1. A circle inscribed in a right angle of an isosceles right triangle divides the hypotenuse into 3 equal parts. Find the radius of the circle if the leg of the triangle is 1.
## Answer: $1 / 3$ or $5/3$.

Solution. It is clear that the radius of the circle inscribed in the right angle $A$ is equal to the distance from its vertex to the point of tangency of the circle with the side. Let $B C$, the hypotenuse of the given triangle $A B C$, intersect the circle at points $D$ and $E$, and $K$ be the point of tangency with the line $A B$ (see figure). By the tangent-secant theorem, $B K^{2}=B D \cdot B E=(2 / 9) B C^{2}=4 / 9$, i.e., $B K=2 / 3$.

![](https://cdn.mathpix.com/cropped/2024_06_04_c790142104ba3f7af959g-19.jpg?height=634&width=608&top_left_y=631&top_left_x=1295)

If the point of tangency lies on the leg $A B$ (point $K$ in the figure), then the desired radius $r=A K=1-B K=1 / 3$.

If the point of tangency lies on the extension of the leg $A B$ (point $K_{1}$ in the figure), then the desired radius $r_{1}=A K_{1}=1+B K_{1}=5 / 3$.