the anti Pythagorean theorem?
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- Опубліковано 29 чер 2024
- an anti Pythagorean theorem. Here is a geometry identity related to Pythagoras from 8th grade math which you may not know about. It relates the sides of a triangle where one angle is 60 degrees. It all starts with a high school problem about triangles and circles and uses the law of cosine from precalculus and leads to a special 8 7 5 triangle with integer sides. Then we'll see if there are other triangles of this form, which involves the quadratic formula. In a surprising turn of events, this is related to ellipses and lines. Special thanks goes to Ian Fowler who has introduced me to this problem. He is a retired high school math teacher and graded math competition problems at the University of Waterloo in Canada
Part 2: • Anti Pythagoras 2: the...
0:00 Introduction
0:30 Problem
3:10 Special Triangles
4:50 Thirty Years Later
6:50 Miracle
Law of Cosines: • Al-Kashi's Mysterious ...
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Hello again Peyam,
All I can say is wow, double wow, n factorial wow!! You have really captured the excitement and the and the joy of the whole experience. I want to thank you again for your wonderful generosity and enthusiasm- you really captured “the moment” and made my day! I can hardly wait for Part 2 so I can send out the links to my close friends who were a part of this. This is just so unbelievably cool. How long do we have to wait for Part 2?
Thanks again - Ian
Thanks so much for the wonderful idea, I genuinely had fun making the video! I think part 2 will be up within a month or so, or I’ll make it available for YT members, once it’s available
I am from the University of Waterloo, feels so nice to see that you've worked with profs over here!
I attended Central Michigan University almost 30 years ago, and twice visited Kitchener for computer programming competitions… "our" regional was held at U Waterloo! That's when I learned that "we" were known as "the other CMU" (LOL) and Canada has a lower drinking age. My favorite memory of that competition: having a blast at "Kaboom!"
@@YBwaibeeYB I was in the Teaching Option and grad in 1977 with a BM and teaching credentials. I started marking Descartes Contests back in the days of grade 13. Still marking the Senior in Dec and Euclid in April. UofW was the best thing that that could have happened to me out of high school. Thanks for the shout out
@@YBwaibeeYB A lot of the markers are current/retired high school math teachers along with many of the profs
I have been lucky to keep those contacts active for so many years.
Folks, Dr. Peyam didn't say it explicitly-he probably should have-but to me, it was clear that he meant solutions that are not simply integer multiples of a base case. He meant that there are an infinite number of solutions _where the sides don't have common factors,_ that is, gcd(a,b,c) = 1.
That would be kinda trivial, wouldn’t it? 😂
@@drpeyam kinda primitive or coprime
I think he didn't say it explicitly because it was too obvious. The sides have to be coprime to count as interesting.
@@chessandmathguy No doubt. Then again, did you notice how many viewers said that you can simply scale the base (coprime) case?
@@nHans I didn't notice, but maybe I didn't look hard enough.
"No, who cares?"
When it comes to mathematics, someone always does
the "who cares" mentality is just the crappiest kind of things to do in mathematics and everywhere. The thing no one cares about at some point in math becomes the next breakthrough discovery that helps other fields of science. And in my opinion it's a sign of low intelligence.
Every knowledge is useful.
exactly. Mathematics deserves to be explored for its own sake. It doesn't need to prove to be immediately applicable
@@themibo899 Yeah
Having watched Dr. Borcherds solve this for Pythagorean triples, I feel confident I can do this.
We want to solve a^2 + c^2 - ac = b^2 over the integers.
We can set x = a/b , y = c/b. Then we have to solve x^2 + y^2 - xy = 1 over the rationals.
This is the equation of the ellipse Dr. Peyam mentioned.
We want a nice way to characterise the rational points on an ellipse. One way is this:
Pick a point on the ellipse (say (1,1) ) and draw a line through it with gradient equal to some rational number t. So, the line has equation y-1 = t(x-1).
When we substitute this into the equation for the ellipse, we'll get a quadratic equation in x or y with rational coefficients. Since one of the solutions is 1, the other must be a rational number too.
Conversely, drawing a line between any rational point on the ellipse and (1,1) will give a line with rational gradient. So, there is a one-to-one correspondence between rational numbers t and points on the ellipse (excluding (1,1) ).
After a bunch of algebra, you can get the following solution for x,y:
x = (t^2 - 2t)/(1 - t + t^2)
y = (1 - 2t)/(1 - t + t^2)
So the primitive solutions for the triangle side lengths are (I've sent t -> -t so that most of the interesting solutions occur for positive t).
a = 2t + t^2
b = 1 + 2t
c = 1 + t + t^2
All other solutions are multiples of these.
(The 120 degree case is left as an exercise)
Very nice!
I think this works for conics in general i.e. quadratic plane curves considered projectively. It shows you can make an inverse stereographic projection of line points (1, t, 0) to the curve by quadratics in t, if you know a point P of the conic. If the conic's coefficients are rational and P and t are rational, so is the projected point.
You have the 1 parameter solution. Well done. So cool.
Yes, the 3,7,5 triangle is a solution for
c^2 = a^2 + b^2 +ab and there an infinite it # on solutions to the 120 deg triangles
@@ianfowler9340 And what about other angles (other than 60 and 120) with rational cosines?
The condition b²=a²+c²-ac (where a
There has been some confusion about the (8,7,5) 60 deg triangle (b^2 = a^2 + c^2 - ac) . The (8,7,5) 60 deg triangle actually spawns another "sibling pair" (8,7,3) with the SAME 60 DEG ANGLE. Here's how to get the sibling. Draw the (8,7,5) roughly to scale. Vertex opp. 8 = C, Vertex opp. 5 = B. Vertex opp. 7 = A. Now rotate BC center B until it intersects AC again at D. We now have BD = 7 - isosceles triangle. Now use the cosine law and factoring the quadratic on triangle BAD to find AD. Factors nicely to get AD = 3 and leaving DC = 2. Now we have the "sibling" (8,7,3) triangle with the same 60 deg triangle. I wish I could show a diagram, but I don't think I can here. This is a very similar set up to the ambiguous case with the sine law (SSA) The 120 deg triangle is another story
@Ian Fowler. Great clarification - I did wonder why (8,7,5) in the start of the video was related to (8,7,3) at the end.
so it was ambiguous case triangle, where you can get many triangles by acccidentally modify the diagram
@@somethingisglowinganditsnotyou There will always be the amb. case since the middle side is always opposite the 60 deg angle (the middle angle)
This is wild.
I think that by just prolonging the shortest side (the 5) until it has the same length as the longest side (the 8), you get an equilateral triangle divided into two unequal parts which are the two sibling triangles. In the end, if you want to get a triangle PQR and you have its sibling PQS, where P is the largest side and S is the smallest, you just need to do R = P - S and you are done.
Your passion and excitement while telling about this subjects is amazing. Congratulations bro, and really amazing math out there!
This was super interesting, thanks! Loved the story telling too!
Thanks Dr Peyam, love your story telling. Waiting for the next video on line intersecting ellipse.
I appreciate the cliff hanger, because it gave me a chance to try and work out the relationship between the identity and the secants.
I didn't get a concrete proof or anything, but I found that one can get Heronian triangles by connecting the foci to the points where certain secants intersect the ellipse. The only thing is, I can't figure out the relationship between those secants, and I'm sure that's where the really interesting part lies.
You’re on the right path!
I enjoyed the whole video but the last part really grabbed me. I've been looking for a way to generate diophantine conics, and as soon as you said "intersection of a line with an ellipse" I predicted I will learn something amazing from the 2nd part.
I love how excited you are about this! It's very cool
Amazing video ! It's nice to see such excitement about simple mathematical problems
Anxious to watch pt 2
this is so amazing! maths is indeed so interesting, and even little observations like this can be so fascinating!
Agreeeeed
The miracle here was more than one person being excited about it. Once in a life time experience. Great story.
I competed in mathematics in high school where we played with fun problems like this.
I haven't had tingles like this in 35 years!
Thank you
Incredible video!!! Now I have to see the next one!!
Hi Prof. Peyam. Thank you for this magnificent video. And for the encouragement: never give up.
I just love how he is so pationate, and enthusiastic about this. It's so rare to see that, and I also wouldn't see my math teacher so excited about this too... Thanks for the amazing content!
Thank youuuu
Looking forward to seeing the sequel with the line intersecting the ellipse Dr. Peyam!
Awesome video! Thank you!
“ Today I want to tell you the greatest math story ever.” Greatest intro ever
It really made my day. Brought a tear to my eye. Math is FUN.
My Körper is super excited to see the next video about the ellipse.
As a former math major at university, and now math tutor for all kinds of grade students, I've always been interested in mathematical concepts like these. That's how I also know that with two exceptions (0 and 3), x^2 - 1 is always composite and I can easily tell you the factors of 323 (1, 17, 19, 323) off of that. I don't remember what it was over, but my math professor even decided to name one of my mathematical epiphanies after me when he used it in class. I've always been fascinated by alternative methods and such in math, so thank you for this fascinating watch!
I really appreciate the enthusiasm :)
I love how you look even more enthusiastic about this than the actual math teacher would be.
Actually George Kitchen, former math teacher at Portage Northern high school, was that way, with his Jacky-wacky and slashy-washy. There’s a lecture series at Kalamazoo College named after him
Well maybe - but I was jumping up and down.
Funny you should bring this up. Six years ago, in Jan 2016, I was toying with 120º triangles with integer sides, and came up with a way to generate infinitely many of them.
After some algebra, the method I came up with boiled down to:
• Choose integers (d,e) s.t. ½e < d < (√3 - 1)e
• Form the triple (α, β, γ) = (2ed-e², e²-d², e²+d²-ed) = ([2d-e]e, [e-d][e+d], [e-d]²+ed)
• Reduce these by their GCD, call it g, to get (α, β, γ)/g = (a,b,c) = sides of a ∆ in which ∠C = 120º, so that
c² = a² + b² - 2ab cosC = a² + ab + b²
The smallest (d,e) that can satisfy the condition of the first line, is (2,3), which produces the ∆ with sides (a,b,c) = (3,5,7). Some other triples are
(5,16,19), (7,33,37), (7,8,13), (26,39,49), (9,56,61), . . .
It's really cool to see where this collection of integer 120º triangles was stumbled across earlier. (Thank you, Dr. Peyam!)
I like to think of that (3,5,7) triangle as the 120º analog of the (3,4,5) rt. ∆.
Having said all this, I now go to the part 2 video to see how it was developed by those guys...
Fred
Great video!
Also, may I point out that -8 is also a reasonable solution for Y (at time 6:10). It's actually drawn on the left already! 8 units on the left, as the other side of the triangle on the left is also 7. Never ceases to amaze me how the equation one sets up gives you all - even those answers you may not have considered or sought to begin with.
Thank you, very helpful explanation 👍
Very interesting topic, thank you Mr. Payam
Dr Peyam, you have the best mathematics channel. Keep Going
I look forward to seeing the geometric interpretation with ellipses. Meanwhile, expressing
b² = c² + a² ± ac
as a quadratic in a and solving yields
a = ½ [ ±c ± √( b² - 3c² ) ]
which has integer solutions a, b, c whenever the discriminant is a perfect square; i.e. that for any integer solution to the Pell equation
p² - 3q² = 1
we set
- for odd p: b = p, c = 2q, and a = ½(c ± 2).\; and
- for even p: b = ½p, c = q, a = ½(c ± 1).
For (p,q) = (2,1) this yields (a,b,c) = (1,1,1).
For (p,q) = (7,4) this yields (a,b,c) = (3,7,8) or (5,7,8).
For (p,q) = (26,15) this yields (a,b,c) = (7,13,15) or (8,13,15).
The double solutions are from the set angle being either 60 or 120 degrees, with opposite sines for cosine, and I'm guessing are distinct triangles from the same cut on the ellipse but going to different foci or vertices.
The additional solutions above are generated from the base solution (2, 1⋅√3) as
(p', q'⋅√3) = (2, 1⋅√3)ᵏ
for any natural number k > 1.
These are only the solutions where the discriminant n² = 1². Other solution families can be obtained through setting c to some even number, and using square-gonomons to then determine b and n. For example:
Setting:
- c = 8 yields n = 11 and (a,b,c) = ( 7,13,8) or (15,13,8)
- c = 10 yields (a,b,c) = (16,19,5) or (21,19,5) for n = 37; and (3,7,8) or (5,7,8) again for n = 11.
p² - 3q² = 1 rearranges to p² - 1 = 3q², (p-1)(p+1) = 3q².
So solve p-1 = q/k, p+1=3qk for arbitrary integers q and k.
I guessed there might be a Pellian somewhere in there. I actually found the method by diagonalising the quadratic form a^2+b^2-ab-c^2, details higher up.
My solution will be different :)
@@drpeyam I count on it. ;-)
Yes that's why if the first int. point has rational co-ordinates and the slope of the line is rational then the second point must have rations co-ordinates.
I tried solving the diophantine b^2=a^2+c^2-ac for integers a, b and c and got the following parametric representation of the three sides.
c=t, a=(1+3t^2)/4, b=(3t^2+2t-1)/4. Putting t = 3 we get the triangle mentioned in the video with sides c=3,a=7 and b=8. putting t=5 we get a triangle with sides c=5,a=19 and b=21. Ive compiled a list of a few more triangles obtained by putting different values of t.
a=19,b=20,c=5
a=27,b=40,c=7
61,65,9
91,96,11
Infact one can show that for all odd values of t, there is an integral solution.
A couple days ago I found a method that finds solutions for all positive integers for the initial input.
Love your videos Dr. Peyam! I must thank you for teaching me so much calculus :)
You’re welcome!!!
Very good, man !
interesting and excellent lesson on Pythagorean Theorem!
These and the normal Pythagoras ones are handy when building special shapes, for example buildings with other than only 90 degrees walls.
can't wait for that video
triangles with a 120 degree angle seem interesting too
I think they work out to be of the form
b^2=a^2+c^2+ac instead of minus
which means for every triple abc for a 60 degree angle, b+2ac is the b for a 120 degree triangle with sides ac.
Isn't it painfully obvious that there are infinitely many triangles with integer sides? Any integer solution to a+b
That’s not what the problem is saying
Wow. just wow. Thanks for your video!
I am so grateful to Dr. P for telling my story. He has a natural talent and enthusiasm that influences everyone in Mathematics Education in such a positive way. So cool!
Part 2 now up: ua-cam.com/video/Sxq2hA7qZYM/v-deo.html
4:02 - wouldn't any integer multiple of the integer side (8,7,5) triangle also be a cool triangle that satisfies it? I'm guessing what you mean is that a (48,43,13) triangle is of a different form? I mean, a (48,42,30) triangle is an (8,7,5) triangle, just 6 times bigger. Later on at 4:37 it's a bit of a tough sell that DB is 5 if CD is 3 the way you've drawn it! CA and CB are equal, so it should be an isosceles triangle with a low peak and wide base, not scalene looking. Maybe use graphics to scale instead of sketches on a white board? Oh, and let's not forget that at 6:33 the (3,7,8) [aka (8,7,3)] triangle miraculously re-appears - even though the one he was talking about earlier was an (8,7,5)...? Something's wrong, or something got missed. Am I seriously the only one to notice this?
I can't wait for the sequel!
I saw a similar problem in Gordin's collection. it was necessary to find the radius of the circle passing through the points of contact of the three circles. So I think I can make a decision. Thanks for the task!
P.s. I am writing all this with the help of a translator, because I am Russian
This is really awesome and a proof to me that mathematics get discovered rather than invented. Quicky going to watch the next episode in this triangle-fiesta.
Also, you can count on students to say things like "that's obivious, its like ... " :-D
The equation can also be presented as b^2=(a-c)^2 + a*c.
The most basic of these threesomes, of course, is (1,1,1)!
The Euclid competition... the Pascal, Descartes, and Fermat competitions were the other grade levels, as I recall.
Also, making circles of various sizes, we can construct many such triangles, of various lengths. Like say if we have radii A, B, and C, then we can make triangles having side lengths A+B, B+C, A+C. The radius sizes just either need to be all positive integers, or all 0.5 less than a positive integer. This would allow you to produce any possible such triangle. I don't know if this relates to his story or if is useful or not...
Please let me in on this cliffhanger. Great video and story 👍
It’s in the description
The angles of triangles that have integer sides have rational cosines.
For any acute angle having a rational cosine, triangles with rational sides can be generated using two integer variables, M and N, where M and N have no common factors. Here are the formulas for finding sides a, b, and c:
a = M^2 - N^2 (for obtuse angle B)
b = 2M(M cos(C) + N)
c = M^2 + N^2 + 2MN cos(C)
a' = N^2 - M^2 + 2b cos(C) (for acute angle B)
These equations can be checked by substitution of a,b,c into the law of cosines.
Notice that these equations also work when Cos(C) = 0 (right angle). Also notice that, except for Cos(C)=0, the solutions come in pairs, where only the length of ‘a’ differs between an obtuse and acute angle B.
If you use the value of Cos(C)=.5, the formula gives the sides of scalene triangles that have angle C = 60 degrees.
I am looking forward to the ellipse video!
This is a great and inspiring story, thanks.
Thank you so much! It was quite a roller coaster ride.
@@ianfowler9340 You are totally very welcome. Having been a math enthusiast all my life, I can understand the excitement of a new discovery: you never know at first how it is going to work out in the end. And being 69 yo, a puzzle that lasts 3 decades adds that much more richness to the experience. And of course, there are few people on Earth that match the enthusiasm level of our Dr. Peyam.
The most brilliant teacher ❤❤
This man is more excited about triangles than is humanly possible.
The condition b²=a²+c²-ac (where I am assuming a
Wow I like this a lot!!
@@drpeyam Thanks! As a matter of curiosity, I came up with this form as a way to make it easier to check solutions to the equation mentally, by analogy with the method I use for checking Pythagorean triples: rewrite a²+b²=c², where a
We can also use this form to easily check Malhar Managoni's parametric formula, rewritten so a
@@MichaelRothwell1 Nice
I have found a geometrical interpretation of my notion of "twin" solutions in terms of the ellipse used in Malhar Managoni's solution: if you take the midpoint of the intercepts of a horizontal line with the ellipse, the locus of such midpoints will be the line y=2x. In particular, this line y=2x intercepts the ellipse at its highest and lowest points (where the tangent is horizontal). This last observation can readily be checked by implicit differentiation of the ellipse equation x²+y²-xy=1.
Here is the explanation of this geometrical interpretation:
In Malhar Managoni's solution, b & c are swapped, giving:
a²+b²-ab=c², which can be rewritten as (b-c)(b+c)=a(b-a).
Dividing the rewritten form by c², and using x=a/c, y=b/c, we get (y-1)(y+1)=x(y-x).
We note that if (x,y) lies on the ellipse, so does (y-x,y). So if we take the intercepts of the horizontal line y=k with the ellipse, the x-coordinates of the two intercepts add up to k, so the midpoint will be at (k/2, k), points whose locus is y=2x.
This geometrical interpretation also allows us to understand why the twin solutions are not obvious from Malhar Managoni's parameter solution: if we let A=(1,1), and B and C the points where y=k intercepts the ellipse, then the parameters t for these solutions correspond to (minus) the gradients of AB and AC, however the relationship between these gradients does not appear to be simple.
As an example, for t=2, we get the solution a,b,c= 8,5,7; for t=4, we get the a multiple of the twin solution a,b,c = 24,9,21 = 3×(8,3,7). Actually, these solutions correspond to the points (8/7,5/7) and (8/7,3/7), which lie on the same vertical line x=8/7. This is because I should really have considered the values t=1/2, t=1/4 (giving the points reflected in y=x, which lie on the same horizontal line y=8/7), but chose the integer values of t for ease of calculation.
I am a math teacher and I am always impressed by mathematicians, who become excited by problems they can't solve.
Thanks!
A few years ago, I discovered algorithms for deriving solutions to these equations. I started with pythagorean triples and found the algorithm for producing the tree of primitive pythagorean triples. Then I moved on to precisely this 60/120 degree formula and found similar algorithms. Unfortunately I do not know who to talk to about such things. I would love to talk to an expert and see if I have discovered anything of worth. Even if I have not, I would at least like to no longer worry about this problem. I would appreciate help pointing me in the direction of someone who can help.
Deriving such identities is a fun exercise, but forming Pythagorean triples has been done before. I believe 3blue1brown at least touched on this. As for the 60/120, that's not commonly discussed. You should make a video about it!
Outside of number theorists and recreational mathematicians, UA-camrs would be interested. Did you do it parametrically or through matrices or another way? I’m interested in figuring it out by linear transformations, if possible.
@@dagordon1 My algorithms are algebraic but I believe they can be translated into matrices relatively easily. If you look at the Wikipedia page for tree of primitive pythagorean triples, you can see that they use matrices but it’s equivalent. I would love to share my work but I’ve always been hesitant because of the possibility for it to be new.
@@zucc4764 Tried 3b1b but he did not respond. His loss.
I have still no idea why this is exciting.
This would be great for estimating distances on a hex grid.
So after the simplest (1,1,1) 60degrees triangle, is there any other smaller solutions than (7,8,3) and (7,8,5)?
There are any number of such triangles. Imagine a circle of radius 2 and a circle of radius 5, edges touching each other. Then say imagine a circle of radius 1, drawn such that its edge touches both of the other circles. So then by the way we constructed it we have a 7, 6, 3 triangle.
Or make a smaller one, use radius 2, radius 3, and radius 1 circles drawn the above way. Then we have a 5, 3, 4 triangle.
Obviously there's any number of triangles that could be made this way. I'm not quite sure how this fits into his explanation.
Didn't we start with a 5, 7, 8 triangle, not a 3,7,8 triangle, which we had at the end? They can't both have a 60deg angle between the 8&3 sides and 8&5 sides and still have opposite sides of 7 or what am I missing?
OK, I drew it out and discovered the answer. Between the 3,7,8 and 5,7,8 triangles there's a 4,sqrt48,8 right angled triangle. and in the same diagram, there's a pair of 1,7,sqrt48 right triangles. So both 3,7,8 and 5,7,8 are correct.
Well said, I've just pointed this out too and just found your comment to assure myself I'm not the only one who noticed!
@@salamander5703 If he'd pointed this out in the video, it would've been the coolest part of it!
So many good parts! And we still haven't got to the _best_ part!
Wait until you see the 2-paramter solution that generates "primitive" solutions. A line intersecting an ellipse!!
I have a feeling I see where this is going by the end of the video. . Can’t wait for the next one!
It’s already up!
@@drpeyam just watched it; the triplets are born!
Hahaha
It's easy to write a computer program that checks all triples up to some limit. Here's all < 100 that are not reflections of others or multiples of smaller triples: {(35, 31, 24), (8, 7, 5), (91, 79, 40), (99, 91, 80), (99, 91, 19), (15, 13, 7), (1, 1, 1), (65, 61, 56), (35, 31, 11), (77, 67, 45), (55, 49, 39), (96, 91, 85), (40, 37, 7), (91, 79, 51), (77, 67, 32), (21, 19, 5), (48, 43, 35), (65, 61, 9), (8, 7, 3), (80, 73, 63), (96, 91, 11), (80, 73, 17), (15, 13, 8), (48, 43, 13), (55, 49, 16), (40, 37, 33), (21, 19, 16)}
Thank you!!!
It would be cool if you calculated the triples with the formula from the nest video and see if it gives all of them
Guys just use
a = t (q² + pq + p²)
b = t (q² - p²)
c = t (q² +2pq)
Where a is the side opposite to the 60° angle.
Amazingly I just found the 8-13-15 triangle by accident!
I copied the problem to try it myself, but made a copying error (I put the 60° at the center of the R=5 circle), and I found R=10.
Hey dude! I won't be able to sleep ever again, this cliffhanger is too intense!
Right??? 😂
06:10: y = -8 means that point B coincides with point A, which is quite cool
True!!
When the margin is too small to include your proof, ellipses come and save the day !!
They will! Just be patient!
Excellent :)
Nice reference to Fermat.
I remember those academic competitions, They were an awesome excuse to get out of class lol.
Hahaha
I had a similar feeling when I realised independently that if the gradient of y=arcsin(x) = 1/SqRt(1-x^2), then the time dilation equation Gamma=1/SqRt(1-(v/c)^2) is also the gradient of y=arcsin(x) when x=v/c
The limit of x is 1 (sine function of any angle cannot be greater than 1) , just as the limit of v/c is 1 (1=100% the speed of light)
Which in turn means that 1/Gamma = cosy = cos(arcsinx), which, if graphed as y= 1/gamma and x= v/c, it produces a unit circle
Which means that all objects move at the speed of light through space and time, and in order to travel faster through space, you must travel slower through time.
My mind turned inside out
I love math videos. But this is about the funniest one I've seen. Huzzah!
I think I showed you this stuff last year, or maybe the year before. These are Eisenstien Triples. And they also have a Mandelbrot Set. There is an entire algebra for that 45 degree angle ellipse. I made some video on the subject.
I never showed this actually
@@drpeyam I just watched your new video on this. I liked it. There is a second set.
g[x]:=(x^2-y^2)/(x^2+x*y+y^2)
g[y]:=(2*x+y)*y/(x^2+x*y+y^2)
I would like to point out this is the second one. If you want to find out the fourth one, just put g[x] and g[y] into the corresponding x's and y's. Then do it again and find the 8th one. You now have the second, fourth, and eighth dual functions. From there you should be able to find the third, fifth, sixth, and seventh dual functions. I haven't done the ellipsoid yet, but I know it exists. I think I did have it at one point, but I could be wrong. Could you make the second triple function with x, y, z using that same process you used in your video?
@@drpeyam the jump to the 3d ellipsoid is completely different. I would have to do a slope for each axis. I think, instead of m/n, I would just put the axis letter in. x would have to be changed to t, I guess. The equation is x^2+y^2+z^2-x*y-x*z=1, maybe. I dunno. It seems to work, but just because it seems to work, doesn't mean it's right. So, I would have to use that, the 3d line thing, and the new slope stuff. I don't think this all goes together nicely. I'm missing something. Is this another one of those 3d is really 2d things, like sin from cross product and cos from dot product?
Now for the 120 deg triangle. Take the( 3,7,8) triangle - 60 deg (vertex D) opp. the 7-side. Vertex A opp. the 3-side, Vertex D opp. the 7-side. Vertex C opp. the 8-side. Now extend AD to B so that CB = 7. We now have triangle CBD with the 120 deg (vertex D) triangle opp. the 7-side. Again solve for DB using the cosine law and the quadratic in DB to get DB = 5. Now we have triangle CBD (3,5,7) with angle 120 deg opp. the 7-side. And we have c^2 = a^2 + b^2 + ab, the identity for a 120 deg triangle with whole number sides. This is the "cousin" to the original (8,7,5) triangle.
Very nice! But the triangle he found after 30 years was different; it was a 3,7,8 triangle unlike the initial triangle which was 5,7,8
You are correct. It was that problem that also got us thinking about the 120 deg triangles. Every 60 deg triangle has a "sibling" 120 deg triangle with +ab in the identity. But it was seeing the 5,7,8 staring me in the face that got me jumping up and down.
@@ianfowler9340 I'm very curious about the relationship with the ellipse!
@@ianfowler9340 There are sets of two acute triangles (3,8,7),(5,8,7) and one obtuse triangle (3,5,7) that are related. The two acutes are “siblings” and the obtuse is the “cousin”, and those are all the same generation
According to the first drawing of the triangle, angle C is 90°. Side AB is the hypotenuse. so by the Pythagorean theorem it should look like AB squared equals AC squared plus BC squared. those. if you want to write BC squared equals, then it should look like BC squared equals AB squared minus AC squared. Isn't it?
If you do (R+5)^2 then
(R+5)^2=8^2 - (R+3)^2 =>
=> R^2 - 8R - 15 = 0
Well, its the same thing as when integer solutions of a^2+b^2=c^2 are associated with rational solutions to x^2+y^2=1 (divide the previous eqn by c^2). We can obtain these solutions by intersecting the unit circle with a line y=kx where k is a rational number.
Here we just have a^2 +b^2 -ab = c^2 which is canonically ellipse (after dividing by c^2).
Interesting!!!
I don't really get this. For k=1 (which is rational), the intersecting point is (1/sqrt(2), 1/sqrt(2)), which are irrational numbers.
@@l_szabi He meant y=kx+1. In the case k=1 you have the point (-1,0)
@@noahtaul Not really. y=kx, but not for every rational k tho. You can find more if you google "Rational points on a unit circle"
@@Grentanksmog Has to pass through the ellipse at a rational point and y = kx will not do the job
What a journey!
"...find R!" Mathematician: "Ooooh, easy, let's break out the cosines..." Engineer: "Let's open the nearest parametric CAD, constrain the angle to 60 degrees... it's 2!"
Well, that is interesting. There is the unsolved problem of calculating the circumference of an elipse - the area is easy enough (a*b*pi).
I met the problem in a slightly different form:
Bottles tend to be round because it gives the smallest circumference (hence material used) for a given area. Problem is that square bottles use shelfspace better. (about 10%) partly because the internal wastage of space (generally occupied by primary shoolteachers) is there, partly because round bottles cannot be compacted around the edge of the shelf (f.i. in a cupboard).
Now I buy a specific brand of olive oil because of their square bottomed flasks (incidentally the cheapest brand) - which is not surprising, as transportation is a major cost factor for something as cheap as olive oil. The other factor is because the flasks are GREEN, which means the content is inpenetrable to ultraviolet light - and ultraviolet light makes an unsaturated fat like olive oil go rancid quicker. Something similar is the case for beer bottles.
This is actually a derivative of the question of WHY iron is the most stable nucleous in the periodic system. I.e. has the lowest ratio of neutrons to positrons. But nobody (that I know of) has ever in quantum physics thought of that.
Iron has the atomic number 26 or 13 alfa particles. And the dodekahedron has 12 vortices; BUT inside a dodekahedron there is space for yet another alfa particle.
This contemplation I have not heard any nuclear physicist ever mention - ok, it takes quite a lot of violence to force the last alfa particle into the center - but never mind, there is enough of that in the center of a star.
Have I made a discovery about the strong nuclear force? Perhaps, but we will never know, as nobody is paying me for the derivation. I might win eternal fame (or somebody will steal the idea - just as Leibnitz stole the idea of calculus from Newton) - and again: What do I care about the admiration of school teachers? The waste of intellectual space.
Well, why buy the cheapest propritary brand of olive oil? Well producers are out to maximize their profit, and they are not interested in smashing their production line with producing olive oil of an inferior quality - they earn their money with the subsidies of the EU for quantum production.
1. There s a formal solution to the ellipse perimeter. It is 4aE(e) where E is the complete elliptic integral of the second kind, e is eccentricity and a is the semimajor axis. Kind of a cheat, because the integral is the one you would get from the geometry. But E has a power series and can be computed very efficiently.
2. Very interesting remark about the dodecahedron and the iron nucleus. I never heard it either
Such an amazing and deep response. Wow. But, you know, N did not steal the Calculus from L. This time the underdog, although it was a close race, really did come in 2nd place.
The only thing I saw that was anti-pythagerous was the click baity image that clearly relies on people adding 25 to 49 and screwing that up and getting 64, instead of 74. Pythagoras would have disapproved of that I'm sure.
Pythagoras told me he loves this video!
If you apply the law of sines to the given data with R = 2, is the law of sines satisfied? It seems that the given data is not consistent.
Dr. Peyam, on the back of your T-shirt, you neglected to mention that c must also be an element of K in order for the associative laws of addition and multiplication, as well as the distributive law, to be true.
Hahaha you want the shirt to be longer? 😂
easy v =(8,7,5) is just a base of the solution so every lambda*v is in the solution space
3,5,7 is not in that space
Oops! I think you have mixed things up a bit. A 5-7-8 triangle is not the same as a 3-7-8 triangle. Nevertheless, there is a strong link between the two. If they are superimposed so that they share the 60-degree angle, the side of length 3 will lie along the side of length 5. Indeed, if we construct an equilateral triangle of side-length 8, and divide each side into segments of length 3, 2 and 3, then line segments joining each division point to the opposite apex of the equilateral triangle will each have a length of 7. The resulting figure will incorporate three 7-2-7 isosceles triangles that overlap one another in the middle, forming a sort of distorted Star of David. Morley's miracle may come to mind, but that's another story. (This very strongly suggests that triangles that satisfy the criteria of one 60-degree angle and three integer-length sides will come in matching pairs. E.g., 15-7-13 and 15-8-13). Better still, if we now scale the figure up by a factor of two, then any perpendicular to a side of the equilateral triangle that passes through a vertex of one of the isosceles triangles will divide the equilateral triangle's side into segments of integer length. The perpendiculars themselves will have lengths that are multiples of √3. If only UA-cam comments allowed insertion of images. I'm looking forward to ellipses and lines.
The condition b²=a²+c²-ac (where I am assuming a
Me upon realising the video ends in a cliffhanger: "NOOOOOOO!!"
(Joking aside, I literally exclaimed "No!" out loud)
Very nice problem I love 😘❤️
Any triangle with a = b = c will fullfill 60° and a² = b² + c² - ac becomes an identity. You have to ask about a not equal to b not equal to c.
The 30-years-later problem could be solved much more simpler geometrically by making an equilateral triangle using C and D and a point E between A and D. Since CDE is constructed to be equilateral, this makes AE being 5 and also makes the angle AEC measuring 120°. Then, the triangle AEC will have measures 5, 3 and 7 with 120° at E, being a perfect mirror image of CDB with 120° at D. Hence DB measures the same as AE and thus it measures 5. So, there is no need for complex algebra nor trigonometry and not even the anti-pythagorean theorem here.
Thank you. I was starting to wonder if anyone else noticed this was an easier geometry problem than myself.
Since you noticed that, I have a question, because I think I missed something major on the premise of this video.
I thought the Pythagorean Theorem was only applicable when you new for sure you had one 90° angle?
I thought in school I was taught you can't use it for non-right triangles.
Reason I ask, because as soon as I saw the numbers, I immediately knew this wasn't a right triangle so you can't use the theorem, which means it can't be anti-pythagorean.
Did I miss something?
@@targetthetank
The Pythagorean Theorem is only for 90° angles, and it is old known formula (c² = a² + b²), where c is opposing the 90° angle.
However, here, Ian Fowler and Peyam are showing the Anti-Pythagorean Theorem which is only for 60° angles instead, and its formula is (c² = a² + b² - ab), where c is opposing the 60° angle.
Both the Pythagorean and the Anti-Pythagorean Theorems are special cases of the formula (c² = a² + b² - 2ab . cos(C)), which you could use for any angle C opposing to the edge c.
The Anti-Pythagorean Theorem itself would be uninteresting and not very useful if was not for the fact that it shows something about a family of triangles that are: (a) escalene, (b) having all sides relating to each other with integer numbers and (c) having a 60° angle and thus, at least one angle not being a right angle nor an irrational fraction of the circle. Such triangles are rare, difficult to conceive and very poorly studied, and that is the entire point of the video.
@@vwss-java Thank you.
I just find it weird to call it anti, since the point is the relationship of the sides to the angles.
I will do more research now.
Thank you again.
Actually Y=-8 makes sense, which is the triangle on the left and Y is in the opposite direction.
Good point
Yes!yes! It means that there are an infinite number of solutions to c^2 = a^2 + b^2 "+" ab where the angle is 120 deg.
triangle
How can the integer of any side of a triangle be negative when its angles add up to 180°? You're no longer looking at a triangle, but rather some other geometric shape. Even if you were to place it in the 2nd or 3rd quadrants on a graph, its sides still require positive integers to complete the shape.
At one point, you solve for y = -8 and y = +5 and you dismiss y = -8... but what if we accept triangles with sides of negative length? Could the topic of your video, as well as other trigonometric and geometric ideas be more fully generalized by accepting negative lengths? (Or negative angles?) Just a thought.
So now we have:
original (8,7,5) 60 deg triangle: c^2 = a^2 + b^2 - ab
sibling (8,7,3) 60 deg triangle: c^2 = a^2 + c^2 - ab
cousin (3,5,7) 120 deg triangle: a^2 = b^2 + c^2 + ac
Very funny and interesting.👌
EDIT: This might be a horrible spoiler, but I very much thank Dr Peyam for showing this problem (and Ian for bringing it to light). Anyway I solved it (I think). I'll be fascinated to see the Waterloo guy's solution.
2ND EDIT: I see Pieter Geerkens found this too.
Here's how to make integer triples a,b,c with c^2 = a^2 + b^2 - ab
Take integers y, k
Make x = (y/k - 3yk) / 2
z = (y/k + 3yk) / 2
a = x +y
b = x - y
c = z
Discard zeros in a, b or c, multiply out denominators, take out common factors, change signs if required and permitted
E.g.
y k a b c
[ 1, 2, -7/4, -15/4, 13/4] a,b,c => 7, 15, 13
[ 1, 3, -10/3, -16/3, 14/3] a,b,c => 5, 8, 7
[ 1, 4, -39/8, -55/8, 49/8] => 39, 55, 49
[ 1, 5, -32/5, -42/5, 38/5] => 16, 21, 19
[ 2, 4, -39/4, -55/4, 49/4] => 39, 55, 49
I found this by diagonalising the given quadratic form, such that it becomes
x^2 + 3 y^2 - z^2 = 0 with x = (a + b)/2, y=(a - b)/2, z=c
Rearrange
z^2 - x^2 = 3 y^2
(z - x)(z + x) = 3 y^2
Put z - x = 3yk, z+x = y/k
:-))
The one I’ll present is different!
Ugh, now i have to watch Anti Pythagorean Theorem: Endgame
"Who cares" is probably the saddest thing someone can ever say in math.
Agreed!
It was more "never heard of it" followed by "indifference". "Who cares" may have been a bit harsh.
Thanku
Liked for the video and story, disliked for the cheeky cliffhanger! >_
me gustó y aquí te dejo dos inquietudes si en una superficie plana trazo dos líneas rectas y las hago formando un ángulo 60° a cuántos segmentos de línea diferentes de la línea recta le puedo calcular su medida con todo el conocimiento que hay hoy referente a la medida del espacio
Y la otra pregunta es si creen posible calcular medidas de ángulos con una regla así como utilizando un transportador
Y les aseguro que si se puede
Atte Jhonny Angarita