Products of Chords in a Circle (math visualization)

In this video, we describe a classic problem about finding the product of the lengths of the chords obtained by placing n equally spaced points on the unit circle and connecting one of the points to each of the others. The answer is surprising and one proof relies on a limiting argument using complex numbers and the nth roots of unity.
Once we have this interesting fact, we then show how to couple the fact with the law of cosines to get some interesting trigonometric identities for strange products of sines, cosines, and tangents.
Finally, we finish with a puzzle about what happens when you do ask the same question about an ellipse. We suggest one particularly interesting ellipse to investigate.
#math #mathvideo #manim #circle #chords #visualproof #trigonometry #sine #cosine #tangent #rootsofunity #complexanalysis #complexnumbers #products #geometry
This animation is based on a famous problem that has been discussed in numerous places. In particular, a great reference for this problem and related ones (with a fantastic bibliography) is the source Chords of an Ellipse, Lucas Polynomials, and Cubic Equations in Issue 8 of the 2020 American Mathematical Monthly (doi.org/10.1080/00029890.2020...) by Ben Blum-Smith and Japheth Wood. You can also find the source here: arxiv.org/abs/1810.00492.
06:09 note that at the end there are rounding error typos for the challenge problem. The first mistake is at 90, which should be 91. The rest have rounding errors most likely.
For more about using manim, see www.manim.community/.
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Music in this video:
I Am Running Down the Long Hallway of Viewmont Elementary by Chris Zabriskie is licensed under a Creative Commons Attribution 4.0 license. creativecommons.org/licenses/...
Source: chriszabriskie.com/honor/
Artist: chriszabriskie.com/

[@phiarchitect]

The more we look, the more fascinating the relationships become.

[@MathVisualProofs]

Definitely!

[@canyoupoop]

I actually encountered a problem where a regular octagone was inscribed in a unit circle and I had to find the product of the dline segments connecting any one fixed vertex with... yeah other all vertex I did that by plotting this on a graph, and using the fact that it reall just represents those standard trig values I found the coordinates for and found the distance, did its product yada yada I then realised this was also true of other n gons but couldn't prove it. The idea of introducting limit here just blows my mind! Actually fantastic!!

[@MathVisualProofs]

👍😀

[@luukaskyronlahti5205]

I just found your channel i think this is paradise :D

[@MathVisualProofs]

Thanks!! Glad you like it!

[@phiarchitect]

The heptadecagon (17-gon) is an interesting case - and probably worth a further look. I started an algebraic construction a while back using my GEOMETOR explorer but have not finished it. Your video is inspiring me to get it done.

[@MathVisualProofs]

That’s a serious one :)

[@phiarchitect]

@@MathVisualProofs Even more serious is the heptagon (7-gon)

[@MathVisualProofs]

@@phiarchitect impossible of course :)

[@phiarchitect]

@@MathVisualProofs unless there is a bridge somewhere :)

[@sehomecanoe]

Awesome. Big fan.

[@lloki2215]

That was excellent. Thank you 😊

[@MathVisualProofs]

Glad you enjoyed it!

[@jinishgaming3240]

It was mind-blowing

[@MathVisualProofs]

:)

[@FromTheMountain]

I'm not sure how to interpret the phrase "equally spaced points" on an ellipse. Does it mean that the angles that the lines connecting those points to the origin are equally spaced, does it mean that the arc length between two consecutive points must be the same, something else? Either way, it seems like it wll be harder to simplify the product like you did with the circle, but I guess there will be a nice trick.

[@MathVisualProofs]

Should not be equal arc length between points. Here’s a hint (divide the nth entry by n, assuming the sequence starts at n=2). (Edited typo)

[@FromTheMountain]

@@MathVisualProofs I did some experimentation in Python, and I think equal arc length is not correct. Using this interpretation, my code gave the sequence (2.00, 5.62, 11.99, 19.88, 33.93, ...). I have been trying to post a link, but my comment keeps getting deleted. However, when I used the same parameterization of the curve as in the code, but then used equally spaced t, I got the sequence (2, 6, 12, 25, 48, 91, 168, 306, 550, ...), which is much closer to what you got, although there are still some small differences. I'm guessing you made a mistake there, especially because your hint seemed to suggest that the nth entry should be divisible by n, which is not the case for the pair (90, 7) for instance. I still need to think about how I can mathematically prove a closed-form solution for this sequence, but at least I know what is asked now.

[@FromTheMountain]

By the way, the parameterization I meant was x(t) = -sqrt(5) * sin(2 pi t), y(t) = cos(2 pi t), with t ranging from 0 to 1.

[@MathVisualProofs]

@@FromTheMountain gah! I must have rounding errors when I went to create the text. 91 is right so I’ve definitely got mistakes. Thanks for catching it. I’ll note it in description. Also, the paper linked in description will have answers if you need them. :)

[@MathVisualProofs]

@@FromTheMountain yes. Again. I had typo in my response. They should not be equal arc length. This parametrization should do it (and I shouldn’t respond using my phone )

[@mad_vegan]

I thought the ellipse problem was just an exercise and thought I was dumb for failing to solve it. Then I gave up and looked at the paper. It's more challenging than I thought. Although I was on the right track. I thought of using some holomorphic transformation to stretch the real and imaginary parts independently. A very simple solution is az + b/z.

[@MathVisualProofs]

It’s a great paper! Also I have a typo in my data (rounding error I think using a computer). Keep on it!

[@mathewlopes5657]

I learnt many things that I didn't learn at college.

[@MathVisualProofs]

👍

[@jeffkevin3]

I have some memory told me that this can be proved elegantly using complex numbers. 😂

[@lgooch]

I don’t understand 3:09. ( how that product is = to |z^7-1| )

[@MathVisualProofs]

Search roots of unity on Wikipedia or check this article : en.m.wikipedia.org/wiki/Cyclotomic_polynomial. Essentially x^n-1 has n roots and they are the nth roots of unity.

[@lgooch]

@@MathVisualProofs thanks!

[@lgooch]

I watched this again and don’t know why I didn’t understand that part lol. I think I forgot that zeta_i are the roots of unity.

[@MathVisualProofs]

@@lgooch No worries. Thanks for checking it out again!

[@omargaber3122]

We want a video showing a solution to the last problem, please

[@MathVisualProofs]

Hmm... well note that I have a computation error in there (check in the description). I bet you can figure it out though...

[@muqtarjamaegal6071]

Please calculus equations

[@MathVisualProofs]

??

[@cheedozer7391]

In the case of an ellipse, isn't it as simple as a change of variable to make it look like a circe and then using the same argument? After that, all one needs to do is change the variable back.

[@MathVisualProofs]

See if you can make it work… :)

[@cheedozer7391]

@@MathVisualProofs It was pretty tricky for me to figure out how to change the variables "nicely", but I made it work in the end!! This exercise was enlightening.

[@columbus8myhw]

Here's a neat thing: suppose instead of multiplying the lengths, we add 1/length^2 over all the lengths. Is there a formula? Before tackling this question, use a computer to get some numerical data, and find the quadratic of best fit. How big are the errors? (Bonus point: use this and a clever limit to find 1+1/2^2+1/3^2+…)

[@MathVisualProofs]

Great stuff here. Thanks!

[@columbus8myhw]

Also Desmos has a "quadratic of best fit" feature built-in, I'm pretty sure!

[@MathVisualProofs]

@@columbus8myhw nice! I play around with desmos from time to time. I’ll check for this