12.395 The ratio of the radius of the circle inscribed in an isosceles triangle to the radius of the circle circumscribed around it is $\boldsymbol{m}$. Find the angles of the triangle and the permissible values of $\boldsymbol{m}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-284.jpg?height=453&width=1018&top_left_y=1251&top_left_x=73)

Let $O$ be the center of the inscribed circle of radius $r$ in the given $\triangle ABC$, $\alpha = \angle A = \angle C$, and $R$ be the radius of the circumcircle of $\triangle ABC$. Then, if $K$ is the midpoint of $AC$, we have $r = OK = AK \cdot \tan \angle OAK = \frac{AC}{2} \cdot \tan \frac{\alpha}{2}$.

By the Law of Sines, $R = \frac{AC}{2 \sin 2\alpha} \Rightarrow$

$\Rightarrow m = \frac{r}{R} = \tan \frac{\alpha}{2} \cdot \sin 2\alpha = \frac{1 - \cos \alpha}{\sin \alpha} \cdot 2 \sin \alpha \cdot \cos \alpha \Rightarrow$

$\Rightarrow 2 \cos^2 \alpha - 2 \cos \alpha + m = 0 \Rightarrow \cos \alpha = \frac{1 \pm \sqrt{1 - 2m}}{2}, \alpha = \arccos \frac{1 \pm \sqrt{1 - 2m}}{2} \Rightarrow$

$\Rightarrow \angle B = 180^\circ - 2\alpha = 2 \arcsin \frac{1 \pm \sqrt{1 - 2m}}{2} \quad (\arcsin \alpha + \arccos \alpha = 90^\circ)$.

Since $0 < \alpha < \frac{\pi}{2}$, we have $0 < \frac{1 \pm \sqrt{1 - 2m}}{2} < 1 \Leftrightarrow 0 < m \leq \frac{1}{2}$.

Answer: $\arccos \frac{1 + \sqrt{1 - 2m}}{2}, 2 \arcsin \frac{1 + \sqrt{1 - 2m}}{2}$ or

$$
\arccos \frac{1 - \sqrt{1 - 2m}}{2}, 2 \arcsin \frac{1 - \sqrt{1 - 2m}}{2} ; 0 < m \leq \frac{1}{2}
$$