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 lim R→0 R -2 ∫ D a(x, y, R) dx dy. Solution
: 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 r 2 = x 2 + R 2 + 2xR cos θ. Hence the length of arc outside the disk is 2R cos -1 ( (r 2 -x 2 -R 2 )/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/R 2 ∫ 2π x 2R cos -1 ( (r 2 -x 2 -R 2 )/2xR) dx. Put x = r - yR, and we get 4πr ∫ 0 1 (1 - y R/r) cos -1 ( (2Rry - (y 2 +1)R 2 )/2R(r - yR) ) dy. As R tends to 0, (1 - y R/r) tends to 1 and (2Rry - (y 2 +1)R 2 )/2R(r - yR) tends to y, so we get 4πr ∫ 0 1 cos -1 y dy = 4πr (y cos -1 y - √(1-y 2 ) )| 0 1 = 4πr. 8th Putnam 1948 © John Scholes jscholes@kalva.demon.co.uk 11 Mar 2002