# A natural function with discontinuities (Part 2)

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