10th Putnam 1950

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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

Answer: π(a + b).

Put b2 = a2(1 - ε). The perimeter is 4 ∫0π/2 (a2sin2t + b2cos2t)1/2 dt = 4a ∫ (1 - ε cos2t)1/2 dt = 4a ∫ ( 1 - 1/2 ε cos2t - 1/8 ε2 cos4t + O(ε4) dt = 2π a ( 1 - ε/4 - 3ε2/64 + O(ε4) ).

b = a(1 - ε)1/2 = a(1 - ε/2 - ε2/8 + O(ε3) ). Hence π (a + b) = 2π a (1 - ε/4 - 4ε2/64 + O(ε3) ). Whilst 2π√(ab) = 2π a (1 - ε/2 - ε2/8 + O(ε3) )1/2 = 2π a (1 - ε/4 - 6ε2/64 + O(ε3) ).

Hence the error in π (a + b) is π a ε2/32 + O(ε3), and the error in 2π√(ab) is 3π a ε2/32 + O(ε3), which is roughly three times as large for small ε.

 


 

10th Putnam 1950

© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002