Problem 2082

If the lengths $a$, $b$ and $c$ of the sides are rational, since $a h_a = b h_b = c h_c = 2A$, where by $h_a$, $h_b$ and $h_c$ we denote the lengths of the altitudes of the triangle, it is enough to find an infinite number of triangles with rational area. From Heron's formula we have
$$A = \sqrt{\frac{3}{2} \left(\frac{3}{2} - a\right) \left(\frac{3}{2} - b\right) \left(\frac{3}{2} - c\right)} = \frac{1}{4} \sqrt{3(3 - 2a)(3 - 2b)(3 - 2c)}.$$
and hence, in order the area be rational for an infinite number of sides, it is enough the quantity under the radical to be square of a rational number. Therefore it is enough to find rational numbers $x$, $y$ and $z$ such that
$$3 - 2a = 3x^2, \quad 3 - 2b = 3y^2, \quad 3 - 2c = 3z^2.$$
This is feasible by putting
$$x = \frac{2uv}{u^2 + v^2 + w^2}, \quad y = \frac{2uw}{u^2 + v^2 + w^2}, \quad z = \frac{-u^2 + v^2 + w^2}{u^2 + v^2 + w^2}.$$
where $u$, $v$ and $w$ are rational. It is easily checked that for these values of $x$, $y$ and $z$ we have $x^2 + y^2 + z^2 = 1$ and therefore, there exists a triangle of side lengths $a$, $b$ and $c$ with perimeter $3$.

Solution 2:
All triangles with side lengths $\frac{3a}{a+b+c}$, $\frac{3b}{a+b+c}$, $\frac{3c}{a+b+c}$ where $a$, $b$ and $c$ are integers such that $a^2 + b^2 = c^2$ satisfy the condition of the problem. Since there are infinitely many right triangles with integer sides no two of which are similar, we are done.