[ Inscribed and circumscribed circles ] [ Pythagorean Theorem (direct and inverse).]

The legs of a right triangle are 36 and 48. Find the distance from the center of the inscribed circle to the height drawn to the hypotenuse.
Calculate the specified height, the radius of the inscribed circle, and the distance from the center of the circle to the vertex of the right angle.

## Solution

## First Method.

Let $r$ be the radius of the circle with center $O$, inscribed in the right triangle $ABC$ with legs $AC=48$ and $BC=36$, $CH$ be the height of triangle $ABC$, $M$ and $K$ be the points of tangency of the circle with sides $AB$ and $AC$ respectively, and $P$ be the projection of the center $O$ onto $CH$. Then

$$
\begin{gathered}
AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{48^{2}+36^{2}}=\sqrt{12^{2}\left(4^{2}+3^{2}\right)}=60 \\
OM=OK=r=\frac{AC+BC-AB}{2}=\frac{48+36-60}{2}=12 \\
CH=AC \cdot \frac{BC}{AB}=48 \cdot \frac{36}{60}=\frac{144}{5} \\
CP=CH-PH=CH-OM=CH-r=\frac{144}{5}-12=\frac{84}{5} \\
OC=\frac{OK}{\sin \angle OCK}=\frac{r}{\sin 45^{\circ}}=r \sqrt{2}=12 \sqrt{2}
\end{gathered}
$$

$$
OP=\sqrt{OC^{2}-CP^{2}}=\sqrt{(12 \sqrt{2})^{2}-\left(\frac{84}{5}\right)^{2}}=12 \sqrt{2-\frac{49}{25}}=\frac{12}{5}
$$

Second Method.

Let $r$ be the radius of the circle with center $O$, inscribed in the right triangle $ABC$ with legs $AC=48$ and $BC=36$, $CH$ be the height of triangle $ABC$, $M$ and $K$ be the points of tangency of the circle with sides $AB$ and $AC$ respectively, and $P$ be the projection of the center $O$ onto $CH$. Then

$$
\begin{gathered}
AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{48^{2}+36^{2}}=\sqrt{12^{2}\left(4^{2}+3^{2}\right)}=60 \\
OM=OK=r=\frac{AC+BC-AB}{2}=\frac{48+36-60}{2}=12 \\
BH=\frac{BC^{2}}{AB}=\frac{36^{2}}{60}=\frac{108}{5} \\
BM=BK=BC-CK=BC-r=36-12=24 \\
OP=MH=BM-BH=24-\frac{108}{5}=\frac{12}{5}
\end{gathered}
$$

## Answer

$\frac{12}{5}$.