In an isosceles triangle $A B C \quad \angle B=\operatorname{arctg} 8 / 15$. A circle of radius 1, inscribed in angle $C$, touches side $C B$ at point $M$ and cuts off a segment $K E$ from the base. It is known that $M B=15 / 8$. Find the area of triangle $K M B$, given that points $A, K, E, B$ follow on the base $A B$ in the specified order.
Prove that the center of the circle lies on the base of the triangle.

## Solution

Let $O$ be the center of the given circle. From the right triangle $OMB$, we find that $\operatorname{tg} \angle MBO = \frac{OM}{MB} = \frac{8}{15} = \operatorname{tg} \angle B$. Therefore, $\angle MBO = \angle B$. Since the rays $BO$ and $BA$ are on the same side of the line $BC$, the center $O$ lies on the side $AB$.

By the Pythagorean theorem, $OB^2 = OM^2 + MB^2 = \left(\frac{17}{8}\right)^2$.

Let $h$ be the height of the triangle $OMB$ (see the figure). Then $OB \cdot h = OM \cdot MB$, from which $h = \frac{15}{17}$. Therefore, $S_{\text{KMB}} = \frac{1}{2}(KO + OB)h = \frac{375}{272}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-18.jpg?height=497&width=468&top_left_y=838&top_left_x=795)

Answer

$\frac{375}{272}$.