Trisecting the Angle/Archimedean Spiral
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Theorem
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of an Archimedean spiral.
Construction
In the figure, the blue line is an Archimedean spiral.
Let $\angle AOB$ be the angle to be trisected, where $OB$ is on the polar axis.
Trisect the segment $OA$ such that $OC = \dfrac 1 3 OA$.
Draw a circle with $O$ as the center and $OC$ as the radius.
Let $D$ be the intersection of the circle and the spiral.
Then $\angle DOB = \dfrac 1 3 \angle AOB$.
Proof
Let the equation of the Archimedean spiral be $r = a \theta$.
Then:
\(\ds \angle DOB\) | \(=\) | \(\ds \frac {OD} a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {OA/3} a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 3 \angle AOB\) |
$\blacksquare$
Also see
Historical Note
Use of the Archimedean spiral to trisect an angle was a standard technique for mathematicians following Archimedes.