Extreme Value with Triangle 2.1

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 minimum when $\angle A$ is maximum. This occurs when vertex $A$ is nearest to $I$ when $\angle A$ subtends the largest arc $BC$ . That is vertex $A$ is at the end nearer to $I$ of the diameter of circumcircle joining $O$ , $I$ and $A$ (see figure).

We note that $\sin \frac A2 = \frac 34$ . Therefore $\cos A = 1 - 2\sin^2 \frac A2 = 1 - 2 \left(\frac 34\right)^2 = - \frac 18 = \boxed{-0.125}$