it can be hard to follow in realtime. I need to come back to it later when I'm dedicated enough to follow along and really think about it. But it has already shown me a relationship between modular arithmetic and complex multiplication that I hadn't considered.
7:36 "only addition, subtraction, multiplication, division and square root" - why only those? Somehow you skipped over the most interesting part. I can guess, that the proof that no other operations are possible with compass and ruler is hard, but you could have said at least something about it.
@@mntrmntr the thing is i believe the proof isn't even that difficult. Since the only objects you have a circles and lines you just need to show that their intersection can be represented as only square roots, and +,-,*,/ of the center of the circle, the radius, and two known points on the line
"Squaring the circle" is an idiom I've heard but I had no idea it was an actual thing in maths. I had to rewind like 8 times to realise you're not using the idiom and I should look up what that means.
Loved the video, it's really well done but I think you made a small mistake. You called the modulus/length the argument, when instead the argument is the angle.
@@mathkiwiI appreciate you calling the modulus the argument as an Easter egg to see who was really paying attention. Nicely done. I love additive things like this. Just a few simple operations, but they hide incredible complexity and diverse application inside those deceptively simple steps. Cheers!
@@deserado11it's about complex numbers. The modulus is the distance from the origin on the complex plane, and the argument is the angle from the origin to the point on the complex plane.
This video went from elementary to mind blowing and iluminating real fast, especially if the viewer knows its fair share of undergraduate math. I love when different branches of math are shown to be just different faces of the same dice.
Here is my explanation of the construction of the pentagon in the video using the Carlyle circle for the appropriate quadratic (which turns out to be x²-(-1)x+(-1)=0), the key aim being the construction of the length cos(2π/5). The 5th roots of 1 are ζ⁰, ζ¹, ζ², ζ³, ζ⁴. As powers of 2 (mod 5) are 2, 4, 3, 1 we can order the powers of ζ according to the powers of 2 (instead of the powers of 3 used for the 17th roots of 1): ζ¹→ζ²→ζ⁴→ζ³ This time, the conjugate of each number is 2 places after the number. By symmetry, ζ⁰+ζ¹+ζ²+ζ³+ζ⁴=0 So ζ¹+ζ²+ζ³+ζ⁴=-ζ⁰=-1 and ζ¹ζ²ζ³ζ⁴=ζ¹⁰=1 Now split the sequence ζ¹→ζ²→ζ⁴→ζ³ in half, taking alternating terms: η₀=ζ¹+ζ⁴. η₁=ζ²+ζ³ Note that η₀, η₁ are real as both are the sum of a 5th root of 1 and its conjugate. In particular, η₀=2Re(ζ)=2cos(2π/5). η₀+η₁=ζ¹+ζ⁴+ζ²+ζ³=-1 η₀η₁=(ζ¹+ζ⁴)(ζ²+ζ³)=ζ³+ζ⁴+ζ¹+ζ²=-1 So η₀, η₁ satisfy x²-(-1)x+(-1)=0 This quadratic can be solved using the Carlyle circle construction (see en.m.wikipedia.org/wiki/Carlyle_circle ), with a circle having as diameter the line segment joining (-1,-1) to (1,0), so centre (-½, 0), and indeed marking this point is the first step in the pentagon construction in the video. The next step is marking the radius of the Carlyle circle, from (-½, 0) to (1,0), followed by marking η₀, the positive root of the quadratic (the roots are the points where the Carlyle circle, with centre and radius just found, intersects the x-axis) which equals 2cos(2π/5), and halving this gives cos(2π/5), the x-coordinate of ζ. Finally we construct the point ζ as the intersection of the perpendicular to the x-axis at η₀=cos(2π/5) with the circle in the 1st quadrant.
For the nonagon, I had to think on it a bit to realize that you can "combine" two polygons like the pendecagon if the constituent polygons' side counts are coprime (since then you dont need to worry about accidentally using the same vertex twice). Great video!
This means that whether there are infinitely many odd-sided constructible polygons or not is still an open question because it's not known whether there are infinitely many fermat primes or not.
This is an amazing video, especially if the viewer is already mathematically minded and/or knowns at least some advanced highschool or undergrad level mathematics! It's challenging at points, but in the way all good maths is. I have a couple of suggestions from a didatic point of view, but feel free to take them or leave them: 1) It would be good if you showed at least a quick graphic method for each operation you talk about or at least the crucial non-obvious step to get there. For example, angle addition - it is really straightforward if you know how to copy angles, but if someone is new to the topic they might not have the intuition/technique necessary to know where to start. 2) I really like the bits where you encourage exploration! For more viewers to engage with these moments it might be worth giving those that are newer to the field a tip or direction about where to start. This can be done a couple moments after you pause, so that it doesn't spoil it for those who don't want tips. Once again, great video! Really surprised to see a video with such production quality and such good content in a channel with relatively few subscribers. Edit: Beautiful title by the way, it perfectly manages to make the video really enticing without feeling clickbait-y in any negative sense!
I agree with all the above. I found your pace somewhat rushed. I know a 20 minute video gets less than half the views of a 10 minute one. But what you're communicating is worth extra time. I think 3B1B has the opposite flaw, he tends to over-repeat things. I wish I could listen as fast as you think I can. ;)
4:25 i don't know if anyone has mentioned it yet, but this animation is incorrect. it implies that the length OB is being conserved to BX while the length OA is conserved to AX, which would result in a kite, not a parallelogram
9:15 zeta^35 would be zeta^1 15:39 it might be nice to explain that 2^k + 1 can't be prime if k has any odd factors, because x+1 divides x^k + 1 when k is odd otherwise, great video!
The problem of constructing with 'compass and straightedge' is indeed easy of we change the definition of 'straightedge' -- hence changing the definition of the problem. Cheers!
@@NoActuallyGo-KCUF-Yourself This works to solve the original problem only if solving the engineering problem helps to solve the original problem. Does it in this case?
The question of what additional regular polygons could be constructed with gaining a method to trisect an angle is very open-ended as the method is not specified. Judging based on the one method I know of making constructable angle trisection possible (allowing for folds, folding points on to each other, and folding points on to lines, which other than drawing a circle or arc, also covers all of the possible constructions with straightedge and compass), though, you would also gain the possibility to solve cubics, meaning that you could have any prime n where n-1=2^a*3^b for some integers a and b. You'd also, of course, have an arbitrary power of 3 as an option in your polygons just like bisection gives you 2.
It turns out that you can solve cubic equations with just straight edge, compass, and angle trisector. Then instead of working with powers of 2 and products of fermat primes, you deal with powers of 2 and 3, and products of pierpont primes (p = 2^m * 3^n + 1) That said, the 11-gon is still impossible to construct.
This is an excellent video and explanation. I have known for a long time which polygons can be constructed with straight edge and compass, but now I finally know why. At least, I understand why the constructable polygons are constructable. However, I'm not sure that I can see why in all other cases the polynomials cannot be reduced to solving successive quadratics.
The main focus was what polygons are constructible. For the other ones, their polynomial can't be reduced to a quadratic one. For n=7, you get degree 7->6->3.
@@mathkiwi I get that. BTW, this was an excellent practical example of field extensions by quadratic roots. I guess kind of heading towards Galois theory...
I actually learned this construction by Gauss when I first learned Geometry in college, and I was so amazed. Next year when I was learning Galois Theory it came up again and had a deeper understanding of it's constructibility. The fact that I couldn't share how amazing this is with my highschool students was the reason I couldn't continue as a teacher anymore 😂
When I was a child I was shown how to draw polygons. Draw a circle, line out the diameter, and draw a line from left edge upwards anywhere so it hit hit perimeter. Divide that line into n segments where n was the number of sides you wanted. (It's easily done with a pencil, ruler and set square). Then you drew a line from the centre through the 'second' mark on the line and where it hits the circumference make a Cord from the left diameter point. You then have your side. Try it 🙂
This appears to be a constructible method for finding a chord on the circle given any n. Since an arbitrary n-gon is not constructible, it's not possible that it always produces the chord for an n-gon. But I was hoping that it produced a close-enough approximation for some range of n, so I wanted to understand the method to see why it might work approximately. Unfortunately, I wasn't able to follow your directions. I tried coming up with an interpretation of "draw a line for left edge upwards anywhere so it hit[s] perimeter", but didn't produce results that are even approximately correct. Here's my (failed) interpretation: Given a circle with center O and point E on the edge, draw the diameter line OE. Pick a point P on segment OE and draw a line perpendicular to OE at P and label an intersection of this line and the circle OE as T, creating segment PT.[*] Find point F on PT that is 2/n of the distance from P to T (so PF:PT :: 2:n). Find point C which is the intersection of the ray OF and the circle OE, making chord EC, which is supposed to be one of the sides of the n-gon inscribed in the circle OE. [*] This sentence corresponds to the part I couldn't interpret. The above fails not only because the choice of P is arbitrary and affects the size of the chord greatly, but because even a "reasonable" choice for P (bisecting OE) produces a chord that is way too short. Can you clarify and correct the above to describe the actual method? In particular: Is the "perimeter" for the circle OE? What is "upwards" mean here? What flexibility does "anywhere" describe? And what does "left edge" reference? NB: I interpreted the "perimeter" as being the perimeter of the circle, "upwards" to mean perpendicular to the diameter OE, "anywhere" to indicate the arbitrary choice for point P where the perpendicular is drawn, and "left edge" to be the left segment of the diameter, which I chose to be segment OE. Thanks
@@mjeffery Sorry about my sloppy language. Also my memory was clearly a bit foggy. I google and found this video of what I so poorly described. I'd be very interested in seeing your results, perhaps a bound on the error? Best Regards ... ruclips.net/video/SBdSHOiMJS4/видео.html
@@alphalunamare OK, interesting. This method takes the diameter of the circle AB and picks point X on 2/n of the way from A to B (so AX:XB :: 2:n). It also picks point P to form equilateral triangle ABP. Point C is formed by intersecting the line PX with the circle so that P and C are on opposite sides of the diameter AB. This is clearly exact for a square, and with a small amount of work can be shown to be exact for a triangle, which immediately implies that at works for a hexagon, too. It does not work for any other regular n-gon. For a pentagon, it undershoots the angle by about .065%. For a heptagon, it overshoots the angle by .17%. For an octagon, it overshoots by .42%. By the 25-gon, it has an error of 4.4% and overshoots enough that it doesn't manage to even get 24 sides. (These errors were found numerically.) As a physical method to draw a good approximation of an n-gon with a small n, it's pretty good, but as expected it isn't a correct general method for constructing one. Nor does it do a good job as n grows larger.
@@mjeffery I am impressed! :-) Here's a question:- If you plot the error as n grows far greater than 25 do you notice a cyclicity in its graph? Another consideration is to make an adjustment to the 'second' point through which the penultimate construction line is drawn, but that smells of epicycles. :-)
Really cool video. The way the roots of unity came up and the recursive quadratics are super reminiscent of FFT . My only wish is that you really dive into why nth roots are not constructible (except for n=2^k). Every time people talk about what constructible numbers are they explain how you can construct+,-,*,/,and sqrt but they don't explain why you can't get for example e^x or cube root(x)
Why do we choose the way to split the zetas into the etas and then the mus? Not quite know where these comes from, and don't know how those summations (the 3^(nk+i) ones) comes about
It is very fascinating that only very recently, the constructive methodology of continuous geometry has been complemented with origami, by which trisection of angle, regural heptagon etc. can be constructed.
I think a video on angle trisection would be a wonderful one for you to make. There’s a plethora of great math there and it can explain why you can construct the heptagon and nonagon and still remain in the complex plane.
I think it would be possible to construct any regular polygon if we were allowed to move the "compass" in three dimensions. it would be similar to the process of dividing a line segment into equal parts. a tridimensional compass should be able to divide any angle in an arbitrary number of equal parts, so through a similar process it could also create any regular polygon
My guess with having trisection would be that it expands to cubics so you can find any n-gon where n is a prime and n-1 is a power of 2 times a power of 3. Also can multiple by 3 in addition to 2 of course.
Yeah, that wasn't really explained well, or at all, on how you can get a 90 degree angle with the tool mentioned in the very beginning. Edit: eventually I was able to make a square. It's a lot of steps though. Maybe I did extra steps, but it's possible.
A Vesica Pisces allows the bisection of any base line at 90° and an identical circle to those used for the Vesica and originating at that intersection of the base line gives four equi-distant points on the circumference of that circle….(from a non-mathematician). Probably irrelevant.
Probably not relevant anymore, but still here is an explanation: (1) At 0:49 we are able to see three points on the screen, all sitting on one line. Let us call the most left one of these points A and the most right one B (both of them also on the circle depicted). Let us call that segment of the line that goes through A and B, wich lies inbetween the points A and B AB. (2) We can draw a circle with center A and with a radius length same as the length of AB. We can also draw a circle with center B and with an radius of of the same length as the segment AB. Let us call these two circles C1 and C2. (3) C1 and C2 intersect in two points. Draw the line, on wich these two points lie. This is the line we where searching for.
I don't know what the rules are, I've somehow managed to go my whole life without being familiar with the basic graphing tools like compasses, and even rulers.
@flameofthephoenix8395 you have a compass and a straightedge. The straightedge let’s you draw a straight line between any two points. The compass let’s you draw a circle centered at a point going through any other point.
I've never seen the neusis construction as any more artificial than the compass and straightedge. Neusis-constructible polygons then include the heptagon, nonagon and tridecagon for example. It's an open problem whether the icosipentagon is neusis-constructible
There were two pints created with green intersections at exacly 2:58 and 2:59. What are they constructed after ? what method was used to obtain these two points?
Those are just the midpoints between the origin and the previous points. Their construction is actually shown at 13:02, and explained (although a bit rushed) in the few seconds before. Cheers!
Probably breaks some rule , but N-secting could be done by making a long strip with a line and compass marking N successive points, then bending the strip into a full circle and transferring the points to a planar circle. The angles could then be copied to the angles needing dividing.
Though then you have introduced a new tool. Also, the mathematical construction would have to first lay your strip on an arc segment of the angle then straighten it out, n-sect the line segment on your strip which is now a line (a rather straight forward process), and then moving the strip back to the arc to n-sect the angle.
This video proves that you can definitely construct a polygon as long as it is a multiple of some power of 2 and some fermat primes. But I don't think it proves that others cannot be constructed for sure, does it? I guess that will require a proof that you can't exponentiate with a rational number unless the denominator is a power of 2. Since |cosine of the angle|
yea that makes sense. Just wanted to point out that it wasn't explicitly stated in the video why cuberoots or higher order roots aside from powers of 2 are not constructable
@@capnabulel It has to do with the 'degree' of the constructed points. The degree of a constructible point is always a power of 2, but the degree of the cube root of 2 is 3. To fully define degree and prove its properties requires a little bit of linear algebra.
The heptagon is not constructible because simply; 7 is not a Fermat prime whereas 17 is. A regular polygon is only constructible if its sides equal 2 to the power of (n), like 4, 8, 16 and so on, or if its sides equal one of the 5 Fermat primes (3, 5, 17, 257, 65537).
You can trisect a line segment or indeed divide it into any number n of equal parts. What you can't do with a straight edge and compass is trisect an arbitrary angle (you can trisect some angles, like 90°).
@@MichaelRothwell1There actually is a way to trisect an angle! You just bisect it, then bysect the section, then you bisect the section of the previous section and so on. Then you just add up the parts like digits in a number until you grt close enough. In fact you are only limited by how much time you are willing to waste and by the compass and paper you use. If you allow for infinite bisections then dividing any angle by any value even including irrational ones immediately become possible at least in theory.
@@Kirillissimus It's true that you can approximate the trisection of an angle as closely as you want, but when one talks about a construction using a straight edge and compass what is meant is an exact construction requiring a finite number of steps (and not the limit of a procedure with an infinite number of steps).
I have a small question which I might know the answer to but I want to confirm. In the end, we concluded that any polygon with (2^k)p_1...p_t sides, where k,t > 0 is constructable. But why is 4 also on this list? My guess at an answer is simply that we don't use the same method of constructing a square than we do for the others? Another question, but that might be because it's 2AM and I'm way too tired: are these numbers the only ones we can construct? Are there others that we may be able to construct in a different way?
Wonderful video, but you might want to replace the auto generated transcript with a hand written one... every time you said Fermat, it guessed something utterly wrong.
@@Kirillissimus sure it does need infinite digits, but those infinite digits are just a pattern of a finite amount of digits repeated infinitely many times
Dear friends, thank you very much. You have opened spaces for me, that I haven't been in, for long time. Theoretical mathematics superb poetry. The lofty flight of our mind. Severall of your posts prove, that the same mind has quantumnes, and that regardless. Srodingers electron, and that there can be more space simultaneously ontologically condensing them. Your animations are perfect. I subscribe you, with pleasure.
What I dislike about current state of math most - is goddam Eta, Theta, and the rest of the Zoo, that for some reason is still kept alive. Let the dinos die out, they are history
3:07 it’s unclear how you determined where these points should lie. Plus if you’re going to talk about constructing polygons, it may be a good idea to show the basic techniques, such as how to construct a perpendicular line. Very few people ever have to construct geometric constructs manually these days, so few will remember it if they’re even taught it at all.
Those points seemed a bit arbitrary to me, too, but when he goes over the proof at the end and where the construction comes from, the construction of those points is actually shown. Steps are still skipped, though.
Nice vid. Just a note: music is a bit loud and distract from what is important: the voice. For intro/outro is ok, but for an entire video hasn't meaning.
There is always a way to construct any polygon. Any number a side can be, it is constructible., the video is wrong. And if so, why haven't you told that the hexagon isn't perfect? Like by 1/10*? Not hating, just saying that you shouldn't assume anything by mathematical proof.
Nach dem Akzent zu urteilen sind Sie Deutscher. Hier ein kleiner Englisch-Tip: „not constructible as well“ ist falsch. Es heißt „not constructible either“.
Using a 90deg angle is kinda cringe. If you want to draw a right angle ABC where point B is the vertex, A and B are the original points on the line, just construct a point D using a compass and a ruler on the the same line as AB but on the other side, then construct two circles with the same radius on A and D with any length greater than AB, then use the two intersections of the circles to draw a line which is 90 degrees with line AB. That way you only need two tools.
@@zapazap Unwatchable to anyone who can't block out the background music, which adds nothing to help the content and plenty to distract from it, Chhers!
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Hi, I hope you liked my video!
Please leave your feedback, questions, etc. in the comments.
Also, please read the description.
it can be hard to follow in realtime. I need to come back to it later when I'm dedicated enough to follow along and really think about it. But it has already shown me a relationship between modular arithmetic and complex multiplication that I hadn't considered.
7:36 "only addition, subtraction, multiplication, division and square root" - why only those? Somehow you skipped over the most interesting part. I can guess, that the proof that no other operations are possible with compass and ruler is hard, but you could have said at least something about it.
You really should use a black background with white lines, like 3Blue1Brown
@@mntrmntr the thing is i believe the proof isn't even that difficult. Since the only objects you have a circles and lines you just need to show that their intersection can be represented as only square roots, and +,-,*,/ of the center of the circle, the radius, and two known points on the line
"Squaring the circle" is an idiom I've heard but I had no idea it was an actual thing in maths. I had to rewind like 8 times to realise you're not using the idiom and I should look up what that means.
Loved the video, it's really well done but I think you made a small mistake. You called the modulus/length the argument, when instead the argument is the angle.
You're totally right
@@mathkiwiI appreciate you calling the modulus the argument as an Easter egg to see who was really paying attention.
Nicely done.
I love additive things like this. Just a few simple operations, but they hide incredible complexity and diverse application inside those deceptively simple steps.
Cheers!
Also I think zeta^35 should be zeta^1
... say what? 🤪 (fastenated by mathematics but can't do it to save my life 😢)
@@deserado11it's about complex numbers. The modulus is the distance from the origin on the complex plane, and the argument is the angle from the origin to the point on the complex plane.
I have a degree in engineering. This is the first time I've had any real appreciation for how classical geometry and algebra can be connected. Thanks!
This video went from elementary to mind blowing and iluminating real fast, especially if the viewer knows its fair share of undergraduate math.
I love when different branches of math are shown to be just different faces of the same dice.
Here is my explanation of the construction of the pentagon in the video using the Carlyle circle for the appropriate quadratic (which turns out to be x²-(-1)x+(-1)=0), the key aim being the construction of the length cos(2π/5).
The 5th roots of 1 are ζ⁰, ζ¹, ζ², ζ³, ζ⁴.
As powers of 2 (mod 5) are 2, 4, 3, 1 we can order the powers of ζ according to the powers of 2 (instead of the powers of 3 used for the 17th roots of 1):
ζ¹→ζ²→ζ⁴→ζ³
This time, the conjugate of each number is 2 places after the number.
By symmetry, ζ⁰+ζ¹+ζ²+ζ³+ζ⁴=0
So ζ¹+ζ²+ζ³+ζ⁴=-ζ⁰=-1
and ζ¹ζ²ζ³ζ⁴=ζ¹⁰=1
Now split the sequence ζ¹→ζ²→ζ⁴→ζ³ in half, taking alternating terms:
η₀=ζ¹+ζ⁴.
η₁=ζ²+ζ³
Note that η₀, η₁ are real as both are the sum of a 5th root of 1 and its conjugate.
In particular, η₀=2Re(ζ)=2cos(2π/5).
η₀+η₁=ζ¹+ζ⁴+ζ²+ζ³=-1
η₀η₁=(ζ¹+ζ⁴)(ζ²+ζ³)=ζ³+ζ⁴+ζ¹+ζ²=-1
So η₀, η₁ satisfy
x²-(-1)x+(-1)=0
This quadratic can be solved using the Carlyle circle construction (see en.m.wikipedia.org/wiki/Carlyle_circle ), with a circle having as diameter the line segment joining (-1,-1) to (1,0), so centre (-½, 0), and indeed marking this point is the first step in the pentagon construction in the video.
The next step is marking the radius of the Carlyle circle, from (-½, 0) to (1,0), followed by marking η₀, the positive root of the quadratic (the roots are the points where the Carlyle circle, with centre and radius just found, intersects the x-axis) which equals 2cos(2π/5), and halving this gives cos(2π/5), the x-coordinate of ζ.
Finally we construct the point ζ as the intersection of the perpendicular to the x-axis at η₀=cos(2π/5) with the circle in the 1st quadrant.
For the nonagon, I had to think on it a bit to realize that you can "combine" two polygons like the pendecagon if the constituent polygons' side counts are coprime (since then you dont need to worry about accidentally using the same vertex twice). Great video!
Thanks!
You can probably extend this logic further and conclude that using n-gram and k-gram you can contruct m-gram where m=LCM(n, k)
This means that whether there are infinitely many odd-sided constructible polygons or not is still an open question because it's not known whether there are infinitely many fermat primes or not.
This is an amazing video, especially if the viewer is already mathematically minded and/or knowns at least some advanced highschool or undergrad level mathematics! It's challenging at points, but in the way all good maths is.
I have a couple of suggestions from a didatic point of view, but feel free to take them or leave them:
1) It would be good if you showed at least a quick graphic method for each operation you talk about or at least the crucial non-obvious step to get there. For example, angle addition - it is really straightforward if you know how to copy angles, but if someone is new to the topic they might not have the intuition/technique necessary to know where to start.
2) I really like the bits where you encourage exploration! For more viewers to engage with these moments it might be worth giving those that are newer to the field a tip or direction about where to start. This can be done a couple moments after you pause, so that it doesn't spoil it for those who don't want tips.
Once again, great video! Really surprised to see a video with such production quality and such good content in a channel with relatively few subscribers.
Edit: Beautiful title by the way, it perfectly manages to make the video really enticing without feeling clickbait-y in any negative sense!
Thank you so much for the detailed feedback!
I agree with all the above. I found your pace somewhat rushed. I know a 20 minute video gets less than half the views of a 10 minute one. But what you're communicating is worth extra time. I think 3B1B has the opposite flaw, he tends to over-repeat things. I wish I could listen as fast as you think I can. ;)
Really nicely animated video! It also reminds of a puzzle game I used to play called Euclidea
4:25 i don't know if anyone has mentioned it yet, but this animation is incorrect. it implies that the length OB is being conserved to BX while the length OA is conserved to AX, which would result in a kite, not a parallelogram
Yes. This error is acknowledged in the video's description.
9:15 zeta^35 would be zeta^1
15:39 it might be nice to explain that 2^k + 1 can't be prime if k has any odd factors, because x+1 divides x^k + 1 when k is odd
otherwise, great video!
Just make a ruler that has transcendental & irrational numbers on it, but the ruler is a fractal so it's not just an appromixation. Easy.
The problem of constructing with 'compass and straightedge' is indeed easy of we change the definition of 'straightedge' -- hence changing the definition of the problem.
Cheers!
Step 1. Turn it into an engineering problem.
Step 2. Solve the engineering problem.
@@NoActuallyGo-KCUF-Yourself This works to solve the original problem only if solving the engineering problem helps to solve the original problem.
Does it in this case?
Well-done video. Speaking of 7 and 17, did you guys notice that the year this video was published: 2023 = 7*17^2
Perfect timing
Oh yes it is
The question of what additional regular polygons could be constructed with gaining a method to trisect an angle is very open-ended as the method is not specified. Judging based on the one method I know of making constructable angle trisection possible (allowing for folds, folding points on to each other, and folding points on to lines, which other than drawing a circle or arc, also covers all of the possible constructions with straightedge and compass), though, you would also gain the possibility to solve cubics, meaning that you could have any prime n where n-1=2^a*3^b for some integers a and b. You'd also, of course, have an arbitrary power of 3 as an option in your polygons just like bisection gives you 2.
It turns out that you can solve cubic equations with just straight edge, compass, and angle trisector. Then instead of working with powers of 2 and products of fermat primes, you deal with powers of 2 and 3, and products of pierpont primes (p = 2^m * 3^n + 1)
That said, the 11-gon is still impossible to construct.
This is an excellent video and explanation.
I have known for a long time which polygons can be constructed with straight edge and compass, but now I finally know why.
At least, I understand why the constructable polygons are constructable.
However, I'm not sure that I can see why in all other cases the polynomials cannot be reduced to solving successive quadratics.
The main focus was what polygons are constructible. For the other ones, their polynomial can't be reduced to a quadratic one. For n=7, you get degree 7->6->3.
@@mathkiwi I get that. BTW, this was an excellent practical example of field extensions by quadratic roots. I guess kind of heading towards Galois theory...
I actually learned this construction by Gauss when I first learned Geometry in college, and I was so amazed.
Next year when I was learning Galois Theory it came up again and had a deeper understanding of it's constructibility.
The fact that I couldn't share how amazing this is with my highschool students was the reason I couldn't continue as a teacher anymore 😂
No te subestimes, de todas formas, no es fácil explicar la construcción de esto:14:53
When I was a child I was shown how to draw polygons. Draw a circle, line out the diameter, and draw a line from left edge upwards anywhere so it hit hit perimeter. Divide that line into n segments where n was the number of sides you wanted. (It's easily done with a pencil, ruler and set square). Then you drew a line from the centre through the 'second' mark on the line and where it hits the circumference make a Cord from the left diameter point. You then have your side. Try it 🙂
This appears to be a constructible method for finding a chord on the circle given any n. Since an arbitrary n-gon is not constructible, it's not possible that it always produces the chord for an n-gon. But I was hoping that it produced a close-enough approximation for some range of n, so I wanted to understand the method to see why it might work approximately.
Unfortunately, I wasn't able to follow your directions. I tried coming up with an interpretation of "draw a line for left edge upwards anywhere so it hit[s] perimeter", but didn't produce results that are even approximately correct. Here's my (failed) interpretation:
Given a circle with center O and point E on the edge, draw the diameter line OE. Pick a point P on segment OE and draw a line perpendicular to OE at P and label an intersection of this line and the circle OE as T, creating segment PT.[*] Find point F on PT that is 2/n of the distance from P to T (so PF:PT :: 2:n). Find point C which is the intersection of the ray OF and the circle OE, making chord EC, which is supposed to be one of the sides of the n-gon inscribed in the circle OE.
[*] This sentence corresponds to the part I couldn't interpret.
The above fails not only because the choice of P is arbitrary and affects the size of the chord greatly, but because even a "reasonable" choice for P (bisecting OE) produces a chord that is way too short.
Can you clarify and correct the above to describe the actual method? In particular: Is the "perimeter" for the circle OE? What is "upwards" mean here? What flexibility does "anywhere" describe? And what does "left edge" reference?
NB: I interpreted the "perimeter" as being the perimeter of the circle, "upwards" to mean perpendicular to the diameter OE, "anywhere" to indicate the arbitrary choice for point P where the perpendicular is drawn, and "left edge" to be the left segment of the diameter, which I chose to be segment OE.
Thanks
@@mjeffery Sorry about my sloppy language. Also my memory was clearly a bit foggy. I google and found this video of what I so poorly described. I'd be very interested in seeing your results, perhaps a bound on the error? Best Regards ... ruclips.net/video/SBdSHOiMJS4/видео.html
@@alphalunamare OK, interesting. This method takes the diameter of the circle AB and picks point X on 2/n of the way from A to B (so AX:XB :: 2:n). It also picks point P to form equilateral triangle ABP. Point C is formed by intersecting the line PX with the circle so that P and C are on opposite sides of the diameter AB.
This is clearly exact for a square, and with a small amount of work can be shown to be exact for a triangle, which immediately implies that at works for a hexagon, too.
It does not work for any other regular n-gon. For a pentagon, it undershoots the angle by about .065%. For a heptagon, it overshoots the angle by .17%. For an octagon, it overshoots by .42%. By the 25-gon, it has an error of 4.4% and overshoots enough that it doesn't manage to even get 24 sides. (These errors were found numerically.)
As a physical method to draw a good approximation of an n-gon with a small n, it's pretty good, but as expected it isn't a correct general method for constructing one. Nor does it do a good job as n grows larger.
@@mjeffery I am impressed! :-) Here's a question:- If you plot the error as n grows far greater than 25 do you notice a cyclicity in its graph? Another consideration is to make an adjustment to the 'second' point through which the penultimate construction line is drawn, but that smells of epicycles. :-)
Really cool video. The way the roots of unity came up and the recursive quadratics are super reminiscent of FFT . My only wish is that you really dive into why nth roots are not constructible (except for n=2^k). Every time people talk about what constructible numbers are they explain how you can construct+,-,*,/,and sqrt but they don't explain why you can't get for example e^x or cube root(x)
Why do we choose the way to split the zetas into the etas and then the mus? Not quite know where these comes from, and don't know how those summations (the 3^(nk+i) ones) comes about
Good video! I just find the music a little distractive, I think it is a little too high in volume
It is very fascinating that only very recently, the constructive methodology of continuous geometry has been complemented with origami, by which trisection of angle, regural heptagon etc. can be constructed.
I think a video on angle trisection would be a wonderful one for you to make. There’s a plethora of great math there and it can explain why you can construct the heptagon and nonagon and still remain in the complex plane.
What a fantastic video. Keep up the great work! ❤️🥳🙌🏼
Ahora, un vídeo de los polígonos que se pueden construir con el método Neusis.
Gauss figured out how to construct a regular 17-gon when he was 19.
I think it would be possible to construct any regular polygon if we were allowed to move the "compass" in three dimensions. it would be similar to the process of dividing a line segment into equal parts. a tridimensional compass should be able to divide any angle in an arbitrary number of equal parts, so through a similar process it could also create any regular polygon
My guess with having trisection would be that it expands to cubics so you can find any n-gon where n is a prime and n-1 is a power of 2 times a power of 3. Also can multiple by 3 in addition to 2 of course.
Beautiful animations!
Thank you for this attractive video. I would like to see the Galois group table of the 16 roots of unity. At least a link to.
I love derivations like this.
12:21 new pokemon lore just dropped
I never realised Mew's name was just a single letter
I'm having a really hard time figuring out how you got your starting point for the line that creates the 90 degree angle for the square. 0:49 .
Yeah, that wasn't really explained well, or at all, on how you can get a 90 degree angle with the tool mentioned in the very beginning.
Edit: eventually I was able to make a square. It's a lot of steps though. Maybe I did extra steps, but it's possible.
A Vesica Pisces allows the bisection of any base line at 90° and an identical circle to those used for the Vesica and originating at that intersection of the base line gives four equi-distant points on the circumference of that circle….(from a non-mathematician). Probably irrelevant.
You can divide a line in half, I think.
Probably not relevant anymore, but still here is an explanation:
(1) At 0:49 we are able to see three points on the screen, all sitting on one line. Let us call the most left one of these points A and the most right one B (both of them also on the circle depicted). Let us call that segment of the line that goes through A and B, wich lies inbetween the points A and B AB.
(2) We can draw a circle with center A and with a radius length same as the length of AB. We can also draw a circle with center B and with an radius of of the same length as the segment AB. Let us call these two circles C1 and C2.
(3) C1 and C2 intersect in two points. Draw the line, on wich these two points lie. This is the line we where searching for.
I don't know what the rules are, I've somehow managed to go my whole life without being familiar with the basic graphing tools like compasses, and even rulers.
@flameofthephoenix8395 you have a compass and a straightedge. The straightedge let’s you draw a straight line between any two points. The compass let’s you draw a circle centered at a point going through any other point.
I've never seen the neusis construction as any more artificial than the compass and straightedge. Neusis-constructible polygons then include the heptagon, nonagon and tridecagon for example. It's an open problem whether the icosipentagon is neusis-constructible
There were two pints created with green intersections at exacly 2:58 and 2:59. What are they constructed after ? what method was used to obtain these two points?
Those are just the midpoints between the origin and the previous points. Their construction is actually shown at 13:02, and explained (although a bit rushed) in the few seconds before.
Cheers!
u can do 7 side polygon. i had to do that sh*t on my engineering drawing class lol O.o
Probably breaks some rule , but N-secting could be done by making a long strip with a line and compass marking N successive points, then bending the strip into a full circle and transferring the points to a planar circle. The angles could then be copied to the angles needing dividing.
Though then you have introduced a new tool. Also, the mathematical construction would have to first lay your strip on an arc segment of the angle then straighten it out, n-sect the line segment on your strip which is now a line (a rather straight forward process), and then moving the strip back to the arc to n-sect the angle.
This video proves that you can definitely construct a polygon as long as it is a multiple of some power of 2 and some fermat primes. But I don't think it proves that others cannot be constructed for sure, does it? I guess that will require a proof that you can't exponentiate with a rational number unless the denominator is a power of 2. Since |cosine of the angle|
The compass solves a quadratic equation and the straightedge a linear one. That might be an answer
yea that makes sense. Just wanted to point out that it wasn't explicitly stated in the video why cuberoots or higher order roots aside from powers of 2 are not constructable
@@capnabulel I think it comes down to we can bisect arbitrary angles but we cannot trisect or more-sect arbitrary angles.
@@capnabulel It has to do with the 'degree' of the constructed points. The degree of a constructible point is always a power of 2, but the degree of the cube root of 2 is 3. To fully define degree and prove its properties requires a little bit of linear algebra.
16:26 - I will try to guess : 2^m * 3^n * p1 * p2 * ... * pt
The heptagon is not constructible because simply; 7 is not a Fermat prime whereas 17 is. A regular polygon is only constructible if its sides equal 2 to the power of (n), like 4, 8, 16 and so on, or if its sides equal one of the 5 Fermat primes (3, 5, 17, 257, 65537).
How do you construct a hexamyriapentachiliapentacosiatriacontaheptagon?
How did you used 3b1b music? I thought that was copyrighted.
Does the 17 sided polygon use the 8-15-17 right triangle? That would explain it
Me parece que utiliza un triángulo rectángulo donde aparece la raíz cuadrada de 68
68=(17)(4)
very good Greek letter pronunciation! (i am Greek)
Regular heptagon can be construct by compass and scale
Underrated video!
Why would you be allowed to bisect a line but not trisect it with this setup?
You can trisect a line segment or indeed divide it into any number n of equal parts. What you can't do with a straight edge and compass is trisect an arbitrary angle (you can trisect some angles, like 90°).
@@MichaelRothwell1There actually is a way to trisect an angle! You just bisect it, then bysect the section, then you bisect the section of the previous section and so on. Then you just add up the parts like digits in a number until you grt close enough. In fact you are only limited by how much time you are willing to waste and by the compass and paper you use. If you allow for infinite bisections then dividing any angle by any value even including irrational ones immediately become possible at least in theory.
@@Kirillissimus It's true that you can approximate the trisection of an angle as closely as you want, but when one talks about a construction using a straight edge and compass what is meant is an exact construction requiring a finite number of steps (and not the limit of a procedure with an infinite number of steps).
@@MichaelRothwell1 Angle trisection is also possible if you allow the paper to be folded during the construction, but that's cheating of course. 🙂
@@X_Baron Yes indeed, using the Beloch fold :)
Contradictions are underrated!
this is how to catch my attention
Lmao I accidentally set the playback speed to 0,5 and i thought he was on drug at the start of the video 😂
Si a la lista admites la trisección del ángulo como método válido, entonces, quedan los polígonos verdaderamente difíciles de construir.
11:29 -- the so-called "Carlyle circle".
The video is really good, but I think you made the music too loud, it seems to be at an equal or higher volule than your voice
Try different speakers or sound settings.
Can't we construct cube roots in 3d space? Trying to use compass in space is awkward in practice, though.
I have a small question which I might know the answer to but I want to confirm.
In the end, we concluded that any polygon with (2^k)p_1...p_t sides, where k,t > 0 is constructable. But why is 4 also on this list?
My guess at an answer is simply that we don't use the same method of constructing a square than we do for the others?
Another question, but that might be because it's 2AM and I'm way too tired: are these numbers the only ones we can construct? Are there others that we may be able to construct in a different way?
4=2^2. And yes, these are the only numbers you can construct. That is because a line solves a linear equation and a circle a quadratic one.
@@mathkiwi but 4 = 2² doesn't fit the pattern of 2ⁿp_1...p_t. I guess it's just an exception?
I think many of your yellow line segments change length during the rotation animation. Just a heads up.
Couldn’t you just rotate the sides with an interior angle of 158.82°?
could you please provide one version of your videos without musique?
Wonderful video, but you might want to replace the auto generated transcript with a hand written one... every time you said Fermat, it guessed something utterly wrong.
that's a nice video but why can't one construct a cube root or pi?
I'm interested as to whether or not 360 is divisible by 17
360/17 has a lot of digits, it's probably irrational
@@xXJ4FARGAMERXx 360/17 is still rational
Examples of irrational numbers are stuff like √3, π and e
@@HatterTobiasEven though 360/17 is rational its precise decimal representation still requires infinite amount of digits.
@@Kirillissimus sure it does need infinite digits, but those infinite digits are just a pattern of a finite amount of digits repeated infinitely many times
isn't there a method to draw any polygon with arbitary sides?
Dear friends, thank you very much.
You have opened spaces for me, that I haven't been in, for long time.
Theoretical mathematics superb poetry. The lofty flight of our mind.
Severall of your posts prove, that the same mind has quantumnes, and that regardless. Srodingers electron, and that there can be more space simultaneously ontologically condensing them. Your animations are perfect. I subscribe you, with pleasure.
Didn't origami artists cracked splitting angle in 3?
What kind of music is played in this video? Who is the band ? Thanks so much
In the description there are some links for them.
Gutes Video mit musik
feedback: the background music is too loud
What I dislike about current state of math most - is goddam Eta, Theta, and the rest of the Zoo, that for some reason is still kept alive. Let the dinos die out, they are history
3:07 it’s unclear how you determined where these points should lie. Plus if you’re going to talk about constructing polygons, it may be a good idea to show the basic techniques, such as how to construct a perpendicular line. Very few people ever have to construct geometric constructs manually these days, so few will remember it if they’re even taught it at all.
Those points seemed a bit arbitrary to me, too, but when he goes over the proof at the end and where the construction comes from, the construction of those points is actually shown. Steps are still skipped, though.
Love the video
Slight pedantic point. With a compass and straight edge, you can also take complex conjugates.
really cool
brasilian schools were not good for me, most of it flew way over my head
This was something
7+10
3+4
I constructed it. Well it is in conatruction.
Nice vid. Just a note: music is a bit loud and distract from what is important: the voice. For intro/outro is ok, but for an entire video hasn't meaning.
There is a method to construct any number of of angles ... with just a compass and a straight edge. What are you talking about?
ruclips.net/video/eaKOyI9Srxc/видео.html
@mathkiwi
ruclips.net/video/3tqwPJsyRqM/видео.html&ab_channel=ArthurGeometry
7 sides = Heptagon
uh i actually managed to construct a heptagon
There is always a way to construct any polygon. Any number a side can be, it is constructible., the video is wrong. And if so, why haven't you told that the hexagon isn't perfect? Like by 1/10*?
Not hating, just saying that you shouldn't assume anything by mathematical proof.
Nach dem Akzent zu urteilen sind Sie Deutscher. Hier ein kleiner Englisch-Tip: „not constructible as well“ ist falsch. Es heißt „not constructible either“.
Very interesting but ain’t understandable
Dude! Please slow down! I'd much rather watch a 30 minute video where you explain every step than a 15 minute video where I get lost every 2 minutes
the f....
Using a 90deg angle is kinda cringe. If you want to draw a right angle ABC where point B is the vertex, A and B are the original points on the line, just construct a point D using a compass and a ruler on the the same line as AB but on the other side, then construct two circles with the same radius on A and D with any length greater than AB, then use the two intersections of the circles to draw a line which is 90 degrees with line AB. That way you only need two tools.
Ripping off 3b1b's music smh
Unwatchable because of the intrusive music
Unwatchable to whom?
Cheers!
@@zapazap Unwatchable to anyone who can't block out the background music, which adds nothing to help the content and plenty to distract from it, Chhers!
@@monitorcomputersystemsltd2375 I a cannot block but I could watch it So again: unwatchable to who?
I agree it is more prominent than it has to be, but I wouldnt call it unwatchable
@@zapazap I refer the honourable gentleman to my previous answer