The hyperbolic spiral as a trisectrix
It is, of course, impossible to subdivide an arbitrary angle into equal thirds using a compass and straightedge alone. But there are many known constructions that rely on other techniques. One of those techniques is the use of a trisectrix, a curve of a special form that is assumed to be already drawn for you, so that with your compass and straightedge you can find its crossing points with lines and circles. More strongly, some curves can be used as a “sectrix” to subdivide an angle into any given number of equal parts; these include the Archimedean spiral, quadratrix of Hippias, and sine curve.
The hyperbolic spiral is closely related to all three of these sectrix curves. It is what you get when you perform a circle inversion of the Archimedean spiral, centered at the center of the spiral. The sine curve is what you get from a helix (the curve of a spiral staircase) by perpendicular projection onto a plane parallel to the axis of the helix, the quadratrix is what you get by slicing the helix by an inclined plane and then projecting it onto a plane perpendicular to the axis, and the hyperbolic spiral is what you get from a perspective projection of the helix onto a plane perpendicular to the axis. Intuitively, it’s what you see when you look straight up from a point centered at the bottom of a spiral staircase.
Because of its close connection to these curves, it should not be a surprise that a hyperbolic spiral can also be a sectrix, but I could not find a reference saying so. Here’s a construction:
The pre-given parts of the diagram are the red hyperbolic spiral, shown with its center point \(O\), its asymptotic line \(AB\) (the lower horizontal line), another horizontal line \(OC\) parallel to the asymptotic line through the center of the spiral, and two perpendicular lines \(AO\) and \(BC\). The horizontal offset between the vertical lines \(AO\) and \(BC\) doesn’t matter for this construction.
Given your angle to be trisected (or \(n\)-sected), place it as \(POS\) at the center of your given hyperbolic spiral. Construct circles with radii \(OP\) and \(OS\) (the inner and outer circles of the figure), and intersect them with line \(OC\) at \(P'\) and \(S'\). Extend the lines \(AP'\) and \(AS'\) to cross line \(BC\) at \(P''\) and \(S''\). Subdivide line segment \(P''S''\) into as many equal-length pieces as desired; here I’ve drawn three pieces, separated at \(Q''\) and \(R''\). Then reverse the construction for \(Q''\) and \(R''\): extend the lines \(AQ''\) and \(AR''\) to cross \(OC\) at \(Q'\) and \(R'\), draw circles with radii \(OQ'\) and \(OR'\), and find the crossing points \(Q\) and \(R\) of these circles with the spiral. The result is that angle \(POS\) is subdivided into three equal angles, \(POQ\), \(QOR\), and \(ROS\).
Why does this work? Let’s go back to the spiral staircase, and the fact that the hyperbolic spiral is its perspective projection. Here’s a view of this taken from Wikipedia, drawn looking down from the top of the spiral staircase rather than up from the bottom:
The right side of the trisection diagram can be thought of as your view up the staircase, with points \(PQRS\) marking four equally-spaced points along the staircase, marking out angles with respect to the center axis of the staircase that are equal from your perspective at the bottom of this axis. Because it’s a spiral staircase, you know that these four points are actually at equally spaced heights with respect to each other, although your view does not let you see these heights. If the staircase winds up the side of a cylindrical tower, the circles of the diagram represent your projected view of circular cross-sections of this tower (such as one of its floors).
The left side of the diagram can be thought of as a cross-section of the staircase as seen from the side. From this view, the staircase itself would appear to be a vertical sine curve, but I left this out of the diagram because it’s not used in the trisection. In this side-on view, point \(A\) is the bottom center point of the staircase (from which you were looking up), and line \(OC\) is a cross-section of the horizontal plane onto which your upward view of the staircase was projected. As seen in a side-on cross-section, the cylindrical tower looks like a rectangle with two vertical sides; vertical line \(BC\) is one of these two sides, and can be thought of as a vertical line on the cylinder itself. Points \(P''\), \(Q''\), \(R''\), and \(S''\) are the points on this vertical line where it crosses the cross-sections of the cylinder that you see are circles. The lines from \(A\) through these four points project them onto the viewing plane seen in cross-section as \(OC\). In short, the left side of the diagram depicts the projection of vertical line \(BC\) from the cylinder onto the viewing plane \(OC\) of the hyperbolic spiral. Through this construction, we correspond the (unprimed) points on the staircase, the (single prime) points in your projected view of the leftmost point of each horizontal cross-section circle, and the (double prime) points, at their actual height on the cylinder in its side-on cross-section view.
Equal angles, on the right side, correspond to equal heights, on the left side. Therefore, a subdivision of an angle into equal parts, on the left, corresponds to a subdivision of a line segment into equal parts, on the right. And subdivisions of a line segment into equal parts are easy to construct.