Sorry, no pictures. Explanation here.
This approximate trisection, due to Avni Pllana, was announced
in a
message
in the geometry.puzzles newsgroup on July 23, 2003.
Scroll to the bottom of that page to view the related discussion thread.
Consider the angle $AOB$ given by the circular arc $AB$ centered at $O$, as shown in the diagram above.
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AON$, respectively. One may verify that: \[ \beta = \arctan \frac {\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}} {\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}} = \frac{1}{3}\alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5) = \frac{1}{3}\alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5). \] Hint: Represent the points as complex numbers in the polar form $re^{i\theta}$.
The error $ \ds e(\alpha) = \frac{\alpha}{3} - \beta $ increases monotonically with $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2)$ = 0.000757 radians = 0.0434 degrees. The worst error on the interval $0 \le \alpha \le \pi$ is $e(\pi)$ = 0.00630 radians = 0.361 degrees. That's quite good for such a simple construction.
The angle $\beta$ constructed by this method coincides exactly with that of Lindberg's construction, where $\beta$ is given as: \[ \beta = \frac{1}{4} \alpha + \arctan \frac{\sin\frac{\alpha}{4}}{2+\cos\frac{\alpha}{4}}. \] One way to verify that the seemingly different expressions for $\beta$ are in fact identical, is to compare their derivatives. In both cases we have: \[ \frac{d\beta}{d\alpha} = \frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}. \]
This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on June 10, 2010.
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