Tag Archives: geometry

Content-free mathematics

All of mathematics is content-free.

That’s a claim that would probably get me lynched in any self-respecting math department, and anyone who has taken a real math class would feel that it isn’t true. How could anything that causes so much pain and confusion be content-free?

In some sense, however, every statement in mathematics is trivial. After all, a proof is no more than a series of logical implications, and every such implication is logically trivial (i.e. correct). The difficulty in understanding a proof is simply a parsing problem. For a technically challenging proof, there may be many definitions and lemmas to parse, and the writing may not be very clear, but all of the logical steps are there. In mathematics, proving is hard, but once a proof has been written, checking the proof requires much less inspiration and genius. In fact, given a proof, it could be written in computer-checkable form, and we all (hopefully) believe that computers aren’t yet sentient.

This isn’t to say that reading and understanding proofs is entirely easy; if this were true, learning math would take no time at all and checking Perelman’s proof of the Poincare Conjecture would not have taken many months and years. Even though I can try to justify a theoretical claim that mathematics is trivial, it is nonetheless true that I struggle to understand things. My brain isn’t a computer; though it’s probably better as being creative, it’s certainly worse at parsing. An argument about the triviality of math because purely philosophical and is itself content-free.

However, there are some parts of math that I do believe to be content-free, not just because all proofs are logically correct, but because they have been generalized to the point of being eaten up by other broader fields. These domains of mathematics become special cases of general statements, and though they may maintain their clever tricks and elegant arguments, work in these fields can contribute little that is not trivially known.

Consider Euclidean geometry as an example. In the time of Euclid and the Greeks, geometry was considered the essence of mathematics; they even used it to prove facts about the primes. Now, however, Euclidean geometry as moved into the realm of the content-free. Every statement in pure Euclidean geometry can be restated in terms of basic building blocks like collinearity and concurrency, and these can be translated into systems of polynomial equations that can be solved by standard techniques (e.g. Gröbner basis). So the ability to solve problems in Euclidean geometry is a corollary of an understanding of systems of equations, and the work of Euclid can now be automated by a computer.

Of course, plenty of people still care about Euclidean geometry. Even if the mathematical statements themselves are no longer important for the development of mathematics as a whole, Euclidean geometry still provides a good place to learn how to think about math, and that is a skill that transcends mere problems or mere fields. In addition, Euclidean geometry is still a great source of problems for contests because of its abundance of elegant results. For me, an argument in pure Euclidean geometry still carries an aesthetic appeal that a system of equations will never be able to match.

Will every mathematical field end up as a subfield of a more general theory? As I see more math, I believe that the answer is yes: There is always a more general framework in which old ideas are simplified and trivialized. There’s no reason that anyone would care about such a general framework, however, unless it can serve to unify seemingly unrelated ideas. The pursuit of generality for generality’s sake feels to me as rather silly; adding needless abstraction to a problem does not make it any more important or more interesting. Instead, I believe in abstraction when mathematics is ready — and not before.

In the end, there will always be problems and puzzles for mathematicians to ponder, and that is what is important. Regardless of whether mathematics is actually content-free (and throughout this discussion, we never defined “content-free”, so this is an subjective opinion), mathematics will always be interesting and worth studying.

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A Converging Rabbit

I’ve decided to make a page of some fun problems. I’m not sure how successful this will be, but in theory, there will eventually be a nice long list of my favorite problems. Here’s the first one:

A rabbit climbs out at its hole, and walks 1 mile in a straight line. Then, the rabbit repeatedly turns \pi/3 radians and walks half of the distance it just walked,
as pictured below.

How far away from the rabbit’s hole is the point at which the rabbit converges?

This is an ARML-style problem: There are lots of ways of solving this, but some methods are cleaner and faster than others. Click to see some solutions.

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Angle Bisectors

Given the sides of a triangle, what is the length of the angle bisector? Consider the following image of triangle ABC, where the sides opposite points A, B, C have lengths a, b, c. We want to find length d.

First, we can apply the Angle Bisector Theorem to see that \frac{a}{x} = \frac{b}{y}, so that x = \frac{ay}{b} and y = \frac{bx}{a}. Plugging these into the equation c = x + y yields c = x + \frac{bx}{a} = \frac{a + b}{a} x and c = \frac{ay}{b} + y = \frac{a + b}{b} y, so that x = \frac{ca}{a+b} and y = \frac{cb}{a+b}.

We can now apply Stewart’s Theorem to see that a^2 y + b^2 x = c(d^2 + xy). Plugging in our expressions for x and y, we see that
ab = \frac{ab (a + b)}{a + b} = \frac{a^2 b}{a + b} + \frac{b^2 a}{a + b} = \frac{a^2 y + b^2 x}{c} = d^2 + xy = d^2 + \frac{c^2 ab}{(a + b)^2}.

Therefore, the length of the angle bisector at vertex C is
\displaystyle \sqrt{ab - \frac{c^2 ab}{(a + b)^2}} = \sqrt{ab \left( 1 - \frac{c^2}{(a + b)^2}\right)}.
The lengths of the other two angle bisectors can be found analogously and are given by a simple permutation of coordinates.

Here’s another way to compute the length of the angle bisector that might be a bit simpler: Let \angle C be 2\alpha, so that the angle bisector separates two angles with measure \alpha. The area of triangle ABC is the sum of the areas of the two smaller triangles, which can be expressed as \frac12 ad \sin \alpha + \frac12 bd \sin \alpha = \frac12 ab \sin 2\alpha = ab \sin \alpha \cos \alpha. Simplifying and rearranging, we see that (a + b) d = 2 ab \cos \alpha, so the length of the angle bisector is therefore
\displaystyle d = \frac{2ab}{a + b} \cos \alpha.
This is simpler than what we had before, though it also involves an angle.

We can use this expression for the length of the angle bisector to show that the sum of lengths of the angle bisectors is less than the perimeter. The length of the angle bisector to angle C is d = \frac{2ab}{a + b} \cos \alpha < \frac{2ab}{a + b}, and the other angle bisectors are similar. Therefore, it is sufficient to show that
\displaystyle \text{sum of angle bisectors} < \frac{2ab}{a + b} + \frac{2ac}{a + c} + \frac{2bc}{b + c} \le a + b + c = \text{perimeter}.

This inequality actually follows as a simple application of the arithmetic mean – harmonic mean (AM-HM) inequality, which states that
\displaystyle \frac{2ab}{a+b} = \frac{2}{\frac1a + \frac1b} \le \frac{a  +b}{2}.
Applying this three times yields
\displaystyle \frac{2ab}{a + b} + \frac{2ac}{a + c} + \frac{2bc}{b + c} \le \frac{a + b}{2} + \frac{a + c}{2} + \frac{b+c}{2} = a + b + c,
which is what we wanted to show. Therefore, we've proven that the sum of lengths of angle bisectors in a triangle is less than the perimeter.