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0|0|0|||||Proof of the Pythagorean Theorem %28and Law of Cosines%29|Don McConnell|1|15:57:56|08/25/2010|I thought you might be interested in %28yet%29 another proof of the Pythagorean Theorem%2C but I wasn%27t sure where to post it. %0D%0A%0D%0AActually%2C what I have is a proof of the Law of Cosines that I believe to be original %28and very cut-the-knot-ty%2C if I do say so myself%29%3A%0D%0A%0D%0Ahttp%3A%2F%2Fdaylateanddollarshort.com%2Fmisc%2Flawofcos.jpg%0D%0A%0D%0ADrop three perpendiculars and let the definition of cosine give the lengths of the sub-divided segments. Then%2C observe that like-colored rectangles have the same area %28computed in slightly different ways%29%2C and the result follows immediately.%0D%0A%0D%0AWhen C is a right angle%2C the blue rectangles vanish and we have the Pythagorean Theorem via what amounts to Proof %235 on Cut-the-Knot%27s Pythagorean Theorem page. %28Note that%2C as mentioned on CtK%2C the use of cosine here doesn%27t amount to an invalid %22trigonometric proof%22.%29 It seems that Proof %2365 may also be a relative.%0D%0A%0D%0AThe picture can be adjusted for obtuse C as well. %28Proof left as an exercise for the reader.%29%0D%0A%0D%0AA final note ... Because the same-colored rectangles have the same area%2C they%27re %22equidecomposable%22 %28aka %22scissors congruent%22%29%3A it%27s possible to cut one into a finite number of polygonal pieces that reassemble to make the other. While there%27s at least one standard procedure for determining how to make the cuts%2C the resulting pieces aren%27t necessarily pretty. Some popular dissection proofs of the Pythagorean Theorem --such as Proof %2336 on Cut-the-Knot-- demonstrate a specific%2C clear pattern for cutting up the figure%27s three squares%2C 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 %22obviously%22 have the same area.%0D%0A%0D%0A%28Another exercise for the reader%2C perhaps%3F %3A%29
1|1|0|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|alexb||07:39:52|08/26/2010|Thank you for that. I am especially moved by the introduction of the %5Bem%5Dcut-the-knot-ty%5B%2Fem%5D category.%0D%0A%0D%0AI have put you write-up at %0D%0A%0D%0Ahttp%3A%2F%2Fwww.cut-the-knot.org%2Fpythagoras%2FDonMcConnell.shtml%0D%0A%0D%0AI have a strong feeling that I have seen something similar. I%27ll check this up in a few days when back from a vacation.
2|1|0|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|C Reineke||08:27:38|08/26/2010|Don%2C%0D%0A%0D%0Aplease take a look at the nice paper by Dr. Uwe Sonnemann%0D%0A%28Alas%2C it%92s written in German%2C but you will immediately understand it%29%3A%0D%0A%0D%0Ahttp%3A%2F%2Fwww.mathe-ok.de%2Fpythagoras%2FPythagoras.pdf%0D%0A%0D%0AOn the pages 12 and 13 %28the diagram%21%29 you will find your proof - or do you %0D%0Asee any difference%3F%0D%0A%0D%0AKind regards%0D%0A%0D%0AChris%0D%0A%0D%0AP.S. There is a typo on page 12. For obtuse %28stumpfwinklige%29 triangles%0D%0A       we have c%5E2%3D a%5E2 %2B b%5E2 %93%2B%94 2ap%0D%0A
3|2|2|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|alexb||09:15:52|08/26/2010|A very nice paper indeed.
4|2|2|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Don McConnell|1|12:25:17|08/26/2010|That does indeed look like my proof %28though I can%27t read German%29. Leave it to the internet to squash my belief of originality. %3A%29%0D%0A%0D%0AI guess the best I can claim now is an %2Aindependently-discovered%2A proof%2C dating back to 1989 or 1990%2C when I was perusing a just-purchased copy of Loomis%27 %22The Pythagorean Proposition%22. I was struck by the author%27s claim that no trigonometric proof of the result was possible when I had just devised one ... of course%2C by then%2C it was a little late to correct the guy.%0D%0A%0D%0AAt least my equation full of boxes is still pretty cool. %3A%29%0D%0A%0D%0A%0D%0AThanks for the reference.%0D%0A%0D%0AI wonder if anyone has worked out a straightforward dissection to prove that the same-colored rectangles have the same area.%0D%0A%0D%0A--Don%0D%0A
5|3|4|4c76c1ca3dfa84cb.jpg||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Bui Quang Tuan||00:25:55|08/26/2010|%3EI wonder if anyone has worked out a straightforward %0D%0A%3Edissection to prove that the same-colored rectangles have %0D%0A%3Ethe same area. %0D%0A%3E%0D%0A%3E--Don%0D%0ADear Don McConnell%2C%0D%0AIt is very clear because two color filled triangles are congruent.%0D%0APlease see my attached image%21%0D%0ABest regards%2C%0D%0ABui Quang Tuan
6|4|5|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Don McConnell|1|14:38:26|08/27/2010|Thank you for your response.%0D%0A%0D%0AYour diagram adapts the key point from the classic %22windmill%22 proof of the Pythagorean Theorem%2C which is very nice because it provides a visual demonstration of the equality of the rectangle areas%2C no cosine formulas required.%0D%0A%0D%0AMy quest%2C though%2C is for something in the spirit of the dissection proofs of the Pythagorean Theorem%3A a way to show that the rectangle areas match by cutting them into a common set of pieces%2C using a single cutting pattern that works in all cases%2C or proof that such a single cutting pattern does not exist.%0D%0A%0D%0ARestricting attention to the Pythagorean case%3A It%27s interesting that in the various dissection proofs%2C the pieces that make up the leg-squares are %22inter-mingled%22 with each other inside the hypotenuse-square. Finding pieces that form two boxes within the hypotenuse-square seems trickier%2C since there%27s not as much room to maneuver. I wouldn%27t be surprised to learn that it can%27t be done. %28I even suspect that.%29
7|5|6|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Bui Quang Tuan||00:17:04|08/27/2010|Thank you for your explanation%21%0D%0ANow we try to find somethings by your directions.%0D%0ABest regards%2C%0D%0ABui Quang Tuan%0D%0A%3EMy quest%2C though%2C is for something in the spirit of the %0D%0A%3Edissection proofs of the Pythagorean Theorem%3A a way to show %0D%0A%3Ethat the rectangle areas match by cutting them into a common %0D%0A%3Eset of pieces%2C using a single cutting pattern that works in %0D%0A%3Eall cases%2C or proof that such a single cutting pattern does %0D%0A%3Enot exist. %0D%0A%3E%0D%0A
8|6|7|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Don McConnell|1|00:34:17|08/28/2010|Given that the problem is likely impossible as described%2C I%27ll revise the challenge%3A%0D%0A%0D%0A%5Bb%5DFind a collection of red%2C green%2C and blue polygons such that the greens and blues alone can be arranged to form the red %28a%5E2%29 square%2C the blues and reds alone can be arranged to form the green %28b%5E2%29 square%2C and the reds and greens alone can be arranged to form the blue %28c%5E2%29 square.%5B%2Fb%5D Different-colored polygons are allowed to %22inter-mingle%22 within a square%3B they don%27t have to be --and probably %2Acannot be%2A-- sub-divisions of the rectangles in my diagram.%0D%0A%0D%0AIdeally%2C the dissection pattern will be %22symmetric%22 in the sense that it shouldn%27t matter which side of the triangle is which. %28This is unlike in a Pythagorean dissection%2C where the leg-squares and hypotenuse-square are treated differently.%29%0D%0A%0D%0ASuch a dissection should degenerate into a Pythagorean dissection as one of the angles approaches 90 degrees%2C with one collection of same-colored polygons shrinking into nothingness%2C leaving single-colored polygons in each leg-square%2C and a mix of those polygons in the hypotenuse-square. This being the case%2C 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 %22symmetric%22 expectation%2C Pythagorean dissections such as CTK%27s %2328 wouldn%27t seem to serve as good starting points%2C since they treat the legs very differently. On the other hand%2C %2336 seems pretty tantalizing%2C though I haven%27t been able to do anything with it myself.
9|3|4|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|jmolokach||15:02:39|09/02/2010|%3EThat does indeed look like my proof %28though I can%27t read %0D%0A%3EGerman%29. Leave it to the internet to squash my belief of %0D%0A%3Eoriginality. %3A%29 %0D%0A%3E%0D%0A%3EI guess the best I can claim now is an %0D%0A%3E%2Aindependently-discovered%2A proof%2C dating back to 1989 or %0D%0A%3E1990%2C when I was perusing a just-purchased copy of Loomis%27 %0D%0A%3E%22The Pythagorean Proposition%22. I was struck by the author%27s %0D%0A%3Eclaim that no trigonometric proof of the result was possible %0D%0A%3Ewhen I had just devised one ... of course%2C by then%2C it was a %0D%0A%3Elittle late to correct the guy. %0D%0A%3E%0D%0A%3EAt least my equation full of boxes is still pretty cool. %3A%29 %0D%0A%3E%0D%0A%3E%0D%0A%3EThanks for the reference. %0D%0A%3E%0D%0A%3EI wonder if anyone has worked out a straightforward %0D%0A%3Edissection to prove that the same-colored rectangles have %0D%0A%3Ethe same area. %0D%0A%3E%0D%0A%3E--Don %0D%0A%0D%0AStill%2C I am rather impressed that you discovered this independently.  I came across proofs 6 and 19 on my own as well%2C and really should have searched before posting%2C but in your case I believe it nearly impossible to know that you duplicated someone else%27s work.  Hats off to you%21%0D%0A
10|4|9|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|C Reineke||07:05:05|09/03/2010|%3CHats off to you%21%3E%0D%0A %0D%0AYes%2C and hats off to you%2C John. My congratulations to you both%21%0D%0A%0D%0ABut I think I have to clarify some things.%0D%0A%0D%0AI%92ve quoted a p a p e r by Dr. Sonnemann and not a p r o o f  by Dr. Sonnemann.%0D%0A%0D%0AHis diagram is just the geometrical interpretation of formulas which you can %0D%0Afind in Euclid%92s %93Elements%94.  %0D%0A%0D%0AI quote %28p.12%2C below the formulas%29%3A%93%85in den Elementen von Euklid%85%93%0D%0A%0D%0AThe formulas%2C a generalization of PT%2C and the diagram %28I first saw it in a book %0D%0Aprinted in Leipzig 1930%21%29 are well-known%2C I think.%0D%0A%0D%0ADon%2C who obviously did not know that%2C rediscovered a very old theorem.%0D%0A%0D%0ABut it%92s a nice approach to the Law of Cosines and his boxes are %93pretty cool%94. %3B-%29%0D%0A%0D%0A%0D%0AAll the best %0D%0A%0D%0AChris%0D%0A
11|1|0|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Scott|1|18:42:48|09/04/2010|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.%0D%0A%0D%0APerhaps it is worth pointing out that the same %22shearing and rotation%22 argument%2C which is used to prove the Pythagorean Theorem %28either with triangles or rectangles%2Fparallelograms%29%2C as in the animation by Jim Morey featured in your proof number 1%2C applies equally in the context of the Law of Cosines -- you apply the shearing-and-rotation to all three pairs of %22adjacent%22 rectangles in the squares on the sides%2C and immediately justify McConnell%27s argument.  The key insight is that the shear-and-rotation process applies to the %22extra%22 rectangles in exactly the same way as it applies to the two rectangles which are used to build the square on the %22hypotenuse%22%0D%0A%0D%0ASee for example the following animated illustration%2C which allows the user to discover this for himself%2C albeit without exquisite accuracy in the drawings%3A%0D%0A%0D%0A%0D%0Ahttp%3A%2F%2Fwww.ies.co.jp%2Fmath%2Fjava%2Ftrig%2Fyogen1%2Fyogen1.html%0D%0A%0D%0AI realize that this is not quite the %22cut and paste%22 argument which McConnell requests%2C but it is too simple and elegant to pass up%21%0D%0A%0D%0AScott
12|2|11|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|mr_homm||22:43:00|09/07/2010|Not as simple as Scott%27s link%2C but perhaps more in the spirit of cut and paste%3A%0D%0A%0D%0Ahttp%3A%2F%2Fhome.comcast.net%2F%7Estuartmanderson%2Fimages%2FLawOfCos.png%0D%0A%0D%0AAL%2C BD%2C and CJ are perpendiculars from each corner through the opposite sides%2C as usual.  First rotate triangle BDA a quarter turn clockwise around B%2C into BEF%2C and likewise rotate BDC a quarter turn counterclockwise around B%2C into BGH.  Of the two small right triangles sharing BM as hypoteneuse%2C note that one is congruent to HKL and the other to FIJ.%0D%0A%0D%0ACut 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 %28by dissection%29 into two parallelograms.  Note that both share a common base BM%2C and their altitudes are BE and BG%2C which are both equal to BD by construction.  Therefore%2C these trapezoids have equal areas.%0D%0A%0D%0AWe are done at this point%2C but of course if you want an explicit procedure for dissecting one rectangle to form the other%2C note that there is a standard method for dissecting a parallelogram into an equivalent rectangle with the same altitude.  If this is done%2C then one has a sequence of steps which turn both rectangles into the same figure%2C hence obviously%2C we now have a sequence of dissections which will convert one rectangle into the other.%0D%0A%0D%0ACan anyone think of a simpler explicit dissection procedure%3F  If so%2C please share%21%0D%0A%0D%0AStuart Anderson
13|3|12|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|Don McConnell|1|06:16:50|09/08/2010|Hello%2C Stuart ...%0D%0A%0D%0AGood thoughts.%0D%0A%0D%0AThe thing about the %22standard%22 methods for dissection %28of parallelograms into rectangles and of rectangles into each other%29 is that they%27re lacking in context-specific insight. If I recall correctly from reading %22Equivalent and Equidecomposable Figures%22 %28Boltyanskii%29 years ago%2C one such method involves making numerous slices a fixed distance apart until space runs out%2C then going from there%3B 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-- %2Aworks%2A%2C of course%2C but the results of the mechanical slice-and-dice aren%27t anywhere near as appealing as the %22famous%22 dissections%2C which use patterns with a fixed number of pieces designed to work in the Pythagorean context.%0D%0A%0D%0AI%27m doubting that a %22fixed%22 pattern can decompose my same-colored rectangles into common pieces%3B almost-certainly%2C the mechanical method %28with arbitrary numbers of pieces%29 is necessary when the aspect ratios of the rectangles are very different. This is why I%27ve shifted my own focus as described in the revised %22challenge%22 described in a previous response in this thread.%0D%0A%0D%0A--Don%0D%0A%0D%0A%0D%0A
14|4|13|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|alexb||08:00:23|09/08/2010|As a challenge%3A%0D%0A%0D%0AIf we combine Pappus%27 generalization with Vecten%27s diagram%2C i.e.%2C apply Pappus%27 proof to the diagram with squares on the sides of an arbitrary triangle%2C then%2C at the side of our choice%2C we%27ll get both a square %28by construction%29 and a parallelogram %28from Pappus%27 proof.%29 They coincide for a right triangle%2C of course. Is there any way to show that the difference between the two is what it should be%3F
15|4|13|||||RE%3A Proof of the Pythagorean Theorem %28and Law of Cosines%29|mr_homm||10:33:24|09/08/2010|Hi Don%2C%0D%0A%0D%0AFurther thoughts%3A  Here is a dissection that looks nice at least some of the time.  First the dissection%2C then a discussion of its behavior%3A%0D%0A%0D%0Ahttp%3A%2F%2Fhome.comcast.net%2F%7Estuartmanderson%2Fimages%2FLawOfCos2.png%0D%0A%0D%0AStarting with the usual figure%2C reflect triangle ABC across side AB%2C so C goes to C%27%2C and likewise across AC so B goes to B%27.  Cut triangle ADC%27 and paste it so AD lies on FG%3B likewise cut triangle AEB%27 and paste it so AE lies on HI.  This converts rectangles ADGF and AEIH into %5Bi%5Dcongruent%5B%2Fi%5D parallelograms.%0D%0A%0D%0AAs long as no altitude is longer than the side it meets%2C this works just fine.  In other cases%2C it is necessary to %22take several strips%22 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.%0D%0A%0D%0ASo this still isn%27t perfect%2C but at least it looks pretty %5Bi%5Dsome%5B%2Fi%5D of the time.%0D%0A%0D%0A--Stuart Anderson