A tetrahedron ABCD has edges of total length 1. The angles at A (BAC etc) are all 90o. Find the maximum volume of the tetrahedron.
Solution
Answer: (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 + √(x2 + y2) + √(y2 + z2) + √(z2 + x2) = 1. By AM/GM we have x + y + z ≥ 3k, where k = (xyz)1/3. Also x2 + y2 ≥ 2xy, so √(x2 + y2) + √(y2 + z2) + √(z2 + x2) ≥ √(2xy) + √(2yx) + √(2zx). By AM/GM that is >= 3√2 k. So we have 1 ≥ 3(1 + √2) k. Hence k3 ≤ (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.
© John Scholes
jscholes@kalva.demon.co.uk
11 Jul 2002