Japan Mathematical Olympiad

Let $H$ be the orthocenter of triangle $ABC$ and $K$ be the common orthocenter of triangle $ABQ$ and triangle $ACP$. Let $D$ be the foot of the perpendicular from $A$ to line $BC$. By definition, $A,D,H,K$ are collinear.

Since $AB=10$, $AC=11$ and $BC>5$, both $\angle B$ and $\angle C$ are less than $90^{\circ }$. Therefore $D$ lies on side $BC$ and $D$ is different from $H$. Assume that $D$ and $K$ coincide, then two points $P$ and $Q$ coincide with $D$. The Pythagorean theorem shows that $A{B}^2-B{P}^2=A{D}^2=A{C}^2-C{Q}^2$, which is a contradiction because $10^2-5^2\ne 11^2-6^2$. It follows that $D$ and $K$ are different points.

Two lines $BH$ and $PK$ are both perpendicular to side $AC$. Therefore these two lines are parallel and $DB:BP=DH:HK$ follows. Similarly, two lines $CH$ and $QK$ are parallel and $DC:CQ=DH:HK$ follows. From the above, it follows that $DB:BP=DC:CQ$ and $DB:DC=BP:CQ=5:6$. Set $DB=5t$, $DC=6t$ where $t$ is a real number, then by using Pythagorean theorem for triangles $ABD$ and $ACD$ we have $10^2-(5t{)}^2=11^2-(6t{)}^2$. Since the required length is $BC=11t$, the answer is $\sqrt{231}$.