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 give info $OA=125$ and $OB=375\sqrt{2}$ a By Pythagoras 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 $\frac{1200\pi}{1600\pi}=\frac{3}{4}$. So the sector is 270 degrees. 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 so the radius through B will divide the circle into two sectors each with measure 135. Draw in $BA$ to create $\triangle{ABO}$. Now by Law of Cosines $AB^{2}=(125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)$ from there $AB=\sqrt{ (125)^{2}+(375\sqrt{2})^{2}-2(125\cdot375\sqrt{2})(cos 135)}=625$

See also

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