Putnam 1967/A5 solution

28th Putnam 1967

Problem A5

K is a convex, finite or infinite, region of the plane, whose boundary is a union of a finite number of straight line segments. Its area is at least π/4. Show that we can find points P, Q in K such that PQ = 1.

Solution

Suppose the result is false, so that the maximum distance is 2d < 1. Take a diameter of K (in other words two points A, B for which AB = 2d). Let the x-axis lie along this diameter with the origin at its midpoint. Take the y-axis perpendicular to the x-axis as usual. Then K lies entirely between x = -d and x = +d. Let t(x) be the top boundary of K (above the x-axis) and -b(x) be the bottom boundary of K (below the x-axis).

The area A of K is the area under the curve t(x) plus the area between b(x) and the x-axis. In other words A = ∫_-d^d t(x) dx + ∫_-d^d b(x) dx = ∫_-d^d t(x) dx + ∫_-d^d b(-x) dx = ∫_-d^d( t(x) + b(-x) ) dx. Now (t(x) + b(-x))² = D² - (2x)², where D is the distance between the two points (x, t(x)) and (-x, b(-x)). Certainly, D < 1, so ∫_-d^d( t(x) + b(-x) ) dx < ∫_-d^d √( 1 - 4x² ) dx.

The indefinite integral is 1/2 √( 1 - 4x²) + (1/4) sin^-1(2x), so integrated between -1/2 and 1/2 it gives π/4. The integrand is positive, so between -d and d it gives a smaller result. In other words we have established that if 2d < 1, then A < π/4. Contradiction.

28th Putnam 1967