### Solving Puzzles with Socrates

In the dialog Meno, Plato famously illustrates Socrates' argument that (based on the fact that a diagonal cuts a square into two equal isosceles triangles) of the two (outlined) squares below

one is twice as large (ie, has twice as big the area as) the other one. This can also be demonstrated by folding corners of the big square towards its center. The small square (standing on a vertex and that might be perceived as a kite) is, in turn, twice as big as one of the four smaller squares obtained by drawing the (dashed) midlines. Thus the same diagram shows how to both double and halve a given square - with paper folding if needs be.

The diagram also serves to explain why the diagonal in a square of size 1 equals without a recourse to the Pythagorean theorem. This is of course a particular case of the theorem; amazingly, it can be used to establish the full blown case.

It could be argued that even more naturally in the diagram below

the triangles on the left are twice as big as the triangles on the right.

Of the two proofs of the Pythagorean theorem - I.47 and VI.31 - included by Euclid in his *Elements*, the former is more famous, the later is almost unknown in elementary mathematics, even though it's by far more natural and elegant. That proof is based on the observation that areas of similar plane figures relate as the squares of the lengths of any of their corresponding components (Euclid VI.19, VI.20).

Indeed, Euclid deals with the areas of polygons, but the principle applies to curvelinear figures as well, circle in particular. The formula for the area of a circle with radius , is one manifestation of the general principle.

This remark helps solved the following problem: in the diagram below, sum the areas of the circles in the two squares; which is larger: the sum of the two areas on the left or that of the four circles on the right? Or, may they per chance be equal?

The answer is obtained without invoking the area of the circle formula. Since two small triangles on the right add up to one triangle on the left, the same holds for their areas and for the areas of the inscribed circles, making the two sums in question equal.

[...] Bogomolny presents a solution to a geometric puzzle from Plato’s dialog Meno in Solving Puzzles with Socrates posted at CTK [...]

October 13th, 2012 at 5:36 am[...] Bogomolny presents a solution to a geometric puzzle from Plato’s dialog Meno in Solving Puzzles with Socrates posted at CTK [...]

July 12th, 2013 at 11:58 pm