Difference between revisions of "2003 USAMO Problems/Problem 4"
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\Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \ | \Longleftrightarrow\angle CBM = \angle CBG = \angle CFG = \angle DCF = \angle DCM \ | ||
\Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}</math> | \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD = MC^{2}</math> | ||
+ | |||
+ | Added diagram: | ||
+ | |||
+ | <asy> | ||
+ | import graph; size(5cm); | ||
+ | real labelscalefactor = 0.5; | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); | ||
+ | pen dotstyle = black; | ||
+ | real xmin = -4.3, xmax = 6.36, ymin = -3.98, ymax = 6.3; | ||
+ | draw((-0.16,3.1)--(-2.48,0.52)); | ||
+ | draw((-2.48,0.52)--(3.78,0.52)); | ||
+ | draw((3.78,0.52)--(-0.16,3.1)); | ||
+ | draw(circle((-0.42,1), 2.12)); | ||
+ | draw((1.04,4.43)--(3.78,0.52)); | ||
+ | draw((1.04,4.43)--(-2.48,0.52)); | ||
+ | draw((1.04,4.43)--(1.64,0.52)); | ||
+ | draw((2.41,2.48)--(-2.48,0.52)); | ||
+ | dot((-0.16,3.1),dotstyle); | ||
+ | label("$A$", (-0.4,3.3), NE * labelscalefactor); | ||
+ | dot((-2.48,0.52),dotstyle); | ||
+ | label("$B$", (-2.84,0.54), SW * labelscalefactor); | ||
+ | dot((3.78,0.52),dotstyle); | ||
+ | label("$C$", (3.86,0.64), NE * labelscalefactor); | ||
+ | dot((1.4,2.08),dotstyle); | ||
+ | label("$D$", (1.48,2.2), NE * labelscalefactor); | ||
+ | dot((1.64,0.52),dotstyle); | ||
+ | label("$E$", (1.72,0.64), NE * labelscalefactor); | ||
+ | dot((1.04,4.43),dotstyle); | ||
+ | label("$F$", (1.12,4.56), NE * labelscalefactor); | ||
+ | dot((2.41,2.48),dotstyle); | ||
+ | label("$M$", (2.48,2.6), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | </asy> | ||
== Resources == | == Resources == |
Revision as of 16:21, 12 April 2012
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:
Added diagram: