O is the circumcenter of the tetrahedron ABCD. The midpoints of BC, CA, AB are L, M, N respectively. AB + BC = AD + CD, CB + CA = BD + AD and CA + AB = CD + BD. Show that ∠LOM = ∠MON = ∠NOL.
Solution
Adding two of the equations and subtracting the third, we get AB = CD, BC = AD, AC = BD. Let L' be the midpoint of AD, M' the midpoint of BD, N' the midpoint of CD. Then L'M' = ½ AB = LM. Also both L'M' and LM are parallel to AB, so they are parallel to each other and hence coplanar. Also L'M = ½ CD = ½ AB = LM, so LML'M' is a rhombus. Let X be the intersection of its diagonals. Then X is the midpoint of LL' and MM', and LL' and MM' are perpendicular. A similar argument shows that the midpoints of LL' and NN' coincide, so X is also the midpoint of NN'. Similarly, LL' and NN' are perpendicular, and MM' and NN' are perpendicular.
NN' is perpendicular to LL' and MM' and hence is normal to the plane containing the rhombus LML'M'. But AB and CD are both parallel to this plane, and hence NN' is perpendicular to AB and CD. But the plane through N perpendicular to AB contains the circumcenter O. So does the plane through N' perpendicular to CD. These planes meet in the line NN', so O lies on NN'. Similarly, it must lie on MM' and LL', so X must be O. Hence ∠LOM = ∠MON = ∠NOL = 90o.
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2003
Last corrected/updated 1 Jan 03