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I'm really curious to know any relationships between the altitude of a tetrahedron and how the foot of this altitude splits the base triangle. For example if you have a tetrahedron PABC with apex P, and then you drop the altitude from P to the opposite base how does it split this base? Is the foot of the altitude the centroid/incenter/orthocenter of the base? Are there any special properties related to this?

Thanks :)

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Are you talking about a regular tetrahedron with equilateral faces, or an arbitrary tetrahedron? If the latter, the foot of the altitude can be anywhere: just move the point $P$ around. –  Rahul Nov 22 '12 at 18:57
    
Hi, I was referring to any tetrahedron. Hmm that's true, but the reason I'm asking this question is that I'm trying to grasp what's so special about the tetrahedron in the following question and how does its altitude split the base triangle: –  user45220 Nov 22 '12 at 20:15
    
    
Damn it, contest problem: mathcomp.leeds.ac.uk –  Will Jagy Nov 22 '12 at 21:50
    
@user45220: Please do not post new questions as answers. –  Eric Naslund Nov 22 '12 at 22:59

1 Answer 1

In a general tetrahedron, you can freely move the apex in a plane parallel to the base, so the foot can be any place in the base triangle. No special relations can be inferred from the general case.

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