Variation of Russian Olympiad

$\textbf{Lemma:}$ $BXMY$ is cyclic.

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$\textbf{Proof:}$ We use barycentric coordinates.

Let $AP = MQ = u$ and $PM = QC = v$ where $u + v = \frac{b}{2}$ . This implies that $\begin{aligned} P & = (b - u : 0 : u) \\ Q & = (v : 0 : b - v) \end{aligned}$ We find the equations of the two circles.

$\begin{aligned} (APQ) & : -a^2yz - b^2xz - c^2xy +(x+y+z)bvz = 0 \\ (BCP) & : -a^2yz - b^2xz - c^2xy +(x+y+z)bux = 0 \end{aligned}$ From here we find $X$ and $Y$ .

$\begin{aligned} X & = (0:bv:a^2-bv) \\ Y & = (c^2 - bu : bu : 0) \end{aligned}$ From here, we can find the equation of the circumcircle of $\triangle BXY$ .

$\triangle BXY : -a^2yz - b^2xz - c^2xy + (x+y+z)(bux +bvz) = 0$

We just need to check if $(1:0:1)$ satisfies the circle equation and it does. We are done.

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Now, by properties of cyclic quadrilaterals, we have $\angle BYM + \angle BXM = 180^{\circ} \implies \angle BXM = 135^{\circ}$