Hippias and his quadratrix

Hippias of Elis (430 BC) was a sophist who invented the quadratrix curve to trisect an angle. The problem of trisecting a given angle was one of the problems that generated a lot of mathematics during this period, and several mathematicians devised methods for solving this problem. Like many other sophists, Hippias was an itinerant teacher who made his living wowing the locals with his knowledge.

Apparently, he did alright, but didn't leave much of a legacy except for the quadratrix.

Definition of the curve

The curve can be described in a few sentences. Let ABCD denote a square. Over a unit time period, allow the top segment of the square to fall at a uniform speed to the bottom of the square. During the same time, allow the left side of the square to rotate clockwise at a uniform speed to the bottom of the square. At each time, the two segments will intersect in a point P. The totality of all these points P is defined as the quadratrix.

Drawing the quadratrix

One can imagine how Hippias might have sketched the quadratrix in the sand, but one can hardly image how he would have made an accurate sketch of it.

First make a square. Color it tan and label the vertices A, B, C, and D.

> sqr :=plots[display](
[plots[textplot]({[0,-.05,`A`],[1,-.05,`B`],[1,1.05,`C`],[0,1.05,`D`]}),
plots[polygonplot]([[0,0],[1,0],[1,1],[0,1]],color=tan)],scaling=constrained,axes=none):
sqr;

As time t goes from 0 to 1, we want the top of the square to fall to the bottom at a constant speed. That means that the y coordinate will be 1-t at time t.

An animation of the top falling to the bottom

a fleeting image of the quadratrix

the quadratrix being traced out

Using the quadratrix to trisect an angle

How did Hippias use the curve to trisect an angle? Very simply. He placed the square containing the curve over the angle with the vertex at the lower left hand corner and the initial ray on the bottom of the square. He then marked the point P on the curve where the terminal ray crossed it and found the point Q at the same height on the left side of the square. Then he constructed the point R on the segment AQ such that AR/AQ = 1/3 and found the point S on the curve whose height is the same as R. By the definition of the curve, the measure of angle SAB is 1/3 the measure of the original angle.

Problems with the quadratrix

1. Find the x-coordinate of the lower end of the quadratrix.

> p(1);

Error, (in p) division by zero

> p(t);

> limit(p(t)[1],t=1)=evalf(limit(p(t)[1],t=1));

This can also be obtained easily with L'Hospitals rule.

2. Find the length of the quadratrix.

> F := int(sqrt(diff(p(t)[1],t)^2+diff(p(t)[2],t)^2),t=0..1);

> evalf(F);

3. Find the area under the quadratrix.

> Int(p(1-t)[1],t=0..1)=int(p(1-t)[1],t=0..1);

> rhs(%) = evalc(rhs(%));

> (rhs(%))=evalf(rhs(%));

>

Using the quadratrix to square the circle.

The quadratrix can be used to square the circle, in fact the name, quadratrix, means 'square making'. Dinostratus (circa 350 BC) was the first to use it for this purpose, according to Pappus (circa 300 AD).