Question

立方体的三个顶点为 $P = (7, 12, 10)$ 、 $Q = (8, 8, 1)$ ，和 $R = (11, 3, 9)$ 。求该立方体的表面积。

Three of the vertices of a cube are $P = (7, 12, 10),$ $Q = (8, 8, 1),$ and $R = (11, 3, 9).$ What is the surface area of the cube?

答案

答案：294

知识点：正方体距离公式表面积

难度评级：2020

解答：

计算距离的平方： $PQ^2 = 1^2 + 4^2 + 9^2 = 98$ ， $QR^2 = 3^2 + 5^2 + 8^2 = 98$ ，且 $RP^2 = 4^2 + 9^2 + 1^2 = 98$ 。所以 $P$ 、 $Q$ 、 $R$ 构成边长为 $\sqrt{98} = 7\sqrt{2}$ 的等边三角形。

立方体中三个两两等距的顶点必由面对角线相连，而边长为 $s$ 的立方体的面对角线长为 $s\sqrt{2}$ 。因此 $s = 7$ ，表面积为 $6 \cdot 7^2 = 294$ 。

Compute the squared distances: $PQ^2 = 1^2 + 4^2 + 9^2 = 98,$ $QR^2 = 3^2 + 5^2 + 8^2 = 98,$ and $RP^2 = 4^2 + 9^2 + 1^2 = 98.$ So $P,$ $Q,$ and $R$ form an equilateral triangle with side $\sqrt{98} = 7\sqrt{2}.$

Three mutually equidistant vertices of a cube must be joined by face diagonals, and a face diagonal of a cube with edge $s$ has length $s\sqrt{2}.$ Thus $s = 7,$ and the surface area is $6 \cdot 7^2 = 294.$

2002 AIME II 第 2 题

答案：294

难度评级：2020

解答：

其他年份的第 2 题