Suppose $ABC$ is a triangle with $M$ the midpoint of $BC$. Suppose that $AM$ intersects the incircle at $K,L$. We draw parallel line from $K$ and $L$ to $BC$ and name their second intersection point with incircle $X$ and $Y$. Suppose that $AX$ and $AY$ intersect $BC$ at $P$ and $Q$. Prove that $BP=CQ$.
Problem
Source: Iran TST 2006
Tags: geometry, geometry proposed
Amir.S
17.04.2006 15:53
Easy problem for IRAN TST. Lemma.let $D$ , $E$ and $F$ be intesection points of incircle with $BC,AC,AB$ respectively. prove that median $AM,DI,EF$ are concurrent. Proof. with sinus law. back to the problem from the lemma we can get that $(A,K,X,L)=-1$ hence the the polur of $X$ is a line($\ell$) parallel to $BC$ and pass from $A$.it's simple from here
Nima Ahmadi Pour
17.04.2006 16:00
Yes. Almost all of the participants will get full mark for this day! (I think about 14 out of 16)
imortal
23.04.2006 20:19
Hi,amir2, please help me, I can't understand your solution. The lemma is OK, actually it helped me solve the problem in my way but I don't know what you mean by (A,K,X,L)=-1; and the polur of X through which inversion, (or maybe this means something else ;? ) Thank you.
darij grinberg
24.04.2006 13:39
also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=80872 . darij
xirti
01.05.2006 21:15
This problem apeared in a romanian competition this year, so I can guess where it comes from