Γ is a plane curve of length 1. Show that we can find a closed rectangle area 1/4 which covers Γ.

**Solution**

Let the endpoints of the curve be A and B. Take lines p, q parallel to AB and on opposite sides of it so that the band between them just covers the curve. Similarly take lines r, s perpendicular to AB so that the band between them just covers the curve (and so that r is nearer to A than to B).

Reflect A in q to get A_{1}. Reflect A_{1} in r to get A_{2}. Similarly, reflect B in p to get B_{1}, then B_{1} in s to get B_{2}. Consider the line segment A_{2}B_{2}. By reflection we can derive from it a curve from A to B consisting of five straight line segments which intersects p, q, r and s. Moreover it is evidently the shortest curve with this property.

Let X be the distance between r and s, Y the distance between p and q, and Z the distance between A and B. We have (X - 2Y)^{2} + (X - Z)^{2} + 2X(X - Z) ≥ 0 (*). Hence (2X - Z)^{2} + (2Y)^{2} ≥ 4XY, which tells us that the square of the length A_{2}B_{2} is greater than or equal to four times the area of the rectangle formed by the intersection of the two bands. Thus if the curve Γ has length 1, then A_{2}B_{2} must have length at least 1 and hence we have found a rectangle with area at most 1/4 which covers Γ. Note that (*) shows the area will be less than 1/4 unless Z = X (so that r goes through A and s through B) and X = 2Y.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002