Geometry problem

Solution: Let S be the set of all points with coordinates of the form (1;1; :::;1)
in n-dimensional space, and for each p 2 S, let Ap be the set of all points in S that
dier from p in exactly one coordinate. Each Ap contains n points, so if we count all of
the points in every Ap as we let p range over all points in B, we will count more than
n 2n+1=n = 2n+1 points. Since there are 2n points in S, by the pigeonhole principle,
we must count some point x at least three times. Thus, there are three points in B
that dier from x in exactly one coordinate. These three points must each dier from
each other in two coordinates, and so they form an equilateral triangle with side length
2p2.