[This is a back-issue of this site’s newsletter]

Saw these online, thought I’d share them with you all –

- Parallel lines have so much in common. It’s a shame they’ll never meet.
- I took the shell off my racing snail, hoping it would make him faster. If anything, though, he just become more sluggish.
- How do you think the unthinkable? Why, with an itheberg, of course!
- I tried to catch some fog yesterday. Mist.

Badum-tsss! And it’s time for Monday Morning Math!

A few weeks ago, I promised some math movies. They’re ready! They’re ready! I’ll show them to you

*very soon*, I promise. (Or, check out my youtube channel for a preview)Last week, we were chatting s complex numbers and pythagorean triangles.

With a pythagorean triangle, if the two sides near the right angle are a and b, the other side is the square root of a

^{2}+b^{2}. On the other hand, the absolute value of a complex number a+bi is the same, it’s the square root of a^{2}+b^{2}. Last week I showed that this “coincidence” can be used to find a formula for pythagorean triples – a triangle with sides P=a^{2}-b^{2}, Q=2ab and R=a^{2}+b^{2}is a right-angled triangle.What about a triangles with other angles, say 60

^{o}? Can we pull off a similar trick?A bit of trigonometry tells us that if a triangle has sides a and b, and in between them is a 60

^{o}angle, the length of the third side is the square root of a^{2}-ab+b^{2}. It would be nice if there was some kind of number system, something like the complex numbers, where the absolute value of a+bi was a^{2}-ab+b^{2}instead of a^{2}+b^{2}.Well, as it happens, there is!

When you learn about complex numbers, you meet a mystery number i, whose square is -1. That is, it’s a root of the polynomial x

^{2}+1, which has no real roots. There’s no particular reason why that polynomial is special, though – we could have used any polynomial with no real roots.Suppose we used x

^{2}+x+1. It’s got no real roots, but let’s invent a root for it, and call it w. We’ll end up with a system of numbers very much like the complex numbers – the elements are a+bw instead of a+bi. Adding them works the way you’d expect, but the rules for multiplying them are slightly different now: simply because w^{2}is -w-1, not -1.If we square a+bw, we get a

^{2}+ 2abw + b^{2}w^{2}, which is a^{2}+ 2abw – b^{2}w – b^{2}, so that (a+bw)^{2}= (a^{2}-b^{2}) + (2ab-b^{2})w.Despite the different multiplication rule, we could use these numbers just like we use complex numbers – we could figure out what the “conjugate” of a+bw is (it’s not a-bw, it turns out). And we could work out the absolute value of a+bw. We could do everything with these numbers that we can do with complex numbers. The reason is simply that these new numbers a+bw are just the complex numbers in disguise.

Now, when we do work out absolute values, it turns out that |a+bw| is not a

^{2}+b^{2}, but a^{2}-ab+b^{2}. And that’s just what we need to solev the triangles puzzle!Because |(a+bw)

^{2}| = |a+bw|^{2}, and (a+bw)^{2}= (a^{2}-b^{2}) + (2ab-b^{2})w, this means |(a^{2}-b^{2}) + (2ab-b^{2})w| = (a^{2}-ab+b^{2}). Let’s simplify things by making P equal a^{2}-b^{2}, Q equal 2ab-b^{2}and R will be a^{2}-ab+b^{2}. Then, we’ve got |P+Qw|=R, that is, P^{2}-PQ-Q^{2}= R^{2}, which is exactly right for P, Q and R are to be the sides of a triangle with and angle of 60^{o}between P and Q.Some examples:

- Choose a=2 and b=1, and you get P = a
^{2}-b^{2}= 3, Q = 2ab-b^{2}= 3, and R = a^{2}-ab+b^{2}= 3. An equilateral triangle! - Choosing other values of a and b give other triangles with a sixty degree angle. If a=3 and b=2, you get P=5, Q=8 and R=7.
- Pick any values of a and b you like, and you can generate as many 60 degree triangles as you like!

There’s a similar trick for generating 120 degree triangles as well. It involves numbers a+bv for which v2=1-v, and |a+bv|=a

^{2}+ab+b^{2}.If you found this email fascinating, here’s a challenge for you. Can you figure out why there’s no formula for generating triangles with whole number edges, and a 45 degree angle?

That’s all for this week! Happy mathing!