Question

如图，外层正方形 $S$ 的边长为 $40$ 。在 $S$ 内部作第二个正方形 $S'$ ，其边长为 $15$ ，中心与 $S$ 相同，并且边与 $S$ 的边平行。从 $S$ 每条边的中点，向 $S'$ 的两个最近顶点各连一条线段。得到一个内接于 $S$ 的四尖星形图案。将这个星形剪下，再折叠成一个以 $S'$ 为底面的棱锥。求这个棱锥的体积。

In the accompanying figure, the outer square $S$ has side length $40.$ A second square $S'$ of side length $15$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S.$ From each midpoint of a side of $S,$ segments are drawn to the two closest vertices of $S'.$ The result is a four-pointed starlike figure inscribed in $S.$ The star figure is cut out and then folded to form a pyramid with base $S'.$ Find the volume of this pyramid.

答案

答案：750

知识点：展开图（立体几何）棱锥体积勾股定理

难度评级：2230

解答：

沿着 $S'$ 的边折叠星形时，四个三角形尖端（也就是 $S$ 各边的中点）会汇合成一个顶点 $V$ 。令 $M$ 为 $S'$ 的中心， $P$ 为它一条边的中点。在平面图中， $P$ 到其三角形尖端的距离为 $20 - \frac{15}{2} = \frac{25}{2}$ ，折叠后这就是斜高 $PV$ 。

三角形 $PMV$ 在 $M$ 处为直角，且 $PM = \frac{15}{2}$ ，所以高为 $\begin{aligned} VM &= \sqrt{\left(\tfrac{25}{2}\right)^2 - \left(\tfrac{15}{2}\right)^2} \\ &= \sqrt{100} = 10. \end{aligned}$ 体积为 $\frac{1}{3} \cdot 15^2 \cdot 10 = 750$ 。

Folding the star along the sides of $S'$ lifts the four triangular points so that their tips (the midpoints of the sides of $S$ ) meet at a single apex $V.$ Let $M$ be the center of $S'$ and $P$ the midpoint of one of its sides. In the flat figure, the distance from $P$ to the tip of its triangle is $20 - \frac{15}{2} = \frac{25}{2},$ and this becomes the slant $PV$ after folding.

Triangle $PMV$ has a right angle at $M,$ with $PM = \frac{15}{2},$ so the height is $\begin{aligned} VM &= \sqrt{\left(\tfrac{25}{2}\right)^2 - \left(\tfrac{15}{2}\right)^2} \\ &= \sqrt{100} = 10. \end{aligned}$ The volume is $\frac{1}{3} \cdot 15^2 \cdot 10 = 750.$

2012 AIME II 第 5 题

答案：750

难度评级：2230

解答：

其他年份的第 5 题