Belarusian Mathematical Olympiad

First solution. Let $\Gamma$ be a circle passing through $M,P,B$. Let $\Gamma$ touch the side $AC$ at $S$, and $T$ be the intersection of the lines $MS$ and $BC$. Since point $S$ lies on the side $AC$, it follows that $T$ lies on the extension of the side $BC$ (over $C$).



Since $AM=MB=MP$ and $MSPB$ is an inscribed quadrilateral, we see that $\angle MBP=\angle MPB=\angle MSB=\beta$, $\angle MBS=\angle ASM=\angle TSC=\phi$. Let $\angle CAB=\alpha$. From the triangle $ASB$ we have $\alpha +\beta +2\phi =\pi$. Since $\angle SMB$ is an external angle of $△ASM$, we have $\angle SMB=\alpha +\phi$. Hence $\angle MTB=\pi -(\alpha +\phi )-\beta =\phi$. Thus, $\angle CTS=\angle CST(=\phi )$, so $CS=CT$. By Menelaus' theorem ($△ACB$ and points $M,S,T$), $$\frac{AM}{MB}\cdot \frac{BT}{TC}\cdot \frac{CS}{SA}=1⟺BT=SA.$$ By the power of point theorem, $B{T}^2=A{S}^2=AM\cdot AB=BM\cdot BA$. It is equivalent to the fact that $BT$ is tangent to ${\Gamma }_1$, passing through $A,M,T$. Let $Q^{\prime }$ be the intersection point of ${\Gamma }_1$ and the line $AC$. Since ${\Gamma }_1$ touches the line $BT$, we see that $Q^{\prime }$ lies on the side $AC$ (not on the extension). Further $$\begin{array}{rl}\sim AM+\sim Q^{\prime }T& =2\angle TSC=2\phi =2\angle STC=\\ & =[TC\text{ is tangent to }{\Gamma }_1]=\sim M{Q}^{\prime }T⟺\\ \sim MA+\sim Q^{\prime }T& =\sim M{Q}^{\prime }T=\sim M{Q}^{\prime }+\sim Q^{\prime }T⟺\sim AM=\sim M{Q}^{\prime }.\end{array}$$

So $AM=M{Q}^{\prime }$ (chords subtending the equal arcs). Therefore, $BM=AM=M{Q}^{\prime }$, which gives that $A{Q}^{\prime }B$ is a right-angled triangle ($\angle A{Q}^{\prime }B=90^{\circ }$). Thus $B{Q}^{\prime }\perp AC$, i.e. points $Q$ and $Q^{\prime }$ coincide, which completes the proof.

Second solution. Let $\Gamma$ be a circle passing through $M,P,B$. Let $\Gamma$ touch the side $AC$ at $S$. Since $APB$ and $AQB$ are right-angled triangles, we have $MA=MB=MQ=MP$, so $A,Q,P,B$ lie on the circle ${\Gamma }_0$ with the center at $M$. Consider the inversion (with $M$ as center) with respect to ${\Gamma }_0$. Since $\Gamma$ passes through the center of the inversion, its image is a line $\ell$, but $B$ and $P$ are fixed points, the line $\ell$ coincide with the line $BC$. Similarly, the image of the line $AC$ ($M\notin AC$) is a circle ${\Gamma }_1$ passing through $A,Q,M$, since $A$ and $Q$ are fixed points. Therefore, $S$ as the tangency point of $\Gamma$ and the side $AC$ is transformed to $S^{\prime }$ which is the tangency point of the line $BC$ and ${\Gamma }_1$. To complete the proof, it suffices to note that $S^{\prime }$ lies on the ray $MS$, i.e. is the intersection point of the ray $MS$ and the line $BC$. Since $S$ lies on the side $AC$, we see that $S^{\prime }$ lies on the extension of the side $BC$ (over $C$).