### 57th Putnam 1996

 A1.  What is the smallest α such that two squares with total area 1 can always be placed inside a rectangle area α with sides parallel to those of the rectangle and with no overlap (of their interiors)? A2.  Two circles have radii 1 and 3 and centers a distance 10 apart. Find the locus of all points which are the midpoint of a segment with one end on each circle. A3.  There are six courses on offer. Each of 20 students chooses some all or none of the courses. Is it true that we can find two courses C and C' and five students S1, S2, S3, S4, S5 such that each Si has chosen C and C', or such that each Si has chosen neither C nor C'? A4.  A is a finite set. S is a set of ordered triples (a, b, c) of distinct elements of A, such that:   (a, b, c) ∈ S iff (b, c, a) ∈ S;   (a, b, c) ∈ S iff (c, b, a) ∉ S;   (a, b, c) and (c, d, a) ∈ S iff (b, c, d) and (d, a, b) ∈ S. Prove that there exists an injection g from A to the reals, such that g(a) < g(b) < g(c) implies (a, b, c) ∈ S. A5.  Let p be a prime ≥ 5. Prove that p2 divides ∑0[2p/3] pCr. A6.  Let R be the reals and k a non-negative real. Find all continuous functions f : R → R such that f(x) = f(x2 + k) for all x. B1.  Let N be the set {1, 2, 3, ... , n}. X is selfish if |X| ∈ X. How many subsets of N are selfish and have no proper selfish subsets. B2.  Let f(n) = ( (2n+1)/e )(2n+1)/2. Show that for n > 0: f(n-1) < 1·3·5 ... (2n-1) < f(n). B3.  (x1, x2, ... , xn) is a permutation of (1, 2, ... , n). What is the maximum of x1x2 + x2x3 + ... + xn-1xn + xnx1? B4.  Let B be the 2 x 2 matrix (bij) with b11 = b22 = 1, b12 = 1996, b21 = 0. Can we find a 2 x 2 matrix A such that sin A = B? [We define sin A by the usual power series: A - A3/3! + A5/5! - ... .] B5.  We call a finite string of the symbols X and O balanced iff every substring of consecutive symbols has a difference of at most 2 between the number of Xs and the number of Os. For example, XOOXOOX is not balanced, because the substring OOXOO has a difference of 3. Find the number of balanced strings of length n. B6.  The origin lies inside a convex polygon whose vertices have coordinates (ai, bi) for i = 1, 2, ... , n. Show that we can find x, y > 0 such that a1xa1yb1 + a2xa2yb2 + ... + anxanybn = 0 and b1xa1yb1 + b2xa2yb2 + ... + bnxanybn = 0.

To avoid possible copyright problems, I have changed the wording, but not the substance of all the problems. The original text of the problems and the official solutions are in American Mathematical Monthly 104 (1997) 746.

Putnam home

© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 1998