2003 AIME I 第 6 题

Question

一个

1

乘

1

乘

1

立方体的八个顶点中，任取三个作为三角形的顶点。所有这类三角形的面积之和为

m + \sqrt{n} + \sqrt{p}

，其中

m

、

n

和

p

是整数。求

m + n + p

。

The sum of the areas of all triangles whose vertices are also vertices of a

1

by

1

by

1

cube is

m + \sqrt{n} + \sqrt{p},

where

m,

n,

and

p

are integers. Find

m + n + p.

答案

答案：348

知识点：正方体三角形面积分类讨论

难度评级：2370

小提示：

每条边都是立方体的棱、长度为

\sqrt{2}

的面对角线或长度为

\sqrt{3}

的体对角线；对

\binom{8}{3} = 56

个三角形分类。

Each side is a cube edge, a face diagonal of length

\sqrt{2},

or a space diagonal of length

\sqrt{3};

classify the

\binom{8}{3} = 56

triangles

大提示：

三种类型分别由两条棱和一条面对角线、三条面对角线，以及一条棱、一条面对角线和一条体对角线组成；数量分别为

24

、

8

和

24

。

The three types are edge-edge-face diagonal, three face diagonals, and edge-face diagonal-space diagonal; there are

24,

8,

and

24

of them

解答：

这类三角形的每条边都是立方体棱、长度为

\sqrt{2}

的面对角线，或长度为

\sqrt{3}

的体对角线。只会出现三种形状。由两条相邻棱和一条面对角线组成的三角形是直角三角形，面积为

\frac{1}{2}

；每个面有

4

个，共

24

个。由三条面对角线组成的三角形是等边三角形，面积为

\frac{\sqrt{3}}{2}

；立方体的

8

个顶点中，每个顶点相邻的三个顶点都确定一个这样的三角形，所以有

8

个。由一条棱、一条面对角线和一条体对角线组成的三角形是直角三角形，直角边为

1

和

\sqrt{2}

，面积为

\frac{\sqrt{2}}{2}

；

4

条体对角线中的每一条，都可与不在该对角线上的

6

个顶点各形成一个，所以有

24

个。（确实

24 + 8 + 24 = \binom{8}{3} = 56

。）

总面积为

\begin{aligned} &24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} \\ &= 12 + 4\sqrt{3} + 12\sqrt{2} \\ &= 12 + \sqrt{48} + \sqrt{288} \end{aligned}

因此

m + n + p

= 12 + 48 + 288

= 348

。

Every side of such a triangle is a cube edge, a face diagonal of length

\sqrt{2},

or a space diagonal of length

\sqrt{3}.

Only three shapes occur. A triangle of two adjacent edges and a face diagonal is right with area

\frac{1}{2};

there are

4

per face, or

24.

A triangle of three face diagonals is equilateral with area

\frac{\sqrt{3}}{2};

each is determined by the three vertices adjacent to one of the

8

cube vertices, so there are

8.

A triangle of an edge, a face diagonal, and a space diagonal is right with legs

1

and

\sqrt{2},

so its area is

\frac{\sqrt{2}}{2};

each of the

4

space diagonals forms one with each of the

6

vertices off that diagonal, so there are

24.

(Indeed

24 + 8 + 24 = \binom{8}{3} = 56.

)

The total area is

\begin{aligned} &24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} \\ &= 12 + 4\sqrt{3} + 12\sqrt{2} \\ &= 12 + \sqrt{48} + \sqrt{288}, \end{aligned}

so

m + n + p

= 12 + 48 + 288

= 348.

2003 AIME I 第 6 题

答案：348

难度评级：2370

小提示：

大提示：

解答：

其他年份的第 6 题