A line parallel to the side $BC$ of a triangle $ABC$ meets the sides $AB$ and $AC$ at points $P$ and $Q$, respectively. A point $M$ is chosen inside the triangle $APQ$. The segments $MB$ and $MC$ meet the segment $PQ$ at points $E$ and $F$, respectively. Let $N$ be the second intersection point of the circumcircles of the triangles $PMF$ and $QME$. Prove that the points $A,M,N$ are collinear.
Problem
Source: Sharygin geometry olympiad 2016, grade 10, Final Round, Problem 1.
Tags: geometry, circumcircle, Sharygin Geometry Olympiad
anantmudgal09
05.08.2016 16:11
Let the circumcircles of the triangles $PMF$ and $QME$ meet the sides $AB$ and $AC$ again at points $P'$ and $Q'$ respectively. We need to prove that $A$ lies on the radical axis of these two circles, hence it is equivalent to showing that $AP'\cdot AP=AQ'\cdot AQ$. Dilating about $A$, with factor $\frac{BC}{PQ}$, it suffices to showing that points $B,P',Q',C$ lie on a circle. However, we notice that $\angle PEM=\angle MCB=\angle PP'M$ and so the point $P'$ lies on the circumcircle of triangle $BMC$. The same holds for $Q'$ and the conclusion holds.
Comment This problem also appeared in India TST 1995.
Favel48
06.10.2016 07:51
Let $S=BM\cap PN$, $T=CM\cap QN$, by angle chasing $SMTN$ is a cyclic quadrilateral and $ST\parallel EF\parallel BC$. Now we prove by Menelaus that $A,M,N$ are collinear using $\triangle BSP$: \[\frac{BM}{MS}\cdot \frac{SN}{NP} \cdot \frac{PA}{AB}=1\]But using triangle similarities: $\triangle MST\sim \triangle MBC$, $\triangle NST \sim \triangle NPQ$, and $\triangle APQ\sim \triangle ABC$, this transforms into: \[\frac{BC}{ST}\cdot \frac{ST}{PQ}\cdot \frac{PQ}{BC}=1\]which is true and ends our proof.
math90
06.10.2016 09:12
Very easy problem, my solution is the same as anantmudgal09's.
khanhnx
17.03.2019 13:49
Here is my solution for this problem Solution Let $T$, $U$ be second intersections of ($PMF$), ($QME$) with $AB$, $AC$ We have: ($TM$; $TP$) $\equiv$ ($FM$; $FP$) $\equiv$ ($CM$; $CB$) (mod $\pi$) and ($UM$; $UQ$) $\equiv$ ($EM$; $EQ$) $\equiv$ ($BM$; $BC$) (mod $\pi$) So: $B$, $T$, $M$, $U$, $C$ lie on a circle But: $PQ$ $\parallel$ $BC$ then: $T$, $P$, $U$, $Q$ lie on a circle Then: $P_{A / (PMF)}$ = $\overline{AP}$ . $\overline{AT}$ = $\overline{AU}$ . $\overline{AQ}$ = $P_{A / (QME)}$ Hence: $A$ lies on radical axis of ($PMF$), ($QME$) or $A$, $M$, $N$ are collinear
WALKERSkeren
13.11.2022 13:55
Favel48 wrote: Let $S=BM\cap PN$, $T=CM\cap QN$, by angle chasing $SMTN$ is a cyclic quadrilateral and $ST\parallel EF\parallel BC$. Now we prove by Menelaus that $A,M,N$ are collinear using $\triangle BSP$: \[\frac{BM}{MS}\cdot \frac{SN}{NP} \cdot \frac{PA}{AB}=1\]But using triangle similarities: $\triangle MST\sim \triangle MBC$, $\triangle NST \sim \triangle NPQ$, and $\triangle APQ\sim \triangle ABC$, this transforms into: \[\frac{BC}{ST}\cdot \frac{ST}{PQ}\cdot \frac{PQ}{BC}=1\]which is true and ends our proof. what a nice solution