Putnam 1996 Problem A1 What is the smallest α such that two squares with total area 1 can always be placed inside a rectangle area α with sides parallel to those of the rectangle and with no overlap (of their interiors)? Solution
: 1/2 (1 + √2). Easy. Let the squares, have sides x ≥ 1/√2 ≥ √(1 - x 2 ). Then the smallest rectangle containing them has sides x, x + √(1 - x 2 ), and hence area f(x) = x 2 + x √(1 - x 2 ). Evidently α is the maximum value of f(x) in the range 1/√2 ≤ x ≤ 1. Evidently x 2 is an increasing function, and x √(1 - x 2 ) a decreasing function, so f(x) has a single maximum where f '(x) = 0. f '(x) = 2x + √(1 - x 2 ) - x 2 /√(1 - x 2 ), so f '(x) = 0 for 8x 4 - 8x 2 + 1 = 0, or x 2 = 1/2 + √2/4 (we want the larger root so that x ≥ 1/√2). Substituting in f(x) gives f(x) = 1/2 (1 + √2). Putnam 1996 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998