Jigsaw-style proofs of the Pythagorean theorem with non-square squares

Question

The two squares on the legs of a right triangle can be chopped up (or "dissected") into several pieces that can be reassembled jigsaw-style into a square congruent to that whose side is the hypotenuse.

If a plane region of some other shape than a square is used, with a side having the length of one of the sides of the triangle, the theorem remains true if the same shape is glued onto all three sides.

If a different shape is used, might this dissection proof become simpler or more comprehensible or more enlightening or otherwise better? If not, can that negative result be made precise and proved?

https://www.facebook.com/MathMunch/posts/1241067935920008?fref=nf — Michael Hardy, Dec 12 '15 at 19:25

score 4 · Accepted Answer · answered Dec 12 '15 at 01:04

4

The following is a minimalist answer. Let $\triangle ABC$ be right-angled at $C$. Drop a perpendicular from $C$ to $P$ on $AB$. This divides $\triangle ABC$ into two similar right triangles which, without any cutting at all, can be reassembled (without any motion) to make the triangle on the hypotenuse.

If we prefer the figures to be erected "outside" the original triangle, reflect the pieces across the sides.

answered Dec 12 '15 at 01:04

André Nicolas

507,029

Amazing! Is this your own argument? ${}\qquad{}$ – Michael Hardy Dec 12 '15 at 01:31
Well, it is a variant of the scaling proof of the Pythagorean Theorem. – André Nicolas Dec 12 '15 at 01:46
What is the scaling proof? ${}\qquad{}$ – Michael Hardy Dec 12 '15 at 02:09
Here is my version. Take our triangle $T$. Let $U$ be similar to $T$ but with unit hypotenuse. Suppose that (for some weird unit of area) $U$ has area $1$ unit. Then the two triangles I split $T$ into have, by scaling, area $a^2$ and $b^2$. But the whole of $T$ has area $c^2$ and by the picture $a^2+b^2=c^2$. – André Nicolas Dec 12 '15 at 02:15
That doesn't appear to be a different proof, but a slightly more complicated way of stating the one in the posted answer. ${}\qquad{}$ – Michael Hardy Dec 12 '15 at 02:20
Well, it is a variant. The one in the comment yields the "formula" explicitly. – André Nicolas Dec 12 '15 at 02:22
Well, maybe the "formula" should be considered a needless complication. ${}\qquad{}$ – Michael Hardy Dec 12 '15 at 02:24

Jigsaw-style proofs of the Pythagorean theorem with non-square squares

1 Answers1

Linked