2005 AMC 12A Problem 25

Below is the professionally curated solution for Problem 25 of the 2005 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

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Concepts:equilateral triangle3D geometrycasework

Difficulty rating: 2640

25.

Let SS be the set of all points with coordinates (x,y,z),(x, y, z), where x,y,x, y, and zz are each chosen from the set {0,1,2}.\{0, 1, 2\}. How many equilateral triangles have all their vertices in S?S?

7272

7676

8080

8484

8888

Solution:

The three equal sides of such a triangle must all have the same length. Checking the possible squared lengths in the 3×3×33 \times 3 \times 3 grid, only three families of side occur.

Face diagonals of a unit cube (length 2\sqrt2): each of the 88 unit cubes contributes 88 triangles, one at each corner, for 88=64.8 \cdot 8 = 64.

Face diagonals of the 2×2×22 \times 2 \times 2 cube (length 222\sqrt2): the three faces meeting at a vertex form one triangle, giving 88 triangles.

Edge-midpoint segments (length 6,\sqrt6, joining midpoints of two edges): each of the 1212 edge midpoints is a vertex of two such triangles, for 1223=8.\dfrac{12 \cdot 2}{3} = 8.

The total is 64+8+8=80.64 + 8 + 8 = 80.

Thus, the correct answer is C.

Problem 25 in Other Years