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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.

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  • $\begingroup$ What is integer in your triangles? Side lengths, coordinates of the vertices, angles in degrees, $\ldots$? $\endgroup$ – Christian Blatter Dec 21 '19 at 14:46
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    $\begingroup$ An integer triangle is a triangle all of whose sides have lengths that are integers. $\endgroup$ – Witold Dec 21 '19 at 14:48
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    $\begingroup$ The formulas for $OH$ and $IH$ are here mathworld.wolfram.com/Orthocenter.html . ($S_{w}$ is Conway triangle notation - mathworld.wolfram.com/ConwayTriangleNotation.html) $\endgroup$ – Witold Dec 21 '19 at 14:58
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    $\begingroup$ @ Jean Marie: Also, distance-between-incentre-and-orthocentre. $\endgroup$ – g.kov Dec 21 '19 at 16:14
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    $\begingroup$ It could be reasonable to limit the variables in given formulas.We can use these relations: $R=\frac{abc}{4s}$, $r=\frac{s}{p}$ and a relation between (abc) and (a+b+c), such that only variable are p and s.Hence $IO=f(p, s)$, $OH=g(p,s)$ and $IH=h(p, s)$. $\endgroup$ – sirous Dec 22 '19 at 12:49
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(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 $IO+IH-HO=6s+3\gt 0$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Witold Dec 27 '19 at 2:03

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