In triangle $ABC$, $AC=1, \angle ABC=30^{\circ}, \angle BAC=60^{\circ}$, and the foot of the altitude from $C$ is $D$. How far apart are the centers of the inscribed circles of triangles $ACD$ and $BCD$?
In the right-angled triangle $\mathrm{Az} ABC$, it is half of a regular triangle with a side length of 2 units, so $AB=2$ and $CB=\sqrt{3}$. From the similarity of triangles $\mathrm{Az} ACD$ and $ABC$, it follows that $CD=\frac{\sqrt{3}}{2}$ (the similarity ratio is $1:2$), and $AD=\frac{1}{2}$ and $BD=\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2a0487fbb1424c42da9bg-1.jpg?height=504&width=466&top_left_y=342&top_left_x=819)

The radius $\varrho$ of the inscribed circle of the triangle is given by the known relationship $\varrho=\frac{T}{s}$, where $T$ is the area and $s$ is the semiperimeter of the triangle. Therefore, the radius of the smaller circle is:

$$
\varrho_{1}=\frac{\frac{1}{2} \cdot \frac{\sqrt{3}}{2}}{\frac{3}{2}+\frac{\sqrt{3}}{2}}=\frac{1}{2} \cdot \frac{\sqrt{3}}{3+\sqrt{3}}=\frac{\sqrt{3}-1}{4}
$$

The radius of the larger circle is:

$$
\varrho_{2}=\frac{\frac{3}{2} \cdot \frac{\sqrt{3}}{2}}{\frac{3}{2}+\frac{3 \sqrt{3}}{2}}=\frac{1}{2} \cdot \frac{\sqrt{3}}{1+\sqrt{3}}=\frac{3-\sqrt{3}}{4}
$$

Let the centers of the inscribed circles be $O_{1}$ and $O_{2}$, and their projections on the side $AC$ be $T_{1}$ and $T_{2}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_2a0487fbb1424c42da9bg-1.jpg?height=247&width=371&top_left_y=1454&top_left_x=866)

In the right-angled trapezoid $T_{1} O_{1} O_{2} T_{2}$, draw a line through $O_{1}$ parallel to $T_{1} T_{2}$. In the resulting right-angled triangle $O_{1} O_{2} P$, $O_{1} P=\varrho_{1}+\varrho_{2}$ and $O_{2} P=\varrho_{2}-\varrho_{1}$.

The distance $O_{1} O_{2}$ we are looking for, according to the Pythagorean theorem, is:

$$
\begin{aligned}
O_{1} O_{2} & =\sqrt{\left(\varrho_{1}+\varrho_{2}\right)^{2}+\left(\varrho_{2}-\varrho_{1}\right)^{2}}=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{2-\sqrt{3}}{2}\right)^{2}}= \\
& =\sqrt{\frac{8-4 \sqrt{3}}{4}}=\sqrt{2-\sqrt{3}} \approx 0.5177 \text{ units. }
\end{aligned}
$$