## [ Right triangle with an angle of $30^{\circ}$ ]

In a right triangle $ABC$ with acute angle $A$ equal to $30^{\circ}$, the bisector $BD$ of the other acute angle is drawn.

Find the distance between the centers of the inscribed circles of triangles $ABD$ and $CBD$, if the shorter leg is 1.
If $O_{1}, O_{2}$ are the centers of the given circles, then $\angle O_{1} D O_{2}=90^{\circ}$.

## Solution

Let $O_{1}$ and $O$ be the centers of the inscribed circles, and $P$ and $Q$ be the points of tangency with side $A C$. Then $A C=\sqrt{3}, \quad D C=\frac{\sqrt{3}}{3}, \quad B D=A D=\frac{2 \sqrt{3}}{3}, A B=2, D P=\frac{1}{2}(A D-A B+B D)=\frac{2-\sqrt{3}}{\sqrt{3}}, D Q=\frac{1}{2}(D C - C B + D B)=\frac{\sqrt{3}-1}{2}$

$D O_{1}=\frac{D P}{\cos 60^{\circ}}=\frac{2(2-\sqrt{3})}{\sqrt{3}}, D O_{2}=\frac{D Q}{\cos 30^{\circ}}=\frac{\sqrt{3}-1}{\sqrt{3}}$.

Since triangle $O_{1} D O_{2}$ is a right triangle, then

$O_{1} O_{2}^{2}=O_{1} D^{2}+O_{2} D^{2}=\left(\frac{2(2-\sqrt{3})}{\sqrt{3}}\right)^{2}+\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)^{2}=\frac{32-18 \sqrt{3}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-37.jpg?height=340&width=746&top_left_y=1282&top_left_x=656)

## Answer

$\frac{\sqrt{96-54 \sqrt{3}}}{3}$