3rd Australian 1982 Problem B1 For what d does a continuous function f: [0, 1] → R with f(0) = f(1) always have a horizontal chord of length d?
d ∈ {1, 1/2, 1/3, 1/4, ... }. Solution We show first that there is always a chord length 1/n. For n = 1, this is obvious, so assume n > 1. Define g(x) = f(x + 1/n) - f(x) for 0 ≤ x ≤ 1 - 1/n. We have to find a zero for g(x). Consider the n values g(0), g(1/n), g(2/n), ... , g((n-1)/n). If any of them are zero we are home. So assume they are all non-zero. Their sum is f(1) - f(0) = 0, so some must be negative and some positive. But g is continuous, so it must assume the intermediate value 0. Conversely, we construct a function which does not have any chords with lengths in the interval (1/(n+1), 1/n). We take a piecewise linear function such as that shown. All segments are in one of two directions. The horizontal distance between adjacent parallel lines of one set is DE, the horizontal distance for the other set is AC. The idea is that there are no horizontal lines with lengths between DE and AC. We choose the slope of the dotted lines so that AB = DE. The result is now clear from the diagram. We need A to be (0,0) and X to be (1,0). It is convenient to take the lines to have slopes ±1. Then the first peak has height AB/2. The second peak is BC/2 lower. So we need AB to be a multiple of BC. Suppose we take AB/BC = n. It is clear from the diagram that AX = nAC, so AC = 1/n. Hence AB = 1/(n+1), and BC = 1/(n(n+1)). Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust, can be ordered by email. 3rd Aus 1982 © John Scholes jscholes@kalva.demon.co.uk 2 January 2004 Last updated/corrected 2 Jan 04