Problem:
The sum of the areas of all triangles whose vertices are also vertices of a by by cube is , where , and are integers. Find .
Solution:
The sides of the triangles may be cube edges, face-diagonals of length , or space-diagonals of length . A triangle can consist of two adjacent edges and a face-diagonal; three face-diagonals; or an edge, a face-diagonal, and a spacediagonal. The first type of triangle is right with area , and there are of them, on each face. The second type of triangle is equilateral with area . There are of these because each of these triangles is uniquely determined by the three vertices adjacent to one of the vertices of the cube. The third type of triangle is right with area . There are of these because there are four space-diagonals and each determines six triangles, one with each cube vertex that is not an endpoint of the diagonal. (Note that there is a total of triangles with the desired vertices, which is consistent with the above results.)
The desired sum is thus , and .
The problems on this page are the property of the MAA's American Mathematics Competitions