8th Putnam 1948

Problem A4

Let D be a disk radius r. Given (x, y) ∈ D, and R > 0, let a(x, y, R) be the length of the arc of the circle center (x, y), radius R, which is outside D. Evaluate limR→0 R-2D a(x, y, R) dx dy.



Answer: 4πr.

Let P be a point in the disk a distance x from its centre. Suppose that the circle centre P radius R cuts the disk perimeter at A and B. Let angle APB be 2θ. We have r2 = x2 + R2 + 2xR cos θ. Hence the length of arc outside the disk is 2R cos-1( (r2-x2-R2)/2xR). This applies for r-R <= x <= r. For smaller x the small circle lies entirely in the disk. Thus we have to evaluate 1/R2 ∫ 2π x 2R cos-1( (r2-x2-R2)/2xR) dx. Put x = r - yR, and we get 4πr ∫01 (1 - y R/r) cos-1( (2Rry - (y2+1)R2)/2R(r - yR) ) dy. As R tends to 0, (1 - y R/r) tends to 1 and (2Rry - (y2+1)R2)/2R(r - yR) tends to y, so we get 4πr ∫01 cos-1y dy = 4πr (y cos-1y - √(1-y2) )|01 = 4πr.



8th Putnam 1948

© John Scholes
11 Mar 2002