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BdMO Online Forum • View topic - USA(J)MO 2017 #3

USA(J)MO 2017 #3

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USA(J)MO 2017 #3

Post Number:#1  Unread postby dshasan » Sat Apr 22, 2017 6:06 pm

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 $\bigtriangleup ABC$
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Re: USA(J)MO 2017 #3

Post Number:#2  Unread postby Thanic Nur Samin » Sat Apr 22, 2017 6:23 pm

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 $ABC$.
Last edited by Thanic Nur Samin on Mon Apr 24, 2017 1:43 pm, edited 1 time in total.
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Re: USA(J)MO 2017 #3

Post Number:#3  Unread postby Zawadx » Sat Apr 22, 2017 6:48 pm

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~
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Re: USA(J)MO 2017 #3

Post Number:#4  Unread postby joydip » Sat Apr 22, 2017 7:04 pm

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 , $${[DFE}]={[EPF]}+{[FPD]}+{[DPE]}$$

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

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

$$=2{[ABC]} $$
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Re: USA(J)MO 2017 #3

Post Number:#5  Unread postby Thamim Zahin » Sat Apr 22, 2017 9:31 pm

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?
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Re: USA(J)MO 2017 #3

Post Number:#6  Unread postby Atonu Roy Chowdhury » Mon Apr 24, 2017 11:26 am

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.
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Re: USA(J)MO 2017 #3

Post Number:#7  Unread postby Thanic Nur Samin » Mon Apr 24, 2017 1:45 pm

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 :cry:

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