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Topic: hard question
Replies: 4   Last Post: Jun 23, 2005 1:47 PM

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Rob Morewood

Posts: 171
Registered: 12/6/04
Re: hard question
Posted: Jun 23, 2005 1:47 PM
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ddginat (ddgiant@connect2.com) wrote:
: I have a circle with unknown diameter.
: I have been provided 28 different cord lengths for the circle.
: All of the cords put together close or complete the circle.

So you have a cyclic 28-gon with known sides.

: Question: Is there a way to calculate the diameter?

Yes, but it's not terribly pretty.
Each chord, of length C_k, will subtend a central angle of measure:

2 * Inverse Sine (C_k / D) , where D is the diameter.

[Note: Inverse Sine is multiple valued! But I shall ignore
that for a few moments and just consider the "principal value"
InvSin() = the value which a calculator will give.]

So you should get the equation in the unknown diameter:

Sum (over the 28 chords) 2*InvSin(C_k/D) = 2*Pi (radians)

or divide out the 2 to get:

Sum (over the 28 chords) InvSin(C_k/D) = Pi (radians)

You could take the Sine of both sides and reduce this to a
polynomial in D, but I'm much to lazy to attempt that!
(There is a "standard" formula for the Sine of a sum of
N angles. I saw it derived in an old College Trigonometry
book last summer - but I don't remember the details.)

If you just want a solution for 'D', feed the InvSin equation
above into a graphing calculator or computer algebra package.
Once you've finished the typing, a good approximation should
be easy to find.

This equation will have a unique solution (when it has any
solution) since, as D increases from C_k towards infinity,
each InvSin(C_k/D) is strictly decreasing towards zero.
So the entire sum is strictly decreasing towards zero
and can only take on the value Pi at most once. And it
WILL take on THAT value if the value, when D equals the
longest chord, is greater than or equal to Pi.


However, remembering that Inverse Sine is multiple valued,
at most one of the central angles can be larger than
Pi radians, and it must be smaller than 2*Pi.
So, for at most one chord, say C_1 (the longest one!),
we can replace InvSin(C_1/D) with Pi-InvSin(C_1/D)
to produce the alternate equation:

InvSin(C_1/D) = Sum (over other chords) InvSin(C_k/D)

Does this equation also have an most one solution?

Robert
|)|\/| || Burnaby South Secondary School
|\| |orewood@olc.ubc.ca || Beautiful British Columbia
Mathematics & Computer Science || (Canada)

A counter-example for those who thought the Perimeter of
the 28-gon divided by Pi ought to give a good approximation
to the diameter. For a REGULAR 28-gon, that procedure is
pretty good, yeilding an error of just over a fifth of a
percent. However, the given polygon is explicitely NOT
regular.

In a circle of diameter ONE, imagine a 28-gon inscribed
with two sides forming an angles of just over zero degrees
on either side of a diameter. The remaining 26 chords will
be vanishingly small. So the perimeter is almost exactly 2.
But 2/Pi is not a terribly good approximation for 1.


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