I'm a newbie in this site. I tried to search if this question was already answered but I'm not sure on how to do it. The problem is: given three distincts points $O,H,I$ namely the circumcenter, the orthocenter and the incenter of a triangle $\triangle ABC$, construct $\triangle ABC$.
What have I tried:
$$\left| \overline{OI} \right| ^2 = R^2 - 2Rr$$
Taking the midpoint $N$ of $\ \overline{OH} $ (center of the nine point circle) we have:
$\left| \overline{NI} \right| = \frac{R-2r}{2}$ So it seems to me we can obtain $R$ and $r$ from these points. I also tried to solve it using complex numbers it looked like this:
Let $a^2,b^2,c^2$ be the vertexes of $\triangle ABC$ in the complex plane in which the origin is the circumcenter let us also define $R=1$ (so $|a|=|b|=|c|=1)$ to simplify. Then we have that the orthocenter $H=a^2+b^2+c^2$ and the incenter $I=ab-ac +bc$ to find a solution to my problem is to find $a^2,b^2,c^2$ in terms of $H$ and $I$ but it was too complex to me. This complex approach indicates that the reflection of the incenter with respect to the circumcenter is important. Another cool distance formula is $\left| \overline{OH} \right|^2 = 9R^2 - (a^2+b^2+c^2)$ with $a,b,c$ being the sides of $\triangle ABC$
I asked the same question to myself a while ago, and now I'm glad to share my conclusion.
To build a triangle given $O,H,I$ is substantially equivalent to solving a general cubic, hence it cannot be done with straightedge and compass only. To prove it, notice that we know $$ (R,r,a^2+b^2+c^2) $$ and we have to find $(a,b,c)$ or, equivalently, the exradii $(r_a,r_b,r_c)$. However, the relations: $$ \sum_{cyc} r_a = 4R+r,\quad \sum_{cyc}\frac{1}{r_a}=\frac{1}{r},$$ $$ \sigma_2 = \sum_{cyc} r_a r_b = \frac{a^2+b^2+c^2}{2}+r(4R+r) $$ give that the exradii are the roots of the polynomial: $$ z^3-(4R+r)z^2+\sigma_2 z - r\sigma_2.$$