Subject: Re: [HM] "President Garfield's Proof"
From: Lambrou Michael (lambrou@itia.math.uch.gr)
Date: Mon Jan 17 2000 - 06:02:26 EST
The following was Clark's kind reply to an HM reader's question.
> An HM reader has asked which proof of the Pythagorean theorem is the
> one attributed to Garfield. The answer may be of general interest.
>
> "Garfield's proof" goes with a figure consisting of three right
> triangles. Let ABC be the given one, with hypotenuse BA and shortest
> side BC. Draw right triangle ABD with length(BD)=length(BA) and D on
> the side of AB that doesn't contain C. Then draw right triangle BDE
> with length(DE)=length(BC) and E on the side of BD that doesn't
> contain A. Triangle BDE is congruent to triangle ABC. Let a,b,c
> denote the common sidelengths. To finish the proof, I'll quote
> Malcolm Graham's article, "President Garfield and the Pythagorean
> Theorem," in The Mathematics Teacher, Dec. 1976 (in a series for the
> American Bicentennial called "Events in the History of American
> Mathematics):
>
> "In figure 1, we see a trapezoid with bases a and b and height (a+b).
> The trapezoid is the union of three right triangles. Hence, the area
> of the trapezoid is equal to the sum of the areas of the three
> triangles.
>
> (a+b)(a+b)/2 = ab/2 + ab/2 + (c^2)/2
> (a+b)(a+b) = ab ab + c^2
> a^2 + 2ab + b^2 = 2ab + c^2
> a^2 + b^2 = c^2 ."
.......
>
> Clark Kimberling
Well, to criticize the proof, I feel that it does not have enough new
ideas to qualify it as a genuinely new proof: Take the classic Indian
proof of a square of sides a+b with four right angled triangles arranged
around it, and the hypotenuses forming an inside square (with corners on
the outer square). The Garfield figure is half of that (just draw the
diagonal of the inside square).
Then instead of considering the area of the square in two different ways,
Garfield takes the trapezoid (=half the square). No wonder his proof
starts with (a+b)(a+b)/2, and then clears the 2 in the denominator, which
2 was put there in the previous step!
Michael Lambrou
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