The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Inactive » k12.ed.math

Topic: hard question
Replies: 4   Last Post: Jun 26, 2005 12:09 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
J. J. Sroka

Posts: 85
Registered: 12/6/04
Re: hard question
Posted: Jun 24, 2005 2:37 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article <>, Robert Morewood
<see.sig@for.real.address> wrote:

> Joseph Sroka-10.2.8 ( wrote:
> : maybe one of the notorious homework solution providers of k.e.m.
> : will spoil your fun by providing a complete solution.
> Hi Joseph,
> I provided much more detailed directions than you did as I'm
> pretty sure this is not homework and there were already two
> incorrect answers posted. It looked like a legitimate topic
> for discussion. Although, when the trig is removed, I get an
> equation with a sum of more than 134 million square roots of
> polynomials in Diameter-Squared, with degrees up to 27. Yikes!
> (OK, I guess removing the trig was a bad idea. It does show
> that the problem is algebraic rather than transcendental...)
> Is this sufficiently complex that only a numerical approximation
> is possible?
> Anyway, I suspect the original poster was interested in this
> problem for something more than extra credit or amusement.
> Perhaps our giant friend will clarify why he is interested.
> : Are the 28 lengths actually guaranteed to be from such an
> : inscribed polygon?
> The original poster implied this is guaranteed in his/her case.
> Of course it's not true in general. Anyone want to take a
> shot at necessary and sufficient conditions? (For exactly
> one solution, two solutions, more?)
> Robert
> |)|\/| || Burnaby South Secondary School
> |\| | || Beautiful British Columbia
> Mathematics & Computer Science || (Canada)


I agree that this is almost certainly not a homework problem.

After I had sent my brief reply to Giant, I wrote up the following. It is
much like your response, even as to the notation used.

This has been an interesting problem, seeing as how I at first had
thought, like Jim, that there would not be a unique solution in general.

Also, your comment about about necessary and sufficient conditions: I
believe that carefully looking at your or my method will show what these

In what I wrote below, I used degrees for values of inverse trig
functions, rather than radians. It really makes no real difference.

Here's the solution that I wrote up:

Every chord must be no longer than the diameter of the circle. Said
another way, the diameter of a circle that has 28 chords which form an
inscribed polygon, must be greater than or equal to the maximum length

Also the diameter cannot be greater than the sum of the chords' lengths,
or even better, the diameter cannot be greater than one half the sum of
the chords' lengths.

All we really need to understand is that possibilities for the length of
the diameter are bounded between two numbers.

Suppose now that we have a circle in which one of the given chords, call
it K, fits.

Then the central angle that this chord spans can be found pretty simply.
Look at the triangle formed by joining the two ends of the chord to the
center of the circle. This is an isosceles triangle with the chord as one
side, and the other two sides having length equal to the radius of the

If you bisect this triangle, with a line from the center of the circle to
the midpoint of the chord, you get two right triangles. Look at the angle
which is half the central angle. Call this angle "theta".

Because it's a right triangle, sin(theta) = opp/hypotenuse. Opp has
length K/2, and hypotenuse = radius = diameter/2, call it D/2.

So, sin(theta) = (K/2)/(D/2) = K/D.

The max that the central angle can be is 180 degrees, so the max that
theta (which is half of that central angle) can be is 90 degrees.

So, without ambiguity, theta = arcsin(K/D).

We have 28 chords, call them K_1, ... , K_28, and 28 half-central angles,
call them theta_1, ... , theta_28.

Add all these half-central-angles together, and you should get...badda
bing! 180 degrees.

So, we should get

180 = arcsin(K_1/D) + ... + arcsin(K_28/D).

Of course, we don't know the value of D, but just start trying values of D
starting with D = max{K_1, ... K_28}. A programmable calculator or the
ability to program, say in BASIC, would make it pretty easy to evaluate

arcsin(K_1/D) + ... + arcsin(K_28/D) for any particular value of D.

Note that the expression arcsin(K_1/D) + ... + arcsin(K_28/D) defines a
function of D, and it's a *decreasing* function of D.

So, if your first try for D results in a value larger than 180, then just
increase D a bit.

The max D to try should be no bigger than one half the sum of the lengths
of the chords, so in a reasonable number of tries for D, you should come
pretty close to 180 degrees. That D will then be pretty close to the


Thanks, Giant, for an interesting problem.

--- Joe

Sent via 10.2.8 at 11:18am PDT, June 23, 2005.

Delete the second "o" to email me.

submissions: post to k12.ed.math or e-mail to
private e-mail to the k12.ed.math moderator:
newsgroup website:
newsgroup charter:

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.