Given n points was incircle have radius 1.
Two points called goog pear if the points have distance more than sqrt(2). Let the circle
have m good pear, prove that 3m

Question

Given n points was incircle have radius 1.
Two points called goog pear if the points have distance more than sqrt(2). Let the circle 
have m good pear, prove that 3m<=n^2

hero · Answer

You should draw a picture of the problem. That would help.

anonymous · Answer

OK. my picture was on... but i dont have idea to solve this problem

anonymous · Answer

This is just Turan's theorem: there are no four-cliques, so the best we can do is to divide the points into three as-equal-as-possible groups.

anonymous · Answer

Define a graph whose vertex set is the  $n$  points, with an edge connecting two points 
if and only if they have distance larger than  $\sqrt{2}$ from each other. This graph contains 
no clique with four vertices...because in the best condition u that for will lie on the perimeter of circle...|dw:1344495070212:dw| so by  Turan's theorem maximum number of edges (its equal to maximum number of good pairs) is $$\frac{r-1}{r} \frac{n^2}{2}$$because we know that Among the n-vertex simple graphs with no (r + 1)-cliques, Turan's graph has the maximum number of edges. here we have  $r=3$ so $$m \le\frac{r-1}{r} \frac{n^2}{2}$$$$m \le\frac{3-1}{3} \frac{n^2}{2}=\frac{n^2}{3}$$$$3m \le n^2$$