APMO 2013 P5

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Atonu Roy Chowdhury
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APMO 2013 P5

Unread post by Atonu Roy Chowdhury » Fri Aug 19, 2016 8:17 pm

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.

joydip
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Re: APMO 2013 P5

Unread post by joydip » Tue Nov 22, 2016 4:10 pm

Let, $G=BD \cap AC,F=DC \cap AQ$ .Then $,(P,G;A,C)= -1 $.So , R,F,G are collinear.Applying pascals theorem on hexagon $AEBDCC$ we get $EB, CC$(tangent at $C),FG$ are concurrent $\Rightarrow B,E,R$ are collinear.
The first principle is that you must not fool yourself and you are the easiest person to fool.

tanmoy
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Re: APMO 2013 P5

Unread post by tanmoy » Tue Nov 22, 2016 10:18 pm

"Questions we can't answer are far better than answers we can't question"

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ahmedittihad
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Re: APMO 2013 P5

Unread post by ahmedittihad » Sun Dec 11, 2016 4:38 pm

Take a projective transformation that fixes ω and sends the pointAC ∩ BD to the center of the circle. Thus ABCD is a rectangle. Because ABCD is harmonic, it must in fact be a square. Thus P is the point at infinity along AB CD and the problem is not very hard now. -Copied from Evan Chen's book
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