Dynamic Geometry: P44

Label the triangle as $ABC$ such that $CA=b=15$ and $BC=a=8$ . Let $A$ be the origin $(0,0)$ of the $xy$ -plane with $AB$ along the $x$ -axis, $O(x,y)$ be the variable circumcenter, and $\angle C = \theta$ .

Then by cosine rule , $c^2 = a^2 + b^2 - 2ab \cos \theta = 289 - 240 \cos \theta$ . Since circumcenter $O$ is where the three perpendicular side dividers meet, $x = \dfrac c2 = \dfrac 12 \sqrt{289-240 \cos \theta}$ .

Let the circumradius be \R). Then $2R = \dfrac c{\sin \theta} \implies R = \dfrac x{\sin \theta}$ . By Pythagorean theorem :

$y^2 = R^2 - x^2 = \frac {x^2}{\sin^2 \theta} - x^2 = x^2 \cot^2 \theta \implies y = \frac {x \cos \theta}{\sin \theta}$

Since $2R = \dfrac b{\sin B}$ , $R$ is minimum, when $B=90^\circ$ (see Dynamic Geometry: P43 ). Then $\theta = \cos^{-1} \dfrac 8{15}$ . Since the area of $\triangle ABC$ is given by $A_\triangle = \dfrac {ab\sin \theta}2$ , $A_\triangle$ is maximum, when $\theta = \dfrac \pi 2$ (see Dynamic Geometry: P42 ). The area bounded by the blue locus of $O$ , the cyan line and orange line is:

$\begin{aligned} A & = \int_{x_a}^{x_b} y \ dx = \int_{x_a}^{x_b} \frac {x \cos \theta}{\sin \theta} \ dx \\ & = \int_{\cos^{-1}\frac 5{18}}^\frac \pi 2 \frac 12 \sqrt{289-240\cos \theta} \cdot \frac {\cos \theta}{\sin \theta} \cdot \frac {240 \sin \theta \ d\theta}{4\sqrt{289-240\cos \theta}} \\ & = \int_{\cos^{-1}\frac 5{18}}^\frac \pi 2 30 \cos \theta \ d\theta \\ & = 30 \sin \theta \ \bigg|_{\cos^{-1}\frac 8{15}}^\frac \pi 2 = 30 - 30 \cdot \frac {\sqrt{161}}{15} = 30 - 2\sqrt{161} \end{aligned}$

Therefore the required answer is $30+2+161 = \boxed{193}$ .