Does this mean that, since division of a length by any positive number is possible using these tools, we can divide the circle into any natural number of equal areas with this method?
@@MathVisualProofs Yep, I just confirmed it. The reason this works is that the area of a semicircle is quadratic in r. Stepping into the world of discrete calculus, the difference between successive terms of this quadratic sequence is a linear function of r. This difference sequence corresponds to the “upper” part of each slice. Because this sequence of areas is linear, we can take the reverse of this sequence (the rotated “lower” semicircle), pair up matching pieces, and each slice now has the same area, because the constant amount the area changes in the upper semicircle is exactly negated by the change in the lower semicircle.
I think it would have been interesting to explain how you can cut the diameter in 7 pieces using the straight edge and compass, but I really like the conciseness of this video
Yes. That would have made the video quite a bit more complicated in the middle and would have possibly confused the issue. There have been enough comments about this that maybe I should do an auxiliary one showing how to do that division. Thanks for the feedback!
While you can construct any rational number length (in particular 1/7), the easier approach is to just take a circle of diameter 7 instead, starting with a given length of 1.
@@MathVisualProofs I also realized that you didnt really dive into why a pentagon was constructible and not a heptagon, which I kind of took for granted because I was taught this a long time ago while I only realized how to multiply and divide recently, even though it's arguably easier. I see why it doesnt make much sense to include it here since explaining everything would change the video completely
I can see how you can create a circle with a diameter of 7x by drawing a line, then using the compass pick a length, and repeat it 7 times on a line. This becomes your diameter. Then bisect your line using the compass giving you a radius where you can complete your circle. But that is backwards. I would also like to see how you divide the diameter into 7. Edit: I saw the solution in a post below. Fantastic. It's funny because the solution is basically doing exactly what I said except creating 2 parallel lines of 7 segments, then you transfer those proportions to the fixed line. Very interesting.
Thanks! But Whoops. I copied and pasted a title and included the # here. I'll remove it because my official SoME submission is a different video. Maybe I should have made it this one...
@@zhinkunakur4751 Yes - and it's fixed. I created the videos in the same time frame and wasn't sure which one to submit to SoME :) It appears maybe I should have submitted this one because people like it better :)
This is brilliant and beautiful! Very nicely done! At first glance, it's not at all obvious how I would divide the diameter into 7 equal segments, but I'll take your word for it that it can be done.
Not obvious at all. But you can divide a line into n parts for any positive integer n with straightedge and compass. Here is the technique: www.mathopenref.com/constdividesegment.html
Step 1: draw a line starting at one of the endpoints (aside from needing to not be parallel with the diameter, it doesn't matter which direction) Step 2: mark seven equally spaced points along that line Step 3: draw a line from the farthest point to the other endpoint Step 4: draw lines parallel to the one drawn in step 3 from the other points
If I'm not mistaken, another construction would be possible by first drawing a line through an offset point C roughly parallel to the original AB (I don't think it's a requirement, but I'm not working with the most rigor here) Then make equidistant marks on the line through C, finishing with point D Then construct line AC and BD so that they intersect at a new point E Then connect each of the points on CD to E, and where they intersect like AB will also be equidistant (like parallel lines running to a vanishing point in a renaissance art piece)
@@AllThingsPhysicsRUclips Less rigorous, more accessible: "make up two parallel lines out of X segments, draw vertical lines between the segments of the two lines, and so any slice of these new rectangles has the vertical lines equally distant. Construct it smart from the beginning so that the line you wanted to divide is one such slice." Because if you slice a metal fence, any angle you cut it, the tips of the posts will be equidistant.
Very nice! You could alternatively mark points on the radius at distances sqrt(1/7), sqrt(2/7), sqrt(3/7), ... and draw concentric circles, but this method is so much more elegant!
@@fejfo6559 you can construct the square root of any given length either the geometric mean theorem ( en.wikipedia.org/wiki/Geometric_mean_theorem ) or the intercept theorem ( en.wikipedia.org/wiki/Intercept_theorem with the segments on the two rays being 1 + s and s + x, respectively). Both are fairly simple methods, personally I lean toward to the GMT for some reasons. Alternatively, you can construct sqrt(7) by a series of right-angled triangles starting with the classical 1-1-sqrt(2) triangle (the half of a unit square) then the first leg of each next triangle is the hypotenuse of the previous one and the second leg is always 1, and then you can divide it by 7. Well you can start the series with the 1-2-sqrt(5) triangle since 2 is the largest whole number of which square is smaller than 7, but this would be much more effort than either of the first two, I think. Still, I like the (mental) image of the resulting "spiral" of triangles. On the other hand, if you start from the 1-1-sqrt(2) triangle, you'll get all the square roots up to 7 which would be useful for scaling the sqrt(1/7) if you want to construct sqrt(2/7), sqrt(3/7), sqrt(4/7) etc as well, like in the original comment suggested.
Create seven end to end line segments of equal length to make a line segment of seven equal sections, with one end positioned at the start of the line segment you want to divide, and point it off to one side so the two are arranged like a V. Add another line segment joining the ends of the v to make a triangle. Parallel to this new line segment, draw another line segment from each of the nodes joining the seven equal segments you drew earlier, such that the new lines intersect the original line segment which was to be divided. The intersection points divide the initial line segment evenly into seven. It's not obvious no :)
Alternatively to dividing a line segment into 7 equal parts is to start with a short line segment, build 6 additional line segments onto the end of it, and construct the circle from these. Technically it's not dividing the circle into 7 parts, but I think this counts.
This technique has been known for quite some time. It is Proposition 9 of Book VI of Euclid's Elements, written c. 300BCE. See aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI9.html On the other hand, the fact that you can't construct a regular heptagon with only a compass and straight edge has only been known since 1837, thanks to the Gauss-Wentzel theorem. This states that a regular n-gon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and any number of distinct Fermat primes (including none). The Fermat Primes are the primes of the form 2^2^n+1, where n is a natural number, such as 3, 5, 17, 257, 65537. These are the only Fermat primes known. So after the regular pentagon, the next few odd-sided regular n-gons you can construct are for n=15, 17, 51, 85, 255 and 257. See en.m.wikipedia.org/wiki/Constructible_polygon
@@sweetcornwhiskey Is true that shows the length 7 is constructible, but it does take a bit more to show that 1/7 is constructible. But you are right that it is perhaps technically correct because I didn't say you had to start with a particular circle :)
Although you've spoiled the answer in the video thumbnail, the idea is so brilliant I had to watch it. Thank you for bring this up. I wonder which areas are constructible given a unit area circle, besides the rationals. Wow!
😀 thanks! Sorry to spoil it early. The RUclips thumbnail game is beyond me and the hardest part about having a channel. I just want to share some visual math ideas :)
@@MathVisualProofs Maybe it was for the best, who knows, without the spoiler I could just think "No." and move on thinking it was just a video about the unconstructability of the heptagon.
Honestly, I like that the thumbnail showed the answer. At a glance, it allows the viewer to see the answer while scrolling through, yet they can still watch the video to learn why it's the answer. In this case too, the squiggly segments can pique the viewer's interest because thet might not make a lot of sense without the proof.
According to wikipedia en.wikipedia.org/wiki/Constructible_polygon#Detailed_results_by_Gauss's_theory, an n-gon is constructible when n is a product of a power of 2 with any combination of the numbers 3, 5, 17, 257, and 65537. All other n-gons are not constructible. So, for example, a 64*5*257-gon is constructible, while a 25-gon is not. If we discover any other “Fermat primes”, those primes will be added to the list with 3,5,etc.
@@MathVisualProofs I remember back in school my teacher tossed out the "trisect an angle" challenge with a compass and a straight edge. I wanted so bad to find a solution :)
@@dougsholly9323 love that! Even though it’s impossible a challenge like that can really inspire. Speaking of, at some point I will have a video that shows how to trisect an angle…. But it needs one extra tool that isn’t classical :)
I was just thinking of perspective vanishing points in art. That stuff is all about preserving angles. These lines that represent length can be vectors. That being said, any vector can be any unit. So, a perspective vector doesn't have to represent length, it can be anything.
Your initial examples of dividing the area of the circle into 2, 3, 4, 5 and 6 equally-sized pieces were a bit misleading, since those pieces were also equally-shaped, whereas in your solution for the division into 7 equally-sized pieces the pieces were not equally-shaped but different when compared to each other. So I was a bit confused first and felt a bit mislead about your solution, since I thought you were solving a different problem (= that the pieces should have to be equally-shaped). Is there a solution for the 7-division where the pieces have the same shape too?
No intent to mislead. That is why I specifically used the phrase "of equal area" instead of equal size. The first examples are the natural thing to do... it is too bad that it fails for lots of values of n. So that's why I went somewhere else with n=7. Your question is excellent. I don't know of such a division... I would guess the answer is no, but I have no reason for that guess yet :) Thanks!
This is super interesting and informative, as well as being very intuitive, but I'm more confused about how other regular polygons can be constructed using classical tools but the heptagon cannot lol. Can I get an explanation as to why?
This is actually quite a deep and challenging problem. You can find constructions of the polygons up to hexagon online. But proving that the heptagon cannot be constructed with classical tools is one of the famous impossibility problems and required a lot of relatively deep mathematics to prove.
@@MathVisualProofs , heh, perhaps. Although, I wasn't trying to say the heptagon was derivative of the pentagon, just that, if I had the appropriate skill level to figure out one, I'd have the skill level to figure out the other.
@@SgtSupaman But even this isn't right :) You can obtain the skill to create the pentagon, but you will never obtain the skill to create a heptagon. And it takes a lot more work to prove that the heptagon is impossible :)
@@MathVisualProofs , only impossible in theory. One can definitely draw a regular heptagon in a circle to a degree of certainty that is realistically covered by the width of the lines used. It's like trying to say it's impossible to ever travel a distance because you first have to travel half that distance, but you then have to first travel half that distance, and so on and so on. But, in reality, you quickly reach a point where the distances are so impossibly small they can't be halved.
Wait, but I still don't undestand why a regular heptagon can't be constructed using a straight edge and compass? Where the areas of those 7 sectors not equal?
It turns out you just can't do that construction with straightedge and compass. You can divide the circle into seven equal areas with the regular heptagon, and there are ways to construct the regular heptagon; just can't do it with the two classical tools.
Would've been nice, if mathematical channels forgot the "straigt edge and compass" rule, and remembered that you can use a marked ruler (neusis) too, or a right angle ruler, with wich you can quite easily construct a heptagon.
@@MathVisualProofs It's just seems ridiculous, that some DIY channels show how you can easily trisect an angle with a ruler, and some other geometrical shtiks up the woodworkers sleeves, but mathematical channels seems to avoid these themes altogether (I just happen to watch both). Would've been nice collaboration actually, if woodworker channel and mathematical channel made a collab, lol.
@@Itoyokofan I’ll look around for someone maybe ;) as to the ruler and straightedge from math - it leads to famous impossibilities and the math that arose out of that is powerful and interesting so I guess that’s why math channels hold on to the classics tools
Really cool proof, but it relies on a lot of assumptions. I'm only vaguely familiar with compass and straightedge problems, so when you say things like "you can construct 3- to 6-sided shapes but not 7-sided" or "go ahead and divide the diameter into 7 equal parts," I end up needing to take your word for it. Not sure how complicated it would have been to fully illustrate the context and process of this proof, maybe that's outside the scope of the video, but I would have liked to have it
Thanks! Yes, those would take some time and would have taken away from the point of the video. Plus there are lots of places showing constructions of the regular n-gons, so I didn't necessarily want to hit those again.
@@MathVisualProofs makes sense. I think the reason you're getting some comments like mine is because this video got picked up by the algorithm, so it has a broader audience with less subject matter familiarity than your other videos.
@@BeefinOut You're probably right :) Funny thing about the algorithm is that most of my videos are broader in the background knowledge required - but it liked this one for some reason :) Thanks!
I've got one in the works! It goes along with my squaring the circle videos: ruclips.net/video/0D3KiCmum90/видео.html and ruclips.net/video/_e4Yn5uGznI/видео.html :)
My approach qould be to just divide 360° into 7, and have each "pizza" slice have that angle. That is one simple way, although the video solution is cool too
Yes, this works but not with a straight edge and compass (one of the famous impossibilities is that you cannot draw 7 equally spaced points on the circle with classical tools).
@@MathVisualProofs I now believe I am ignorant (as in lacked knowledge) of what exactly using a straight edge and compass means. Afterlooking it up and understanding it further, I now understand the premise of the video lol. Thanks for the reply! it opened up my understanding more
Hang on, how do you divide up the diameter of the semicircle into 7 equal parts? Edit, I don't know, but I rid realize it's easy to just make a line segment and repeat it 7 times. Find the mind point of the middle one, make a circle from that.
The way I'd tackle this problem would be to first construct a line segment with length root 7. From there one would construct concentric circles, one with radius 1 and one with radius root 7. The space between the outer circle and the inner circle can then be divided in six leaving one circle and six arches of equal area.
Without showing how you would do the hexagon it's not very enlightening to say you can't do a heptagon. What is the technique that works for the other regular shapes that doesn't work for the 7 sided one?
I found out this idea was extended from constructing Yin-Yang symbol, especially the one used in South Korean flag "Taegeukgi (태극기, 🇰🇷)"! How amazing it is...!
There is a lot of modern mathematics that was developed to prove that certain things are impossible. So the limitation is a nod to that I think. The idea of proving the impossibility of certain tasks really was a fascinating change in thinking.
You didn't show that making the regular heptagon was impossible. I would assume one could generate one from a seven-side star, easily and inscribe a circle within said star. I like your method more though. The swooshes look cool. Though I'm pretty sure as a kid I tooled around with a similar method already I just didn't do any proof of anything, it was just drawing.
Showing that a regular heptagon isn't constructible is not easy, but the claim is true. Yes, you could generate one easily if you had a regular seven-sided star, but that shape also cannot be constructed.
i think the issue is with dividing 900 degrees by 7, as you end up with a repeating number that you cannot guarantee you marked exactly. Never mind that, that needs a protractor anyways.
This demonstration is not limited to 7 sections. I wrote some functions in desmos and both plotted and calculated the areas of m sections where m is a positive integer, and it holds true for any value of m (odd, even, prime, nonprime, etc.) its an interesting bit of work but i don't think its worth saving as a unique file, so if you want to view it yourself, copy and paste the following code into desmos (make sure to limit m to integers by setting step to 1): m=1 a_{1}=\operatorname{round}\left(\frac{2\pi}{m^{2}}\left(\left[1...m ight]^{2}-\left[0...m-1 ight]^{2} ight),4 ight) a_{2}=\operatorname{round}\left(a_{1}\left[\operatorname{length}\left(a_{1} ight)...1 ight],4 ight) a_{3}=\operatorname{round}\left(\left(a_{1}+a_{2} ight),3 ight) A=\sum_{n=1}^{m}a_{3}\left[n ight] c=\frac{\left[1...m ight]}{m} c_{0}=\left[0...m-1 ight] c_{1}=\left[1...m ight] y=\sqrt{c^{2}-\left(x-c ight)^{2}} y=-\sqrt{c^{2}-\left(x+c-2 ight)^{2}} If you want to display some numerical data, copy/paste these three lines in and label them with the following information (desmos doesn't seem to support pasting label, color, or formatting information so I cant just paste it all in one nice block) \left(2c-\frac{1}{m},.05 ight) \left(2c-\frac{1}{m},-.05 ight) \left(2c-\frac{1}{m},-1 ight) {a1} {a2} {a3}
@@MathVisualProofs oh duh. Still, would have been nice to mention it. But turns out that it didn't actually matter at all what their areas were, because of the symmetry in the final step, so you could have even skipped that part!
![Natanael Cano - Amor Eterno [Official Video]](http://i.ytimg.com/vi/EO4prCs6WpM/mqdefault.jpg)
Does this mean that, since division of a length by any positive number is possible using these tools, we can divide the circle into any natural number of equal areas with this method?
Give it a try with 8 pieces and see what you get :)
@@MathVisualProofs Yep, I just confirmed it. The reason this works is that the area of a semicircle is quadratic in r. Stepping into the world of discrete calculus, the difference between successive terms of this quadratic sequence is a linear function of r. This difference sequence corresponds to the “upper” part of each slice. Because this sequence of areas is linear, we can take the reverse of this sequence (the rotated “lower” semicircle), pair up matching pieces, and each slice now has the same area, because the constant amount the area changes in the upper semicircle is exactly negated by the change in the lower semicircle.
@@jakobr_ 😀👍
@@jakobr_ Isn't it a linear function of r^2?
@@jamesparsley5796 The difference between successive terms in a quadratic (in r) sequence is linear (in r)
oh thats nice.
Thanks!
Real nice!
@@SuperDydx thanks!
69th like.... nice to your nice.
I think it would have been interesting to explain how you can cut the diameter in 7 pieces using the straight edge and compass, but I really like the conciseness of this video
Yes. That would have made the video quite a bit more complicated in the middle and would have possibly confused the issue. There have been enough comments about this that maybe I should do an auxiliary one showing how to do that division. Thanks for the feedback!
While you can construct any rational number length (in particular 1/7), the easier approach is to just take a circle of diameter 7 instead, starting with a given length of 1.
@@MathVisualProofs I also realized that you didnt really dive into why a pentagon was constructible and not a heptagon, which I kind of took for granted because I was taught this a long time ago while I only realized how to multiply and divide recently, even though it's arguably easier. I see why it doesnt make much sense to include it here since explaining everything would change the video completely
@@jercki72 Yes. That is a good idea though. Maybe I can do a construction of the pentagon and a dividing into n pieces video. Thanks!
I can see how you can create a circle with a diameter of 7x by drawing a line, then using the compass pick a length, and repeat it 7 times on a line. This becomes your diameter. Then bisect your line using the compass giving you a radius where you can complete your circle. But that is backwards. I would also like to see how you divide the diameter into 7.
Edit: I saw the solution in a post below. Fantastic. It's funny because the solution is basically doing exactly what I said except creating 2 parallel lines of 7 segments, then you transfer those proportions to the fixed line. Very interesting.
But the real question is: can you also make a NON-regular square?
Haha! I should have fixed that. But left it. Re-recording audio isn’t my fave :)
@@MathVisualProofs wdym, don't you know about the _other_ squares?
@@AJMansfield1 If you meet the other squares, let me know :)
In some surfaces yes, a quadrilateral may have all sides with the same length, all angles right, and still have no vertice transitivity.
i think times square is not a regular square
I think all of the #some2 videos have been great, but there is just something so nice about a short, but elegant video like this
Thanks! But Whoops. I copied and pasted a title and included the # here. I'll remove it because my official SoME submission is a different video. Maybe I should have made it this one...
@@MathVisualProofs was that really a mistake? ; )
@@zhinkunakur4751 Yes - and it's fixed. I created the videos in the same time frame and wasn't sure which one to submit to SoME :) It appears maybe I should have submitted this one because people like it better :)
Pmmpm
This is brilliant and beautiful! Very nicely done! At first glance, it's not at all obvious how I would divide the diameter into 7 equal segments, but I'll take your word for it that it can be done.
Not obvious at all. But you can divide a line into n parts for any positive integer n with straightedge and compass. Here is the technique: www.mathopenref.com/constdividesegment.html
Step 1: draw a line starting at one of the endpoints (aside from needing to not be parallel with the diameter, it doesn't matter which direction)
Step 2: mark seven equally spaced points along that line
Step 3: draw a line from the farthest point to the other endpoint
Step 4: draw lines parallel to the one drawn in step 3 from the other points
@@MathVisualProofs Got it...not obvious, but also not that hard to understand.
If I'm not mistaken, another construction would be possible by first drawing a line through an offset point C roughly parallel to the original AB (I don't think it's a requirement, but I'm not working with the most rigor here)
Then make equidistant marks on the line through C, finishing with point D
Then construct line AC and BD so that they intersect at a new point E
Then connect each of the points on CD to E, and where they intersect like AB will also be equidistant (like parallel lines running to a vanishing point in a renaissance art piece)
@@AllThingsPhysicsRUclips Less rigorous, more accessible: "make up two parallel lines out of X segments, draw vertical lines between the segments of the two lines, and so any slice of these new rectangles has the vertical lines equally distant. Construct it smart from the beginning so that the line you wanted to divide is one such slice."
Because if you slice a metal fence, any angle you cut it, the tips of the posts will be equidistant.
Very nice! You could alternatively mark points on the radius at distances sqrt(1/7), sqrt(2/7), sqrt(3/7), ... and draw concentric circles, but this method is so much more elegant!
That's what I was thinking the pizza with 7 slices bahaha
but how do you construct sqrt(1/7) with straight edge and compas?
@@fejfo6559 pythagorean theorem is pretty good for this
@@fejfo6559 you can construct the square root of any given length either the geometric mean theorem ( en.wikipedia.org/wiki/Geometric_mean_theorem ) or the intercept theorem ( en.wikipedia.org/wiki/Intercept_theorem with the segments on the two rays being 1 + s and s + x, respectively). Both are fairly simple methods, personally I lean toward to the GMT for some reasons.
Alternatively, you can construct sqrt(7) by a series of right-angled triangles starting with the classical 1-1-sqrt(2) triangle (the half of a unit square) then the first leg of each next triangle is the hypotenuse of the previous one and the second leg is always 1, and then you can divide it by 7. Well you can start the series with the 1-2-sqrt(5) triangle since 2 is the largest whole number of which square is smaller than 7, but this would be much more effort than either of the first two, I think. Still, I like the (mental) image of the resulting "spiral" of triangles.
On the other hand, if you start from the 1-1-sqrt(2) triangle, you'll get all the square roots up to 7 which would be useful for scaling the sqrt(1/7) if you want to construct sqrt(2/7), sqrt(3/7), sqrt(4/7) etc as well, like in the original comment suggested.
But how do you divide a line segment into seven equal parts with a straight edge and a compass? Am I missing something obvious?
It’s not obvious but here is a source that gives the idea: www.mathopenref.com/constdividesegment.html
Create seven end to end line segments of equal length to make a line segment of seven equal sections, with one end positioned at the start of the line segment you want to divide, and point it off to one side so the two are arranged like a V. Add another line segment joining the ends of the v to make a triangle. Parallel to this new line segment, draw another line segment from each of the nodes joining the seven equal segments you drew earlier, such that the new lines intersect the original line segment which was to be divided. The intersection points divide the initial line segment evenly into seven. It's not obvious no :)
Alternatively to dividing a line segment into 7 equal parts is to start with a short line segment, build 6 additional line segments onto the end of it, and construct the circle from these. Technically it's not dividing the circle into 7 parts, but I think this counts.
This technique has been known for quite some time. It is Proposition 9 of Book VI of Euclid's Elements, written c. 300BCE. See
aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI9.html
On the other hand, the fact that you can't construct a regular heptagon with only a compass and straight edge has only been known since 1837, thanks to the Gauss-Wentzel theorem. This states that a regular n-gon can be constructed with compass and straightedge if and only if n is the product of a power of 2 and any number of distinct Fermat primes (including none). The Fermat Primes are the primes of the form 2^2^n+1, where n is a natural number, such as 3, 5, 17, 257, 65537. These are the only Fermat primes known. So after the regular pentagon, the next few odd-sided regular n-gons you can construct are for n=15, 17, 51, 85, 255 and 257. See en.m.wikipedia.org/wiki/Constructible_polygon
@@sweetcornwhiskey Is true that shows the length 7 is constructible, but it does take a bit more to show that 1/7 is constructible. But you are right that it is perhaps technically correct because I didn't say you had to start with a particular circle :)
I wish I had seen this when I had a customer who wanted their pizza in 7 slices lmao
My 6 friends are going to love it when I pull this trick out to cut a pizza for us.
I’d like to see it done!
Stay away from my pizza.
Although you've spoiled the answer in the video thumbnail, the idea is so brilliant I had to watch it. Thank you for bring this up. I wonder which areas are constructible given a unit area circle, besides the rationals. Wow!
😀 thanks! Sorry to spoil it early. The RUclips thumbnail game is beyond me and the hardest part about having a channel. I just want to share some visual math ideas :)
@@MathVisualProofs Maybe it was for the best, who knows, without the spoiler I could just think "No." and move on thinking it was just a video about the unconstructability of the heptagon.
@@sachs6 good point!
Honestly, I like that the thumbnail showed the answer. At a glance, it allows the viewer to see the answer while scrolling through, yet they can still watch the video to learn why it's the answer. In this case too, the squiggly segments can pique the viewer's interest because thet might not make a lot of sense without the proof.
@@RainShinotsu Thanks!
After being separated those don't even look like they should fit together to make a circle but the proof is all there, incredible
Right? Pretty cool.
Well, they do, but sure.
Amazing. What other regular polygons can we NOT make using the classical tools?
That’s a good question! Give a search for constructible regular polygons and you’ll find out. There are lots that aren’t. :)
According to wikipedia en.wikipedia.org/wiki/Constructible_polygon#Detailed_results_by_Gauss's_theory, an n-gon is constructible when n is a product of a power of 2 with any combination of the numbers 3, 5, 17, 257, and 65537. All other n-gons are not constructible.
So, for example, a 64*5*257-gon is constructible, while a 25-gon is not.
If we discover any other “Fermat primes”, those primes will be added to the list with 3,5,etc.
@@jakobr_ exactly! Makes the Fermat primes even more intriguing :)
Great visual proof!
Thanks!
Very interesting solution, but by your first few examples, I (incorrectly) assumed the final shape would be a pie piece. Tricky :)
Yes. Didn’t mean to mislead. If you could do a pie shape I think you could do the impossible and construct the heptagon :)
@@MathVisualProofs I remember back in school my teacher tossed out the "trisect an angle" challenge with a compass and a straight edge. I wanted so bad to find a solution :)
@@dougsholly9323 love that! Even though it’s impossible a challenge like that can really inspire. Speaking of, at some point I will have a video that shows how to trisect an angle…. But it needs one extra tool that isn’t classical :)
I was just thinking of perspective vanishing points in art. That stuff is all about preserving angles. These lines that represent length can be vectors. That being said, any vector can be any unit. So, a perspective vector doesn't have to represent length, it can be anything.
👍
Thats actually beautiful
😀
Wow, love the solution 😀 I was always into math/physics, so it’s kinda fun 😉 Hope for more videos and keep up a good work 👍
Thanks! Check out my catalog for more visual proofs/visualizations. :)
This helped me when cutting the pizza for me and my six friends, thank you very much 😊
😀
I wish I was shown this when I was in junior high/ high school
man, this would have been awesome
:)
Extremely Satisfying.
:)
This is beautiful, for years I watch 3blue1brown videos and your videos give me the same vibe. I love it keep up the great videos
Thanks! I am using his amazing software so it makes sense it would give that vibe :)
That's gorgeous!
😄
Beautiful!
Thanks! 😀
Damn those shapes look so cool!
Right? 👍
Very intriguing! I didn't know it was possible, but you made the problem and proof easy to understand.
:) Thanks for the comment! Glad you enjoyed the video.
Mind Blown ! Subscribed
Thanks! Glad you liked this one.
Holy shit..this was a goldmine!
😀
This is really cool
Thanks!
Your initial examples of dividing the area of the circle into 2, 3, 4, 5 and 6 equally-sized pieces were a bit misleading, since those pieces were also equally-shaped, whereas in your solution for the division into 7 equally-sized pieces the pieces were not equally-shaped but different when compared to each other. So I was a bit confused first and felt a bit mislead about your solution, since I thought you were solving a different problem (= that the pieces should have to be equally-shaped). Is there a solution for the 7-division where the pieces have the same shape too?
No intent to mislead. That is why I specifically used the phrase "of equal area" instead of equal size. The first examples are the natural thing to do... it is too bad that it fails for lots of values of n. So that's why I went somewhere else with n=7. Your question is excellent. I don't know of such a division... I would guess the answer is no, but I have no reason for that guess yet :) Thanks!
This is super interesting and informative, as well as being very intuitive, but I'm more confused about how other regular polygons can be constructed using classical tools but the heptagon cannot lol. Can I get an explanation as to why?
This is actually quite a deep and challenging problem. You can find constructions of the polygons up to hexagon online. But proving that the heptagon cannot be constructed with classical tools is one of the famous impossibility problems and required a lot of relatively deep mathematics to prove.
@@MathVisualProofs thank you. This would have been a nice bit to add at the beginning.
it's a nice trick but I would like to point that, although all seven areas have the same size, they aren't equal. they have different shape.
Yes. Most times I said "of equal area" but in one place I said "equal parts" (should have inserted the word "area" maybe).
This was an awesome video! Straight to the point and shows an amazingly simple solution to a complicated problem
Thanks!
I'm pretty sure if I could do a pentagon, I could do a heptagon. I can't do either, but you get my point.
But you’d be wrong :). Knowing how to do a pentagon gives no strategy for the heptagon. It’s impossible to construct a heptagon with classics tools.
@@MathVisualProofs , heh, perhaps. Although, I wasn't trying to say the heptagon was derivative of the pentagon, just that, if I had the appropriate skill level to figure out one, I'd have the skill level to figure out the other.
@@SgtSupaman But even this isn't right :) You can obtain the skill to create the pentagon, but you will never obtain the skill to create a heptagon. And it takes a lot more work to prove that the heptagon is impossible :)
@@MathVisualProofs , only impossible in theory. One can definitely draw a regular heptagon in a circle to a degree of certainty that is realistically covered by the width of the lines used. It's like trying to say it's impossible to ever travel a distance because you first have to travel half that distance, but you then have to first travel half that distance, and so on and so on. But, in reality, you quickly reach a point where the distances are so impossibly small they can't be halved.
Thanks for another great visual proof.
Thanks for watching!
Practically works for any number of divisions
Yep!
This needed to be done on IRL paper with the tools mentioned for added effect 😅
Some day maybe ;)
woah that is super cool!
👍 thanks!
Finally, a way to cut pizza for 7-people party
😀
This is an awesome visual proof
Thanks!
“bro just take a pizza slice like a normal person”
😀
This is so cool !!
😀
Wait, but I still don't undestand why a regular heptagon can't be constructed using a straight edge and compass? Where the areas of those 7 sectors not equal?
It turns out you just can't do that construction with straightedge and compass. You can divide the circle into seven equal areas with the regular heptagon, and there are ways to construct the regular heptagon; just can't do it with the two classical tools.
I'm sorry. But, you fell short on this exercise in math and geometry. Take a closer look. It's fascinating.
Can you explain or clarify?
Yes, but if I divide the cake in this way among 7 guests, only one will get a piece with a cherry
Save that sweet center piece for yourself :)
POV: ur Voldemort and ur stuck on trying to figure out how to split ur soul into seven pieces for beginners
Beautiful!!!
😀
this is so cool thank you.
Thanks for watching!
That was brilliantly beautiful, sir!
Thanks!!
Brilliant!
👍
Nice tutorial-5 Bro!
:) Thanks!
nice. A question how do you construct the regular pentagon with a straight edge and compass?
Here’s one possible source: www.mathopenref.com/constinpentagon.html
Would've been nice, if mathematical channels forgot the "straigt edge and compass" rule, and remembered that you can use a marked ruler (neusis) too, or a right angle ruler, with wich you can quite easily construct a heptagon.
Sometimes I don’t use the classical requirement. I have two videos where I square the circle and so I have to use other tools :)
@@MathVisualProofs It's just seems ridiculous, that some DIY channels show how you can easily trisect an angle with a ruler, and some other geometrical shtiks up the woodworkers sleeves, but mathematical channels seems to avoid these themes altogether (I just happen to watch both).
Would've been nice collaboration actually, if woodworker channel and mathematical channel made a collab, lol.
@@Itoyokofan I’ll look around for someone maybe ;) as to the ruler and straightedge from math - it leads to famous impossibilities and the math that arose out of that is powerful and interesting so I guess that’s why math channels hold on to the classics tools
now i just gotta wait for an opportunity to show that off,
the problem ist that german schools couldn't care less about problem solving
Great Idea!
Thanks! :)
Very good - very satisfying.
Thanks ! 😀
Interesting. Now i wonder if there's a way so that 7 dwarfs can also simultaneously share the same amount of crust🤔
👍
Beautiful
😃
So it's the same concept of a ying-yang shape (where n=2 in that case), but expanding it further. Neat
👍
that is so friggen cool lmao. thanks for making this video!
Thanks for watching this video! :)
Really cool proof, but it relies on a lot of assumptions. I'm only vaguely familiar with compass and straightedge problems, so when you say things like "you can construct 3- to 6-sided shapes but not 7-sided" or "go ahead and divide the diameter into 7 equal parts," I end up needing to take your word for it. Not sure how complicated it would have been to fully illustrate the context and process of this proof, maybe that's outside the scope of the video, but I would have liked to have it
Thanks! Yes, those would take some time and would have taken away from the point of the video. Plus there are lots of places showing constructions of the regular n-gons, so I didn't necessarily want to hit those again.
@@MathVisualProofs makes sense. I think the reason you're getting some comments like mine is because this video got picked up by the algorithm, so it has a broader audience with less subject matter familiarity than your other videos.
@@BeefinOut You're probably right :) Funny thing about the algorithm is that most of my videos are broader in the background knowledge required - but it liked this one for some reason :) Thanks!
Simply beautiful
Thank you! 😊
Trisection of an angle next!
I've got one in the works! It goes along with my squaring the circle videos: ruclips.net/video/0D3KiCmum90/видео.html and ruclips.net/video/_e4Yn5uGznI/видео.html :)
Generalized version of the yin and yang
ngl, I nearly lost it
in a good way
my mind was blown at the end
then I said, "that was so cool"😀🤤
😀Thanks for sharing!
great stuff
Thanks!
My approach qould be to just divide 360° into 7, and have each "pizza" slice have that angle. That is one simple way, although the video solution is cool too
Yes, this works but not with a straight edge and compass (one of the famous impossibilities is that you cannot draw 7 equally spaced points on the circle with classical tools).
@@MathVisualProofs I now believe I am ignorant (as in lacked knowledge) of what exactly using a straight edge and compass means. Afterlooking it up and understanding it further, I now understand the premise of the video lol. Thanks for the reply! it opened up my understanding more
@@lmarsh5407 No worries! It is a bit of a niche idea in mathematics :) Glad you checked it out, though! It leads to some cool mathematics.
cant wait to do this to a pizza
😀
Pepsi Logo:
*heavy breathing*
Haha
Also, who discovered this gem of a proof?
Not sure exactly I cited Roger Nelson’s book Icons of Math because that’s where I learned it.
Next time I get a pizza I will order it cut just like that.
I’d like to see it!
Wow! If it was a logo, I'll be happy!
my six friends gonna hate me next pizza party
:)
this video is great
Thanks!
But the pie crust allocation is far from equal which could lead to negative outcomes
Good point. :)
If you can't make a heptagon using a staight edge and compass, then use a protractor. 360/7=51.42.
Still only an approximation to 360/7.
Subscribed
Thanks!
(1) In "Edison" it would be possible to create and play Loops in Any Order we like. Example. Loops=(2, 6, 8, 1) or (12, 1, 3, 6.10,) and
?
But WHY can't we construct a regular heptagon using a straight edge and compass? Can you prove it can't be done?
It is provable. But not in my format
Can’t wait to do thos with my next pizza order :D
Want to see this done :)
I wonder if this is how those swirley marbles are made assuming this works with a three dimensional sphere
Interesting thought! I have no clue about that :)
That was completely pointless and completely awesome!
👍
How am I still into this kinda stuff?
This reminds me of a puzzle.
With more slices it could be a cool one to have :)
LS149 Marble Thirds
Hang on, how do you divide up the diameter of the semicircle into 7 equal parts?
Edit, I don't know, but I rid realize it's easy to just make a line segment and repeat it 7 times. Find the mind point of the middle one, make a circle from that.
Far from obvious. Here is a link that shows how to cut any line into n pieces for n a positive integer : www.mathopenref.com/constdividesegment.html
Well this could be an interesting way to divide up a pizza.
Definitely want to see it :)
The way I'd tackle this problem would be to first construct a line segment with length root 7. From there one would construct concentric circles, one with radius 1 and one with radius root 7. The space between the outer circle and the inner circle can then be divided in six leaving one circle and six arches of equal area.
Is a good way!
Without showing how you would do the hexagon it's not very enlightening to say you can't do a heptagon. What is the technique that works for the other regular shapes that doesn't work for the 7 sided one?
There are many resources showing how to do classical constructions so I figured I’d let interested people seek those out if intrigued.
Nice!
😀
Wow wow wow, how did you divide the segment in 7 without measuring? Soo much jumping
See this follow up: ruclips.net/video/afZiAGaKEY4/видео.html. Thought I added a mid-video card for it but it’s not showing up. I’ll fix that.
I found out this idea was extended from constructing Yin-Yang symbol, especially the one used in South Korean flag "Taegeukgi (태극기, 🇰🇷)"! How amazing it is...!
Very cool!
Nice sir
Thanks!
Is there any modern import to proofs or operations limited to classical tools? Or is it simply a limitation for its own sake (fun)?
There is a lot of modern mathematics that was developed to prove that certain things are impossible. So the limitation is a nod to that I think. The idea of proving the impossibility of certain tasks really was a fascinating change in thinking.
you skipped the major nontrivial part of the construction, dividing the diameter into 7 parts.
I disagree. That is done many places. It would have made this video too complicated looking.
You didn't show that making the regular heptagon was impossible. I would assume one could generate one from a seven-side star, easily and inscribe a circle within said star. I like your method more though. The swooshes look cool. Though I'm pretty sure as a kid I tooled around with a similar method already I just didn't do any proof of anything, it was just drawing.
Showing that a regular heptagon isn't constructible is not easy, but the claim is true. Yes, you could generate one easily if you had a regular seven-sided star, but that shape also cannot be constructed.
This is right. It’s much harder to show this. And a visual proof would be even harder :) thanks for your reply!
i think the issue is with dividing 900 degrees by 7, as you end up with a repeating number that you cannot guarantee you marked exactly. Never mind that, that needs a protractor anyways.
This demonstration is not limited to 7 sections. I wrote some functions in desmos and both plotted and calculated the areas of m sections where m is a positive integer, and it holds true for any value of m (odd, even, prime, nonprime, etc.) its an interesting bit of work but i don't think its worth saving as a unique file, so if you want to view it yourself, copy and paste the following code into desmos (make sure to limit m to integers by setting step to 1):
m=1
a_{1}=\operatorname{round}\left(\frac{2\pi}{m^{2}}\left(\left[1...m
ight]^{2}-\left[0...m-1
ight]^{2}
ight),4
ight)
a_{2}=\operatorname{round}\left(a_{1}\left[\operatorname{length}\left(a_{1}
ight)...1
ight],4
ight)
a_{3}=\operatorname{round}\left(\left(a_{1}+a_{2}
ight),3
ight)
A=\sum_{n=1}^{m}a_{3}\left[n
ight]
c=\frac{\left[1...m
ight]}{m}
c_{0}=\left[0...m-1
ight]
c_{1}=\left[1...m
ight]
y=\sqrt{c^{2}-\left(x-c
ight)^{2}}
y=-\sqrt{c^{2}-\left(x+c-2
ight)^{2}}
If you want to display some numerical data, copy/paste these three lines in and label them with the following information (desmos doesn't seem to support pasting label, color, or formatting information so I cant just paste it all in one nice block)
\left(2c-\frac{1}{m},.05
ight)
\left(2c-\frac{1}{m},-.05
ight)
\left(2c-\frac{1}{m},-1
ight)
{a1}
{a2}
{a3}
:) nice work
Please explain for us how you got to the area of each semi circle section. You just flew right over that.
Just used the circle area formula. So semi circle is 1/2 pi times radius squared.
@@MathVisualProofs oh duh. Still, would have been nice to mention it. But turns out that it didn't actually matter at all what their areas were, because of the symmetry in the final step, so you could have even skipped that part!
All hail the Math Pepsi.
I could also say this would have made a SICK sports equipment logo from the 70s-80s.
@@CathodeRayKobold :)
Yin, yang, and uh... yong, yum, yanney, yan, and yern.
Hah! :)