Dear ,
Everything has changed over this past month. Due to the COVID-19 pandemic, the state of New York is under a state of emergency, and schools are closed. The impact on the math enrichment community is significant, although small compared to the gravity of the public health threat.
It seems like everything has been canceled: NYSML, which was planned for this past weekend, did not take place. ARML, scheduled for the end of May, will not be held. The AMC contests, including AIME II, the USA(J)MO and MOP, have been suspended.
But there is some good news, and there are some opportunities.
For those students who take the initiative to learn the tools and to acquire the skills to be part of an online team, the Purple Comet Math Meet, which is easily adapted to a completely online format, is still on at the end of April. And ARML Local has been rescheduled for the weekend of May 30th and 31st, and there might be an online option. Also, ARML is offering their 2009–2014 Contest Book as a free download.
What does this mean for the Upstate New York Math Team? To look at things optimistically, this situation gives us an additional year to prepare for our return to ARML. And given the possibility of ARML Local participation, we have modified and published the team application form (click the button below).
There is another important initiative that we are starting. We would like to incorporate the Upstate New York Math Team as a nonprofit entity to ensure its existence into the future. We are especially interested in involving math team coaches, team alumni, those with legal training, fundraising prowess, web skills, and most of all, those with the know-how and drive to get things done. Please email me directly if you are interested in making this happen.
And, as always, a huge thanks to the Problem Squad. May their efforts in developing the practice problem sets challenge you, frustrate you, and give you many sleepless nights this coming month!
– Japheth Wood
P.S.: As usual, I ask you to share the newsletter sign up form with students, parents, math teachers and math team coaches. We have a year to connect to the high school math enrichment community across the state, and return to ARML in 2021!
Problems of the Month - March
Prepared by The Problem Squad (Matthew Babbitt and Brendan Caseria)
Each month we offer a set of problems to help you prepare for ARML. We'll send you the solutions a month later. If you solve a problem and want to share your solution, please email us!
We hope you enjoy these challenge and learn some unexpected mathematics in the process. Solve these with your team or on your own!
NEW! April 2020 Problems
March 2020 Problems
February 2020 Problems
January 2020 Problems
December 2019 Problems
November 2019 Problems
NEW! March 2020 Solutions
February 2020 Solutions
January 2020 Solutions
December 2019 Solutions
November 2019 Solutions
We received solutions to March Problems from student Soumyadeep Bhattacharjee, and from Bard College undergraduates Tina Giorgadze, Woochan Hwang, and Julia Sheffler. Thank you!
We'd love to see your solutions to the April Problems. If you prepare them carefully in LaTeX, we may be able to include them in the official solutions.
Did you know that ARML expects your solution to be in a particular format?
Please read the ARML conventions here: ARML Conventions
Solution of the March Bonus Problem
The bonus problem for March was offered by Jessie Tan, the president of the Cornell Math Club.
Suppose that each of the elementary symmetric polynomials S1, ..., Sn in n given real numbers x1, ..., xn is non-negative. Prove or disprove that each number xi is non-negative.
Definitions: The elementary symmetric polynomial Sk in n variables {x1, …, xn} is the sum of products of the xi, taken k at a time. For example, the elementary symmetric polynomials of three variables {a,b,c} are S1=a+b+c, S2 = ab+ac+bc, and S3 = abc. And the elementary symmetric polynomials in n variables {x1, …, xn} are S1 = x1 + x2 + ⋯ + xn, S2 = x1x2 + x1x3 + ⋯ + xn-1xn, ⋯, and Sn = x1x2⋯xn. (And, of course, "non-negative" means "greater than or equal to zero".)
Solution: Let P(x) = (x + x1)(x + x2) … (x + xn). The roots of this polynomial are the values −x1, −x2, …, −xn. When expanded, we have that P(x) = xn + S1xn−1 + S2xn−2 + … + Sn−1x + Sn, by Vieta's Theorem.
Suppose, for the sake of contradiction, that some xi is a negative number. Therefore, there exists a positive root, specifically −xi, of P(x). However, when we substitute −xi for x in the expanded form of P(x), we get:
P(−xi) = (−xi)n + S1(−xi)n−1 + S2(−xi)n−2 + … + Sn−1(−xi) + Sn
Since −xi > 0, (−xi)n is clearly positive. All other terms are at least 0, since each is a non-negative symmetric sum multiplied by a power of −xi. So, P(−xi) > 0, which contradicts the notion that −xi is a root of P(x). Therefore, since it cannot be true that some xi is negative, all xi must be non-negative. ∎
If you are reading this, and you have a math problem that you want to share, send an email to Matthew Babbitt! Include the problem, an answer, and a full solution to your problem.We will include it in a future newsletter, so the other readers have a chance to try it out.
