NMO Selection Tests for BMO and IMO

For convenience, let $\Omega$ and $\gamma$ denote the circles $AEFPQ$ and $MPQ$, respectively. Letting the line $AD$ cross $\omega$ again at $T$, we will show that $\gamma$ and $\omega$ are tangent at $T$.

We first prove that the points $M, P, Q, T$ are concyclic, so $T$ lies on $\gamma$. Let $A'$ be the center of $\omega$, and notice that $AA'$ is a diameter of $\Omega$, since $A'E$ and $A'F$ are perpendicular to $AE$ and $AF$, respectively. Let $N$ be the midpoint of the line segment $DT$. Since $A'D = A'T$, the angle $A'NA$ is right, so $N$ also lies on $\Omega$, and therefore $DP \cdot DQ = DA \cdot DN$; and since $DA = 2DM$ and $DT = 2DN$, it follows that $DP \cdot DQ = DM \cdot DT$, showing that $M, P, Q, T$ are indeed concyclic.

If $EF$ and $BC$ are parallel, then the triangle $ABC$ is isosceles with apex at $A$ and the conclusion follows by symmetry.

Otherwise, the lines $PQ$ and $EF$ cross at a point $R$; these lines are the radical axes of $\Omega, \gamma$ and $\Omega, \omega$, respectively, so $R$ is the radical center of the three circles, and $RT$ is the radical axis of $\gamma$ and $\omega$.

To conclude the proof, it is therefore sufficient to show that $RT$ is tangent to $\omega$. With respect to this circle, the line $DR$ is the polar of $D$, and the line $EF$ is the polar of $A$, so $R = DR \cap EF$ is the pole of the line $ADT$. Consequently, $RT$ is indeed tangent to $\omega$.