Question

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.

Answer

Solution:

The slant height is $\sqrt{600^2 + (200\sqrt{7})^2} = \sqrt{360000 + 280000} = 800.$ Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius $800$ whose arc has the base circumference $2\pi \cdot 600 = 1200\pi;$ since a full circle of radius $800$ has circumference $1600\pi,$ the central angle is $\frac{3}{4} \cdot 360^\circ = 270^\circ.$ A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is $135^\circ$ away.

The shortest crawl is the straight segment between the two points, at radii $125$ and $375\sqrt{2}$ with a $135^\circ$ angle between them. By the law of cosines, $d^2 = 125^2 + (375\sqrt{2})^2 - 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ = 15625 + 281250 + 93750 = 390625.$

Thus $d = 625.$

2004 AIME II Problem 11

Solution:

Problem 11 in Other Years