A natural function with discontinuities (Part 3)

This post concludes this series about a curious function:

discontinuousIn the previous post, I derived three of the four parts of this function. Today, I’ll consider the last part (90^\circ \le \theta \le 180^\circ).

obtuseangleThe circle that encloses the grey region must have the points (R,0) and (R\cos \theta, R \sin \theta) on its circumference; the distance between these points will be 2r, where r is the radius of the enclosing circle. Unlike the case of \theta < 90^\circ, we no longer have to worry about the origin, which will be safely inside the enclosing circle.

Furthermore, this line segment will be perpendicular to the angle bisector (the dashed line above), and the center of the enclosing circle must be on the angle bisector. Using trigonometry,

\sin \displaystyle \frac{\theta}{2} = \frac{r}{R},


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

We see from this derivation the unfortunate typo in the above Monthly article.

One thought on “A natural function with discontinuities (Part 3)

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 )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.