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Topic: hard question
Replies: 2   Last Post: Jun 28, 2005 11:48 AM

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J. J. Sroka

Posts: 85
Registered: 12/6/04
Re: hard question
Posted: Jun 28, 2005 11:48 AM
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(Not again...the earlier posted version of this corrected two typos --
and I used the character "pi". Some newsreaders display the Greek letter,
others may display the fraction one-fourth. Here's a version that just
says "pi", not the Greek letter itself; also some M's and D's have been
put in their proper places. Sorry 'bout that.)

Robert Morewood asked: Anyone want to take a shot at necessary and
sufficient conditions?

So...

Here's a statement of a necessary and sufficient condition that a set of
line segments can form an inscribed polygon in a circle.

Let n >= 3, and let K_1, ...,K_n be the lengths of n line segments.

Let M = max{K_1, ...,K_n}.

In order for the n line segments to form an inscribed polygon of a circle,
it is necessary and sufficient that

arcsin(K_1/M) + ... + arcsin(K_n/M) >= pi

where arcsin is the principal value of the inverse sine, that is, the
range of arcsin = [-pi/2, pi/2].

The unique diameter, D, of that circle is then given by D such that D >= M and

arcsin(K_1/D) + ... + arcsin(K_n/D) = pi.

*********************

Note that the *diameter* is unique, but the n segments can form more than
one non-congruent inscribed polygon, all within the same circle. So maybe
this is the non-uniqueness which Jim Spriggs and I had previously had in
mind.

Anyone care to find a formula for the number of non-congruent inscribed
polygons in the case that the n segments are all of different lengths?

--- Joe (Sent via 10.2.8 at 2:02am PDT, June 28, 2005).

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