12.397 In a segment with a central angle $\alpha$, a regular triangle is inscribed such that one of its vertices coincides with the midpoint of the segment's chord, and the other two lie on the arc of the segment. The height of the triangle is $h$. Find the radius of the arc of the segment.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-285.jpg?height=295&width=416&top_left_y=1448&top_left_x=203)

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Let $O$ be the center of the circle with radius $x$ and the given segment $P Q, \alpha=\angle P O Q, B$ be the midpoint of the chord $P Q, \triangle A B C$ be an equilateral triangle with height $h$, inscribed in the segment $P Q$. Extend $O B$ to intersect $A C$ at point $K$.

$\triangle A O B=\triangle C O B \Rightarrow O K$ is the bisector

$\angle A O C \Rightarrow O K$ is the altitude of $\triangle A O C \Rightarrow$

$$
\Rightarrow x^{2}=K C^{2}+O K^{2}=\left(\frac{h}{\sqrt{3}}\right)^{2}+\left(h+x \cdot \cos \frac{\alpha}{2}\right)^{2}
$$

(since $K C=\frac{h}{\sqrt{3}}, B K=h, O B=x \cdot \cos \frac{\alpha}{2} \Rightarrow x^{2} \sin ^{2} \frac{\alpha}{2}-2 x h \cos \frac{\alpha}{2}-\frac{4}{3} h^{2}=0$.

Solving this equation, we ultimately get the answer.

Answer: $x=\frac{h}{\sin ^{2} \frac{\alpha}{2}} \cdot\left(\cos \frac{\alpha}{2}+\sqrt{1+\frac{1}{3} \sin ^{2} \frac{\alpha}{2}}\right)$.