2017 USAJMO #3

Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PA$ and $BC$ intersect at $D$, $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$.

Solution 1: 

Note that $$\angle{DPF}=\angle{FPE}=\angle{EPD}=120^{\circ}.$$ Now, using the sin area formula, we get $$[DEF]=[DPF]+[FPE]+[EPD]=\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right).$$We also have that $$[ABC]=BC^2 \cdot \frac{\sqrt{3}}{4}.$$ So, it suffices to prove $$\frac{\sqrt{3}}{4}\left(DP \cdot FP+FP \cdot EP+EP \cdot DP\right)=2 \cdot \frac{BC^2\sqrt{3}}{4} \implies DP \cdot FP+FP \cdot EP+EP \cdot DP=2BC^2.$$By the Law of Cosines on $\triangle{BPC}$, we obtain $$BC^2=b^2+c^2-2 \cdot b \cdot c \cdot \cos{120^{\circ}}=b^2+c^2+bc,$$ and Ptolemy's Theorem on quadrilateral $ABPC$ gives$$AP \cdot BC=BP \cdot AC+CP \cdot AB \implies AP=b+c.$$From some angle chasing, we know that $\triangle{ACD} \sim \triangle{APC}$, and therefore $$\frac{AC}{AP}=\frac{CD}{PC}=\frac{AD}{AC} \implies AC^2=AD \cdot AP=AD(b+c) \implies AD=\frac{b^2+bc+c^2}{b+c}.$$So, we have $$PD=AP-AD=b+c-\left(\frac{b^2+bc+c^2}{b+c}\right)=\frac{bc}{b+c}.$$Now, from angle chasing again, we see that $$\triangle{FBP} \sim \triangle{ACP} \implies \frac{FB}{AC}=\frac{BP}{CP}=\frac{FP}{AP} \implies FP=\frac{BP}{CP} \cdot AP=\frac{b(b+c)}{c},$$ and similarly $EP=\frac{c(b+c)}{b}.$ Thus, we finally have 

$$DP \cdot FP+FP \cdot EP+EP \cdot DP=\frac{bc}{b+c} \cdot \frac{b(b+c)}{c}+\frac{b(b+c)}{c} \cdot\frac{c(b+c)}{b}+\frac{c(b+c)}{b} \cdot \frac{bc}{b+c}$$

$$=b^2+(b+c)^2+c^2=2b^2+2bc+2c^2=2(b^2+bc+c^2)=2 \cdot BC^2,$$ as desired. $\square$

Solution 2: 

There is a solution with barycentric coordinates, which should be pretty obvious if you know the technique. Regardless, I will upload this solution when I have time.