Orthocentric Tetrahedron

Date: 11/30/2002 at 03:19:08
From: Mrs. D
Subject: Orthocentric tetrahedron

I am a high school math teacher and a student came to me with this 
problem:

Recall that the opposite edges of an orthocentric tetrahedron are 
perpendicular. Let ABCD be an orthocentric tetrahedron. Show that 
AB^2 + CD^2 = AD^2 +BC^2

I really want to help this student (who is studying for a math 
competition) but I am pretty stuck on this one.  Please help!

- Mrs. D

Date: 12/01/2002 at 13:01:21
From: Doctor Floor
Subject: Re: Orthocentric tetrahedron

Hi,

Thanks for your question.

Let's consider an orthocentric tetrahedron and put it into a 3D grid. 
We may take A(0,0,0) and B(1,0,0). Then we may rotate around AB in 
such a way that CD is parallel to the y-axis. So we have C(f,g,h) and 
D(f,g+t,h) for some parameter t.

We know that AB^2 + CD^2 = 1 + t^2................[1]

We know that AD and BC are perpendicular, so the vectors 
v(A,D) = (f,g+t,h) and v(B,C) = (f-1,g,t) have inproduct equal to 0.
This leads to 

  f*(f-1)+(g+t)*g + h^2 = 0
  f^2 + g^2 + h^2 + tg - f = 0....................[2]

And now we can use this to simplify AD^2 + BC^2:

  AD^2 = f^2 + g^2 + 2tg + t^2 + h^2
  BC^2 = f^2 -2f + 1 + g^2 + h^2
  ---------------------------------- +
  AD^2 + BC^2 = 2*(f^2 + g^2 + h^2 + tg - f) + 1 + t^2  (use [2])
              = 1 + t^2...........................[3]

And [1] and [3] show the correctness of our theorem.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/