EXCERPT FROM THE NEW YORK TIMES
URL:
http://wordplay.blogs.nytimes.com/2016/07/18/imo-2016/?_r=0The competition, held July 6-16 in Hong Kong, included teams from over 100 countries. The winning U.S. team score was 214 out of a possible 252, ahead of the Republic of Korea (207) and China (204). Rounding out the top ten were Singapore (196), Taiwan (175), North Korea (168), Russian Federation (165), United Kingdom (165), Hong Kong (161) and Japan (156).
Congratulations to the U.S. team: students Ankan Bhattacharya, Michael Kural, Allen Liu, Junyao Peng, Ashwin Sah, and Yuan Yao, head coach Po-Shen Loh, and deputy coach Razvan Gelca.
This week we’ll be featuring two of the problems from this year’s six-problem test, with our discussion moderated by Po-Shen Loh himself.
Here are the two problems — our challenges for this week.The first challenge is IMO 2016 Problem 2:
Find all positive integers n for which each cell of an n×n table can be filled with one of the letters I, M and O in such a way that:
• in each row and each column, one third of the entries are I, one third are M and one third are O; and
• in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third are I, one third are M, and one third are O.Our second challenge is a bit more difficult. Here’s an introduction by Dr. Loh:
This year’s IMO featured an unusually large number of non-standard problems which combined multiple areas of mathematics into the same investigation. The most challenging problem turned out to be #3, which was a fusion of algebra, geometry, and number theory. On that question, the USA achieved the highest total score among all countries, ultimately contributing to its overall victory — a historic repeat #1 finish (2015 + 2016), definitively breaking the 21-year drought since the last #1 finish in 1994, and the first consecutive #1 finish in the USA’s record.
Let’s give it a try. Here’s IMO 2016 Problem 3:
Let P = A1 A2 … Ak be a convex polygon on the plane. The vertices A1, A2, …, Ak have integral coordinates and lie on a circle. Let S be the area of P. An odd positive integer n is given such that the squares of the side lengths of P are integers divisible by n. Prove that 2S is an integer divisible by n.That concludes this week’s challenges. If you’d like to try your hand at the remaining four problems in this year’s IMO exam, you’ll find the complete test here. To obtain a perfect score for the test, you must successfully provide the correct answer for each problem, and also prove that each answer is correct.