6th USAMO 1977

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Problem 4

ABCD is a tetrahedron. The midpoint of AB is M and the midpoint of CD is N. Show that MN is perpendicular to AB and CD iff AC = BD and AD = BC.

 

Solution

Use vectors. Take any origin O and write the vector OX as X. Then MN perpendicular to AB and CD is equivalent to:
(A + B - C - D).(A - B) = 0 and
(A + B - C - D).(C - D) = 0.
Expanding and adding the two equations gives (A - D)2 = (B - C)2 or AD = BC. Subtracting gives AC = BD.

Conversely, AD = BC and AC = BD gives:
(A - D)2 = (B - C)2 and
(A - C)2 = (B - D)2.
Adding gives (A + B - C - D).(A - B) = 0, so MN is perpendicular to AB. Subtracting gives MN perpendicular to CD.

 


 

6th USAMO 1977

© John Scholes
jscholes@kalva.demon.co.uk
20 Aug 2002