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### 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.

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/
```
Associated Topics:
College Polyhedra
High School Polyhedra

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