12.403 The ratio of the radius of the circle circumscribed around a trapezoid to the radius of the circle inscribed in it is $\boldsymbol{k}$. Find the angles of the trapezoid and the permissible values of $k$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-289.jpg?height=229&width=346&top_left_y=276&top_left_x=149)

Let $r$ be the radius of the circle inscribed in the given trapezoid $ABCD (BC \| AD)$, $R$ be the radius of the circle circumscribed around $ABCD$, $CK$ be the height of the trapezoid, and $\alpha = \angle ADC$. Then $ABCD$ is isosceles, $CK = 2r$, $CD = \frac{2r}{\sin \alpha}$, $AB = \frac{BC + AD}{2} \Rightarrow AB = AK = \frac{2r}{\sin \alpha}$ (Theorem 2). By the Law of Sines for $\triangle ACD$, $AC = 2R \sin \alpha \Rightarrow$

\[
\begin{aligned}
& \Rightarrow AC^2 = 4R^2 \sin^2 \alpha = AK^2 + CK^2 = 4r^2 \left( \frac{1}{\sin^2 \alpha} + 1 \right) \Rightarrow k^2 \sin^4 \alpha - \sin^2 \alpha - 1 = 0 \Rightarrow \\
& \Rightarrow \sin \alpha = \frac{1}{k} \sqrt{\frac{1 + \sqrt{1 + 4k^2}}{2}}, \text{ with } 0 < \sin \alpha < 1 \Leftrightarrow \left\{ \begin{array}{l}
k > \frac{1}{\sqrt{2}} \\
1 + 4k^2 > 0
\end{array} \right.
\end{aligned}
\]

Answer: $\arcsin \frac{1}{k} \sqrt{\frac{1 + \sqrt{1 + 4k^2}}{2}}, \pi - \arcsin \frac{1}{k} \sqrt{\frac{1 + \sqrt{1 + 4k^2}}{2}}$ for $k > \sqrt{2}$.