Let $ABC$ be an acute triangle, $M$ be the midpoint of $BC$ and $P$ be a point on line segment $AM$. Lines $BP$ and $CP$ meet the circumcircle of $ABC$ again at $X$ and $Y$ , respectively, and sides $AC$ at $D$ and $AB$ at $E$, respectively. Prove that the circumcircles of $AXD$ and $AYE$ have a common point $T \ne A$ on line $AM$.
Problem
Source: 2013 Saudi Arabia IMO TST III p3
Tags: geometry, circumcircle
Ricochet
24.07.2020 06:02
We need to prove that $A, T, P$ are collinear, see that $AT$ is radical axis of $(AXD), (AYE)$, but also by Cheva, we have $DE \| BC$ so by Reim's $DEYX$ is cyclic, and $P$ is also on the radical axis of $(AXD), (AYE)$
nabodorbuco
25.07.2020 23:28
Hi there! I made a video about it. https://youtu.be/HBL2GMdYMVs
dgreenb801
26.07.2020 03:21
By Ceva's Theorem, $\frac{BE}{EA} \cdot \frac{AD}{DC} = 1 \implies DE \parallel BC$. Let $AM$ intersect $ED$ at $T$, then since $BM=MC$, $ET=TD$. Also, $\angle AYC = \angle ABC = \angle AED$, so the circumcircle of $\triangle AYE$ is tangent to line $DE$. Similarly, the circumcircle of $\triangle AXD$ is tangent to the line $DE$. Then the power of $T$ with respect to both circle is the same (it is $TE^2=TD^2$). Thus, line $AT$ is the radical axis of the two circles, so the second intersection point of the circles lies on $AT$.