A circle with its center on the side $AB$ of an isosceles triangle $ABC$ ($AB = BC$) touches the segment $AC$ at point $F$, intersects the segment $BC$ at point $G$, passes through point $B$, and intersects the segment $AB$ at point $E$, such that $AE = a$, and $\angle BFG = \gamma$. Find the radius of the circle.
Let $R$ be the radius of the given circle, and $O$ its center. The inscribed angles $B E G$ and $B F G$ subtend the same arc, so $\angle B E G = \angle B F G = \gamma$. The inscribed angle $B G E$ subtends the diameter $B E$, so $\angle B G E = \frac{\pi}{2}$. From the right triangle $B G E$, we find that $\angle E B G = \frac{\pi}{2} - \angle B E G = \frac{\pi}{2} - \gamma$. Therefore,

$$
\angle B A C = \frac{\pi - \angle A B C}{2} = \frac{\pi - \left(\frac{\pi}{2} - \gamma\right)}{2} = \frac{\pi}{4} + \frac{\gamma}{2}
$$

as the angle at the base of the isosceles triangle $A B C$. Since $O F \perp A C$ (as the radius of the circle drawn to the point of tangency), triangle $A O F$ is a right triangle, so

$$
O F = A O \sin \angle O A F = (A E + E O) \sin \angle B A C
$$

or

$$
R = (a + R) \sin \left(\frac{\pi}{4} + \frac{\gamma}{2}\right)
$$

From this equation, we find that $R = \frac{a \sin \left(\frac{\pi}{4} + \frac{\gamma}{2}\right)}{1 - \sin \left(\frac{\pi}{4} + \frac{\gamma}{2}\right)}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-49.jpg?height=955&width=1358&top_left_y=0&top_left_x=390)

## Answer

$\frac{a \sin \left(\frac{\pi}{4} + \frac{\gamma}{2}\right)}{1 - \sin \left(\frac{\pi}{4} + \frac{\gamma}{2}\right)}$

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-49.jpg?height=54&width=197&top_left_y=1231&top_left_x=16)

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