# Circumradius of a triangle inscribed in an elllipse

Let ABC be a triangle with interior angles $\alpha, \beta$ and $\gamma.$ If this triangle inscribed in an ellipse with semi-major axis $a$ and semi-minor axis $b,$ how can we find the radius of the circumcircle of ABC?

I am thinking of this problem for a long time, but could not think of a way to solve it. Any idea?

It is simple: you cannot solve it, because the given parameters do not fix $R$. Given three non-collinear points in the plane, here it is a set of ellipses with aspect ratio $2:1$ going through such points:
This can be drawn by picking a direction, scaling with respect to the orthogonal direction, drawing the circumcircle of the scaled triangle and scaling it back to have an ellipse with aspect ratio $2:1$ going through the vertices of the original triangle. In particular, $\alpha,\beta,\gamma,a,b$ do not fix $R$. The next animation is simply derived from the previous one:
• @Bumblebee: I believe this problem cannot be forced to have a unique solution in any reasonable way. The curves described by the main centers of the rotating triangles are algebraic curves with high degree (sort of skew Lissajous curves), so even if you constrain the circumcenter of $ABC$ to lie on the major/minor axis of the ellipse you have a couple of distinct solutions. – Jack D'Aurizio Nov 19 '17 at 10:17
• @Bumblebee: $R$ can take any value in the range between the minimum circumradius and the maximum one. To find an explicit solution to this minimization/maximization problem should be time-consuming but doable. It is simpler to tackle the dual problem, i.e. to fix a triangle and find the largest/smallest circumscribed ellipse with fixed aspect ratio (the situation of the first animation). – Jack D'Aurizio Nov 19 '17 at 18:42