Question

一个以 $ABCD$ 为底面、 $E$ 为顶点的正方形棱锥有八条长为 $4$ 的边。一个平面经过 $\overline{AE}$ ， $\overline{BC}$ ，和 $\overline{CD}$ 的中点。该平面与棱锥的截面面积可表示为 $\sqrt{p}$ 。求 $p$ 。

A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4.$ A plane passes through the midpoints of $\overline{AE},$ $\overline{BC},$ and $\overline{CD}.$ The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}.$ Find $p.$

答案

答案：80

知识点：立体几何棱锥坐标几何面积分割

难度评级：3060

解答：

将底面放在 $A = (0,0,0)$ ， $B = (4,0,0)$ ， $C = (4,4,0)$ ， $D = (0,4,0)$ ；则顶点为 $E = (2, 2, 2\sqrt{2})$ ，因为 $2^2 + 2^2 + 8 = 16$ 。给定的三个中点为 $R = (1, 1, \sqrt{2})$ ， $S = (4, 2, 0)$ ，和 $T = (2, 4, 0)$ ，它们都满足 $x + y + 2\sqrt{2}\,z = 6$ ，这是切割平面的方程。

对边 $\overline{BE}$ 和 $\overline{DE}$ 参数化，可得平面与它们分别交于 $U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ 和 $V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right)$ 。截面是五边形 $RUSTV$ ，其中 $RU = RV = \sqrt{7}$ ， $US = VT = \sqrt{3}$ ， $ST = 2\sqrt{2}$ ，对角线 $UV = 3\sqrt{2}$ 。

沿 $\overline{UV}$ 分割五边形。等腰三角形 $RUV$ 的高为 $\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}}$ ，面积为 $\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}$ 。等腰梯形 $USTV$ 的高为 $\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}}$ ，面积为 $\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}$ 。总面积为 $4\sqrt{5} = \sqrt{80}$ ，所以 $p = 80$ 。

Place the base at $A = (0,0,0),$ $B = (4,0,0),$ $C = (4,4,0),$ $D = (0,4,0);$ the apex is then $E = (2, 2, 2\sqrt{2}),$ since $2^2 + 2^2 + 8 = 16.$ The given midpoints are $R = (1, 1, \sqrt{2}),$ $S = (4, 2, 0),$ and $T = (2, 4, 0),$ and all three satisfy $x + y + 2\sqrt{2}\,z = 6,$ the equation of the cutting plane.

Parametrizing edges $\overline{BE}$ and $\overline{DE}$ shows the plane meets them at $U = \left(\frac{7}{2}, \frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ and $V = \left(\frac{1}{2}, \frac{7}{2}, \frac{\sqrt{2}}{2}\right).$ The cross-section is the pentagon $RUSTV$ with $RU = RV = \sqrt{7},$ $US = VT = \sqrt{3},$ $ST = 2\sqrt{2},$ and diagonal $UV = 3\sqrt{2}.$

Split the pentagon along $\overline{UV}.$ Isosceles triangle $RUV$ has height $\sqrt{7 - \frac{9}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2} \cdot 3\sqrt{2} \cdot \sqrt{\frac{5}{2}} = \frac{3\sqrt{5}}{2}.$ Isosceles trapezoid $USTV$ has height $\sqrt{3 - \frac{1}{2}} = \sqrt{\frac{5}{2}}$ and area $\frac{1}{2}(3\sqrt{2} + 2\sqrt{2})\sqrt{\frac{5}{2}} = \frac{5\sqrt{5}}{2}.$ The total is $4\sqrt{5} = \sqrt{80},$ so $p = 80.$

2007 AIME I 第 13 题

答案：80

难度评级：3060

解答：

其他年份的第 13 题