APMO 2013 Problem 5

Discussion on Asian Pacific Mathematical Olympiad (APMO)
User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:
APMO 2013 Problem 5

Unread post by Phlembac Adib Hasan » Thu May 09, 2013 11:16 am

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.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
SANZEED
Posts:550
Joined:Wed Dec 28, 2011 6:45 pm
Location:Mymensingh, Bangladesh

Re: APMO 2013 Problem 5

Unread post by SANZEED » Fri May 10, 2013 12:59 am

(I got the idea to use Harmonic Division in this problem from Nadim Ul Abrar vai after the exam. This problem actually made me learn Harmonic Division)
Let $BR\cap \omega=E', CE'\cap AR=S$.
Now by the first lemma of Zhao, $AC$ is a symmedian of $\triangle ABD$. Thus $(C,A;D,B)=-1$.
Thus $E'(C,A;D,B)\cap AR\Rightarrow (S,A;D,R)=-1$.
Now $C(S,A;D,R)\cap \omega \Rightarrow (E',A;D,C)=-1$.
This means that $AE'$ passes through the intersection point of the tangents to $\omega$ at $C,D$.
Thus we get that $E\equiv E'$ and $B,E,R$ collinear.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

Post Reply