Let $\mathcal{R}$ be the region consisting of the set of points in the [coordinate plane](https://artofproblemsolving.com/wiki/index.php/Coordinate_plane) that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the [line](https://artofproblemsolving.com/wiki/index.php/Line) whose equation is $3y - x = 15$, the [volume](https://artofproblemsolving.com/wiki/index.php/Volume) of the resulting [solid](https://artofproblemsolving.com/wiki/index.php/Solid) is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime), and $p$ is not divisible by the square of any prime. Find $m + n + p$.
The [inequalities](https://artofproblemsolving.com/wiki/index.php/Inequality) are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)$. The region $\mathcal{R}$ is a [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle), as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right [cones](https://artofproblemsolving.com/wiki/index.php/Cone) that share the same base and axis. 


Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\frac 13 Bh$; since both cones share the same base, then the desired volume is $\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)$. The [distance](https://artofproblemsolving.com/wiki/index.php/Distance) from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}$. The distance between $\left(\frac 92, \frac {13}2\right)$ and $\left(\frac{39}{4}, \frac{33}{4}\right)$ is given by $h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}$. Thus, the answer is $\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}$. (Note to MAA: Is it a coincidence that this is the number of days in a non-leap year?)