The bad news is I'm having trouble sleeping tonight; the good news is that I came up with another proof of the Pythagorean theorem, to add to yesterday's.

Given a right triangle with sides a and b and hypotenuse c, construct the figure shown below.

By similarity, we will have (i) x/(y+b) = a/c, (ii) y/x = a/c, and (iii) x/a = c/b. From (iii) we have x = ac/b, so from (ii) we have y = ax/c = a^2/b, so from (i) we have (ac/b)/(a^2/b+b) = a/c, which simplifies to a^2 = b^2 = c^2. QED.

This approach, of course, turned out to be an oldie. Wikipedia presents a similar proof - which, however, takes the "outer" triangle as the object of study, rather than "tacking on" a triangle as I have done. The similar triangles approach is sometimes ascribed to Legendre, but is almost certainly older (e.g. Bhaskara).

Truth be told, I came up with this proof by a circuitous route. I realized that the usual cosine addition identity cos(m-n) = cos(m)cos(n) + sin(m)sin(n) can be derived independently of the Pythagorean theorem, and that if we put m=n we get cos(0) = cos^2(m) + sin^2(m). If we grant that cos(0)=1, we have the Pythagorean identity cos^2(m) + sin^2(m) = 1, from which the Pythagorean theorem obviously follows.

With this in mind, I found a nice proof of the cosine addition identity at MathWorld; see Equations (49)-(52). To make my proof of the Pythagorean theorem trig-free, I modified the MathWorld diagram by letting h approach x. This explains why my picture ended up looking different than the classic similar triangles approach.

Many people claim that trigonometric proofs of the Pythagorean theorem are necessarily circular (see e.g. Wikipedia). Based on the path I took to a valid proof, I think the situation is more subtle.

## Wednesday, February 25, 2009

## Tuesday, February 24, 2009

### A Physicist's Proof of the Pythagorean Theorem

One night last week, as I lay in bed thinking about the challenges of the economy, terrorism, and global climate change, I thought to myself,

This pleasant diversion helped me to fall asleep for a few nights in a row, at least until I came up with the following approach. I call it a physicist's proof not only because it's less than mathematically rigorous, but also because the idea is to begin with a triangle of zero size and derive its law of growth. This is a more "dynamical" way of understanding the theorem than the traditional dissection methods.

Without further ado, here is the argument:

We label the sides by

I believe this can be made rigorous using (sin q)' = cos q and (cos q)' = -sin q (relations which can be proved

Anyway, having found a solution, I poked around the web to compare it to the extant proofs. As I expected, my approach was not new. The Wikipedia entry on the Pythagorean theorem features this same basic argument, as does this page's amazing list. The idea is attributed to Michael Hardy of the University of Toledo, from 1988. Let this post serve to celebrate the idea's 20th Anniversary.

As long as I'm being unoriginal, let me also link to a cool math page from the website of Contra Costa Community College. Yesterday I was on this site by chance and happened to see two of my recent triangle puzzles!

*What this world really needs is another proof of the Pythagorean theorem*.This pleasant diversion helped me to fall asleep for a few nights in a row, at least until I came up with the following approach. I call it a physicist's proof not only because it's less than mathematically rigorous, but also because the idea is to begin with a triangle of zero size and derive its law of growth. This is a more "dynamical" way of understanding the theorem than the traditional dissection methods.

Without further ado, here is the argument:

We label the sides by

*x*and*y*, so the hypotenuse is some function*f*(*x*,*y*). We assume*f*(0,0) = 0. Now let*y*increase by an infinitesimal amount d*y*; then the small triangle is similar to the large right triangle, so that d*f*/d*y*=*y*/*f*. Separating, we find*f*^2 =*y*^2 +*C*(*x*). Repeating the argument with the horizontal leg, we find*f*^2 =*x*^2 + D(*y*). Thus*C*(*x*) -*x*^2 = D(*y*) -*y*^2 =*E*, a constant, whence*f*^2 =*y*^2 +*x*^2 +*E*. But*f*(0,0) = 0 implies*E*=0, so*f*^2 =*y*^2 +*x*^2. QED.I believe this can be made rigorous using (sin q)' = cos q and (cos q)' = -sin q (relations which can be proved

*without*recourse to the Pythagorean theorem!)Anyway, having found a solution, I poked around the web to compare it to the extant proofs. As I expected, my approach was not new. The Wikipedia entry on the Pythagorean theorem features this same basic argument, as does this page's amazing list. The idea is attributed to Michael Hardy of the University of Toledo, from 1988. Let this post serve to celebrate the idea's 20th Anniversary.

As long as I'm being unoriginal, let me also link to a cool math page from the website of Contra Costa Community College. Yesterday I was on this site by chance and happened to see two of my recent triangle puzzles!

## Saturday, February 7, 2009

### Math Moment at Au Bon Pain

Last week I was in Washington, D.C., for a series of meetings. During a break, I went across the street to a bakery/cafe called Au Bon Pain. There I sat at my table, watching an employee as she restocked paper cups and plates. At one point, I saw the woman, about 25 years old, waving her hands over a tray loaded with paper cups arranged in nested stacks. Thinking she must have been counting the cups, I said, "How many?" She said, "No, I didn't count them, I was just wondering would they fit on this shelf."

A minute later I said, "OK, I can't resist, I have to know how many there are." So I stood up and counted: "16..." (the number of cups in one stack) "...times...7" (the number of stacks) "...so...that's 112."

"Wow!" she said. "I never could have figured that out."

***

A very informative report has just been published by the National Governors' Association, the Council of Chief State School Officers, and Achieve, Inc. The full report, titled "Ensuring U.S. Students Receive a World-Class Education," is here. Of particular interest are pages 20-21, which address many of the myths about why other countries outperform the U.S. on international exams.

In an earlier post, I wrote about the academic underperformance of poor children in this country. Figure 15 from the Achieve report is worth looking at in this connection. Click to enlarge:

(n.b. From Sweden on up, the country's performance is measurably better than the U.S. as a whole; from Italy on down, the country's performance is measurably worse than the U.S. as a whole.)

A minute later I said, "OK, I can't resist, I have to know how many there are." So I stood up and counted: "16..." (the number of cups in one stack) "...times...7" (the number of stacks) "...so...that's 112."

"Wow!" she said. "I never could have figured that out."

***

A very informative report has just been published by the National Governors' Association, the Council of Chief State School Officers, and Achieve, Inc. The full report, titled "Ensuring U.S. Students Receive a World-Class Education," is here. Of particular interest are pages 20-21, which address many of the myths about why other countries outperform the U.S. on international exams.

In an earlier post, I wrote about the academic underperformance of poor children in this country. Figure 15 from the Achieve report is worth looking at in this connection. Click to enlarge:

(n.b. From Sweden on up, the country's performance is measurably better than the U.S. as a whole; from Italy on down, the country's performance is measurably worse than the U.S. as a whole.)

### A Trio of Triangle Puzzles

Last week I was daydreaming out the window and thought of a few puzzles. The first one in particular doesn't require any specialized math knowledge.

(Note, a triangle is said to be "isosceles" when at least two of its sides have equal lengths. Or to put it another way, a triangle is isosceles when not all of the side lengths are different.)

The next one is a little more mathy:

It makes a nice image if you carefully draw the answers to (1) and (2) on the same sheet of paper.

I'm using a new laptop that doesn't have any drawing software on it, so I'll have to post the answers later. Or, feel free to upload your answer to the Web someplace and link to it in the comments!

1. Given two distinct points A and B in the plane, find all points C so that triangle ABC is isosceles.

(Note, a triangle is said to be "isosceles" when at least two of its sides have equal lengths. Or to put it another way, a triangle is isosceles when not all of the side lengths are different.)

The next one is a little more mathy:

2. Given two distinct points A and B in the plane, find all points C so that triangle ABC is a right triangle.

It makes a nice image if you carefully draw the answers to (1) and (2) on the same sheet of paper.

3. What would be the answers to (1) and (2) if the two points A and B were given in three-dimensional space, instead of a plane?

I'm using a new laptop that doesn't have any drawing software on it, so I'll have to post the answers later. Or, feel free to upload your answer to the Web someplace and link to it in the comments!

Subscribe to:
Posts (Atom)