A natural function with discontinuities (Part 2)

Yesterday, I began a short series motivated by the following article from the American Mathematical Monthly.

discontinuous

Today, I’d like to talk about the how this function was obtained.

If 180^\circ \le latex \theta \le 360^\circ, then clearly r = R. The original circle of radius R clearly works. Furthermore, any circle that inscribes the grey circular region (centered at the origin) must include the points (-R,0) and (R,0), and the distance between these two points is 2R. Therefore, the diameter of any circle that works must be at least 2R, so a smaller circle can’t work.

reflexangle

The other extreme is also easy: if \theta =0^\circ, then the “circular region” is really just a single point.

Let’s now take a look at the case 0 < \theta \le 90^\circ. The smallest circle that encloses the grey region must have the points (0,0), (R,0), and (R \cos \theta, R \sin \theta) on its circumference, and so the center of the circle will be equidistant from these three points.

acuteangle

The center must be on the angle bisector (the dashed line depicted in the figure) since the bisector is the locus of points equidistant from (R,0) and (R \cos \theta, R \sin \theta). Therefore, we must find the point on the bisector that is equidistant from (0,0) and (R,0). This point forms an isosceles triangle, and so the distance r can be found using trigonometry:

\cos \displaystyle \frac{\theta}{2} = \displaystyle \frac{R/2}{r},

or

r = \displaystyle \frac{R}{2} \sec \frac{\theta}{2}.

This logic works up until \theta = 90^\circ, when the isosceles triangle will be a 45-45-90 triangle. However, when \theta > 90^\circ, a different picture will be needed. I’ll consider this in tomorrow’s post.

Leave a comment

1 Comment

  1. A Natural Function with Discontinuities: Index | Mean Green Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: