A cube has 8 vertices. If each vertex is cut off to form a triangle, the new solid will have 3 x 8 = 24 vertices. If each of these vertices is then connected directly to each of the others via a straight line segment, how many of these segments will go through the body of the solid, rather than along its surface?

From each triangular face pair that is adjacent to two common octagonal faces created from the truncation, a straight line connected from the two furthest vertices of these triangles will pass through the solid. There are six octagonal faces, each with eight vertices, thus there are 48 of these lines.

From each triangular face pair that share only one common octagonal face created from the truncation, there are three straight lines that connect to the furthest vertex that will pass through the solid. With six octagonal faces/2, there are 36 of these lines.

From each triangular face pair that share no common octagonal face created from the truncation, there are three straight lines from each vertex that pass through the solid. With each triangle having three vertices, there are nine of these lines. As there are four pairs these triangles, the number of these straight lines of this type is 36.

The total number of straight lines that pass through the solid, then, is 48+36+36 = 120.

A second approach is to take one vertex. Along the surface/edge of the truncated hexahedron, it will connect to 13 vertices, leaving 10 vertices of the 24 it must, then, connect via passage through the solid.  As these are pairs, we multiply by the number of remaining vertices and divide by 2. 
24 x 10 / 2 = 120. Confirming that there are 120 line segments connecting vertices that pass through the body of the object. 

Edited on February 14, 2008, 11:08 pm

Posted by Dej Mar on 2008-02-14 12:46:26