Coordinates of a triangle given orthocenter, circumcenter and circumradius

Question

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?

Answer 1

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.

Answer 2

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 $R=\tfrac12$.