I am looking for information regarding which triangles $ABC$ satisfy $BC^2+AC^2=n\cdot AB^2$ for $n=1,2,3,...$. I'm sure that work has already been done in this area since it is a fairly simple question to ask.
As an example, I have just proved that $n=2$ is true if and only if $\angle LAC =\angle ABM$ or $\angle ANC=\angle ALB$, where $L,M,N$ are midpoints of sides $BC,CA,AB$ respectively. (This means that in any triangle $ABC$ we have $\angle LAC=\angle ABM\iff \angle ANC=\angle ALB$, which is a problem by David Monk.)
What happens for $n=3,4,5,...$?
Thanks.
In terms of numbers, you want $a^2+b^2 = nc^2$ where $a+b > c, a+c > b, b+c > a$.
Assume $a \ge b$.
If $a = b$ then $2a^2 = nc^2$ or $a = c\sqrt{n/2}$. Since $c < 2a$, $a < 2a\sqrt{n/2}$ or $1 < 2n$ or $n > 1/2$. So every positive integer will work. These correspond to isosceles triangles that get narrower as $n$ gets larger.
If $a > b$, once you have $a^2+b^2 = c^2$, just multiply $a$ and $b$ by $\sqrt{n}$ to get $a^2+b^2=nc^2$.
Don't know if this helps or if this is what you want.