2005 AMC 12A Problem 25

Question

Let

S

be the set of all points with coordinates

(x, y, z),

where

x,

y,

and

z

are each chosen from the set

\{0, 1, 2\}.

How many equilateral triangles have all their vertices in

S?

72

76

80

84

88

Answer: C

Concepts:equilateral triangle 3D geometry casework

Difficulty rating: 2640

Small hint:

An equilateral triangle in this grid must have all three sides equal, so each side is a face diagonal or a specific slanted segment

Big hint:

Count separately the triangles whose sides are unit-cube face diagonals, big-cube face diagonals, and edge-midpoint-to-edge-midpoint segments of length

\sqrt{6}

Solution:

The three equal sides of such a triangle must all have the same length. Checking the possible squared lengths in the

3 \times 3 \times 3

grid, only three families of side occur.

Face diagonals of a unit cube (length

\sqrt2

): each of the

8

unit cubes contributes

8

triangles, one at each corner, for

8 \cdot 8 = 64.

Face diagonals of the $2 \times 2 \times 2$ cube (length

2\sqrt2

): the three faces meeting at a vertex form one triangle, giving

8

triangles.

Edge-midpoint segments (length

\sqrt6,

joining midpoints of two edges): each of the

12

edge midpoints is a vertex of two such triangles, for

\dfrac{12 \cdot 2}{3} = 8.

The total is

64 + 8 + 8 = 80.

Thus, the correct answer is C.