pizza riddle: cut all three in half, arrange them to form a triangle with the cut sides. If the triangle is acute, the small pizzas is the best deal. If it is obtuse, the large pizza is the best deal ...or just count the number of pepperonis on each
You raise an excellent point. We haven't defined the "inherent value" of any given pizza to the customer; it will differ from person to person, and total surface area isn't necessarily what it will be. It could just be the total amount of toppings, if you don't care much about the cheese, the sauce, and the crust. Fred
I never got to algebra in school. Never made it to college either. All that well over 40+ years ago. But I enjoy these videos like you wouldn't believe. I feel like I'm learning via osmosis. Wish we had this back in the day. No telling where I'd be today. I do work these problems and am understanding algebra a bit. So please continue making these and please keep in mind, some of us are old dogs but we are learning new tricks. Thanks guys.
I am 54, I hear ya! There are a lot of university lectures here on TH-cam as well, Yale and Stanford and many others have their own channels with courses. Here's the link to Yale where you can see on the side links to other universities, if you're interested. th-cam.com/users/YaleCourses Click their "Playlists" to find something of interest.
I am from Ireland and algebra is taught to students from 11 years up. As youngster, I had studied mechanical drawing at school and that has hugely helped me with mathematics. For example, it taught me to understand the theorems of circles, squares, rectangles, etc.
Agreed. The observation that if the sum of the areas of two similar figures attached to the sides of a right triangle equals the area of a third similar figure attached to the hypotenuse, then Pythagoras follows is clever. And the one-line construction that yields these similar figures is truly elegant. I am now convinced that for sheer utility and richness of results, Pythagoras's Theorem beats Euler's Identity as the most beautiful result in Mathematics. I can't wait to share this proof, as well as de Gua's theorem and the quadrature sum of the areas of a right pyramid, with all my friends! Also looking forward to more fun and games when I get my hands on "A dingo ate my math book." Thanks Mathologer for making these entertaining and informative videos.
Based on the name, I suppose there's a connection with solutions to those Diophantine equations and Eisenstein integers (i.e. complex numbers of the form a + b\omega + c\omega^2, where \omega is a primitive third root of unity and a, b, c are integers) similar to 3b1b's video on Pythagorean theorem and Gaussian integers.
Tehom It would be better to remove all multiples and duplicates, and simply list the primitives. e.g. (3,5,7) (7,8,13) (1,1,1) (3,7,8) (5,7,8) (7,13,15) (8,13,15) based on what you provided
Usually just a quick google of whatever is on the t-shirt will get you there :) By popular demand I've also put some that I designed myself here shop.spreadshirt.com.au/1035536/
Sleves sqaured plus hat squared equals trousers hoses squared? Sleeves volumes squared plus hat volume squared plus trousers hoses squared equals muscle shirt volume squared? IDK I'm a flatlander with poor imagination.
I´m German and wondered, why i do understand perfectly the Englisch of Burkard. Well, now I know, Burkard is German ... Really nice proofs an even better animation !!!
11:39 the angles were selected a bit uncomfortable making the left corner of the red triangle appear as if the outer lines meeting there would be straight. but the "thesis" that all orange triangles are same shape just different scale and that the upper two triangles sum in their area to the lower one is fine. thus the definite conclusion drawing a square around them will also make the areas of the upper two squares sump to the area of the lower squares - is definitely neat.
It took me a while, but I think I finally got it. I just need more words to guide my mind through an understanding of it. Whatever same shape figures are attached to the sides (each same shape triangle has the right angle away from the center) will scale proportionally. Could be semi-circles (half of each pizze in his challenge) or smiley faces. So, pretend that figure is a square. We know that the cut triangles came from the original, so the summation of their areas equal the whole, which relation must hold for any chosen figure. Therefore it must be true that a^2 + b^2 = c^2. Thanks.
My answer would be to cut the pizza's in half and put them together into a triangle. Putting the small and medium at a 90 degree angle to eachother and then fitting the large slice inbetween the remaining 2 vertices. If the diameter of the bigger pizza slice is smaller than the distance between the vertices, then it is a bad deal, if it is larger then it is a good deal. If it fits exactly then both are a great deal.
FINALLY, I got the proof of that 1/A²+1/B²=1/D² stuff. Consider the area of the given triangle. Since it is a right triangle, the area (I call it F) can be calculated with F = 1/2*A*B. Since D is the height on C, the Area can be evaluated with F = 1/2*C*D aswell. Combining these two equations we get: C*D=A*B. Squaring both sides give us C²*D²=A²*B². But C²=A²+B². So we get the final equation: D²(A²+B²)=A²*B². Rearranging will give us what we initially wanted to show. q.e.d.
Nice! I solved it in a more pedestrian way :-(. I called the the two intervals into which D dissects C C1 and C2. Then we have three equations: Two P-theorems: (1) C1²+D²=A² (2) C2²+D²=B² And, as the two internal triangles have the same proportions: (3) C1/D = D/C2, which can be reerranged to obtain: (3') C1*C2=D² Express C1 and C2 from (1) and (2), plug them into (3') and after rearrangement you get the expression. I prefer yours though.
For 19:13, make the three lengths from the right angle a, b, and c. How we find the sum of the areas and use Heron’s formula for D^2. After some simplification the two expressions become the same.
My own favorite proof is based of that shape :: 4 identical copies of a right triangle of sides a, b and c organized in a square shape.Exterior square has side s = a+b and its area is s^2 = a^2 + b^2 + 2ab . But the area is also the sum of the four triangles plus the inner square of side c.So Area = 4* (ab/2) + c^2 = 2ab + c^2 = s^2 = a^2 + b^2 + 2ab. And, therefore a^2 + b^2 = c^2. Done.
OK, here's my answer to the pizza problem: Cut all three pizzas in half and form a triangle with one half from each pizza--the cut edge (diameter) being used as each side. If the triangle is right, both deals are equal. If it's an acute triangle, the two smaller pizzas are a better deal, and if it's an obtuse triangle, the larger pizza is the best deal. Great video; planning to watch it several times over.
+John Chessant Cut it in half by "eyeballing" it. A small error in the angle of the cut would result in a much smaller error (percentage-wise) in the length of the cut. +letao12 You can look at it and see if it's close to a right triangle; if so, you can consider the deals equal within a few pennies. To both: If I can't make these assumptions, I don't think a solution exists.
I love your sense of humor so much, man... You are a delight. Thank you for doing these - my world is enriched with you in it, in so many ways. ...Okay, I may have a bit of a crush.
I proved the 3D counterpart (square areas) when I was in high school. When I went to uni, I showed it to a classmate who then proved a 4D version. He showed it to one of his computer science tutors, Carroll Morgan, who then proved it for all dimensions, but I don't know if he ever published it. A few months later, I was stunned when I was browsing at the library and by chance I opened up a journal and saw another proof. That was around the late 1970s.
@@T.R.-TRUTH.REASON more than that, people proved an nD version. the 3D version is the sum of the squares of the areas of the three right-angular sides of a tetrahedron equals the square of the area of the largest side.
@@T.R.-TRUTH.REASON Yes, diagonally slice a tetrahedron from the corner of a cube. sum of squares of areas of the three rectangles in the corner equals square of area of diagonal side. For 4D, it's squares of volumes of tetrahedra. And so on for higher dimensions.
8:40 Well; that also happens with the irrationality proofs of roots: I would argue that it’s much easier to prove that n√k ≠ a/b, where b > 1; for all the infinitely many positive integer values of k, n, a, and b; via the Fundamental Theorem of Arithmetic; than it is, to prove the irrationality of any individual number, like: √2, or: √3. 🙂
*UPDATE:* In fact; I recently noticed that you don’t even need the F.T.Arithmetic, for that proof. Simply knowing that all roots and powers of 1 are equal to 1, is enough. 🙂
cut the pizzas in half arrange them into a triangle if the triangle is acute take the two smaller pizzas if the triangle is obtuse take the large pizza if its a right triangle then take either
I think that works. But the smaller two diameters don't have to add up too the diameter of the larger one. If that's the case they don't make a triangle. Which obviously means the bigger one has more area than the smaller combined. Nice one.
Proof of 1/A^2 + 1/B^2 = 1/D^2: Scale the right triangle by a factor of 1/BD so that it has hypotenuse H = 1/D and left side 1/B. Call the bottom side E. Since the left triangle is similar to this triangle we know A/H = D/E, which tells us A = 1/E. Thus by Pythagoras we're done. I also have a geometric proof of X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) that uses similar ideas to the proof of the cosine formula presented in the video, but it's hard to explain in a youtube comment.
Great video, simple and very fun! Reminds me a bit of Byrne's edition of Euclid due to the sheer artistry of the whole ''coloured shapes manipulation'', while this may not be the most rigorous or useful way to do geometry, darn is it esthetically beautiful. Have to check out that book of 371 proofs of pythagoras, sounds like recreational heaven!
Yes, Byrne is great, here is an online version www.math.ubc.ca/~cass/euclid/byrne.html I've also provided a link to the book of proofs in the description. Have a look :)
When angle = 60*. 5^2 + 8^2 - (5*8*) = 49 = 7^2. When angle = 120*. 7^2 + 8^2 + (7*8) = 169 = 13^2. Thanks you Mathologer, because of you I find some really interesting new stuff.
I was asked to prove Pythagoras during a university interview back in 1983. I used the first proof with the big square in it, which seemed to go down well. However, I think the scaling proof is definitely my new favourite!
9:10 The fraction of the semicircle occupying a square can be written as: the area of the semicircle/the area of the square Since both the semicircle's diameter and the square side lengths is equal, their lengths can be given by a. the area of the semicircle will simply be = π/2•(a/2)^2 = π/8•a^2. The fraction can then be written as = (π/8•a^2)/a^2 and since a^2 are like terms this simplifies to π/8 which is roughly equal to 0.39.
Sorry, my eyesight much be failing me. Of course, you were after cubes and not squares. Here you go 1³+6³+8³=9³ . These is an example of a cubic quadruple.
Mathologer Oh great. Thanks! I wonder if you can prove it, like you can prove Pythagoras, but using cubes and some sort of 3D shapes. Does it have any applications in geometry?
Hey mathologer, I just wanted to say that i am a big fan of your videos and have been a fan for over a year and a half.Recently my uncle gave me a book for my birthday and coincidentally it was written by you.It was Sciencia and your book (QED Beauty in mathematical proofs) was written very well and was informative.I just wanted to say thank you for spreading math on youtube and also writing great books.
It's also pretty simple to prove in general Hilbert spaces: ||a+b||² = = + + + = ||a||² + 2 Re + ||b||². So if a and b are orthogonal, i.e. = 0, we have ||a+b||² = ||a||² + ||b||².
Yes, but Hilbert spaces all use the L2 metric, which is practically the same thing as taking the Pythagorean theorem as an axiom. Of course it's easy to prove, it's the very thing powering the entire space!
My mind exploded when you showed the "make three copies of the ABC triangle and scale each by A, B, and C" proof; that's my new favorite proof. Though, oddly enough, I already knew about the "square to parallelogram to rectangle" proof, but I had no good way of describing what I knew. I also knew about how you can extend the Pythagorean Theorem to ANY triangle and it involved adding or subtracting something multiplied by a trig function to compensate; I just happen to stumble upon it while watching TV, and that's how I first learned about trigonometry as early as 6th grade.
That simplest proof actually blew my mind the moment I saw it. It was like Pythagoras just became as immediately obvious as common sense in an instant. Wow.
at 6:48 is this what you meant? BB=B times B BA=B times A BC=B times C 11:23 is this one using the 6:48's BB BC BA logic? or how do i find what A^2 abd B^2 are? all i can think was 1/2 A * B = 1/2 C *h, where h is the middle line that you divided it. then I got stuck how to come up with a2 +b2 = c2 by using h
12:15 - cut each in half across the diameter and arrange the small and medium at right angles and compare the large to the hypotenuse made by small and medium.
i think I'd use the pizza cutter as a tachometer wheel and measure the circumference of the two smaller pizzas and if the square of both is greater than that of the square of the larger, I'd go for the two smaller :)
Proof for 16:00 both triangles are similiar: A/X = B/D Pythagoras: X = sqrt(A^2-D^2) =>A/sqrt(A^2-D^2) = B/D => A^2/(A^2-D^2) = B^2/D^2 =>A^2/A^2 - D^2/A^2 = D^2/B^2 => 1-D^2/A^2 = D^2/B^2 =>1 = D^2/A^2+D^2/B^2 =>1/D^2 = 1/B^2+1/A^2 Proof for 16:38 Using the cosine rule we can determine the following(the three angles of the red triangle are named alpha,beta,gamma): A^2 = B^2+C^2 -2BC*cos(alpha) B^2 = A^2+C^2 -2AC*cos(beta) C^2 = A^2+B^2 -2AB*cos(gamma) => A^2+B^2+C^2 = 2*(A^2+B^2+C^2) - 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) => 0 = A^2+B^2+C^2 - 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) => A^2+B^2+C^2 = 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) We know that the angles of the green triangles that are next to the angles of the red triangle are 180° - the corresponding angle of the red triangle, because the sum of the angle of the red triangle, the green triangle and 180° from the two right angles from the blue squares must equal 360° So: X^2 = B^2+C^2 -2BC*cos(180° - alpha) Y^2 = A^2+C^2 -2AC*cos(180°-beta) Z^2 = A^2+B^2 -2AB*cos(180° - gamma) => X^2+Y^2+Z^2 = 2*(A^2+B^2+C^2) - 2 * (AB*cos(180° - gamma)+AC*cos(180° - beta)+BC*cos(180 ° - alpha)) since -cos(180-x) = cos(x) => X^2+Y^2+Z^2 = 2*(A^2+B^2+C^2) + 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) we can substitute 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) with our previous term => X^2+Y^2+Z^2 = 3*(A^2+B^2+C^2)
Yep the super-theory is actually wrong "for all shapes". It needs to be "for all 2 dimensional shapes" Fractals have fractional dimensions and don't scale the way one might normally expect.
Fractals don't always have to have fractional dimension. A filled-in koch snowflake has a fractal dimension of two, even though its boundary hasn't. The mandelbrot set even has a two-dimensional boundary and is still considered a fractal.
Jorge. I didn't forget the question mark, I intentionally didn't include it. I already figured that in the case of fractals it's not going to work when the Hausdorff dimension doesn't equal it's topological dimension, and that we'd need to replace "area" for something like "the sum of the area of all 2-faces" if we want to include shapes with a topological dimension not equal to 2. So while it's not proper grammar I decided not adding a question mark would eliminate some confusion that I was actually asking "What about fractals?", my point was just to be cheeky and pedantic.
Fanstastic video! It's amazing to see how beautiful and versatile the Pythagorean theorem is and how many generalizations it allows. One could even make the case that the Pythagorean theorem was in a sense the foundation for mathematics! I just think it's unfortunate that Pythagoras got too much credit for it, since he probably wasn't the first to prove it, but I guess that's how history goes.
16:40 That proof is pretty easy actually. The green triangle on the top has sides a and b just like the red one, and the angle between sides a and b is supplementary to its vertical angle since the other two are right angles by virtue of being in squares. That means the vertical angles have the same sines. Because the area of a triangle is a times b times the sine of the angle between them and each of those factors are the same, they have the same area.
That Trump joke was VERY disrespectful, Burkard. Don't you know that Trump invented ALL the proofs of this theorem thousands, millions, billions of years ago??? So ignorant.
JustOneAsbesto Yeah, and he made the babylonians pay for them. He used them to build the numerals great wall of china and crooked pythagoras stole them. Then the christians stole alternate facts and named them lies and orange was outlawed in europe by fake news. He fled to america.aa
2:36 can’t you write that as: (a+b)^2 = 4ab/2+c^2 And then if you do the maths, and simplify it, you get: a^2 + 2ab + b^2 = 2ab + c^2 - 2ab. - 2ab a^2 + b^2 = c^2
I would like to see how all Trump detractors also post a self discovered proof of Pythagora's theorem, seeing that all of them are much smarter than him.
They are measuring him on another former president, not on themselves. Of course, they are judging him on a matter that hasn't got much to do with what a President needs to do, and I doubt he would be any more pleasent if he was more intelligent, but it seems like this was a joke.
12:09 the combo Proof: ( C = circumference of the pizzas, b = circumference of small pizza, sp = small pizza, mp = medium pizza, lp = large pizza) let us say that the circumference of the pizzas are the sides of a right triangle (the large pizza is the hypotenuse) Note: C of sp = b, C of mp = 2b, and C of lp = b√5 so the Pythagorean theorem works Then, b + 2b = 3b so, 3b > b√5 √5 is between 2 & 3
Well, more like pi - tha - gO - ras The intonation is in the "ό" letter (πυθαγόρας) Here is how we pronounce it in modern Greek translate.google.com/#auto/en/%CF%80%CF%85%CE%B8%CE%B1%CE%B3%CF%8C%CF%81%CE%B1%CF%82 I wonder how the actual pronounciation of his name, was in ancient Greek...
Animation is a really powerful tool for explaining these things. I have seen several of these proofs before, understood them, and promptly forgotten them. I think I will remember two of these.
I enjoy the spontaneous reactions from the guy behind the camera: to me he kind of represents all of us. He is also kind of Burkard’s ‘Greek Chorus’. Their interactions are real, relaxed and charming.
It blows my mind that nobody has pointed out that your shirt is actually even funnier when you realize that Einstein's equation E =mc2 is actually a very specific application of pythagorean's theorem. Because the full form of it is E^2 = (mc^2)^2 + (pc)^2 The version we normally see is for a stationary object with Rest Mass. So p is equal to 0, canceling the second term.
I had to spend another hour or two studying the triangle at 11:22 (also 15:50) and I believe I have successfully calculated the ratios of all the edges using arithmetic. I had been assuming by seeing it that it was the famous 3:4:5 triangle, but it wasn't. The length of B is A*SQRT(3) and the length of C is 2*A. What really surprised me was spotting an isosceles triangle of 120 degrees in there. Did you see that too?
Thanks....thanks...thanks... When I first saw this video title I thought it was something trivial that I was supposed to know ....but after watching it...I'm still in awe !....the elegant universe of mathematics....😎
The problem on minute 16:37 for the area of red and green triangles is actually easy. Just rotate 90° the red triangle around any of its vertices. Then one of its sides will fit along the side a green triangle, and we find that both triangles have the same base and the same height, and henceforth the same area.
At 16.42: extend the sides of the green triangles to form a new medium sized triangle. That triangle is right angled if and only if the original (ie red) triangle is isosceles. At 16.49: connect the corners of adjacent pairs of the yellow/orange squares and extend those line segments to the points where they intersect, forming a new large triangle. That triangle is always similar to the original right triangle, with side lengths equal to the following multiple of the original side lengths: (2+A/B)*(2+B/A) where A and B are the sides that form the right angle.
My favourite proof of Pythagoras is the one that uses calculus, because it's so very 'sledgehammer to crack a nut'. It's not complicated to do, it's just that calculus is so much more sophisticated, and came so much later than Pythagoras.
You have a right triangle with sides A, B and C. You extend the line A slightly (dA), which creates a new right triangle with dA as its hypotenuse, and dC as one of the sides. This new triangle, when the sides are arbitrarily small, is similar to the original such that dC/dA = A/C. Rearrange into CdC = AdA, and integrate to get C^2 = A^2 + [constant]. To find the constant, let A tend to 0 on your original triangle, which gives you that [constant] = B, and therefore that A^2 + B^2 = C^2. Some of the steps are cut for the sake of brevity, so it's not as rigorous as it could be, but hopefully this should feel right. It's available with diagrams at greater length on the Wikipedia page for the Pythagoras theorem, and I'm sure many other places.
12:00 - buy all 3 pizzas, cut them all in half, arrange them in a triangle. If the angle formed by the small and medium pizzas is a right angle, there is no better deal. If it is an obtuse angle, I think then the large pizza is better to buy, and if it is an acute angle, get the small and medium.
The implication that it worked for all shapes, not just squares was mind blowing. Oh, and as a consequence, my answer to the pizza question is that they are the same amount of pizza for the same price. Brilliant!
Slice each pizza into n equal slices and arrange in an alternating up-down pattern, then take the limit as n goes to infinity. This converts every circular pizza into a rectangular pizza. Next you submerge the rectangular pizzas in a bathtub and mark the height of the water level. You will then know which pizza combination is largest by the change in mass of the bathtub.
Pythagoras' theorem and the fact that it shows areas scaled to each side all add so that the two smaller areas equals the larger one seems to follow from the fact that the surface area of a sphere is proportional to the square of its radius. You can think of a slide projector at some distance R, from the screen. The area of a projected image at one distance Rc, can be made to equal the sum of the areas of images projected at two shorter distances, Ra and Rb. It can easily be shown that for that condition to exist, Rc^2 = Ra^2 + Rb^2 and that Ra,Rb and Rc must combine together to form a right triangle.
The proof for 16:36 X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) : Just apply the law of cosines to all 4 triangles and use the cosine of the supplement is the negative of the cosine.
@ 14:40 You asked for some non-trivial solutions. I give you infinite :D Choose the ones you like! Solutions of a^2+b^2-ab=c^2 are given by (a, b, c)=(4kxy, k(3x^2 + 2xy - y^2), k(3x^2 + y^2)) Solutions of a^2+b^2+ab=c^2 are given by (a, b, c)=(4kxy, k(3x^2 - 2xy - y^2), k(3x^2 + y^2)) Edit: Obviously keep in mind that a,b,c should be positive, so choose your x,y accordingly!
Solution to the pizza problem. You cut each pizza in half. You form a triangle with pizzas. If the angel between small pizzas is greater than 90 degrees you should take the big pizza. If it's less than 90 degrees you should take the small pizzas.
For the very first proof, if you take into account that the sum of the angles of a triangle must be 180 degrees, it follows that you can form the needed blue squares with any right triangle. I loved that proof and it made me crazy that one couldn't easily see that the squares are effectively squares, the proof is visually obvious except by that detail.
That shirt is more brilliant than it lets on, because the extended form of E=mc^2 basically turns mass and momentum into the lesser sides of a right triangle, with E being the hypotenuse!
Brilliant.I hope some of my ex-pupils were watching and some of the proofs and ideas come back to haunt them.I also was made aware of so much more.Thank you
10:31 - it's not that amazing at all - it's just a consequence of area having unit of length squared. We could *change our units* and see the same "constancy." And our choice of units can't affect any real results.
Let a, b, and c be three side then from the triangle inequality we can easily say--- a+b>c Which indeed States that, the offer of " Small + Medium> Large"
Außergewöhnlich! Even I as a nonmathematician have now a grasp (a slight one perhaps -- and probably very imperfect at that) of Pythagoras & the wider implications. Vielen Dank, Herr Professor Mathologer!
E=m(a²+b²).
:)
My Life has been a lie...
(fixed)
E²=m²c⁴+p²c² !
e^(i*pi) = -m(a^2 + b^2)/E
Did anyone notice that the Mathologer logo is a proof for the Pythagorean theorem?
I did right before I noticed you point it out.
I did then I scrolled, and there was the first comment pointing it out.
@@topherthe11th23 You're welcome.
Thanks
@@jamesolatunji5 you bet
pizza riddle: cut all three in half, arrange them to form a triangle with the cut sides. If the triangle is acute, the small pizzas is the best deal. If it is obtuse, the large pizza is the best deal
...or just count the number of pepperonis on each
You raise an excellent point. We haven't defined the "inherent value" of any given pizza to the customer; it will differ from person to person, and total surface area isn't necessarily what it will be. It could just be the total amount of toppings, if you don't care much about the cheese, the sauce, and the crust.
Fred
26+36 vs 81
Unfortunately you still have to pay for all three pizzas after cutting them up and playing with them :P
But yeah, I came to the same technique.
No. Big one is obviously the best, as it's perimeter is the smallest, so less wasted space for unnecessary crust.b
haha yes, and thats why pizza have a circle shape, to maximize the surface/perimeter ratio.
Pizzaiolo are really smart people.
I never got to algebra in school. Never made it to college either. All that well over 40+ years ago. But I enjoy these videos like you wouldn't believe. I feel like I'm learning via osmosis. Wish we had this back in the day. No telling where I'd be today. I do work these problems and am understanding algebra a bit. So please continue making these and please keep in mind, some of us are old dogs but we are learning new tricks. Thanks guys.
And please continue to be interested :) It's always a great inspiration for us youngsters to see that curiosity is a quality you can have at any age !
Big Nasty If you have time just try learning from Khan academy; it has videos and workouts
This is geometry though
I am 54, I hear ya! There are a lot of university lectures here on TH-cam as well, Yale and Stanford and many others have their own channels with courses.
Here's the link to Yale where you can see on the side links to other universities, if you're interested.
th-cam.com/users/YaleCourses
Click their "Playlists" to find something of interest.
I am from Ireland and algebra is taught to students from 11 years up. As youngster, I had studied mechanical drawing at school and that has hugely helped me with mathematics. For example, it taught me to understand the theorems of circles, squares, rectangles, etc.
The scaling proof is absolutely beautiful
by A, B and C? That was my favorite one
Ikr its the best one imo
Wow. The simplicity blew the hell out of me.
I agree. The general shape proof, that he favored was merely consistent but not conclusive
Agreed. The observation that if the sum of the areas of two similar figures attached to the sides of a right triangle equals the area of a third similar figure attached to the hypotenuse, then Pythagoras follows is clever. And the one-line construction that yields these similar figures is truly elegant. I am now convinced that for sheer utility and richness of results, Pythagoras's Theorem beats Euler's Identity as the most beautiful result in Mathematics.
I can't wait to share this proof, as well as de Gua's theorem and the quadrature sum of the areas of a right pyramid, with all my friends! Also looking forward to more fun and games when I get my hands on "A dingo ate my math book." Thanks Mathologer for making these entertaining and informative videos.
Integer-triangles with an angle of 60 degrees or 120 degrees are called "Eisenstein triples":
60 degrees: (3, 8, 7); (5, 8, 7); (5, 21, 19); (7, 40, 37); ...
120 degrees: (3, 5, 7); (7, 8, 13); (5, 16, 19); ...
Based on the name, I suppose there's a connection with solutions to those Diophantine equations and Eisenstein integers (i.e. complex numbers of the form a + b\omega + c\omega^2, where \omega is a primitive third root of unity and a, b, c are integers) similar to 3b1b's video on Pythagorean theorem and Gaussian integers.
simpler for 60deg: 1, 1, 1 (or any other Number, they just have to be equal) :D :P
UniTrader he says „nontrivial“ in the video
1 1 0 is a trivial solution. 1 1 1 is OK.
This man always has the best shirts
Jonathan Hernandez your profile pic is great
6:15 scaling of triangles. Absolutely beautiful
Oh my god it made me laugh out loud
Some other-Pythagorean triples:
120 degree triples: (a,b,c) = (3,5,7), (5,3,7), (6,10,14) (10,6,14) (7,8,13) (8,7,13)
60 degree triples: (a,b,c) = (1,1,1) (2,2,2) (n,n,n) etc (3,8,7) (8,3,7)(5,8,7)(8,5,7)(6,16,14)(16,6,14)(7,15,13)(15,7,13)(8,15,13)(15,8,13)(10,16,14)(16,10,14)
Tehom
It would be better to remove all multiples and duplicates, and simply list the primitives.
e.g.
(3,5,7) (7,8,13)
(1,1,1) (3,7,8) (5,7,8) (7,13,15) (8,13,15)
based on what you provided
In general,
(a² - b²)² + (2ab - b²)² - (a² - b²)(2ab - b²) = (a² - ab + b²)²
(a² - b²)² + (2ab + b²)² + (a² - b²)(2ab + b²) = (a² + ab + b²)²
I would consider (n,n,n) to be trivial, since you just get the equality n²=n².
I absolutely love your T-shirts, can we get them anywhere?
icestork.com/product/pythagoras-vs-einstein-c2-shirt/ might get it myself.
Usually just a quick google of whatever is on the t-shirt will get you there :) By popular demand I've also put some that I designed myself here shop.spreadshirt.com.au/1035536/
I will not be surprised if they are soon out of stock.
shouldn't Pythagoras be wearing trousers?
Sleves sqaured plus hat squared equals trousers hoses squared?
Sleeves volumes squared plus hat volume squared plus trousers hoses squared equals muscle shirt volume squared? IDK I'm a flatlander with poor imagination.
I´m German and wondered, why i do understand perfectly the Englisch of Burkard. Well, now I know, Burkard is German ...
Really nice proofs an even better animation !!!
You couldn't tell from the accent?
11:39
the angles were selected a bit uncomfortable making the left corner of the red triangle appear as if the outer lines meeting there would be straight.
but the "thesis" that all orange triangles are same shape just different scale and that the upper two triangles sum in their area to the lower one is fine.
thus the definite conclusion drawing a square around them will also make the areas of the upper two squares sump to the area of the lower squares - is definitely neat.
It took me a while, but I think I finally got it. I just need more words to guide my mind through an understanding of it. Whatever same shape figures are attached to the sides (each same shape triangle has the right angle away from the center) will scale proportionally. Could be semi-circles (half of each pizze in his challenge) or smiley faces. So, pretend that figure is a square. We know that the cut triangles came from the original, so the summation of their areas equal the whole, which relation must hold for any chosen figure. Therefore it must be true that a^2 + b^2 = c^2. Thanks.
My answer would be to cut the pizza's in half and put them together into a triangle. Putting the small and medium at a 90 degree angle to eachother and then fitting the large slice inbetween the remaining 2 vertices. If the diameter of the bigger pizza slice is smaller than the distance between the vertices, then it is a bad deal, if it is larger then it is a good deal. If it fits exactly then both are a great deal.
Jelle I'm not sure if this would work since you ONLY have a pizza knife...
Jelle (not then again, it's a math problem)
Jelle How do you cut the pizzas exactly in half?
We have to assume we can do at LEAST that, with a mathematical pizza knife.
mushookie man I suppose so. I asked because finding the center of a given circle is a common compass-and-straightedge exercise.
FINALLY, I got the proof of that 1/A²+1/B²=1/D² stuff.
Consider the area of the given triangle. Since it is a right triangle, the area (I call it F) can be calculated with F = 1/2*A*B. Since D is the height on C, the Area can be evaluated with F = 1/2*C*D aswell. Combining these two equations we get: C*D=A*B. Squaring both sides give us C²*D²=A²*B². But C²=A²+B². So we get the final equation: D²(A²+B²)=A²*B². Rearranging will give us what we initially wanted to show. q.e.d.
Good job, I just found out it is called the "inverse Pythagorean theorem"
Thanks man
Put together the same proof today. Using the equivalent areas is the key.
angeluomo exactly 👍
Nice! I solved it in a more pedestrian way :-(. I called the the two intervals into which D dissects C C1 and C2. Then we have three equations:
Two P-theorems:
(1) C1²+D²=A²
(2) C2²+D²=B²
And, as the two internal triangles have the same proportions:
(3) C1/D = D/C2,
which can be reerranged to obtain:
(3') C1*C2=D²
Express C1 and C2 from (1) and (2), plug them into (3') and after rearrangement you get the expression. I prefer yours though.
For 19:13, make the three lengths from the right angle a, b, and c. How we find the sum of the areas and use Heron’s formula for D^2. After some simplification the two expressions become the same.
The Mathologer logo is the first proof of Pythagoras's theorem. ;)
That's it. In fact, I used 3-4-5 triangles for the triangles in my logo :)
Its a new logo
No, the channel logo.
Nvm false memory
My own favorite proof is based of that shape :: 4 identical copies of a right triangle of sides a, b and c organized in a square shape.Exterior square has side s = a+b and its area is s^2 = a^2 + b^2 + 2ab . But the area is also the sum of the four triangles plus the inner square of side c.So Area = 4* (ab/2) + c^2 = 2ab + c^2 = s^2 = a^2 + b^2 + 2ab. And, therefore a^2 + b^2 = c^2. Done.
OK, here's my answer to the pizza problem:
Cut all three pizzas in half and form a triangle with one half from each pizza--the cut edge (diameter) being used as each side. If the triangle is right, both deals are equal. If it's an acute triangle, the two smaller pizzas are a better deal, and if it's an obtuse triangle, the larger pizza is the best deal.
Great video; planning to watch it several times over.
Ken Haley How do you cut the pizza exactly in half?
Also, how do you measure angles if we've only given a pizza knife?
+John Chessant Cut it in half by "eyeballing" it. A small error in the angle of the cut would result in a much smaller error (percentage-wise) in the length of the cut.
+letao12 You can look at it and see if it's close to a right triangle; if so, you can consider the deals equal within a few pennies.
To both: If I can't make these assumptions, I don't think a solution exists.
I love your sense of humor so much, man... You are a delight.
Thank you for doing these - my world is enriched with you in it, in so many ways. ...Okay, I may have a bit of a crush.
I proved the 3D counterpart (square areas) when I was in high school. When I went to uni, I showed it to a classmate who then proved a 4D version. He showed it to one of his computer science tutors, Carroll Morgan, who then proved it for all dimensions, but I don't know if he ever published it. A few months later, I was stunned when I was browsing at the library and by chance I opened up a journal and saw another proof. That was around the late 1970s.
Is it a 3D version of the Pythagorean theorem?
@@T.R.-TRUTH.REASON more than that, people proved an nD version.
the 3D version is the sum of the squares of the areas of the three right-angular sides of a tetrahedron equals the square of the area of the largest side.
@@MusicalRaichuI think you're talking about a cuboid or a cube, right?
@@T.R.-TRUTH.REASON Yes, diagonally slice a tetrahedron from the corner of a cube. sum of squares of areas of the three rectangles in the corner equals square of area of diagonal side.
For 4D, it's squares of volumes of tetrahedra. And so on for higher dimensions.
@@MusicalRaichu Yes，because it is a square.I have got it.
The proof at 6:30 - 6:50 blew my mind. So neat and beautiful.
The last six minutes were mind-blowing!!
8:40 Well; that also happens with the irrationality proofs of roots: I would argue that it’s much easier to prove that n√k ≠ a/b, where b > 1; for all the infinitely many positive integer values of k, n, a, and b; via the Fundamental Theorem of Arithmetic; than it is, to prove the irrationality of any individual number, like: √2, or: √3. 🙂
*UPDATE:* In fact; I recently noticed that you don’t even need the F.T.Arithmetic, for that proof. Simply knowing that all roots and powers of 1 are equal to 1, is enough. 🙂
cut the pizzas in half
arrange them into a triangle
if the triangle is acute take the two smaller pizzas
if the triangle is obtuse take the large pizza
if its a right triangle then take either
Michael P it's*
I think that works. But the smaller two diameters don't have to add up too the diameter of the larger one. If that's the case they don't make a triangle.
Which obviously means the bigger one has more area than the smaller combined.
Nice one.
Awesome, a fellow intellectual who watches Rick and Morty with a IQ of 300
one scenario you forgot to mention is if the triangle couldn't be made. In that case again get the large pizza.
I'm only eating at pizza places that give the square root of the price from now on.
Proof of 1/A^2 + 1/B^2 = 1/D^2: Scale the right triangle by a factor of 1/BD so that it has hypotenuse H = 1/D and left side 1/B. Call the bottom side E. Since the left triangle is similar to this triangle we know A/H = D/E, which tells us A = 1/E. Thus by Pythagoras we're done.
I also have a geometric proof of X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) that uses similar ideas to the proof of the cosine formula presented in the video, but it's hard to explain in a youtube comment.
Great video, simple and very fun! Reminds me a bit of Byrne's edition of Euclid due to the sheer artistry of the whole ''coloured shapes manipulation'', while this may not be the most rigorous or useful way to do geometry, darn is it esthetically beautiful. Have to check out that book of 371 proofs of pythagoras, sounds like recreational heaven!
Yes, Byrne is great, here is an online version www.math.ubc.ca/~cass/euclid/byrne.html
I've also provided a link to the book of proofs in the description. Have a look :)
When angle = 60*. 5^2 + 8^2 - (5*8*) = 49 = 7^2. When angle = 120*. 7^2 + 8^2 + (7*8) = 169 = 13^2. Thanks you Mathologer, because of you I find some really interesting new stuff.
I was asked to prove Pythagoras during a university interview back in 1983. I used the first proof with the big square in it, which seemed to go down well. However, I think the scaling proof is definitely my new favourite!
9:10
The fraction of the semicircle occupying a square can be written as: the area of the semicircle/the area of the square
Since both the semicircle's diameter and the square side lengths is equal, their lengths can be given by a.
the area of the semicircle will simply be = π/2•(a/2)^2 = π/8•a^2.
The fraction can then be written as = (π/8•a^2)/a^2 and since a^2 are like terms this simplifies to π/8 which is roughly equal to 0.39.
What about A³+B³+C³=D³? Does this have a positive integer solution?
Yes, for example, 1³+2³+2³=3³ . Have a look at this en.wikipedia.org/wiki/Pythagorean_quadruple
Mathologer Umm, I mean its cubed, not squared. Sorry about that...
Sorry, my eyesight much be failing me. Of course, you were after cubes and not squares. Here you go 1³+6³+8³=9³ . These is an example of a cubic quadruple.
Mathologer Oh great. Thanks! I wonder if you can prove it, like you can prove Pythagoras, but using cubes and some sort of 3D shapes. Does it have any applications in geometry?
Hey mathologer, I just wanted to say that i am a big fan of your videos and have been a fan for over a year and a half.Recently my uncle gave me a book for my birthday and coincidentally it was written by you.It was Sciencia and your book (QED Beauty in mathematical proofs) was written very well and was informative.I just wanted to say thank you for spreading math on youtube and also writing great books.
Glad you like what I do and thank you very much saying so :)
Thank you!!
It's also pretty simple to prove in general Hilbert spaces:
||a+b||² = = + + + = ||a||² + 2 Re + ||b||². So if a and b are orthogonal, i.e. = 0, we have ||a+b||² = ||a||² + ||b||².
Yes, but Hilbert spaces all use the L2 metric, which is practically the same thing as taking the Pythagorean theorem as an axiom. Of course it's easy to prove, it's the very thing powering the entire space!
What are Hibert Spaces?
+tamara ciocan A generalized version of Euclidean vector spaces that allows for infinite dimensions and complex vectors.
a52Productions exactly!
My mind exploded when you showed the "make three copies of the ABC triangle and scale each by A, B, and C" proof; that's my new favorite proof. Though, oddly enough, I already knew about the "square to parallelogram to rectangle" proof, but I had no good way of describing what I knew.
I also knew about how you can extend the Pythagorean Theorem to ANY triangle and it involved adding or subtracting something multiplied by a trig function to compensate; I just happen to stumble upon it while watching TV, and that's how I first learned about trigonometry as early as 6th grade.
Omg Pythagoras and Einstein fighting over c squared, and that's the first thing I see...
I want that T shirt
6:46 OHHHH WOW!!! That has got to be absolutely the most wonderful proof ever!
I absolutely appreciate the amazing visualizations in each of the videos here. Great work!
That simplest proof actually blew my mind the moment I saw it.
It was like Pythagoras just became as immediately obvious as common sense in an instant.
Wow.
The text at 10:04 says: "Any shape with non zero area!!!" :)
at 6:48
is this what you meant?
BB=B times B
BA=B times A
BC=B times C
11:23
is this one using the 6:48's BB BC BA logic? or how do i find what A^2 abd B^2 are?
all i can think was 1/2 A * B = 1/2 C *h, where h is the middle line that you divided it. then I got stuck how to come up with a2 +b2 = c2 by using h
Giving credit to the real inventors of the theorem. You earned yourself a subscriber.
Thank you so much... I have shared this with many kids in lower school grades they have found this to be very helpful!!!
Proving Pythagoras theorem by proving fermat's last theorem
This.
12:15 - cut each in half across the diameter and arrange the small and medium at right angles and compare the large to the hypotenuse made by small and medium.
I agree. Also, the question was poorly conveyed
i think I'd use the pizza cutter as a tachometer wheel and measure the circumference of the two smaller pizzas and if the square of both is greater than that of the square of the larger, I'd go for the two smaller :)
Proof for 16:00
both triangles are similiar:
A/X = B/D
Pythagoras: X = sqrt(A^2-D^2)
=>A/sqrt(A^2-D^2) = B/D
=> A^2/(A^2-D^2) = B^2/D^2
=>A^2/A^2 - D^2/A^2 = D^2/B^2
=> 1-D^2/A^2 = D^2/B^2
=>1 = D^2/A^2+D^2/B^2 =>1/D^2 = 1/B^2+1/A^2
Proof for 16:38
Using the cosine rule we can determine the following(the three angles of the red triangle are named alpha,beta,gamma):
A^2 = B^2+C^2 -2BC*cos(alpha)
B^2 = A^2+C^2 -2AC*cos(beta)
C^2 = A^2+B^2 -2AB*cos(gamma)
=> A^2+B^2+C^2 = 2*(A^2+B^2+C^2) - 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha))
=> 0 = A^2+B^2+C^2 - 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha))
=> A^2+B^2+C^2 = 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha))
We know that the angles of the green triangles that are next to the angles of the red triangle are 180° - the corresponding angle of the red triangle, because the sum of the angle of the red triangle, the green triangle and 180° from the two right angles from the blue squares must equal 360°
So:
X^2 = B^2+C^2 -2BC*cos(180° - alpha)
Y^2 = A^2+C^2 -2AC*cos(180°-beta)
Z^2 = A^2+B^2 -2AB*cos(180° - gamma)
=> X^2+Y^2+Z^2 = 2*(A^2+B^2+C^2) - 2 * (AB*cos(180° - gamma)+AC*cos(180° - beta)+BC*cos(180 ° - alpha))
since -cos(180-x) = cos(x)
=> X^2+Y^2+Z^2 = 2*(A^2+B^2+C^2) + 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha))
we can substitute 2 * (AB*cos(gamma)+AC*cos(beta)+BC*cos(alpha)) with our previous term
=> X^2+Y^2+Z^2 = 3*(A^2+B^2+C^2)
9:53, What about fractals :p
Yep the super-theory is actually wrong "for all shapes". It needs to be "for all 2 dimensional shapes"
Fractals have fractional dimensions and don't scale the way one might normally expect.
8bitpineapple you forgot the question mark
Fractals don't always have to have fractional dimension. A filled-in koch snowflake has a fractal dimension of two, even though its boundary hasn't. The mandelbrot set even has a two-dimensional boundary and is still considered a fractal.
Jorge. I didn't forget the question mark, I intentionally didn't include it. I already figured that in the case of fractals it's not going to work when the Hausdorff dimension doesn't equal it's topological dimension, and that we'd need to replace "area" for something like "the sum of the area of all 2-faces" if we want to include shapes with a topological dimension not equal to 2.
So while it's not proper grammar I decided not adding a question mark would eliminate some confusion that I was actually asking "What about fractals?", my point was just to be cheeky and pedantic.
I've never found Pythagoras so entertaining! A fascinating, fun and educational video! Many thanks, Mathologer!
Geometers are always in love with Pythagoras...
14:21 And as the 1000th commenter, I would like to say:
Hexagons are the bestagons
For the pizza problem:
Go to the pizzaman with the knife, cut him half, see whether it creates an 90 degree triangle and then eat the pizzas
Why can't I stop smiling while watching a math video? Beautiful
ayyye sunday evening is saved :D
Fanstastic video! It's amazing to see how beautiful and versatile the Pythagorean theorem is and how many generalizations it allows. One could even make the case that the Pythagorean theorem was in a sense the foundation for mathematics! I just think it's unfortunate that Pythagoras got too much credit for it, since he probably wasn't the first to prove it, but I guess that's how history goes.
I wonder if Trump has ever heard of Pythagoras. He probably thinks that Pythagoras is a refugee who wants to steal all his money :P
16:40 That proof is pretty easy actually. The green triangle on the top has sides a and b just like the red one, and the angle between sides a and b is supplementary to its vertical angle since the other two are right angles by virtue of being in squares. That means the vertical angles have the same sines. Because the area of a triangle is a times b times the sine of the angle between them and each of those factors are the same, they have the same area.
Nice video before watching it
You are like the best math teacher I never had
That Trump joke was VERY disrespectful, Burkard.
Don't you know that Trump invented ALL the proofs of this theorem thousands, millions, billions of years ago???
So ignorant.
JustOneAsbesto Yeah, and he made the babylonians pay for them. He used them to build the numerals great wall of china and crooked pythagoras stole them. Then the christians stole alternate facts and named them lies and orange was outlawed in europe by fake news. He fled to america.aa
Less is more.
I hate the fuckiiiin tao. Modern symbol sucks!
2:36 can’t you write that as:
(a+b)^2 = 4ab/2+c^2
And then if you do the maths, and simplify it, you get:
a^2 + 2ab + b^2 = 2ab + c^2
- 2ab. - 2ab
a^2 + b^2 = c^2
I would also like to know Donald Trump's tweeted proof of Pythagoras's theorem, immediately.
I would like to see how all Trump detractors also post a self discovered proof of Pythagora's theorem, seeing that all of them are much smarter than him.
John ... well we aren't the president, are we?
It begins "First, construct a wall W..."
"I know proofs, I have the best proofs, believe me..."
They are measuring him on another former president, not on themselves. Of course, they are judging him on a matter that hasn't got much to do with what a President needs to do, and I doubt he would be any more pleasent if he was more intelligent, but it seems like this was a joke.
This is video made me open my mouth in awe more than once.
It made me pause and run to my notes,
It made me think,
It was so beautiful. Thank you
Your t-shirt 😓 suuu goooddd make me 😳laugh so badly 🤤
12:09 the combo
Proof:
( C = circumference of the pizzas, b = circumference of small pizza, sp = small pizza, mp = medium pizza, lp = large pizza)
let us say that the circumference of the pizzas are the sides of a right triangle (the large pizza is the hypotenuse)
Note: C of sp = b, C of mp = 2b, and C of lp = b√5 so the Pythagorean theorem works
Then, b + 2b = 3b so,
3b > b√5
√5 is between 2 & 3
Pi ta go Ra s
Well, more like pi - tha - gO - ras
The intonation is in the "ό" letter (πυθαγόρας)
Here is how we pronounce it in modern Greek
translate.google.com/#auto/en/%CF%80%CF%85%CE%B8%CE%B1%CE%B3%CF%8C%CF%81%CE%B1%CF%82
I wonder how the actual pronounciation of his name, was in ancient Greek...
You mean Pi Ta Go Ra Su I Cchi, right?
Animation is a really powerful tool for explaining these things. I have seen several of these proofs before, understood them, and promptly forgotten them. I think I will remember two of these.
That's great, mission accomplished then as far as you are concerned :)
The guy in the background sounds a bit like a sitcom laugh track - I think he adds nothing to an already great video
okarakoo That guy is the cameraman... He's just doing unscripted friendly interaction.
erm ... no.
The guy in the background in Marty Ross, one of Burkhard's collaborators.
I enjoy the spontaneous reactions from the guy behind the camera: to me he kind of represents all of us. He is also kind of Burkard’s ‘Greek Chorus’. Their interactions are real, relaxed and charming.
The best math channel ever.
I hope you never stop making videos!
i just realized i conjectured de gua's theorem when i was 15.
great video as always, congrats on the book
r/iamverysmart
Another glorious romp in geometry led by the ever-enthusiastical Mathologer! ABC-Delicious!
It blows my mind that nobody has pointed out that your shirt is actually even funnier when you realize that Einstein's equation E =mc2 is actually a very specific application of pythagorean's theorem.
Because the full form of it is E^2 = (mc^2)^2 + (pc)^2
The version we normally see is for a stationary object with Rest Mass. So p is equal to 0, canceling the second term.
Mathologer is really amazing and I love it. Nice T-shirt!
Omg what a roller coaster of a video, I had no clue on how many levels was Pythagoras cool!
I had to spend another hour or two studying the triangle at 11:22 (also 15:50) and I believe I have successfully calculated the ratios of all the edges using arithmetic. I had been assuming by seeing it that it was the famous 3:4:5 triangle, but it wasn't. The length of B is A*SQRT(3) and the length of C is 2*A. What really surprised me was spotting an isosceles triangle of 120 degrees in there. Did you see that too?
Thanks....thanks...thanks...
When I first saw this video title I thought it was something trivial that I was supposed to know ....but after watching it...I'm still in awe !....the elegant universe of mathematics....😎
The problem on minute 16:37 for the area of red and green triangles is actually easy. Just rotate 90° the red triangle around any of its vertices. Then one of its sides will fit along the side a green triangle, and we find that both triangles have the same base and the same height, and henceforth the same area.
At 16.42: extend the sides of the green triangles to form a
new medium sized triangle. That triangle is right angled if and only if the
original (ie red) triangle is isosceles.
At 16.49: connect the corners of adjacent pairs of the
yellow/orange squares and extend those line segments to the points where they
intersect, forming a new large triangle.
That triangle is always similar to the original right
triangle, with side lengths equal to the following multiple of the original side
lengths: (2+A/B)*(2+B/A) where A and B are the sides that form the right angle.
The best teacher and the best material on youtube ... thanks
My favourite proof of Pythagoras is the one that uses calculus, because it's so very 'sledgehammer to crack a nut'. It's not complicated to do, it's just that calculus is so much more sophisticated, and came so much later than Pythagoras.
You have a right triangle with sides A, B and C. You extend the line A slightly (dA), which creates a new right triangle with dA as its hypotenuse, and dC as one of the sides. This new triangle, when the sides are arbitrarily small, is similar to the original such that dC/dA = A/C. Rearrange into CdC = AdA, and integrate to get C^2 = A^2 + [constant]. To find the constant, let A tend to 0 on your original triangle, which gives you that [constant] = B, and therefore that A^2 + B^2 = C^2.
Some of the steps are cut for the sake of brevity, so it's not as rigorous as it could be, but hopefully this should feel right. It's available with diagrams at greater length on the Wikipedia page for the Pythagoras theorem, and I'm sure many other places.
Oh wow, the proof with the scaled triangles is so beautiful.
12:00 - buy all 3 pizzas, cut them all in half, arrange them in a triangle. If the angle formed by the small and medium pizzas is a right angle, there is no better deal. If it is an obtuse angle, I think then the large pizza is better to buy, and if it is an acute angle, get the small and medium.
The implication that it worked for all shapes, not just squares was mind blowing. Oh, and as a consequence, my answer to the pizza question is that they are the same amount of pizza for the same price. Brilliant!
Slice each pizza into n equal slices and arrange in an alternating up-down pattern, then take the limit as n goes to infinity. This converts every circular pizza into a rectangular pizza. Next you submerge the rectangular pizzas in a bathtub and mark the height of the water level. You will then know which pizza combination is largest by the change in mass of the bathtub.
If you love pepperoni, choose the large pizza , but if you love the crust, choose the small+ medium pizza
Pythagoras' theorem and the fact that it shows areas scaled to each side all add so that the two smaller areas equals the larger one seems to follow from the fact that the surface area of a sphere is proportional to the square of its radius. You can think of a slide projector at some distance R, from the screen. The area of a projected image at one distance Rc, can be made to equal the sum of the areas of images projected at two shorter distances, Ra and Rb. It can easily be shown that for that condition to exist, Rc^2 = Ra^2 + Rb^2 and that Ra,Rb and Rc must combine together to form a right triangle.
The proof for 16:36 X^2 + Y^2 + Z^2 = 3(A^2 + B^2 + C^2) : Just apply the law of cosines to all 4 triangles and use the cosine of the supplement is the negative of the cosine.
@ 14:40 You asked for some non-trivial solutions. I give you infinite :D Choose the ones you like!
Solutions of a^2+b^2-ab=c^2 are given by (a, b, c)=(4kxy, k(3x^2 + 2xy - y^2), k(3x^2 + y^2))
Solutions of a^2+b^2+ab=c^2 are given by (a, b, c)=(4kxy, k(3x^2 - 2xy - y^2), k(3x^2 + y^2))
Edit: Obviously keep in mind that a,b,c should be positive, so choose your x,y accordingly!
I really enjoy these amazing gems of math wisdom! Keep it up! It is so enjoyable to learn and relearn these interesting facts!!!
Thank you!
Solution to the pizza problem. You cut each pizza in half. You form a triangle with pizzas. If the angel between small pizzas is greater than 90 degrees you should take the big pizza. If it's less than 90 degrees you should take the small pizzas.
For the very first proof, if you take into account that the sum of the angles of a triangle must be 180 degrees, it follows that you can form the needed blue squares with any right triangle. I loved that proof and it made me crazy that one couldn't easily see that the squares are effectively squares, the proof is visually obvious except by that detail.
That shirt is more brilliant than it lets on, because the extended form of E=mc^2 basically turns mass and momentum into the lesser sides of a right triangle, with E being the hypotenuse!
Happy birthday, Dr. Polster!
Brilliant.I hope some of my ex-pupils were watching and some of the proofs and ideas come back to haunt them.I also
was made aware of so much more.Thank you
I guarantee it, this video will become one of the Mathologer classics
(also 1000th comment!)
10:31 - it's not that amazing at all - it's just a consequence of area having unit of length squared. We could *change our units* and see the same "constancy." And our choice of units can't affect any real results.
Let a, b, and c be three side then from the triangle inequality we can easily say---
a+b>c
Which indeed States that, the offer of " Small + Medium> Large"
That subtle morph at 10:34 literally hurt my brain.
14:51
Sure!
For the second equation: (8,5,7), (15,7,13), (21,16,19),...
For the third equation: (3,5,7), (8,7,13), (5,16,19),...
In general,
(a² - b²)² + (2ab - b²)² - (a² - b²)(2ab - b²) = (a² - ab + b²)²
(a² - b²)² + (2ab + b²)² + (a² - b²)(2ab + b²) = (a² + ab + b²)²
Infinite triples generator.
Außergewöhnlich!
Even I as a nonmathematician have now a grasp (a slight one perhaps -- and probably very imperfect at that) of Pythagoras & the wider implications.
Vielen Dank, Herr Professor Mathologer!
That's great :)
Hey, bist du auch Deutscher?
I dont catch the so-called 'simplest', at 10:45... Dont see any square... Help !
Is there a website which proves the theorem at 16:39 ?