|The Universe of Discourse|
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Thu, 29 Jan 2009
A simple trigonometric identity
Here I put A at (0,0), B at (4,0), and 1 at (2,2). This is clear enough, but I wished that it were more nearly equilateral.
So that night as I was waiting to fall asleep, I thought about the problem of finding lattice points that are at the vertices of an equilateral triangle. This is a sort of two-dimensional variation on the problem of finding rational approximations to surds, which is a topic that has turned up here many times over the years.
Or rather, I wanted to find lattice points that are almost at the vertices of an equilateral triangle, because I was pretty sure that there were no equilateral lattice triangles. But at the time I could not remember a proof. I started doing some calculations based on the law of cosines, which was a mistake, because nobody but John Von Neumann can do calculations like that in their head as they wait to fall asleep, and I am not John Von Neumann, in case you hadn't noticed.
A simple proof that there are no equilateral lattice triangles has just now occurred to me, though, and I am really pleased with it, so we are about to have a digression.
The area A of an equilateral triangle is s√3/2, where s is the length of the side. And s has the form √t because of the Pythagorean theorem, so A = √(3t)/2, where t is a sum of two squares, because the endpoints of the side are lattice points.
By Pick's theorem, the area of any lattice triangle is a half-integer. So 3t is a perfect square, and thus there are an odd number of threes in t's prime factorization.
But t is a sum of two squares, and by the sum of two squares theorem, its prime factorization must have an even number of threes. We now have a contradiction, so there was no such triangle.
Wasn't that excellent? That is just the sort of thing that I could have thought up while waiting to fall asleep, so it proves even more conclusively that starting with the law of cosines was a mistake.
Okay, end of digression. Back to the law of cosines. We have a triangle with sides a, b, and c, and opposite angles A, B, and C, and you no doubt recall from high school that c2 = a2 + b2 - 2ab cos C. We'll call this "law C".
Before I fell alseep, it occurred to me that you could take the analogous law B, which is b2 = a2 + c2 - 2ac cos B, and substitute the right-hand side for the b2 term in law C. Then a bunch of stuff will cancel out and you should either get something interesting or something tautological. Von Neumann would have known right away which it was, but I needed paper.
So today I got out the paper and did the thing, and came up with the very simple relation that:
c = a cos B + b cos AWhich holds in any triangle. But somehow I had never seen this before, or, if I had, I had completely forgotten it.
The thing is so simple that I thought that it must be wrong, or I would have known it already. But no, it checked out for the easy cases (right triangles, equilateral triangles, trivial triangles) and the geometric proof is easy: Just drop a perpendicular from C. The foot of the perpendicular divides the base c into two segments, which, by the simplest possible trigonometry, have lengths a cos B and b cos A, respectively. QED.
Perhaps that was anticlimactic. Have I mentioned that I have a sign on the door of my office that says "Penn Institute of Lower Mathematics"? This is the kind of thing I'm talking about.
I will let you all know if I come up with anything about the almost-equilateral lattice triangles. Clearly, you can approximate the equilateral triangle as closely as you like by making the lattice coordinates sufficiently large, just as you can approximate √3 as closely as you like with rationals by making the numerator and denominator sufficiently large. Proof: Your computer draws equilateral-seeming triangles on the screen all the time.
I note also that it is important that the lattice is two-dimensional. In three or more dimensions the triangle (1,0,0,0...), (0,1,0,0...), (0,0,1,0...) is a perfectly equilateral lattice triangle with side √2.
[ Addendum 20090130: Vilhelm Sjöberg points out that the area of an equilateral triangle is s2√3/4, not s√3/2. Whoops. This spoils my lovely proof, because the theorem now follows immediately from Pick's: s2 is an integer by Pythagoras, so the area is irrational rather than a half-integer as Pick's theorem requires. ]