The legs of a right triangle are 3 and 4. Find the distance between the centers of the circumscribed and inscribed circles.

#
If $M$ is the point of tangency of the inscribed circle of triangle $ABC$ with side $AB$, and $p$ is the semiperimeter of the triangle, then $AM = p - BC$.

## Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-12.jpg?height=640&width=534&top_left_y=0&top_left_x=765)

## Answer

$\frac{\sqrt{5}}{2}$. Let $M$ be the point of tangency of the inscribed circle with the hypotenuse $AB$ of the given right triangle $ABC$, where $BC = 4$, $AC = 3$. $O$ is the midpoint of the hypotenuse (the center of the circumscribed circle), $Q$ is the center of the inscribed circle, $r$ is its radius, and $p$ is the semiperimeter of the triangle. Then,

$$
\begin{gathered}
AB = \sqrt{3^2 + 4^2} = 5, \quad r = \frac{3 + 4 - 5}{2} = 1, \quad AM = p - BC = 6 - 4 = 2 \\
OM = AO - AM = \frac{5}{2} - 2 = \frac{1}{2}
\end{gathered}
$$

From the right triangle $OMQ$, we find that

$$
OQ = \sqrt{OM^2 + QM^2} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2}
$$