Thank you BPRP!!! Big honor for me to have your attention. You know, in math, an odd question is even to everyone. I always ask you on my strange thinking, because I envy the Giraffe! I am trying so hard to be diligent in learning to catch it!
The angle sum formula can be derived from the circle proof at 10:45. For every vertex, the number of corresponding arcs separated by other vertices is p-2q (since u have q arcs on either side of that vertex). Thus when u sum up the angles of the corresponding arcs u get (360)p-2q, divide by 2 u get 180(p-2q).
I wondered if the angle sum formula shouldn't be with absolute values: 180°|p-2q| Because without them you can't use the q of the equivalent stars. For example you can use {7,2} but you can't use {7,5}, wich is equivalent but will give you a negative angle.
22:27 NOTE III: actually, the gcd(p, q) determines how man paths you have to draw. So, the path count, Pc = gcd(p, q) Try it for yourself with 9 points and a spead of 3, or 21 points and a spread of 9.
@@aram8832 floor means rounding down to an integer floor(3.26) = 3 floor(7.95)=7 floor(2)=2 floor(-1.3)= -2 I guess it's the greatest integer below the input? so maybe.
Note II proof: Consider a {p/q} star inscribed in a circle, p vertices define p arcs of circumference. Each angle insist on some of the p arcs, exactly p-2q arcs because from the total of p arcs we subtract the q arcs that have as a rope the first segment that define one of the angle and another q subtract for the others q arcs that have as a rope the second segment that define an angle. Naming the p arcs A1, A2,…,AP We can say that the first angle a1=1/2 (A1+A2+…+A(P-2Q)) the second angle a2 = 1/2(A2+A3+…) … and the final angle ap= 1/2(Ap+A1+…). So the sum a1+a2+…= 1/2( a lot of arcs) we can easily convince ourselves that “a lot of arcs”=(P-2Q)(A1+…+AP)= 360°(P-2Q) => Angles sum= 1/2 360° (P-2Q) = 180° (P-2Q) (Sorry for my bad English)
As a fellow German, I wouldn't presume the Swiss pronunciation of the name :) I'm pretty sure my own pronunciation of it would be ridiculed by Swiss people.
For the angle sum, the p-2q factor shows how many arcs there are opposite a given point of the star. Using the same angle versus arc argument, yields p*2q times the coverage of the circle, thus half that many times 360° angle sum.
This is just a simple observation: I noticed that the Schläfl notation on 7-pointed stars or the concept of 7 pointed stars could leads us also in understanding simple notations and concepts in elementary mathematics, like fractions. Noticing that as q in {p/q}, where q is the turning number and p are the points of the star, could lead us somehow in understanding the value of fractions (what’s larger and what’s smaller). As the turning value (q) increases, the star gets little and little area and space around it. From 7/2 to 7/4, the star’s shape, area, region, and angle relation had become little. Into understanding the value of fractions, taking also 7/2 to 7/4, this would also state that the lesser the denominator (in my case), the bigger the value is. How big the area, the region, and the visual appearance of the star (I think angle could also do), would relate to the value of fractions on which one’s larger (namely 7/2 => 3.5, 7/3 => 2.33, and 7/4 => 1.75, thus also proving that the fraction with the least denominator, with the same numerator cuz that’s always a problem, works with this concept). I suddenly remember the A&T - McDonald’s Burger Problem back in the 1980s concerning people choosing Mc’s 1/4 pound burger over A&T’s 1/3 pound burger, because people thought the 1/4’s bigger due to its denominator. (Funny story tho) Although a bit obvious, it could at least help students have this concept concerning the value of fractions and using a bit Schläfl notation / 7 pointed stars on calculations. No more calculators to calculate what value is it compared to that or doing some mental scratches (for students who are not prone to mental calculation). :)
I remember doing this in my college geometry class. We said if p & q are relatively prime instead of the GCD(p,q)=1 (they are the same thing, though coprime is sometimes used also). We did some other stuff with modulo sets essentially proving it a different way. And then when we did non-euclidean geometries we did some other fun stuff. In the hyperbolic space you get a strange answer to the number of degrees in a circle using this as a basis to start with (it's not 360°). For my senior project in college I used permutation groups and this as an example to demonstrate that finite permutation groups are isomorphic to any finite set of the same size.
When I was in high school and college I would always draw these. It me a while to find information on them before but if you look up "Star Polygons" you can find a lot more info about them. A very important note that may be good for another video is that the star polygons relate a lot to cyclic groups in abstract algebra. Since I drew these stars all the time, I had a visual to relate cyclic groups to when looking at isomorphisms. Because of this abstract algebra was my favorite course in college
I'll introduce some my notation related to base in number theory. Whenever you talking base of a number, like base 2 (binary), you have enough digits in numeric notation as from 0,1,2,3,4,5,6,7,8,9. So that means from base 1 to base 10, we have enough digits from numeric digits so that we don't need to invent some symbols. But then, when we talk about some high base like base 100 or base 256, we've to accepted that there isn't enough symbols to categorized all the digits. So what do we do? We introduced a notation called: Base-Divide notation (or you can called it as Base-Power notation for yourself) The notation looks like this: (|1|2|3|4|)13 with 13 is subscript decode what the "number" in notation should be in base n So what's useful? When we derive a base, we need convert it to decimal for our human purpose reading. So the way we'll convert the Base-Divide Notation to Base 10 is very easy (just sort of general formula to convert base n to base 10) (|n1|n2|n3|n4|...n(k)|)m Where k is the "digit" of the notation and m is the base. Note that the value contains in n1, n2, n3, n4 should not be larger or equal to the base that they have set to (kind of 1023, there is no single digit that is larger than the base "10") General formula to convert the Base-Divide notation to Base 10: Examples: (|3|6|7|9|)11 = 3*11^3 + 6*11^2 + 7*11^1 + 9*11^0 = 4805 in base 10 (|1|2|3|4|5|)15 = 1*15^4 + 2*15^3 + 3*15^2 + 4*15^1 + 5*15^0 = 58115 Note to know: You cannot write the notation like this: (|12|32|11|5|)19 where there is/are number that's larger than the base You need to rewrite whenever you write "wrong" Base-Divide Notation (Technically, you can write the wrong notation for yourself if you don't publish it in paper or share this notation to another human than you know) How to rewrite the notation when you write it wrong: Example: (|34|12|55|14|)23 Compute 14/23 = 0*23 + 14 Compute 55/23 = 2*23 + 9 (the 2 here is the result when divide so in next step we add the result to the remainder) Compute 12/23 = 0*23 +12+2 (it's like carry sums from the long multiply that learn in elementary school) Compute 34/23 = 1*23+11+0 (actually, you can +0 if you want becuase +0 means "no carry before". But it's meaningless) Then the result are: (|34|12|55|14|)23 -> (|1|11|14|9|14|)23 So we done here... Not yet The thing that you can do with this notation is plug any kind of formula that you can imagine, quadratic, cubic, sigma, pi, e, definite integral, taking a derivative value... Anything Take this as example: You can write like this: (|13|21|4|n|)35 where n is a real number from 0 (|1|2|3|6|0|)7 (Note that we doesn't use the equal sign when rewriting) Or like this: (|n+2|n+1|n+3|)12 where n must be smaller than 9 (Real numbers plug in base? Use the base convert formula to solve)
This might be irrelivant but Ive heard from somewhere that (5^a - 3^b)/(2^c) has a solution for every whole number over 0 where a,b and c are also whole number. I dont know if it is true and why it is the case.
assuming you mean for every whole number n>0, there are whole numbers a,b,c such that (5^a - 3^b)/(2^c) = n, this is not the case. Let n be any positive multiple of 3, and suppose (5^a - 3^b)/(2^c) = n for some a,b,c. Then 5^a = (2^c)n + 3^b, but unless b = 0, the right hand side is divisible by 3, and so 3 divides 5^a, which is impossible. Thus the only choice for b is 0, so we have 5^a = (2^c)n + 1. Furthermore, if we also take n to be divisible by 5 (e.g. n=15), we have 5|(5^a - (2^c)n) and so 5|1, a contradiction. Answer is no, it does not hold for every whole number n over 0
PS: Both the {7/3} and the {8/3} stars contain the other type of star for their number of vertices ({7/2} and {8/2} respectively) within them, as well as the respective polygon. This pattern seems to continue: {9/4} contains {9/3}, {9/2} and {9/1} as well. I postulate that this might be the case for all {n/ceil(0.5n)-1} stars.
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To figure out the angle sum, you can also imagine a rod sliding back and forth along the lines. How much you rotate it is the angle sum.
There's an easy way to find the angle sum for any star and it's the same method as you used for the {7/3} star. The arc covered after two points in the star will be p-2q turns of the circle when added together, which you can divide in half to find the total angle. That's why it's 180*(p-2q).
is it necessarry though for the vertices to be on a circle? seems not so for 5 and 6, how about 7 or more ... oops: I see someone asked same question and @jursamaj replied
I am feeling smarter now, that I can say that I found all these myself when I was in 10th class. I also found a method to check how many such stars exist for n points.
That angle sum formula can be proved quite readily using the last proof you gave for the {7/3} star! The fraction of the circle that each angle subtends is given by (p-2q)/p. This can be seen because there are p many segments between the vertices, and q are not included clockwise, and q are not included anticlockwise! Then there are p of these, giving us a total of p-2q fractions of the circle, but with the half correspondence this becomes 180(p-2q)
I like that the formula for the angle sum deprecates into the polygon formula as well. If q=1 it's really a polygon rather than a star, but the formula turns into the polygon angle sum formula: 180(p-2)
Cool! Just ran into a situation where I wanted to know the sum of angles on a 5 pointed star. I kinda asummed it would've been 360 intuitively but it appeared my intuition was just dead wrong so then I got interested in the general case and stumbled right into a trusted old channel. Genuinely baffled at how much more complex the issue is than I initially thought it'd be, but I knew I could rely on you to clarify the matter :p
Hello mentor, I made several videos trying to convince people that square root of any positive real number is POSITIVE, but people still argue. Could you please make a video to show the whole world the difference between y=√4 and y²=4
The first illustrated 7-point star looks like an expanded version of the 5-pointed star. Haha. By the way do some videos about the Pascal’s triangle and other different interesting patterns in mathematics (like Fibonacci, and others). :)
2:34 . . . Yes it is indeed possible to finish the star in one go. Method (1) Start from any one of the six intersection points (A) of the two triangles, draw one of the triangles and return to A. Immediately draw the other triangle. Method (2) Start from any one of the six intersection points (A) of the two triangles, draw the hexagon and return to A. Immediately draw the outer triangles. Voila, star drawn easily in one go!
I have my doubts about the 6 pointed star because every other star you‘ve drawn was drawn in a continuous line and had p points only if p was relatively prime to q but the six pointed star has p = 6 and q = 2 which are obviously not relatively prime, thus you had to draw to independent triangles. Furthermore I would argue that there is no way to draw a six pointed star because every number you choose that is less than 6 is either not relatively prime to 6 or it is congruent 1/-1 mod 6 so it would just form a hexagon. But I liked the video anyways ;)
2:33 you can do it in one go it is called the Star of David and it is in the flag of my country (israel) I will happily send you how to draw this in one go if you want
I think the rule is we should start with a vertex and then rotate q times (like how I demonstrated in the video). I didn’t think harder on drawing that star in one go if I could start anywhere I wanted. But I see it now. 😆
The 2 inner red equals to the 2 red in the quad, and the 2 red in the quad is already considered as sum of angles inside a quadrilateral, so you can ignore the inner 2 red then all you are left with is the triangle because the inner 2 red angles which are not included in the triangle are now included in the quadrilateral. Hope that helps
There are four natural numbers N > 1 such that you cannot draw an N-pointed star in one go: they are 2, 3, 4, and 6. There are four natural numbers N > 1 such that you can tesselate the plane with tiles of rotational symmetry N: they are 2, 3, 4, and 6. The first fact I can account for given the kind of reasoning in this video. The second I keep _feeling_ like I can do the same, but when I try and articulate the reasoning it runs away on me.
quick question, are there non flat, or non euclidean surfaces were u can draw the 6 pointed star without lifting your pen? i am a dumb person and i will try, but i also want somebody more educated to look/explain.
Can u explain the angle sum formula? :) Its so elegant but i dont know how to derive it. I know that for odd n stars that when q is (p-1)/2 that the sum is 180° because i found the angle sum in another manner. I found it by turning the seven point star into a five point star with out changing the angle sum with the exact same trick u turned a five point star into a triangle. But all the other sums are still a mystery to me.
thanks for ur AMAZING CONTENT!!!!!! i am so glad my dad told me about this channel in 2018!! but then sometimes i didnt watch u. but now i do! i had already watched this causer it was in a playlist, while unlisted. but best best luck for your future!!!!
If they're relatively prime, then you can draw it without lifting your pen. But We can still draw {6/2} even though they aren't relatively prime. If {p/q} is a valid star I think the rule is that you can draw it in gcd(p,q) parts.
Remember if you are comfortable in algebra more than geometry then complex numbers are there for you. I will prove that the sum of angles of 5/2 is 180: Let w be the 5th root of unity, then the vertices of the star are 1,w,w^2,w^3,w^4. Now the sum of angles is Sigma arg((w^k+3-w^k)/(w^k+2-w^k)) = Sigma arg((w^3-1)/(w^2-1)) = 5.arg((w^3-1)/(w^2-1)) And one can show that arg((w^3-1)/(w^2-1))= pi/5 Thus the sum of angles is 5.pi/5 =pi And so we are done! (Moral: if you are not comfortable with Geo, learn complex numbers 😉)
Actually, you can. Start from the star in the circle. Now, move any point, while keeping it on one of its lines. The sum of the angles at either end of that line will remain the same. Using this method repeatedly, you can deform the circular star to any arbitrary star, keeping the angle sum the same at each step.
Thank you BPRP!!! Big honor for me to have your attention. You know, in math, an odd question is even to everyone. I always ask you on my strange thinking, because I envy the Giraffe! I am trying so hard to be diligent in learning to catch it!
Here’s the man!!! Thanks again!! 😃
What do you mean the giraffe?
@@colleen9493 He has a giraffe very good at calculus. Lol
@@hotlatte1222 LOL
@@hotlatte1222 Are you 90 on a freeway?
Also note that the angle sum formula 180°(p-2q) holds for the case q=1, where the star {p/1} is just a polygon and has angle sum 180°(p-2) :D
ya
Also holds for the {6, 3} "cutting a pizza" one.
You're a fantastic teacher and a very patient human being. Thank you, sir.
The angle sum formula can be derived from the circle proof at 10:45. For every vertex, the number of corresponding arcs separated by other vertices is p-2q (since u have q arcs on either side of that vertex). Thus when u sum up the angles of the corresponding arcs u get (360)p-2q, divide by 2 u get 180(p-2q).
I wondered if the angle sum formula shouldn't be with absolute values: 180°|p-2q|
Because without them you can't use the q of the equivalent stars. For example you can use {7,2} but you can't use {7,5}, wich is equivalent but will give you a negative angle.
22:27 NOTE III: actually, the gcd(p, q) determines how man paths you have to draw.
So, the path count, Pc = gcd(p, q)
Try it for yourself with 9 points and a spead of 3, or 21 points and a spread of 9.
(p-3)/2 when odd and (p-4)/2 when even can be simplified to floor[(p-3)/2]
Yes and that is better!
Is floor means greatest integer function?
@@aram8832 floor means rounding down to an integer
floor(3.26) = 3
floor(7.95)=7
floor(2)=2
floor(-1.3)= -2
I guess it's the greatest integer below the input? so maybe.
@@BigDBrian ohh, thanks for clearing that up!
@@BigDBrian at or below
When you said factorial at 22:04 I laughed out loud. I'm gonna start saying that from now on!
Thanks for mentioning that. I missed it! Yes, it made me smile.
FACTOREO
Note II proof:
Consider a {p/q} star inscribed in a circle, p vertices define p arcs of circumference. Each angle insist on some of the p arcs, exactly p-2q arcs because from the total of p arcs we subtract the q arcs that have as a rope the first segment that define one of the angle and another q subtract for the others q arcs that have as a rope the second segment that define an angle. Naming the p arcs A1, A2,…,AP
We can say that the first angle a1=1/2 (A1+A2+…+A(P-2Q)) the second angle a2 = 1/2(A2+A3+…) … and the final angle ap= 1/2(Ap+A1+…). So the sum a1+a2+…= 1/2( a lot of arcs) we can easily convince ourselves that “a lot of arcs”=(P-2Q)(A1+…+AP)= 360°(P-2Q) =>
Angles sum= 1/2 360° (P-2Q) = 180° (P-2Q)
(Sorry for my bad English)
As a german, the way you pronounce "Schläfli" is really cute haha
Not cute but false!
As a fellow German, I wouldn't presume the Swiss pronunciation of the name :) I'm pretty sure my own pronunciation of it would be ridiculed by Swiss people.
Thanks BPRP for this wonderful video. Explained really well.
Yes, you can draw the 6-point star in one go. Draw a hexagon and then the triangles on top. ;)
For the angle sum, the p-2q factor shows how many arcs there are opposite a given point of the star. Using the same angle versus arc argument, yields p*2q times the coverage of the circle, thus half that many times 360° angle sum.
Seems like there are more and more geometry recently. Excellent content as always!
This is just a simple observation: I noticed that the Schläfl notation on 7-pointed stars or the concept of 7 pointed stars could leads us also in understanding simple notations and concepts in elementary mathematics, like fractions.
Noticing that as q in {p/q}, where q is the turning number and p are the points of the star, could lead us somehow in understanding the value of fractions (what’s larger and what’s smaller). As the turning value (q) increases, the star gets little and little area and space around it. From 7/2 to 7/4, the star’s shape, area, region, and angle relation had become little. Into understanding the value of fractions, taking also 7/2 to 7/4, this would also state that the lesser the denominator (in my case), the bigger the value is. How big the area, the region, and the visual appearance of the star (I think angle could also do), would relate to the value of fractions on which one’s larger (namely 7/2 => 3.5, 7/3 => 2.33, and 7/4 => 1.75, thus also proving that the fraction with the least denominator, with the same numerator cuz that’s always a problem, works with this concept). I suddenly remember the A&T - McDonald’s Burger Problem back in the 1980s concerning people choosing Mc’s 1/4 pound burger over A&T’s 1/3 pound burger, because people thought the 1/4’s bigger due to its denominator. (Funny story tho)
Although a bit obvious, it could at least help students have this concept concerning the value of fractions and using a bit Schläfl notation / 7 pointed stars on calculations. No more calculators to calculate what value is it compared to that or doing some mental scratches (for students who are not prone to mental calculation). :)
I remember doing this in my college geometry class. We said if p & q are relatively prime instead of the GCD(p,q)=1 (they are the same thing, though coprime is sometimes used also). We did some other stuff with modulo sets essentially proving it a different way. And then when we did non-euclidean geometries we did some other fun stuff. In the hyperbolic space you get a strange answer to the number of degrees in a circle using this as a basis to start with (it's not 360°). For my senior project in college I used permutation groups and this as an example to demonstrate that finite permutation groups are isomorphic to any finite set of the same size.
When I was in high school and college I would always draw these. It me a while to find information on them before but if you look up "Star Polygons" you can find a lot more info about them. A very important note that may be good for another video is that the star polygons relate a lot to cyclic groups in abstract algebra. Since I drew these stars all the time, I had a visual to relate cyclic groups to when looking at isomorphisms. Because of this abstract algebra was my favorite course in college
The math teacher i wanted, but never had.
Same feeling...🤧
Same here
You guys should be happy to have the opportunity to watch this freely on RUclips
@@guillaume5313 yah.. that's a good point though
💯
Same here 🙁😔
I'll introduce some my notation related to base in number theory. Whenever you talking base of a number, like base 2 (binary), you have enough digits in numeric notation as from 0,1,2,3,4,5,6,7,8,9. So that means from base 1 to base 10, we have enough digits from numeric digits so that we don't need to invent some symbols. But then, when we talk about some high base like base 100 or base 256, we've to accepted that there isn't enough symbols to categorized all the digits. So what do we do?
We introduced a notation called:
Base-Divide notation (or you can called it as Base-Power notation for yourself)
The notation looks like this:
(|1|2|3|4|)13 with 13 is subscript decode what the "number" in notation should be in base n
So what's useful?
When we derive a base, we need convert it to decimal for our human purpose reading. So the way we'll convert the Base-Divide Notation to Base 10 is very easy (just sort of general formula to convert base n to base 10)
(|n1|n2|n3|n4|...n(k)|)m
Where k is the "digit" of the notation and m is the base. Note that the value contains in n1, n2, n3, n4 should not be larger or equal to the base that they have set to (kind of 1023, there is no single digit that is larger than the base "10")
General formula to convert the Base-Divide notation to Base 10:
Examples:
(|3|6|7|9|)11 = 3*11^3 + 6*11^2 + 7*11^1 + 9*11^0 = 4805 in base 10
(|1|2|3|4|5|)15
= 1*15^4 + 2*15^3 + 3*15^2 + 4*15^1 + 5*15^0
= 58115
Note to know:
You cannot write the notation like this:
(|12|32|11|5|)19 where there is/are number that's larger than the base
You need to rewrite whenever you write "wrong" Base-Divide Notation (Technically, you can write the wrong notation for yourself if you don't publish it in paper or share this notation to another human than you know)
How to rewrite the notation when you write it wrong:
Example:
(|34|12|55|14|)23
Compute 14/23 = 0*23 + 14
Compute 55/23 = 2*23 + 9 (the 2 here is the result when divide so in next step we add the result to the remainder)
Compute 12/23 = 0*23 +12+2 (it's like carry sums from the long multiply that learn in elementary
school)
Compute 34/23 = 1*23+11+0 (actually, you can +0 if you want becuase +0 means "no carry before". But it's meaningless)
Then the result are:
(|34|12|55|14|)23 -> (|1|11|14|9|14|)23
So we done here... Not yet
The thing that you can do with this notation is plug any kind of formula that you can imagine, quadratic, cubic, sigma, pi, e, definite integral, taking a derivative value... Anything
Take this as example:
You can write like this:
(|13|21|4|n|)35 where n is a real number from 0 (|1|2|3|6|0|)7
(Note that we doesn't use the equal sign when rewriting)
Or like this:
(|n+2|n+1|n+3|)12 where n must be smaller than 9
(Real numbers plug in base? Use the base convert formula to solve)
This might be irrelivant but Ive heard from somewhere that (5^a - 3^b)/(2^c) has a solution for every whole number over 0 where a,b and c are also whole number. I dont know if it is true and why it is the case.
assuming you mean for every whole number n>0, there are whole numbers a,b,c such that (5^a - 3^b)/(2^c) = n, this is not the case. Let n be any positive multiple of 3, and suppose (5^a - 3^b)/(2^c) = n for some a,b,c. Then 5^a = (2^c)n + 3^b, but unless b = 0, the right hand side is divisible by 3, and so 3 divides 5^a, which is impossible. Thus the only choice for b is 0, so we have 5^a = (2^c)n + 1. Furthermore, if we also take n to be divisible by 5 (e.g. n=15), we have 5|(5^a - (2^c)n) and so 5|1, a contradiction. Answer is no, it does not hold for every whole number n over 0
PS: Both the {7/3} and the {8/3} stars contain the other type of star for their number of vertices ({7/2} and {8/2} respectively) within them, as well as the respective polygon.
This pattern seems to continue: {9/4} contains {9/3}, {9/2} and {9/1} as well. I postulate that this might be the case for all {n/ceil(0.5n)-1} stars.
To figure out the angle sum, you can also imagine a rod sliding back and forth along the lines. How much you rotate it is the angle sum.
There's an easy way to find the angle sum for any star and it's the same method as you used for the {7/3} star. The arc covered after two points in the star will be p-2q turns of the circle when added together, which you can divide in half to find the total angle. That's why it's 180*(p-2q).
This is very nice!!
is it necessarry though for the vertices to be on a circle? seems not so for 5 and 6, how about 7 or more ... oops: I see someone asked same question and @jursamaj replied
@@blackpenredpen plise arccos(x) =6
You can also draw a different 7 point star by alternating how many points you skip. Edit: sorry, I was thinking of a 8 point star
I am feeling smarter now, that I can say that I found all these myself when I was in 10th class. I also found a method to check how many such stars exist for n points.
Stars(n) = [(n-3)/2]
Where [x] is the floor function
Love your content as always !!! thanks for sharing your passion and keep it up !
That angle sum formula can be proved quite readily using the last proof you gave for the {7/3} star! The fraction of the circle that each angle subtends is given by (p-2q)/p. This can be seen because there are p many segments between the vertices, and q are not included clockwise, and q are not included anticlockwise! Then there are p of these, giving us a total of p-2q fractions of the circle, but with the half correspondence this becomes 180(p-2q)
I like that the formula for the angle sum deprecates into the polygon formula as well. If q=1 it's really a polygon rather than a star, but the formula turns into the polygon angle sum formula: 180(p-2)
You probably mean "degenerates" or "is valid also in the particular case" ;-)
Love geometry vids like this! Videos like yours inspire me to share my own maths content as well!
Cool! Just ran into a situation where I wanted to know the sum of angles on a 5 pointed star. I kinda asummed it would've been 360 intuitively but it appeared my intuition was just dead wrong so then I got interested in the general case and stumbled right into a trusted old channel. Genuinely baffled at how much more complex the issue is than I initially thought it'd be, but I knew I could rely on you to clarify the matter :p
Hello mentor,
I made several videos trying to convince people that square root of any positive real number is POSITIVE, but people still argue.
Could you please make a video to show the whole world the difference between y=√4 and y²=4
your cool reasoning for the 180° angle sum of the 7 point star could also have been used for the 5 point star ;)
The first illustrated 7-point star looks like an expanded version of the 5-pointed star. Haha.
By the way do some videos about the Pascal’s triangle and other different interesting patterns in mathematics (like Fibonacci, and others). :)
2:34 . . . Yes it is indeed possible to finish the star in one go.
Method (1)
Start from any one of the six intersection points (A) of the two triangles,
draw one of the triangles and return to A.
Immediately draw the other triangle.
Method (2)
Start from any one of the six intersection points (A) of the two triangles,
draw the hexagon and return to A.
Immediately draw the outer triangles.
Voila, star drawn easily in one go!
en.wikipedia.org/wiki/Unicursal_hexagram
I have my doubts about the 6 pointed star because every other star you‘ve drawn was drawn in a continuous line and had p points only if p was relatively prime to q but the six pointed star has p = 6 and q = 2 which are obviously not relatively prime, thus you had to draw to independent triangles. Furthermore I would argue that there is no way to draw a six pointed star because every number you choose that is less than 6 is either not relatively prime to 6 or it is congruent 1/-1 mod 6 so it would just form a hexagon. But I liked the video anyways ;)
In general, if d==gcd(p,q), the star can be drawn with _d_ continuous lines
@@giacomolanza1726 underrated
You've mentioned not being into geometry before, it's good to see you explore it
2:33 you can do it in one go it is called the Star of David and it is in the flag of my country (israel) I will happily send you how to draw this in one go if you want
I think the rule is we should start with a vertex and then rotate q times (like how I demonstrated in the video). I didn’t think harder on drawing that star in one go if I could start anywhere I wanted. But I see it now. 😆
Just wanted to point out : en.wikipedia.org/wiki/Unicursal_hexagram
That is really cool!
This is one of the things that got me interested in math back in the day lol.
Super Fantastic , Wonderful Amazing
I was lost at 9:44. The part where the 2 inner red angles equals to the other 2 red angles and how the inner are already considered
The 2 inner red equals to the 2 red in the quad, and the 2 red in the quad is already considered as sum of angles inside a quadrilateral, so you can ignore the inner 2 red
then all you are left with is the triangle because the inner 2 red angles which are not included in the triangle are now included in the quadrilateral. Hope that helps
Good explanation
@blackpenredpen can you solve the deceptively simple ( e^sin x = csc x ) ?
三角形，四角形。
五星形，由這個開始出現跳異點，
六星形，…
其實很好去區別，
先以單雙數區別
單數組3、5、7、9、11、13、15、17、19…
雙數組2、4、6、8、10、12、16、18、20…
因為星形所以先將234放一邊，
所以單數首即是5，雙數首即時6，
再來觀察六星與五星角的排列特性得出暫時的兩種可能A、B，
之後用途，n角星是不是趨近圓，解答嗎？
是趨近的。「但是不是實圓是虛實圓(在數學上)」,
然在現實世界中確實是圓，
主要差異在於現實中無法將極龐大的物體和及渺小的物體進行外觀上(比較計算)，
以上存屬瞎扯蛋抱歉了！
你用中文 所以我來幫自己打個廣告 我有個中文版頻道叫 黑筆紅筆 😃
Can you make a video where you find the angle of curvature of a circumference? Maybe thinking it as a regular poligon with infinitely many sides?
Here before someone comments right after me: "I watched the whole 23 min video in 3 mins! It was cool!"
Cooll..
But how?
If that were to happen, I'll applaud them heartily on their interdimensional time travel. Beats everything I've got.
Interesting 😊
There are four natural numbers N > 1 such that you cannot draw an N-pointed star in one go: they are 2, 3, 4, and 6.
There are four natural numbers N > 1 such that you can tesselate the plane with tiles of rotational symmetry N: they are 2, 3, 4, and 6.
The first fact I can account for given the kind of reasoning in this video. The second I keep _feeling_ like I can do the same, but when I try and articulate the reasoning it runs away on me.
So, it's that mean
Sum of angel {p/q}
= Sum of angel {p/p-q} ?
absolutely speaking... yes. same value, just opposite direction ;)
1:56 This shape is actually a kind of Jewish symbol which called in Hebrew "Mgaen David" (shield of David). You can see it on the Israeli flag.
Guardaré esta canal para cuando aprenda por completo hablar inglés
I really had to wonder if he didn't know he drew a Star of David at the beginning, and the (often inevitable) heavy implications of it, lol.
Couldn’t you say that q < p/2? I noticed that when you have a 7 point star, you only have q = 2, 3, and 7/2 = 3.5, and 3 < 3.5
quick question, are there non flat, or non euclidean surfaces were u can draw the 6 pointed star without lifting your pen? i am a dumb person and i will try, but i also want somebody more educated to look/explain.
Can u explain the angle sum formula? :)
Its so elegant but i dont know how to derive it. I know that for odd n stars that when q is (p-1)/2 that the sum is 180° because i found the angle sum in another manner. I found it by turning the seven point star into a five point star with out changing the angle sum with the exact same trick u turned a five point star into a triangle. But all the other sums are still a mystery to me.
Nevermind found it
Shout out to Stanley
Hello BRPR! I was wondering if you programmed a 100 Limits in one take video. I believe it would be as useful as the other 100 in one take! :D
Nice content sir👌👌
Subscribed because I know I'ma need that math sooner or later
Has anyone taken it to the third dimension? I have memories of the "Stella Octangula"; presumably an
intersection of planes in a sphere/ball.
gcd(p,q) = 1 is possible in one go because gcd(p,q) is equal to the amount of goes you have to go
I think it is possible to draw the 6 point star in one go because each vertex of the 6 point star is of even degree
Can find a formula for oscillating sequnce 0,1,0,1,1,0,1,0,0,1,0,1,1,0,1,0....
How to measure the angle of an arc
1;56 star of David
21:37
Dang I thought he was gonna write "coprime" :(
Hello can you explain gcd use calculus
That's so nice!
How about 3 dimensional stars?
Nice
This time i don't see Dr. Peyam 's comment 1 week ago
Sir is it mandatory to show face in the short video about ourselves..
Or is it fine even if not too?
No
@@blackpenredpen ok sir
Its Blackpen, Redpen, Bluepen
me after seeing the angle sum formula: 21:49
Do you like trigonometry???
Sir, I live in Bangladesh.. I like your video so much...
You are quite cool!
Some people just deserve a life, and others a like. - Yuansheng Zhong
ahh okok
thanks for ur AMAZING CONTENT!!!!!! i am so glad my dad told me about this channel in 2018!! but then sometimes i didnt watch u. but now i do! i had already watched this causer it was in a playlist, while unlisted. but best best luck for your future!!!!
I think that if we can draw {p/q} then it means that p should be relatively prime to q
If they're relatively prime, then you can draw it without lifting your pen. But We can still draw {6/2} even though they aren't relatively prime. If {p/q} is a valid star I think the rule is that you can draw it in gcd(p,q) parts.
next video: how to make a philosopher stone
In Discrete Math ,We call em Hamiltonians if you can draw em without liftin a hand or repeatin the vertices
Oh, hi Steve!
Hi Mike!
excillent class
simply ossom ! :))
What if you make 3 turns each time to draw a 6-point star
13:12
"Point stars" I saw something else!
Right after he made a video about how he doesn’t like geometry😂
I have concluded that blackpenredpen cut Dr Peyam =/0
Group theory???
Remember if you are comfortable in algebra more than geometry then complex numbers are there for you.
I will prove that the sum of angles of 5/2 is 180:
Let w be the 5th root of unity, then the vertices of the star are 1,w,w^2,w^3,w^4.
Now the sum of angles is
Sigma arg((w^k+3-w^k)/(w^k+2-w^k))
= Sigma arg((w^3-1)/(w^2-1))
= 5.arg((w^3-1)/(w^2-1))
And one can show that
arg((w^3-1)/(w^2-1))= pi/5
Thus the sum of angles is 5.pi/5 =pi
And so we are done!
(Moral: if you are not comfortable with Geo, learn complex numbers 😉)
@Complētus Ineptus Yes, you are right! I wrote that in a hurry, luckily it was still true 😂
I will edit it
Nice
there's a problem: your 7 pointed star was inside a circle, if it wasn't you couldn't prove the sum to be 180deg
Actually, you can. Start from the star in the circle. Now, move any point, while keeping it on one of its lines. The sum of the angles at either end of that line will remain the same. Using this method repeatedly, you can deform the circular star to any arbitrary star, keeping the angle sum the same at each step.
you have a PokeBall!!
You missed -1 point Star :/
Because of this video I think you should rename your channel as " Blackpen Redpen Bluepen".
Anyways nice video and very informative. Thank you sir. 😊
Me, too lol
Am I the only who knew about these already
Infinite star is circle
Why you deleted my comment suggesting you to hang the 6 point star on the gallows ?
?
I never saw such a comment.
@@blackpenredpen Maybe guys from youtube considered it as antisemitic
What da beard doin ?