na. u just missed think again.
but i say u missed vey narrowly
Given a regular 2007-gon. Find the minimal number k such that: Among every k vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
we need to ensure that 4 adjacent vertices are always formed
yes the answer is coming out to be 1506.
As celestine said we must ensure that 4 adjacent vertices are aways formed.So, let us try to find the maximum number of points we can choose so, that we can avoid 4 adjacent vertices.
so, we start choosing all the alternative vertices, and so we have chosen : \frac{2007-1}{2}=1003 vertices.
Now, in order to maximise the number of choices so that 4 adjacent vertices are not formed we choose all the vertices in the alternative gaps in between the first choice, so number of points chosen in the second round is = \frac{2007+1}{4}=502
so, total number of choices = 1503+502=1505
This ends our quest and if we choose one more than this then we are bound to have 4 adjacent vertices.
so, our answer is 1506