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### USA(J)MO 2017 #3

Posted: **Sat Apr 22, 2017 6:06 pm**

by **dshasan**

Let $ABC$ be an equilateral triangle, and point $P$ on it's 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 $\bigtriangleup DEF$ is twice the area of $\bigtriangleup ABC$

### Re: USA(J)MO 2017 #3

Posted: **Sat Apr 22, 2017 6:23 pm**

by **Thanic Nur Samin**

We use barycentric coordinates.

Let $P\equiv (p:q:r)$. Now, we know that $pq+qr+rp=0$ [The equation of circumcircle for equilateral triangles].

Now, $D\equiv (0:q:r), E\equiv (p:0:r), F\equiv (p:q:0)$.

So, the area of $\triangle DEF$ divided by the area of $\triangle ABC$ is:

$$\dfrac{1}{(p+q)(q+r)(r+p)} \times \begin{vmatrix}

0 & q & r\\

p & 0 & r\\

p & q & 0

\end{vmatrix}$$

$$=\dfrac{2pqr}{(p+q+r)(pq+qr+rp)-pqr}$$

$$=\dfrac{2pqr}{-pqr}=-2$$.

The reason of negativity is that we took signed area.

Therefore the area of $DEF$ is twice the area of $ABC$.

### Re: USA(J)MO 2017 #3

Posted: **Sat Apr 22, 2017 6:48 pm**

by **Zawadx**

Thanic Nur Samin wrote:We use barycentric coordinates.

Let $P\equiv (p:q:r)$. Now, we know that $pq+qr+rp=0$ [The equation of circumcircle for equilateral triangles].

Now, $D\equiv (0:q:r), E\equiv (p:0:r), F\equiv (p:q:0)$.

So, the area of $\triangle DEF$ divided by the area of $\triangle ABC$ is:

$$\dfrac{1}{(p+q)(q+r)(r+p)} \times \begin{vmatrix}

0 & q & p\\

p & 0 & r\\

p & q & 0

\end{vmatrix}$$

$$=\dfrac{2pqr}{(p+q+r)(pq+qr+rp)-pqr}$$

$$=\dfrac{2pqr}{-pqr}=-2$$.

The reason of negativity is that we took signed area.

Therefore the area of $DEF$ is twice the area of $ABC$.

There's a typo in the determinant: zero for you~

### Re: USA(J)MO 2017 #3

Posted: **Sat Apr 22, 2017 7:04 pm**

by **joydip**

For those who loves synthetic geometry

Throughout the proof signed area will be used.

**Lemma**: Let $ABC$ be an equilateral triangle, and point $P$ on its circumcircle. Let $PB$ and $AC$ intersect at $E$, and $PC$ and $AB$ intersect at $F$.Then $ {[EPF]}={[ABPC]}$

**Proof:** Let the tangent to $(ABC)$ at $A$ meet $BP$ at $J$ .Then applying pascal's theorem on hexagon $AACPBB$ we get $JF \parallel BB \parallel AC$ . So

$${[EPF]}={[ECF]}-{[ECP]}={[ECJ]}-{[ECP]}={[PCJ]}={[PCB]}+{[BCJ]}={[PCB]}+{[BCA]}={[BPCA]}={[ABPC]}$$.

**Problem :** So , $${[DFE}]={[EPF]}+{[FPD]}+{[DPE]}$$

$$={[ABPC]}+{[BCPA]}+{[CAPB]} $$

$$ =\{ {[BPA]}+{[APC]} \}+\{ {[ABC]}-{[APC]} \} + \{ {[ABC]}-{[BPA]} \} $$

$$=2{[ABC]} $$

### Re: USA(J)MO 2017 #3

Posted: **Sat Apr 22, 2017 9:31 pm**

by **Thamim Zahin**

joydip wrote:

**Proof:** Let the tangent to $(ABC)$ at $A$ meet $BP$ at $J$ .Then applying pascal's theorem on hexagon $AACPBB$ we get $JF \parallel BB \parallel AC$ . So

How did you get that idea?

### Re: USA(J)MO 2017 #3

Posted: **Mon Apr 24, 2017 11:26 am**

by **Atonu Roy Chowdhury**

WLOG $P$ lies on the shorter arc $BC$ . So, $[DEF]=[AEF]-[ABC]-[BDF]-[CDE]$

$\angle BAD = \alpha $

Use Sine Law to find $BD$,$DC$,$BF$,$CE$ in terms of $a$ and sine of $\alpha$ and $60-\alpha$, where $a$ is the length of the sides of $\triangle ABC$ . Then we'll use these lengths to find $[AEF]$,$[BDF]$ and $[CDE]$ . We've to prove $[DEF] =\frac{\sqrt3}{2} a^2$

After some simplification, we get

$\frac{(\sin^2 \alpha + \sin^2 (60 - \alpha) )( \sin \alpha + \sin (60-\alpha) ) - \sin^3 \alpha -\sin^3(60-\alpha)}{\sin \alpha \sin (60-\alpha)(\sin \alpha + \sin (60-\alpha))}=1$ which is obviously true, and so we are done.

### Re: USA(J)MO 2017 #3

Posted: **Mon Apr 24, 2017 1:45 pm**

by **Thanic Nur Samin**

Zawadx wrote:
There's a typo in the determinant: zero for you~

Edited. Latexing a determinant is a pain in the first place, locating these typos are difficult

It was correct in my paper, so if I had submitted, it wouldn't have been a zero, rather a seven.