Visualising Pythagoras: ultimate proofs and crazy contortions

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Finally, a Mathologer video about Pythagoras. Featuring some of the most beautiful and simplest proofs of THE theorem of theorems plus an intro to lots of the most visually stunning Pythagoranish facts and theorems from off the beaten track: the Pythagoras Pythagoras (two words :), 60 and 120 degree Pythagoras, de Gua's theorem, etc.

Things to check out:

As usual thank you very much to Marty and Danil for all their help with this video.

Enjoy!

[KekusMagnus]

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

[CrepitusRex]

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.

[DanielGonzalezL]

The scaling proof is absolutely beautiful

[Makebuildmodify]

Did anyone notice that the Mathologer logo is a proof for the Pythagorean theorem?

[ethancheung]

6:15 scaling of triangles. Absolutely beautiful

[johnchessant]

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); ...

[Tehom]

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)

[RealtermDe]

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 !!!

[Jelle_NL]

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.

[dominikstepien]

I absolutely love your T-shirts, can we get them anywhere?

[SuperDreamliner]

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.

[michaelp]

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

[kenhaley]

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.

[johnchessant]

The Mathologer logo is the first proof of Pythagoras's theorem. ;)

[ljfaag]

It's also pretty simple to prove in general Hilbert spaces:
||a+b||² = <a+b, a+b> = <a, a> + <a, b> + <b, a> + <b, b> = ||a||² + 2 Re <a, b> + ||b||². So if a and b are orthogonal, i.e. <a, b> = 0, we have ||a+b||² = ||a||² + ||b||².

[macronencer]

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!

[lovaaaa]

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!

[MusicalRaichu]

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.

[dl]

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.

[bloomface]

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.