A Rational Triangle Idea

Last week Sol at "Wild About Math" posted a link to the Wolfram Demo site related to a novel relationship that is not well known to students/teachers. For every rational number,q, on the open interval from 0 to 1, there is a Pythagorean Triple.

Actually the demo may oversell the idea a little, or perhaps I don't completely understand it, but I don't think it enumerates all the Pythagorean triples, but it is still a nice idea to introduce to students because many of the ideas can be proved with nothing beyond a good grasp of Alg II (for non-US folks, that's about a 14-16 year old students grasp of Algebra).

I had written about the relation back at the end of 2009. If you haven't read that one, and are not familiar with the idea, you might want to start there.

Anyway, Sol's post got me thinking again about how I would approach this idea with students. The first big idea is that if you pick a point on the y-axis between y=0 and y=1, with a rational y-coefficient, then the line through that point and the point (-1,0) will cross the unit circle in the first quadrant in a point (p,r) whose x and y coordinates are both rational. Since the point is on the unit circle, that means that p²+r² = 1. But since both p and r are rational, they can be written with a common denominator, Wolfram uses (a/c, b/c) and so (a/c)²+(b/c)²=1. If we multiply both sides of the expression by c², we get a²+b²= c² and so {a,b,c} is a Pythagorean triple.

So what might we ask our young charges to do with this? Well, first, do we believe that ANY rational on the positive y axis REALLY will produce a rational intersection on the unit circle? The proof requires no more than the solution of a quadratic. If the students are not up to challenging an open proof, have them start by confirming that some given rational y-intercepts will produce a rational intersection on the unit circle, and then go on to find the Pythagorean triple associated with that rational. For example, Wolfram's demo seems to associate the rational 1/2 with a (3,4,5) triangle... can they show why? Can you find the triple associated with 3/7?

I mentioned above that I didn't think this could produce ALL the Pythagorean triples... and we might want to ask our students if they think it could. One of my concerns about the Wolfram presentation is that the point (0, 1/3) is associated with the triple (8,6,10). Why not (4,3,5)??? how did they arrive at the non-primitive triple they chose, and not, for instance, (12,9,15) or ??? pick your favorite. I knew that they had used (3,4,5) in the first example (0, 1/2)...

In fact, would (9,12,15) show up later if I plowed through enough rational fractions? I suspected it wouldn't. First, the symmetry of the situation seemed to play against it. In simplest form, the x coordinate for any version of a (3,4,5) triple must be either 3/5 or 4/5; or at least so it seemed to me.

I think I would challenge my students to work backwards from a given triple and see if they could find the related rational y-intercept. Since we know (see my earlier blog) that the x-coordinate was (1-t²)/(1+t²), we could take any primitive triple, say (5,12,13) and set either the x value = 5/13 or 12/13 and get the two (and only two) values that would generate appropriate intersections on the unit circle. As we do this, we discover that the rational always turns out to be
where H is the hypotenuse of the triple, and L is the leg we chose to use for the x-coordinate ratio. For (5,12,13) we get the two rational values 2/3 (this is number 3 on the Wolfram demo) and 1/5 (number 7, but labeled (24, 10, 26) ) ... Bright students should be challenged to prove that any Primitive Pythagorean triple, will also be such that the ratio (H-L)/(H+L) give will produce a square of a rational number.

So maybe the Wolfram demo is just good enough (and just bad enough) to motivate our students, after all, they love exposing our flaws.

Another Approach to Pythagorean Triples

Click on image to enlarge

I wrote a couple of posts a while back on the Barning Tree method of finding Pythagorean triples using matrices, and then a followup. I recently came across another approach to Pythagorean triples that involves a clever relationship between points on the positive y-axis, and points on the unit circle. (ok, maybe I should have known this, but I didn't.. and I think it is really a neat idea)

On the unit circle, x² + y ² = 1, if we draw a secant from the point (-1,0) through (0,t) on the y-axis, it turns out that if t is a rational number, then the coordinates of P=(x,y) where the secant intersects the circle, will also be rational. Since the slope of the line is also t, the equation is y=tx+t ... and so and t²= (1-x²)/(1+x)². That means t = y/(x+1) which leads to x= (1-t^2)/(1+t^2) and y= ^(2t)/_(1+t^2) If we pick some rational number to be t, say t=2/7,
then x= ⁴⁵/₅₃ and y= ²⁸/₅₃.... Then by similar triangles, there must be a circle with radius 53 and a point on the circle would be (28, 45) and in fact 28²+45²=53²... and any such rational point will produce another Image of unit circle

Even More on Pythagorean Triples and Barning Trees

A while back I wrote a couple of blogs on Primitive Pythagorean triples and Matrices and Barning Trees of Primitive Pythagorean triples. Shortly afterward I got a nice note from H. Lee Price with a link to a paper he wrote on "The Pythagorean Tree: A New Species". A couple of things I learned from it were too good to keep to myself, so here are some of the things I thought were amazing.. or read the paper for yourself to find your own favorites.
One of the clever things that I learned was a way of creating an informative 2x2 matrix for any triple that holds some interesting information. I will use the 5,12,13 triangle as an example, but any of them produce the same sorts of information. Take one leg, and write it as a ratio to the sum of the hypotenuse and other leg, then simplify. Using 5 over 12+13 gives ⁵/₂₅ or ¹/₅. Using the other leg we get ¹²/₁₈ = ²/₃. One of the first amazing things is that there will always be one, and only one even number in the two fractions created. Put the numerator and denominator of the fraction with the even number in the left column of a 2x2 matrix, and the other fraction makes the right column. For the 5, 12, 13 Pythagorean triple we get:
"SO WHAT?", you ask. Well, amazingly, you can use these to quickly find the radii of the in-circle, and all three ex-circles of the triangle. If you multiply across the two rows the two products formed will give you the radius of the in-circle and also the radius of the ex-circle on the hypotenuse of the triangle. Then if you multiply the two diagonals, you get the other two ex-circles radii. In the case of the 5,12, 13 triangle, the in-circle has a radius of 2, and the three ex-circles have radii of 3, 10, and 15 units respectively. I had never observed until it was pointed out to me, that the radius of the ex-circle on the hypotenuse is always the sum of the radii of the in-circle and the other two ex-circles...2 + 3 + 10 = 15....
Finally, these in-circle and ex-circle relations are tied back to the Barning tree. Remember the diagram that shows that each of the Pythagorean triples produces three offspring in the tree. Mr. Price points out that in each case, the new Triangle has an in-circle that is one of the three ex-circles of the parent triangle. The 3,4,5 triangle has ex-circle radii of 2, 3 and 6 units. The off spring triangles are the 5, 12, 13 triangle with in-center radius of 2, the 8, 15, 17 triangle with in-center radius of 3, and the 20,21,29 triangle with in-center radius of 6.
Somehow, all that geometry packed into a simple matrix seems incredible. By the way, the original two fractions we used to form the matrix, they are the tangents of 1/2 the two acute angles of the triangle. The focus of Mr. Price's paper was to point out that from these 2x2 matrices, you can easily build a different tree to produce all the Primitive Pythagorean triples as well. I will return to that when time permits, and I feel I understand it well enough to do it justice.

Pythagoras and Matrices

You can click on any image to expand it and make it sharper

It’s been a good weekend for Geometry for me. Several notes from folks telling me about geometry stuff I never knew… While I’m waiting on permission from the author of one, I wanted to tell you about the other.. The graph (tree) below shows a set of Primitive Pythagorean triples… All the ones with hypotenuse less than 100. Primitive Pythagorean triples are right triangles that have all sides as integers and none of them have a common factor.

What was new (to me) was that any one of them (and all the others not shown here) can be found as transformations of (4,3,5) using only three transformations. Let me make that clearer. If you think of a primitive Pythagorean triple as a point in three-space, then any other primitive Pythagorean triple is a point in three space that is just a transformation of this one… but there are only three transformations needed to get ALL of them. The tree shows which ones are generated by which ones, but you need to crank out the calcualtor and do some of these to see how neat it really is.

This type of graph is called a Barning-Tree because it seems to have first been discovered by F. J. M. Barning, “On Pythagorean and quasi-Pythagorean triangles and a generation process with the help of unimodular matrices, (Dutch) Math. Centrum Amsterdam Afd. Zuivere Wisk, ZW-011 (1963) 37 pp..

If you take the three matrices below, and write any of the primitive triples as a column vector, then multiplying by any of the matrices will give you another unique triple.. they never duplicate one, and they don’t leave any out (OK, I’m taking that on faith as I haven’t proven it for myself yet). I think that is kind of wild, and am totally impressed with people who can notice stuff like that.

Here are the three transformations