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




Re: hard question
Posted:
Jun 28, 2005 11:48 AM


(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 onefourth. 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 noncongruent inscribed polygon, all within the same circle. So maybe this is the nonuniqueness which Jim Spriggs and I had previously had in mind.
Anyone care to find a formula for the number of noncongruent 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).
  Delete the second "o" to email me.
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