Ex. 108. A circle with center at point $O$ is inscribed in triangle $A B C$. A second circle with center at point $P$ touches rays $A B$ and $A C$ and side $B C$ externally. It is known that angle $B A C$ is equal to $\alpha$, and the sum of the distances from point $O$ to vertices $B$ and $C$ is $t$. Find the sum of the distances from point $P$ to the same vertices.
Ex. 108. Answer: $m \operatorname{ctg} \frac{\pi-\alpha}{4}$. Solution. Let the distance $O P$ be denoted by $d$. Angles $O C P$ and $O B P$ are right angles (see the figure), hence the quadrilateral $O C P B$ can be inscribed in a circle, in which $O P=d$ is the diameter.

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-39.jpg?height=368&width=583&top_left_y=364&top_left_x=742)

Let $\angle A B C=\beta, \angle A C B=\gamma$. Then $\angle O P C=\beta / 2, \angle O P B=\gamma / 2, \angle C O P=$ $\pi / 2-\beta / 2, \angle B O P=\pi / 2-\gamma / 2$. Therefore, $O C+O B=d\left(\sin \frac{\beta}{2}+\sin \frac{\gamma}{2}\right)=m, \quad P C+$ $P B=d\left(\cos \frac{\beta}{2}+\cos \frac{\gamma}{2}\right) \Rightarrow \frac{P C+P B}{O C+O B}=\frac{2 \cos \frac{\beta+\gamma}{4} \cos \frac{\beta-\gamma}{4}}{2 \sin \frac{\beta+\gamma}{4} \cos \frac{\beta-\gamma}{4}}=\operatorname{ctg} \frac{\beta+\gamma}{4}=\operatorname{ctg} \frac{\pi-\alpha}{4}$.