APMO 2013 P5
- Atonu Roy Chowdhury
- Posts:64
- Joined:Fri Aug 05, 2016 7:57 pm
- Location:Chittagong, Bangladesh
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.
Re: APMO 2013 P5
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: APMO 2013 P5
Already posted here:
viewtopic.php?f=15&t=2758&p=13832&hilit ... 2F5#p13832
viewtopic.php?f=15&t=2758&p=13832&hilit ... 2F5#p13832
"Questions we can't answer are far better than answers we can't question"
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: APMO 2013 P5
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
Frankly, my dear, I don't give a damn.