## Task B-2.6.

In a right-angled triangle, the lengths of all sides are two-digit natural numbers. If we swap the tens and units digits of the length of one of the legs, we will get the length of the hypotenuse of the triangle. Determine the lengths of the sides of the triangle.
## Solution.

Let $\overline{x y}=10 x+y$ be the length of the leg $a$. Then the length of the hypotenuse $c=\overline{y x}=10 y+x$, and we have:

$$
\begin{array}{rlrl}
(10 x+y)^{2}+b^{2} & =(10 y+x)^{2}, & 2 \text { points } \\
100 x^{2}+20 x y+y^{2}+b^{2} & =100 y^{2}+20 x y+x^{2}, & \\
b^{2}=99 y^{2}-99 x^{2}, & 1 \text { point } \\
b^{2}=3^{2} \cdot 11(y-x)(y+x) . & 1 \text { point }
\end{array}
$$

Since the left side is the square of a natural number, and $x$ and $y$ are single-digit numbers, it must be that $y+x=11$, while $y-x$ must also be the square of a natural number. This leads to the following cases:

1. $y+x=11, y-x=9$

The solution to this system is $y=10, x=1$, which cannot be a solution to the problem because $y$ is a single-digit number (digit).
2. $y+x=11, y-x=4$

The solution to this system is $y=\frac{15}{2}, x=\frac{7}{2}$, which cannot be a solution to the problem because $x$ and $y$ are natural numbers (digits).
3. $y+x=11, y-x=1$

The solution to this system is $y=6, x=5$.

Now, $b^{2}=3^{2} \cdot 11^{2} \cdot 1$, so $b=33$.

Therefore, this is a right triangle with legs of length 56 and 33, and a hypotenuse of length 65.