of a Circle

Jennifer Roth

The answer to this question has been investigated by many mathematicians,
and it has been found that with a compass and straightedge construction
this is impossible.

Hippocrates found that there are certain regions with curved boundaries
that are squarable (called lunes).

A construction similar to his for one squarable lune follows using
GSP.

You start with an isosceles right triangle, and then semicircles
are constructed on the three sides as shown in the following picture.

Consider the ratio of the areas of the semicircles on AB and AC. You get

Now since ABC is an isosceles triangle with AB the hypotenuse,

Therefore

So, Area of the semicircle on AB has twice the area of the semicircle
on AC.

One example of this follows.

Click here to view the GSP file that
demonstrates this.

Therefore the area of a quarter circle from AB is the same area
as the semicircle on AC.

Now, if the area of the segment of the quarter circle is the commen area of the quarter circle on AB and the semicircle on AC

Therefore the area of the lune on AC and the area of triangle AGC are equal. Since AGC is an isosceles right triangle, its area is

Thus the area of the lune is the same as a square of side of length

.

An example of this follows

Click here to view the gsp file that
demonstrates this.

This, however is not a solution to squaring the circle.

Every lune cannot be squared. In fact this particular lune is
one of 5 that can be squared.

The Quadratrix was invented by Hippias to trisect an angle.

The following is an example of a quadratrix. (Click
here to view the GSP file that demonstrates this)

See how the three proportions are equal

The final proportion is (arc BED)/(arc ED)

While Hippias used the Quadratrix to trisect an angle, Pappus
was able to demonstrate how it can be used to find a square equal
in area to a given circle.

If it is assumed that point G (see Quadratrix above) can be found,
then the following can be proved:

(arc BED)/AB = AB/AG

One example of this follows: Click here to view the GSP file that demonstrates this.

Pappus just assumed you would then be able to arrive at a square
equal in area to a given circle since 2 square lengths were used
to describe an arc length.

If you have a circle of radius r, then a line segment of length
s can be constructed in which

(C/4)/r = r/s. This is the general formula for the above formula
(arc BED)/AB = AB/AG.

arc BED was 1/4 of the original circle of radius AB. And the length
s obtained was AG.

Therefore we know C/4 = (r)(r)/s

Let q = (r)(r)/s=C/4

The area of a circle is (pi)(r)(r), and the Circumference is 2(pi)(r),
therefore the the area is (C/2)(r).

**References
**Burton, David M. Burton's History of Mathematics: An
Introduction