Please be nice in the comments - all comments discrediting or disparaging the youngsters will be deleted. It's an area of current research without a consensus by the mathematical community. Please don't treat your own opinion as fact, no matter how smart or qualified you think you are. Let's just celebrate young women in mathematics! Also note that Pythagoras wasn't the first to invent the theorem, but that's what the western world calls it so we're stuck with that.
Minor point about the Guardian: it is rather shameful to claim the Pythagorean theorem is 2000 years old. Pythagoras lived in the 6th century BC :P ^^ Cool video, of course.
the problem is that i dont understand how these teens are being given this credit. I have seen this proof before, how can this be considered new knowledge? Like, i understand its cool and all to have bright young women in maths, but why are they being given credit for something that has been known for decades? What surprises me the most is the claim that it was thought impossible to prove pythagoras only through trig, when it is essential in all classes to make comparisons between trig and pythagoras for students who are getting started to understand and better visualize the concepts. Never have i ever heard that claim, quite on the contrary, all my teachers and my current lecturer have always said that the geometric concepts associated with vector math can always be expressed in a trigonometric form... and i cant exactly expect these young women to be the first people on planet earth to figure out that the modulus of a vector is the same as pythagoras, so yeah, please, enlighten me as to why are they being credited for something that was already known. Why is this suddenly such a big deal?
@@AlFredo-sx2yyEXACTLY! Even I had also used this kind exact proof in high school and am very confused as to what did they discover? I mean I used trigo in many theorems and I can vouch for that my classmates and I did solve the Pythagoras with trigo just like this.
@@DeclanMBrennan: So you're admitting that you were actually expecting them to be morons, and every single mathematician who had looked at it so far morons as well? Maybe you should check in the mirror, I bet you'll find your moron.
Thanks for sharing. It felt really cruel that I couldn't find their proof anywhere. Even if this somehow isn't exactly what they did, it's still awesome and soothes the part of me that was tired of waiting to see it
@@c64116 the news report exists but not the actual proof. I think the students submitted their proof to journal peer review so they are waiting on a journal to publish it.
@@c64116 They don't mean their story, they mean the actual proof the girls wrote. It was presented at the American Mathematical Society's conference, but hasn't been published yet.
The a=b case can also be handled in this proof. You don't have to extend the triangle pair in this case, since the initial two copies make a larger similar right triangle. Then sin α = a/c =c/2a, which gives us 2a^2=c^2, just as in Pythagoras.
@@keriddunk1520 the beta angle and the alpha angle are the same in this case. So the a/c case is the obvious one and the c/2a case if you use the angle labelled beta (same as alpha cus we assumed similar sides) in the diagram in the video taking the mirrored triangle into account.
What's impressive is that there's no circular reasoning here. None of their method's depend on Pythagoras as you said, which is very hard to do in trigonometry. The geometric series is also a bonus.
Hi, my reply is probably too late to get an answer, I would like an explanation why the other proofs are circular. I am most familiar with the proof where you draw a square with sides (a+b), and an embedded square with sides c (tilted). You can then easily proof that the area of the embedded square is (a+b)^2 - 4((a+b)/2) = a^2 + b^2. How is that proof circular? Or is it not considered trigonometric? Thanks.
@@ZeerobZeerob It's not other proofs in general, just ones that use trigonometry. There are a lot of theorems in trigonometry which are proven using the Pythagorean theorem.
Jackson's proof is an exact copy of a proof originally made by B. F. Yanney and J. A. Calderhead in 1896 which can be found at cut-the-knot, proof 60. Johnson's proof can be found in many Calculus textbooks and is a "geometric series" that was already proof of the Pythagorean theorem. She just added an extra triangle.
@Akupara Jesus is God in flesh! He is the most known historical fact by Christian and non Christian historians. All of the apostles and his followers martyred in the 1st and 2nd Century cause they knew him, no one would die for a liar! Seek Jesus before it’s too late, he is the only way truth and life!
I have found in the news videos hints for two other proofs besides this "waffle cone" proof, and there is also a thrid diagram that might be the start of finding a formula for sin(beta - alpha). The diagrams shown are: 1) One showing a square of side length c attached to the hypotenuse of the abc right triangle. Another square is constructed from perpendiculars to sides a and b from the center of the c^2 square. 2) Another showing a right triangle is placed inside a circle with its hypotenuse being a diameter of the circle. A perpendicular bisector of the diameter is made forming a smaller triangle with one cide being c/2. The lengths of the sides are then found. I created proofs of the Pythgorean Theorem from these two diagrams. Proof 1) is a geometric area proof. Proof 2) uses the Law of Sines, and also the formula for sin(beta - alpha) mentioned above. Disclaimer: These ideas come from examining the screenshots taken from two different news stories and they might vary from the actual proofs the students presented. EDITED TO ADD: There is another proof using what I call the student's "sin( beta - alpha)" diagram. It is possible to modify the diagram and derive two expressions for "sin(beta - alpha)." Equating the two expressions gives the Pythagorean Theorem.
I hope that, because they are just highschool kids, someone within their social sphere is helping them to get this published for peer review. This is one of those things that needs be widely published so the two who came up with it don't get lost in the shuffle!
Thank you for explaining this!!! While I use applied mathematics in my occupational field (business/finance), I haven't taken trig in 25 years, so based on the snippet I saw of their presentation in that news report, I was trying to figure out how they used the law of sines. Brilliant work ladies!
Very nice proof. Very impressive that high school students are able to come up with it. Congratulations to J&J and congratulations to their math teacher and school that is teaching actual math!
What a sensational story -- such elegance, creativity and utter brilliance in the proof itself. Truly inspiring -- it underscores the point that innovation is not the preserve of established experts in any field.
@Intel Bro You said "the method of validation with trigonometry has been used over and over many times" Do you know any references in regards to that? Like the names of the math books that have proven Pythagorean theorem using trigonometry? I want to know.
It's an interesting proof, but it seems pretty silly to say it's based on trigonometry instead of just geometry, since it only uses the geometric ratio (sine) between sides of similar right triangles. You could do the exact same proof (in substance) without ever mentioning the sine of any angle.
I don't agree. At 5:10 you see that the proof is relying on the law of sines, which is pretty hard to express "without ever mentioning the sine of any angle"... Sure trig can be done away in 90% of the arguments of the proof, but they definitely use it in a crucial key step. Proofs are often like that, where they depend on something powerful that is only used exactly once.
@@LuisPereira-bn8jq It uses the law of sines, but it doesn't actually rely on it. Note that the law of sines itself can be derived without using trig, which means the same proof could be done without using trig.
@@alanjones4358 Sorry to say, but I don't think you're making any sense. For clarity, if a triangle has sides A,B,C with opposing angles a,b,c, the law of sines says that A/sin a = B/sin b = C/sin c. And the argument on this proof most definitely relies on this identity, and does not carry through without it. And not only is the sine law not a simple geometric fact (it's certainly far more complex than similarity of triangles), how would you even write the sine law without using sines?
@@LuisPereira-bn8jq I think you misunderstood my comment. I'm only saying that since the law of sines can be proven without trig, then a proof using the law of sines logically doesn't rely on trig. There's a big difference between a proof using trig and a proof relying on trig. But I did mispeak: I should have said you can do an equivalent proof without trig instead of "the exact same proof without trig". But I don't want to correct my original comment after your replies. I'm not sure what you mean by "how to write the sine law without using sines", since the sine of any angle of a triangle can easily be expressed as just a ratio of lengths, even if the lengths are not the sides of the triangle in question. It would be less convenient, but not particularly difficult.
@@alanjones4358 "I'm only saying that since the law of sines can be proven without trig, then a proof using the law of sines logically doesn't rely on trig." Again, this makes no sense. What you are effectively implying is that, since sin and cos can be described geometrically, no argument whatsoever counts as trigonometric. Might as well say that trigonometry doesn't really exist while you're at it. "even if the lengths are not the sides of the triangle in question. It would be less convenient, but not particularly difficult." "Less convenient" is a ridiculous understatement. What makes the law of sines valuable is that it is expressed using only the elements that constitute a triangle, namely the 3 sides and the 3 angles. Call it "not difficult" all you want, but to express the law of sines without mentioning sines (and thus the angles), you'd need to tack on 3 arbitrary triangles to the initial one. But hey, since it's "not particularly difficult", how about you do it and present the world with your superior version of the sine law? I'm sure everyone would love to see it...
I clearly remember the high school assignment of writing a proof of the Theorem, only to have mine rejected as "verification, not proof". This video certainly dusted off some very rusty brain cells! :) Congratulations to the two of them!
The right angle triangles can only be isosceles if a=b, and other posters have given a proof for this case. This is an easy case and they surely would have considered it.
Once you get expression for X and Z you can easily show that Z^2 - X^2 = c^2 and thus proving Pythagoras identity for the large triangle cXZ. No trigonometric functions are needed. Also dimensions of smaller triangles can be obtained w/o invoking trigonometric functions explicitly, just ratios of sides in similar triangles should be used. The explicit insight why does the proof works is still kind of missing. It must have something to do with the fact that the large cXY triangle was constructed and that triangle is not similar to small triangle abc (different angles) but X and Z can be expressed by a, b and c.
It's most simple to use the law of cosines (LoC) directly to prove the Pythagorean theorem, if someone wants a purely trigonometric way. LoC itself has a trigonometric proof itself, check the Wikipedia article for it: the proof does not use the sin²x+cos²x=1 identity anywhere, so it is not circular. Why they thought then trigonometric proofs were impossible? Or am I missing something? Someone please enlighten me!
As long as a is not equal to b you can assume that b>a without loss of generality. And the case where they are equal is pretty trivial and easy to prove from the fact that these extended lines would be parallel.
Amazing insight by these students but I'm a bit confused by the rules of the proof game. What is a 'trigonometric proof' as distinct from a geometric proof? All trig functions are just simple statements about basic geometry relationships?
that's a very neat proof! one part I don't understand (and maybe someone here could help) is the notion of it being circular or not; is there a set of well defined axioms for planar geometry? Because as far as I understand that's what you need to talk about independence of one theorem from another - is there a framework to talk about these theorems rigorously? I tried turning to analytic geometry, but I don't think that's valid - most of the stuff is just defined so that it works as it should (e.g. the euclidean metric on R^2). Does anybody know a reference on any rigorous foundations for planar geometry?
Looks valid. But another trigonometric proof exists for a long time using the compound angle formulas for sine and cosine. Is that considered invalid and why ?
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.
One of the student's presentation slides stated that "The Inscribed Angle Theorem is independent of Pythagoras." How might they have used that theorem in their proof?
The first responses to this article I saw were from Twitter… and as you can probably guess already… the responses were immensely negative, derisive, and unfortunately, bigoted. It’s refreshing to see you and the commenters here just excited that we possibly have another trig proof lol. This proof was quite clever.
This is a simple, clever expression of Pythagorean incommensurability of an unequal line being applied to a triangle. By the time of deducing the third successive triangle it is effectively redundant. As, this ought go on forever and the expressions become infinitely convoluted and yield no further proof. Fairly certain I’d already seen this exact same rationale be applied to an equilateral. Specifically, the Pythagorean cult pentacle 🤷♂️
The construction fails if the initial triangle is 45/45/90 as the two extended lines won’t meet. So need an extra step to prove that case. But otherwise a very clever approach.
There is absolutely no need of trigonometry in this proof. The orthogonal rectangle with one side being 2a is homotethic to the original one with scale 2a/b so no need to use any trigonometry. Then once you have X and Z you can see Z^2 = X^2+c^2 proving the Pythagorean theorem without using the sin(2 alpha).
Jackson's proof is an exact copy of a proof originally made by B. F. Yanney and J. A. Calderhead in 1896 which can be found at cut-the-knot, proof 60. Johnsons's proof can be found in many Calculus textbooks and is a "geometric series" that was already proof of the pythagorean theorem. She just added an extra triangle.
Nice! And well done reconstruction. This got me to think that with sin' = cos and cos' = -sin, one can easily proves sin² + cos² = 1. Then the derivatives are proved from the classic trigonometric identities combined with sin(x)/x going to 1 as x goes to zero. From the top of my head that has a geometric demonstration without Pythagoras, using only the area of some triangles, which is also how the sine law used in that new demonstration is usually derived. Therefore I think that a trigonometric derivation has long existed but the novelty here is that it is also a purely geometric demonstration then. Did I err somewhere?
No, actually, only sin x/x is needed. Indeed sin(x+h) - sin x = 2 cos(x+h/2)sin h/2 and cos(x+h) - cos(x) = -2 sin(x+h/2) sin h/2, so by continuity of sin and cos, we only need the limit of sin x/x to conclude, and that can be done geometrically (en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Inequalities). The trigonometric identities used above too (same page) and none of those demonstrations uses Pythagoras. Thus I think my conclusion was correct: the novelty of Johnson's and Jackson's demonstrations is not that it uses only trigonometry but actually that it does not use calculus at all.
Looks like a nice new approach. While the fractal structure of the "wafle cone" is interesting, the infinite sum (the by far heaviest machinery in this proof) is not necessary: the intersection point of sides x and z can be computed using e.g. vectors. Thus I would say that the mirrored triangle, the double angle and its sine are the key ideas.
That doesn't sound right. What exactly do you mean by "the intersection point of sides x and z can be computed using e.g. vectors" ? Seems to me that you'd be finding the coordinates of the intersection point, and thus the components of the vectors along each side of the triangle. But the proof doesn't want coordinates/components, it wants lengths. And calculating the length of a vector from it's components is literally Pythagoras' Theorem...
@@LuisPereira-bn8jq Yes, first the coordinates of the intersection point are computed. But since the side marked as 'x' in the video has angle alpha to the x-axis, its length can be computed from similarity/proportions. Let v denote the vector along side x (and v.x its horizontal component), then length(v):v.x = c:b
Got it now, I think you're right. The only subtlety I'd notice is that for your method to work you need the formulas for the lengths to have the form c*q(a,b), where q is a rational function, but this should be exactly what you get since the intersection point is computed only from a,b. So another way of saying it is that you extend both sides a,b so as to form two new right triangles that have the intersection point as a vertex, so as to get a figure with 4 similar triangles, then do a bunch of similarity analysis. Looked at from that point of view, this does feel like a clunkier version of the proof that just draws the height relative to the hypotenuse to subdivide the triangle into 2 similar ones, then does similarity analysis.
Am I naive here? If we can use calculus, why not just define e^z = sum z^n/n!, prove the addition formula, define sin and cos by e^ix = cos x + i sin x, and then it follows from e^ix*e^-ix = e^0 = 1. Or maybe the point is that the geometric series is easier than power series?
Hi. It's great to such amazing proof. And I really love it! Can I add Chinese caption to the video and upload it to a Chinese video platform called BiliBili? I just want to share it and let more people know about how amazing math is. I will put all your RUclips account information and credit for sure. Hope to see your reply. Thank you.
May be I am missing something but how do you know that the convergent geometric series actually converge to X and Z and not something else? Still, it's a very nice proof.
@@goombacraft dont you just get two more isosceles right triangles when you bisect the original one? that doesnt help. but there are a few comments addressing that case which you can find here. it's even simpler, you dont have to draw the extensions.
Hello Mathtrain, my question might be dumb but how to demonstrate properly that a^2/b^2< 1 in your geometric serie. It seems obvious since it is a right triangle, but how to demonstrate that in a right triangle the hypothenuse is the longer side without using pythagorean theorem.
For a^2/b^2 to be less than 1, we just need a to be less than b. As long as the legs are not equal, we can always define a to he the shorter one. If they are equal length, there's a separate case I didn't cover in the video.
The approach itself is interesting. But I'd say, it is not only trig. We have triangle congruence and series here. There is simple and known proof via trig/congruence without double angle when we put height to hypothenuse and consider congruent triangles.
First, this proof assumes that a != b. Second, they also need to prove sin(2 alpha) = 2 ab / (a^2 + b^2) WITHOUT using Pythagoras, otherwise the proof is circular. That means the proof is incomplete, IMHO.
Here’s a video by polymathematics that also proves sin(2α) = 2ab / (a^2 + b^2) ruclips.net/video/p6j2nZKwf20/видео.html However the above video doesn’t address the case where a=b. The again, the comment by grpthry in this video shows how the proof can be remedied for a=b.
@@adiaphoros6842 That second video proves sin(2α) = 2ab / (a^2 + b^2) by assuming the sine law. The sine law assumes Pythagorean. IMHO that breaks the proof.
@@kaicheung5916 “The sine law assumes Pythagorean.” Wrong, the law of sines uses only the definition of the sine function: opposite / hypotenuse. Here’s a proof: Consider △ABC, where the its sides are labelled as such: AC = b BC = a AB = c and its angles are named the same as its vertices. Draw the altitude h from vertex A to side a From there: sin(B) = h / c h = c sin(B) sin(C) = h / b h = b sin(C) Therefore c sin(B) = b sin(C) sin(B) / b = sin(C) / c Using an altitude from vertex B to side b, you can obtain a similar equation for sin(A) / a As you can see, nowhere in the proof was the Pythagorean theorem used. The Pythagorean theorem only applies to right triangles, and the triangle with side lengths c, c, and 2a at the start is assumed to be not a right triangle.
@@adiaphoros6842 Ay, there's the rub: your definition of the sine function: opposite / hypotenuse, is only valid on euclidean plane. (e.g. on the surface of a sphere, sum of interior angle is > 180 degrees, which violates that definition). Since Pythagoras is equivalent to the parallel postulate, therefore the acceptance of that definition is equivalent to assuming the parallel postulate and the Pythagoras theorem. Therefore, It is circular.
@@kaicheung5916 There is a uniform proof of the law of sines in Euclidean, Spherical, and Hyperbolic geometries in Elementary Differential Geometry (pages 201-209) by Christian Bär. So you don’t need the parallel postulate. But this is going beyond classical geometry and into Riemannian geometry. Also equivalence doesn’t mean dependence. For example the law of cosines is equivalent to but independent from the Pythagorean theorem.
@5:25 The only derivation that I have seen of the Law of Sines employs the Pythagorean Theorem. Does anyone have a link to something that derives it without the Pythagorean Theorem?
Draw a triangle ABC with opposite sides a, b, c respectively. The altitude from C can be computed as either asin(B) or bsin(A). Since those are equal, rearrange to get the law of sines
@@MathTrain1 Thanks! I think I just glitched and was thinking of the Law of Cosines derivation but maybe that can be derived w/o the Pythagorean Theorem, too?
Since we start the series from r + r^2 + ... and not from 1 + r + ..., we multiply the whole result by r (since this is just a term wise multiplication by r). The formula when you do that is r/(1-r) instead of 1/(1-r)
I was able to find this on the site "dirzon". I don't expect youtube to let me link to a pdf so I won't even try. The name of the site should suffice to find the correct file with any search engine. Any way, the shape is similar indeed but quite different when you look at it more closely. It consists entirely of right angle triangles without the initial "pyramid", and the right angles occur along the sides of the big shape. The high school students created right angles that never involve the lines "X and Z" (EDIT: or "U and V" in another video about this topic). The exercise is also about calculating the total length of all the zig-zagging lines inside the shape, not its (out)sides. I think it's safe to say this text book exercise would not have helped the students come up with their proof even if they saw it. But for the record, I do think it's good to be suspicious when highschoolers come up with something "new" in any field.
@@forasago I know that's why I said very similar, but not the exact same. However, now that you mention it, I wonder if the design in the Stewart Calc text could also lead to a proof of the Pythagorean theorem if you follow similar steps to what the highschoolers did.🤔 As there is also an infinite series in the problem.
No lol. Most modern mathematical "discoveries" are just people flexing their PhD. I know these are high school kids, notice the qualifier "most." I have 100000 times more respect and appreciation for discoveries in science than math.
Jaykee the Wolf: No. They used calculus to find out the limit results for u and v. Luis: true. but good work, still. Jaykee the Wolf: not really. just like einstein ripped off maxwell's theory to shape the theory of special relativity, a little of the same is going on here.
Not if you define them as the ratios between sides of a right triangle. It might take Pythagoras' Theorem to calculate them, but not to define or use them in the way presented.
Completely blown away with their proof and if your interpretation is different from them then we might have 🕝🕑. Hope you interpreted them as those geniuses conceptualized.
Thank you. I'd been able to find only their abstract and the same article words, copied from publication to publication. If this is their proof, they need to cover separately the special case of an isosceles right triangle (easily done), so they can ensure that b > a.
Honestly had forgotten to cover that case! But if so then you're right the a=b case is easy. There's a bunch of ways to do it, but I wonder how they'd incorporate trig. Or maybe their proof is just for the a not equal to b case
@explosion612 Plato has Socrates give one in his dialogue Meno. It essentially consists of using a floor tiled with isosceles right triangles and showing that two of them make a square with side length equal to the triangle's leg, while four of them make a square with side length equal to its hypotenuse.
Pff, no need : At least, introduce a vector space just to mathematically deal with concept of "vertic-/horizontality" in order to rearrange ("if needed") how to construct an isosceles "trigone" ("trigono-") to "measure" ("-metry") from a shared/common *longest* side; or labelling "b" whatever side's actually the longest
Well, that's another for the list of proofs On another, more serious note, the fact high schoolers were able to figure this out on their own is insane, and while there are simpler proofs, this one is very nice and elegant. I love how everything comes together in the end to form the all-known a²+b²=c², very satisfying
Nicely done! Jackson & Johnson have explored and discovered a clever use of the math they have learned in high school - exactly the kind of students every teacher would hope for! It turns out that the proof uses the law of sines, triangle geometry and calculus. The convergence of a geometric series for 0 < r < 1 is a topic learned in beginning calculus but not proved. In order to prove this, one needs to show that 1/(1-r) is the least upper bound of the sequence of partial sums of the series. This topic is usually addressed in advanced calculus where the axioms and foundations of the real number system are rigorously discussed.
@@jpharnad My posts describe what is mathematically required of a completely rigorous proof from first principles. Sorry you don't fancy such a proof. Using ellipsis ... or invoking the existence of a limit of an infinite sequence to claim that a real number exists, no matter how intuitively obvious it may seem to you, does not substitute for a rigorous mathematical foundation of these ideas. It's about what counts for a rigorous proof. See for example Walter Rudin's classic "Principles of Mathematical Analysis."
@@jpharnad It requires what it requires and the 'anything' (convergent sequences and series) in your "hardly requires anything" is clearly the key novel ingredient in the current proof that makes it work. Also, as you can see in my posts above the mathematics of convergent sequences and series, already on its own, implies the Pythagorean Theorem.
👍 Amazing result and excellent video. I like the pencil animation technique. I too am curious to see a peer reviewed publication. Seems purely trigonometric, but I'd like more qualified nerds to weigh in on that.
![Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]](http://i.ytimg.com/vi/p6j2nZKwf20/mqdefault.jpg)
Please be nice in the comments - all comments discrediting or disparaging the youngsters will be deleted. It's an area of current research without a consensus by the mathematical community. Please don't treat your own opinion as fact, no matter how smart or qualified you think you are. Let's just celebrate young women in mathematics! Also note that Pythagoras wasn't the first to invent the theorem, but that's what the western world calls it so we're stuck with that.
@@joelleblanc8829 you couldn't disprove it though, but you have no credibility so that tracks.
Minor point about the Guardian: it is rather shameful to claim the Pythagorean theorem is 2000 years old. Pythagoras lived in the 6th century BC :P ^^ Cool video, of course.
Subscribed 👏🏼👏🏼
the problem is that i dont understand how these teens are being given this credit. I have seen this proof before, how can this be considered new knowledge? Like, i understand its cool and all to have bright young women in maths, but why are they being given credit for something that has been known for decades? What surprises me the most is the claim that it was thought impossible to prove pythagoras only through trig, when it is essential in all classes to make comparisons between trig and pythagoras for students who are getting started to understand and better visualize the concepts. Never have i ever heard that claim, quite on the contrary, all my teachers and my current lecturer have always said that the geometric concepts associated with vector math can always be expressed in a trigonometric form... and i cant exactly expect these young women to be the first people on planet earth to figure out that the modulus of a vector is the same as pythagoras, so yeah, please, enlighten me as to why are they being credited for something that was already known. Why is this suddenly such a big deal?
@@AlFredo-sx2yyEXACTLY! Even I had also used this kind exact proof in high school and am very confused as to what did they discover? I mean I used trigo in many theorems and I can vouch for that my classmates and I did solve the Pythagoras with trigo just like this.
I was expecting to see a Sin² + Cos² = 1 somewhere but this is way more impressive. What a beautiful proof.
That is called the Pythagorean identity… so you can kinda understand why they weren’t able to include that in the proof
@@jacob2713 Indeed
@@DeclanMBrennan:
So you're admitting that you were actually expecting them to be morons, and every single mathematician who had looked at it so far morons as well? Maybe you should check in the mirror, I bet you'll find your moron.
It would be circular reasoning.
@@ShaunakDesaiPiano how would it be circular? You can derive that identity using only trig, so then youve proven pythagoras?
Thanks for sharing. It felt really cruel that I couldn't find their proof anywhere. Even if this somehow isn't exactly what they did, it's still awesome and soothes the part of me that was tired of waiting to see it
damn, weird, maybe it has to do with your personalization settings on your devices. ive seen this story blasted EVERYWHERE
@@c64116 the news report exists but not the actual proof. I think the students submitted their proof to journal peer review so they are waiting on a journal to publish it.
@@c64116 They don't mean their story, they mean the actual proof the girls wrote. It was presented at the American Mathematical Society's conference, but hasn't been published yet.
@@c64116 The story is everywhere, but not the proof itself.
Yeah, I couldn't find the proof so assumed it was BS
Wonderful video. No one is talking about the excellent pencil animations. Amazing work.
Thank you very much!
The a=b case can also be handled in this proof. You don't have to extend the triangle pair in this case, since the initial two copies make a larger similar right triangle. Then sin α = a/c =c/2a, which gives us 2a^2=c^2, just as in Pythagoras.
I worked this out as well. This was a fun exercise.
Sin(alpha) = c/2a where did you get that from ?
Oh damn you're right
@@keriddunk1520 the beta angle and the alpha angle are the same in this case. So the a/c case is the obvious one and the c/2a case if you use the angle labelled beta (same as alpha cus we assumed similar sides) in the diagram in the video taking the mirrored triangle into account.
thank the ancestors the Law of Sines is independent from the Pythagorean Theorem. If it were dependent on the latter, this whole proof falls flat.
What's impressive is that there's no circular reasoning here. None of their method's depend on Pythagoras as you said, which is very hard to do in trigonometry. The geometric series is also a bonus.
Hi, my reply is probably too late to get an answer, I would like an explanation why the other proofs are circular. I am most familiar with the proof where you draw a square with sides (a+b), and an embedded square with sides c (tilted). You can then easily proof that the area of the embedded square is (a+b)^2 - 4((a+b)/2) = a^2 + b^2. How is that proof circular? Or is it not considered trigonometric? Thanks.
@@ZeerobZeerob It's not other proofs in general, just ones that use trigonometry. There are a lot of theorems in trigonometry which are proven using the Pythagorean theorem.
Jackson's proof is an exact copy of a proof originally made by B. F. Yanney and J. A. Calderhead in 1896 which can be found at cut-the-knot, proof 60.
Johnson's proof can be found in many Calculus textbooks and is a "geometric series" that was already proof of the Pythagorean theorem. She just added an extra triangle.
math really is beautiful...
God has designed it that way.
@@zackwies1493 which one?
@@lapatria100 Lord Jesus Christ!
@@lapatria100Jesus Christ!
@Akupara Jesus is God in flesh! He is the most known historical fact by Christian and non Christian historians. All of the apostles and his followers martyred in the 1st and 2nd Century cause they knew him, no one would die for a liar! Seek Jesus before it’s too late, he is the only way truth and life!
Thank you for explaining this so much better than 99.9% of journalists did 💚🤓
Journalists are not mathematicians. They took journalism in the first place because they are bad at math.
99.9% of journalists don't even know what trigonometry is! 😂
I have found in the news videos hints for two other proofs besides this "waffle cone" proof, and there is also a thrid diagram that might be the start of finding a formula for sin(beta - alpha).
The diagrams shown are:
1) One showing a square of side length c attached to the hypotenuse of the abc right triangle. Another square is constructed from perpendiculars to sides a and b from the center of the c^2 square.
2) Another showing a right triangle is placed inside a circle with its hypotenuse being a diameter of the circle. A perpendicular bisector of the diameter is made forming a smaller triangle with one cide being c/2. The lengths of the sides are then found.
I created proofs of the Pythgorean Theorem from these two diagrams.
Proof 1) is a geometric area proof.
Proof 2) uses the Law of Sines, and also the formula for sin(beta - alpha) mentioned above.
Disclaimer: These ideas come from examining the screenshots taken from two different news stories and they might vary from the actual proofs the students presented.
EDITED TO ADD:
There is another proof using what I call the student's "sin( beta - alpha)" diagram.
It is possible to modify the diagram and derive two expressions for "sin(beta - alpha)." Equating the two expressions gives the Pythagorean Theorem.
It's amazing to see that these young people can still make incredible discoveries!!!
It's really cool.
I hope that, because they are just highschool kids, someone within their social sphere is helping them to get this published for peer review. This is one of those things that needs be widely published so the two who came up with it don't get lost in the shuffle!
This is in no way a result suitable for publication in a peer-reviewed journal
Thank you for explaining this!!! While I use applied mathematics in my occupational field (business/finance), I haven't taken trig in 25 years, so based on the snippet I saw of their presentation in that news report, I was trying to figure out how they used the law of sines. Brilliant work ladies!
Thank you, both to you and Jackson and Johnson
Very nice proof. Very impressive that high school students are able to come up with it. Congratulations to J&J and congratulations to their math teacher and school that is teaching actual math!
What a sensational story -- such elegance, creativity and utter brilliance in the proof itself. Truly inspiring -- it underscores the point that innovation is not the preserve of established experts in any field.
@Intel Bro You said "the method of validation with trigonometry has been used over and over many times" Do you know any references in regards to that? Like the names of the math books that have proven Pythagorean theorem using trigonometry? I want to know.
Such an elegant and geometrically beautiful proof. Those young women have so much future ahead of them 👏👏
To summarize in three words - ingenious, simple and elegant - the highest level in mathematics.
I also once saw something similar on r/math. (A high-schooler solving a theorem or something along the lines of that...)
One of the easiest proofs of this to understand. Wicked talent
It's an interesting proof, but it seems pretty silly to say it's based on trigonometry instead of just geometry, since it only uses the geometric ratio (sine) between sides of similar right triangles. You could do the exact same proof (in substance) without ever mentioning the sine of any angle.
I don't agree. At 5:10 you see that the proof is relying on the law of sines, which is pretty hard to express "without ever mentioning the sine of any angle"...
Sure trig can be done away in 90% of the arguments of the proof, but they definitely use it in a crucial key step.
Proofs are often like that, where they depend on something powerful that is only used exactly once.
@@LuisPereira-bn8jq It uses the law of sines, but it doesn't actually rely on it. Note that the law of sines itself can be derived without using trig, which means the same proof could be done without using trig.
@@alanjones4358 Sorry to say, but I don't think you're making any sense.
For clarity, if a triangle has sides A,B,C with opposing angles a,b,c, the law of sines says that
A/sin a = B/sin b = C/sin c.
And the argument on this proof most definitely relies on this identity, and does not carry through without it.
And not only is the sine law not a simple geometric fact (it's certainly far more complex than similarity of triangles), how would you even write the sine law without using sines?
@@LuisPereira-bn8jq I think you misunderstood my comment. I'm only saying that since the law of sines can be proven without trig, then a proof using the law of sines logically doesn't rely on trig. There's a big difference between a proof using trig and a proof relying on trig. But I did mispeak: I should have said you can do an equivalent proof without trig instead of "the exact same proof without trig". But I don't want to correct my original comment after your replies.
I'm not sure what you mean by "how to write the sine law without using sines", since the sine of any angle of a triangle can easily be expressed as just a ratio of lengths, even if the lengths are not the sides of the triangle in question. It would be less convenient, but not particularly difficult.
@@alanjones4358
"I'm only saying that since the law of sines can be proven without trig, then a proof using the law of sines logically doesn't rely on trig."
Again, this makes no sense. What you are effectively implying is that, since sin and cos can be described geometrically, no argument whatsoever counts as trigonometric. Might as well say that trigonometry doesn't really exist while you're at it.
"even if the lengths are not the sides of the triangle in question. It would be less convenient, but not particularly difficult."
"Less convenient" is a ridiculous understatement. What makes the law of sines valuable is that it is expressed using only the elements that constitute a triangle, namely the 3 sides and the 3 angles.
Call it "not difficult" all you want, but to express the law of sines without mentioning sines (and thus the angles), you'd need to tack on 3 arbitrary triangles to the initial one.
But hey, since it's "not particularly difficult", how about you do it and present the world with your superior version of the sine law? I'm sure everyone would love to see it...
I clearly remember the high school assignment of writing a proof of the Theorem, only to have mine rejected as "verification, not proof". This video certainly dusted off some very rusty brain cells! :) Congratulations to the two of them!
It kills me that they are making headlines but haven't released the entire proof. What better "peer review" do you need?
Cool
If the right triangle is also isosceles then the extended sides don't meet, are we covering this case separately then?
Honestly not sure! Their proof will hopefully cover this case. Though it's still impressive if they prove it only for a not equal to b.
The right angle triangles can only be isosceles if a=b, and other posters have given a proof for this case. This is an easy case and they surely would have considered it.
All lines meet at infinity their lines will meet instantly It's a square then a equals b.
@@zisuzlabs then they can use the geometric series upto infinite value as it meet at infinity
Once you get expression for X and Z you can easily show that Z^2 - X^2 = c^2 and thus proving Pythagoras identity for the large triangle cXZ. No trigonometric functions are needed. Also dimensions of smaller triangles can be obtained w/o invoking trigonometric functions explicitly, just ratios of sides in similar triangles should be used.
The explicit insight why does the proof works is still kind of missing. It must have something to do with the fact that the large cXY triangle was constructed and that triangle is not similar to small triangle abc (different angles) but X and Z can be expressed by a, b and c.
It's most simple to use the law of cosines (LoC) directly to prove the Pythagorean theorem, if someone wants a purely trigonometric way. LoC itself has a trigonometric proof itself, check the Wikipedia article for it: the proof does not use the sin²x+cos²x=1 identity anywhere, so it is not circular.
Why they thought then trigonometric proofs were impossible? Or am I missing something? Someone please enlighten me!
law of cosines is basically a generalization of the Pythagorean theorem, so i don't think it counts
Suppose angle A and angle B are not known, then angle A = tan^-1(a/b) and angle B = tan^-1(b/a). So, then, c^2 = 2ab/sin2A = 2ab/sin2B.
Very good. The logic behind the process is wonderful/impressive. Well done to the kids
This triangle can be formed if and only if b side is greater than a side. Thus, the geometric series also converges.
As long as a is not equal to b you can assume that b>a without loss of generality. And the case where they are equal is pretty trivial and easy to prove from the fact that these extended lines would be parallel.
Very clever proof... But, I'm not sure the usage of the geometric series makes it a purely trigonometric solution.
Amazing stuff....opened a whole new world of using series to prove trig identities
The limit of the geometric sequence in r equal to 1/(1-r) is valid only if r is small : the Pythagoras theorem is valid for any a and b.
thank you for the video
Amazing insight by these students but I'm a bit confused by the rules of the proof game. What is a 'trigonometric proof' as distinct from a geometric proof? All trig functions are just simple statements about basic geometry relationships?
that's a very neat proof! one part I don't understand (and maybe someone here could help) is the notion of it being circular or not; is there a set of well defined axioms for planar geometry? Because as far as I understand that's what you need to talk about independence of one theorem from another - is there a framework to talk about these theorems rigorously? I tried turning to analytic geometry, but I don't think that's valid - most of the stuff is just defined so that it works as it should (e.g. the euclidean metric on R^2). Does anybody know a reference on any rigorous foundations for planar geometry?
School Mathematics Geometry from the mid 1960s has a pretty good set of axioms.
Aside from Euclid himself, try Hilbert and then Tarski
Looks valid. But another trigonometric proof exists for a long time using the compound angle formulas for sine and cosine. Is that considered invalid and why ?
FYI: They each came up with 2 separate trigonometric proofs on their own.
Thank you for posting this - good work.
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.
One of the student's presentation slides stated that "The Inscribed Angle Theorem is independent of Pythagoras."
How might they have used that theorem in their proof?
I noticed that too, and I couldn't figure it out. We'll just have to wait and see
@@MathTrain1 Maybe they also demonstrate the theorem of sines, to show that it is independent of Pythagoras' theorem.
The first responses to this article I saw were from Twitter… and as you can probably guess already… the responses were immensely negative, derisive, and unfortunately, bigoted.
It’s refreshing to see you and the commenters here just excited that we possibly have another trig proof lol. This proof was quite clever.
This is a simple, clever expression of Pythagorean incommensurability of an unequal line being applied to a triangle. By the time of deducing the third successive triangle it is effectively redundant. As, this ought go on forever and the expressions become infinitely convoluted and yield no further proof. Fairly certain I’d already seen this exact same rationale be applied to an equilateral. Specifically, the Pythagorean cult pentacle 🤷♂️
4:18 i'm curious about how you could prove that r is always less than 1 for all right triangles.
As long as aa, just swap them. If they're equal, you have to handle that in a separate case not covered in this video
The construction fails if the initial triangle is 45/45/90 as the two extended lines won’t meet. So need an extra step to prove that case. But otherwise a very clever approach.
There is absolutely no need of trigonometry in this proof. The orthogonal rectangle with one side being 2a is homotethic to the original one with scale 2a/b so no need to use any trigonometry. Then once you have X and Z you can see Z^2 = X^2+c^2 proving the Pythagorean theorem without using the sin(2 alpha).
Jackson's proof is an exact copy of a proof originally made by B. F. Yanney and J. A. Calderhead in 1896 which can be found at cut-the-knot, proof 60.
Johnsons's proof can be found in many Calculus textbooks and is a "geometric series" that was already proof of the pythagorean theorem. She just added an extra triangle.
I love it, although this looks rather more like a demonstration than a proof?
But won’t it fail if a and b are equal? Because then a/b is 1 then the geometric part will fail cause it only works between -1 to 1 exclusive?
Yes there is another case of the same idea you can do for the a=b case but I didn't cover it
My question is if that special case is provable by trig itself, because if so they don't need to worry about it?
Yes you can use the law of sines
Why hasn’t their paper been released yet?
Nice!
Nice! And well done reconstruction. This got me to think that with sin' = cos and cos' = -sin, one can easily proves sin² + cos² = 1. Then the derivatives are proved from the classic trigonometric identities combined with sin(x)/x going to 1 as x goes to zero. From the top of my head that has a geometric demonstration without Pythagoras, using only the area of some triangles, which is also how the sine law used in that new demonstration is usually derived. Therefore I think that a trigonometric derivation has long existed but the novelty here is that it is also a purely geometric demonstration then. Did I err somewhere?
But then we also need (cos x - 1)/x when x goes to zero, and I can't remember a geometric demonstration for that one! I don't have time to ponder now…
No, actually, only sin x/x is needed. Indeed sin(x+h) - sin x = 2 cos(x+h/2)sin h/2 and cos(x+h) - cos(x) = -2 sin(x+h/2) sin h/2, so by continuity of sin and cos, we only need the limit of sin x/x to conclude, and that can be done geometrically (en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Inequalities). The trigonometric identities used above too (same page) and none of those demonstrations uses Pythagoras. Thus I think my conclusion was correct: the novelty of Johnson's and Jackson's demonstrations is not that it uses only trigonometry but actually that it does not use calculus at all.
Looks like a nice new approach. While the fractal structure of the "wafle cone" is interesting, the infinite sum (the by far heaviest machinery in this proof) is not necessary: the intersection point of sides x and z can be computed using e.g. vectors. Thus I would say that the mirrored triangle, the double angle and its sine are the key ideas.
That doesn't sound right.
What exactly do you mean by "the intersection point of sides x and z can be computed using e.g. vectors" ?
Seems to me that you'd be finding the coordinates of the intersection point, and thus the components of the vectors along each side of the triangle.
But the proof doesn't want coordinates/components, it wants lengths. And calculating the length of a vector from it's components is literally Pythagoras' Theorem...
@@LuisPereira-bn8jq Yes, first the coordinates of the intersection point are computed. But since the side marked as 'x' in the video has angle alpha to the x-axis, its length can be computed from similarity/proportions. Let v denote the vector along side x (and v.x its horizontal component), then length(v):v.x = c:b
Got it now, I think you're right.
The only subtlety I'd notice is that for your method to work you need the formulas for the lengths to have the form c*q(a,b), where q is a rational function, but this should be exactly what you get since the intersection point is computed only from a,b.
So another way of saying it is that you extend both sides a,b so as to form two new right triangles that have the intersection point as a vertex, so as to get a figure with 4 similar triangles, then do a bunch of similarity analysis.
Looked at from that point of view, this does feel like a clunkier version of the proof that just draws the height relative to the hypotenuse to subdivide the triangle into 2 similar ones, then does similarity analysis.
Limit theory...sum to infinity...."using only trig!" Hardly.
Am I naive here? If we can use calculus, why not just define e^z = sum z^n/n!, prove the addition formula, define sin and cos by e^ix = cos x + i sin x, and then it follows from e^ix*e^-ix = e^0 = 1. Or maybe the point is that the geometric series is easier than power series?
@@jimhabegger3712 Yeah, maybe it's the construction that's more interesting than the proof itself.
Hi. It's great to such amazing proof. And I really love it! Can I add Chinese caption to the video and upload it to a Chinese video platform called BiliBili? I just want to share it and let more people know about how amazing math is. I will put all your RUclips account information and credit for sure. Hope to see your reply. Thank you.
That is good with me! Thanks for offering. Can you put the link here when you've uploaded?
@@MathTrain1 Sure. I will see if I can find some free time to finish it and then I will put it in comments at once! Thanks a lot!
May be I am missing something but how do you know that the convergent geometric series actually converge to X and Z and not something else? Still, it's a very nice proof.
Elegant!
Wouldn't considering cos(A-B) = cos(A)cos(B)+sin(A)sin(B) at A=B yeild Pythagoras Theorem using just the ratio definitions of sin and cos as well?
Well... That's neat, and easy to understand
brilliant!
That’s pure genius ❤
Nice. I wonder if they address the case where α = β, in which case the sides marked X and Z are actually parallel.
I guess then this proof isn't valid for α=β, but I suppose you might be able to solve this issue by bisecting that triangle.
@@goombacraft dont you just get two more isosceles right triangles when you bisect the original one? that doesnt help. but there are a few comments addressing that case which you can find here. it's even simpler, you dont have to draw the extensions.
thanks for sharing this
Hello Mathtrain,
my question might be dumb but how to demonstrate properly that a^2/b^2< 1 in your geometric serie.
It seems obvious since it is a right triangle, but how to demonstrate that in a right triangle the hypothenuse is the longer side without using pythagorean theorem.
For a^2/b^2 to be less than 1, we just need a to be less than b. As long as the legs are not equal, we can always define a to he the shorter one. If they are equal length, there's a separate case I didn't cover in the video.
The approach itself is interesting. But I'd say, it is not only trig. We have triangle congruence and series here.
There is simple and known proof via trig/congruence without double angle when we put height to hypothenuse and consider congruent triangles.
First, this proof assumes that a != b. Second, they also need to prove sin(2 alpha) = 2 ab / (a^2 + b^2) WITHOUT using Pythagoras, otherwise the proof is circular. That means the proof is incomplete, IMHO.
Here’s a video by polymathematics that also proves sin(2α) = 2ab / (a^2 + b^2)
ruclips.net/video/p6j2nZKwf20/видео.html
However the above video doesn’t address the case where a=b. The again, the comment by grpthry in this video shows how the proof can be remedied for a=b.
@@adiaphoros6842 That second video proves sin(2α) = 2ab / (a^2 + b^2) by assuming the sine law. The sine law assumes Pythagorean. IMHO that breaks the proof.
@@kaicheung5916 “The sine law assumes Pythagorean.”
Wrong, the law of sines uses only the definition of the sine function: opposite / hypotenuse.
Here’s a proof:
Consider △ABC, where the its sides are labelled as such:
AC = b
BC = a
AB = c
and its angles are named the same as its vertices.
Draw the altitude h from vertex A to side a
From there:
sin(B) = h / c
h = c sin(B)
sin(C) = h / b
h = b sin(C)
Therefore
c sin(B) = b sin(C)
sin(B) / b = sin(C) / c
Using an altitude from vertex B to side b, you can obtain a similar equation for sin(A) / a
As you can see, nowhere in the proof was the Pythagorean theorem used. The Pythagorean theorem only applies to right triangles, and the triangle with side lengths c, c, and 2a at the start is assumed to be not a right triangle.
@@adiaphoros6842 Ay, there's the rub: your definition of the sine function: opposite / hypotenuse, is only valid on euclidean plane. (e.g. on the surface of a sphere, sum of interior angle is > 180 degrees, which violates that definition). Since Pythagoras is equivalent to the parallel postulate, therefore the acceptance of that definition is equivalent to assuming the parallel postulate and the Pythagoras theorem.
Therefore, It is circular.
@@kaicheung5916 There is a uniform proof of the law of sines in Euclidean, Spherical, and Hyperbolic geometries in Elementary Differential Geometry (pages 201-209) by Christian Bär. So you don’t need the parallel postulate. But this is going beyond classical geometry and into Riemannian geometry.
Also equivalence doesn’t mean dependence. For example the law of cosines is equivalent to but independent from the Pythagorean theorem.
I saw that news report and simply decided to wait for RUclips to recommend me a video a few days later and they did it in 3 days.
@5:25 The only derivation that I have seen of the Law of Sines employs the Pythagorean Theorem. Does anyone have a link to something that derives it without the Pythagorean Theorem?
Draw a triangle ABC with opposite sides a, b, c respectively. The altitude from C can be computed as either asin(B) or bsin(A). Since those are equal, rearrange to get the law of sines
@@MathTrain1 Thanks! I think I just glitched and was thinking of the Law of Cosines derivation but maybe that can be derived w/o the Pythagorean Theorem, too?
Im trying to do the proof myself, where did the r in the numerator come from in Z = c+(2cr)/1-r?
Since we start the series from r + r^2 + ... and not from 1 + r + ..., we multiply the whole result by r (since this is just a term wise multiplication by r). The formula when you do that is r/(1-r) instead of 1/(1-r)
That "waffle" cone structure is very similar to a problem in James Stewarts' Calculus textbook. It's on page 721, problem #56, 5th ed.
I was able to find this on the site "dirzon". I don't expect youtube to let me link to a pdf so I won't even try. The name of the site should suffice to find the correct file with any search engine. Any way, the shape is similar indeed but quite different when you look at it more closely. It consists entirely of right angle triangles without the initial "pyramid", and the right angles occur along the sides of the big shape. The high school students created right angles that never involve the lines "X and Z" (EDIT: or "U and V" in another video about this topic). The exercise is also about calculating the total length of all the zig-zagging lines inside the shape, not its (out)sides. I think it's safe to say this text book exercise would not have helped the students come up with their proof even if they saw it. But for the record, I do think it's good to be suspicious when highschoolers come up with something "new" in any field.
@@forasago I know that's why I said very similar, but not the exact same. However, now that you mention it, I wonder if the design in the Stewart Calc text could also lead to a proof of the Pythagorean theorem if you follow similar steps to what the highschoolers did.🤔 As there is also an infinite series in the problem.
Is there a benefit to this method? Its a long way to get there and requires other known methods. Super cool that they were able to do this
No lol. Most modern mathematical "discoveries" are just people flexing their PhD. I know these are high school kids, notice the qualifier "most."
I have 100000 times more respect and appreciation for discoveries in science than math.
Not only trigonometry, there's calculus too
Crystal clear, thank you... It's a beautiful construction
They can spare the trigs if they divide the original angle alpha instead of forming another triangle with the angle 2alpha.
If the pathagorean theorm is proved, does it then become the law of Pythagoras?
Awesome
Absolutely geniuses.
.Yay! Thanks for doing this. It took me two passes through, but I get it. These gals claim not to be math whizzes, but clearly they are.
Well done 👏
They used infinite sums, so not only trigonometry.
Nice💯💯💯
Not completely trigonometric, as it relies on convergence of two infinite sums. But bravo nevertheless.
well i'm glad i watched, but now my brain hurts, now i'm off to watch some asmr
this is actually pretty cool
Goosebumps.
Awesome.
God Bless those two kids!
Erdós would be happy
interesting
Jaykee the Wolf: No. They used calculus to find out the limit results for u and v.
Luis: true. but good work, still.
Jaykee the Wolf: not really. just like einstein ripped off maxwell's theory to shape the theory of special relativity, a little of the same is going on here.
Genius
Sine and cosine are a direct result of Pythagoras' theorem
Not if you define them as the ratios between sides of a right triangle. It might take Pythagoras' Theorem to calculate them, but not to define or use them in the way presented.
I feel that this is weird but my math teacher proved this in the class about 8 months ago
please kids stop comparing me with sheep and rats.
Completely blown away with their proof and if your interpretation is different from them then we might have 🕝🕑. Hope you interpreted them as those geniuses conceptualized.
Thank you. I'd been able to find only their abstract and the same article words, copied from publication to publication.
If this is their proof, they need to cover separately the special case of an isosceles right triangle (easily done), so they can ensure that b > a.
Honestly had forgotten to cover that case! But if so then you're right the a=b case is easy. There's a bunch of ways to do it, but I wonder how they'd incorporate trig. Or maybe their proof is just for the a not equal to b case
@explosion612 Plato has Socrates give one in his dialogue Meno. It essentially consists of using a floor tiled with isosceles right triangles and showing that two of them make a square with side length equal to the triangle's leg, while four of them make a square with side length equal to its hypotenuse.
You are never going to see any proof.
The proof is "Dey blak!" and if you don't just believe them, then "You a racis!".
@@MathTrain1 the a=b case can be done with only the first two triangles.
Pff, no need : At least, introduce a vector space just to mathematically deal with concept of "vertic-/horizontality" in order to rearrange ("if needed") how to construct an isosceles "trigone" ("trigono-") to "measure" ("-metry") from a shared/common *longest* side; or labelling "b" whatever side's actually the longest
Well, that's another for the list of proofs
On another, more serious note, the fact high schoolers were able to figure this out on their own is insane, and while there are simpler proofs, this one is very nice and elegant.
I love how everything comes together in the end to form the all-known a²+b²=c², very satisfying
Nicely done! Jackson & Johnson have explored and discovered a clever use of the math they have learned in high school - exactly the kind of students every teacher would hope for! It turns out that the proof uses the law of sines, triangle geometry and calculus. The convergence of a geometric series for 0 < r < 1 is a topic learned in beginning calculus but not proved. In order to prove this, one needs to show that 1/(1-r) is the least upper bound of the sequence of partial sums of the series. This topic is usually addressed in advanced calculus where the axioms and foundations of the real number system are rigorously discussed.
But the proof if the limit never uses the pythagorean theorem
@@jpharnad My posts describe what is mathematically required of a completely rigorous proof from first principles. Sorry you don't fancy such a proof. Using ellipsis ... or invoking the existence of a limit of an infinite sequence to claim that a real number exists, no matter how intuitively obvious it may seem to you, does not substitute for a rigorous mathematical foundation of these ideas. It's about what counts for a rigorous proof. See for example Walter Rudin's classic "Principles of Mathematical Analysis."
@@jpharnad It requires what it requires and the 'anything' (convergent sequences and series) in your "hardly requires anything" is clearly the key novel ingredient in the current proof that makes it work. Also, as you can see in my posts above the mathematics of convergent sequences and series, already on its own, implies the Pythagorean Theorem.
great vid bro, well explained. Congrats to Calcea and Ne'Kiya on such an acheivement so early in their careers
Those be triangles alright
what a fun proof! thanks for sharing it with us ❤️
Beautiful work.
👍 Amazing result and excellent video. I like the pencil animation technique. I too am curious to see a peer reviewed publication. Seems purely trigonometric, but I'd like more qualified nerds to weigh in on that.
Adding the perpendiculars, etc. was geometric.
@@Paul-222 Good point. Curious to see what real mathematicians make of it.
The pencil animation technique has another name - “writing”.