## Task A-4.7. (10 points)

In a right-angled triangle, the median and the angle bisector of the right angle divide the hypotenuse into three segments whose lengths, in some order, form an arithmetic sequence. Determine all possible ratios of the lengths of the legs of this triangle.

Three numbers form an arithmetic sequence if the sum of the smallest and the largest is equal to twice the middle number.
## First Solution.

Let $d$ denote the length of the perpendiculars dropped from the intersection of the hypotenuse and the angle bisector of the right angle to the legs. These perpendiculars are the legs of two smaller right triangles that are similar to the original triangle.

(1 point)

Due to similarity and the Pythagorean theorem, we have

$$
\frac{d}{\sqrt{\left(x_{1}+x_{2}\right)^{2}-d^{2}}}=\frac{\sqrt{x_{3}^{2}-d^{2}}}{d}=\frac{a}{b}
$$

For simplicity, assume $c=6$. We distinguish two cases:

1. Case: Let $x_{3}<x_{2}<x_{1}$. Then from $x_{1}+x_{3}=2 x_{2}$, a simple calculation gives $x_{2}=\frac{c}{3}=2$ and $x_{3}=\frac{c}{6}=1$.

(2 points)

From this, we get

$$
\frac{d}{\sqrt{25-d^{2}}}=\frac{\sqrt{1-d^{2}}}{d}
$$

hence $d=\frac{5}{\sqrt{26}}$ and $a: b=1: 5$.

2. Case: Let $x_{2}<x_{3}<x_{1}$. Then from $x_{1}+x_{3}=2 x_{2}$, a simple calculation gives $x_{2}=\frac{c}{6}=1$ and $x_{3}=\frac{c}{3}=2$.

(2 points)

From this, we get

$$
\frac{d}{\sqrt{16-d^{2}}}=\frac{\sqrt{4-d^{2}}}{d}
$$

hence $d=\frac{4}{\sqrt{5}}$ and $a: b=1: 2$.