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There is a beautiful fact:

If you take a regular N-sided polygon, which is inscribed in the unit circle and find the product of all its diagonals (including two sides) carried out from one corner you will get N exactly:

$A_1A_2\cdot A_1A_3\cdot ...\cdot A_1A_N = N$

enter image description here

For example, for a square we have $\sqrt{2}\cdot 2\cdot \sqrt{2} = 4$.

I know there is some prove, which is based on complex numbers. But the result is so simple that I wonder is there much more simple prove, which you can explain to a school boy easily?

P.S. Please, use spoiler tag >! in your answers.

Martin Sleziak
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klm123
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  • I'm not sure what you mean with your last comment. You want to know that there is a solution, but you'd also like to find it yourself? In that case, are you eventually going to answer this question yourself? – Joe Z. Jun 15 '14 at 19:46
  • @Joe Z., 1. yes. 2. I would like to, but I am not sure whether I can:) – klm123 Jun 15 '14 at 19:51
  • Also the result is false for diagonals - for $N = 4$, the only diagonal is of length $2$, not $4$. It works again if you use all chords from a point instead of all diagonals of the polygon. – Joe Z. Jun 15 '14 at 19:55
  • @JoeZ., the two sides attached to the corner are also counted as diagonals. Please, see the formula. – klm123 Jun 15 '14 at 20:01
  • What I mean is that you're using the terminology "diagonals" differently from how it's normally used, and that can be confusing. – Joe Z. Jun 15 '14 at 20:03
  • I would start by letting $A_1=1$, all the rest be the complex roots of $1$ and expressing it all in $\cos \theta$ and $\sin \theta$. No guarantees. – Ross Millikan Jun 16 '14 at 14:40
  • @RossMillikan, straightforward, but this makes things only more complicated. Even expression with complex numbers looks much simpler. I would expect to find some analogy, which will make result obvious... – klm123 Jun 16 '14 at 14:55
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    One can show fairly easily that the length of the kth diagonal is 2 sin(k pi/n). But this is just a small step forward. – berkeleychocolate Jun 17 '14 at 17:42
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    One reason to maybe not expect a 'pretty' proof is the awkward dimensionality of the whole thing - the result isn't just $n$, it's (e.g.) $n (\mathrm{meters}^{n-1})$. So any proof will have a hard time expressing the result in terms of geometric quantities, since it won't scale similarly to anything that you can 'point to' on the polygon diagram - it doesn't scale as a length or an area. – Steven Stadnicki Jun 29 '14 at 19:52
  • If you let the circle have a radius of $R$ instead of $1$, then the product of the diagonals is $nR^{n-1}$, which is dimensionally correct. – JimmyK4542 Jun 29 '14 at 20:40
  • @JimmyK4542 Of course, but the point is that that's a much less elegant formula (and still difficult to express in terms of lower-dimensional quantities of interest). – Steven Stadnicki Jun 30 '14 at 04:13

4 Answers4

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Here is an extremely simple proof but it requires working in the field $\mathbb{C}$.

For any natural $n > 0$:

  Let $w = e^{i\frac{2\pi}{n}}$

  $\prod_{k=1}^n (z-w^k) = z^n-1$ because both are monic polynomials of degree $n$ with the same $n$ roots

  Thus $\prod_{k=1}^{n-1} (z-w^k) = \lim_{t \to z} \frac{t^n-1}{t-1} = \sum_{k=0}^{n-1} z^k$

  In particular $\prod_{k=1}^{n-1} (1-w^k) = \sum_{k=0}^{n-1} 1^k = n$

  Therefore the product of the diagonals desired is $|n| = n$

user21820
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  • I'm giving this answer because I'm not sure what is meant by "complex arithmetics". If it means complex calculations, then this isn't complex at all and in my opinion most beautiful. – user21820 Jun 30 '14 at 05:41
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    I believe "complex arithmetics" means using complex numbers instead of complicated calculations. Of course, I had the same interpretation of "complex arithmetics" that you did when I first saw this question. – JimmyK4542 Jun 30 '14 at 05:48
  • @JimmyK4542: Haha. Anyway do you think there is a way to convert this into one that doesn't use $\mathbb{C}$? I would doubt so. – user21820 Jun 30 '14 at 05:52
  • I tried using Google, but all the results prove it using complex numbers. I don't think there is a better solution. – JimmyK4542 Jun 30 '14 at 05:54
  • @JimmyK4542: If I remember correctly I once saw a complicated method that didn't use complex numbers but that was many years ago and I don't remember anything about it. – user21820 Jun 30 '14 at 05:55
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Consider the complex solutions $z_0, \ldots , z_{n-1}$ to $(z+1)^n-1 = 0$. These solutions are evenly spaced on the circle centered at $z = -1$ with radius $1$. Thus, they form a regular $n$-gon.

Since $0 = (z+1)^n-1 = z^n + \cdots + nz+1-1 = z(z^{n-1}+\cdots +n)$, we know that $z_0 = 0$ is one solution, and the product of the other $n-1$ solutions is $z_1 \cdots z_{n-1} = (-1)^{n-1}n$.

Then, the product of the distances from $z_0$ to each of the $z_i$'s is $|z_1-z_0| \cdots |z_{n-1}-z_0| = |z_1| \cdots |z_{n-1}| = |z_1 \cdots z_{n-1}| = |(-1)^{n-1}n| = n$, as desired.

JimmyK4542
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4

Not a proof, but an interesting visualization (using $n=7$) ...

enter image description here

Here, $|\overline{P_0P_1}| = 1$, and each $\triangle P_0 P_k P_{k+1}$ is isosceles with vertex angle $\angle P_k = \frac{2\pi k}{n}$ (or $2\pi-\frac{2\pi k}{n}$). Thus, $$|\overline{P_0 P_{k+1}}| \;=\; 2\;|\overline{P_0 P_{k}}|\;\sin \frac{1}{2}\angle P_{k} \;=\; 2\;|\overline{P_0 P_{k}}|\;\sin\frac{\pi k}{n}$$ so that, ultimately, (with $A_i$ as in your figure) $$|\overline{P_0 P_n}| = 2 \sin\frac{\pi}{n} \cdot 2\sin\frac{2\pi}{n}\cdot\;\cdots\;\cdot 2 \sin\frac{(n-1)\pi}{n} = |\overline{A_1 A_2}| \cdot |\overline{A_1 A_3}| \cdot\;\cdots\;\cdot |\overline{A_1 A_n}|$$

As the diagram makes clear, $P_0$, $P_1$, and $P_n$ are certainly collinear (as are, more generally, points $P_0$, $P_k$, $P_{n-k+1}$ for any $k$). However, while the diagram appears to show that $\overline{P_0 P_7} = 7$ (as it should, because we know that to be the case), this doesn't qualify as a proof.

OP notes that there exist proofs of the relation using "complex arithmetics". This answer, for instance, essentially gives one. (All we need to do is multiply the formula there by $|\overline{A_1A_2}|\cdot|\overline{A_1A_n}| = 2^2 \sin^2\frac{\pi}{n}$ to include the "diagonals" that are actually sides of the polygon.)

In reference to @StevenStadnicki's comment, it does seem that this ...

$$|\overline{P_0 P_n}| = |\overline{A_1 A_2}| \cdot |\overline{A_1 A_3}| \cdot\;\cdots\;\cdot |\overline{A_1 A_n}|$$

... asserts equality between a one-dimensional quantity on the left, and an $(n-1)$-dimensional quantity on the right. Perhaps we should instead write the dimensionless relation ... $$\frac{|\overline{P_0 P_n}|}{|\overline{P_0 P_1}|} = \frac{|\overline{A_1 A_2}|}{|\overline{A_0 A_1}|} \cdot \frac{|\overline{A_1 A_3}|}{|\overline{A_0 A_1}|} \cdot\;\cdots\;\cdot \frac{|\overline{A_1 A_n}|}{|\overline{A_0 A_1}|}$$ ... where $A_0$ is the center of the $A$-polygon's circumcircle. (This is effectively @JimmyK4542's suggestion.)

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4

Here is my attempt at an elementary proof. After hours of staring, I believe (at minimum) a few trigonometric identities would be required in any elementary demonstration.

Given our circle with $n$ nodes going around the circle, it is clear that each edge of interest, $a_k$, $1\le k\le n-1$, is part of an isosceles triangle with node at the center, sides, $1,1,a_k$, and inner angle $(2\pi)/n$. Thus, using law of cosines and a product to some formula, we gain,

$$a_k^2=1^2+1^2-2\cos\left(\frac{2\pi k}{n}\right)=4\sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{(n-k)\pi}{n}\right).$$ We now have, using this (formula 24),

$$a_1^2\cdots a_{n-1}^2 = \prod_{k=1}^{n-1}\left(4 \sin\left(\frac{k\pi}{n}\right)\sin\left(\frac{(n-k)\pi}{n}\right)\right) =\\ 4^{n-1} \left(\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right)\right)^2 = 4^{n-1}\left(\frac{n}{2^{n-1}}\right)^2 = n^2.$$ Hence, our theorem holds, $$a_1\cdots a_{n-1}=n.$$

Bobby Ocean
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    do you have an idea of how to prove that formula? I'd rather not post a question if there's a simple trick I'm missing. – genepeer Jul 01 '14 at 21:32
  • @genepeer At the moment, I honestly don't know how to prove that particular formula (there is a paper referenced, but I haven't read it). However, the difficulty of that little formula seems to lie at the heart of why this seemingly simple problem has a difficult proof. Who would have guessed that a trigonometric product has an explicit rational value? – Bobby Ocean Jul 02 '14 at 07:25
  • Mathworld only references personal communication. Did you find a paper with it? Anyway, I've asked the question here. – genepeer Jul 02 '14 at 12:25
  • I think it is even more complicated solution... are you sure that the formula is not proven similarly, using complex numbers? – klm123 Jul 03 '14 at 10:11
  • @klm123: That's exactly what I would say. I think any proof of that formula can be rephrased more simply using complex numbers. – user21820 Jul 03 '14 at 10:13
  • Besides, your method has an extraneous square, since the desired theorem would have been proven directly with the identity for the product of those $\sin$s. – user21820 Jul 03 '14 at 10:15
  • @klm123 I have found a proof for the formula that does not use complex numbers. I'll be posting it on my question when I finish the writing it up. It's not surprising that it is even more complicated. If the popular proofs use complex numbers, then sticking to real numbers usually makes the problem harder. – genepeer Jul 03 '14 at 10:52
  • @genepeer, you probably mean sticking to trigonometric solution. That's for sure, I didn't expect a trigonometric solution be an answer to my question. But can you show your prove somewhere? – klm123 Jul 03 '14 at 10:54
  • @klm123 the gist of it is in the comments to my question. As per your revised question, I don't think there's a simple solution to this problem that you can explain to a schoolboy. – genepeer Jul 03 '14 at 11:04
  • I've posted a complete proof of the formula. – genepeer Jul 03 '14 at 15:22