2010 AIME I Problem 11

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11.

Let R\mathcal{R} be the region consisting of the set of points in the coordinate plane that satisfy both 8x+y10|8 - x| + y \le 10 and 3yx15.3y - x \ge 15. When R\mathcal{R} is revolved around the line whose equation is 3yx=15,3y - x = 15, the volume of the resulting solid is mπnp,\frac{m\pi}{n\sqrt{p}}, where m,m, n,n, and pp are positive integers, mm and nn are relatively prime, and pp is not divisible by the square of any prime. Find m+n+p.m + n + p.

Answer: 365
Concepts:volumeconecoordinate geometry
Difficulty rating: 2920
Solution:

The condition 8x+y10|8 - x| + y \le 10 means yx+2y \le x + 2 for x8x \le 8 and y18xy \le 18 - x for x8.x \ge 8. Intersecting with the half-plane 3yx153y - x \ge 15 leaves the triangle with vertices A=(92,132)A = \left(\frac{9}{2}, \frac{13}{2}\right) and B=(394,334)B = \left(\frac{39}{4}, \frac{33}{4}\right) on the line 3yx=15,3y - x = 15, and apex C=(8,10).C = (8, 10).

Side ABAB lies on the axis of revolution, and the foot DD of the perpendicular from CC to the line, namely (8.7,7.9),(8.7, 7.9), lies between AA and B.B. So the solid is two cones sharing a base of radius CDCD with heights summing to AB,AB, and its volume is 13πCD2AB.\frac{1}{3}\pi \cdot CD^2 \cdot AB. Here CD=31081510=710,AB=(214)2+(74)2=7104. \begin{aligned} CD &= \frac{|3 \cdot 10 - 8 - 15|}{\sqrt{10}} = \frac{7}{\sqrt{10}}, \\ AB &= \sqrt{\left(\tfrac{21}{4}\right)^2 + \left(\tfrac{7}{4}\right)^2} \\ &= \frac{7\sqrt{10}}{4}. \end{aligned}

The volume is 13π49107104=343π1210,\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+10m + n + p = 343 + 12 + 10 =365.= 365.

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