5th USAMO 1976 Problem 4 A tetrahedron ABCD has edges of total length 1. The angles at A (BAC etc) are all 90 o . Find the maximum volume of the tetrahedron. Solution
: (5√2 - 7)/162. Let the edges at A have lengths x, y, z. Then the volume is xyz/6 and the perimeter is x + y + z + √(x 2 + y 2 ) + √(y 2 + z 2 ) + √(z 2 + x 2 ) = 1. By AM/GM we have x + y + z ≥ 3k, where k = (xyz) 1/3 . Also x 2 + y 2 ≥ 2xy, so √(x 2 + y 2 ) + √(y 2 + z 2 ) + √(z 2 + x 2 ) ≥ √(2xy) + √(2yx) + √(2zx). By AM/GM that is >= 3√2 k. So we have 1 ≥ 3(1 + √2) k. Hence k 3 ≤ (5√2 - 7)/27. Hence the volume is at most (5√2 - 7)/162. This is achieved if x = y = z = (√2 - 1)/3. Then the other three sides are (2 - √2)/3 and the perimeter is 1. 5th USAMO 1976 © John Scholes jscholes@kalva.demon.co.uk 11 Jul 2002