Integer triangle $ABC$ such that $IHO$ is also an integer triangle. An infinite number of such non-similar triangles $ABC$. 
Let $I, H, O$ be the incenter, orthocenter and circumcenter of non-isosceles triangle $ABC$ respectively. Prove that there are infinitely many integer triangles $ABC$, none of which are similar, suchs that for each of them the triangle $IHO$ is also an integer triangle. (An integer triangle  is a triangle all of whose sides have lengths that are integers)

My work. $$IO=\sqrt{R^2-2rR}$$ $$OH=\sqrt{9R^2-(a^2+b^2+c^2)}$$ $$IH=\sqrt{2r^2+4R^2-\frac{1}{2}(a^2+b^2+c^2)}$$
where $R$, $r$ are circumradius and inradius of triangle $ABC$ respectively; $a$, $b$, $c$ are the lengths of the sides of  triangle $ABC$. Unfortunately, these formulas are very inconvenient for analysis.
 A: (Some small examples that I found satisfy $IO=IH$. So...) 
Let us consider $\triangle{ABC}$ satisfying $IO=IH$. We have
$$IO=IH\iff (a^2 - a b + b^2 - c^2) (a^2 - b^2 + b c - c^2) (a^2 - a c - b^2 + c^2)=0$$
So, in the following, let us consider $\triangle{ABC}$ such that $$a^2=b^2-bc+c^2\tag1$$
(which means that $\angle A=60^\circ$.)
Here, it is known that 
$$a=m^2+mn+n^2,\quad b=m^2+2mn,\quad c=m^2-n^2$$
satisfy $(1)$. Then, we get
$$IO=IH=n\sqrt{m^2+mn+n^2},\qquad HO=2mn+n^2\in\mathbb Z$$
Also, it is known that 
$$m=s^2-1,\quad n=2s+1,\quad z=s^2+s+1$$
satisfy $z^2=m^2+mn+n^2$, so if 
$$\begin{cases}a=(s^2+s+1)^2
\\\\b=(s - 1) (s + 1) (s^2 + 4 s  + 1)
\\\\c=s (s + 2) (s^2 - 2 s - 2)\tag2\end{cases}$$
where $s\ (\ge 3)$ is an integer, then we get $$IO=IH=(2s+1)(s^2+s+1)\in\mathbb Z$$

Let us prove that if $(2)$, then $\triangle{ABC}$ are not isosceles triangles.
Proof : 
We have
$$b-a=(2 s + 1) (s^2 - 2 s - 2)\gt 0$$
$$a-c=(2 s + 1) (s^2 + 4 s + 1)\gt 0$$
$$a+c-b=(s - 1) (s + 1) (s^2 - 2 s - 2)\gt 0\qquad\square$$

Next, let us prove that if $(2)$, then no two are similar.
Proof : 
Suppose that a triangle with $(2)$ and a triangle with
$$\begin{cases}a=(t^2+t+1)^2
\\\\b=(t - 1) (t + 1) (t^2 + 4 t  + 1)
\\\\c=t (t + 2) (t^2 - 2 t - 2)\end{cases}$$
where $t\not=s\ (t\ge 3)$ are similar. Then, we have
$$\frac{(s^2+s+1)^2}{(s - 1) (s + 1) (s^2 + 4 s  + 1)}=\frac{(t^2+t+1)^2}{(t - 1) (t + 1) (t^2 + 4 t  + 1)}$$
$$\iff (s - t) (2 s t + s + t + 2) (s^2 t^2 - 2 s^2 t - 2 s^2 - 2 s t^2 - 8 s t - 2 s - 2 t^2 - 2 t + 1) = 0$$
$$\iff s^2 t^2 - 2 s^2 t - 2 s^2 - 2 s t^2 - 8 s t - 2 s - 2 t^2 - 2 t + 1 = 0$$
$$\iff s = \frac{2 (t^2 + 4 t + 1) ± 2(t^2+t+1)\sqrt{3}}{2 (t^2 - 2 t - 2) }$$
which is impossible since the RHS is irrational.$\qquad\square$

Conclusion : 
For $$\begin{cases}a=(s^2+s+1)^2
\\\\b=(s - 1) (s + 1) (s^2 + 4 s  + 1)
\\\\c=s (s + 2) (s^2 - 2 s - 2)\end{cases}$$
where $s\ (\ge 3)$ is an integer, $\triangle{ABC}$ are not isosceles triangles and no two are similar, with $$IO=IH=(2s+1)(s^2+s+1),\qquad HO=(2 s + 1) (2 s^2 + 2 s - 1)$$
where $IO+IH-HO=6s+3\gt 0$.
