## Task B-1.4.

Let $A B C$ be an isosceles right triangle. A circle with its center on one of the legs passes through the vertex of the right angle $C$ and is tangent to the hypotenuse of the triangle $A B C$. Determine the length of the hypotenuse of the triangle $A B C$ if the radius of the given circle is 1.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_4b7629e40479ae2c2357g-03.jpg?height=663&width=873&top_left_y=865&top_left_x=523)

Sketch
1 point

Since the circle is tangent to the hypotenuse of the right triangle, the hypotenuse is a tangent to the circle, and the radius at the point of tangency $D$ is perpendicular to the hypotenuse.

Since triangle $ABC$ is an isosceles right triangle, the angle at vertex $B$ is $45^{\circ}$, and triangle $BSD$ is also an isosceles right triangle.

Therefore, $|BD|=1$, $|BS|=\sqrt{2}$, and $|AC|=|BC|=1+\sqrt{2}$. 1 point

The hypotenuse $|AB|$ of triangle $ABC$ is $|AB|=\sqrt{2(1+\sqrt{2})^{2}}=\sqrt{2}+2$.

Note: Instead of using the Pythagorean theorem, the student can conclude that point $S$ is equidistant from the legs of angle $CAD$, meaning it lies on the angle bisector of that angle, and $|AD|=|AC|=1+\sqrt{2}$, so the hypotenuse is $|AD|+|BD|=2+\sqrt{2}$.

This is worth 2 points.