An angle trisection

From page 26 of
Underwood Dudley, The Trisectors, 2nd edition, 1996.

Drag the point $B$ to change the angle $AOB$ (but keep it less than 90 degrees).
Press “r” to reset the diagram to its initial state.
The red line, $OT$, is an approximate trisector of the angle $AOB$.

Construction

We wish to trisect the given angle $AOB$. Assume the angle is less than 90 degrees; see the diagram above.

  1. Draw a line through $B$ parallel to $OA$.
  2. Extend $OA$ and mark off $AC$ and $CD$ along it, each equal to $OA$.
  3. Draw the arc $DE$ with center $C$ and radius $CD$.
  4. Drop a perpendicular from $E$ to $OD$ and let $F$ be the foot of the perpendicular.
  5. Draw the arc $FT$ with center $O$ and radius $OF$ (shown in green).
The line $OT$ is an approximate trisector of the angle $AOB$.

Error Analysis

Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $FOT$, respectively. It is straightforward to show that \[ \beta = \frac{\sin \alpha}{2+\cos \alpha} = \frac{\alpha}{3} - \frac{1}{2^2 \cdot 3^3 \cdot 5} \alpha^5 + O(\alpha^7) = \frac{\alpha}{3} - \frac{1}{540} \alpha^5 + O(\alpha^7). \] The error $ \ds e(\alpha) = \frac{\alpha}{3} - \beta $ is monotonically increasing in $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2) =$ 0.0236 radians = 1.352 degrees.

It is interesting that the error is $O(\alpha^5)$ rather than $O(\alpha^3)$ as one might have expected. Despite this, the method's accuracy is not particularly remarkable for angles that are not very close to zero.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on July 26, 2002.
Cosmetic revisions on June 6, 2010.

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