To show that an isosceles right triangle with side length 1 has hypotenuse sqrt(2), you can put 4 copies of it together corner-to-corner, which gives you a square, and since each triangle has area 1/2 the square has area 2. So the side length must be sqrt(2).

Attempting to generalize this pretty quickly leads you to the "Behold!" proof.

Finding distance between two points in R^(2).

Here is my favorite math video, and my favorite proof of the Pythagorean theorem, courtesy of Barry Mazur. The argument is unforgettable.

https://www.youtube.com/watch?v=ItiFO5y36kw

As soon as I learned of the statement of the theorem, I noticed it is obvious that a right triangle is the union of two smaller right triangles and all three are similar. Therefore the areas satisfy A+B=C. The area is proportional to the hypotenuse squared hence a^2 + b^2 = c^2 where a,b,c are the corresponding hyotenuses.

 

I later heard that people attribute this proof to this or that famous thinker, it seems obvious that one area is the sum of the other two and all three are similar though.

I don't really see what this has to do with giftedness. What is your point? I don't think the ability to see the proof has anything to do with the ability to hold opposing ideas in your mind at the same time and still retain the ability to function. Divergent thinking would be involved in coming up with this proof, but understanding it is straightforward for someone with a mathematics background imo.

(And apparently I'm not gifted enough to get the image to show in the preview instead of a link. ☹️)

A nice piece of work on someone's part.

I love your way of thinking <3

This was the first proof of Pythagoras identity that I understood.

A better way to visualize is to take the random paths of A and B and overlay them over each other. You will get the random paths of C. There is no fixed starting or end point, they are pure displacements.

Two high school students from Louisiana have just found a new proof of the Pythagorean Theorem. I reverse engineer the proof from a couple of pictures we have from their conference talk. It's fun!

What about this? https://www.cut-the-knot.org/pythagoras/TrigProof.shtml

is it in here? https://www.cut-the-knot.org/pythagoras/

A proof I saw somewhere: if you integrate 2sin(x)cos(x) in two different ways (using a u-substitution with either u = sin(x) or u = cos(x)), you get sin^(2)(x)+C1 on one hand and -cos^(2)(x)+C2 on the other, which shows that sin^(2)(x) + cos^(2)(x) = constant. By plugging in x = 0 we see that the constant must be 1.

So here is one that I came up with myself: you can use the fact that the alternating binomial coefficients add to zero to show that the power series of sin^2 x and cos^2 x have only one non-vanishing term when added together.

I have discovered a truly complicated proof of Pythagoras' theorem, which this margin is too narrow to contain.

There’s a really interesting proof of the Pythagorean theorem using dimensional analysis and Buckingham Pi theorem

We first observe that any quantum field theory that hopes to eventually incorporate general relativity must have equations of motions that are infinitesimally Poincare invariant. We know prove that up to scaling there is exactly one norm induced metric which suffices this, namely the one where if a und b are orthogonal vectors with length |a| and |b|, then a+b has length

|a+b|^2 = |a|^2 + |b|^2 .

The issue is that your staircase curve doesn't approximate the hypotenuse, not in the way you would need to for the arc length to match. It is always going the wrong direction, 45 degrees off. It does stay close to the hypotenuse, but where your curve is has little bearing on how long it is.

It works if you approximate with segments that have both endpoints on the curve you are trying to approximate, or that are tangent to the curve you are trying to approximate. More generally it works if your curve not only converges to the limit curve, but its derivatives also converge to the derivative of the limit curve, since arc length is computed from that derivative.

The same phenomenon appears in other settings as well. Look up the "Schwarz lantern" for a version involving surface area instead of length.

https://www.reddit.com/r/math/comments/1iesw9g/why_doesnt_the_manhattan_distance_approximate_the/

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how do you prove Brahmagupta's formula? It is usually proven by cosine law which is proven by Pythagorean theorem

Looks nice. I like it. I have two questions though. Why 2a, 2b and 2c? What would go wrong with a, b and c? And when you drop perpendiculars to FC, doesnt it create a right angle a E and D?

Looks great! I don’t know anything about its novelty, but I can offer a couple items to clean up for the proof:

1. In the “Construction and Diagram” section, remove the comma after “two parallel bases. You have a list of two items.
2. In the “Construction and Diagram” section, change “AF = BC” to “AF and BC.” This makes it consistent with the earlier statement in the “Construction and Diagram” section and removes redundancy with the “Proof” section.
3. In the “Proof” section, in the line “On the other hand,” rewrite ABCF as Area_ABCF. This promotes consistency with your previous statement and removes ambiguity.

It’s basically Al kashi applied in a right triangle

I haven't followed every step of your argument. I did notice, though, that at one point you use the formula for the sine of the sum; I mention this to say that it is not entirely clear which hypotheses you are using. You are clearly not using "just Euclid's axioms and the definition of the trigonometric functions".

More importantly, though, the argument is unnecessarily convoluted. Already from the graph in your second page (second circle in the first picture) you know from the big triangle that (calling the angle t, I have formatting troubles with 𝜃)

sec t / (tan t + cot t) = sin t.

Simplifying this you get directly that sin^(2) t + cos^(2) t = 1.

This is the place to mention that the whole "trigonometric proof of the Pythagorean Theorem" thing was absolutely blown out of proportion by the press, with totally fallacious titles like "The proof that mathematicians considered impossible". It was nice that those two students came up with some ideas, but it was not a big thing at all.

I did not see in your notes there "therefore a^2+b^2=c^2" which I thought was what you were proving.

However, to cut to the chase of my critique on this ...

at which point does this proof fail for non euclidean geometry?

Since the general form of this is actually 𝛼a^2 + 𝛽b^2 + 𝛾ab = c^2, clearly there must be some point in here were you explicitly use a Euclidean axiom. I believe that you still have not been clear about your axioms and I strongly suspect implicit use of intuition from one of the diagrams. But, the answer to the above two concerns would go a long way to making me happy about it. It would show me that you have a handle on the structure and context of your proof.

I skimmed it. But, I am not able at this time to spend the time to untangle it. If you understood the proof and context, you could answer the question quickly. My question was aimed at discovering whether you actually understood what you were doing. At this point, it seems that you do not. If you can point out how the proof would fail for non Euclidean geometry, say on the surface of a sphere, this would go a long way to convincing me that further effort would be worth my time. Rhetoric will not get my attention.

Note: by "skimmed" I don't mean "glanced at without thought". I gave serious attention looking at the upper left for clearly stated axioms, but did not feel that I saw that. I looked at the bottom right for a clearly stated conclusion. I did not see that either. As an academic, I have good experience checking proofs. My first reaction was that this seems to be not a good proof - but not wishing to merely dismiss it, I asked a pertinant question that will help me to understand what you are are are not achieving. I am seriously spending time and effort on this because I believe in giving people a chance. I hope that clarifies.

This is the first time i have written down everything from axioms, assumptions to end result lol. I am still a high school student and i am not experienced with "real proof writing", so please if you still find any genuine mistakes or flaw in my reasoning, please correct me. This is, In my opinion a novel and rigorous proof, but if you still do find any mistakes or flaws, tell me.

This proof is literally only based on the axioms of the euclidean geometry and the definitions of trigonometric functions, which were defined as the ratio of sides of triangle. If i have to summarize what makes this proof possible, like why does this work? then my answer would be, "because cotx = 1/tanx".

If you are confused about the scaling, then i suggest you draw a circle of radius c, and then do the exact same thing that i did on the 3rd page, you'd get a result with a scaling factor of c, and now just carefully do the manipulations i did at the last line of the 3rd page, and you'd get the result you might be familiar with

a^(2) + b^(2) = c^(2)

Somebody needs to TeX it. Lovely handwriting, but it feels like the whole thing might profit from being spread across 5-10 pages instead of three.

Or at least markdown with LaTeΧ equations

Is there a simpler way of doing this? Draw a square of side length L x L. Divide each side L into lengths a and b, flipping back and forth between a and b as you move around the square.

Draw another square from the points where a and b meet along the length of L. This square will have side length c.

Now you have 4 identical right triangles with side lengths a and b and hypotenuse c, and a square in the middle with side length c. The area of the whole thing, LxL = (a+b)(a+b) = a^2 + 2ab + b^2. The area of the 4 triangles equals ab/2 x 4 = 2ab. The area of the central square equals c^2.

Subtracting the 4 triangles from the larger square should give you the area of the smaller square. So a^2 + 2ab + b^2 - 2ab = c^2 simplifies to a^2 + b^2 = c^2.

I will always support creative mathing. I want to encourage your continued studies in the field.

Cool proof. Just one feedback I'd really want to give: please have your final proof a little neater, having these scribbles on random parts of the proof may confuse people. But aside from that, cool work!

interesting

Proof seems flawless. Good job!