23rd USAMO 1994

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1.  a1, a2, a3, ... are positive integers such that an > an-1 + 1. Put bn = a1 + a2 + ... + an. Show that there is always a square in the range bn, bn+1, bn+2, ... , bn+1-1.
2.  The sequence a1, a2, ... , a99 has a1 = a3 = a5 = ... = a97 = 1, a2 = a4 = a6 = ... = a98 = 2, and a99 = 3. We interpret subscripts greater than 99 by subtracting 99, so that a100 means a1 etc. An allowed move is to change the value of any one of the an to another member of {1, 2, 3} different from its two neighbors, an-1 and an+1. Is there a sequence of allowed moves which results in am = am+2 = ... = am+96 = 1, am+1 = am+3 = ... = am+95 = 2, am+97 = 3, an+98 = 2 for some m? [So if m = 1, we have just interchanged the values of a98 and a99.]
3.  The hexagon ABCDEF has the following properties: (1) its vertices lie on a circle; (2) AB = CD = EF; and (3) the diagonals AD, BE, CF meet at a point. Let X be the intersection of AD and CE. Show that CX/XE = (AC/CE)2.
4.  xi is a infinite sequence of positive reals such that for all n, x1 + x2 + ... + xn ≥ √n. Show that x12 + x22 + ... + xn2 > (1 + 1/2 + 1/3 + ... + 1/n) / 4 for all n.
5.  X is a set of n positive integers with sum s and product p. Show for any integer N ≥ s, ∑( parity(Y) (N - sum(Y))Cs ) = p, where aCb is the binomial coefficient a!/(b! (a-b)! ), the sum is taken over all subsets Y of X, parity(Y) = 1 if Y is empty or has an even number of elements, -1 if Y has an odd number of elements, and sum(Y) is the sum of the elements in Y.

To avoid possible copyright problems, I have changed the wording, but not the substance, of the problems.

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© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002