2003 AIME I Problem 6

Question

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.

Answer

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

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

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.

Problem 6 in other years