### 24th Putnam 1963

Problem A5

R is the reals. f : [0, π] → R is continuous and ∫0π f(x) sin x dx = ∫0π f(x) cos x dx = 0. Show that f is zero for at least two points in (0, π). Hence or otherwise, show that the centroid of any bounded convex open region of the plane is the midpoint of at least three distinct chords of its boundary.

Solution

sin x > 0 for all x in (0, π), so if f did not change sign, then we could not have ∫0π f(x) sin x dx = 0. If there was only one sign change, at k say, then sin(x - k) would also have only one sign change and f(x) sin(x - k) would not change sign in the interval, so ∫0π f(x) sin(x - k) dx = cos k ∫0π f(x) sin x dx - sin k ∫0π f(x) cos x dx would be non-zero. Contradiction. So f must have at least two sign changes and hence at least two zeros in the interval.

Take polar coordinates with the centroid as origin. Let the boundary be f(θ) for 0 ≤ θ ≤ π and -g(θ) for 0 > -θ > -π.

Taking moments about the x-axis gives ∫0π f(θ)sin θ dθ = ∫0π g(θ) sin θ dθ = ∫0π h(θ) sin θ dθ, where h(θ) = g(π - θ). So ∫ (f(θ) - h(θ) ) sin θ dθ = 0.

Similarly, taking moments about the y-axis gives ∫ f(θ) cos θ dθ + ∫ g(θ) cos θ dθ = 0 or ∫ (f(θ) - h(θ) ) dθ = 0. So the result proved in the first part gives that f(θ) = g(π-θ) for at least two values in (0, π). In other words, the centroid is the midpoint of at least two chords. But we could take the x-axis to be one of these chords and then repeat the result to get two more, giving three in total.