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WCSTombs 4 hours ago

There are thousands of different proofs of the Pythagorean theorem, and some of them are really cool. The purely trigonometric proof that was found by some high school students recently is a great one. However, I think the greatest proof of all is this little gem that has been attributed to Einstein [1].

Take any right triangle. You can divide it into two non-overlapping right triangles that are both similar to the original triangle by dropping a perpendicular from the right angle to the hypotenuse. To see that the triangles are similar, you just compare interior angles. (It's better to leave that as an exercise than to describe it in words, but in any case, this is a very commonly known construction.) The areas of the two small triangles add up to the area of the big triangle, but the two small triangles have the two legs of the big triangle as their respective hypotenuses. Because area scales as the square of the similarity ratio (which I think is intuitively obvious), it follows that the squares of the legs' lengths must add up to the square of the hypotenuse's length, QED.

It's really a perfect proof: it's simple, intuitive, as direct as possible, and it's pretty much impossible to forget.

[1] https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99...

JanisErdmanis a few seconds ago | parent | next [-]

Indeed, a wonderful proof. It does, though, make one implicit assumption that if one stretches the fabric by the same amount, all holes in it stretch by the same amount. In particular, it assumes that triangle stretching is size-independent. Perhaps there are fabrics where that is not true...

zeroonetwothree 2 hours ago | parent | prev | next [-]

This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles.

This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction.

To me that truth isn’t necessarily any less fundamental than the Pythagorean theorem itself. But to each their own.

BTW Terrence Tao has a write up of this proof as well: https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem...

fiso64 13 minutes ago | parent [-]

I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.

bryanrasmussen 3 hours ago | parent | prev | next [-]

unfortunately doesn't work for me because of difficulty visualizing things, so I suppose there are probably a good number of people with the same problem.

So I guess for one particular subset of the population it is difficult, impossible to understand, and because it cannot be understood it will not be remembered.

Not complaining just noting the amusing thing that different explanations may have all sorts of problems with it.

Although if there was a video of it I guess I would understand it then. Not sure if everyone with visualization issues would though.

dataflow 2 hours ago | parent | prev | next [-]

> it follows that...

"Now just draw the rest of the owl."

Does it not feel like you skipped something here? The areas add up and area scales quadratically, therefore... Pythagorean Theorem? It definitely is not clear how this follows, even after the questionable assumption that it's obvious area scales quadratically.

codethief 5 minutes ago | parent [-]

He didn't skip anything but he left the details to the reader:

Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C.

The area scaling quadratically with the similarity ratio means that

A = (a/c)² C, and B = (b/c)² C.

Now, plug this into A + B = C, cancel C, rearrange.

card_zero 39 minutes ago | parent | prev | next [-]

Einstein's proof relies on the fact that the theorem works with any shape, not just squares, such as pentagons: https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentag...

Or any arbitrary vector graphics, like Einstein's face. So in the proof, the shape on the hypotenuse is the same as the original triangle, and on the other two sides there are two smaller versions of it, which when joined have the same area (and shape) as the big one.

Fair enough. However, none of the hundreds or thousands of proofs explain it. They all prove it, like by saying "this goes here, that goes there, this is the same as that, therefore logically you're stupid," but it still seems like weird magic to me. Some explanation is missing.

zeroonetwothree 2 hours ago | parent | prev [-]

I think proof #6 on this page is easier to follow and uses the same similar triangles. But then it’s just some basic algebra without assuming anything about areas of similar triangles :)

https://www.cut-the-knot.org/pythagoras/index.shtml#6