POLYA025:   Maximum sum of arithmetic progressions (proposed by Paul Yiu, 12/21/01).

This is a problem from the 1999 Chinese National High School Competition (translation from an article in the Chinese Mathematical Bulletin, 2001, September issue).

Given a positive integer n and a positive number M, find the greatest possible sum a_{n+1} + a_{n+2} + ... + a_{2n+1} of arithmetic progressions a_1, a_2, a_3, .... satisfying (a_1)^2 + (a_{n+1})^2 <= M.

Discussion. [JPE, 1/11/02]: Putting x = a1, y = a2 - a1, we have to find the maximum value of (n+1)(x +3 n y/2) for x^2 + (x+n y)^2 <= M.
We get this maximum S when the line (n+1)(x + 3 n y/2) =S is tangent to the ellipse x^2 + (x+n y)^2 = M.
Computing the discriminant, we get S = (n+1) root(5 M/2) and the corresponding values x = -root(M/10), y = 1/n root(8 M/5).
 

Bibliography.

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