Is it a unique solution?

Choosing a coordinate system so that the circumcentre is at the origin, if the position vectors of $A,B,C$ are $\mathbf{a},\mathbf{b},\mathbf{c}$ respectively, then we know $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$ , and so we know $\mathbf{b}+\mathbf{c}$ . Since $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}|= R$ is known, $\mathbf{b}-\mathbf{c}$ must be perpendicular to $\mathbf{b}+\mathbf{c}$ . Since $A$ is not the orthocentre, $\mathbf{b} + \mathbf{c} \neq \mathbf{0}$ .

If $\mathbf{e}$ is a unit vector perpendicular to $\mathbf{b}+\mathbf{c}$ ( $\mathbf{e}$ is unique to within a choice of sign), then $\mathbf{b} \; = \; \tfrac12(\mathbf{b}+\mathbf{c}) + \lambda\mathbf{e} \hspace{2cm} \mathbf{c} \; = \; \tfrac12(\mathbf{b}+\mathbf{c}) - \lambda\mathbf{e}$ for some $\lambda$ . Since $\mathbf{b}$ and $\mathbf{c}$ both have length $R$ , we deduce that $\tfrac14|\mathbf{b}+\mathbf{c}|^2 + \lambda^2 \; = \; R^2$ which determines $\lambda$ to within a sign. Thus $B$ and $C$ are determined to within labelling (which point is $B$ and which $C$ ). The triangle is uniquely determined.

Let O denote circumcenter, R denote circumradius and H denote orthocenter of ∆ABC. Given OA, we can uniquely draw the circumcicle. AH is well known to be (2RcosA). Given AH and R, cos A has a unique value and therefore Angle A has unique value. Corresponding to this angle, the LENGTH of chord that subtends it (BC) is unique. Now AH must be perpendicular to BC.This leaves only one unique position for BC and hence the ∆ABC is uniquely defined.

Consider the following two cases:

Case 1. The triangle is scalene.

The side opposite the given vertex is perpendicular to the line through the orthocenter and the given vertex, and also to the line through the circumcenter and the midpoint of this side. The median through this side's midpoint passes through the given vertex and so it suffices to say that the the triangle is uniquely defined.

Case 2: The triangle is isosceles.

Here, the side opposite the given vertex is perpendicular to the Euler line, but there is an infinite set of lines that are perpendicular to the Euler line. With only the orthocenter and the circumcenter, there is not much more that can be said. If however, we consider the centroid of the triangle, then we see that only one pair of points satisfy the conditions that the line segment from one to the other is bisected perpendicularly by the Euler line, and the average of their coordinates with those of the given vertex results in the coordinates of the centroid. Equilateral triangles have circumcenters and orthocenters at the same point so there is no need for a third case.

Hence, any triangle is uniquely defined by these three points.

Any three non collinear points determine a triangle