## Task B-2.3.

In a right-angled triangle, the ratio of the height and the median drawn from the vertex of the right angle is equal to $\frac{12}{37}$. Determine the ratio of the lengths of the legs of this triangle.
## Solution.

Let $A, B, C$ be the vertices of the given triangle, and $N$ the foot of the altitude from the right angle vertex $C$.

![](https://cdn.mathpix.com/cropped/2024_05_30_3d5a33db4db92485fe9eg-07.jpg?height=499&width=988&top_left_y=984&top_left_x=474)

With the labels as in the figure, from $v_{c}: t_{c}=12: 37$ it follows that $v_{c}=12 k$ and $t_{c}=37 k$, where $k>0$ is a real number.

Since in a right triangle the center of the circumscribed circle is at the midpoint of the hypotenuse, we have $\frac{c}{2}=t_{c}=37 k$, from which we get $c=74 k$.

Applying the Pythagorean theorem to the right triangle $C N E$, we get

$x^{2}=t_{c}^{2}-v_{c}^{2}=(37 k)^{2}-(12 k)^{2}=(37 k-12 k)(37 k+12 k)=25 k \cdot 49 k$, from which it follows that $x=35 k$.

Since $\triangle A N C \sim \triangle C N B$, we have:

$\frac{a}{v_{c}}=\frac{b}{\frac{c}{2}-x}$

$\frac{a}{b}=\frac{v_{c}}{\frac{c}{2}-x}=\frac{12 k}{37 k-35 k}=6$ or $\frac{b}{a}=\frac{1}{6}$.

Note: Instead of similarity, students can use the tangent of the angle at vertex $A$ or $B$ in the last two steps of the solution.

Similarly, instead of similarity, students can apply the Pythagorean theorem to triangle $A N C$ to get

$b=\sqrt{v_{c}^{2}+\left(\frac{c}{2}-x\right)^{2}}=\sqrt{(12 k)^{2}+(37 k-35 k)^{2}}=\sqrt{148 k^{2}}=2 k \sqrt{37}$

and analogously apply the Pythagorean theorem to triangle $C N B$

$a=\sqrt{v_{c}^{2}+\left(\frac{c}{2}+x\right)^{2}}=\sqrt{(12 k)^{2}+(37 k+35 k)^{2}}=\sqrt{144 k^{2}+5184 k^{2}}=\sqrt{5328 k^{2}}=$ $12 k \sqrt{37}$.