## Task 13/70

We are looking for a right-angled triangle whose side lengths $a, b, c$ are all integers and for the smallest possible $a$ satisfy the relation $4a = \frac{b+c}{2}$.

Here, $a$ denotes a cathetus.
Without loss of generality, it can be assumed that $c$ is the hypotenuse. Then, according to the Pythagorean theorem, $a^{2}=c^{2}-b^{2}$.

On the other hand, from $4 a=\frac{b+c}{2}$, the equation $8 a=c+b$ follows. Now, $a^{2}=8 a \cdot \frac{1}{8} a=c^{2}-b^{2}=(c+b)(c-b)$. This results in the system of equations

$$
c+b=8 a \quad ; \quad c-b=\frac{1}{8} a
$$

which has the solutions $b=\frac{63}{16} a$ and $c=\frac{65}{16} a$. Based on the problem statement, it follows that $a=16 \mathrm{LE}$, $b=63 \mathrm{LE}, c=65 \mathrm{LE}$.