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I've recently become acquainted with Buckingham's Pi theorem for the first time . Then I've found an excercise that says:

Use dimensional analysis to prove the Pythagoras theorem. [Hint: Drop a perpendicular to the hypotenuse of a right-angle triangle and consider the resulting similar triangles.]

Any ideas? Thanks.

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6  
it's a rather well known proof: the perpendicular divides the triangle to 2 triangles, and they are all similar (having two equal angles); the sum of the areas of the small triangles is equal to the area of the original triangle; the area of the square over a side of the original triangle is a constant times the area of one of the 3 triangles (of the one whose hypotenuse is the chosen side) –  user8268 Jun 5 '11 at 14:21
    
@user8268: and this qualifies as a proof by dimensional analysis? –  becko Jun 5 '11 at 16:21
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well, hardly:) I guess the fact that the area of a right-angle triangle is a constant times hypotenuse${}^2$ (where the constant is a constant if we only consider similar triangles) can be seen as dimensional analysis. –  user8268 Jun 5 '11 at 18:21
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I like to call this the one-line proof of the Pythagoras Theorem, the one line being the perpendicular to the hypotenuse. –  Gerry Myerson Jun 6 '11 at 7:33

1 Answer 1

up vote 3 down vote accepted

Here is one formulation of this argument; it is the same as the one suggested by user8268 in the above comments, but perhaps this formulation will make it clearer why this is a proof by dimensional analysis:

  • You want to prove that the sum of the squares on each of the non-hypotenuse sides equals the square on the hypotenuse.

  • You generalize, and instead prove that for any shape, if you scale it by $a$, and then by $b$, the sum of the resulting areas is the area of the shape scaled by $c$. (We began with the case of the unit square.)

  • By thinking about how areas scale, it suffices to check for one particular shape.

  • We check it by taking the shape to be the original triangle (to be pedantic: scaled so that its hypotenuse has length one). This case is clear: just drop a perpendicular from the vertex opposite the hypotenuse to the hypotenuse, and see note that the triangle with hypotenuse length $c$ is the sum of two similar triangle of hypotenuse lengths $a$ and $b$.

The dimensional analysis is in the third step. The point is in the final equality that is proved, i.e. in the final proof of $a^2 + b^2 = c^2$, these quantities are not the area of any particular shape, but rather are the scaling factors for the areas of the original triangle after scaling its lengths by $a$, $b$, and $c$. This is why it is a proof by dimensional analysis.

[I originally posted this here, and you can see the comments there for some historical background on this particular argument.]

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