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................ 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.................... 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 ) = 1 + t^2........................... And  and  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/
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