I recently came across a note on an Annual Meeting of the Rocky Mountain Section of the MAA in 1923. Among the list of presentations was one by W. J. Hazzard, Professor at the Colorado School of Mines on the topic of "Parabolic Grouping of Pythagorean triangles."
I was a little familiar with Prof. Hazard as I had leapfrogged off one of his old posts in the Mathematics Teacher on methods of solving a quadratic equation to write a little about the history of solving quadratics in twenty or so different ways, which I hope someday to reduce to blog posts, but not today.
I even had a copy of one of the good Professor's books in my collection of old math books, but I had not read, nor was I aware of the idea he spoke of. With a few words of guidance from a "very" brief coverage in the article, I was able to extract at least a little that may be of interest to anyone who enjoys Pythagorean relations, and especially if you teach high school math.
If you put one acute vertex of a right triangle at the origin and lay it out so that the shorter leg lies along the positive x-axis, the other vertex will be at the point (a,b) as determined by the two legs of the triangle. In the graph I have shown the position of a 3-4-5 triangle and a 5-12-13 triangle to make my meaning clear.
A natural question is, "So What?" But if we look at several of the points determined by the upper vertex, and select out only some "related" Pythagorean triples, we notice a pattern. In the image at right the points represent the set of triples 3-4-5; 5-12-13; 7-24-25; and 9-40-41. (Any teacher or student who is not aware, there is a simple trick to find an infinite number of these triangles with a longer side one less than the hypotenuse. Just take any odd number to be the short leg, square it, and then divide by two and round up to the next whole for the hypotenuse. For example, 11 is a good odd number, and its square is 121. If we divide 121 by 2 we get 60.5, which is between 60 and 61, and 11, 60, 61 is a Pythagorean triple.)
All the points lie on a parabola y= 1/2 x2 - 1/2 . Since the focal length is 1/(4A), with A = 1/2, the focus must be a distance of 1/2 unit above the vertex, making the focal point at the origin. If we think about the definition of a parabola as the set of points equally distant from a focus and directrix, we realize the line of the directrix must be the line y = -1 so that, for instance the point (3,4) which is 5 units from the origin/focus will also be 5 units away from the directrix.
Admittedly that is a pretty small (although infinity large) sub-set of the Pythagorean triples. What would happen if we plotted other triangles like 8-15-17? It turns out they are not on the parabola drawn... they are on another one. In fact, all the triangles which have a longer leg two less than the hypotenuse will also have a focus at the origin, and the directrix will be ... yeah you knew it would be, y = - 2. That makes the focus at (0,-1). You can write the equation with ease for the parabola passing through any of these Pythagorean vertices, and all the ones with a common difference between the longer leg and hypotenuse share a parabola.
All the triples I've picked so far have been primitive triples. A good question to ask is what would happen if we picked, say, a 6-8-10 triangle. Will it fall on the same parabola as the 3-4-5, or on the ones with a difference of two?
The image below shows parabolas for differences of 1, 2, and 8 between the longer side and hypotenuse, and point D is the 6-8-10 triangle, right there with 8-15-17 and others like it.
I'm not sure you can swap this information for bread or ale at the local inn, but it's pretty interesting stuff.
3 comments:
Hi Pat! Just wanted to let you know the 'the history of solving quadratics in twenty or so different ways" link doesn't go anywhere. Some searching brought up interesting links, but not sure if those are what you were linking to. Thank you!
Thanks, Will try to correct.
Thanks, Will try to correct. https://pballew.blogspot.com/2024/01/solving-quadratic-equations-by-analytic.html
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