A circle touches one of the legs of an isosceles right triangle and passes through the vertex of the opposite acute angle. Find the radius of the circle if its center lies on the hypotenuse of the triangle, and the leg of the triangle is equal to $a$.

#
Express the distance from the center of the circle to the vertices of the acute angles of the given triangle in terms of the desired radius.

## Solution

Let the given circle have center $O$ on the hypotenuse $A B$, touch the leg $B C$ at point $K$, and pass through vertex $A$. Denote the radius of this circle by $x$. Then, in triangle $OKB$, it is known that

$$
\angle B=45^{\circ}, O K=x, O B=a \sqrt{2}-x
$$

Therefore, $a \sqrt{2}-x=x \sqrt{2}$. From this, we find that $x=a(2-\sqrt{2})$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-18.jpg?height=414&width=584&top_left_y=782&top_left_x=699)

Answer

$a(2-\sqrt{2})$