TETRAHEDRA INSIDE OF A REGULAR PARALLELEPIPED

In the latest Wall Street Journal Puzzle Page of September 17-18, 2016 their first question is rather elementary and deals with the volume of a tetrahedron cut out of a cube. We toughen the problem somewhat by a modification where the cube is replaced by a parallelepiped of volume abc and ask the question what is the volume of a tetrahedron where three edges coincide with the parallelepiped as shown-

The vertices of the entire tetrahedron lie at [a,0,0],[0,b,0], [0,0,c] and [0,0,0] .

The simplest way to solve for its volume to take the product of its base area ab/2 and multiply it by one-third of the height c. This immediately produces the answer-

1 ab abc Vol= ( )c  3 2 6

which represents one-sixth of the parallelpiped volume. This also yields the answer posed by the Journal for a cube when we set a=b=c-1. Note that one could cut two identical tetrahedra of the parallelpiped leaving a residue slice of volume 2abc/3.

A more mathematically advanced solution for the volume is to employ calculus. One has-

a b(1x / a ) Vol  dx c[1 (x / a)  ( y / b)]dy x0 y0

cb a x abc   (1 )2 dx  2 x0 a 6

It is always a little surprising that the volume of six of these tetrahedra constitute the total volume of the parallelepiped. One would expect the volume of the tterahedra would be a larger fraction of abc. Other tetrahedra can also be cut out of the parallelepiped. One can make the interesting observation that this tetrahedron has six edges (E=6), four sides (S=4, and four vertices (V=4).It means that-

V+F-E=2

This equality is the famous Euler formula valid for all polyhedra. For example, a cube has V=8, E=12, and F=6 .

An interesting variation of this problem is to see what happens when we cut a total of six pyramids out of the parallelepiped by defining their edges via eight lines drawn from each of the eight vertices of the parallelepiped to its center at [a/2,b/2,c/2]. The resultant volume of each of resulting pyramids (four side faces and one base) have the same value of abc/6. However, the resultant twin pairs of pyramids have three different heights.

It was already known to the ancient Egyptian pyramid builders that the frustum of a square base pyramid of area a2 finished to a height h such that the top surface equals a smaller square of area b2 is-

h Vol  [a 2  ab  b2 ] 3

This solution is given in the Moscow Papyrus and is about three thousand years old. Here is how we get this answer today.

We know that the volume of the frustum of an unfinished pyramid equals its volume of a2c/3 minus the pyramid tip above the frustum of b2(c-h)/3. This yields-

1 1 Vol  a 2c  b2 (c  h) 3 3

But from the geometry we also have (c-h)=bc/a. Using this to eliminate c yields –

h a 3  b3  h Vol     [a 2  ab  b2 ] 3 (a  b) 3

Kudos to the ancient engineer who was able to achieve this result in pre-analytic geometry and calculus days. Once the pyramid reaches half-height, 7/8th of the stone laying will already have been done.

A second more difficult problem posed in this week’s WSJ puzzle page is to find the volume of a 3D solid formed by connecting the six intersection points of a sphere of radius R centered on one of the edges of a tetrahedron of side-length 2R by straight lines. The tetrahedron has four equal area equilateral triangle faces of area sqrt(3)R2 each and a volume of Vol=2sqrt(2)R3/3. A schematic of the problem follows-

At first glance it seems like a difficult task to find the volume corresponding to the shaded area. However a little thought and having played with blocks as a child, it is clear that the shaded polyhedron can be broken up into two smaller tetrahedra of volume sqrt(2)R3/12 each plus a pyramids of base R2 and height h=R/sqrt(2).. The volume of the pyramid is R3/[(3sqrt(2)] or twice the volume of the small tetrahedra. Adding things together produces the following volume for the 3D structure-

2 1 2 Vol  {2[ ] [ ]}R3  R3 3D 12 3 2 3

This represents just one-half the volume of the large tertrahedron. I have constructed two identical cardboard models of this 3D structure. When the square sides are brought together and one is rotated by ninety degrees the full tetrahedron is recovered. Thus cutting a large tetrahedron in exactly one half produces two identical 3D solids. Note the 3D structure has nine edges(E=9), six vertices(V=6), and five faces(F=5).So Euler’s formula is again confirmed since-

F+V-E=5+6-9=2

A side-view of the structure resembles the covering of a Conestoga wagon used by American pioneers moving west along the Oregon trail in the 1840s.

September 19, 2016