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From: bhuber@sjuvax.UUCP
Newsgroups: net.puzzle
Subject: Re: Circles and chords (Reply to Hint from author)
Message-ID: <2876@sjuvax.UUCP>
Date: Sat, 8-Mar-86 10:20:29 EST
Article-I.D.: sjuvax.2876
Posted: Sat Mar  8 10:20:29 1986
Date-Received: Wed, 12-Mar-86 03:46:11 EST
References: <965@h-sc1.UUCP> <2083@jhunix.UUCP> <974@h-sc1.UUCP>
Reply-To: bhuber@sjuvax.UUCP (B. Huber)
Distribution: net
Organization: St. Joseph's University, Phila. PA.
Lines: 50
Keywords: Probability, measure, random, circle
Summary: Any answer is possible

In article <974@h-sc1.UUCP> shields@h-sc1.UUCP writes:
>The original puzzle: draw random chords of a circle of radius 2r,
>what percentage of them will intersect the concentric circle of
>radius r? (This is how it was posed to us)
>
>> So depending upon which method we choose to generate random chords
>> we get an answer of 1/2, 1/3, or 1/4.  When stating that something
>> is to be chosen at random, it is important to state HOW it is to
>> be chosen at random. 
>> 
>>                                                         Dwight S. Wilson
>
>I stand corrected. The problem was posed to us as intentionally ambiguous,
>but my real question for you puzzlers is to find as many different methods
>of drawing random chords that generate different percentages as you can.
>         						 It would be very
>interesting to see what others can come up with.
>
>                                         - Tom Shields
>                                           Harvard University

The problem is quite uninteresting, unless one assumes that the probabilities
involved are invariant under rotations of the circle (at least).  Even so,
it's easy to construct such distributions of chords which yield any answer
you please, between 0 and 1.

Let a be an angular coordinate on the circle, which we orient.  Then
almost every chord (the diameters excepted) is uniquely determined by, and
uniquely determines, the angle b that it makes with a radius drawn to its 
first endpoint, where b ranges between 0 and pi/2, excluding pi/2.  (To
avoid to much pedantry, I let you define 'first endpoint'.  This is why 
one orients the circle.)

"Drawing random chords" in a way that respects the symmetry of the circle
means choosing a normalized product measure on  [0, 2pi[  x  [0, pi/2[ 
(I'm fudging with the diameters) which is uniformly distributed on the
first parameter.

So choose the uniform measure   da/(2pi)  for the endpoint of the chord, 
and let m be any normalized measure for the angle to the radius.  The
measure  da X m  is normalized and respects the symmetry of the circle.  By
choosing m to have support in [pi/6, pi/2], the answer is 0.  By choosing
m to have support in [0, pi/6] the answer is 1.  A linear combination of
such m's will give any desired answer between 0 and 1.

Either more clarification of this problem needs to be given, or else it 
isn't a problem at all.

				bill huber