Solving Triangles - Part 6

Place $A$ and $B$ on the $x$ -axis so that its midpoint is on the origin. Then its coordinates are $A(-\frac{1}{2}c, 0)$ and $B(\frac{1}{2}c, 0)$ . Since $AG \perp BG$ , by Thales' Theorem the locus of points for $G(p, q)$ is a circle, so $p^2 + q^2 = \frac{1}{4}c^2$ .

Now suppose that $C$ has coordinates $C(r, s)$ . By the centroid equation, $\frac{1}{3}(-\frac{1}{2}c + r + \frac{1}{2}c) = p$ and $\frac{1}{3}(0 + s + 0) = q$ , so that $p = \frac{1}{3}r$ and $q = \frac{1}{3}s$ .

Combining $p^2 + q^2 = \frac{1}{4}c^2$ with $p = \frac{1}{3}r$ and $q = \frac{1}{3}s$ gives $r^2 + s^2 = \frac{9}{4}c^2$ , which means the locus of points for $C(r, s)$ is a circle with radius of $\frac{3}{2}c$ , but since $\triangle ABC$ is acute, $-\frac{1}{2}c < r < \frac{1}{2}c$ .

By the distance formula, $b^2 = (r + \frac{1}{2})^2 + s^2$ , and since $r^2 + s^2 = \frac{9}{4}c^2$ , this simplifies to $b^2 = \frac{5}{2}c^2 + rc$ . Similarly, $a^2 = \frac{5}{2}c^2 - rc$ . Therefore, by the law of cosines, $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{\frac{5}{2}c^2 - rc + \frac{5}{2}c^2 + rc - c^2}{2\sqrt{\frac{5}{2}c^2 - rc}\sqrt{\frac{5}{2}c^2 + rc}} = \frac{4c}{\sqrt{25c^2 - 4r^2}}$ , which has a maximum when $r = 0$ (when $C$ is on the $y$ -axis) for a value of $\cos C = \frac{4}{5}$ , and has a minimum when $r^2$ is as large as possible, one place being when $r$ approaches $\frac{1}{2}c$ , for a value of $\cos C$ approaching $\frac{\sqrt{6}}{3}$ .

Therefore, the range of $\cos C$ is $[\frac{4}{5}, \frac{\sqrt{6}}{3})$ , so that $m = 4$ , $n = 5$ , $p = 6$ , and $q = 3$ , and $mnpq = \boxed{360}$ .