## USAMO 2009 #5 July 19, 2009

Posted by Damien Jiang in Problem-solving, Uncategorized.
Tags: ,

I like Olympiad geometry. Therefore, I will give my solution to this year’s USAMO #5; I was rather happy with my solution.

5. Trapezoid ${ABCD}$, with ${\overline{AB}||\overline{CD}}$, is inscribed in circle ${\omega}$ and point ${G}$ lies inside triangle ${BCD}$. Rays ${AG}$ and ${BG}$ meet ${\omega}$ again at points ${P}$ and ${Q}$, respectively. Let the line through ${G}$ parallel to ${\overline{AB}}$ intersects ${\overline{BD}}$ and ${\overline{BC}}$ at points ${R}$ and ${S}$, respectively. Prove that quadrilateral ${PQRS}$ is cyclic if and only if ${\overline{BG}}$ bisects ${\angle CBD}$.

Proof.

First, note that since the homothety centered at ${B}$ that takes ${R,S}$ to ${D,C}$ takes their circumcircles to each other (and therefore their centers), their tangents at ${B}$ are the same. I.e., the circumcircles of ${\triangle BRS}$ and ${\triangle BDC}$ are internally tangent.

If direction:

Let ${\angle CBQ = \alpha, \angle DBQ = \alpha', \angle PAB = \beta, \angle BCD = \theta}$. Then we know ${\alpha = \alpha'}$. By cyclic quadrilaterals, ${\angle QPG = \angle QPA = \angle QBA = \angle QBD + \angle DBA = \alpha + \theta = \angle CBQ + \angle DCB = \angle CBQ + \angle CBX' = \angle QBX' = \angle GBX'}$ where ${X'}$ is the intersection of the common tangent at ${B}$ and ${PQ}$. Hence ${\angle GPX' + \angle GBX' = 180}$, or ${G, P, X', B}$ are cyclic.

This implies ${\angle X'GP = \angle X'BP = \angle PAB}$, i.e. ${X'G || AB}$. But then since ${R,S}$ lie on ${X'G}$, we obtain, by Power of a Point, ${(RX')(SX') = BX'^2 = (PX')(QX')}$, or ${PQRS}$ is cyclic as desired.

Only if direction:

Let ${X}$ be the radical center of the circles of ${BRS, BCD, PQRS}$. (There is no degenerate case here, which was a problem in #1 of this year, because we have an isosceles trapezoid!) Since ${X, R, S, G}$ are collinear, ${XG||AB \implies \angle XGP = \angle BAP = \angle XBP}$, so ${XGBP}$ is cyclic. Hence, ${\angle XBG + \angle XPG = 180 \implies \angle XBG = \angle GPQ = \angle APQ = \angle ABQ}$. But ${\angle XBG = \angle XBQ = \alpha' + \theta}$ and ${\angle ABQ = \alpha + \theta}$, so ${\alpha = \alpha'}$ as desired.