[ipince]

Cool 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.

[camjw]

When a = b then the proof of the Pythagorean theorem is trivial so this is the sort of nit-picking that actual mathematicians don't care about.

[ipince]

As a mathematician myself, I beg to differ. Proofs should be rigorous, and at the very least it should warrant a mention of this case.

[chengiz]

How is it trivial?

[camjw]

Commented this on another thread:

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...

[chengiz]

You're using geoemetrical construction not dissimilar from proving the theorem for a != b. So it's not in the spirit of this new method. No one disputes there are easier methods to prove the theorem.

[tzs]

Note that when a = b we have an equilateral triangle, with area a^2/2.

Draw a line from the 90 degree angle to side c, bisecting the 90 degree angle into two 45 degree angles. This divides the original triangle into two smaller triangles.

From the fact that the sum of the interior angles of a triangle is 180 degrees, it is not hard to see that the two smaller triangles are both equilateral, with sides of a, c/2, and c/2, and the angle between their two c/2 sides is 90 degrees.

That gives c^2/8 for the area of each of the smaller triangles, or c^2/4 for the area of the original triangle which we know to be a^2/2. So c^2/4 = a^2/2 or c^2 = 2 a^2 = a^ + a^2.

[ipince]

Ok, but those triangles aren't equilateral but isosceles :)

[cobertos]

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

[scotty79]

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.

[cobertos]

On the partial sums, okay, I see.

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