Difference between revisions of "2003 USAMO Problems/Problem 4"
(New page: == Problem == Let <math>ABC</math> be a triangle. A circle passing through <math>A</math> and <math>B</math> intersects segments <math>AC</math> and <math>BC</math> at <math>D</math> and ...) |
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Line 4: | Line 4: | ||
== Solution == | == Solution == | ||
− | {{ | + | by April |
+ | |||
+ | Take <math>G\in BD: \,FG\parallel CD</math>. We have: | ||
+ | |||
+ | <math>MF = MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram} \ | ||
+ | \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA = \angle GCD\Longleftrightarrow\angle FDA + \angle CGF = 180^\circ \ | ||
+ | \Longleftrightarrow \angle ABE + \angle CGF = 180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic} \ | ||
+ | \Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \ | ||
+ | \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}</math> | ||
== Resources == | == Resources == |
Revision as of 18:54, 18 December 2008
Problem
Let be a triangle. A circle passing through and intersects segments and at and , respectively. Lines and intersect at , while lines and intersect at . Prove that if and only if .
Solution
by April
Take . We have: