Coordinates of a triangle given orthocenter, circumcenter and circumradius I need to figure out a way to find the coordinates of a scalene triangle given its circumradius and the coordinates of its circumcenter and orthocenter, or show that there isn’t sufficient info to do so. Any ideas?
 A: Let $O$ and $R$ be the center
and the radius of the circumscribed circle
of the scalene $\triangle ABC$
with sidelengths $a,b,c$,
angles $\alpha,\beta,\gamma$,
orthocenter $H$,
incenter $O_i$.
Then it is known that
\begin{align}
a^2+b^2+c^2&=(3R)^2-|OH|^2
\tag{1}\label{1}
.
\end{align}  
To simplify the question without loss of generality,
given $O$, $R$ and $H$,
let's 
move the origin of the coordinate system to $O$
and rotate $OH$
to put $H$ to the right of $O$,
and scale by $(2R)^{-1}$.
Then the side lengths can be replaced as
\begin{align}
a&=\sin\alpha,\quad
b=\sin\beta,\quad
c=\sin\gamma
,
\end{align}
and \eqref{1} becomes
\begin{align}
\sin^2\alpha+\sin^2\beta+\sin^2\gamma
&=(\tfrac32)^2-\frac{|OH|^2}{(2R)^2}
,\\
\sin^2\alpha+\sin^2\beta+\sin^2(\alpha+\beta)
&=(\tfrac32)^2-\frac{|OH|^2}{(2R)^2}
\tag{2}\label{2}
,
\end{align}  
which gives the condition on $OH$ and $R$
\begin{align}
|OH|<3\,R
.
\end{align} 
Thus, we have this simplified picture:
$O$ at the origin, $R=\tfrac12$
and $H$ located somewhere between $O$
and $(\tfrac32,0)$:

Any pair of angles $\alpha,\beta$, 
which agree with \eqref{2}
will define a suitable triangle.
An interesting case is if we add
an extra condition that one angle is $90^\circ$.
In this case solutions are limited to the cases
\begin{align}
\frac{|OH|^2}{(2R)^2}&=\tfrac94-2=\tfrac14
,\\
|OH|&=R
,
\end{align}
and any triangle with one vertex at $H$
and the other two at the ends of the diameter through $O$
will have the orthocenter at $H$, the circumcenter at $O$
and the circumradius $R=\tfrac12$.

A: Given circumcenter $O$ and orthocenter $H$ we can also construct the centroid $G$ of the triangle, because $OG=OH/3$. Pick then any point $A$ on the circumcircle and produce line $AG$ to $M$ such that $GM=AG/2$: point $M$ is then the midpoint of the side $BC$ of the triangle opposite to $A$. As $BC$ is perpendicular to $OM$, it can be readily constructed.
This construction works as long as $OM$ is less than the circumradius.

