Problem 3

Let

ABC

be a scalene triangle with circumcircle

\Omega

. Let

M

be the midpoint of

BC

. A variable point

P

is selected on segment

AM

. The circumcircles of triangles

BPM

and

CPM

meet

\Omega

again at points

D

and

E

, respectively. The lines

DP

and

EP

meet the circumcircles of triangles

CPM

and

BPM

again at points

X

and

Y

, respectively. Prove that, as

P

varies, the circumcircle of triangle

AXY

passes through a fixed point

T

distinct from

A

Step 6 of 9: Fixed points Q, R carry P onto two chords

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CQ\parallel BR\parallel AM\ (Q,R\in\Omega);\quad \angle QDB=\angle QCB=\angle AMB=\angle PDB\ \Longrightarrow\ D,P,Q\ \text{collinear (similarly }E,P,R)

Detailed analysis

Let $Q$ be the second intersection of $\Omega$ with the line through $C$ parallel to $AM$ , and $R$ the second intersection of $\Omega$ with the line through $B$ parallel to $AM$ ; both are fixed, since they depend only on the fixed direction $AM$ and the fixed triangle. Since $Q,D,B,C\in\Omega$ , $\angle QDB=\angle QCB$ ; since $CQ\parallel AM$ , $\angle QCB=\angle AMB$ (transversal $CB$ ); since $P\in AM$ , $\angle AMB=\angle PMB$ ; since $B,P,M,D$ are concyclic, $\angle PMB=\angle PDB$ . Chaining these gives $\angle QDB=\angle PDB$ , so rays $DQ$ and $DP$ coincide, i.e. $D,P,Q$ are collinear. The symmetric argument with $R$ gives $E,P,R$ collinear.