12.198. A circle is inscribed in a rhombus. In the resulting curvilinear triangle (with an acute angle), another circle is inscribed. Find its radius if the height of the rhombus is \( h \), and the acute angle is \( \alpha \).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0945.jpg?height=425&width=969&top_left_y=62&top_left_x=178)

Fig. 12.65
## Solution.

Let in rhombus $A B C D$ (Fig. 12.65) $\angle B A D=\alpha, 0^{\circ}<\alpha<90^{\circ}, O$ - the point of intersection of the diagonals - the center of the inscribed circle, $K$ - the center of the circle inscribed in the curvilinear triangle.

$N$ and $M$ are the points of tangency of the constructed circles with the side $A B$ of the rhombus.

Then $K M \perp A B, O N \perp A B, O N=\frac{h}{2}$.

Let $M K=r$. Draw $K L \perp O N$.

Then $K L \| A B, \angle O K L=\angle O A B=\frac{\alpha}{2}, O L=O N-K M=\frac{h}{2}-r, O K=\frac{h}{2}+r$.

In $\triangle O L K\left(\angle O L K=90^{\circ}\right): \sin \angle O K L=\frac{O L}{O K}$.

$$
\begin{aligned}
& \text { Then } \sin \frac{\alpha}{2}=\frac{\frac{h}{2}-r}{\frac{h}{2}+r} \Leftrightarrow\left(\frac{h}{2}+r\right) \sin \frac{\alpha}{2}=\frac{h}{2}-r \Leftrightarrow r\left(1+\sin \frac{\alpha}{2}\right)=\frac{h}{2}\left(1-\sin \frac{\alpha}{2}\right) \Leftrightarrow \\
& \Leftrightarrow r=\frac{h}{2} \cdot \frac{1-\sin \frac{\alpha}{2}}{1+\sin \frac{\alpha}{2}}=\frac{h}{2} \cdot \frac{1-\cos \left(90^{\circ}-\frac{\alpha}{2}\right)}{1+\cos \left(90^{\circ}-\frac{\alpha}{2}\right)}=\frac{h}{2} \operatorname{tg}^{2}\left(45^{\circ}-\frac{\alpha}{4}\right)
\end{aligned}
$$

Answer: $\frac{h}{2} \operatorname{tg}^{2}\left(45^{\circ}-\frac{\alpha}{4}\right)$.