3.22. In the segment, the arc of which contains $\alpha^{\circ}$, a regular triangle is inscribed such that one of its vertices coincides with the midpoint of the arc, and the other two lie on the chord. The area of the triangle is $S$. Find the radius of the arc of the segment.
3.22. We have $\cup A C=\cup C B=\frac{\alpha}{2}$ (Fig. 3.24), $C D=C E=D E=a, S_{\triangle D C E}=$ $=S, O B=O C=R$ - the desired value.

$$
\text { From the equality } S=\frac{a^{2} \sqrt{3}}{4} \text { (see formula }
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-512.jpg?height=300&width=468&top_left_y=1461&top_left_x=685)
(1.18)), we find $a=2 \sqrt{\frac{S}{\sqrt{3}}}$. Then,

Fig. 3.24

$C F=\frac{a \sqrt{3}}{2}=2 \sqrt{\frac{S}{\sqrt{3}}} \cdot \frac{\sqrt{3}}{2}=\sqrt{S \sqrt{3}} . \quad$ Further, in $\quad \triangle O F B \quad$ we have $O F=O B \cos \frac{\alpha}{2}=R \cos \frac{\alpha}{2}$. Since $O F+C F=R$, then $R \cos \frac{\alpha}{2}+\sqrt{S \sqrt{3}}=$ $=R$, from which

$$
R=\frac{\sqrt{S \sqrt{3}}}{1-\cos \frac{\alpha}{2}}=\frac{\sqrt{S \sqrt{3}}}{2 \sin ^{2} \frac{\alpha}{4}}
$$