Extreme Value with Triangle 2.2

Let the circumcenter and incenter of $\triangle ABC$ be $O$ and $I$ respectively, and $R=8$ and $r=3$ . By Euler's theorem for a triangle , $\overline{OI}^2 = R(R-2r) = 8(8-6) = 16$ $\implies \overline{OI} = 4$ .

Since $\cos \theta$ decreases with $\theta$ , $\cos A$ is maximum when $\angle A$ is minimum. This occurs when vertex $A$ is farthest away from $I$ when $\angle A$ subtends the smallest arc $BC$ . That is vertex $A$ is at the end farther away from $I$ of the diameter of circumcircle joining $O$ , $I$ and $A$ (see figure).

We note that $\sin \frac A2 = \frac 3{12} = \frac 14$ . Therefore $\cos A = 1 - 2\sin^2 \frac A2 = 1 - 2 \left(\frac 14\right)^2 = \frac 78 = \boxed{0.875}$