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 1

The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$ -axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$ . The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800$ . Setting $\theta R=C\implies 800\theta=1200\pi\implies\theta=\frac{3\pi}{2}$ .

If the starting point $A$ is on the positive $x$ -axis at $(125,0)$ then we can take the end point $B$ on $\theta$ 's bisector at $\frac{3\pi}{4}$ radians along the $y=-x$ line in the second quadrant. Using the distance from the vertex puts $B$ at $(-375,-375)$ . Thus the shortest distance for the fly to travel is along segment $AB$ in the sector, which gives a distance $\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}$ .

Solution 2

To find the shortest length from the red to blue points, the net of the side of the cone could be drawn.

The angle $YVX$ is equal to $360^\circ \cdot \frac{1200\pi}{1600\pi} \cdot \frac{1}{2}$ , or $135^\circ$ . Therefore, the law of cosines could be utilized. $AB = \sqrt{(375\sqrt{2})^2 + 125^2 - 2 \cdot (375\sqrt{2})(125)(\cos 135^\circ)} = \boxed{625}$

~Diagram and Solution by MaPhyCom

2004 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2004 AIME II Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

See also