In time of course one will realize that Pythagoras is a theorem, but only within Euclidean geometry and it follows naturally from how we define distance in this system. Any proof of the theorem must, at some level, assume or derive Euclidean distance, meaning that Pythagoras is more of a consequence than something that needs proving. The real foundational assumption is how we define space and distance and Pythagoras follows from that.

Our beloved physicists have already proved to us that space is not entirely Euclidean (a fascinating topic on its own), but it is in actual fact curved (general relativity). Thus Euclidean geometry is an approximation of reality, which, wait for it, also makes Pythagoras (which is a measure of distance in Euclidean space) an approximation 🤯. This is as far down the rabbit hole as we'll go lol.

I always appreciate a good trip down a rabbit hole. That's where the fun lies! On the other hand, keeping things Euclidean makes it a mere mathematical exercise which is also fun. I don't quite understand why you say defining sine and cosine as ratios of sides makes this circular. In my view, it takes the whole trig aspect out of it and leaves us with a proof based on similar triangles, like many others. I wish I had presented the video this way (as NOT a trigonometric proof) instead of falling for the hype that was going around when this proof first came out.

Δεν καταλαβαίνω την ερώτησή σου

ΤΑ SPORTEX SPANOU ΕΙΝΑΙ ΚΑΘΕΤΑ ΣΤΑ ΓΕΝΕΙΑ ΤΟΥ...

Fewer distractions in ancient times, perhaps. 🙂

@ probably!

you can use the formula for range but is there an intuitive way?

It's a great question, and if I had an answer I would have put it in the video. When I was solving this problem, I had already observed from numerical computation that the change in the velocity's angle was 90 degrees for maximum distance. While trying to understand why that is true, I realized that the area of the triangle would also be maximized, and that led me to notice the connection between the formulas for area and for range. I have since tried to develop some intuition for why that should be the case, but I haven't come up with anything concrete.

@@MathyJaphy this is a nice method that may help - apply sin law in that triangle made by vo vf and gt vectors this will give costheta (theta being the angle b/w initial velocity and horizontal) and put that in the expression for range we will finaly get range=vovfsinphi/g where phi is the angle b/w vo and vf vectors

Yes, the law of sines gives a nice way of showing that the range is maximized when sin(phi) is maximized, so phi is 90 degrees. In fact I originally solved it this way before realizing that the area of the triangle is also maximized. I decided that using the area equation would make a better explanation in the video.

@@MathyJaphy how did you think of that solution? because i'm in highschool and my teacher gave this problem and that solution was very wild to me

Glad you enjoyed them. I'll make more when life affords me more time to work on them.

Thanks for you kind words! So glad you appreciate the vids. I needed to take a break, but I hope to be back someday.

@@MathyJaphy It's alright, there are ups and downs in everyone's life and it's great to take a break from the real life. Btw, these animations make me love math even more the textbooks and I appreciate hardwork that you put into the making of this videos, just love it man. Even though I can't do much but my best wishes are with you. Hope you have a great day.

You're right, it's not impossible. Good job proving me wrong! That wording came from my own inability to do that algebra at the time I encountered this problem (long before CAS calculators!). I should have said "seems impossible". Other commenters have shown me how to do it by hand. BTW, it looks as though you missed a 1/2 in front of the v^2. If you put that in and multiply the two sqrt's together, you get the answer I worked out in the end.

Thanks for referring that paper, I hadn't seen it before. I agree completely that this proof uses trig only as terminology for ratios of sides of similar triangles. So, calling it a "trigonometric" proof is a bit weak. The paper is thoughtful and well written, and its only flaw is that it mentions two KZbin videos, neither of which is mine. :-)

I dont get it can you please elaborate

@@sainiyashraj Is a Trigonometric Proof Possible for the Theorem of Pythagoras? Michael de Villiers RUMEUS, University of Stellenbosch CONCLUDING COMMENTS To get back to the original question of whether a trigonometric proof for the theorem of Pythagoras is possible, the answer is unfortunately twofold: yes and no. 1) Yes, if we restrict the domain to positive acute angles, any valid similarity proof can be translated into a corresponding trigonometric one, or alternatively, we could use an approach like that of Zimba (2009) or Luzia (2015). 2) No, if we strictly adhere to the unit circle definitions of the trigonometric ratios as analytic functions, since that would lead to a circularity.

Sorry to hear that. I hope you will remain proud of your accomplishment anyway, and that you will continue to share it with others who can appreciate it.

I used Desmos Graphing Calculator ( www.desmos.com/calculator ) to create the animations, my Mac's built-in screen capture to turn it into a video, and iMovie to edit it to what you see here.

Thank you!

My goodness, that's quite a challenge! Dr. Peyam does about as good a job as I can imagine for rigorousness, but you're right, it is hard to follow. Mathologer does a better job on that front, but he does a bit more hand-waving. I'm afraid I couldn't possibly do any better than those two examples, especially since my specialty is high-school-level graphical proofs. I appreciate that you asked me, though!

Agreed! My assertion that this proof is "based on trigonometry rather than geometry" was poorly worded, if that's what you're referring to.

Point taken. And there’s still some geometry in there too.

Thanks for asking! The descriptions of all my videos give credit and links to the software I use. The text animation is done in Apple Keynote. The graphical animations are done with Desmos Graphing Calculator. The resulting video clips are put together with Apple iMovie. It's not trivial, but at least all the software is free and popular, so there are online tutorials that can help you learn how to use each package individually. Making the pieces work together to create a video presentation is where most of the creativity and ingenuity is required. Yes, take a class that teaches how to use video editing software, and any math class that introduces you to Desmos.

@@MathyJaphy Thank you so much for this. I'm a professional tutor and still training to be a college professor. I would like to utilize technology like this (visual) to teach advanced students better

Thanks for pointing that out. What a cool proof! Definitely similar, but I think the geometry of how the similar triangles are constructed and partitioned is sufficiently different to call it a different proof. Well, I may just have to do this proof #100 as the next in my Pythagorean Theorem series!

I don't know about that. From what I've heard, they worked together on this one proof. Apparently, they've come up with more proofs since then.

That's true, and geometry is used as well!

Yes! Using the area formulas makes for a cleaner proof. One could also argue that trigonometry isn't needed even with the proof as presented here. It's all just ratios of side lengths of similar triangles which we happen to call "sin". This caused me to question what it even means to be a "trigonometric" proof. Even Jason Zimba's proof using the double-angle formula could be questioned (see video description for links if interested). After all, the double-angle formula comes from comparing analogous side lengths in an arrangement of similar triangles!

Thank you! I had fun making those animations. And you're right, my derivation of X/Y is complicated. I like how the complexity dissolves away by the end, leaving the Pythagorean simplicity. However, a good proof remains straightforward throughout. I think your Variation #1 is the winner. Much prettier than mine! By the way, thanks for the mention in your latest Variation video. I am honored!

@@MathyJaphy Overall, I think that Jason Zimba's proof is the best trig proof. It relies on the difference angle formulae. It does not rely on a limiting process.

In a triangle, the sum of any two side lengths is greater than the length of the third side. So your presumption, c=a+b, is incorrect.

@@MathyJaphy The Pythagorean calculation does not include the area even if I assign c' = a + b. The point is that the sides of the triangle are affine without the area; one leg can be on earth, one inthe middle of the 'andromeda cluster, and one at the bottom of the ocean. The equaation c^2 = a^2 + b^2 obtains only if one omits the product ab (i.e., multiplication, where the area of the triangle is A = 2ab. The can only be obtained by using imaginary number where the product +/- iab is eliminatd by complex conjugation psipsi*

# = a+b #^2 = [a^2 + b^2] +[2ab] # = 7 = 3 + 4 #^2 = 7^2 = 49 = [25] +[24] <> 25 (count is preserved under multiplication) That is, c:= a + ib c* = a - ib cc* = [a^2 + b^2] + a(ib) -a(ib) (b is imaginary) <> (a+b) ^2 = [a^2 + b^2] +[2ab] (binomial expansion); Fermat's Last Theorem for the case n=2

@@MathyJaphy That is, Fermat's Last Theorem is valid for the case n=2 for all positive real numbers c^2 <> a^2 + b^2 since in second order (I repeat, sigh. ad infinitum, ad nauseam) c= a + b c^2 = [a^2 + b^2] + [2ab] (Binomial Expansion, proved by Newton) <> [a^2 + b^2] (why) figure it out and you will be enlightened....

@@MathyJaphy 2nd order equations relate areas, not lengths,

Sure, I'm happy to take suggestions. You can email me at MathyJaphy @ gmail.com.

I would be happy to, but I don't understand the question. What do you mean by "solve the factor of..."?

@@MathyJaphy it would be helpful for my project if you could provide the measurement of the sides of the triangle..

​@@SabrinaHoq Yes, you're referring to the part where I scale each triangle by different factors. But I don't know what you mean by "solve". Some commenters have questioned whether it's okay to multiply a length by a length because the units become squared. But I'm not multiplying by a length, just a value that happens to be the same as the length of one of the other sides. You can always scale the sides of a triangle by a constant value without changing its angles. Does that answer your question?

@@MathyJaphy Can I use the same constant value every time,??

@@SabrinaHoq The lengths of the triangles' sides (A, B, C, D, E and F) are defined by the quadrilateral you start with. A, B, C and D are the lengths of the quadrilateral's sides. E and F are the lengths of the diagonals. For the quadrilateral that I used as a demo in the video, the lengths are approximately (A=1.49, B=1.11, C=3.34, D=1.66, E=1.98, F=1.94). Is this helpful?

I made this with Desmos graphing calculator. It's a web-based application anyone can use. Here is a link to an example to get you started. It takes some learning if you're not already familiar with it, but documentation is available online. Try this link and click on the little metronome symbol in the upper left area, below the little plus sign, to start it going. www.desmos.com/calculator/1g1w3gcrkf