3. In a circle with radius $R$, determine the central angle corresponding to a circular segment with the following property: the chord of the segment is equal to the perimeter of the circle with the largest radius that is inscribed in it.
Solution. Let $A M B$ be the circular arc segment and let $\measuredangle A O B=\alpha$. It is clear that $\measuredangle A O M=\frac{\alpha}{2}$. From the power of a point with respect to a circle, we have

$$
\overline{M N} \cdot \overline{N F}=\overline{A N} \cdot \overline{N B}
$$

Let $r$ be the radius of the circle inscribed in the segment. Then $l=2 r \pi, \overline{A N}=\overline{N B}=\pi r$ and $\overline{M N}=2 r, \overline{N F}=2 R-2 r$. Since $\frac{\pi r}{R}=\sin \frac{\alpha}{2}$, we get $R=\frac{\pi r}{\sin \frac{\alpha}{2}}$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_d0c212a3407f50323a50g-07.jpg?height=430&width=392&top_left_y=1054&top_left_x=1068)

$$
\overline{N F}=2 \frac{\pi r}{\sin \frac{\alpha}{2}}-2 r=2 r\left(\frac{\pi r}{\sin \frac{\alpha}{2}}-1\right)
$$

If we substitute in (1), we get

$$
\pi^{2} r^{2}=4 r^{2}\left(\frac{\pi r}{\sin \frac{\alpha}{2}}-1\right), \text { i.e. } \sin \frac{\alpha}{2}=\frac{4 \pi}{4+\pi^{2}}
$$

Finally, $\alpha=2 \arcsin \frac{4 \pi}{4+\pi^{2}}$.