Iranian Mathematical Olympiad

Let in triangle $ABC$, points $I_a$, $H$, and $D$ be the $A$-excenter, foot of the altitude from $A$, and foot of the angle bisector from $A$, respectively. Since $(AD,I{I}_a)=-1$ and $\angle AHD=90^{\circ }$, we have $\angle IHD=\angle DH{I}_a$. Further, $AIHQ$ is cyclic, So $\angle QAI=\angle IHD$, hence $\angle QAI=\angle DH{I}_a$ and $△QAI\sim △I_{a}HE$. We know $\angle AHE\sim \angle FAI$, therefore $\angle QIF\sim \angle I_{a}EA$ and we have $$\angle FQI=\angle A{I}_{a}E=\angle IQB.$$ Now let $\omega$ be the circle centered at $M$, has the radius $MB=MC$, and $X$ and $Y$ be the intersections of line $FQ$ with $AB$ and $AC$, respectively. Point $I$ lies on the angle bisector of $\angle QYC$, since $I$ lies on the angle bisector of $\angle YQB$ and $\angle YCQ$. Therefore $Y$, $I$, and $O_c$ are collinear, where $O_c$ is the center of ${\omega }_c$. Now we have $$\begin{array}{rl}\angle I{O}_{c}C=90^{\circ }-\frac{1}{2}\angle QYC& =\frac{1}{2}(\angle YQC+\angle YCQ)=\angle IQB+\angle ICB\\ & =\frac{1}{2}(\angle B-\angle C)+\frac{1}{2}\angle C=\frac{1}{2}\angle B\end{array}$$ Therefore $O_c$ lies on $\omega$. Similarly we can show that $O_b$ lies on $\omega$, where $O_b$ is the center of ${\omega }_b$. Notice that $I$ is the $A$-excenter of triangle $AXY$ so $$\angle O_{b}I{O}_c=\angle YIX=90-\frac{1}{2}\angle A=\angle B{I}_{a}C,$$ hence $B{O}_b=C{O}_c$. Finally $$P_{{\omega }_b}^M=M{O}_b^2-O_b{B}^2=M{O}_c^2-O_c{C}^2=P_{{\omega }_c}^M,$$ we are done.