Trisecting the Angle/Trisectrix of Maclaurin

Theorem

Let $\alpha$ be an angle which is to be trisected.

This can be achieved by means of a trisectrix of Maclaurin.


Construction

In the figure, the blue line is a trisectrix of Maclaurin.

Let us denote the trisectrix of Maclaurin by $\TT$.


Let $3 \theta$ be the angle to be trisected.

Position the vertex of $3 \theta$ at the point $\tuple {2 a, 0}$ with one arm along the $x$-axis.

Let the other arm of $3 \theta$ intersect the loop of $\TT$ at $P'$.

Let $OP'$ be constructed, where $O$ denotes the origin of the Cartesian plane in which $\TT$ is embedded.


Then:

$\angle POP' = \theta$

and hence $3 \theta$ has been trisected.


Proof


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Sources

  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): trisectrix
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): trisectrix
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