Difference between revisions of "2017 USAJMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
+ | 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 BCE \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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+ | ------------------------------------------------------------------------------------------------------ | ||
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+ | Lemma 1: If <math>BF = k</math>, then <math>CE = \frac 1k</math>. | ||
+ | This lemma results directly from the fact that <math>\triangle FBC</math> ~ <math>\triangle BCE</math>; <math>\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}</math>, or <math>CE = \frac{1}{BF}</math>. | ||
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+ | ------------------------------------------------------------------------------------------------------ | ||
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+ | Lemma 2: <math>[AEF] = (k+\frac 1k + 2)(X+Y)</math>. | ||
+ | We see that <math>[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)</math>, as desired. | ||
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+ | ------------------------------------------------------------------------------------------------------ | ||
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+ | Lemma 3: <math>\frac{X}{Y} = k</math>. | ||
+ | We see that | ||
+ | <cmath>\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.</cmath> | ||
+ | However, after some angle chasing and by the Law of Sines in <math>\triangle BCF</math>, we have <math>\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}</math>, or <math>k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}</math>, which implies the result. | ||
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+ | ------------------------------------------------------------------------------------------------------ | ||
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+ | By the area lemma, we have <math>[BDF] = kX</math> and <math>[CDF] = \frac{Y}{k}</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. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:51, 19 April 2017
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
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.
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 |