In the tetrahedron ABCD, angle BDC = 90o and the foot of the perpendicular from D to ABC is the intersection of the altitudes of ABC. Prove that:
(AB + BC + CA)2 ≤ 6(AD2 + BD2 + CD2).
When do we have equality?
Solution
The first step is to show that angles ADB and ADC are also 90o. Let H be the intersection of the altitudes of ABC and let CH meet AB at X. Planes CED and ABC are perpendicular and AB is perpendicular to the line of intersection CE. Hence AB is perpendicular to the plane CDE and hence to ED. So BD2 = DE2 + BE2. Also CB2 = CE2 + BE2. Subtracting: CB2 - BD2 = CE2 - DE2. But CB2 - BD2 = CD2, so CE2 = CD2 + DE2, so angle CDE = 90o. But angle CDB = 90o, so CD is perpendicular to the plane DAB, and hence angle CDA = 90o. Similarly, angle ADB = 90o.
Hence AB2 + BC2 + CA2 = 2(DA2 + DB2 + DC2). But now we are done, because Cauchy's inequality gives (AB + BC + CA)2 ≤ 3(AB2 + BC2 + CA2).
We have equality iff we have equality in Cauchy's inequality, which means AB = BC = CA.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 1998
Last updated/corrected 14 Mar 03