Difference between revisions of "2017 USAJMO Problems/Problem 3"
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WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. | WLOG, let <math>AB = 1</math>. Let <math>[ABD] = X, [ACD] = Y</math>, and <math>\angle BAD = \theta</math>. After some angle chasing, we find that <math>\angle BCF \cong \angle BEC \cong \theta</math> and <math>\angle FBC \cong \angle BCE \cong 120^{\circ}</math>. Therefore, <math>\triangle FBC</math> ~ <math>\triangle BCE</math>. | ||
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We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven. | We see that <math>[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk</math>. Thus, it suffices to show that <math>X + Y + \frac Xk + Yk = 2X + 2Y</math>, or <math>\frac Xk + Yk = X + Y</math>. Rearranging, we find this to be equivalent to <math>\frac XY = k</math>, which is Lemma 3, so the result has been proven. | ||
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+ | ==Solution 2== | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:58, 5 May 2017
Contents
Problem
() Let be an equilateral triangle and let be a point on its circumcircle. Let lines and intersect at ; let lines and intersect at ; and let lines and intersect at . Prove that the area of triangle is twice that of triangle .
Solution 1
WLOG, let . Let , and . After some angle chasing, we find that and . Therefore, ~ .
Lemma 1: If , then . This lemma results directly from the fact that ~ ; , or .
Lemma 2: . We see that , as desired.
Lemma 3: . We see that However, after some angle chasing and by the Law of Sines in , we have , or , which implies the result.
By the area lemma, we have and .
We see that . Thus, it suffices to show that , or . Rearranging, we find this to be equivalent to , which is Lemma 3, so the result has been proven.
Solution 2
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |