Revision as of 12:33, 19 July 2011 by Teorth
This is the wiki page for the mini-polymath3 project, which seeks solutions to Question 2 from the 2011 International Mathematical Olympiad.
This project will follow the usual polymath rules. In particular:
- Everyone is welcome to participate, though people who have already seen an external solution to the problem should refrain from giving spoilers throughout the experiment.
- This is a team effort, not a race between individuals. Rather than work for extended periods of time in isolation from the rest of the project, the idea is to come up with short observations (or to carry an observation of another participant further) and then report back what one gets to the rest of the team. Partial results or even failures can be worth reporting.
- Participants are encouraged to update the wiki, or to summarise progress within threads, for the benefit of others. (In particular, linking between the wiki and specific comments on the blogs is highly encouraged.)
- Mini-polymath 3: 2011 IMO question, June 9 2011.
- Mini-polymath3 discussion thread, July 19 2011.
- Minipolymath3 project: 2011 IMO, July 19 2011
- Problem 2. Let [math]S[/math] be a finite set of at least two points in the plane. Assume that no three points of [math]S[/math] are collinear. A windmill is a process that starts with a line [math]\ell[/math] going through a single point [math]P \in S[/math]. The line rotates clockwise about the pivot [math]P[/math] until the first time that the line meets some other point [math]Q[/math] belonging to [math]S[/math]. This point [math]Q[/math] takes over as the new pivot, and the line now rotates clockwise about [math]Q[/math], until it next meets a point of [math]S[/math]. This process continues indefinitely.
- Show that we can choose a point [math]P[/math] in [math]S[/math] and a line [math]\ell[/math] going through [math]P[/math] such that the resulting windmill uses each point of [math]S[/math] as a pivot infinitely many times.
Observations and partial results
- Trivial observation: If we start with a point on the convex hull of S and a line l that is “tangent” to the convex hull then we will only iterate over the points from the convex hull.
- One can start with any point (since every point of S should be pivot infinitely often), the direction of line that one starts with however matters!
- If the points form a convex polygon, it is easy.
- Say there are four points: an equilateral triangle, and then one point in the center of the triangle. No three points are collinear. If you start in M, you first hit say A, then C, then M, then B, then A.
- Maybe the strategy should be to take out the convex hull of S from consideration; follow it up by induction on removing successive convex hulls.
- A possible line of enquiry might be to consider how adding additional points effects an already established “windmill” this might lead to an inductive solution if adding new points can be related to already established cases.
- None yet.