Question

一个半径为 $4$ 英尺、高为 $10$ 英尺的圆柱形桶装满了水。把一个边长为 $8$ 英尺的实心立方体放入桶中，使立方体的一条对角线竖直。由此排开的水的体积为 $v$ 立方英尺。求 $v^2$ 。

A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2.$

答案

答案：384

知识点：立体几何体积棱锥

难度评级：2760

解答：

排开水的体积等于立方体位于桶口平面以下的部分体积。由对称性，该区域是从立方体底角切出的四面体：沿立方体棱方向有三条两两垂直且等长的棱，记其长度为 $\ell$ ，顶面是桶口平面中的一个等边三角形。这个等边截面内接于半径为 $4$ 的桶口圆，因此边长为 $4\sqrt{3}$ ，所以 $\ell = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$ 。

取一个等腰直角面为底面，则体积为 $\begin{aligned} &\frac{1}{3}\left(\frac{1}{2}\ell^2\right)\ell = \frac{\ell^3}{6} \\ &= \frac{(2\sqrt{6})^3}{6} \\ &= \frac{48\sqrt{6}}{6} = 8\sqrt{6}. \end{aligned}$

因此 $v = 8\sqrt{6}$ ，所以 $v^2 = 64 \cdot 6 = 384$ 。

The displaced volume equals the volume of the part of the cube lying below the plane of the barrel's rim. By symmetry that region is a tetrahedron cut from the bottom corner of the cube: three mutually perpendicular edges of equal length $\ell$ along the cube's edges, capped by an equilateral triangle in the rim plane. The equilateral cross-section is inscribed in the rim circle of radius $4,$ so its side length is $4\sqrt{3},$ and therefore $\ell = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}.$

Taking one of the right isosceles faces as the base, the volume is $\begin{aligned} &\frac{1}{3}\left(\frac{1}{2}\ell^2\right)\ell = \frac{\ell^3}{6} \\ &= \frac{(2\sqrt{6})^3}{6} \\ &= \frac{48\sqrt{6}}{6} = 8\sqrt{6}. \end{aligned}$

Thus $v = 8\sqrt{6}$ and $v^2 = 64 \cdot 6 = 384.$

2015 AIME II 第 9 题

答案：384

难度评级：2760

解答：

其他年份的第 9 题