# Minipolymath4 Project Now Open!

As we talked about here, there was plans afoot to group-solve online an interesting and problem from the IMO.  The online discussion going on right now (unless, of course, you’re reading this later).  They decided to work on Question 3.  Here it is:

Problem 3.   The liar’s guessing game is a game played between two players $A$ and $B$.  The rules of the game depend on two positive integers $k$ and $n$ which are known to both players.

At the start of the game, $A$ chooses two integers $x$ and $N$ with $1 \leq x \leq N$.  Player $A$ keeps $x$ secret, and truthfully tells $N$ to player $B$.  Player $B$ now tries to obtain information about $x$ by asking player A questions as follows.  Each question consists of $B$ specifying an arbitrary set $S$ of positive integers (possibly one specified in a previous question), and asking $A$ whether $x$ belongs to $S$.  Player $B$ may ask as many such questions as he wishes.  After each question, player $A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wishes; the only restriction is that, among any $k+1$ consecutive answers, at least one answer must be truthful.

After $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers.  If $x$ belongs to $X$, then $B$ wins; otherwise, he loses.  Prove that:

1. If $n \geq 2^k$, then $B$ can guarantee a win.
2. For all sufficiently large $k$, there exists an integer $n \geq 1.99^k$ such that $B$ cannot guarantee a win.

Check out (and join in on) the discussion here.

# MultiPolymaths

As we wrote about 3+ years ago, Tim Gowers ran an experiment on his blog which he called Polymath.  The idea was to have an open collaboration on the internet where anyone who was interested could work together on a (hard!) research problem in combinatorics.  It worked out spectacularly well and a solution was obtained quicker than anyone expected.

The results were written up in a research paper and recently appeared in the Annals (which, as we talked about here, is the highest level math journal out there).  Amusingly, the journal agreed to publish the paper with author “D. H. J. Polymath”!

Two new polymath projects have started up.

As has recently been the tradition, Terry Tao is organizing a Mini Polymath based on the current year’s IMO questions.  This year’s IMO is this July 10-11.  It will be held in beautiful Argentina.  The questions are notoriously challenging/fun!  The mini-polymath will kick off on Thursday, July 12th at 1:00 am (OK time).   To join in, or just to lurk as people collaborate on the IMO problems, go here to the offical page of the mini-polymath.

In the meanwhile, a full scale polymath project has started up to try and prove the “Hot Spots Conjecture for Acute Triangles”.  It was proposed by Chris Evans and they are working on it here.  It is a more geometric/analytic/physics based question this time.  Dr. Evans describes the problem like this:

Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain, is given an initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then given enough time, the hottest point on the metal will lie on its boundary.

– from Chris Evans’s Polymath proposal

A picture of the Hot Spots Conjecture (have they tried applying talcum powder to the problem?)

Also on the polymathprojects.org website you can find several other important unsolved problems in math which might tickle your fancy.  Check it out!

# Mini Polymath

Once again it’s time for the International Math OlympiadAs we talked about last year, that’s an international competition for some of the best (high school aged) math problem solvers in the world.  It’s like the Putnam, but even more extreme!

As last year, Terence Tao is hosting a mini polymath group.  That’s a online group discussion to solve one of the IMO problems.  It started yesterday at 2pm Oklahoma time and already has 130+ comments!  If you’re interested in joining in, go here.  This year’s question is geometric:

Problem 2.  Let $S$ be a finite set of at least two points in the plane. Assume that no three points of $S$ are collinear. A windmill is a process that starts with a line $\ell$ going through a single point $P \in S$. The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point $Q$ belonging to $S$. This point $Q$ takes over as the new pivot, and the line now rotates clockwise about $Q$, until it next meets a point of $S$. This process continues indefinitely.
Show that we can choose a point $P$ in $S$ and a line $\ell$ going through $P$ such that the resulting windmill uses each point of $S$ as a pivot infinitely many times.
For the full list of IMO questions, go here.

# Mini Polymath

IMO Logo

Some time ago we posted about an online math research collaboration project initiated by Timothy Gowers which you, your dog, and everyone else on the Internet was welcome to participate in.  It turned out to be remarkably successful.   Not only was the original problem solved but various other results were obtained as well.

They are currently working on writing up the results of that project as a research paper for publication.  The author of the paper is D. H. J. Polymath.  Timothy Gowers is planning to run another polymath project this fall (keep an eye out for it!).

In the meanwhile, Terence Tao is running a mini-polymath project. The problem is the sixth (and hardest) problem from the 2009 International Math Olympiad.    The problem is:

Problem 6. Let $a_1, a_2, \ldots, a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s = a_1 +a_2 +\ldots+a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \ldots , a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.

The good news is that the IMO is an international competition for high school students.  This means the above problem is meant to be solvable by students of this age and, in particular, is known to have a solution (which doesn’t happen when you’re doing research!).  The bad news is that the IMO is for the best of the best of high school math students from around the world and only 6 or so out of 500 were able to solve it.

If you’re feeling good about your math kung-fu, take a crack at the problem.  If not so much, then swing on by Terence Tao’s blog and join in with the crowd working on solving it right now.