Consider the following diagram made of 16 right triangles of sides and four squares of side . This being a tiling of the outer square (of side length ), we have
hence implying
thus proving the Pythagoras' Theorem.
![]() |
Outer square of side |
This is similar to previously known proofs (eg. see proof #3 and #116 here), but it is hard to say whether this exact tiling appears anywhere before.
Now, consider the following diagram made up of right triangles of sides , (pink) sqaures of side , (yellow) squares of side , and (green) squares of side . The reader is welcome to check that this also gives a completely valid proof of the Pythagoras' Theorem.
![]() |
Outer square of side |
Some of the readers might have started asking the right question. And the answer is yes, it can be generalised! With the outer sqaure of side , we get of the red and yellow squares, of the green squares, and of the triangles. This gives
hence proving the Pythagoras' Theorem. For your viewing pleasure, here's the big square. Feel free to print it out and put it on your wall.
![]() |
Outer square of side |
Of course, the keen ones among you would ask whether this is truly a tiling for any values of
![]() |
Outer square of side |
The keener ones would of course ask for a proof that this is indeed a tiling. For that, one only needs to note that the diagram is made of four spiky triangles (made of a layer of blue triangles followed by a layer of yellow and a layer of pink squares followed by a number of layers of blue triangles) - this is almost a triangle except two spikes on two sides. All one needs to convince oneself is that these spiky triangles fit with each other properly and leave a space in the middle which can be nicely filled by the green squares.
No comments:
Post a Comment