Question

如下图标记的正方体 $ABCDEFGH$ 边长为 $1$ ，被一个经过顶点 $D$ 和两个点 $M$ 、 $N$ 的平面切开；这两个点分别是 $\overline{AB}$ 和 $\overline{CG}$ 的中点。该平面把正方体分成两个立体。较大立体的体积可写成 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG},$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：89

知识点：正方体体积棱锥相似

难度评级：2740

小提示：

将切割平面延伸，使其与直线 $BC$ 在 $B$ 外侧的点 $K$ 相交：由于 $MB \parallel DC$ 且 $MB = \frac{1}{2}DC$ ，点 $B$ 是 $\overline{CK}$ 的中点

Extend the cutting plane to meet line $BC$ beyond $B$ at $K:$ since $MB \parallel DC$ and $MB = \frac{1}{2}DC,$ point $B$ is the midpoint of $\overline{CK}$

大提示：

较小部分是棱锥 $KDCN$ 减去棱锥 $KMBP$ ，后者与前者相似，相似比为 $\frac{1}{2}$

The smaller piece is pyramid $KDCN$ minus pyramid $KMBP,$ which is similar to it with ratio $\frac{1}{2}$

解答：

延伸切割平面。在底面中，直线 $DM$ 与直线 $CB$ 在 $B$ 外侧的延长线相交于点 $K$ ；由于 $MB \parallel DC$ 且 $MB = \frac{1}{2}DC$ ，线段 $MB$ 是三角形 $KDC$ 的中位线，所以 $B$ 是 $\overline{CK}$ 的中点，且 $CK = 2$ 。该平面还与棱 $BF$ 相交于点 $P$ ；而正方体在平面外被切掉的部分，是棱锥 $KDCN$ 去掉小棱锥 $KMBP$ 后剩下的部分。

棱锥 $KDCN$ 的底面 $DCN$ 是直角三角形，两条直角边为 $DC = 1$ 和 $CN = \frac{1}{2}$ ，顶点 $K$ 到该底面所在平面的距离为 $CK = 2$ ，所以体积为 $\frac{1}{3} \cdot \frac{1}{4} \cdot 2 = \frac{1}{6}$ 。棱锥 $KMBP$ 与 $KDCN$ 相似，相似比为 $\frac{KB}{KC} = \frac{1}{2}$ ，所以其体积为 $\frac{1}{8} \cdot \frac{1}{6} = \frac{1}{48}$ 。

因此较小部分体积为 $\frac{1}{6} - \frac{1}{48} = \frac{7}{48}$ ，较大部分体积为 $1 - \frac{7}{48} = \frac{41}{48}$ ，所以 $p + q = 41 + 48 = 89$ 。

Extend the cutting plane. In the bottom face, line $DM$ meets line $CB$ extended beyond $B$ at a point $K;$ since $MB \parallel DC$ and $MB = \frac{1}{2}DC,$ segment $MB$ is a midline of triangle $KDC,$ so $B$ is the midpoint of $\overline{CK}$ and $CK = 2.$ The plane also cuts edge $BF$ at a point $P,$ and the piece of the cube cut off past the plane is the pyramid $KDCN$ with the small pyramid $KMBP$ sliced away.

Pyramid $KDCN$ has base $DCN,$ a right triangle with legs $DC = 1$ and $CN = \frac{1}{2},$ and its apex $K$ is at distance $CK = 2$ from the plane of that base, so its volume is $\frac{1}{3} \cdot \frac{1}{4} \cdot 2 = \frac{1}{6}.$ Pyramid $KMBP$ is similar to $KDCN$ with ratio $\frac{KB}{KC} = \frac{1}{2},$ so its volume is $\frac{1}{8} \cdot \frac{1}{6} = \frac{1}{48}.$

The smaller piece therefore has volume $\frac{1}{6} - \frac{1}{48} = \frac{7}{48},$ and the larger piece has volume $1 - \frac{7}{48} = \frac{41}{48},$ giving $p + q = 41 + 48 = 89.$

2012 AIME I 第 8 题

答案：89

难度评级：2740

小提示：

大提示：

解答：

其他年份的第 8 题