-----Original Message----- From: John McCarthy [mailto:jmc@steam.Stanford.EDU] Sent: Friday, December 13, 2002 3:10 PM Subject: Re: [math-fun] Mathematical Presidents
I recall it as a proof by Garfield. It was in my high school geometry book. Spoilsports claim the proof is older.
Of course, it's entirely possible that Garfield independently rediscovered an existing proof. I'll use this as an excuse to share my favorite proof of the Pythagorean Theorem with the list. It's very simple, and deserves to be better known. Consider a right triangle with legs A and B, and hypotenuse C, with lengths a, b, and c. The claim that a^2 + b^2 = c^2 is equivalent to the statement that the area of a square of side A, plus the area of a square of side B, is equal to the area of a square of side C. But that's not the only geometrical statement equivalent to this equation. The area of any similar plane figures is proportional to the lengths of corresponding sides, so we can use any set of similar figures we choose to prove the Pythagorean theorem. For example, if we could show that the area of a semicircle with diameter A, plus the area of a semicircle with diameter B, was equal to the area of a semicircle with diameter C, that would suffice to show that a^2 + b^2 = c^2. The similar figures I will use in the proof are triangles similar to ABC itself, with the hypotenuse the length of the given side. Proof; draw the altitude of ABC from the vertex opposite C. Observe that the two smaller triangles are similar to the original triangle, and sum to it in area. The hypotenuses of these three triangles are A, B, and C. QED This proof is so simple I was able to give it without a diagram. When I first saw this proof (in the mathematics exhibit in a museum), it was given with diagrams, but no words! There were 5 pictures. The first was the classical picture of a right triangle, with a square erected on each side. The second was the same right triangle, with a semicircle erected on each side. The third was the same right triangle, with irregular but similar figures erected on each side. The fourth was the same right triangle, with a similar right triangle erected on each side, in the orientation that would allow these to be "folded" into the original triangle. The fifth illustration was the same as the third, with the single line described in the proof added. If you understand that the areas of similar figures are proportional to the lengths of corresponding sides, you can't look at these diagrams without being immediately convinced of the Pythagorean Theorem, even if you have never seen it before. Most short or geometrical proofs of the theorem that I've seen make it seem amazing and magical. This one makes it seem obvious. Andy Latto andy.latto@pobox.com