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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}.


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.


1. william - August 8, 2010


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