Proof Problem #1 Solution

pythag proof

IMG_20150717_073010 copy5

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2 thoughts on “Proof Problem #1 Solution

  1. There is another route via an area/dimension argument. Within a family of similar triangles, where a, b, and c denote the lengths of the corresponding sides in three different triangles, there will be a single parameter k (non-zero) such that the areas of the triangles are ka^2, kb^2, and kc^2. Of course, making this completely rigorous depends on a careful work defining area and dimension.

    Apply this to the three right triangles in the diagram using their hypotenuses AB, BC, AC as the corresponding sides with lengths a, b, c, respectively.

    Since the larger triangle’s area is the sum of the smaller two triangles’ areas, we have kc^2 = ka^2+kb^2.

    Barry Mazur gives his own take on this version in a nice numberphile video: Math Fable.

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