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From: ins_ampm@jhunix.UUCP (Michael P McKenna)
Newsgroups: net.puzzle
Subject: Re: Circles and chords (Hint from author)
Message-ID: <2083@jhunix.UUCP>
Date: Tue, 4-Mar-86 13:25:21 EST
Article-I.D.: jhunix.2083
Posted: Tue Mar  4 13:25:21 1986
Date-Received: Fri, 7-Mar-86 05:47:31 EST
References: <965@h-sc1.UUCP>
Reply-To: ins_ampm@jhunix.ARPA (Michael P McKenna)
Distribution: net
Organization: Johns Hopkins Univ. Computing Ctr.
Lines: 73

In article <965@h-sc1.UUCP> shields@h-sc1.UUCP writes:
>
>     Here is a hint to help you on your way to finding solutions to
>the previously posed puzzle: any chord of a circle can be represented by
>its midpoint. Therefore this helps in two ways: it is possible to generate
>random chords by generating random points, and a simple check for
>intersection with the inner circle is to check if the midpoint of the chord
>is in the inner circle. This should help considerably, and make evident
>yet another solution to the puzzle. However, we found more solutions based
>on this property also, see if you can find them.
>
>                                        Happy puzzling,
>
>                                             - Tom Shields
>                                               Harvard University


I don't seem to have received the puzzle, but if it is what I 
think it is you want to be careful about random generation.  This
is discussed in one of Martin Gardner's columns, reprinted in
_The 2nd Scientific American Mathematical Puzzles & Diversions_

The problem (which sounds like it might be your puzzle), is to
determine the probability that a random chord drawn in a circle
is longer than the side of an equilateral triangle inscribed in
the circle.  It is easy to show that this probability depends on
the method of generating the random chord.

The first suggested method is to pick a point A on the circle.  This is
one endpoint of the chord.  The other endpoint is can range uniformly
over the rest of the circle.  By symmetry it doesn't matter where A
is for computing the probability, so assume that A is at the one of
the vertices of the triangle (or just inscribe the triangle so that
one of its vertices is at A if this makes you feel uncomfortable).
Anyway clearly the chord is longer than the side of the triangle, if
and only if it cuts across the triangle.  The angle at the vertex
is 60 degrees out of a possible 180 degrees so the probability is
1/3.

2nd method:  Any chord must be perpendicular to one of the diameters.
Pick a diameter, position the triangle so that a vertex is at one of
the endpoints.  Now the set of chords ranges over the length of the
diameter.  The diameter intersects the triangle at the midpoint of
one side, label this point A.  Label the analagous point on the
opposite side of the diameter B.  Clearly only the chords between
A and B are longer than the triangle's side.  This length is equal
to 1/2 the diameter.  Therefore the probability is 1/2

3rd method:  The one suggested above.  Let the midpoint of the chord
range uniformly over the area of the circle.  It is easy to show
that chords longer that the triangle side must have their midpoints
inside a circle with same center, and 1/2 the radius of the original
circle.  This inner circle therefore has area 1/4 the original circle,
therfore the probability of the midpoint being inside it is 1/4.

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. 

Interestingly the articly states that humans do not seem to use any
of the above methods when asked to draw a random chord.  The probability
of a subject drawing a chord longer than the side of the triangle is
much better than 1/2.

                                                        Dwight S. Wilson

Disclaimer:  Why should I disclaim these views?  They're mathematically
             correct.

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