Pythagoras Would Be Proud: High School Students' New Proof of the Pythagorean Theorem [TRIGONOMETRY]

it's Tuesday new proof of the Pythagorean theorem just dropped I'm very excited today to share a new proof of the Pythagorean theorem and in addition to it being cool just that there's another proof of the Pythagorean theorem out there this one is distinctive for two other reasons first of all it was discovered and presented by two high school students Kelsey Johnson and Nakia Jackson at a regional meeting of the AMS so that right there super cool well done ladies but second of all it is a trigonometric proof of the Pythagorean theorem generally speaking pure trigonometric proofs of the Pythagorean theorem for a long time were not thought to be possible at all because ultimately it was believed that any such proof would come to rely on What's called the Pythagorean identity sine squared X plus cosine squared x equals one as you might imagine based on its name the Pythagorean identity is derived from the Pythagorean theorem so any trigonometric proof that ends up relying on the Pythagorean identity is implicitly a circular proof now although for a long time people say said that there was no such proof possible a little over a decade ago Jason zimba did in fact prove the Pythagorean theorem purely using trigonometry I think he used the sum and difference identities to do it which do not rely on the Pythagorean identity either way though this is still super cool and I want to get into it the first thing that Johnson and Jackson did is they start with just your standard right triangles some right triangle with side lengths A and B with hypotenuse length C and then of course across from those side lengths we have angles in this case complementary angles that we're going to name Alpha and beta I'm not drawing it in here but everything that looks like a right triangle in this picture is a right triangle what we're going to try to do is to get to this statement right here a squared plus b squared of course equals c squared but equals c squared by way of this particular ratio 2 a b over sine 2 Alpha this essentially is going to come from the combination of our basic definition of the sine ratio and the Law of Sines itself neither of which depends in any way on the Pythagorean identity so for your basic sine ratio of course we can say things like sine of alpha is equal to a over C that is the sine of some angle is equal to the ratio of the length of the side opposite that angle divided by the length of the hypotenuse similarly we could also say sine of beta is equal to B over C and in fact that's the one that we really need a little bit more so we're just going to keep that up here sine beta equals B over C but you might notice in the statement that we're aiming for we have a two alpha and there is no angle measure to Alpha in this diagram right now so Miss Johnson and Miss Jackson very cleverly reflected this right triangle across and got us that angle of two alpha now what that created was an isosceles triangle no longer a right triangle with a vertex angle 2 Alpha that's the one that we're going to need for this statement down here and then base angles beta and beta the angle 2 Alpha was across from a side length of 2A and then each of the angles beta are now across from what was a moment ago our hypotenuse length of C if we express this in terms of the Law of Sines we can say that sine of 2 Alpha over 2A is equal to sine of beta over C combining this with our piece of information from a moment ago we can actually go ahead and substitute and say well we knew that sine beta was actually equal to B over C and so with a bit more simplification we can say that sine 2 Alpha over 2A is the same thing as B over c squared if we divide both sides by B here and then use the reciprocal identity flip everything upside down what we're going to get is this statement right here 2A B over sine 2 Alpha is the same thing as c squared so let's also store that piece of information over here for safe keeping now the question at this point should be well okay that's great but where does the a squared plus b squared get involved here how on Earth do we get that without the Pythagorean theorem itself the answer that the high school students came up with very clever was to develop this shape they called the waffle cone the waffle cone is a series of infinitely many similar right triangles to our original ABC right triangle you can tell what that means is the waffle cone is itself a very large right triangle because it has these two complementary angles Alpha and beta in that left-hand Corner because it's a right triangle we can make another statement about sine of two alpha we can go back to our original sine ratio definition and say that sine of two alpha must be the same thing as U this base of the larger right triangle divided by V the hypotenuse of that same right triangle meaning if we can figure out what the lengths are for u and v we can restate what sine of 2 Alpha is and hopefully connected to this statement over here about c squared now this is where the proof is no longer really purely trigonometric because the way that we're going to figure out these lengths u and v is we're going to take advantage of this infinite series of similar right triangles and taking advantage of infinite series is really more of a calculus style thing to do not really a trigonometric style thing to do again all of these are similar they all use the same Alpha and beta angles along with of course the right angle and so all of their side lengths have to be in this ratio A to B to C but then in this first triangle right here the larger leg which in the original triangle was length B has to be length to a so to get to that first triangle we must be multiplying by a ratio to a divided by B the divided by B part is what cancels out B and then of course the 2A is what gives us that length of 2A that we know it's equal to in this larger right triangle using that same ratio that means that the smaller side must be 2A squared over B and the hypotenuse must be 2ac over B so if we wanted to we could label that here so that's 2ac over B and again of course this side here is going to be 2A and then the shorter side on that right triangle is going to be 2A squared over B that shorter side 2A squared over B is actually the same as the longer side of the next triangle down so we can now take this and put it in place of the longer leg 2A and to get from 2A to 2A squared over B we can see we have changed the ratio we're now multiplying by A over B and in fact that is the same ratio we're going to be using for all of these right triangles all the way toward you know the infinitely small tip of that larger right triangle this means the next hypotenuse down is going to be 2A squared C over B squared that is that length right there 2A squared C over B squared and then the shorter side is going to be 2A squared over B times A over B so that ends up being 2A cubed over B squared but in fact all we care about all the way down on both sides are hypotenuses and now that we have the ratio we're going to use the entire time we can can skip ahead and only state those hypotenuse lengths for example this next hypotenuse down is 2A cubed C over B Cubed and then the one down after that is 2A to the fifth C over B to the fifth because in each case we are using two of the ratios to get to that next triangle down on the side that we care about similarly this hypotenuse here is going to be 2A to the fourth C over B to the fourth and this next one down will be 2A to the sixth C over B to the sixth and so on and so on all the way down this gives us a sum for both u and v for U for example we can see it's 2ac over B plus 2A to the third C over B to the third plus two a to the fifth C over B to the fifth on and on forever and that's a pure infinite series we could rewrite that as the sum from 1 to Infinity of 2 AC over B that is the initial term times the ratio a squared over B squared to the N minus 1 power because of course we don't use the ratio on the first term but then we use it to get every subsequent term after that using the formula for the value of a convergent infinite series we can say that U is equal to that initial value 2ac over B divided by 1 minus the ratio a squared over B squared if we multiply everything on top and bottom by B squared we're going to get the statement that U is equal to 2 A B C over B squared minus a squared and that's something that we will go ahead and store over here on the right V is a little bit tougher to parse out because it does have this initial portion C that's not going to end up being part of our infinite series but we'll deal with that in a second we can say that V is equal to C plus 2A squared C over B squared plus 2A to the fourth C over B to the fourth plus to a to the sixth C over B to the sixth on and on forever so in fact we can restate that as C plus the infinite series again from 1 to Infinity of 2A squared C over B squared times a squared over B squared that ratio just like before to the N minus 1 power using our same formula before V equals c plus our initial value two a squared C over B squared divided by 1 minus that same ratio a squared over B squared and just like before we're going to multiply everything in that fraction by B squared over B squared to further simplify this into C plus two a squared C over B squared minus a squared now this one unfortunately takes a little bit more simplification because we basically need to get a common denominator to combine the C plus the 2 2A squared C over B squared minus a squared so another way to think about that is c times B squared minus a squared over itself B squared minus a squared plus that quantity 2A squared C over the same common denominator B squared minus a squared as we distribute that c on top we get B squared C minus a squared C plus the 2A squared C that we had over here all over the common denominator of B squared minus a squared well negative a squared and positive 2A squared are like terms so in fact we can write v as a squared C plus b squared C the a squared C coming from negative 1 a squared C plus two a squared C all over B squared minus a squared and then furthermore we can factor out C and we can say V is equal to c times a squared plus b squared all over over B squared minus a squared and you should be getting excited because you can see we have an a squared plus b squared involved finally unsurprisingly this is the last piece of information we need so let's store this over here with our other series of amazing boxes uh actually that's going to be behind me you're not even going to be able to see that anymore I guess I'll just put it right there all right armed with all of this let's remember where we're trying to go a squared plus b squared equals c squared at this point I don't think we need sine of beta anymore so we can get rid of that but we do want to rewrite this sine of 2 Alpha we're going to say sine of 2 Alpha is equal to U over V where U is this expression here 2 A B C over B squared minus a squared and then the denominator V is this expression here c times the quantity a squared plus b squared also over B squared minus a squared because those use the same denial denominators those denominators actually cancel out for that matter these C's cancel out here and so you can tell that gives us a new statement for sine of two alpha we can say sine 2 Alpha is equal to 2 a b over a squared plus b squared with some fancy proportions we could tell that's the same thing as saying 2 a b over sine of 2 Alpha is equal to a squared plus b squared and that's what we needed we have done it we have shown that a squared plus b squared must be the same thing as 2 a b over sine 2 Alpha must be the same thing as c squared itself the Pythagorean theorem is proven don't worry folks it's definitely true again just a really great job by Kelsey Johnson and Nakia Jackson high school students from New Orleans who have proved the Pythagorean theorem proved it in a mostly trigonometric way using your sine ratio definition and your law of science theorem neither of which depend on the Pythagorean identity there was a little calculus thrown in there for good measure but to me that's all the better they were just using tools across mathematics and they proved the Pythagorean theorem what an amazing accomplishment for those two girls you know I had to do it I loaded this up into Desmos so I've got a little thing that you can play with there you get the UV labels you get the angle labels I will put the link to the Desmos graph in the description I will also link to that Jason zimbaproof that trigonometric proof of the Pythagorean theorem that came out a little over 10 years ago along with some other trigonometric and infinite series style proofs that I found on the cut the knot website so check out all those links in the description that's all I've got for you today thank you for watching I will see you all next time thank you [Music]

Channel: polymathematic

Views: 886,038

Keywords: high school, high school math, new orleans, math, mathematics, maths, mathematical proof, pythagoras, pythagorean theorem, proof, trigonometry, trigonometric proof, law of sines, sine, ratio, trigonometric ratio, calculus, infinite series, convergence, convergent infinite series, geometry, proving pythagorean theorem, new proof, impossible, impossible proof, high school nicki minaj, high school musical, nicki minaj high school, high school math teacher, polymathematic, st marys new orleans

Id: p6j2nZKwf20

Length: 13min 57sec (837 seconds)

Published: Tue Apr 04 2023