2002 AIME II 第 2 题

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

小提示：

计算三条两两距离

PQ

、

QR

和

RP

。

Compute the three pairwise distances

PQ,

QR,

and

RP

大提示：

立方体中三个两两等距的顶点由面对角线相连；边长为

s

的立方体，其面对角线长为

s\sqrt{2}

。

Three mutually equidistant vertices of a cube are joined by face diagonals, and a face diagonal of a cube with edge

s

has length

s\sqrt{2}

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

小提示：

大提示：

解答：

其他年份的第 2 题