Another way to see what Joshua seemed to recall, besides by the theorem that Fred quoted elsewhere, is via the ----- Lemma: Given a nondegenerate tetrahedron T, two of its altitudes (to, say, apices A and B) are intersecting <=> the edge AB of T, and its opposite edge, are perpendicular. (Here "perpendicular" means the two skew lines containing these two edges -- after orthogonal projection to any plane parallel to both of them -- are perpendicular. Or equivalently the isometry group of the union of these skew lines is of size 8.) Proof: Left to the reader. ----- Since the perpendicularity condition is symmetrical with respect to the two edges involved, it must imply that the other two altitudes intersect each other as well. --Dan Joshua wrote: << Fred wrote: << Richard's paper mentions mentions that the 4 altitude lines of a tetrahedron, through vertices and perpendicular to faces, do not necessarily concur in a common point.

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I seem to recall that if one pair of altitudes is intersecting, then the other pair is as well. Does anyone know an easy proof or counterexample?

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"Things are seldom what they seem." --W.S. Gilbert