[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 a2+b2. On the other hand, the absolute value of a complex number a+bi is the same, it’s the square root of a2+b2. Last week I showed that this “coincidence” can be used to find a formula for pythagorean triples – a triangle with sides P=a2-b2, Q=2ab and R=a2+b2 is a right-angled triangle.
What about a triangles with other angles, say 60o? 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 60o angle, the length of the third side is the square root of a2-ab+b2. 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 a2-ab+b2 instead of a2+b2.
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 x2+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 x2+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 w2 is -w-1, not -1.
If we square a+bw, we get a2 + 2abw + b2w2, which is a2 + 2abw – b2w – b2, so that (a+bw)2 = (a2-b2) + (2ab-b2)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 a2+b2, but a2-ab+b2. And that’s just what we need to solev the triangles puzzle!
Because |(a+bw)2| = |a+bw|2, and (a+bw)2 = (a2-b2) + (2ab-b2)w, this means |(a2-b2) + (2ab-b2)w| = (a2-ab+b2). Let’s simplify things by making P equal a2-b2, Q equal 2ab-b2 and R will be a2-ab+b2. Then, we’ve got |P+Qw|=R, that is, P2-PQ-Q2 = R2, which is exactly right for P, Q and R are to be the sides of a triangle with and angle of 60o between P and Q.
- Choose a=2 and b=1, and you get P = a2-b2 = 3, Q = 2ab-b2 = 3, and R = a2-ab+b2 = 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|=a2+ab+b2.
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!