A right-angled triangle has side lengths that are integers, and the measurement of its perimeter is equal to the measurement of its area. Which triangle is this?
Let the lengths of the sides be denoted by $x, y, z$. According to the Pythagorean theorem, $x^{2}+y^{2}=z^{2}$. From this,

$$
x^{2}=z^{2}-y^{2}=(z+y)(z-y)
$$

The second condition states that $2(x+y+z)=x y$. From this,

$$
2(y+z)=x(y-2)
$$

Substituting this into twice the (1) equation,

$$
2 x^{2}=2(y+z)(z-y)=x(y-2)(z-y)
$$

and since no side of the triangle can have a length of 0,

$$
2 x=(y-2)(z-y)
$$

Substituting this into twice the (2) equation,

$$
4(y+z)=(y-2)^{2}(z-y)
$$

We transform the left side,

$$
4(y+z)=8 y+4(z-y)
$$

Thus,

$$
8 y=(z-y)\left[(y-2)^{2}-4\right]=y(y-4)(z-y)
$$

and since

$$
y \neq 0, \quad(z-y)(y-4)=8
$$

Since $y$ and $z$ are integers, only

$$
z-y=8, \quad y-4=1 ; \quad z-y=2, \quad y-4=4
$$

and

$$
z-y=4, \quad y-4=2
$$

are possible. From these, $x, y, z$ are respectively $5,12,13,12,5,13 ; 6,8,10$ and $8,6,10$.