6

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.

a06e
  • 6,665
  • 7
    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 05 '11 at 14:21
  • @user8268: and this qualifies as a proof by dimensional analysis? – a06e Jun 05 '11 at 16:21
  • 1
    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 05 '11 at 18:21
  • 10
    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 06 '11 at 07:33

2 Answers2

3

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.]

Community
  • 1
Matt E
  • 123,735
2

I recently came across a proof of Pythagoras Theorem via dimensional analysis in a book by Paul J Nahin called Mrs. Perkins's Electric Quilt, which goes as follows.

Let there be a right triangle with sides $a, b, c$, with $c$ the hypotenuse and let $\phi$ be one acute angle of this triangle. Since area has dimensions of length squared, and given $c, \phi$ we can uniquely determine the triangle, the area must be $$\Delta = c^2 f(\phi)$$ Now, drop the altitude on $c$. We get two right triangles with hypotenuses $a, b$ and one acute angle $\phi$. Hence, there areas are $$\Delta_1 = a^2 f(\phi) \qquad \Delta_2 = b^2 f(\phi)$$ But, $$\Delta = \Delta_1 + \Delta_2 \implies c^2 f(\phi) = a^2 f(\phi) + b^2 f(\phi)$$ $$\implies \boxed{c^2 = a^2 + b^2}$$

Aaratrick
  • 573