# 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 (No Trig/Bash)

Extend $DP$ to hit $EF$ at $K$. Then note that $[DEF]\cdot\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=[ABC].$ Letting $BF=x$ and $PF=y$, we have that $\frac{x+AB}y=\frac{y+PC}x=\frac{AC}{BP}.$ Solving and simplifying using LoC on $\triangle BPC$ gives $\frac{AB}{AF}=\frac{PC}{PB+PC}.$ Similarly, $\frac{AC}{AE}=\frac{PB}{PB+PC}.$

Now we find $\frac{AK}{DK}.$ Note that $\frac{AD}{DP}=\frac{AD}{BD}\cdot\frac{BD}{DP}=\frac{AC}{PB}\cdot\frac{AB}{PC}=\frac{AB^2}{PB\cdot PC}.$ Now let $E'=DE\cap AF$ and $F'=DF\cap AE$. Then by an area/concurrence theorem, we have that $\frac{DK}{AK}+\frac{DE'}{EE'}+\frac{DF'}{FF'}=1,$ or $\frac{DK}{AK}+(1-\frac{DP}{AP}-\frac{DC}{BC})+(1-\frac{DP}{AP}-\frac{BD}{BC})=1.$ Thus we have that $\frac{DK}{AK}=2\cdot\frac{DP}{AP}.$

Manipulating these gives $\frac{AK}{DK}=\frac{(PB+PC)^2}{2\cdot PB\cdot PC}.$ Thus $\frac{AK}{DK}\cdot\frac{AB}{AF}\cdot\frac{AC}{AE}=\frac12,$ and we are done.

~cocohearts

## 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

We will use barycentric coordinates and vectors. Let $\vec{X}$ be the position vector of a point $X.$ The point $(x, y, z)$ in barycentric coordinates denotes the point $x\vec{A} + y\vec{B} + z\vec{C}.$ For all points in the plane of $\triangle{ABC},$ we have $x + y + z = 1.$ It is clear that $A = (1, 0, 0)$; $B = (0, 1, 0)$; and $C = (0, 0, 1).$

Define the point $P$ as $P = \left(x_P, y_P, z_P\right).$ 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 the condition $x_P + y_P + z_P = 1$ inherent to barycentric coordinates, gives us $x_Py_P + y_Pz_P + z_Px_P = 0.$

We can write the equations of the following lines: $$BC: x = 0$$ $$CA: y = 0$$ $$AB: z = 0$$ $$PA: \frac{y}{y_P} = \frac{z}{z_P}$$ $$PB: \frac{x}{x_P} = \frac{z}{z_P}$$ $$PC: \frac{x}{x_P} = \frac{y}{y_P}.$$

We can then solve for the points $D, E, F$: $$D = \left(0, \frac{y_P}{y_P + z_P}, \frac{z_P}{y_P + z_P}\right)$$ $$E = \left(\frac{x_P}{x_P + z_P}, 0, \frac{z_P}{x_P + z_P}\right)$$ $$F = \left(\frac{x_P}{x_P + y_P}, \frac{y_P}{x_P + y_P}, 0\right).$$

The area of an arbitrary triangle $XYZ$ is: $$[XYZ] = \frac{1}{2}|\vec{XY}\times\vec{XZ}|$$ $$[XYZ] = \frac{1}{2}|(\vec{X}\times\vec{Y}) + (\vec{Y}\times\vec{Z}) + (\vec{Z}\times\vec{X})|.$$

To calculate $[DEF],$ we wish to compute $(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}).$ After a lot of computation, we obtain the following: $$(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = \frac{2x_Py_Pz_P}{(x_P + y_P)(y_P + z_P)(z_P + x_P)}[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].$$

Evaluating the denominator, $$(x_P + y_P)(y_P + z_P)(z_P + x_P) = (1 - z_P)(1 - y_P)(1 - x_P)$$ $$(x_P + y_P)(y_P + z_P)(z_P + x_P) = 1 - (x_P + y_P + z_P) + (x_Py_P + y_Pz_P + z_Px_P) - x_Py_Pz_P.$$

Since $x_P + y_P + z_P = 1$ and $x_Py_P + y_Pz_P + z_Px_P = 0,$ it follows that: $$(x_P + y_P)(y_P + z_P)(z_P + x_P) = -x_Py_Pz_P.$$

We thus conclude that: $$(\vec{D}\times\vec{E}) + (\vec{E}\times\vec{F}) + (\vec{F}\times\vec{D}) = -2[(\vec{A}\times\vec{B}) + (\vec{B}\times\vec{C}) + (\vec{C}\times\vec{A})].$$

From this, it follows that $[DEF] = 2[ABC],$ and we are done.

## Solution 3

$[asy] import cse5; import graph; import olympiad; size(3inch); pair A = (0, 3sqrt(3)), B = (-3,0), C = (3,0), O = (0, sqrt(3)); pair P = (-1, -sqrt(11)+sqrt(3)); path circle = Circle(O, 2sqrt(3)); pair D = extension(A,P,B,C); pair E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label("A", A, N); label("B", B, W); label("C", C, E); label("P", P, S); label("D", D, NW); label("E", E1, SE); label("F", F, SW); dot("O", O, SE); [/asy]$

We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\sqrt{3})$. Using the facts $\triangle{CBP} \sim \triangle{CFB}$ and $\triangle{BCP} \sim \triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \frac{1}{2x}$, and $E = (\frac{1}{x}+1,-\frac{\sqrt{3}}{x})$.

The slope of $FE$ is $$k = \frac{-\frac{\sqrt{3}}{x} + x\sqrt{3}}{2+\frac{1}{x}+x}$$ It is well-known that $\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\frac{1}{k}$. Since $O=(0,\frac{1}{\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \frac{k}{\sqrt{3}}$, i.e., $D = (\frac{k}{\sqrt{3}},0)$. Using shoelace, $$2[\triangle{FDE}] = \frac{k}{\sqrt{3}}(-x\sqrt{3})+(-x-1)(-\frac{\sqrt{3}}{x})- (-x\sqrt{3})(\frac{1}{x}+1) - (-\frac{\sqrt{3}}{x})\frac{k}{\sqrt{3}}$$ $$= 2\sqrt{3} + \sqrt{3}(\frac{1}{x}+x) + k(\frac{1}{x} - x)$$ $$= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+(\frac{1}{x}+x)^2-(\frac{1}{x}-x)^2} {2+\frac{1}{x}+x})$$ $$= 2\sqrt{3} + \sqrt{3}(\frac{2(\frac{1}{x}+x)+4}{2+\frac{1}{x}+x})$$ $$= 4\sqrt{3}$$ So $[\triangle{FDE}] = 2\sqrt{3} = 2[\triangle{ABC}]$. Q.E.D

By Mathdummy.

## Solution 4 Without the nasty computations

Note that $\angle{APB}=\angle{FPB}=\angle{EPC}=\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths.

Let $BP = x$ and $CP = y$. Then, From Law of Cosine, $BC^2 = x^2 + y^2 + xy$.

From Ptolemy's theorem, $AP BC = x AC + y AB$, so $AP = x + y$.

Lemma 1: In Triangle ABC with side lengths $a,b,c$ and $\angle A =120^o$, the length of the angle bisector of $A$ is $$d = \frac{bc}{b+c}$$ This can be easily proved with Stewart's and Law of Cosine.

Using Lemma 1, we have $$PD = \frac{xy}{x+y}$$ $$x = \frac{FP AP}{FP + AP}$$ $$y = \frac{EP AP}{EP + AP}$$ Plug in $AP=x+y$, we get: $$PD = \frac{xy}{x+y}$$ $$FP = \frac{x(x+y)}{y}$$ $$EP = \frac{y(x+y)}{x}$$ Then $$[\triangle{FDE}] = \frac{1}{2}\sin(120)(PDFP + FPEP + EPPD)$$ $$= \frac{\sqrt{3}}{4}(x^2 + (x+y)^2 + y^2)$$ $$= \frac{\sqrt{3}}{2}(x^2 + xy + y^2)$$ $$= \frac{\sqrt{3}}{2} BC^2$$ $$= 2 [\triangle{ABC}]$$

By Mathdummy.