2005 AMC 12A Problem 25

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.