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.

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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
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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

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 $|n| = n$

  • 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 5:41
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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 5:48
  • @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 5:52
  • 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 5:54
  • @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 5:55

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.

  • The OP is specifically interested in proofs that avoid "complex arithmetics". – Blue Jun 29 '14 at 20:35
  • Ahh sorry, I incorrectly interpreted "complex" as "complicated". – JimmyK4542 Jun 29 '14 at 20:37

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.)

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, $$a_1\cdots a_{n-1}=n.$$

  • 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 1 '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 2 '14 at 7:25
  • Mathworld only references personal communication. Did you find a paper with it? Anyway, I've asked the question here. – genepeer Jul 2 '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 3 '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 3 '14 at 10:13

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