AoPS Academy
satisfy 
. Prove that
Also, find the values of for which the equality happens.
with no three collinear. Prove that one can relabel the points with the labels 
so that for each 
the segments 
and 
do not intersect and the following inequality holds.![\[ B_1 B_2 + B_3 B_4 + \dots + B_{2n-1} B_{2n} \ge \frac{2}{\pi} (A_1 A_2 + A_3 A_4 + \dots + A_{2n-1} A_{2n}) \]](http://latex.artofproblemsolving.com/b/5/0/b50de6dbc93a4b82cf02de5606253f5c7fe0eb2b.png)
has expected projection length equal to 
on a random line 
. Thus by an averaging argument, there exists some line such that the projection lengths of 
add up to at least the RHS quantity. Now suppose that the points 
project to 
on , where are listed from left to right, and 
need not correspond to 
. Then the sum of the projection lengths of is at most 
for any permutation 
of 
.
be the points in 
that correspond to the projections 
respectively. Let 
be the remaining points (that correspond to projections 
). Then for any permutation we have
such that the line segments 
are non-intersecting. We claim that such a matching between the "
-points" and the "
-points" exists whenever there is a line 
separating the two groups of points. Here we can take to be a line perpendicular to separating 
from 
. To prove the claim, pick the line segment 
whose intersection with is leftmost. Then all the other points 
must lie toward the right of the line determined by . This means will not intersect with any other 
, so we can safely "match" these two points. Repeating the argument for the remaining points establishes the previous claim, and the problem is solved!
-points" and the "-points" exists whenever there is a line separating the two groups of points.
and 
.
