10th Putnam 1950 Problem B2 An ellipse with semi-axes a and b has perimeter length p(a, b). For b/a near 1, is π(a + b) or 2π √(ab) the better approximation to p(a, b)? Solution
: π(a + b). Put b 2 = a 2 (1 - ε). The perimeter is 4 ∫ 0 π/2 (a 2 sin 2 t + b 2 cos 2 t) 1/2 dt = 4a ∫ (1 - ε cos 2 t) 1/2 dt = 4a ∫ ( 1 - 1/2 ε cos 2 t - 1/8 ε 2 cos 4 t + O(ε 4 ) dt = 2π a ( 1 - ε/4 - 3ε 2 /64 + O(ε 4 ) ). b = a(1 - ε) 1/2 = a(1 - ε/2 - ε 2 /8 + O(ε 3 ) ). Hence π (a + b) = 2π a (1 - ε/4 - 4ε 2 /64 + O(ε 3 ) ). Whilst 2π√(ab) = 2π a (1 - ε/2 - ε 2 /8 + O(ε 3 ) ) 1/2 = 2π a (1 - ε/4 - 6ε 2 /64 + O(ε 3 ) ). Hence the error in π (a + b) is π a ε 2 /32 + O(ε 3 ), and the error in 2π√(ab) is 3π a ε 2 /32 + O(ε 3 ), which is roughly three times as large for small ε. 10th Putnam 1950 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002