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

Answer: 1/2 (1 + √2).

Easy.

Let the squares, have sides x ≥ 1/√2 ≥ √(1 - x2). Then the smallest rectangle containing them has sides x, x + √(1 - x2), and hence area f(x) = x2 + x √(1 - x2). Evidently α is the maximum value of f(x) in the range 1/√2 ≤ x ≤ 1.

Evidently x2 is an increasing function, and x √(1 - x2) a decreasing function, so f(x) has a single maximum where f '(x) = 0. f '(x) = 2x + √(1 - x2) - x2/√(1 - x2), so f '(x) = 0 for 8x4 - 8x2 + 1 = 0, or x2 = 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