Wrapping the Real Numbers Around a Circle… of Radius 1

I’d like to thank my amazing Calculus III instructor, Professor Wolfson, for showing me this. Undoubtedly will always be my favorite professor and I only wish to someday be a quarter as knowledgeable as him. Anyhow! Let’s jump right in.

What if I told you that you can wrap the entire real number line, every point on it, around a circle of radius 1? Go ahead, choose whether I’m a loony or not now before continuing to read.

Did you choose? Alright. Now look at the animated graph below.

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Circle has radius 1, line is in the form y=mx+1 with -10<m<10.

I’m going to try to motivate you to sort of “prove” this in your head before I get into the mathematics behind it. While looking at the gif, just note the following:

  1. Consider the very top point of the circle, the point (0,1), as undefined.
  2. Note that as the slope changes, where the line intersects the circle and the x-axis changes as well. In a single snapshot of the gif, let’s call the point it intersects on the x-axis as (x_0, 0) or just x_0 for short. Call the point it intersects on the circle (x_c,y_c)
  3. Now try to imagine this: in that random snapshot of the gif (where the line is held still) map (“assign”) the point x_0 to (x_c,y_c). Now go to the next snapshot and do it again, and again, …

Now let the slope instead of being anything in the range (-10, 10), be anything in ( - \infty, \infty). What follows is every single point on the real number line (x-axis) gets assigned to a point on the circle! With room to spare even, that point on the top of the circle never got assigned anywhere after all.

Let’s get down and dirty in some math now.

Err, I’m sorry, I won’t say that again. Sounded a lot cooler in my head.

Finding the Function:
We can actually find the exact function f that tells us where our point x_0 goes on the circle. It’ll be a function mapping one number to two, or more explicitly, f: \mathbb{R} \to \mathbb{S}, where \mathbb{S} = \{ x \in \mathbb{R}^2 | x^2+y^2 =1 \}.

To get started we need to know what we have to work with. Without loss of generality let’s let the radius be 1 (that is, you can make the radius whatever you please, I just wanted to make it look pretty in the end). What we have is

\begin{cases} x^2+y^2=1 \\ y=mx+1 \text{ where } m = -\frac{1}{x_0} \end{cases}

which is actually all we need. Let’s leave the m in there for now though to avoid having to deal with messy equations, and just remember that our circle doesn’t include the point (0,1).

From here it’s a matter of some plug and chug!

x^2+(mx+1)^2=1
x^2+m^2x^2+2mx=0
(1+m^2)x^2+2mx=0
\displaystyle{ x=\frac{-2m \pm \sqrt{4m^2}}{2(1+m^2)}}
\implies x=\displaystyle{\frac{-2m}{1+m^2}}

Great! Now we complete the same procedure for the other coordinate.

\displaystyle{\left(\frac{y-1}{m} \right) ^2 + y^2 = 1}
y^2-2y+r^2+m^2y^2=m^2
(1+m^2)y^2-2y+(1-m^2)=0
y=\displaystyle{\frac{2 \pm \sqrt{4-4(1+m^2)(1-m^2)}}{2(1+m^2)}}
y= \displaystyle{\frac{1 \pm m^2}{1+m^2}}
Here we must choose 1-m^2 for the numerator because otherwise things would simplify to y=1, and remember, (0,1) wasn’t defined on our circle.
\displaystyle{ \implies y=\frac{1-m^2}{1+m^2}}

The Function:
All this, in summary, provides the following function:
f(x_0)=\displaystyle{ \left< \frac{-2m}{1+m^2}, \frac{1-m^2}{1+m^2} \right>}
which, after plugging in the fact that m=-\frac{1}{x_0} provides…

f(x_0)=\displaystyle{ \left< \frac{2x_0}{{x_0}^2+1}, \frac{{x_0}^2-1}{{x_0}^2+1} \right>}

And there we have it! Any point x_0 on the real number line can be mapped to the point f(x_0) on our circle of radius 1! Go ahead and try a few points and mess around if you want. Let’s end on some final notes regarding this:

  • It’s pretty cool looking at how “dense” the circle is packed, for instance, the bottom half of the circle is mapped only points from [0,1] but what about perhaps the upper-left tiny little 1/1000 segment of the circle? Oh dear…
  • Thing about the difference in the arc-length of a circle over its entire domain [0, 2\pi] versus the arclength of x=0 over entires domain (-\infty, \infty). What in the world is going on here?
  • Explicitly calculate the inverse of f(x_0), that is, if I gave you a point (x_c, y_c) on the circle, can you tell me what x_0 it was assigned?
    • By finding the explicit inverse function (showing it exists), we showed that the circle and the real numbers are isomorphic (have the “same size” set of points!!).

Edit: Part 2 of this is up!

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Area of a Triangle on a Sphere

My good friend who runs the wonderful blog In The Margins gave me this problem quite some time ago, and I honestly enjoyed completing it quite a bit. Thought I’d go ahead and share it with the world!

Problem:
“Find the area of a triangle on a sphere without using Calculus, given just its angles \mathbf{\alpha, \ \beta, \ \text{and } \gamma}.”

I recommend pausing here and trying this yourself first, it’s quite fun!

Solution:great-circle-figure
First, make sure you know what a great circle is. It’s the intersection of a plane (“paper”) and a sphere (“orange?”) where the plane goes through the center of the sphere. Look at the image to the right for an example (for instance, the Earth’s equator is a great circle in a sense!)

Begin by first drawing a triangle on a sphere with arbitrary angles \alpha, \ \beta, \ \text{and } \gamma. Afterwards, extend the sides of the triangle to form great circles. In doing this, you should notice that the same triangle appears on the back of the sphere.

Now denote the side between \gamma and \alpha as 1, \alpha and \beta as 2, and \beta and \gamma as 3. Now only looking at side 1, shade in the hemisphere not containing the triangle, and the repeat the process with the other two sides.

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The numbers in circles represent how many times that particular partition was shaded over.

Let’s denote the following:
A_1 := Set of Shaded Points from The Hemisphere of Side 1
A_2 := Set of Shaded Points from The Hemisphere of Side 2
A_3 := Set of Shaded Points from The Hemisphere of Side 3

Note then, that this follows:
|A_1| = |A_2| = |A_3| = 2 \pi r^2
|A_1 \cap A_2 \cap A_3|=A_T (where A_T is the area of the triangle)
|A_1 \cup A_2 \cup A_3| = 4 \pi r^2 - A_T
|A_1 \cap A_2| = 2 \alpha r^2 (this, along with the following, are called antipodal digons)
|A_2 \cap A_3| = 2 \beta r^2
|A_3 \cap A_1| = 2 \gamma r^2

Looks like we’re in a good situation to employ the Inclusion-Exclusion Principle!

Inclusion-Exclusion Principle for 3 Finite Sets:
Let A, B, and C be three finite sets. Then
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C| .

Which means we must have
|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|

\implies 4 \pi r^2 - A_T= 6 \pi r^2 - 2r^2(\alpha + \beta + \gamma) + A_T
\implies A_T = r^2(\alpha + \beta + \gamma - \pi)

Thus we have shown the following:

Proposition:
The area of a triangle A_T on a sphere of radius r with angles \alpha, \ \beta, \ \gamma is A_T = r^2(\alpha + \beta + \gamma - \pi) .

And there we have it! There’s our area. Oh! And also!

Corollary:
The sum of the interior angles of a triangle on a sphere must exceed \pi.

Isn’t that so cool? Being used all your life to the sum of the interior angles of a triangle being precisely 180 degrees — how do you feel seeing that now it’s mandatory it exceed 180 degrees?