## Posts Tagged ‘area’

### Solution: Halving the Triangle

November 2, 2010

Suppose you have an equilateral triangle and you want to cut it in half using a pizza cutter (which can cut curves, not just straight lines). What is the shortest cut you can make?

Two commenters got the correct answer. The cut is one sixth of a circle whose center is at a vertex of the triangle.

It’s not immediately obvious to me why this is true. However, the following picture from Mahajan’s book makes it pretty clear.

Assuming the cut goes from one side of the triangle to another, we could tile six triangles around to make a hexagon, and find a minimum-length cut to divide all the triangles in half. As long as you believe a circle has the minimum perimeter for a fixed area, this should be pretty convincing.

### The Root of the Issue

September 29, 2009

I … regard it as mysterious that an object moving in an inverse square force law traces out a conic section. There are lots of ways to prove it, of course. Newton did it using Euclidean geometry. My homework problems above give two other ways. The one using the Runge-Lenz vector is pretty… but I’m still looking for the truly beautiful way, where you leave the room saying: “Inverse square force law… conic sections… of course! Now the connection is obvious!” – John Baez, Mysteries of The Gravitational 2-Body Problem

I’m not ready to going to give the beautiful proof Baez has been searching for throughout a long career in mathematical physics. I’d like to, but I’m not brilliant enough to be that easy to understand. Instead, I’m going to talk about some middle school math.

I’ve found that in most circumstances, if I’m willing to sit and be confused for long enough, I can normally understand what people are trying to tell me about math. That’s nice. It doesn’t work that way for everyone, and I’m glad I was lucky enough to be able to enjoy it, to the extent I can. But more than knowing the answer to a difficult problem, we humans like to know where the answer comes from.

Suppose, for example, I tell you that the solution to

$x^2 + ax = c$

is

$x = -a/2 + \sqrt{b + a^2/4}$.

It’s easy to verify – just plug the purported solution back into the original equation and you’ll see that it works. But if you have a soul you’ll wonder where in Hell I got that from. (Which is the sort of thinking that, ironically, will cost you your soul due to blasphemy.) Actually where I got it from was my seventh grade math teacher. But I should have gotten it by drawing a picture.

We start with $x^2$. It looks like this:

We'll represent the values on each side of the equation by the area of a figure. This is how we start.

Next is $+ ax$. Where is $ax$ in that picture? $x$ is the side of the square, so if we add on to that with a rectangle of width $a$, we’ll have a region $x^2 + ax$, which is the left hand side of the original equation.

The total area is now x^2 + ax. I know the picture says bx where it should say ax. This is the internet. You get what you pay for.

But that is not really fair to the other side of the square, who didn’t get any region added on. Let’s be symmetric about this, and add $ax/2$ to both sides.

That’s the left hand side of the equation. The right hand side is just $c$.

Here's a picture of the entire equation: x^2 + ax = c.

The thing on the left, though, is so damn close to being tractable. It’s almost a square. Once we’ve drawn this picture, it’s natural to “complete the square” by adding in that little region that’s $a/2 * a/2 = a^2/4$.

We add a little region to both sides, completing the square.

Now let’s do some morphing and stretch that right hand side out into whatever shape we want. Also a square, perhaps?

The equation discusses area, not shape, so we can make the shape on the right into any shape we want, as long as we give it the same area. We'll merge the red and green of the previous image to make purple, and form a square to give each side of the equation the same shape.

Great. Now we have two squares that are the same size. Therefore, the lengths of their edges are the same. And the length of the edge on the right hand side is just the square root of the area $c + a^2/4$. The length of the edge of the square on the left is $x + a/2$, so equating them gives

$x + a/2 = \sqrt{a^2/4 + c}$

or

$x = -a/2 + \sqrt{a^2/4 + c}$.

Cool.

Now let $a=14$ and $b=-23$. Then $x = -7 + \sqrt{72} = -7 + 6\sqrt{2}$.

Aaaaaaaaaaaw PHOOEY! I don’t know what the square root of two is. I am a Pythagorean and lived thousands of years ago. I believe that everything is ratios of integers and that you can’t eat beans because they resemble a fetus! But wait, I do have a little trick to get good rational approximations of root two. The trick is to make a table with two columns, and use a rule to generate new entries as you go down. It looks like this:

 1 1 2 3 5 7 12 17 29 41

The rule is, to put a new entry in the left hand column, add the two entries in the row above it. To put a new entry in the right hand column, again go to the row above, but this time add twice the left hand column to the right hand column. Just try; you’ll see.

The miraculous part comes when you divide the right hand column by the left. Let’s do this again.

 X Y Y/X 1 1 1 2 3 1.5 5 7 1.4 12 17 1.417 29 41 1.4138

It’s getting suspiciously close to root two. It’s not hard to verify that the table isn’t going to make a mistake; it will get closer and closer to root two. First we’ll write down formulas explaining the rule. Call the entry in the left column and $n^{th}$ row $X_n$ and right column $Y_n$. Then the rule is

$X_{n+1} = X_n + Y_n$

$Y_{n+1} = 2X_n + Y_n$.

Because we think the ratio is going towards $\sqrt{2}$, let $Y_n = \sqrt{2}X_n + \epsilon$. Then algebra tells you (on simplification)

$Y_{n+1} = \sqrt{2}X_{n+1} - \epsilon(\sqrt{2} - 1)$

from which we see that the ratio $Y_n/X_n$ converges to root two, oscillating above and below it each iteration.

The method is a general one for extracting square roots. Simply let $Y_n = cX_n + Y_n$ and you’ll extract the square root of $c$. It’s similar to an even-simpler method for finding rational approximations to the golden ratio – take successive terms of the Fibonacci sequence, which converges to the exponential of the golden ratio plus a constant. But I don’t see where it comes from.

I suppose we could write the iteration equations as a matrix, and talk about having $(1, \sqrt{2})$ as an eigenvector, etc. But the only reason I think to do that is that linear algebra and calculus are about the only math I know, so I try to use them for everything. If we start by assuming we’re going to find some rule that lets us make a table to find rational approximations of root two, it’s not hard to believe someone would come upon the rule. But who would get this idea of the table (or some similar equivalent idea) in the first place? I don’t know. But I’m happy people told me about it.

And this is only one of many ways to extract a square root. We could guess-and-check at will, trying first 1.5, then 1.4, then 1.45, then 1.42, then 1.41, then 1.415, etc. That’s a very intuitive rule, although a poor one from an efficiency standpoint. Or we could use some other rule to generate a new guess. For example, first take a guess at root two. Then average your guess $g$ with $2/g$. This generates a new, better guess. Repeating the process creates a sequence of rationals that converges to root two, just like the original method did. This new guess, though, has a wonderfully-intuitive explanation, and is almost obvious. It’s Newton’s method. I’ve seen one or two other methods that were more complicated – enough that I didn’t care to work through them in great detail when, after all, I can tell Wolfram Alpha to do whatever I want. (Well, not quite. Just now I told it to make me a sandwich, and it replied that it was assuming that “sandwich” was a city in Massachusetts).

So sometimes things make a lot of sense. Sometimes they only make a medium amount. Bit by bit, as I learn, the sphere of things that makes sense gradually expands. On the other hand, the sphere of things I heard about but don’t make sense expands even faster, and on top of that, the sphere of things I’ve heard about but understand so poorly I can’t even claim the results don’t make sense to me, since I don’t know what the results are saying, outstrips them all.

### New Problem: Transformation of Volumes

October 30, 2008

Here’s a real life problem that came up when I was looking for dark matter (I checked under my bed like six times, but I didn’t find any.  When I presented this research at group meeting, everyone got really quiet.  I think they were in awe.)

In the problem below, I use “volume” where you might think “area” makes more sense.  2D volume is area, so same thing.  But I use the term “volume” to suggest that you might want to generalize the problem to higher dimensions.

Suppose you have some 2-D volume in the plane.  You then do some sort of pointwise transformation, which maps this into a new volume, like so:

A super-spooky example of a smooth pointwise transformation from R2 to R2

You can describe this transformation by

$x' = f(x,y)$

$y' = g(x,y)$

where $(x',y')$ is the coordinate to which you map the point $(x,y)$.  Assume that $f$ and $g$ are smooth and invertible.  Points inside the region stay inside the region, and things behave nicely, if weirdly.  Straight lines map onto curvy lines (or straight lines, if they want), but not broken up segments or points.  The neighborhood of a point transforms into the neighborhood of the transformed point.  Different points stay different (never wind up on top of each other).  Photographs of your head become distorted, but your eye (if you have one, this is an equal-opportunity blog) is still next to your nose.

Question: What is the area of the transformed region, in terms of the old region and the transformation equations?  (see tags for a hint or two)