Given $100$ coplanar points, no three collinear, prove that at most $70\%$ of the triangles formed by the points have all angles acute.
Problem
Source: IMO 1970, Day 2, Problem 6
Tags: acute, Triangle, point set, counting, combinatorial geometry, IMO, IMO 1970
Peter
13.11.2005 23:52
Does 'at most' imply giving an equality case as well? As I seem unable to find one...
Ravi B
14.11.2005 00:02
No, equality is not achieved. For example, in the thread why 70%, I showed that the fraction of acute triangles is at most $66.7\%$.
antimonyarsenide
01.04.2012 02:18
I'm a bit confused about this: Is this statement valid? : "Since among the triangles formed by any 5 points at most 70% are acute, at most 70% of all the triangles are acute." I can't seem to wrap my head around the statement, because it seems to be applying pigeonhole to a set of 3-element sets...
feel
01.04.2012 09:48
It doesn't work, because when you take two subsets of 5 points, they can both contain the same triangle.
SCP
16.05.2012 21:29
It does work, cause each triangle is counted the same number. showing the meaning one one person writing so easy is bad at this site, but I know it is difficult and made hundres of mistakes too
Legend-crush
12.08.2014 00:33
Is there any solution ? Moderator says: yes, follow the link given by Ravi B.