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 Subject: "Proof of the Pythagorean Theorem (and Law of Cosines)" Previous Topic | Next Topic
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Don McConnell
guest
Aug-25-10, 03:57 PM (EST)

"Proof of the Pythagorean Theorem (and Law of Cosines)"

 I thought you might be interested in (yet) another proof of the Pythagorean Theorem, but I wasn't sure where to post it. Actually, what I have is a proof of the Law of Cosines that I believe to be original (and very cut-the-knot-ty, if I do say so myself):Drop three perpendiculars and let the definition of cosine give the lengths of the sub-divided segments. Then, observe that like-colored rectangles have the same area (computed in slightly different ways), and the result follows immediately.When C is a right angle, the blue rectangles vanish and we have the Pythagorean Theorem via what amounts to Proof #5 on Cut-the-Knot's Pythagorean Theorem page. (Note that, as mentioned on CtK, the use of cosine here doesn't amount to an invalid "trigonometric proof".) It'seems that Proof #65 may also be a relative.The picture can be adjusted for obtuse C as well. (Proof left as an exercise for the reader.)A final note ... Because the same-colored rectangles have the same area, they're "equidecomposable" (aka "scissors congruent"): it's possible to cut one into a finite number of polygonal pieces that reassemble to make the other. While there's at least one standard procedure for determining how to make the cuts, the resulting pieces aren't necessarily pretty. Some popular dissection proofs of the Pythagorean Theorem --such as Proof #36 on Cut-the-Knot-- demonstrate a specific, clear pattern for cutting up the figure's three squares, a pattern that applies to all right triangles. I have yet to find a similarly straightforward cutting pattern that would apply to all triangles and show that my same-colored rectangles "obviously" have the same area.(Another exercise for the reader, perhaps? :)

Subject     Author     Message Date     ID
Proof of the Pythagorean Theorem (and Law of Cosines) Don McConnell Aug-25-10 TOP
RE: Proof of the Pythagorean Theorem (and Law of Cosines) alexb Aug-26-10 1
RE: Proof of the Pythagorean Theorem (and Law of Cosines) C Reineke Aug-26-10 2
RE: Proof of the Pythagorean Theorem (and Law of Cosines) alexb Aug-26-10 3
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Don McConnell Aug-26-10 4
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Bui Quang Tuan Aug-26-10 5
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Don McConnell Aug-27-10 6
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Bui Quang Tuan Aug-27-10 7
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Don McConnell Aug-28-10 8
RE: Proof of the Pythagorean Theorem (and Law of Cosines) jmolokach Sep-02-10 9
RE: Proof of the Pythagorean Theorem (and Law of Cosines) C Reineke Sep-03-10 10
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Scott Sep-04-10 11
RE: Proof of the Pythagorean Theorem (and Law of Cosines) mr_homm Sep-07-10 12
RE: Proof of the Pythagorean Theorem (and Law of Cosines) Don McConnell Sep-08-10 13
RE: Proof of the Pythagorean Theorem (and Law of Cosines) alexb Sep-08-10 14
RE: Proof of the Pythagorean Theorem (and Law of Cosines) mr_homm Sep-08-10 15

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alexb
Charter Member
2595 posts
Aug-26-10, 07:39 AM (EST)

1. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #0

 Thank you for that. I am especially moved by the introduction of the cut-the-knot-ty category.I have put you write-up at https://www.cut-the-knot.org/pythagoras/DonMcConnell.shtmlI have a strong feeling that I have seen something similar. I'll check this up in a few days when back from a vacation.

C Reineke
Member since Jul-9-10
Aug-26-10, 08:27 AM (EST)

2. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #0

 Don,please take a look at the nice paper by Dr. Uwe Sonnemann(Alas, it's written in German, but you will immediately understand it):https://www.mathe-ok.de/pythagoras/Pythagoras.pdfOn the pages 12 and 13 (the diagram!) you will find your proof - or do you see any difference?Kind regardsChrisP.S. There is a typo on page 12. For obtuse (stumpfwinklige) triangles we have c^2= a^2 + b^2 “+” 2ap

alexb
Charter Member
2595 posts
Aug-26-10, 09:15 AM (EST)

3. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #2

 A very nice paper indeed.

Don McConnell
guest
Aug-26-10, 12:25 PM (EST)

4. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #2

 That does indeed look like my proof (though I can't read German). Leave it to the internet to squash my belief of originality. :)I guess the best I can claim now is an *independently-discovered* proof, dating back to 1989 or 1990, when I was perusing a just-purchased copy of Loomis' "The Pythagorean Proposition". I was struck by the author's claim that no trigonometric proof of the result was possible when I had just devised one ... of course, by then, it was a little late to correct the guy.At least my equation full of boxes is still pretty cool. :)Thanks for the reference.I wonder if anyone has worked out a straightforward dissection to prove that the same-colored rectangles have the same area.--Don

Bui Quang Tuan
Member since Jun-23-07
Aug-26-10, 00:25 AM (EST)

5. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #4

 >I wonder if anyone has worked out a straightforward >dissection to prove that the same-colored rectangles have >the same area. >>--DonDear Don McConnell,It is very clear because two color filled triangles are congruent.Please see my attached image!Best regards,Bui Quang Tuan
Don McConnell
guest
Aug-27-10, 02:38 PM (EST)

6. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #5

 Thank you for your response.Your diagram adapts the key point from the classic "windmill" proof of the Pythagorean Theorem, which is very nice because it provides a visual demonstration of the equality of the rectangle areas, no cosine formulas required.My quest, though, is for something in the spirit of the dissection proofs of the Pythagorean Theorem: a way to show that the rectangle areas match by cutting them into a common set of pieces, using a single cutting pattern that works in all cases, or proof that such a single cutting pattern does not exist.Restricting attention to the Pythagorean case: It's interesting that in the various dissection proofs, the pieces that make up the leg-squares are "inter-mingled" with each other inside the hypotenuse-square. Finding pieces that form two boxes within the hypotenuse-square seems trickier, since there's not as much room to maneuver. I wouldn't be surprised to learn that it can't be done. (I even suspect that.)

Bui Quang Tuan
Member since Jun-23-07
Aug-27-10, 00:17 AM (EST)

7. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #6

 Thank you for your explanation!Now we try to find somethings by your directions.Best regards,Bui Quang Tuan>My quest, though, is for something in the spirit of the >dissection proofs of the Pythagorean Theorem: a way to show >that the rectangle areas match by cutting them into a common >set of pieces, using a single cutting pattern that works in >all cases, or proof that such a single cutting pattern does >not exist. >

Don McConnell
guest
Aug-28-10, 00:34 AM (EST)

8. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #7

 Given that the problem is likely impossible as described, I'll revise the challenge:Find a collection of red, green, and blue polygons such that the greens and blues alone can be arranged to form the red (a^2) square, the blues and reds alone can be arranged to form the green (b^2) square, and the reds and greens alone can be arranged to form the blue (c^2) square. Different-colored polygons are allowed to "inter-mingle" within a square; they don't have to be --and probably *cannot be*-- sub-divisions of the rectangles in my diagram.Ideally, the dissection pattern will be "symmetric" in the sense that it'shouldn't matter which side of the triangle is which. (This is unlike in a Pythagorean dissection, where the leg-squares and hypotenuse-square are treated differently.)Such a dissection should degenerate into a Pythagorean dissection as one of the angles approaches 90 degrees, with one collection of same-colored polygons shrinking into nothingness, leaving single-colored polygons in each leg-square, and a mix of those polygons in the hypotenuse-square. This being the case, one way to proceed is to look for Pythagorean dissections where the cutting patterns for the legs can be viewed as degenerated versions of the cutting pattern in the hypotenuse. With the "symmetric" expectation, Pythagorean dissections such as CTK's #28 wouldn't seem to serve as good starting points, since they treat the legs very differently. On the other hand, #36 seems pretty tantalizing, though I haven't been able to do anything with it myself.

jmolokach
Member since Aug-17-10
Sep-02-10, 03:02 PM (EST)

9. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #4

 >That does indeed look like my proof (though I can't read >German). Leave it to the internet to squash my belief of >originality. :) >>I guess the best I can claim now is an >*independently-discovered* proof, dating back to 1989 or >1990, when I was perusing a just-purchased copy of Loomis' >"The Pythagorean Proposition". I was struck by the author's >claim that no trigonometric proof of the result was possible >when I had just devised one ... of course, by then, it was a >little late to correct the guy. >>At least my equation full of boxes is still pretty cool. :) >>>Thanks for the reference. >>I wonder if anyone has worked out a straightforward >dissection to prove that the same-colored rectangles have >the same area. >>--Don Still, I am rather impressed that you discovered this independently. I came across proofs 6 and 19 on my own as well, and really should have searched before posting, but in your case I believe it nearly impossible to know that you duplicated someone else's work. Hats off to you!Success is getting up one more time than you fall down...

C Reineke
Member since Jul-9-10
Sep-03-10, 07:05 AM (EST)

10. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #9

 Yes, and hats off to you, John. My congratulations to you both!But I think I have to clarify some things.I’ve quoted a p a p e r by Dr. Sonnemann and not a p r o o f by Dr. Sonnemann.His diagram is just the geometrical interpretation of formulas which you can find in Euclid’s “Elements”. I quote (p.12, below the formulas):“…in den Elementen von Euklid…“The formulas, a generalization of PT, and the diagram (I first saw it in a book printed in Leipzig 1930!) are well-known, I think.Don, who obviously did not know that, rediscovered a very old theorem.But it's a nice approach to the Law of Cosines and his boxes are “pretty cool”. ;-)All the best Chris

Scott
guest
Sep-04-10, 06:42 PM (EST)

11. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #0

 I enjoyed the recent page by Don McConnell on the Law of Cosines. He asks for a simple demonstration of the equality of the areas of the same-colored rectangles.Perhaps it is worth pointing out that the same "shearing and rotation" argument, which is used to prove the Pythagorean Theorem (either with triangles or rectangles/parallelograms), as in the animation by Jim Morey featured in your proof number 1, applies equally in the context of the Law of Cosines -- you apply the shearing-and-rotation to all three pairs of "adjacent" rectangles in the squares on the sides, and immediately justify McConnell's argument. The key insight is that the shear-and-rotation process applies to the "extra" rectangles in exactly the same way as it applies to the two rectangles which are used to build the square on the "hypotenuse"See for example the following animated illustration, which allows the user to discover this for himself, albeit without exquisite accuracy in the drawings:I realize that this is not quite the "cut and paste" argument which McConnell requests, but it is too simple and elegant to pass up!Scott

mr_homm
Member since May-22-05
Sep-07-10, 10:43 PM (EST)

12. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #11

 Not as simple as Scott's link, but perhaps more in the spirit of cut and paste:AL, BD, and CJ are perpendiculars from each corner through the opposite sides, as usual. First rotate triangle BDA a quarter turn clockwise around B, into BEF, and likewise rotate BDC a quarter turn counterclockwise around B, into BGH. Of the two small right triangles sharing BM as hypoteneuse, note that one is congruent to HKL and the other to FIJ.Cut FIJ and HKL off the ends of rectangles BFJ and BHL respectively and paste them onto the congruent triangles below B. This converts the two rectangles (by dissection) into two parallelograms. Note that both share a common base BM, and their altitudes are BE and BG, which are both equal to BD by construction. Therefore, these trapezoids have equal areas.We are done at this point, but of course if you want an explicit procedure for dissecting one rectangle to form the other, note that there is a standard method for dissecting a parallelogram into an equivalent rectangle with the same altitude. If this is done, then one has a sequence of steps which turn both rectangles into the same figure, hence obviously, we now have a sequence of dissections which will convert one rectangle into the other.Can anyone think of a simpler explicit dissection procedure? If so, please share!Stuart Anderson

Don McConnell
guest
Sep-08-10, 06:16 AM (EST)

13. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #12

 Hello, Stuart ...Good thoughts.The thing about the "standard" methods for dissection (of parallelograms into rectangles and of rectangles into each other) is that they're lacking in context-specific insight. If I recall correctly from reading "Equivalent and Equidecomposable Figures" (Boltyanskii) years ago, one such method involves making numerous slices a fixed distance apart until space runs out, then going from there; the number of pieces involved depends upon the relative lengths of the rectangle sides. Applying that process to the Pythagorean case --decomposing the longer-leg-square and hypotenuse-square to have the same height as the shorter-leg-square-- *works*, of course, but the results of the mechanical slice-and-dice aren't anywhere near as appealing as the "famous" dissections, which use patterns with a fixed number of pieces designed to work in the Pythagorean context.I'm doubting that a "fixed" pattern can decompose my same-colored rectangles into common pieces; almost-certainly, the mechanical method (with arbitrary numbers of pieces) is necessary when the aspect ratios of the rectangles are very different. This is why I've shifted my own focus as described in the revised "challenge" described in a previous response in this thread.--Don

alexb
Charter Member
2595 posts
Sep-08-10, 08:00 AM (EST)

14. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #13

 As a challenge:If we combine Pappus' generalization with Vecten's diagram, i.e., apply Pappus' proof to the diagram with squares on the sides of an arbitrary triangle, then, at the side of our choice, we'll get both a square (by construction) and a parallelogram (from Pappus' proof.) They coincide for a right triangle, of course. Is there any way to show that the difference between the two is what it'should be?

mr_homm
Member since May-22-05
Sep-08-10, 10:33 AM (EST)

15. "RE: Proof of the Pythagorean Theorem (and Law of Cosines)"
In response to message #13

 Hi Don,Further thoughts: Here is a dissection that looks nice at least some of the time. First the dissection, then a discussion of its behavior:Starting with the usual figure, reflect triangle ABC across side AB, so C goes to C', and likewise across AC so B goes to B'. Cut triangle ADC' and paste it'so AD lies on FG; likewise cut triangle AEB' and paste it so AE lies on HI. This converts rectangles ADGF and AEIH into congruent parallelograms.As long as no altitude is longer than the side it meets, this works just fine. In other cases, it is necessary to "take several strips" to convert the rectangles into parallelograms. The behavior is worst when the triangle ABC is very long and thin. and one is building parallelograms from the rectangles on the shortest side.So this still isn't perfect, but at least it looks pretty some of the time.--Stuart Anderson