### Problem Description

Archimedes made significant contribution to mathematics when he developed an
iterative method for calculating the value of Pi with greater
precision than then existed. This program satisfies my curiosity about how
he did it.

### Background & Techniques

Here is an illustrated description of Archimedes' method for estimating Pi
developed a couple of hundred years BC. Starting with super-scribed and
inscribed hexagons for a given circle, he calculated the perimeters and used the
ratio of each perimeter to the diameter of the given circle as an estimate of
Pi. Since the circumscribed perimeter will be larger than the circumference of
the circle, the derived estimate of Pi will be too large. Similarly the
inscribed perimeter will be too small and produce a low estimate. As the number
of sides increases, the two estimates converge towards Pi.

Archimedes started with what was probably the largest number of sides for which
he knew the actual ratio of the side lengths to the diameter, a hexagon. He then
cleverly derived the ratios for 12 sided regular polygons and

epeated the same logic for 24, 48, and 96 sided polygons. Most current
explanations of Archimedes procedure, cheat and use trigonometry but here we'll
do it the way Archimedes did it using only algebra and geometry. The English
translation of Proposition 3 in "The Measure of a Circle" used here is from "The
Works of Archimedes" translated from Greek by T.L. Heath and published in 1897.
A free download is available from the link below.

After 4 iterations, Archimedes determined that the true value of Pi was greater
than 3 10/70 and less that 3 10/71. Decimal terminology had not yet been
developed so he used rational fraction estimates for irrational values, starting
with estimates for the square root of 3, but making sure that the error
introduced only made his estimate worse than if he could have used the true
value. Pretty smart fellow!

In the tabs above, I'll try to describe the "clever" part in getting from the
known perimeter of the hexagon to the perimeter of the 12 sided polygon.
Actually, he only needed to concentrate on the length of one side and what
happens to side lengths if we bisect N sides to create a polygon with 2N sides.
The technique for interior and exterior side lengths both use Euclid's Angle
Bisector Theorem: "The angle bisector any angle in a triangle divides the
opposite side in the same ratio as the sides adjacent to the angle." It is from
Euclid's Book VI, Proposition 3 and, as heath did, we'll refer to it as EUCL
VI.3 in the following explanations. Archimedes tended to omit intermediate steps
and explanations in his proofs. Heath added some explanatory text in his
translation and I added a bit more where the rational was not clear to a
jack-leg "mathematician" such as myself :>).

Each click of the "Next Polygon" button will show the next doubling of the
original 6 sided circumscribed and inscribed polygons here. Two additional pages
will describe the logic used to derive the interior and exterior side lengths.

##### Circumscribed Detail

On this page, I'll follow Archimedes' derivation of the Pi estimate for
circumscribed polygons. Heath placed his explanations in [ ] brackets
which I have tried to retain. I've added more explanation for the parts that
weren't as obvious to me as they must have been to Archimedes and Heath.

For the exterior hexagon, he considers the triangle formed by connecting point
O, the circle's center to point A, the point of tangency of a side, to point C,
the end of that side. He knows that angle AOC is 1/3 of a right angle (30
degrees to us). He also knows that, for such a triangle (which we commonly call
a 30-60-90 triangle),

Archimedes then says:

(1) the ratio of OA:AC [= sqrt(3):1] >265:152 and

(2) the OC:AC ratio i[= 2:1] = 306:153.

Me: From this we can derive an estimate of Pi as the perimeter divided by the
diameter (twice the length of OA ) or 12*AC/(2*OA) = 6*(AC/OA)=6*1/sqrt(3)=3.464
or, for Archimedes since decimal notation did not yet exist, six times the ratio
of the radius to the semi-perimeter = 6*265/153. We'll use this same concept to
track the values for the rest of the exercise.

Archimedes: First, draw OD bisecting the angle AOC and meeting AC in D.

Me: Draw line OD bisecting angle AOC at point D {so that now AD represents half
the length of one side of a 12 sided polygon.

Archimedes:

Now CO:OA=CD:DA [By EUCL VI.3]

So that [(CO+OA):OA = CA:DA or]

(CO+OA):CA = OA:DA

Me: I had to step through it this way:

CO/OA=CD:DA

CO/OA+1 = CD/DA +1

==> CO/OA+OA/OA = CD/DA + DA/DA

==> (CO+OA)/OA = (CD+DA)/DA

==> [(CO+OA)/OA = CA/DA]

Multiply both sides of the above by OA/CA to get

==> (CO+OA):CA = OA:DA

Archmedes: Therefore [by (1) and (2)] OD/DA > 571/153 (3)

Me: The estimate for Pi, for the 12 sided polygon, if A. had bothered to record
it, is

Pi < 12*AD/OA=12*153/571 = 3.215

Archimedes:

Hence sqr(OD)/sqr(AD) [= (sqr(OA) + sqr(AD))/sqr(AD)] > (sqr(571 )+ sqr(153)) /
sqr(153)

> 349350/23409,

So that OD/DA > 591 1/8 / 153 (4)

Me: We now have a ratio for the triangle AOD which we can bisect again, giving
Archimedes' point E for the 24 sided polygon and giving an OA/AE ratio of
1162.125/153 and a Pi estimate of 24*153/1162.125=3.160.

Similarly Archimedes uses AOE to calculate AOF for the 48 sided case, he gets
2334 1/4 : 153 and we can calculate Pi < 153*48/2334.25 = 3.1462.

An finally, from AOG, the 96 side case, he gets 4673 1/2 : 153 and Pi <
153*96/4673.5 = 3.1428 which he expresses as 3 1/7.

Inscribed Detail

Calculating Pi estimates for the inscribed polygons seemed trickier than for
circumscribed. Again, I'll expand the Archimedes proof to add my explanations of
things which may not be obvious to us mere mortals.

Archimedes: Let AB be the diameter of a circle and let AC meeting the circle at
C make the angle CAB equal to one third of a right angle. Join BC.

Then AC:CB [sqrt(3):1)] < 1351 :780.

Me: This time the estimate of sqrt(3) is less than the true value so we can be
sure that it does not make the estimate, which will always be low, closer than
the rest of the mathematics indicate.

We can make our first estimate of Pi from the known ratios From the given
triangle; we know that the ratio of the diameter (AB) to one side of the hexagon
(CB) is 1/2 or so our initial (low) estimate of the value of Pi is 6/2 or 3.00.

Archimedes: First let AD bisect the angle BAC and meet BC in d and the circle in
D. Join BD.

Then angle BAD = angle dAC = angle dBD and the angles at D and C are both right
angles.

It follows that the triangles ADB [ACd], and BDd are similar.

Me: BAD = dAC by construction . Angles at D and C are right angles because they
are angles inscribed in a semicircle which are always right angles. (Search Web
for "angle inscribed semicircle" for many proofs). The vertical angles at d are
equal. So angle dAC = 180-90-Cda = 180-90-DdB = DBd, The three triangles are
simlar all have the same three angle values so are similar by definition.

Archimedes:

Therefore AD : DB = BD : Dd [=AC : Cd] =AB : Bd [Eucl. VI.3] = AB + AC : Bd + Cd
= AB + AC : BC or BA + AC : BC = AD : DB

Me: It might be obvious to geometers, but it wasn't clear to me that if AC:Cd=AB:Bd
then it also true that AC:Cd = (AB+AC):(Cd+Bd). Here's my proof:

AC/Cd=AB/Bd = some constant, k. Then AC= k*Cd and AB = k*Bd. Adding these
equations, we get AC+AB=k*Cd+k*Bd, so AC+AB= k*(Cd+Bd) and (AC+AB)/(Cd+Bd)=k.
===> (AC+AB) : (Cd+Bd) = k = AC:Cd

Archimedes: But AC:CB < 1351:750 from above while BA:BC = 2:1 = 1550:780

Therefore AD:DB < 2911 : 780 (1) [Hence sqr(AB):sqr(BD) < sqr(2911) + sqr(780) :
sqr(780) < 9082321:608400] Thus AB:BD < 3013 3/4: 780 (2)

Me: This is enough to give us our estimate for the 12 sided incribed polygons.
Because AB is the hypotenuse of right triangle ABD, AB:DB less than the
calculated value 3013 3/4:780 and Pi > 12* (780/3013.75) = 3.1057

Now the hard thinking is done and with the help of a calculator or computer,
even most of the work is done. Archimedes bisects DAB with AE to get the 24
sided ratios. AB:BE < 1838 9/11: 240 (4) giving the 24 sided case estimate: Pi >
24*(240/1838.818) = 3.1325. Similarly

48 side case: AB:BF < 1009 1/6: 66 and Pi > 48*66/1009.167 = 3.1392 and

96 side case: AB:BG < 2017 1/6: 66 and Pi > 96*66/2017.167 - 3.1410

Non-programmers are welcome to read on, but may want to jump to bottom of
this page to download the executable program now.

### Programmer's Notes:

### Running/Exploring the Program