Sunday, November 19, 2006

Birthday present

Here's the first of a pair of birthday presents for BA, the Marine Corps, and all you other Scorpios, both courtesy of George Polya. The second comes in a following post.

First, a proof of the Pythagorean theorem that I'd never seen before:

(1) Instead of drawing squares on the sides of the triangle, draw regular hexagons.

(2) Even though the area of the hexagon isn't the square of the side, it's proportional to it. A = kS².
If this is obvious, fine. If not, try drawing radii, in your mind, for the hexagon, and consider what happens to the areas of the six triangles as you grow the hexagon in your mind: double the side, quadruple the areas.
From this, you can conclude that if the Pythagorean theorem, c² = a² + b², is true then so is another flavor of the same thing: the hexagon on the hypotenuse is equal to the sum of the hexagons on the other two sides.
kc² = ka² +kb²
(3) This same argument holds for any similar polygons, regular or irregular. (After all, you can decompose any polygon into triangles.) The Pythagorean theorem is just a special case.

(4) Run the argument in step (2) backwards. If you can prove an analogue of the Pythagorean theorem for some kind of polygon, the general theorem in (3) follows, and the Pythagorean theorem is also proved.

(5) Imagine a right triangle. Flip it around so it rests on its hypotenuse, and the vertex of the other two sides is sticking way up in the air. Drop a line down from that vertex, to meet the hypotenuse at a nice, right angle. (The base of the triangle is now the hypotenuse, and the line you drew is the altitude.)

The two new little triangles you just created are each similar to the original triangle (same angles) and to one another. They sum to the whole area.

Quod erat demonstrandum.

2 Comments:

Anonymous B A said...

Dammmmm - that's shiny! Now I guess I'll have to buy a copy of György's book - or two or three.

3:54 AM  
Blogger A said...

I wouldn't go that far :)

1:27 AM  

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