Balkan Mathematical Olympiad Shortlist

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 $$3-2a=3x^2,3-2b=3y^2,3-2c=3z^2.$$ This is feasible by putting $$x=\frac{2uv}{u^2+v^2+w^2},y=\frac{2uw}{u^2+v^2+w^2},z=\frac{-u^2+v^2+w^2}{u^2+v^2+w^2}.$$ 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.