2008 AIME I Problem 5

Attempt Problem 5 of the 2008 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AIME I solutions, or check the answer key.

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5.

A right circular cone has base radius rr and height h.h. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 1717 complete rotations. The value of h/rh/r can be written in the form mn,m\sqrt{n}, where mm and nn are positive integers and nn is not divisible by the square of any prime. Find m+n.m + n.

Answer: 14
Concepts:conecircumferencePythagorean Theorem
Difficulty rating: 2300
Solution:

The contact point of the base stays at distance =r2+h2\ell = \sqrt{r^2 + h^2} (the slant height) from the fixed vertex, so it traces a circle of radius .\ell. Rolling without slipping, the cone makes one rotation for each base circumference of arc, so returning after exactly 1717 rotations means 2πr2+h2=172πr,2\pi\sqrt{r^2 + h^2} = 17 \cdot 2\pi r, i.e. r2+h2=17r.\sqrt{r^2 + h^2} = 17r.

Squaring gives h2=288r2,h^2 = 288r^2, so h/r=288=122,h/r = \sqrt{288} = 12\sqrt{2}, and m+n=12+2=14.m + n = 12 + 2 = 14.

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