From Euclid to Spigotry
Approximating Pi by Angle Bisection
Last modified: September 9, 2004
References:
Beckmann, A History of Pi Chapter 6
Doerrie, 100 Great Problems of Elementary Mathematics, pages 184-188 
Lecture topic 1: Inscribed and circumscribed polygons
Start with a circle of radius r, and create an inscribed polygon and a cirumscribed polygon,
both with n sides.

Show that the perimeter of the inscribed polygon is pn = 2nr sin(p/n)

while that of the circumscribed polygon is Pn = 2nr tan(p/n)

Show that the area of the inscribed polygon is an = nr2/2 sin(2p/n)

while that of the circumscribed polygon is An = nr2 tan(p/n)

Evaluate these quantities numerically for the case of a square (n = 4) and a hexagon (n = 6)
Here is the diagram that you will need (illustrated for n = 4).

Lecture topic 2 : Archimedes' recursive formulas
Let pn denote the perimeter of a polygon of n sides inscribed within a circle, and let Pn
denote the perimeter of a polygon circumscribed about the same circle.  Prove by geometry that

= 
 = 

The proof as presented by Archimedes is almost impossible to follow, since numerical approximations are brought in each time an angle is bisected.  The proof based on Figure 30 in Doerrie (reproduced above) is a good old-fashioned similar-triangles argument that will lead you to the Archimedes recurrence relations without any trigonometry. Right triangles AOC and AZM are similar, as are isosceles triangles COM and CMD.

Lecture topic 3: Archimedes by trigonometry

This approach can be found on pp. 65 and 66 of Beckmann.

If you are willing to use trigonometry life is easy.  Just let  denote the angle p/n radians.

  Then the perimeter of the inscribed polygon of n sides (in a circle of unit radius)  is

 as proved in topic 1

while the perimeter of the circumscribed polygon is
 as proved in topic 1

The first Archimedes recurrence, = , translates into
= 
which can be rewritten as the standard half-angle formula
= 
The second one,  = , translates into
= 
Square this, and cancel out a factor of  to obtain
=
Divide both sides of this by and it becomes
 = 

Lecture topic 4: Biography of Archimedes
There is a chapter on Archimedes in Beckmann, but there is lots more biographical material in books in Cabot and on the Web.  Archimedes was a great military engineer as well as a great mathematician.  Present a 10-minute biography. 

Lecture topic 5: Archimedes in Mathematica

Archimedes had to struggle to complete the calculations for a polygon of 96 sides, but
you can replicate his results in a few minutes in Mathematica.

Start, as Archimedes did, with a hexagon of radius 1/2 and diameter 1.
For the inscribed hexagon, p6 = 3
For the circumscribed hexagon, P6 = 2
Now you can just use the formulas of Archimedes over and over again to recalculate
P12 and then p12, P24, and then p24.  Remember to use //N to get a numerical
value.

Email your finished .nb file to bamberg@tiac.net and it will be made available on the Web site.
Lecture topic 6: An important limit
In the diagram below, if the angle x is measured in radians, the circular sector has area A = 
The smaller triangle, inside the sector, has area  while the larger one, outside the sector
has area 
Conclude that for any positive x,
sin x < x < tan x
or, inverting everything and reversing the inequality, that
csc x > 1/x > cot x
Now multiply by sin x to get
1 > >cos x
Since cos x approaches 1 as x approaches 0, the quantity , which is "squeezed" between
cos x and 1, must also approach a limit of 1 as x approaches 0.

Lecture topic 7: Constructing the quadratrix of Hippias
This is explained on pp. 40-44 of Beckmann, but once he gets into trigonometry it is
easy to lose sight of the fact that infinitely many points on the quadratrix can be
constructed with compass and straightedge.  Show first how to construct the point
that corresponds to q = p/4, then the points that correspond to q = p/8 and q = 3p/8.
At this point it should be clear how to construct the point that corresponds to
q = 
A French curve will be available in class so that people can draw a smooth curve through the points that they have constructed.  This, however, is not a legal operation according to Euclid's rules!
Lecture topic 8: Using the quadratrix for angle trisection
This is not mentioned in Beckmann, but it is a classic use of the quadratrix.
We know that an angle of p/3 (60 degrees) cannot be constructed with compass and straightedge.
The point on the quadratrix that corresponds to q = p/3 can be constructed -- show how.
The point that corresponds to q = p/9 (20 degrees) cannot we constructed directly.
However, is is easy to divide the segment on the base line (Beckmann's OA) into three equal
parts.  Drawing vertical lines from the division points to intersect the quadratrix then gives angles
of 20 degrees and 40 degrees!
More generally, once you have a quadratrix, the problem of dividing an angle into n equal parts
is reduced to the easier problem of dividing a line segment into n equal parts. 
Lecture topic 9: Using the quadratrix to "square the circle"
Well, not exactly, but we can use it to find two lines segment, one of which is p times as long as the other.
The explanation in Beckmann is correct, but it can be made much simpler.  Try using the diagram
below.

For a circle of radius r, a point
L on the quadratrix is defined
by the intersection of two lines
where = 
or q = 
Now y = x cot q
so the equation of the quadratrix
is y =  as on page 43
of Beckmann.
Take the limit as x approaches 0.
Since the limit of is 1,
the limit of y is . 

So the diameter AD is p times longer than the segment OQ.

Lecture topic 10: Vieta's infinite product by geometry
This approach is presented on pp. 92-93 of Beckmann, A History of Pi
Show (again) that the area of an inscribed polygon of n sides in a circle of radius r is
A(n) = n r2/2 sin(2p/n).  Then show that 
Starting with n = 4 (a square) show that 
Explain how to calculate all the necessary cosines by extracting square roots.
Lecture topic 11: Vieta's product by trigonometry and induction
This is a more "modern" version of the same argument that never mentions geometry explicitly.
Start with any angle x.  By using sin x = 2 sin (x/2) cos (x/2) ,
prove by induction that for any value of n,
sin x = 2n 
Remember, this means that you may assume the formula is true for n in showing that it is true for n+1.
Show that when n is very large,  is well approximated by x
Now choose x = p/2 to get Vieta's formula
p = 
Lecture topic 12: Biography of Vieta.
For a Huguenot in 16th century France, "publish or perish" had to to be taken fairly literally!  Report
on the mathematical and political careers of Vieta in the context of the Reformation and religious
strife in France.
Lecture topic 13: Vieta in Mathematica
Use Mathematica to get a good approximation to p by Vieta's formula.  Since really good trig 
tables did not exist in the 16th century, you should compute the cosines by using a half-angle formula.
It's OK just to compute the first few approximate values, but if you have any experience in simple
programming you may be able to use the Mathematica function For[] to automate the process.
If you are willing to use Mathematica's Cosine[] function you can also use the function Product[]
to get the answer for arbitrary n.