2004 AIME II Problem 11

Attempt Problem 11 of the 2004 AIME II 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 2004 AIME II solutions, or check the answer key.

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11.

A right circular cone has a base with radius 600600 and height 2007.200\sqrt{7}. A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125,125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is 3752.375\sqrt{2}. Find the least distance that the fly could have crawled.

Answer: 625
Concepts:conenet (3D geometry)law of cosines
Difficulty rating: 2990
Solution:

The slant height is 6002+(2007)2\sqrt{600^2 + (200\sqrt{7})^2} =360000+280000= \sqrt{360000 + 280000} =800.= 800. Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius 800800 whose arc has the base circumference 2π600=1200π;2\pi \cdot 600 = 1200\pi; since a full circle of radius 800800 has circumference 1600π,1600\pi, the central angle is 34360=270.\frac{3}{4} \cdot 360^\circ = 270^\circ. A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is 135135^\circ away.

The shortest crawl is the straight segment between the two points, at radii 125125 and 3752375\sqrt{2} with a 135135^\circ angle between them. By the law of cosines, d2=1252+(3752)221253752cos135=15625+281250+93750=390625. \begin{aligned} d^2 &= 125^2 + (375\sqrt{2})^2 \\ &\quad {}- 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ \\ &= 15625 + 281250 + 93750 \\ &= 390625. \end{aligned}

Thus d=625.d = 625.

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