The Largest Circle Within the Largest Inellipse

The Steiner ellipse is the maximum inellipse of a triangle, and its semi-major and semi-minor axes are given by $\frac{1}{3}\sqrt{a^2 + b^2 + c^2 \pm 2Z}$ , were $Z = \sqrt{a^4 + b^4 + c^4 -a^2b^2 - b^2c^2 - a^2c^2}$ .

In this problem $a = b = 1$ , so the semi-axes are $\frac{1}{3}\sqrt{1^2 + 1^2 + c^2 \pm 2\sqrt{1^4 + 1^4 + c^4 -1^21^2 - 1^2c^2 - 1^2c^2}}$ , which simplify to $\frac{1}{6}\sqrt{3}c$ and $\frac{1}{6}\sqrt{4 - c^2}$ .

The largest circle in an ellipse would be when its radius is the same as the semi-minor axis, so the one with a maximum radius would be the largest minimum of $\frac{1}{6}\sqrt{3}c$ and $\frac{1}{6}\sqrt{4 - c^2}$ at a given value of $c$ . Since $\frac{1}{6}\sqrt{3}c$ is an increasing function and $\frac{1}{6}\sqrt{4 - c^2}$ is a decreasing function (for $c > 0$ ), the largest minimum of these two would be when $\frac{1}{6}\sqrt{3}c = \frac{1}{6}\sqrt{4 - c^2}$ , which solves to $c = 1$ (an equilateral triangle), for semi-axes values (and a radius value) of $r = \frac{1}{6}\sqrt{3}$ .

Therefore, $A = 1$ , $B = 6$ , $C = 3$ , and $A + B + C = \boxed{10}$ .