### Fun with right triangles

Last year my son enjoyed learning the pythagorean theorem. He was fascinated by a picture containing a proof and being able to calculate the length of one of the sides from the other two.

On vacation a couple weeks ago, we found a fun (well, for geeks anyway) driving game, which we called “Iron Chef Triangles”. The game is very simple: given a number, construct a pythagorean triple containing it – that is, find a right triangle whose sides are whole numbers one of which matches the given number. For example, if I say “5”, you might respond “3, 4, 5” (because 3 squared plus 4 squared equals 5 squared, so you can make a right triangle with sides of length 3, 4 and 5) or you might say “5, 12, 13” (because 5 squared plus 12 squared is 13 squared). If I say 8 you might say “6, 8, 10” or you might say “8, 15, 17”.

Try a few: how about 11? How about 20? See any patterns?

Once you’ve figured out how to easily do these in your head, here are two harder problems.

Hard: given right triangle whose sides aren’t necessarily integers, can you make a right triangle whose sides are integers with approximately the same angles? How close can you get?

Observation: the product of the three parts of a pythagorean triple is always a multiple of 60.

Unfairly hard problem: can you find two different pythagorean triples whose product is the same?

Have fun,

— Max

Of course, finding Pythagorean triples with legs equal to 1 or 2 is a bit of a problem. As a kid, I got fascinated with the problem of finding Pythagorean triples where the legs differ by 1. 3,4,5 is the first; 20,21,29 is the second; 119,120,169 the third; and so on. Decades later I was able to figure out a recurrence that allows you to start with 3,4,5 and generate all the following solutions in order. The nature of the recurrence allows you to compute a closed form formula in n (in the same style as for Fibonacci numbers) that computes the nth solution. All in all, good stuff. Interesting post!

Yes fun stuff. Along the way you probably discovered that these approximations of an isosceles triangle match right up with continued fractions expansion of the square root of two [at least the ones with an odd denominator anyway, if you think about it a little the even denominators are doomed to futility in this exercise].

I think it’s quite amusing to see that rational points are not only infinite but dense on the curve x^2 + y^2 = 1 but nonexistent with a higher exponent. First part of that comes semi-easily, the second part is the real bear 🙂

Actually, I was unaware of the continued fraction correspondence, but I must check that out. I suppose it’s somewhat unsurprising, given the Pell’s equation relevance. I was also unaware about the rationals being dense on the unit circle, but I have an idea for the proof of that. Given that Fermat’s last problem has been solved (right?), the “second part” thing is not so hard. 🙂

Anyway, thanks again for sharing these tidbits of fun! I’ll have to look into the parts of which I was unaware.