2017 USAJMO Problems/Problem 3

Problem

( $*$ ) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ ; let lines $PB$ and $CA$ intersect at $E$ ; and let lines $PC$ and $AB$ intersect at $F$ . Prove that the area of triangle $DEF$ is twice that of triangle $ABC$ .

$[asy] size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), P = (0, -sqrt(3)), D = (0, 0), E1 = (6, -3sqrt(3)), F = (-6, -3sqrt(3)), O = (0, sqrt(3)); draw(Circle(O, 2sqrt(3)), black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); label("A", A, N); label("B", B, W); label("C", C, E); label("P", P, S); label("D", D, NW); label("E", E1, SE); label("F", F, SW); [/asy]$

Solution 1

WLOG, let $AB = 1$ . Let $[ABD] = X, [ACD] = Y$ , and $\angle BAD = \theta$ . After some angle chasing, we find that $\angle BCF \cong \angle BEC \cong \theta$ and $\angle FBC \cong \angle BCE \cong 120^{\circ}$ . Therefore, $\triangle FBC$ ~ $\triangle BCE$ .

Lemma 1: If $BF = k$ , then $CE = \frac 1k$ . This lemma results directly from the fact that $\triangle FBC$ ~ $\triangle BCE$ ; $\frac{BF}{BC} = \frac{BF}{1} = \frac{BC}{CE} = \frac{1}{CE}$ , or $CE = \frac{1}{BF}$ .

Lemma 2: $[AEF] = (k+\frac 1k + 2)(X+Y)$ . We see that $[AEF] = (X+Y) \frac{[AEF]}{[ABC]} = (k+1)(1+\frac 1k)(X+Y) = (k + \frac 1k + 2)(X+Y)$ , as desired.

Lemma 3: $\frac{X}{Y} = k$ . We see that $\frac XY = \frac{\frac 12 (AB)(AD) \sin(\theta)}{\frac 12 (AC)(AD) \sin (60^{\circ} - \theta)} = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}.$ However, after some angle chasing and by the Law of Sines in $\triangle BCF$ , we have $\frac{k}{\sin(\theta)} = \frac{1}{\sin(60^{\circ} - \theta)}$ , or $k = \frac{\sin(\theta)}{\sin (60^{\circ} - \theta)}$ , which implies the result.

By the area lemma, we have $[BDF] = kX$ and $[CDF] = \frac{Y}{k}$ .

We see that $[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \frac Xk + \frac Yk + 2X + 2Y - X - Y - Xk - \frac Yk = X + Y + \frac Xk + yk$ . Thus, it suffices to show that $X + Y + \frac Xk + Yk = 2X + 2Y$ , or $\frac Xk + Yk = X + Y$ . Rearranging, we find this to be equivalent to $\frac XY = k$ , which is Lemma 3, so the result has been proven.

Solution 2

$A = (1, 0, 0)$ $B = (0, 1, 0)$ $C = (0, 0, 1)$ $P = \left(x_P, y_P, z_P\right)$ $D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)$ $E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)$ $F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right)$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

2017 USAJMO (Problems • Resources)
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2017 USAJMO Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

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