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
[hide]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 |