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From: eklhad@ihnet.UUCP (K. A. Dahlke)
Newsgroups: net.puzzle,net.math
Subject: Re: Another neat problem from Putnam Exam
Message-ID: <170@ihnet.UUCP>
Date: Tue, 23-Oct-84 10:32:18 EDT
Article-I.D.: ihnet.170
Posted: Tue Oct 23 10:32:18 1984
Date-Received: Wed, 24-Oct-84 04:00:20 EDT
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Organization: AT&T Bell Labs, Naperville, IL
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< x --- y >

This proof is not what you had in mind, since it does not use
the triangular inequality, but you might find it interesting.
Mine is an inductive proof, the first case being trivially true
(one line segment connecting the point in x to the point in y).
For the larger sets, we shall divide and conquer.
Chose the point farthest to the right (the lower if two such points exist).
Label the sets so this point (p) is in x, and draw a line through p
(a virtually vertical line)
such that x and y lie entirely on the half-plane bounded by this line.
Rotate the line around p, and increment a counter when you cross a point
in x, and decrement the counter when you cross a point in y.
Begin the counter at 1 (since the line has already hit p (in the set x)).
The counter must reach 0 after a rotation of 180 degrees (if not sooner).
Since no three points are collinear, the counter presents a continuous function
(in a discrete sort of way).
It begins positive (at 1), wonders around up there, and becomes 0 upon 
crossing some point in y.
It may become 0 several more times, but we don't care.
Draw the line connecting p to the point in y which first makes the counter 0.
There are an equal number of x and y points on either side of this line,
and both half-planes can be "segmented" independently.

I hope someone posts the "triangular" proof, i would be very interested.
-- 

Karl Dahlke    ihnp4!ihnet!eklhad