Question

Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15.$ When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15,$ the volume of the resulting solid is $\frac{m\pi}{n\sqrt{p}},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p.$

Answer

Solution:

The condition $|8 - x| + y \le 10$ means $y \le x + 2$ for $x \le 8$ and $y \le 18 - x$ for $x \ge 8.$ Intersecting with the half-plane $3y - x \ge 15$ leaves the triangle with vertices $A = \left(\frac{9}{2}, \frac{13}{2}\right)$ and $B = \left(\frac{39}{4}, \frac{33}{4}\right)$ on the line $3y - x = 15,$ and apex $C = (8, 10).$

Side $AB$ lies on the axis of revolution, and the foot $D$ of the perpendicular from $C$ to the line, namely $(8.7, 7.9),$ lies between $A$ and $B.$ So the solid is two cones sharing a base of radius $CD$ with heights summing to $AB,$ and its volume is $\frac{1}{3}\pi \cdot CD^2 \cdot AB.$ Here $CD = \frac{|3 \cdot 10 - 8 - 15|}{\sqrt{10}} = \frac{7}{\sqrt{10}}, \qquad AB = \sqrt{\left(\tfrac{21}{4}\right)^2 + \left(\tfrac{7}{4}\right)^2} = \frac{7\sqrt{10}}{4}.$

The volume is $\frac{1}{3}\pi \cdot \frac{49}{10} \cdot \frac{7\sqrt{10}}{4} = \frac{343\pi}{12\sqrt{10}},$ so $m + n + p = 343 + 12 + 10 = 365.$

2010 AIME I Problem 11

Solution:

Problem 11 in Other Years