12.162. The perimeter of a sector is equal to $l$. Find the distance from the vertex of the central angle of the sector to the center of the circle inscribed in this sector, if the radius of the arc of the sector is $R$.
## Solution.

Let $E$ be the point of tangency of the inscribed circle and the radius $O_{1} A$ of the sector, the center $O$ of the circle lies on the bisector $O_{1} C$ of the angle $A O_{1} B$ (Fig. 12.34). If $\angle A O_{1} B=\alpha$, then the length of the arc $A C B$ is $\alpha R$ and

$$
\begin{aligned}
& l=2 R+R \alpha, \alpha=\frac{l-2 R}{R} . \text { Let } O O_{1}=x . \mathrm{B} \quad \triangle O E O_{1}\left(\angle O E O_{1}=\frac{\pi}{2}\right) \\
& O E=x \sin \frac{\alpha}{2} . O C=O E=x \sin \frac{\alpha}{2} . O O_{1} C=O O_{1}+O C, R=x+x \sin \frac{\alpha}{2} \\
& x=\frac{R}{1+\sin \frac{\alpha}{2}}=\frac{R}{1+\cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}=\frac{R}{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}=\frac{R}{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{l-2 R}{4 R}\right)}= \\
& =\frac{R}{2 \cos ^{2} \frac{(\pi+2) R-l}{4 R}} \\
& \text { Answer: } \frac{R}{2 \cos ^{2} \frac{(\pi+2) R-l}{4 R}}
\end{aligned}
$$