8,9 The angle at the base of an isosceles triangle is equal to $\varphi$. Find the ratio of the radius of the inscribed circle to the radius of the circumscribed circle of this triangle.
Express the radii of the circles in terms of the base of the triangle.

## Solution

Let the base $BC$ of the isosceles triangle $ABC$ be denoted by $a$, the radii of the inscribed and circumscribed circles by $r$ and $R$ respectively, the center of the inscribed circle by $O$, and the midpoint of $BC$ by $M$. Then,

\[
\begin{gathered}
R=\frac{BC}{2 \sin \angle BAC}=\frac{a}{2 \sin (180^\circ - 2\varphi)}=\frac{a}{2 \sin 2\varphi} \\
r=OM=BM \operatorname{tg} \frac{1}{2} \angle B=\frac{a}{2} \operatorname{tg} \frac{\varphi}{2}
\end{gathered}
\]

Therefore,

\[
\frac{r}{R}=\operatorname{tg} \frac{\varphi}{2} \sin 2\varphi
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-22.jpg?height=466&width=700&top_left_y=825&top_left_x=679)

## Answer

$\operatorname{tg} \frac{\varphi}{2} \sin 2\varphi$