Task B-2.8. The base $\overline{B C}$ of the isosceles triangle $A B C$ is of length $a$, and the angle opposite the base is $\alpha$. The circle $k$ touches the legs $A B$ and $A C$ and the circumcircle of the triangle $A B C$. Determine the radius of the circle $k$ (in terms of $a$ and $\alpha$).
Solution. Let the desired radius be denoted by $r$, and the radius of the circumcircle of triangle $ABC$ by $R$.

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-13.jpg?height=900&width=811&top_left_y=515&top_left_x=619)

Sketch...

(2 points)

We will calculate the radius $r$ from the right triangle $DAS$. It holds that

$$
|AS| = 2R - r, \quad \sin \frac{\alpha}{2} = \frac{r}{2R - r}
$$

(2 points)

We will calculate the radius $R$ from the right triangle $FOB$. Since $\alpha$ is an inscribed angle over the chord $\overline{BC}$, the angle $FOB$ is also $\alpha$ (half of the central angle over $\overline{BC}$).

(1 point)

Then, $\sin \alpha = \frac{a}{2R}$ or $2R = \frac{a}{\sin \alpha}$.

From $\left({ }^{*}\right) \sin \frac{\alpha}{2} = \frac{r}{\frac{a}{\sin \alpha} - r}$, we have

$$
\begin{aligned}
& \sin \frac{\alpha}{2} \left( \frac{a}{\sin \alpha} - r \right) = r \\
& r + r \sin \frac{\alpha}{2} = \frac{a \sin \frac{\alpha}{2}}{\sin \alpha} \\
& r = \frac{a \sin \frac{\alpha}{2}}{\sin \alpha \left(1 + \sin \frac{\alpha}{2}\right)}
\end{aligned}
$$

Note. The radius $R$ can also be calculated from the right triangle $ABE$ (angle $ABE$ is an inscribed angle over the diameter) or from the formula for the area $\left(P = \frac{abc}{4R}\right)$. Any method of calculation is worth 3 points.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

February 28, 2011.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THAT METHOD AND SCORE IT APPROPRIATELY.