Computing the length of one side of a triangle from distance between orthocenter and the midpoint of another side

Question

$\triangle{ABC}$ is an acute triangle, $\overline{AB} > \overline{BC}$, and $M$ is the midpoint of $\overline{AC}$. $H$ is the orthocenter of the triangle, and $P$ is the point on the minor arc $BC$ that is the intersection of the line through $H$ and $M$ and the circle circumscribing the triangle. If $\vert \overline{HM}\vert = 5$ and $\vert \overline{HP}\vert = 16$, and if $\angle{ABP}$ is a right angle, compute the length of side $\overline{BC}$.

Answer 1

Solution

As the figure shows, denote the circumcenter as $O$, and the intersection point of $MP$ and $BC$ as $N$.

Accoring to the given conditions, it's clear that $A,O,P$ are collinear. Notice that $PB \perp AB$ and $CH \perp AB$. Hence $PB // CH$. Similarily, $PC \perp AC$ and $BH \perp AC $. Hence $PC // BC$. As a result, $BPCH$ is a parallelogram.

Thus, $N$ is the midpoint of $BC$. Hence, $MN // AB$. But $CH \perp AB$, therefore $CH \perp MP$.

Notice that $CH$ is the altitude to the hypotenuse $MP$ in the right triangle $\triangle MPC$. Hence,$$CH=\sqrt{HM \cdot HP}=4\sqrt{5}.$$ Moreover,$$HN=\frac{1}{2}HP=8.$$ Thus, $$CN=\sqrt{CH^2+HN^2}=12.$$ It follows that $$BC=2CN=24.$$