TITLE: Isosceles triangles with integer side lengths whose vertices are lattice points
QUESTION [1 upvotes]: I was wondering if such a triangle exists and I indeed found one: the triangle whose vertices are
$(0,0), (-33,56), (63,16)$. But is there a way to somehow "generate" these triangles? By the way, I'm only interested in isosceles triangles that don't have horizontal or vertical sides.
Here's what I've done so far:  To make things easier, I made the origin the vertex. The other vertices are denoted as $(a,b)$ and $(c,d)$, and $$a^2+b^2=c^2+d^2=r^2$$$$(a-c)^2+(b-d)^2=n^2$$ where $a, b, c, d, r, $ and $n$ are integers. The second equation becomes $$a^2-2ac+c^2+b^2-2bd+d^2=2r^2-2ac-2bd$$$$2(r^2-(ac+bd))=n^2$$
I don't know how to proceed from this. Any ideas?

REPLY [1 votes]: Here is one method. Given two Pythagorean triples: $\,a^2+b^2=c^2\,$
and $\,d^2+e^2=f^2,\,$ use the first one to create an isosceles triangle with
vertices $\,(-a,0),(a,0),(0,b)\,$ and sides $2a,c,c.\,$ Regard each vertex
as a point in the complex plane and multiply each point by $\,d+ei\,$ to
get the triangle with vertices $\,(-ad,-ae),(ad,ae),(-be,bd)\,$ and sides multiplied by $\,f.\,$ This triangle will have horizontal or vertical
sides iff $\,\,a/b=d/e\,$ or $\, a/b=e/d.\,$
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