Question

一个直圆锥的底面半径为 $600$ ，高为 $200\sqrt{7}$ 。一只苍蝇从圆锥表面上一点出发，该点到圆锥顶点的距离为 $125$ ，并沿圆锥表面爬到圆锥正对面的一个点，该点到顶点的距离为 $375\sqrt{2}$ 。求苍蝇可能爬行的最短距离。

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.

答案

答案：625

知识点：圆锥展开图（立体几何）余弦定理

难度评级：2990

解答：

圆锥斜高为 $\sqrt{600^2 + (200\sqrt{7})^2}$ $= \sqrt{360000 + 280000}$ $= 800$ 。沿经过起点的母线剪开并展开，得到半径为 $800$ 的扇形，其弧长等于底面周长 $2\pi \cdot 600 = 1200\pi$ 。半径为 $800$ 的整圆周长为 $1600\pi$ ，所以扇形的圆心角为 $\frac{3}{4} \cdot 360^\circ = 270^\circ$ 。圆锥正对面的点相当于绕底面半圈，在展开扇形中相隔 $135^\circ$ 。

最短路径是展开图中两个点之间的线段，这两个点到顶点的距离分别为 $125$ 和 $375\sqrt{2}$ ，夹角为 $135^\circ$ 。由余弦定理， $\begin{aligned} d^2 &= 125^2 + (375\sqrt{2})^2 \\ &\quad {}- 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ \\ &= 15625 + 281250 + 93750 \\ &= 390625. \end{aligned}$

因此 $d = 625$ 。

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 $d = 625.$

2004 AIME II 第 11 题

答案：625

难度评级：2990

解答：

其他年份的第 11 题