In $\triangle PBC$, $\angle PBC=60^{o}$, the tangent at point $P$ to the circumcircle$g$ of $\triangle PBC$ intersects with line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and $g$ respectively, satisfying $\angle DBE=90^{o}$, $PD=PE$. $BE$ and $PC$ meet at $F$. It is known that lines $AF,BP,CD$ are concurrent. a) Prove that $BF$ bisect $\angle PBC$ b) Find $\tan \angle PCB$
Problem
Source: Chinese Western Mathematical Olympiad 2006, Problem 3
Tags: geometry, circumcircle, trigonometry, geometry unsolved
Jumbler
11.11.2006 08:58
Anybody solve it?
yetti
12.11.2006 03:02
First, switch the notation of the points P, A. Let G be the concurrency point of PF, BA, CD and let BD meet CA at Q. By Menelaus' theorem for $\triangle ABC$ cut by the transversal PGF, $\frac{PB}{PC}\cdot \frac{FC}{FA}\cdot \frac{GA}{GB}= 1$ and by Ceva's theorem for the same triangle with the concurrent cevians AD, BD, CD, $\frac{PB}{PC}\cdot \frac{QC}{QA}\cdot \frac{GA}{GB}= 1.$ Combing, $\frac{QC}{QA}= \frac{PC}{PB}\cdot \frac{GB}{GA}= \frac{FC}{FA}$ Thus F, Q divide the segment CA harmonically. Since $BF \equiv BE \perp BD \equiv BQ,$ BF is the internal and BQ the external bisector of the angle $\angle B.$ Note that for this, we did not use the conditions that AP is a tangent to the circumcircle of the $\triangle ABC$ at A, $\angle B = 60^\circ$ and AE = AD. Using these conditions now, E is the midpoint of the circumcircle arc CA opposite to B, $\triangle AOE$ is equilateral and AE = R, where O is the circumcenter and R the circumradius. On the other hand, using the sine theorem for the $\triangle ADB,$ $\frac{AD}{AB}= \frac{\sin \widehat{ABD}}{\sin \widehat{ADB}}$ $\angle ABD = 90^\circ-\frac{\angle B}{2}= 60^\circ$ and $\angle DAB = \angle C$ (because AD is the circumcircle tangent at A), so that $\angle ADB = 120^\circ-\angle C.$ Therefore, $R = AE = AD = \frac{c\ \sin 60^\circ}{\sin(120^\circ-C)}= \frac{2R\ \sin C \sin 60^\circ}{\sin (120^\circ-C)}$ $\frac{\sqrt{3}}{2}\cos C+\frac{1}{2}\sin C = \sqrt{3}\sin C$ $\tan C = \frac{\sqrt{3}}{2 \sqrt{3}-1}= \frac{6+\sqrt{3}}{11}$ Urrrgh !!!