Foxmaths! 2.0

June 24, 2008

Perimeter = Area

Filed under: Maths — Tags: , , — Fox @ 7:20 pm

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.

Regular Hexagon
(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,

P = n * s

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,

A = \frac{1}{2} * s * a * n

Settering perimeter equal to area,

P = A

n * s = \frac{1}{2} * s * a * n

1 = \frac{1}{2} * a

a = 2

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

Filed under: Maths — Tags: , , — Fox @ 6:39 pm

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

Fox Graph

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

Filed under: Maths — Tags: , , — Fox @ 6:05 am

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

USA - Blank

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.
(more…)

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