Pythagorean Theorem in Three Dimensions

Date: 05/18/2001 at 00:15:28
From: Erin Avery
Subject: Tetrahedron with trirectangular vertex Pythagorean proof

Suppose a tetrahedron has a trirectangular vertex S. Let A, B and C be 
the areas of the 3 faces that meet at S, and D be the area of the face 
opposite S. Prove that
 
     D^2 = A^2 + B^2 + C^2

(This is a three-dimensional version of the Pythagorean theorem.)

Date: 05/18/2001 at 08:55:09
From: Doctor Rick
Subject: Re: Tetrahedron with trirectangular vertex Pythagorean proof

Hi, Erin.

I haven't heard the term "trirectangular" before, but I can guess that 
it means a vertex such that each of the three faces meeting at the 
vertex has a right angle at the vertex. Thus we can place this 
tetrahedron with S at the origin of a Cartesian coordinate system, and 
the three edges meeting at S lie along the x-, y- and z-axes.

One way to prove the theorem follows these lines. I will take face A 
to be in the y-z plane; B in the x-z plane; and C in the x-y plane. 
The lengths of the edges along the three coordinate axes will be 
called x, y and z.

Express the areas A, B, and C in terms of x, y and z. Now you can 
express A^2 + B^2 + C^2 in terms of x, y and z.

Now you can find the lengths of the other three edges of the 
tetrahedron in terms of x, y and z, using the Pythagorean theorem. 
Finally, use Heron's formula for the area of a triangle in terms of 
its sides, to get the area D; and show that this expression is equal 
to the expression for A^2 + B^2 + C^2. You can find Heron's formula in 
our Dr. Math FAQ (under Formulas), if you don't know it.

You might have trouble simplifying the area D. I suggest that you look 
for a way to write Heron's formula in terms of the SQUARES of the 
three sides; it can be done, using the factorization of a difference 
of squares in reverse. Then you won't have to think at all about 
square roots.

This is a nice theorem, isn't it? When you get such a simple result, 
you wonder if it's necessary to go through this much algebra along the 
way. Maybe there's a more elegant solution. Let us know if you learn 
one.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/