Angles on an Angle (A Mathematical Tangent)

I guaranteed there would be a post, and this is proof of that guarantee.

I like math. I wrote a post a while back about Fermat and how he is a super math prankster. This post isn’t going to be biographic. This post is experimental. Buckle up and get ready for the paint-application-level images.

I’ve always had a design in my head that is kind of inspired from taking the limit of a regular shape as its sides approach infinity. In basic terms, if you keep adding sides to a square such that all the sides are equal length once that side is added, at infinity you’ll get really dang close to making a circle.

So here is my example time. I was up one night doing this because it seemed fun and I was particularly fixated on figuring this out.

So this is a 90 degree angle.

If you put a single point somewhere in that quadrant, you’ll get another 90 degree angle if you draw perpendicular lines from the axes. Like the one below.

Now that’s a square and all the angles equal 90 degrees. But what if we add another point out in that space, such that we maintain the perpendicular angles on the axes, and make their angles out in the quadrant equivalent. It will look something like the thing down below.

The question that remains, what would the angles of the intersections have to be in order for them to be equal to each other?

What you are looking at is one fourth of an octagon. Also you are looking at two points, both with 135 degree angles in the quadrant. It is pretty easy to figure out that it is 135 since you can reason that they are perpendicular lines, splitting another 90 degree angle between them, forming two 135 degree angles.

So let’s add 3 more points for the fun of it, putting a total of 5 points out in the quadrant. You’ll see the result (or my best paint of it) down below.

I won’t lie, I eyeballed that figure so they probably won’t all be perfect 162 degree angles. How did I come up with that number? This one you can’t really just eyeball 162 degrees, but by the time I got to this many points that night, I already had a formula in hand (more about that later).

One thing that you should notice though is how much more like a circle it already looks than the 90 degree figure. That’s because all of the angles are getting closer to 180 degrees, which every point on a circle should essentially be (for our sake). And here is where the fun and elegant part comes in.

Like I said before, there is a formula to figure out what angle each intersection would have to be in order to have equivalent angles to each other, I couldn’t find one, but I made one and it looks like this.

[90(1+2x)]/(1+x) x being the amount of points added minus 1.

So as a chart
1 point = [90(1+2(0))]/(1+0) = 90 degree angle
2 points = [90(1+2(1))]/(1+1) = (90*3)/2 = 135 degree angles
3 points = [90(1+2(2))]/(1+2) = (90*5)/3 = 150 degree angles
4 points = [90(1+2(3))]/(1+3) = (90*7)/4 = 157.5 degree angles
5 points = [90(1+2(4))]/(1+4) = (90*9)/5 = 162 degree angles
…   skip a few
2000 points = [90(1+2(2000))]/(1+2000) = (90*4001)/2001 = 179.955 degree angles

So if you took the limit of this equation as you take the number of points to infinity you’ll get a nice crisp 180 degrees. This verifies the assertion that the limit actually creates a circle. Since, although a circle doesn’t have a 180 degree angle for every single point, it has something infinitely close to 180 degrees without ever actually being 180 degrees.

Limits are pretty cool and I might write about Euler’s number one of these days, or even the Monty Hall problem. For right now though, I hope this post satisfied your mathematical interests until the next one comes along.

Also feel free to test the formula, it should undoubtedly work and looks awesome when you decide to choose a large number of points.


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