Tetrahedron vs Parallelepiped
How few equal volume tetrahedrons(connecting different sides of the parallepiped) can fit in a parallelepiped. In other words, what's the smallest number of times the equal volumes of tetrahedrons can equal to the volume of the parallelepiped?
Please try to post different solutions using vectors or geometry.
The answer is 6.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
The question should read "What's the minimum number of tetrahedrons of equal volume that can combine to form a parallelepiped?" Obviously, if 6 can "fit in a box", and since any box can be further subdivided to as many boxes one wants, then the number is just about anything you want over 6.
An [irregular] tetrahedron by definition has 4 triangular faces. A parallelepiped has 6 parallelogram faces. which means 12 triangular faces. That's 4 irregular tetrahedrons of equal volumes. Since that makes up 2/3 of the volume of the parallelepiped, the remaining irregular tetrahedron that has no face that's part of the surface of the parallelepiped has 1/3 of the volume, and thus can be divided in half into two more irregular tetrahedrons of equal volume.