\alpha +30^{\circ })$$ and $\angle BED=\angle BEP-15^{\circ }$, we have $$\cos (\alpha -15^{\circ })=\frac{BE}{DE}=\frac{2\sin (\alpha +30^{\circ })}{2\cos 15^{\circ }},$$ $$\cos (\alpha -15^{\circ })\cos 15^{\circ }=\sin (\alpha +30^{\circ }),$$ $$\cos \alpha +\cos (\alpha -30^{\circ })=2\sin (\alpha +30^{\circ }),$$ $$\cos \alpha +\cos \alpha \cos 30^{\circ }+\sin \alpha \sin 30^{\circ }=\sqrt{3}\sin \alpha +\cos \alpha ,$$ $$1+\frac{\sqrt{3}}{2}+\frac{1}{2}\tan \alpha =\sqrt{3}\tan \alpha +1.$$ So $$\tan \alpha =\frac{6+\sqrt{3}}{11}.$$