> By all accounts, these two teenage math students are the exact opposite of the majority of the math establishment. They are female, they are African-American, and they come from an area which is not particularly renowned for producing high academic achievers. This is just an awesome turn of events and one which should inspire anyone — no matter what their ethnic, gender or socio-demographic background — that excell…
How would it be remiss to omit these factors, several of which seen irrelevant. Are you suggesting that their proof be discounted because it might be divinely inspired?
New Orleans teenagers found a new proof of the Pythagorean Theorem
191–200 of 287 posts
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#192Nice trick, but this text only handles the case of “When our extended lines from steps 2 and 3 meet” . What if they don’t, that is, what if α = β = π/4, and the triangle is isosceles and rectangular? I haven’t seen the original text, but this proof may be incomplete.
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#193Cool proof, though it doesn't consider the case where a=b. If so, the geometric series is non-converging since the ratio isn't less than 1. Geometrically, the construction wouldn't work because the sides A and C of the large "triangle" would be parallel to each other.
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#194Nice trick, but this text only handles the case of “When our extended lines from steps 2 and 3 meet” . What if they don’t, that is, what if α = β = π/4, and the triangle is isosceles and rectangular? I haven’t seen the original text, but this proof may be incomplete.
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#195Earlier quoted context omitted.
Geometrically, this happens when alpha is 45. The two lines in the diagram from the article will be parallel and never converge. 2*alpha = beta+alpha I was waiting for them to break this out into a special case or something but the article never did. Can't find any other material on this proof that mentions it
You can divide partial sums first and then take the limit at infinity instead summing first and then dividing. Not sure how much that helps with "infinite triangle" with two 90deg and one 0deg angles. That's btw how you get sin90=1 which doesn't have any geometrical sense when we consider finite triangles. Or in case of triangle with 45, 45, 90 maybe you could just pick different angle than 90 to be 2alpha.
Though, still for the 45-45-90 I don't think you can pick a different angle? At least for alpha > 45 (because it also doesn't work for this, the lines diverge), you can always swap it so beta is > 45 for those cases. If you pick something other than 90 to be 2 alpha, the reflection mentioned in step 1 can't be done
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#196Nice trick, but this text only handles the case of “When our extended lines from steps 2 and 3 meet” . What if they don’t, that is, what if α = β = π/4, and the triangle is isosceles and rectangular? I haven’t seen the original text, but this proof may be incomplete.
I mean this special case is also trivial, so it seems pretty reasonable to omit it. Feels quite uncharitable to describe this proof as a nice trick and then claim its incomplete because of such a simple special case. Mathematicians wouldn't consider this incomplete when the "missing" case can be solved almost by looking at it.
Can you educate me?
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#197Nice trick, but this text only handles the case of “When our extended lines from steps 2 and 3 meet” . What if they don’t, that is, what if α = β = π/4, and the triangle is isosceles and rectangular? I haven’t seen the original text, but this proof may be incomplete.
Then you go to the opposite direction, and the math still holds, but now you have inverses. And those inverses, at proportionality final formulae in the text, still gives you the Pythagorean formulae. I suggest you do see the original text.
If I could find it, I would have. This discussion mentions https://meetings.ams.org/math/spring2023se/meetingapp.cgi/Pa..., but I can’t find the paper there. Do I overlook something on that page?
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#198Earlier quoted context omitted.
I think any new proof of PT is pretty cool. it doesn't even have to be "elegant" or better than previous proofs. Just a new one. Will there at some point be a proof that all possible proofs of Pythagorean Theorem have been found. Or for any theorem? Is it possible to prove that all possible proofs of something have been found?
I wonder, are there an infinite number of proofs of the theorem, each more complex than the last? Can I rephrase what they did, make it more convoluted, and call it a new proof?
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#199Earlier quoted context omitted.
You can define sine and cosine together using the functional equations S(X)C(Y)+C(X)S(Y)=S(X+Y) C(X)C(Y)−S(X)S(Y)=C(X+Y) The only solutions to this are the constant 0 functions and the sine-cosine pair.
This is super cool, I've never seen it before! Do you know what this is called so I can look up a proof/theorem on it?
Re: New Orleans teenagers found a new proof of the Pythagorean Theorem
#200Earlier quoted context omitted.
I mean this special case is also trivial, so it seems pretty reasonable to omit it. Feels quite uncharitable to describe this proof as a nice trick and then claim its incomplete because of such a simple special case. Mathematicians wouldn't consider this incomplete when the "missing" case can be solved almost by looking at it.
I don’t see it being trivial. Of course, ‘everybody’ knows the diagonal of the unit square has length √2, but don’t we know that because of the Pythagorean theorem? Can you educate me?
Okay so take the triangle made by taking the diagonal of the unit square. This has side lengths 1, 1, and c and has area 1/2.
Now, take four of these and arrange them in a square with the side length being c. It would be easier to draw this... basically you stick the right angles in the center. If this isn't clear I can draw a diagram.
Anyway, you just made a square with side length c but since its made of four of those original triangles we know that the area of it is 4 * (1/2) = c^2 so c^2 = 2.
EDIT: made an excalidraw to explain this construction - maybe helpful https://excalidraw.com/#room=2298a8fd232d5f58e8ca,HmUwSqOt6J...