Japan competition problems, 2022

Let the radius of the circumcircles of triangle $ABC$ and $BHC$ be $R$ and $R'$, respectively. From the sine theorem, we have $\frac{BC}{\sin \angle BAC} = 2R$ and $\frac{BC}{\sin \angle BHC} = 2R'$. Also, by $\angle BHC = \angle BAC + \angle ABH + \angle ACH = \angle BAC + 2 \cdot (90^\circ - \angle BAC) = 180^\circ - \angle BAC$, we have $\sin \angle BAC = \sin \angle BHC$. Therefore, $R = R'$.

Denote the intersection of the circumcircle of triangle $BHC$ and line $PM$ other than $P$ by $Q$. Then we have $\angle PBQ = \angle PBC + \angle CBQ = \angle PBC + \angle CPQ = \angle PBC + \angle ABP = \angle ABC$. By the sine theorem, we get $\frac{AC}{\sin \angle ABC} = 2R = 2R' = \frac{PQ}{\sin \angle PBQ}$, which implies $PQ = AC = 10$. Therefore, we obtain $QM = PQ - PM = 7$, and by the power of a point theorem, we have $BM \cdot CM = PM \cdot QM = 21$. Since $BM = CM$, we obtain $BM = \sqrt{21}$, and the answer is $BC = 2BM = 2\sqrt{21}$.