7 ms·

I have read about this proof for a bit and this is the first write-up that gives the slightest details. The phrase "using trigonometry" is confusing. What they

I have read about this proof for a bit and this is the first write-up that gives the slightest details. The phrase "using trigonometry" is confusing. What they do is assume functions sine and cosine exist, as normally defined, as ratios of triangle values, without assuming these have the various Pythagorean-theorem derived properties. They then construct an infinite series of nested triangles and use the formula for the sum of geometric series' to derive the length of the original triangle's hypotenuse. It certainly seems clever. I'm still confused what axioms they're effectively using relative to the usual Pythagorean theorem proofs - most of these use the formula for area of a right triangle and this seemingly doesn't. On the other hand, it seems an infinite construct would require things like the axiom of induction, which may or may not be included in axiom of axiomatic geometry.

scythe 4y ago
Induction isn't so bad. I'm not sure how they're getting away with the Law of Sines, though. The usual proof of LoS that I know is dependent on the existence of the circumcircle. But I don't know how to prove the existence of the circumcircle without dragging in a lot of geometry. Or you can use the area formula, which makes the proof similar to other arguments that use the area formula.
- SamReidHughes 4y ago
  There is a very straightforward proof that involves dropping an altitude. https://www.mathopenref.com/lawofsinesproof.html https://www.mathopenref.com/lawofsinesproof.html You could use a similar technique to make this proof not reference trigonometric functions.
  - crdrost 4y ago
    For that matter the easiest proof of Pythag that I know of involves dropping an altitude: Look at the right triangle the normal way up, clearly the area of the triangle is k c² (k = ½ sin α sin β if you like, but it just matters that it's the same nonzero k for all similar triangles). Now roll it onto its hypotenuse, drop an altitude, and observe that both subtriangles are similar to the first one, kc² = ka² + kb². The diagram that's a bit involved is the angle sum diagram, you start with a right triangle (a,b,c) with some angle α, extend it to a new triangle (a,b', c') with angle α+β, then make the new triangle with angle β that you stacked on top of the original triangle into a right triangle with angle β (c', d, e) by extending the hypotenuse of the (a,b,c) triangle to a point P, basically until the angle with the hypotenuse c' is 90°. Drop a dotted line to the x-axis from P and you can work out that the dotted line is at x=cos α cos β, and its distance to a is sin α sin β. Similarly the y-coordinates give sin α cos β + cos α sin β. As you say, you can do all of this without angles except for defining the first triangle with angle α+β, which you might not even need... We just need it here for sin(2 α) which is something like reflecting the same triangle about its hypotenuse?
    - alex_smart 4y ago
      For this proof to be a proof, you first have to define what area is and why it should scale as kc² and why the sum of areas of the two the smaller triangles should equal the area of the larger triangle.
- contravariant 4y ago
  Is there not a proof using similar triangles somehow? Because the whole 'sine' part seems like a bit of a red herring, they're basically just considering a couple of ratios between different lengths, they do not use any properties of the sine function as such (in particular it does not look like they're using sin(x)^2 + cos(x)^2 = 1, which would make the proof trivial)
  - chr1 4y ago
    Yes, if they droped height h to the side of isoscales triangle, then from similar triangles they would have h/A=c/C and h/b=2a/c and would get the result without mention of sine rule or areas.
jonahx 4y ago
It's a very clever proof. Agreed "using trigonometry" is potentially misleading. After reading the proof, the only 2 senses in which "trignometry" is being used are: 1. The term "sin a" is used to denote the ratio opposite / hypotenuse. But this can be considered a purely notational convenience. They could have called it "foo a" and nothing would change, or they could have inlined the referred-to ratio everywhere. 2. The law of sines is required. But the proof of this law [1] also boils down to nothing more than the ratio definition and some algebra. So afaict no circular logic is being used, but at the same time it doesn't seem to be doing anything previously thought to be impossible, unless there was a previous belief that the law of sines could not be used in a proof, which would be a strange belief to hold. I see it simply as a creative, unexpected proof. [1] https://www.youtube.com/watch?v=4xO8xqLyEbA https://www.youtube.com/watch?v=4xO8xqLyEbA
- deleted 4y ago
  [deleted]
- fnordpiglet 4y ago
  Is there more to trigonometry? I’m not a abstract math person, so forgive the ignorance, but my understanding was all trigonometric functions derive from ratios of angles and lengths of triangles so in the end each occurrence of a trigonometric function can be replaced by the corresponding ratios in some triangle. There are other ways to construct things, such as power series representations, etc, but even these must necessarily be replaceable by the ratio of angles and lengths of some triangle. What am I missing?
  - jonahx 4y ago
    > What am I missing? Nearly all the nontrivial results of trigonometry do in fact rest on the pythagorean theorem. The trig identities you learned in high school, as well as more advanced results like power series, etc. These results would be inadmissible. So the “uses trigonometry” part of this story feels like an attempt to manufacture mystery and hype. Which is a shame, because the geometric series construction is imo the interesting part, and can stand on its own merits.
deleted 4y ago
[deleted]
deleted 4y ago
[deleted]
xiphias2 4y ago
,,In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M.'' It would be interesting to see if the original proofs work with non-complete metric spaces or not, as probably this proof doesn't. https://en.wikipedia.org/wiki/Complete_metric_space https://en.wikipedia.org/wiki/Complete_metric_space https://sharegpt.com/c/qRsum0c https://sharegpt.com/c/qRsum0c
silentguy 4y ago
I don't see any Cosine being mentioned in the proof. I don't see any circular logic here.
soegaard 4y ago
Here is a proof of Pythagoras using a geometric series: https://www.cut-the-knot.org/pythagoras/Proof100.shtml https://www.cut-the-knot.org/pythagoras/Proof100.shtml The proof is by John Arioni and also features an infinite number of similar triangles.

scythe 4y ago

Induction isn't so bad. I'm not sure how they're getting away with the Law of Sines, though. The usual proof of LoS that I know is dependent on the existence of the circumcircle. But I don't know how to prove the existence of the circumcircle without dragging in a lot of geometry. Or you can use the area formula, which makes the proof similar to other arguments that use the area formula.

SamReidHughes 4y ago
There is a very straightforward proof that involves dropping an altitude. https://www.mathopenref.com/lawofsinesproof.html https://www.mathopenref.com/lawofsinesproof.html You could use a similar technique to make this proof not reference trigonometric functions.
- crdrost 4y ago
  For that matter the easiest proof of Pythag that I know of involves dropping an altitude: Look at the right triangle the normal way up, clearly the area of the triangle is k c² (k = ½ sin α sin β if you like, but it just matters that it's the same nonzero k for all similar triangles). Now roll it onto its hypotenuse, drop an altitude, and observe that both subtriangles are similar to the first one, kc² = ka² + kb². The diagram that's a bit involved is the angle sum diagram, you start with a right triangle (a,b,c) with some angle α, extend it to a new triangle (a,b', c') with angle α+β, then make the new triangle with angle β that you stacked on top of the original triangle into a right triangle with angle β (c', d, e) by extending the hypotenuse of the (a,b,c) triangle to a point P, basically until the angle with the hypotenuse c' is 90°. Drop a dotted line to the x-axis from P and you can work out that the dotted line is at x=cos α cos β, and its distance to a is sin α sin β. Similarly the y-coordinates give sin α cos β + cos α sin β. As you say, you can do all of this without angles except for defining the first triangle with angle α+β, which you might not even need... We just need it here for sin(2 α) which is something like reflecting the same triangle about its hypotenuse?
  - alex_smart 4y ago
    For this proof to be a proof, you first have to define what area is and why it should scale as kc² and why the sum of areas of the two the smaller triangles should equal the area of the larger triangle.
contravariant 4y ago
Is there not a proof using similar triangles somehow? Because the whole 'sine' part seems like a bit of a red herring, they're basically just considering a couple of ratios between different lengths, they do not use any properties of the sine function as such (in particular it does not look like they're using sin(x)^2 + cos(x)^2 = 1, which would make the proof trivial)
- chr1 4y ago
  Yes, if they droped height h to the side of isoscales triangle, then from similar triangles they would have h/A=c/C and h/b=2a/c and would get the result without mention of sine rule or areas.

jonahx 4y ago

It's a very clever proof. Agreed "using trigonometry" is potentially misleading. After reading the proof, the only 2 senses in which "trignometry" is being used are: 1. The term "sin a" is used to denote the ratio opposite / hypotenuse. But this can be considered a purely notational convenience. They could have called it "foo a" and nothing would change, or they could have inlined the referred-to ratio everywhere. 2. The law of sines is required. But the proof of this law [1] also boils down to nothing more than the ratio definition and some algebra. So afaict no circular logic is being used, but at the same time it doesn't seem to be doing anything previously thought to be impossible, unless there was a previous belief that the law of sines could not be used in a proof, which would be a strange belief to hold. I see it simply as a creative, unexpected proof. [1] https://www.youtube.com/watch?v=4xO8xqLyEbA https://www.youtube.com/watch?v=4xO8xqLyEbA

deleted 4y ago
[deleted]
fnordpiglet 4y ago
Is there more to trigonometry? I’m not a abstract math person, so forgive the ignorance, but my understanding was all trigonometric functions derive from ratios of angles and lengths of triangles so in the end each occurrence of a trigonometric function can be replaced by the corresponding ratios in some triangle. There are other ways to construct things, such as power series representations, etc, but even these must necessarily be replaceable by the ratio of angles and lengths of some triangle. What am I missing?
- jonahx 4y ago
  > What am I missing? Nearly all the nontrivial results of trigonometry do in fact rest on the pythagorean theorem. The trig identities you learned in high school, as well as more advanced results like power series, etc. These results would be inadmissible. So the “uses trigonometry” part of this story feels like an attempt to manufacture mystery and hype. Which is a shame, because the geometric series construction is imo the interesting part, and can stand on its own merits.

deleted 4y ago

[deleted]

deleted 4y ago

[deleted]

xiphias2 4y ago

,,In mathematical analysis, a metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M.'' It would be interesting to see if the original proofs work with non-complete metric spaces or not, as probably this proof doesn't. https://en.wikipedia.org/wiki/Complete_metric_space https://en.wikipedia.org/wiki/Complete_metric_space https://sharegpt.com/c/qRsum0c https://sharegpt.com/c/qRsum0c

silentguy 4y ago

I don't see any Cosine being mentioned in the proof. I don't see any circular logic here.

soegaard 4y ago

Here is a proof of Pythagoras using a geometric series: https://www.cut-the-knot.org/pythagoras/Proof100.shtml https://www.cut-the-knot.org/pythagoras/Proof100.shtml The proof is by John Arioni and also features an infinite number of similar triangles.