Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcircles of triangles $BEN$ and $CDN$. $l_1$ and $l_2$ are perpendicular lines to $PM$ and $PT$ in points $M$ and $T$ respectively. Prove that lines $l_1$ and $l_2$ intersect on the circumcircle of triangle $ABC$. Proposed by Nima Hamidi
Problem
Source: Iran 3rd round 2012-Geometry exam-P2
Tags: geometry, circumcircle, geometric transformation, geometry proposed
tobash_co
20.09.2012 14:19
If incircle touches $AB,AC$ at $X,Y$ then $BE=AX=AY=CD$. Since $P$ is the center of spiral similarity sending $BE$ to $DC$, we immediately get $\bigtriangleup PEB\equiv \bigtriangleup PCD$. So $\angle EMP=\angle CTP$, so $A,M,P,T$ are concyclic together with $G$. Thus if $\l_1,\l_2$ intersect at $G$ then $GM=GT$, $BM=CT$ and $\angle GMB=\angle GTC$. Thus $\bigtriangleup GMB\equiv \bigtriangleup GTC$, and so \[\angle ABG=\angle AMG+\angle BGM=\angle ATG+\angle CGT=\angle ACG\], so $A,B,C,G$ are indeed concyclic.
ACCCGS8
22.09.2012 07:42
Observe that $M$ is the Miquel point of complete quadrilateral consisting of points $A, E, B, C, D, N$ so $ABPD$ and $AEPC$ are cyclic. Thus $\angle BEP = \angle DCP$ and $\angle EBP = \angle CDP$. Also $BE=AF=AG=DC$ so $\triangle BEP$ is congruent to $\triangle DCP$. Now $PM, PT$ are corresponding medians, so $\angle PME =\angle PTC$ so $AMPT$ is cyclic. Thus $AMPTX$ is cyclic so $\angle MXT = \angle BAC$. By RHS, $\triangle XMP$ is congruent to $\triangle XTP$ so $XM=XT$. Also, $MB=TC$ and since $AMTX$ is cyclic, $\angle XMB= \angle XTC$ so by SAS $\triangle XMB$ is congruent to $\triangle XTC$. Thus $\angle MXB=\angle TXC$ so $\angle BXC=\angle MXT=\angle BAC$ so $l_1, l_2$ intersect on the circumcircle of $\triangle ABC$.