30th Putnam 1969

Problem B4

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 A1. Reflect A1 in r to get A2. Similarly, reflect B in p to get B1, then B1 in s to get B2. Consider the line segment A2B2. 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 A2B2 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 A2B2 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.