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

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.)

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.

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!

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

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

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

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?

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