5. An isosceles triangle \(ABC\) with vertex \(B\) is inscribed in a circle of radius \(R\). A circle of radius \(r\) is inscribed in it. Another circle of radius \(r\) is inscribed in the segment bounded by the chord \(AB\) and the smaller arc. Find the ratio \(R: r\).

For a complete solution to each problem, 7 points are awarded.

The maximum total score is 35.
5. Answer: $\frac{8}{3}$.

Let $a=AB=BC, AC=2b$. Then the height dropped to the base is $h=\sqrt{a^{2}-b^{2}}$. The area is $S=bh$, and the semiperimeter is $p=a+b$. Therefore, $r=\frac{S}{p}=\frac{b \sqrt{a^{2}-b^{2}}}{a+b}$.

By the Law of Sines, $a=2R \sin \hat{A}=\frac{2Rh}{a}$, from which $R=\frac{a^{2}}{2 \sqrt{a^{2}-b^{2}}}$. Finally, the circle inscribed in the segment touches the segment $AB$ at its midpoint. The height of the segment is $2r$, so the distance from the center $O$ of the circumscribed circle to the midpoint $D$ of the segment $AB$ is $R-2r$. From the similarity of right triangles $OBD$ and $CBK$, where $K$ is the midpoint of $AC$, we have $\frac{R-2r}{R}=\frac{b}{a}$. Thus, $\frac{2r}{R}=\frac{a-b}{a}$. Using the expressions for $r$ and $R$ in terms of $a, b$, we have the equation

$$
\frac{2r}{R}=\frac{4b(a^{2}-b^{2})}{a^{2}(a+b)}=\frac{4b(a-b)}{a^{2}}=\frac{a-b}{a}
$$

Simplifying, we get $a=4b$. Therefore, $r=\frac{b \sqrt{15}}{5}$ and $R=\frac{8b}{\sqrt{15}}$, so $R: r=\frac{8}{3}$.