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$\triangle{ABC}$ 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 $\overline{BC}$.

user74973
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Solution

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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 $$BC=2CN=24.$$

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mengdie1982
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    By the Inscribed-Angle Theorem, $\overline{AP}$ is a diameter of the circumscribing circle, and $A$, $O$, and $P$ are collinear. Since $\overline{AP}$ is a diameter, $\angle{ACP}$ is a right angle. – user74973 Jul 19 '18 at 20:16
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    Line segment $\overline{CH}$ is constructed; it is perpendicular to $\overline{AB}$ because $H$ is the orthocenter of $\triangle{ABC}$. – user74973 Jul 19 '18 at 20:16
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    Two line segments perpendicular to the same line segment are parallel to each other. So, $\overline{BP}$ and $\overline{CH}$ are parallel to each other. – user74973 Jul 19 '18 at 20:17
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    Likewise, $\overline{BH}$ and $\overline{CP}$ are parallel to each other. – user74973 Jul 19 '18 at 20:17
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    Quadrilateral $BPCH$ is a parallelogram. – user74973 Jul 19 '18 at 20:20
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    The diagonals of a parallelogram bisect each other. $N$ denotes this intersection. $N$ is the midpoint of $\overline{BC}$. – user74973 Jul 19 '18 at 20:37
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    So, $\overline{MN}$ is a line segment between the midpoints of two sides of $\triangle{ABC}$; it is parallel to the third side $\overline{AB}$. – user74973 Jul 19 '18 at 20:37
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    Since $\overline{CH}$ is perpendicular to $\overline{AB}$, it is perpendicular to $\overline{MN} \subsetneq \overline{MP}$. – user74973 Jul 19 '18 at 20:41