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2004 AIME II Problems/Problem 11

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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.

size(200);import three; pointpen = black; pathpen = black + linewidth(0.7); currentprojection = perspective(0,-20,5);triple O...     pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0), A=(-125,0), B=375*2^.5*expi(pi/4); D(arc(O,800,180,-90)); D...

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})}=\boxed{625}.

See also

2004 AIME II (ProblemsResources)
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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