To participate in the notorious supernational Putnam Mathematics Competition contact Prof. Steven Miller and/or show up at 9:45 am Saturday in Bronfman 106. Last year the Williams team of Carlos Dominguez, Jared Hallett, and Wei Sun finished 19th in the nation on the infamous William Lowell Putnam Mathematical Competition Saturday, December 1, 2012. An amazing 23 (over 1% of all 2181 Williams students) took the exam, including many first year students.
2012 Putnam Exam solutions have appeared in the February Mathematics Magazine and at Kedlaya’s website.
If you’re interested in math puzzles and competitions, email Professor Miller and feel free to come to the math puzzle nights.
Proposition. Let f be a continuous function on R^2 with average 0 on every rectangle of area 1. Then f = 0.
Proof. A variational argument (moving two sides, preserving area), shows that the average of f over every side is the same. Suppose there is such a function other than 0. By translation and multiplying by a constant, we may assume that f(0,0) = 2. Since f is continuous, for some e > 0, for 0 ≤ y ≤ 2e, f(0,y) > 1. Consider rectangles of height e and width 1/e with left side on the initial 2e interval of the y-axis. Since the average of f over the left side exceeds 1, so does the average over the bottom side. Hence the average over the rectangle [0,1/e]x[0,e] exceeds 1, contradiction.
Putnam Problem B2. Given a polyhedron, prove that there is a constant c > 0 such that if n balls cover the surface of P, then n > c/V^2.
Solution sketch (Jared Hallet). Choose K>0 such that Sigma r^2 is at least K and choose C such that C^2 = K/2. Given covering by n balls, consider the “large” balls of radius at least Cn^-.5. The sum of the other r^2s is then at most n (Cn^-.5)^2 = C^2. Hence V sqrt n ~ sqrt n Sigma r^3 > sqrt n Sum of large r^3s > C Sum of largest r^2s > C (K – C^2) > CK/2.
Putnam Problem B3. In a round-robin tournament each of 2n teams plays every other team over 2n-1 days. Can one necessarily choose one winning team from each day without choosing any team more than once?
Solution. Yes, each day can choose or “marry” one of its winners. Indeed, by Hall’s Marriage Theorem, it suffices to show that for every collection of k days, there are at least k winners. Given a loser, the k teams it loses to provide the requisite k winners.
Last year the Williams team (Carlos Dominguez, Jared Hallett, and Liyang Zhang) placed in the top ten and received an Honorable Mention out of 460 teams from 572 participating colleges and universities. Account in Williams Record. Check out the problems and solutions. President Falk won honorable mention on the 1985 Putnam Exam, the same year that Martin Hildebrand ’86 scored in the (unranked) top five. (See Amer. Math. Monthly 93 (1986), 620-626.)