Triangles having integer sides and integer area. Find all non-right angled dissimilar triangles having integer sides and integer area simuntaneously. Are there infinitely many such triangle? 
 A: Say you have a triangle with integer sides and area. Let the origin $O$ be one vertex of your triangle, and place another vertex $A$ at $(a, 0)$ (with $a$ positive integer). It will be convenient if none of these are obtuse angles of the triangle, so for the argument's sake, assume $OA$ is the longest side of the triangle.
Any third point $B$ with coordinates $(x, y)$ would yield a triangle $OAB$ with integer area as long as $y$ is an integer (base times height divided by $2$) (if $a$ is odd, $y$ needs to be even as well). So we need to figure out what such points yield integer sides $AB$ and $OB$. We have:
$$
|OB| ^2 = x^2 + y^2 \quad \quad |AB|^2 = (a-x)^2 + y^2 = a^2 + 2ax + x^2 + y^2
$$
The first equation says that $x^2$ needs to be an integer, since $y^2$ is, so $x$ is the square root of an integer. But then the second equation says that $2ax$ needs to be an integer, so $x$ needs to be rational also. The only rational numbers with integer squares are integers, so $x$ needs to be an integer.
This means that the triangle defined by $O$, $B$ and $(x, 0)$ is a Pythagorean triangle. This also goes for the triangle defined by $B$, $A$ and $(x, 0)$. So any time you have a triangle with integer sides and area it can be divided into two Pythagorean triangles sharing a leg. And any time you have two Pythagorean triangles sharing a leg length and either both having half-integer or both having integer area, you can glue them together to form a triangle with integer sides and integer area.
Classifying such Pythagorean triangles and taking similarity into account is beyond the time I have available, but letting the two triangles be equal gives you enough isosceles examples to conclude that there are infinitely many.
Edit After some thought I realized I missed an important point: $y$ doesn't have to be an integer. In that case, it still needs to be a rational number, since the area $ay/2$ is an integer. Since $y$ is a rational number and we're looking for dissimilar triangles, some scaled-up version of the triangle (e.g. the one with all measures increased by a factor of $a/2$) will have integer $y$-coordinate for point $B$ and thus by this argument be dividable into two Pythagorean triangles.
A: $A=(0,0)$, $B=(n^2-1,0)$, $C=(0,2n)$ with $n\ge 2$.
