Task B-2.5. In a right-angled triangle, a height and a median are drawn from the vertex of the right angle. The ratio of their lengths is 12 : 13. Determine the ratio of the legs of this triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6e5f8ba1616e2c4b87f4g-07.jpg?height=382&width=596&top_left_y=2243&top_left_x=735)

Since $v: t=12: 13$, we can write $v=12 k, t=13 k, k \in \mathbb{N}$.

(1 point)

In a right triangle, the median to the hypotenuse is equal to the radius of the circumscribed circle, which is half the length of the hypotenuse.

(1 point)

Therefore, $|A D|=|D B|=t=13 k$. From the right triangle $\triangle C D E$, we get

\[
\begin{aligned}
& |D E|=\sqrt{(13 k)^{2}-(12 k)^{2}}=5 k \\
& |E B|=13 k-5 k=8 k
\end{aligned}
\]

We have

\[
\begin{aligned}
& b^{2}=v^{2}+|A E|^{2}=(12 k)^{2}+(18 k)^{2}=468 k^{2} \\
& a^{2}=v^{2}+|B E|^{2}=(12 k)^{2}+(8 k)^{2}=208 k^{2}
\end{aligned}
\]

Then the ratio of the legs is

\[
\begin{aligned}
& \left(\frac{b}{a}\right)^{2}=\frac{468 k^{2}}{208 k^{2}}=\frac{9}{4} \\
& \frac{b}{a}=\frac{3}{2} \text{ or } b: a=3: 2
\end{aligned}
\]