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Difference between revisions of "2017 USAJMO Problems/Problem 3"

(→‎Solution 2)
(→‎Solution 2)
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<cmath>C = (0, 0, 1)</cmath>
 
<cmath>C = (0, 0, 1)</cmath>
 
<cmath>P = \left(x_P, y_P, z_P\right)</cmath>
 
<cmath>P = \left(x_P, y_P, z_P\right)</cmath>
 +
The fact that <math>P</math> lies on the circumcircle of <math>\triangle{ABC}</math> gives us <math>x^2_P + y^2_P + z^2_P = 1.</math> This, along with <math>x_P + y_P + z_P = 1,</math> gives us <math>x_Py_P + y_Pz_P + z_Px_P = 0.</math>
 +
 
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath>
 
<cmath>D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)</cmath>
 
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath>
 
<cmath>E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)</cmath>

Revision as of 22:06, 5 May 2017

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

\[A = (1, 0, 0)\] \[B = (0, 1, 0)\] \[C = (0, 0, 1)\] \[P = \left(x_P, y_P, z_P\right)\] The fact that $P$ lies on the circumcircle of $\triangle{ABC}$ gives us $x^2_P + y^2_P + z^2_P = 1.$ This, along with $x_P + y_P + z_P = 1,$ gives us $x_Py_P + y_Pz_P + z_Px_P = 0.$

\[D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)\] \[E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)\] \[F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right)\]

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

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
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