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Given any 6 points inside a circle of radius 1, some two of the 6 points are within 1 of each other.
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I'll give two proofs.
Divide the circle into six equal circular sectors. If necessary, rotate the sectors
around the center until one of the points (say, A) falls on one of the radii. Note that, since each sector
is a portion of the Reuleaux triangle of width 1, no points in a sector are father from each other than 1.
If in one of the two sectors to which A belongs there is another point of the set, these two
will be within 1 of each other. Otherwise, we are left with 4 sectors and 5 points and, thus,
are able to claim that one of the sectors contains at least two points of the set. These two then
satisfy the requirements.
The second proof is by William A McWorter Jr..
We may assume that no point is at the center O of the circle and
that no two of the points are on a radius; for then we are done. Radii through O
and each of the six points partition 360o into six parts, at least
one of which is less or equal 60o. Let A and B be two of the points
such that the AOB is less or equal 60o. AO and BO, being on
radii, have length less or equal 1. But also, one of these segments is
opposite an angle greater or equal 60o in the AOB. Hence the length
of AB is less or equal that of one of AO or BO, both of which are less or
equal 1.

Copyright © 1996-2009 Alexander Bogomolny
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