2004 AIME II Problem 11

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

The slant height is

\sqrt{600^2 + (200\sqrt{7})^2} = 800;

unroll the cone into a circular sector of radius

800

The sector’s central angle is

270^\circ,

so exactly opposite points are

135^\circ

apart in the sector; apply the law of cosines

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,

\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.

Problem 11 in other years