An angle trisection

Construction by Wayne Baker

Sorry, no pictures. Explanation here.

The Basic Construction

Here is a very simple straightedge and compass construction of an approximate angle trisector due to Wayne Baker.

Let us represent the angle by the circular arc $AB$ centered at $O$; see the diagram above. The angle's size may be anything from 0 to 180 degrees. To trisect, do:

  1. Quadrisect the angle $AOB$, that is, divide it into four equal parts. The arc $AP$ in the diagram above represents one quarter of the original arc $AB$. Let $L$ be the length of the chord $AP$ (shown in green).
  2. Draw a circular arc (shown in orange) centered at $O$ and radius 3/4 of $OA$. Mark $A'$ and $B'$ its intersections with the rays $OA$ and $OB$, respectively.
  3. Swing an arc (not shown) of radius $L$ centered at $A'$ and mark $P'$ its intersection with the arc $A'B'$, as shown.

The line $OP'$ is an approximate trisector of the angle $AOB$.

Note added October 2024

Wayne as provided this javascrip application which provides an interactive interface to his construction. Thanks, Wayne!

Error Analysis

Let $\alpha$ and $\beta=\tau(\alpha)$ be the sizes of the angles $AOB$ and $A'OP'$, respectively. It is straightforward to show that \[ \beta = 2 \arcsin\big(\frac{4}{3}\sin\frac{\alpha}{8}\big) = \frac{\alpha}{3} + \frac{7}{2^7\cdot3^4}\alpha^3 + O(\alpha^5) = \frac{\alpha}{3} + \frac{7}{10368}\alpha^3 + O(\alpha^5). \]

The error $ \ds e(\alpha) = \tau(\alpha) - \frac{\alpha}{3} $ is monotonically increasing in $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2)$ = 0.002695 radians = 0.154 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.0237 radians = 1.360 degrees.

Iterative Improvement

As we see in the asymptotic expansion shown above, the angle $\tau(\alpha)$ is slightly larger than the target value of $\alpha/3$. Making three copies of the constructed angle, and putting them end-to-end as in arcs $A'P'$, $P'P''$, and $P''P'''$ shown in the diagram below, we arrive at the endpoint $P'''$ which is very slightly off the point $B'$, and just outside the arc $A'B'$. The constructible angle $B'OP'''$ is exactly three times the error $e(\alpha)$. If we were able to trisect $B'OP'''$ exactly, then we would know the error, and consequently will have achieved the exact trisection of the original angle. Of course the exact trisection of $B'OP'''$ is impossible in general, but we may repeat the method outlined in the Basic Construction above to obtain an approximate trisection of $B'OP'''$, which will yield $ \tau\big(3\tau(\alpha) - \alpha\big) $, and consequently an improved trisection $\tau_{\mathrm{improved}}(\alpha)$ of the original angle: \[ \tau_{\mathrm{improved}}(\alpha) = \tau(\alpha) - \tau\big(3\tau(\alpha) - \alpha\big) = \frac{\alpha}{3} - \frac{7^4}{2^{28}\cdot3^{13}} \alpha^9 + O(\alpha^{11}). \] The error $ \ds e_{\mathrm{improved}}(\alpha) = \frac{\alpha}{3} - \tau_{\mathrm{improved}}(\alpha)$ is monotonically increasing in $\alpha$. In particular, $e_{\mathrm{improved}}(\pi/2) = 1.5\times 10^{-9}$ radians $ = 8.6\times10^{-8}$ degrees.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on May 31, 2010.

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