[ Inscribed and Circumscribed Circles ] [ Trigonometric Ratios in a Right Triangle ]

In a right triangle $A B C$, the legs $A B$ and $A C$ are 4 and 3, respectively. Point $D$ bisects the hypotenuse $B C$. Find the distance between the centers of the circles inscribed in triangles $A D C$ and $A B D$.

#
If $O_{1}$ and $O_{2}$ are the centers of the given circles, then triangle $O_{1} D O_{2}$ is a right triangle.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles inscribed in triangles $A D C$ and $A B D$ respectively, and let $P$ and $Q$ be their points of tangency with side $B C$. Denote $\angle A D B=\alpha$.

From the isosceles triangle $A D B$, we find that

$$
\begin{gathered}
\sin \frac{\alpha}{2}=\frac{4}{5}, \cos \frac{\alpha}{2}=\frac{3}{5} \\
D Q=\frac{D B+A D+A B)}{2}-A B=\frac{1}{2}
\end{gathered}
$$

Similarly, we find that $D P=1$. Then

$$
\begin{gathered}
D O_{2}=\frac{D Q}{\cos \frac{\alpha}{2}}=\frac{5}{6} \\
D O_{1}=\frac{D P}{\cos \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{D P}{\sin \frac{\alpha}{2}}=\frac{5}{4} \\
O_{1} O_{2}^{2}=D O^{2}{ }_{1}+D O^{2}{ }_{2}=\left(\frac{5}{4}\right)^{2}+\left(\frac{5}{6}\right)^{2}=\frac{325}{144}
\end{gathered}
$$

Therefore,

$$
O_{1} O_{2}=\frac{5 \sqrt{13}}{12}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-17.jpg?height=343&width=866&top_left_y=0&top_left_x=593)

## Answer

$\frac{5 \sqrt{13}}{12}$.