26th Putnam 1965

α and β are positive real numbers such that 1/α + 1/β = 1. Prove that the line mx + ny = 1 with m, n positive reals is tangent to the curve x^α + y^α = 1 in the first quadrant (x, y ≥ 0) iff m^β + n^β = 1.

Solution

Suppose mx + ny = 1 is tangent to the curve. Suppose it touches at (a, b). Differentiating, we see that the tangent at (a, b) is a^α-1x + b^β-1 = 1, so m = a^α-1, n = b^β-1. Hence, using αβ - β = α, we have that m^β + n^β = a^α + b^α = 1.

Conversely, suppose that m^β + n^β = 1. Take a = m^β/α, b = n^β/α. Then a^α + b^α = 1, so (a, b) lies on the curve in the first quadrant. Its tangent is Mx + Ny = 1, where M = a^α-1, N = b^β-1. But a = m^β/α and β/α (α - 1)= 1, so M = m. Similarly, N = n. Thus we have established that mx + ny = 1 is tangent to the curve in the first quadrant as required.

26th Putnam 1965

© John Scholes
jscholes@kalva.demon.co.uk
25 Jan 2002