>The n-1 numbers then satisfy
>
>1 + (z+1) + (z+1)² + ... + z^n-1 = 0.
>
>If you multiply through, then the free term will be the
>product of the n-1 (complex) chords and equal to n.

However, upon reflection, I see that you mean to use the fact that the product of the roots of a polynomial is equal to its constant term. The z values are of course the roots, and the constant term is n, as is clear when you multiply out the terms. From there on, everything is obvious.

This an interesting trick. Thank you.

--Stuart Anderson

Strange. I was sure that it is common to call the constant term "free", meaning "free of the variable." Now that this confused you, I made a search on the web and could not find anybody who shared the same view. I have then admit it is probably a translation from Russian that no one else uses. Thanks for bringing this up.

>>The n-1 numbers then satisfy
>>
>>1 + (z+1) + (z+1)² + ... + z^n-1 = 0.
>>
>>If you multiply through, then the free term will be the
>>product of the n-1 (complex) chords and equal to n.
>
>However, upon reflection, I see that you mean to use the
>fact that the product of the roots of a polynomial is equal
>to its constant term.

(plus or minus.)

>The z values are of course the roots,
>and the constant term is n, as is clear when you multiply
>out the terms. From there on, everything is obvious.

When it comes to computing the diagonals of a regular polygon, you can't avoid using trigonometry, which is often simplified from embedding into complex plane.

I had surmised that this must be the meaning, but was not sure.

>> the product of the roots of a polynomial is equal
>>to its constant term.
>
>(plus or minus.)

Yes, I meant to say "monic polynomial," but forgot to insert the qualifier.

I showed my brother this solution, and he thought it was quite neat. He is also interested in mathematics, and we are one day going to write a short note about an interesting approach to Heron's formula which we thought of while talking late one evening last year, if we can both find the time.

--Stuart Anderson

For monic polynomials the sign of the product is (-1)ⁿ times the constant term.

-- Just had to make this clarification lest a potential reader gets confused.

>I showed my brother this solution, and he thought it was
>quite neat. He is also interested in mathematics, and we
>are one day going to write a short note about an interesting
>approach to Heron's formula which we thought of while
>talking late one evening last year, if we can both find the
>time.

So you were not alone in that remote place where you grew up. Or is he much younger?

Thanks. I should be more careful, but since magnitudes would be taken later, I simply ignored the sign issue. You are right, of course, that to someone else reading this thread, my post was potentially misleading.

>So you were not alone in that remote place where you grew
>up. Or is he much younger?

I have a brother 2 years younger and another 4 years younger. Both have mathematical talent, but they were not interested in math as children. I was fascinated by it, back to my earliest memories, while they were bored with it in school (if was of course not very challenging for them) and turned to other interests, mostly music.

When they started college, they encountered mathematical subjects such as physics, realized that there was much more to mathematics than they had thought, and began taking mathematics courses. At that point they becamse interested in it, but until then, I was more or less alone as far as mathematical interests were concerned.

At present, Steve (2 years younger, with whom I was discussing Heron's formula) works with me at the University of Washington Instructional Center, while John (4 years younger) is a mechanical engineer at the Kenworth truck company, their chief North American specialist in modeling vibrations in systems of very many degrees of freedom. So, things have worked out well for us after all.

--Stuart Anderson

Thank you for sharing. Enjoy the rest of it.

Alex