Perimeter = Area

Over at JD2718, jd2718 asked readers to consider a square such that the area of the square equalled the perimeter. Of course, the solution is relatively uninteresting – consider a square of side length 4. It will have a perimeter of 4 + 4 + 4 + 4 = 16, and a side length of 4*4 = 16.

Given that uninteresting result, he asks whether or not there is an interesting problem that can be gleaned from this.

Well, an interesting problem, I don’t know – but an interesting result? Absolutely.

Consider the figure below. I realize, having six sides, it is in fact a hexagon, but we’re going to assume it has n many sides.

(Credit where credit is due, this hexagon was lifted from Wikipedia)

Being a regular polygon, it has n many sides of equal length – we’ll call this length s. The important thing here is that regular polygons can be divided radially into n many triangles of equal size and shape.

The other thing of note is the green line, extending from the center to meet perpendicularly with the side at the midpoint. This distance is known as the apothem of the figure. We’ll call this length a.

As such, the perimeter of the figure is easily expressed as the length of the side times the number of sides,

Area is also easily expressed. The area of a triangle is simple to compute, as one half the base width times the height. The base of each triangle in the figure is a side length, s, and the height of each triangle is the apothem, a. The area of the polygon is merely the area of one of the triangles times the number of triangles,

Settering perimeter equal to area,

Thus we get that the apothem is of length 2. Notice that n, the number of sides of the polygon, cancelled out completely – the apothem of any polygon such that perimeter equals area will be of length 2.

This strikes me as a disturbingly general result ^^ Very neat.

A Graph Of Note

This was moved up from the previous post, because of some changes I made, and I think it warrants individual consideration.

Consider the graph above, 11 nodes with connections between some of the nodes. We would like to color the nodes of the graph such that no two nodes that are connected have the same color. That’s a very general question, but it has a very interesting result as it applies to the graph above.

Consider just five nodes connected in a ring. Including all rotations, reflections, and color transpositions, there are 240 different ways of coloring those nodes. Now, when you start adding nodes internally, you would assume that the colors of the internal nodes would influence the coloring on the boundary. If there were one node connected radially from the center to all five nodes on the outside, the boundary coloring can not include whatever color the inside node is colored. Boundary colorings in the 240 that include the color of the inside node are therefore eliminated.

What’s fascinating to me about the above graph is that there are enough ways to four color the whole graph such that any one of the 240 possible boundary colorings will occur. Any one of them. The internal graph might as well not exist, given how it influences the coloring on the boundary. Very, very interesting.

Of course, I checked that by hand, and I might’ve made a mistake so feel free to check me : )

Four Colors and a Theorem

(Ed: Some of the original post has been removed. )

There is a small Theorem of Note that says, in short, that four colors are sufficient to color any map such that no two regions that share a border have the same color.

There are of course provisions, for example the regions must be contiguous. The United States would generally be disqualified because of Michigan, but it turns out not to matter in that particular case.

This idea fascinated and puzzled me for a long time after I first heard it. It seems so shockingly general. Given the infinitude of possible border shapes and geometries and arrangments, how could you even begin to approach the problem in a sensible way? Of course, you know how Mathematicians are – always up to something.

The below is a rather long post, encompassing a bit of history of the problem, -a lot- of background and introduction to the problem, a watered down description of old proofs, some analysis of the problem, and current areas of interest with regard to the problem.

I Like to … Kick Them When They’re Down

So, above I have charted the IMDB rating of each of M. Night Shyamalan’s movies, in chronological order. He was the rising genius behind The Sixth Sense, all those years ago. Since then, as you can see, there has been a slide. Notice though that even with the slide, he’s still better now than when he started!

Part of the mystery though, is just what to attribute this slide to. His abilities as a director? A writer? A producer? Actor? Difficult to say, as he has performed in each of those roles in his latest movies. Perhaps he’s merely stretched too thin…

And maybe it’s just too easy to mock easy targets.

Only as Good As …

So, as I’m waiting to get the cable sorted out, I’m fiddling with Mathematica. I ask it to, for example, factor 2⁵⁰⁰+1. Big number. 151 digits, which I won’t reproduce here. But, factored,

17 * 401 * 4001 * 61681 * 340801 * 1074001 * 2020001 * 2787601 * 22624001 * 3173389601 * 1481124532001 * 8877945148742945001146041439025147034098690503591013177336356694416517527310181938001

Simple enough. What’s interesting is the time it took to do it. Mathematica has a function, Timing[expr] which returns the time it takes to calculate the value of the expression expr. To factor 2⁵⁰⁰+1, according to the timing function, took 0.313 seconds.

All right, maybe not so interesting. But then let’s look at 2⁵⁰⁰-1.

Down for the Count

A storm on Thursday took out my cable, hence the silence. Everything is fine now, but I really couldn’t resist posting a pic like that : )

Representations: Something to Think About

Discussion prompt. Hopefully I’ll get to the real post tomorrow.

(And no bonus points for nit-picking my geography.)

The Wheel Of Theodorus and Complex Numbers

The Wheel Of Theodorus, or Square Root Spiral, is a rather simple geometric construction that allows you to construct line segments with length of the square root of any integer. Observe the figure below.

You’ll notice I’ve stolen this image from the Figures Speak weblog. I’m currently without my aged copy of the Geometer’s Sketchpad, so crudely wrought and stolen images will have to suffice.

The construction is fairly simple. Starting with a 1-1-root 2 right triangle, use the hypotenuse as one leg, and add a perpendicular segment of length 1 to create another right triangle. Pythathagoras says that the resulting triangle has a hypotenuse of root 3. And so it goes. And it creates this very nice spiral effect.

The construction is very simple, but I would like a way to express this spiral figure itself mathematically. For example, a smooth parameterization would be very nice. Unfortunately I’m not sure one exists. A parameterization exists, but it sort of cheats, and will not be discussed here because it distracts from my main point.

And that main point is that complex numbers provide a very convenient and very elegant way to express the geometric relationships in the figure.

How Not To Play Dice

The question is as follows: Can two loaded dice be constructed such that, rolling the pair gives the numbers 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 with equal probability?

Now, a couple of warnings going in. First and foremost, I don’t really know how you go about loading dice. I’m a mathematician, not a dicetition, so this takes a very abstract approach to loading dice. For each die, we’re basically choosing the probability with which each number occurs in a roll.

Second, as little as I know about loading dice, I’m even less sure about my abilities with probability. Honestly, I avoid doing probability as often as I possibly can. Well, I don’t go out of my way to find probability to avoid, but it does make me very uncomfortable. So if I make any mistakes, feel free to comment. The only reason I’m doing this problem at all is that it involves a neat algebraic jump.

So here we go.

What Is Math? (Baby Don’t Hurt Me, No More)

Today’s XKCD

Now, I generally avoid philosophy at all costs. All costs. For many reasons. Any good thought has likely been thought and better expressed before. It’s hard to separate philosophy from the pretensiousness of many who like to discuss it. Many reasons. But I am a mathematician, and it’s hard not to think about what it is I’m actually doing.

So, the comic expresses the traditional stacking of sciences. Biological, Chemical, Physical, Mathematical. And somewhere, burning in some deep, dark pit of hell, Computational Linguistics. It’s an easy stack to make. As in the quantum physics course I just took, it’s often difficult to tell which came first, the physics of quantum mechanics, or the math of it. Indeed, it often appears as though quantization in physics, is a result of a necessity of the math. Relativity, I have argued, is a logical necessity – from the math. Sometimes, it’s even more difficult to separate the math from ‘reality’. So it begs the question, I think so at least, what is math?

To which I would like to make a proposal.