3. In a right triangle $ABC\left(\measuredangle C=90^{\circ}\right)$, the medians $\overline{A B}=m$ and $\overline{B N}=n$ are given. Let $O$ be the center of the inscribed circle in triangle $ABC$. Express the radius $R$ of the circumscribed circle around triangle $ABO$ in terms of $m$ and $n$.
Solution. From the right triangles $A M C$ and $B N C$ we have

$$
m^{2}=b^{2}+\frac{a^{2}}{4} \text { and } n^{2}=a^{2}+\frac{b^{2}}{4},
$$

so

$$
m^{2}+n^{2}=\frac{5}{4}\left(a^{2}+b^{2}\right)=\frac{5}{4} c^{2}
$$

i.e., $c^{2}=\frac{4}{5}\left(m^{2}+n^{2}\right)$. Let $O$ be the center of the inscribed circle of $\triangle A B C$ and $S$ be the center of the circumscribed circle around $\triangle A B O$. Then

$$
\measuredangle B A O+\measuredangle A B O=\frac{\alpha}{2}+\frac{\beta}{2}=45^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_8dd66bca97b851071799g-2.jpg?height=517&width=598&top_left_y=417&top_left_x=862)

Furthermore, the central angle is twice the inscribed angle subtending the same arc, so

$$
\measuredangle A S O=2 \measuredangle A B O=2 \frac{\beta}{2}=\beta \text { and } \measuredangle B S O=2 \measuredangle B A O=2 \frac{\alpha}{2}=\alpha .
$$

Thus,

$$
\measuredangle A S B=\measuredangle A S O+\measuredangle B S O=\beta+\alpha=90^{\circ}
$$

i.e., $\triangle A B S$ is an isosceles right triangle. Therefore,

$$
R=\frac{\overline{A B}}{\sqrt{2}}=\frac{c}{\sqrt{2}}=\sqrt{\frac{2}{5}\left(m^{2}+n^{2}\right)}
$$