Find a set of 8 points with no three collinear such that no subset of 5 forms a convex pentagon.

here is an idea I've had for a real solution.  Drew it out and it looks like it works but can't quite prove it yet, perhaps somebody with more geometry skill than myself can do so.

Start with an 3 non co-linear points.  Draw the triangle fromed by the three points.  Next draw the 3 medians of this triangle and choose their common intersection as the 4th point.  Now this 4th point along with the 3 points where the medians intersect the 3 sides of the triangle break the triangle up into 6 sub-triangles.  Now pick any one of these triangles.  Pick a point inside this triangle as the 5th point.  Now if you draw a line from the 5th point through the 4th point it will pass through another sub-triangle.  Pick a point in this sub-triangle as the 6th point.  Now draw the angle from 5th to 4th to 6th point.  on the side of this angle that is greater than 180 degrees there should be two more sub-triangles.  Pick a point in one of these 2 sub-triangles as the 7th point.  Finally draw a line from the 7th point through the 4th point and pick a point in the sub-triangle that it passes through as the 8th point.

From what I can see this should always give 8 points no 3 of which are collinear and no 5 of which form a convex pentagon.  But I have as yet to prove it.

Posted by Daniel on 2009-10-09 13:37:53