For a set $ P$ of five points in the plane, no three of them being collinear, let $ s(P)$ be the numbers of acute triangles formed by vertices in $ P$. Find the maximum value of $ s(P)$ over all such sets $ P$.
Problem
Source: MEMO Team Competition, Question 6
Tags: combinatorics proposed, combinatorics
mszew
25.09.2007 15:04
The regular pentagon has $ 5$ acute triangles and there are $ C_{5,3}=10$ triangles but the sum of all the interior angles of any convex pentagon is $ 540$ so at least one angle is bigger than $ 90$ same for a non-convex pentagon thus: $ 5\leq s(P)\leq 9$
patham
25.09.2007 19:40
Is there actually a set $ P$ having $ s(P) = 9$ ? I myself feel that maximum value of $ s(P)$ is between $ 6$ and $ 7$ (although I have no proof).
Ravi B
25.09.2007 20:44
In the following link, I showed that $ s(P)\le 7$ for all sets $ P$ of 5 points: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14796. Actually, looking back, all I did was link to someone else's solution to Problem 6 from IMO 1970. It's not too hard to come up with a convex pentagon showing that $ s(P) = 7$ can be achieved.
patham
26.09.2007 11:03
Can you post the figure for $ s(P)=7$ ? The maximum I have got is $ 6$.
