In a right triangle $A B C$ with hypotenuse $A C$, equal to 2, medians $A M$ and $C N$ are drawn. A circle can be circumscribed around quadrilateral $A N M C$. Find the radius of this circle.
Prove that triangle $ABC$ is isosceles.

## Solution

From the theorem of the tangent and the secant, it follows that

$$
BA \cdot BN = BC \cdot BM, \text{ or } 2 BN^2 = 2 BM^2.
$$

Therefore, $BN = BM$ and $AB = BC = \sqrt{2}$, i.e., triangle $ABC$ is isosceles.

$$
CN = \sqrt{BC^2 + BN^2} = \sqrt{(\sqrt{2})^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{5}{2}}
$$

If $R$ is the desired radius, then

$$
CN = 2R \sin \angle BAC = 2R \sin 45^\circ = R \sqrt{2}
$$

$$
R = \frac{CN}{\sqrt{2}} = \frac{\sqrt{5}}{2}.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-30.jpg?height=405&width=529&top_left_y=232&top_left_x=770)

## Answer

$\frac{\sqrt{5}}{2}$.