2004 AIME II Problems/Problem 11

Problem

A right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$ , 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 $375\sqrt{2}.$ Find the least distance that the fly could have crawled.

Solution

Label the starting point of the fly as $A$ and the ending as $B$ and the vertex of the cone as $O$ . With the given information, $OA=125$ and $OB=375\sqrt{2}$ . By the Pythagorean Theorem, the slant height can be calculated by: $200\sqrt{7}^{2} + 600^2=640000$ , so the slant height of the cone is $800$ . The base of the cone has a circumference of $1200\pi$ , so if we cut the cone along its slant height and through $A$ , we get a sector of a circle $O$ with radius $800$ . Now the sector is $\frac{1200\pi}{1600\pi}=\frac{3}{4}$ of the entire circle. So the degree measure of the sector is $270^\circ$ . Now we know that $A$ and $B$ are on opposite sides. Therefore, since $A$ lies on a radius of the circle that is the "side" of a 270 degree sector, $B$ will lie exactly halfway between. Thus, the radius through $B$ will divide the circle into two sectors, each with measure $135^\circ$ . Draw in $BA$ to create $\triangle{ABO}$ . Now, by the Law of Cosines, $AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})$ . From there we have $AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(\cos {135})}=625$ .

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2004 AIME II Problems/Problem 11

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