An angle trisection

Construction by cdsmith

Sorry, no pictures. Explanation here.

Construction

To trisect the angle $AOB$ (with $OA=OB$), do:

1. Find the midpoint $M$ of the line segment $AB$.
2. Draw circles centered at $A$, $M$, and $B$, each of radius $\frac{1}{2}AB$, and mark their intersection points $C$ and $D$, as shown in the diagram above.

Error Analysis

Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOC$, respectively. One may verify that \[ \beta = \arctan\bigg( \frac{ \sin\frac{\alpha}{2} \sin\big(\frac{\pi}{6}+\frac{\alpha}{2}\big) }{ 1 + \sin\frac{\alpha}{2} \cos\big(\frac{\pi}{6}+\frac{\alpha}{2}\big) } \bigg) = \frac{1}{4}\alpha + \frac{\sqrt{3}}{16} \alpha^2 - \frac{1}{16} \alpha^3 + O(\alpha^4). \]

This says that $\ds \beta \approx \frac{1}{4}\alpha$ when $\alpha$ is small, so small angles are quadrisected, rather than trisected! (This is clearly visible in the interactive diagram above.) For not-so-small angles, the method works reasonably well. In fact, it produces exact trisection for angles $\alpha=\pi/2$ and $\alpha=\pi$.

The worst error in the range $0 \le \alpha \le \pi$ is 0.0214 radians = 1.23 degrees. This occurs at $\alpha=2\arctan(\sqrt{3}\pm\sqrt{2})$ which corresponds to angles of approximately 35 degrees and 145 degrees.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on June 14, 2010.

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