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Anti Pythagoras 2: the Line, Ellipse, and Infinite Triplets
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- Опубликовано: 30 июл 2024
- Anti Pythagoras 2: the Line, the Ellipse, and the Infinity. How to generate infinitely many anti Pythagorean triplets of the form b^2 = a^2 + c^2 - ac The triplets relate the sides of a triangle where one angle is 60 degrees. It is a variation of the Pythagorean theorem from 8th grade math. It has to do with intersecting an ellipse with a line of rational slope and performing some clever algebra like expanding quadratic polynomials and quadratic forms and comparing numerator and denominators of fractions with integers. Then we give an example of a triangle with such a triplet.
Special thanks goes to Ian Fowler who has introduced me to this problem. Ian would like to give a big shout out to Emidio Iacobucci and Frank Cirone for their part in the adventure.
0:00 Introduction
1:40 Rational Slopes
2:40 Finding Nemo
6:00 The triplets are born
Anti Pythagorean theorem: • the anti Pythagorean t...
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I want to take a moment here and personally thank everyone who watched and/or commented on the 2 videos. Your comments have given me some cool insights and extensions that I did not see before. Many of you have an expertise well above my pay-grade and I found your input both generous and helpful. Collaboration adds so much to the experience.
What was the name of the student?
Very impressive. It's a shame you're not more enthusiastic about it (joke). I marvel at your "child on Christmas" excitement. Thank you for making this Math analysis so intriguing.
Consider the line passing through P(0,-1) and intersecting the ellipse at the 2nd point. Now rotate that line about center P and imagine that as it sweeps around it will intersect the ellipse in a set of continuous real valued co-ordinates. Now each time the second point of intersection happens to have rational co-ordinates that will correspond to a unique values of m and n which leads to a unique primitive solution to :
b^2 = a^2 + b^2 - ab
A 1-1 correspondence.
The analogous identity for Pythagorean triplets (m² - n²)² + (2mn)² = (m² + n²)² shows that there are an infinite number of them (in terms of coprime integers).
I already knew that, so I had correctly guessed that the anti-Pythagorean triplets also would have a similar equation. Unfortunately I didn't succeed in finding it. So I'm very happy to see its final form here, after Dr. Peyam did all the hard work!
There we go! Yes, excellent observation Niranjan! Dr. Peyam's parametrization is not the same but I was wondering why the expressions looked familiar!
Also interesting to note that replacing M by M-N gives the same triplet in the opposite order (e.g. M=5, N=2 gives (16,19,21), while M=5, N=3 gives (21,19,16)). Which can be proven directly by doing that replacement in the three formulas.
Indeed!
Very interesting, Dr. Peyam and Ian! In the past I worked on finding dissections using Pythagorean triplets (I found various very interesting and beautiful infinite series), this video inspired me to think about using these anti-Pythagorean triplets for dissections as well. This caused me to notice that for every valid triplet a,b,c with a^2 + c^2 - ac = b^2 that also (c-a),b,c is a valid triplet. It is very easy to prove this. The 5,7,8 and the 3,7,8 that are both shown in this video, clearly are each others complements this way.
Indeed!
Thanks Dr. Peyams Körper for producing this video!
Very well explanation of anti-Pythagorean theorem. And i see it's easier to achieve general formulas for a-P triplets than Pythagorean where you need to start from stereographic projection. Found the formulas on "Topology of numbers" by Allen Hatcher.
Love the way u enjoy maths .
Thank you very much sir 🙏
Nice video, thank you! This kind of idea is also used in algebraic geometry. For instance, the rational parametrization of the circle yields all pythagorean triples (see en.wikipedia.org/wiki/Birational_geometry#Birational_equivalence_of_a_plane_conic); indeed, the parametrization in question defines a bijection between the rational numbers and the rational points of the unit circle (except one, which corresponds to the "point at infinity"). For this reason, I believe that the rational parametrization of the ellipse considered here also yields all anti-pythagorean triples.
wow thank you for introducing! i think that can be a good motivation for birational geometry
Nice to see there's another class of Pythagorean triplets Dr. Peyam! Is there a reason why we can not use the ordered pairs (x,y) = {(1,0), (-1,0) and (0,1)} as inputs to make another possible linear parametrization?
Perfectly fine to use any of those.
@@ianfowler9340 Yo! The man himself!
Very curious to see what comes out of those new formulas and how they relate to the triples in this video. Could they not hypothetically relate to other triangle constructions?
it is already the complete set of solutions. because x,y have to be rationals for every solution - and that implies a rational M/N. (the starting point doesn't matter as long as it has rational coordinates. but (0,-1) gives the nice constraint M>N, because the ellipse has slope 1 at that point)
i love this! it seems like generalization of pythagorian triplets
I like it,and i have already subscribed to your great channel.
Very nice cliffhanger. I really like your videos. Please go crazy :)
Will do!!! Love your profile pic btw 🏳️🌈
@@drpeyam thanks. Colour codes for resistors :)
Nifty. I’ve never seen it before and don’t think it was in the literature even of Ancient India. So it opens the door for some weird factorization exercises because, if you equate the a formula with a number you wish to factor, call it c, you can consider N and 2M-N factors. Then b is a number you can guess by adding another parameter on a, thus factoring a in afew steps using the derived system. Call the parameter k, then k=-1 gives you seven. Add the two and get m^2+mn= 2a-1=15, factoring fifteen, noting m divides fifteen and by construction m
An equilateral triangle with side length 3 is the simplest of this infinite group of Anti- Pythagorean family of triangles.corresponding to M=2 and N =1 :-)
M=1, N=0 gives you (a,b,c) = (0,1,1). In the original problem with the three tangent circles, the one circle collapses to a point.
Thanks Doc! This video helped me make progress on my Diophantine conic project without using number theory algorithms. For example, picking a line with slope 7/5 and using the "5 points define an ellipse" method, leads me to the implicit function x^2-x y+y^2=1521, which is guaranteed to have integer solutions (18 in this case).
thanks so much!!! Good luck
Wow; that is exiting!
Loved it! NB a, b, and c must all be positive integers such that M>N or a and b would be zero or negative.
Bonne Explicitation Dr Peyam. Greeeeat
Beautiful
If we're talking about integer length sides of triangles, you could build a complete list using by starting with sides of length one and employing the triangle inequality to produce all triangles with those two sides.
(1, 1) > (1, 1, 1)
(1, 2) > (1, 2, 2)
(1, 3) > (1, 3, 3)
(2, 3) > (2, 3, 2), (2, 3, 3), (2, 3, 4)
(3, 4) > (3, 4, 2), (3, 4, 3), (3, 4, 4), (3, 4, 5), (3, 4, 6)
and so on.
For (8,7) you can have triangles with sides of length 8 and 7 with the third being 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, or 14.
The "anti-pythagoras" triples' construction is: a=2mn-n^2, b=m^2-mn+n^2, c=m^2-n^2
The pythagorean triples' construction is: a=m^2-n^2, b= 2mn, c=m^2+n^2
So, the only difference is one of the triangles' sides is offset by that correction term of mn
Perhaps there are other similar constructions to derive the rest of the integer triplets.
I am not sure if it is just how he speaks, but this man is really excited about triangles.
i believe this solution generates all possible solutions if we go through all (m,n) pairs. the reason is that any solution would be a rational point on the ellipse, and hence, the slope of the line connecting it to the point (0,-1) would have rational slope equal to m/n.
Thanks!
This is making me recall another video on how there's no simple equation for the perimeter of an ellipse. For some reason my brain is connecting the two concepts, but i cannot fathom how they intersect.
The real maths are the friends we made along the way
Awwwww
Nice
There are some interresting facts about the ellipse x^2+y^2-xy=1.
1. It intersects the x- and y-axis in the same points as the unit circle.
2. If one intersects it with the unit hyperbola x^2-y^2=1, then the x-coordinate of that point is equal to the eccentricity of the ellipse.
3. The semi major axis of the ellipse is equal to the eccentricity of the hyperbola.
Wow that is interesting!
Nice
2. There are 4 points of intersection: +-(2/sqrt(3),1/sqrt(3)) and (+-1,0) (of which we want the former). More importantly, what you call eccentricity is better known as linear eccentricity (distance from center to a focus), which is different from what's usually referred to as eccentricity (linear eccentricity divided by semi-major axis).
3. I was about to say something similar here, but then realized that this hyperbola's eccentricity and linear eccentricity are equal....
Very nice! I just wish someone would figure out whether there are finitely many or infinitely many twin primes 🙁
sorry for my english
1. why is it anti pythagorian if ac is always > 0?!
2. how do you get the (3,7,8) triangle with this formula? or does this formula just proves that there are infinitely many triplets, but doesn't find all of them?
3. for every (integer) triplet there is an "adjacent" (integer) one. like (3,7,8)&(5,7,8) and (13,43,48)&(35,43,48). if(x,y,z) is an ordered triplet (when x
thanks
What are the properties on the back of your shirt for? They kind of look like the properties of a linear (or vector) space, but there are some stuff I don’t recognize (this is my first year of Linear Algebra actually.)
It’s the properties of a field, like R or C
An ellipse is just a circle viewed off center. Imagine an angled slice through a cone (funnel). Now continue to turn the view till the plane of the circle(ellipse) is on edge. Viewed this way in two dimensions you have a triangle. I commented on a video a couple years ago when the presenter was doing some tortured calculations trying to determine the area of an ellipse. Seems they can understand they are barking up the wrong tree. The area is always going to be a circle but what you want to solve for is the viewing angle in 3 dimensions like you have drawn on your grid.
Dr Norman Wildberger of the University of New South Wales has many videos on RUclips describing the laws of his rational trigonometry, which parameterizes the circle with rationals and avoids irrational square roots and the concept of angle.
This is great. Thank you.
I did find some duplicates for coprime m and n. A stronger constraint on m, n than just being coprime seems necessary to prove an infinity of solutions (Leaning on the Pell equation, I believe n^2 - 3 m^2 = 1 would do; but possibly n^2 - 3 m^2 = k^2 for some k might do as well and include a broader class of such triangles. (Of course the ellipse x^2 - xy + y^2 = 1 is just the ellipse x^2/2 + 3y^2/2 = 1 rotated by 45 degrees CCW.)
Duplicates:
- (n,m) = (1,3), (2,3) both produce (5,8,7), permuted
- (n,m) = (1,4), (3,4) both produce (7,15,13), permuted
- (n,m) = (1,5), (4,5) both produce (3,8,7), permuted
- (n,m) = (1,6), (5,6) both produce (11,35,31) permuted
- (n,m) = (1,7), (6,7) both produce (13,48,43) permuted
- (n,m) = (2,5), (3,5) both produce (16,21,19) permuted
- (n,m) = (2,7), (5,7) both produce (8,15,13), scaled
- (n,m) = (3,7) and (4,7) both produce (33,40,37) permuted
- (n,m) = (3,8) and (5,8) both produce (39,55,49) permuted
It is easy to show there are infinite possibilities: take n=2 and let m be any odd number larger than 1.
Then you get the triplets 8,7,5; 16,19,21; 24,39,45; 32,67,77 etc,
or with for some positive integer p: the triplets 8*p, 4*p^2 + 3, 4*p^2 + 4*n - 3.
for comparison P triples… (a, b, c) = [ (m2 − n2); (2mn); (m2 + n2)]
Ellipses are so cool
And ellipsoids are even cooler
I want that German pun shirt!
It’s in my teespring store
If you were to take those definitions of a, b, & c, and substitute them into the 3 versions of the law of cosines, you should get true statements, right?
You say that the line y = ( m / n )x - 1 goes through 45 degrees for an angle. However, a slope that corresponds to a 45-degree angle is 1.
No I don’t say that the line has a 45 degree angle, I say that the ellipse is tilted by 45 degrees
Thanks Michael Empeigne - I think we have misconception here but to be honest I wouldn't have noticed it if you didn't point it out. But yeah as Dr. Peyam says the whole ellipse is tilted by 45 degrees not the line has 45 degree angle. If someone can clarify the difference between these two that would be nice. Great work Dr. Peyam!
He is talking about all lines (with rational slopes) that intersect the ellipse at (0, 1). Those lines do not have to have a 45 degree slope.
Does that mean, that there are infinately many ways to construct an angle of 60 degrees by using whole numbered sidelengths?
Provided that the converse of the anti Pythagorean theorem holds
Cool T-shirt
Okay, so it’s parameterized. What if it’s different from (0,-1)? What about b2 = a2 + c2 + ac ? Looking forward to the next video
The +ab applies to 120 deg triangles. Just be careful which of the 3 side lengths is the longest. In the -ac version b has to be the middle length since it is opposite the 60 deg angle. In the +ab version the c^2 is the longest side since it is opposite the 120 deg angle.
The research note below shows that 60, 90 and 120 degree angles are the only possible ones for triangles where the cosine is also rational. So these infinite sets of whole number triangles with 60 and 120 degree angles are more like partners to pythagorean triangles: semi-pythagorean instead of anti-pythagorean. It was really fun to learn this new bit of math! Definitely worth writing up somewhere.
www.uni-math.gwdg.de/jahnel/Preprints/cos2.pdf
For c=5, the solution give by your equations is (a,b,c)=(8,7,5) but there is other solution (a,b,c)=(21,19,5) not given by your equations and also the trivial solution (a,b,c)=(5,5,5)
That's not how you're supposed to use the equations. You don't pick values for a, b, or c. You pick values for M and N (coprime, with M > N). Also, scalar multiples of a solution are considered the same, so e.g. (5,5,5) and (1,1,1) are the same solution.
M = 2, N = 1 gives (3,3,3) = 3*(1,1,1).
M = 3, N = 2 gives (8,7,5).
M = 8, N = 7 gives (63,57,15) = 3*(21,19,5).
@@cQunc ok, i agree but in pythagorean integers triplet the formulas with m and n give all solutions, here your formulas don't
@@rachidiksi994 No, you don't get all Pythagorean triplets, e.g. not (9,12,15). Just like the formulas here, they're only guaranteed to give you all solutions up to common factors. It's a good point, though, that you can't always get the coprime version of anit-Pythagorean triplets, like (21,19,5). For Pythagorean triplets, I'm pretty sure I've proven to myself that you can.
Why is the correction term ac? Is this the definition of antipythagorean triplet?
Yes
There is a flaw in your derivation. One angle can be bigger than 90°. Therefore a^2 = b^2 +c^2 +bc is possible too. If you derive formulars with your method you get addional solutions. For exsample you get (16,19,21) but not (5,19,21). In both cases the 60° angle is opposite of 19.
Your identity works for 120 deg triangles
@@ianfowler9340 No. (5, 19,21) has one 60° angle. You don't get this solution from the other formula. There ar solutions with one angle between 90° and 120° and one with the difference to 120° und of course one 60°. The method using a rational point of a curve, a line intersecting this point and calculate the other intersection point is right, but there are 4 points. And there is a second curve.
I calculated triangles with both sets of formulas and I calculted the angles. I did it with excel. The angles of each triangle were calculated with die cosine law. Therefore there is no space for a mistake. I varied M and N much more then in the video suggested. Just to learn wether the first set of formulas give all possible answers by extending the range of M and N. You get impossible solutions with sum of angles more than 180° using N bigger than M or 0 or negative. So it is not possible to get all possible solutions with the first set. But with the second set of formulars one can use negative N's but they must be bigger than -M/2. But it doesn't deliver all solutions either.
I made a python program to calculate pythagorean triples. The input is a number and the program delivers every possible Pythagorean triple with the smallest angle. I want to make one für antipythagorean triples too. At the moment I struggle with the formulas.
I keep on investigating this problem. Some results: The smallest non-trivial antipythogorean triple is not (5,7,8) with. It is (3,7,8). The 60° angle is opposite to 7 - in both cases. Smallest edge of both solutions add up to the longest vertice. That is no coincidence!
On the other day, Dr. Peyam did mention the last term with cosine.
@@demophilo1 Since you are squaring a and c and multiplying ac you can negate both a and c without changing the values of a^2, c^2 and ac.
b = m^2 + n^2 - mn
a = n^2 - 2mn
c = n^2 - m^2
also works.
m = 2 and n= -3 yields b = 19, a = 21, c = 5
m = -2 and n = 3 also yields b = 19, a = 21, c = 5
from
b = m^2 + n^2 - mn
a = 2mn - n^2
c = m^2 - n^2
m= 5 and n = 3 yields b = 19, a = 21, c = 16
(m = 5 and n = 2 flips a and c)
Also each 60 deg triangle is a side,side,angle (ambiguous case) and produces 2 triangles. Your (5,19,21) pairs up with the (16,19,21).
This is also why (5,7,8) and (3,7,8) are paired together. They both come from the fact that a side,side,angle triangle with angle = 60 deg has 2 triangles that satisfy the SSA. We used to call this the Ambiguous Case in teaching the Sine Law. So you are correct - it's no coincidence! I like to call them "sibling" solutions.
I should also add that one of m,n being negative means the line intersects the ellipse in the 2nd quadrant.
This reminds me of the Minkowski metric
if x=a/b, and at the same time x=c/d, this does not give us the right to consider a=c and b=d.
for example, according to your logic x=1/2 and x=2/4 then 1=2 and 2=4
It’s ok actually, I never said I found all the solutions, just infinitely many of them
X*Y=a/b*c/b=ac/b^2......i think there is wrong here ??
No?
i knew this already
Good for you?
2:05 now i know why there r *straight* lines .. in contrary to -krumm- .. _just_ lines" ^v^
Show the anti pythagorean pyramid.
So it took the guy 30+ years and a bunch of university colleagues to solve a math problem he'd been giving to grade schoolers? Cool.
No? I think you misunderstood the context
@@drpeyam I was exaggerating the truth for humor.
I don't like the name "anti-Pythagorean" theorem, that is a discourtesy to Pythagoras. Why no call it "The non-standard theorem of Pythagoras." or the "elliptical theorem of Pythagoras" , or, better still, "the Peyam-Fowler theorem"
The standard theorem of Pythagoras relates to circles and this one relates to ellipses. Perhaps there are other ones that relate to to other conic sections, that deserve to be investigated.
Fun fact: This guy is left handed.
That's "M divided by N", not "M over N". "M over N" would be M! / [(M - N)! N!].
No? M over N is M/N
What you’re referring to is M choose N
How many neoplatonists are here because of the click-baity title?
Eminem?
wasnt it fucking 8,7,3 ???
x^2+y^2±x*y=1
area= (2/3)*Pi*sqrt(3)
perimeter=4*sqrt(2)*EllipticE((1/3)*sqrt(6))=7.1343450992015408312
EllipticE(k)=sum of 0 to 1 (sqrt(-k^2*t^2+1)/sqrt(-t^2+1))
Ellipse Positive Set, x^2+x*y+y^2=1, :
[x/sqrt(x^2+x*y+y^2), y/sqrt(x^2+x*y+y^2)]
[(x^2-y^2)/(x^2+x*y+y^2), y*(y+2*x)/(x^2+x*y+y^2)]
[(x^3-3*x*y^2-y^3)/(x^2+x*y+y^2)^(3/2), 3*y*x*(x+y)/(x^2+x*y+y^2)^(3/2)]
[(x^4-6*x^2*y^2-4*x*y^3)/(x^2+x*y+y^2)^2, (4*x^3*y+6*x^2*y^2-y^4)/(x^2+x*y+y^2)^2]
[(x^5-10*x^3*y^2-10*x^2*y^3+y^5)/(x^2+x*y+y^2)^(5/2), y*(5*x^4+10*x^3*y-5*x*y^3-y^4)/(x^2+x*y+y^2)^(5/2)]
Ellipse Negative Set, x^2-x*y+y^2=1, :
[x/sqrt(x^2-x*y+y^2), y/sqrt(x^2-x*y+y^2)]
[(x^2-y^2)/(x^2-x*y+y^2), y*(-y+2*x)/(x^2-x*y+y^2)]
[(x^3-3*x*y^2+y^3)/(x^2-x*y+y^2)^(3/2), 3*y*x*(x-y)/(x^2-x*y+y^2)^(3/2)]
[(x^4-6*x^2*y^2+4*x*y^3)/(x^2-x*y+y^2)^2, (4*x^3*y-6*x^2*y^2+y^4)/(x^2-x*y+y^2)^2]
[(x^5-10*x^3*y^2+10*x^2*y^3-y^5)/(x^2-x*y+y^2)^(5/2), y*(5*x^4-10*x^3*y+5*x*y^3-y^4)/(x^2-x*y+y^2)^(5/2)]
Nice