In an isosceles triangle $ABC (AB = BC)$, the ratio of the distances from the center of the inscribed circle of triangle $ABC$ to the vertices of angles $B$ and $C$ is $k$. Find the angles of triangle $ABC$. What are the possible values of $k$?
Let $O$ be the center of the circle inscribed in triangle $ABC$. Denote $\angle BAC = \alpha$, express the angles of triangle $BOC$ in terms of $\alpha$, and apply the Law of Sines to it.

## Solution

Let $O$ be the center of the circle inscribed in triangle $ABC$. Denote $\angle BAC = \angle BCA = \alpha, OC = x$. Then

$$
OB = kx, \angle BCO = \frac{\alpha}{2}, \angle OBC = 90^\circ - \alpha
$$

(since $O$ is the intersection point of the angle bisectors of triangle $ABC$). Applying the Law of Sines to triangle $BOC$, we get the equation

$$
\frac{BO}{\sin \frac{\alpha}{2}} = \frac{OC}{\sin (90^\circ - \alpha)}, \text{ or } \frac{kx}{\sin \frac{\alpha}{2}} = \frac{x}{\cos \alpha}, \text{ or } k \cos \alpha = \sin \frac{\alpha}{2}
$$

The last equation can be rewritten as

$$
2k \sin^2 \frac{\alpha}{2} + \sin \frac{\alpha}{2} - k = 0
$$

Since $0 < \alpha < 180^\circ$, we have $0 < \sin \frac{\alpha}{2} < 1$. Therefore, we need to check that

$$
\frac{\sqrt{1 + 8k^2} - 1}{4k} < 1 \text{ for } k > 0
$$

Thus,

$$
\begin{gathered}
\angle BAC = \angle ACB = \alpha = 2 \arcsin \frac{\sqrt{8k^2 + 1} - 1}{4k} \\
\angle ABC = 180^\circ - 2\alpha = 180^\circ - 4 \arcsin \frac{\sqrt{8k^2 + 1} - 1}{4k}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-24.jpg?height=535&width=554&top_left_y=979&top_left_x=755)

Answer

$\angle A = \angle C = 2 \arcsin \frac{\sqrt{8k^2 + 1} - 1}{4k}, \angle B = \pi - 2 \angle A, k > 0$.