What can be built using polygons?

[@NStripleseven]

There’s actually many irrational collections that are buildable. For instance, two lines, one rotated 1/sqrt(5) of the circle away from the other, form a clearly buildable, yet irrational collection. This works because for any p and I sub p you can pick, its average contains the opposite point, and so works out to be the center.

[@TheGrayCuber]

Yes, this is a great point! I think ‘most’ buildable collections are irrational. Assuming finite points and equivalence up to rotation, there is a countable amount of rational buildable collections. But even just considering 4-points irrational buildables as you pointed out - there are an uncountable amount.

[@minamagdy4126]

What was proved around 7-8 minutes in is that it's impossible to have a buildable set of points with a point that has no other point a rational portion away from it. A buildable set must always have all points be partnered up with others such that the partner is a rational distance from the original. However, there can absolutely be sets of points that are buildable where there are points an irrational distance apart. The simplest example is the two opposite-point pairs which are irrational distances apart. I believe that a buildable set must have all rational closures (sets of points that are all rational distances apart) are centered, although arbitrary unions are possible from there. The converse (edit: watched more of the video) can also be proven (that all such sets are centered).

[@NStripleseven]

@@minamagdy4126 true, though that’s not what was said. I’m just clarifying.

[@BritishBeachcomber]

It's all about how to balance a centrifuge.

[@7177YT]

Those collections you chose study are neat, the questions you posed quite interesting, and your investigation intellectually satisfying.

[@levav8]

galois theory, done in a way a highschooler can understand. amazingly well done!

[@zalibecquerel3463]

I like the graphics and the layout ofyour presentation. What tool or software did you use?

[@TheGrayCuber]

Thanks! This was all made in Google Slides

[@pridepotato314]

@@TheGrayCuberwoah

[@octag0nist]

I hit the like button too early in the video because now we're talking about building polynomials and I'm even more excited then before but don't have any way to easily express it; and thus have been forced to use the treacherous comment function.

[@samueldeandrade8535]

Wait a second ... THIS IS VERY COOL!!!

[@graf_paper]

Oh I am so eager to investigate centered collections on the surface of a sphere! Trivially all of the properties of centered collection in 2D generalize to 3D, but I don't have any intuition about what more interesting configurations might exist!

[@guilhermegondin151]

This is so amazing, it let me wondering if there is such a way to make all the contructible shapes instead of regular polygons, maybe that could give some insight into transcendental numbers.

[@younscrafter7372]

0:01 My guess: the points are evenly spaced in the sense that they consist of several sets of points, each of which is rotationally symmetrical. This also means that if these points had mass, the center of mass would be in the middle of the circle. I'm assuming this based on a video about balancing 7 tubes in a 12 slot centrifuge

[@shy_dodecahedron]

Yes. There's a non-bailable centered collection. Consider putting two vertical lines πr/4 from the centre where r is the radius. The dots from intersection are centered, but not buildable

[@TheGrayCuber]

That collection will form a rectangle, so we can build it using two pairs of opposite points.

[@shy_dodecahedron]

@@TheGrayCuber but how one could go about getting those non-algebraic coordinates?

[@konraddapper7764]

N=5 is the Minimum number of Points for a non buildable collection brause N=1 does Not Work AS Seen in the Video N=2 call the Points A=(1,0) (roate if requierd) given It is centerd B has to bee (-1,0) thus it is constructeble N=3 Let A =(1,0 ) -> B=(bx.by) C=(cx cy) With by+cy=0 and bx +by=1 -> by= -cy =y Given B,C are on the circle WE get that bx= +-sqrt(1-y^2) =x and the Same for cx Obviusly cx hav to have the Same sign in oder to add Not add to Zero and x=-1/2 else they would Not add to one -> B=(-1/2, sqrt(3/4)) similar for C Thus all cases are perfekt triangels this buildable. N=4 similar results in only perfekt rectangels. All of them are buildable As a Set of two balanced points. And N=5 is possible as shown by the example in the Video.

[@dominiquelaurain6427]

Nice video. It's related to Centrifuge problem (Numberphile has a video with Hannah Fry explaining it).

[@timofeysobolev7498]

24:06 I think Answer is 3 points, because we can distribute points in such way that them will be in one half of circle

[@jamesm8136]

This is so interesting!

[@m9l0m6nmelkior7]

I love this ! Allowing for negative points seems like it makes this a vector space over R/2piR ? or R/4piR ? idk

[@nadyanabahi8259]

bless this channel, did you find this yourself?

[@TheGrayCuber]

Yeah! The division property and indentity formulas of cyclotomics are well known, but this problem of building collections of points is something I thought of

[@Ragecardo]

Thank you for sharing your research

[@felixstrider]

Another commenter stated "A buildable set must always have all points be partnered up with others such that the partner is a rational distance from the original." Why impose this arbitrary limit? If we allow non-rational rotation of polygon's points, and allow the removal of balanced, polygon-based subsets of points, wouldn't all centered sets of points be buildable using a potentially infinite number of shapes and point nullification?

[@felixstrider]

i now notice the thumbnail in the top right of my screen stating "anything is buildable". lol

[@miruten4628]

Just to nitpick (or because I like edge cases) I think Phi_1(z) = z - 1 should be interpreted as the empty collection, i.e. not having any points at all. Since in this case z = 1, so so it's a positive and a negative point at position 1 cancelling each other out. (Btw, Phi_1 doesn't actually matter for the proof, since we're only building collections of ≥2 points. Really, the base case could just be the primes.)

[@debblez]

So we have seen that the roots of cyclotomic polynomials can get you any rational/buildable collection. if we allow ourselves to use the roots of any (integer) polynomial can we then make any centered collection? Or are there even transcendental collections??

[@CYXXYC]

as a programmer it kinda stuck out to me that the "negative" points or polygons could be nicely replaced by a XOR (addition modulo 2) operation, so when two points match, 1 XOR 1 = 0

[@TheGrayCuber]

That would handle a negative meeting a positive, but it only allows one point at each location. I intentionally called these ‘collections’ instead of sets to allow for duplicate points. There doesn’t technically need to be negative points - they could just be rotated 180 degrees to become positive, but I found it easier for visualization to have a negative and positive point at the same spot cancel as opposed to having two opposite positive points which get removed by subtracting a 2-gon

[@CYXXYC]

@@TheGrayCuber oh yeah, getting rid of a point that, for example, a 2gon and a 3gon would share would uncenter the resulting collection

[@angelmendez-rivera351]

@@TheGrayCuber The formal term you are looking for to describe the concept is a multiset. But yes, describing them vaguely as "collections" is perfectly fine for the purposes of this video.

[@TheGrayCuber]

Thank you, that is great to know!

[@localidiot4078]

3 points can only be balanced im almost sure. as for 4, i suspect not, and it probably has something to do with not being able to find an equation for a polynomial of degree 5. you need to be able to "split" the points of either a line or a triangle in a way that i dont think you can do

[@quentind1924]

3 points can easiely not be balanced. Place all 3 near the top of the circle and it doesn’t work for example

[@DinoMomPlays]

Is this investigation published in a journal anywhere? It certainly seems publishable and it just has that flavor of novel mathematics that will have unexpected utility somewhere down the line.

[@TheGrayCuber]

I haven't published this is a journal.

[@alex_zetsu]

You started with a series of points that are not evenly spread in the "traditional" sense but not only do those points all lie on a circle but their center of mass is the circle's central point. Is there a special term for a series of points arranged in this way? Or is "series of points whose center of mass is a point equidistant to all the points" is the only way to call it? I mean we have a special term for a polygon whose vertices lie on a circle, that is they are cyclic polygons.

[@KookyPiranha]

tip: set playback speed to 1.5x

[@johnydl]

1 isn't possible cause it's not centred 2 isn't possible because any deviation from the opposite positions isn't centred 3 isnt' possible because any deviation from the equilateral triangle isn't centred 4 may have the necessary degrees of freedom, but I cannot produce a counter-example, starting from an arbitrary 3 point collection that doesn't contain a pair in opposition generates a balancing point off the circle. I don't believe 4 points is possible either but I can't prove it. There may be a pathological singular counter example I cannot think of. 5 is the first arrangement where there is definitively the necessary degrees of freedom and you can generate an infinite number of counter examples. Start with a point at (0,1) and then for every point (x,y) there's another point (-x,y) for every point (p,q) where q

[@ntuneric]

i think 4 is always 2 pairs of opposite points

[@TheGrayCuber]

For the problem of 4 points, it helps to fix one of the points as some p. This reduces the problem to finding three points that sum to -p with none of them being -p, or otherwise showing that this is impossible.

[@ntuneric]

1st point can be fixed at 1 without loss of generality

[@johnydl]

@@TheGrayCuber Draw a circle centred at O with unit radius, point A at (1,0), for point B on the circle (not A') move point C around the circle and plot the possible locations for D (as the result of O-(a+b+c) where a, b, c are the vectors to A, B, C). The result is a ghost circle which intersects the unit circle at A' and B', when D is at A' then C is at B' and vice versa. I can show this result through trial and error I just have difficulty proving it (I last did proofs 15 years ago) and this doesn't fully prove no pathological cases exist.

[@Zicrus]

4 has to be a rectangle, which is obviously buildable: Choose 2 of the points arbitrarily. Assume WLOG that they are not opposites (if no such pair exists, it is clearly buildable). Rotate the circle (with center at 0) in the complex plane, so one point is the complex conjugate of the other. Now, for it to be centered, the two remaining points must also form a conjugate pair (otherwise, the average would not have a 0i component). Furthermore, their real part must be negative the real part of the first pair (since both pairs are conjugate pairs, and the average of their real parts must be 0 to be centered). Since the second pair must lie on the circle, they only have one valid pair of positions with this real value, which forms a rectangle with the first pair.

[@alanspyglass3304]

There are still irrational numbers that when summed they can be rational. Does that change how you define the I_p collection?

[@dressedmolerat]

Go hyperlegible!!

[@TheGrayCuber]

en soweli sitelen!

[@Ykulvaarlck]

the 5 points thing doesn't have anything to do with galois, right...?

[@TheGrayCuber]

Not directly. Galois theory also uses cyclotomic polynomials, but the result about fifth degree formula is focused on the symmetry of the five points evenly spaced around the circle. Those five points are buildable in the sense of this video, but their symmetry is not 'solvable' which means they cannot be written in a closed form using only arithmetic and nth roots

[@liamismath1]

15:18 the profile picture

[@izme1000]

I'm having trouble with "remove the triangle of points". Removing is not the same as adding, and can skew the results.

[@TheGrayCuber]

If you begin with a collection of points that sum to 0, and remove points that also sum to 0, the sum of the result must be 0 - 0 = 0. In order to do this, we either need to require that the removed points already existed in the beginning set, or otherwise allow ‘negative’ points, but either of these conditions will allow subtraction as a valid operation

[@izme1000]

​@@TheGrayCuber I went back and looked at your example. The issue that was bothering me was the line was no longer centered. None of the polygons are individually centered anymore, but the collection as a whole is. It just messes with your brain visually.

[@blockshift758]

7:10 off center day ruined 😔 literally unwatch able. (Great video actually❤)