1) The document explains that while π is proven to be transcendental, it can theoretically be approximated to any desired accuracy using only a compass and ruler by repeatedly bisecting basic angles and using trigonometric functions of those angles. 2) Squaring the circle can also be approximated by drawing two concentric circles with one 1.25 times the radius of the other and connecting the points of intersection, resulting in a square with an error of only 0.26%. 3) Determining the "best" approximation of π or finding an exact expression using combinations of basic operations is conjectured to be an NP-hard problem similar to the subset sum problem.

1. Elsevier Editorial System(tm) for Journal of Mathematical Analysis and Applications
Manuscript Draft
Manuscript Number:
Title: Approximations in drawing π and squaring the circle
Article Type: Regular Article
Section/Category: Miscellaneous
Keywords: Lindemann-Weierstrass, pi (π), squaring, circle, subset-sum problem, NP-complete, NP-
hard
Corresponding Author: Mr. Chris De Corte,
Corresponding Author's Institution: KAIZY BVBA
First Author: Chris De Corte
Order of Authors: Chris De Corte
Abstract: In this document, I will explain how one can theoretically draw π (pi), up to any desired
accuracy, using only a ruler and a compass.
Afterward, I will show the easiest way to approximate the squaring of a circle.

2. Approximations in
drawing π and squaring the circle
By: Chris De Corte
Cover Letter

3. Approximations in drawing π and squaring the circle
Abstract
In this document, I will explain how one can theoretically draw π (pi), up to any
desired accuracy, using only a ruler and a compass.
Afterward, I will show the easiest way to approximate the squaring of a circle.
I am well aware that “Carl Louis Ferdinand von Lindemann (April 12, 1852 –
March 6, 1939) published in 1882 his proof that π (pi) is a transcendental
number, i.e., it is not a root of any polynomial with rational coefficients”.
The reader is warned that following this, it should not be possible to draw π
hence to square the circle.
Key-words
Lindemann–Weierstrass, pi (π), squaring, circle, subset-sum problem, NP-
complete, NP-hard
Introduction
In the first part of this document, I will explain how to approximate pi (π).
Afterwards, I will approximate the squaring of the circle by ruler and compass
only.
Methods & Techniques
1. Approximating pi (π)
Using simple geometry, it is easy to draw the following basic starting angles: 72°
(pentagram), 60°, 45°. Let's call them αi.
The above basic angles can be split n times in 2, each time using a simple
bisection, resulting in n bisections to the new angle αi /2n
.
From all these angles, it is also easy to project the radius of a unit circle onto the
x- or y-axis to obtain the cosine, sinus and even the tangent.
*Manuscript

4. So, I will list a limited list of all these possible angles in a spreadsheet software
and also calculate the respective trigonometric values of them. Then I will sort
the values from high to low and manually start selecting the best fit values to
construct our approximation of π up to the desired number of digits accuracy.
The result can be seen in the below snapshot of a spreadsheet.
I only attempted to construct 3 different approximations of π up to 8 digits
accuracy.
Because of possible rounding inaccuracies in the spreadsheet software, I didn't
make attempts of higher accuracy than 8 digits but this should not be a problem
with the right software.

6. Out of the above spreadsheet, we can deduct following approximations for π:
Approximation 1:
Approximation 2:
Approximation 3:

7. 2. Squaring the circle
Using the above method to determine and draw π, it is clear that there will be an
infinite number of ways to square any circle up to a desired accuracy.
However, during my attempts, I discovered a particularly easy way to
approximate this by using the following method:
Draw a first circle of size unity (1.00) as a start.
Using two bisections of this agreed unity to obtain the length of 0.25, draw also a
second circle of unity times 1.25 from the same origin.
Draw a +45° and -45° line through the origin of both the circles.
The 4 intersections with the second circle will serve as the corners of the square
that will be the approximation of the squared first circle.
The error will only be 0.26% (=(1.2533141-1.25)/1.2533141) which will be as
good as unnoticeable by the naked eye. This is because to be correct, the
second circle should have had radius sqrt(π/2) which is 1.2533141 while we
approximated it to be 1.25.

8. Results
Drawing π up to any agreed number of digits is theoretically possible using only a
compass and a ruler.
Squaring a circle can be easily approximated with the shown method.
Discussions:
The examples for approximating π only uses simple angle bisections on a limited
list of basic angles followed by addition and subtraction of projections. No
multiplication, division, squares, roots or nested roots are used of the projections.
So, many more complicated scenario's are possible.
I think that it is clear from the above spreadsheet and from the simple examples
that:
1. approximations to any more digits accuracy is theoretically possible. For
each extra digit in accuracy that is wanted in π, we will always find a
combination of tan, cosine or sine (no matter close to 0° or 90° angles)
that will make us able to construct the desired accuracy of π.
2. An infinite number of combinations will be possible to approximate π to the
desired accuracy in this way (as we can always increase one of the
coefficients of the last approximation and recalculate the other coefficients
so that the new approximation again achieves the desired accuracy).

9. Two open question remain with regard to the approximation of π:
1. Considering that π is transcendental. But also most of the used sine,
cosine and tangents are also transcendental. Does the proof of Ferdinand
von Lindemann also include the use of (x=) transcendentals to come to a
new transcendental (π)? I wouldn't think so because otherwise, the
polynomial π = 1 * π would proof the contrary. Suppose this reasoning is
correct, how could one be sure that there does not exist a lucky
combination that will work out true exactly?
2. If the answer to the above question that there is definitively no solution
that results in an exact value for π remains valid, then, which formula (with
combinations of addition, subtraction, division and multiplication of nested
roots and squares of basic angles and powers of bisected angles) will
come most close to π?
I think the finding of the answer to either of the above 2 questions has similarities
to the subset-sum problem (given a set of integers, is there a non-empty subset
whose sum is zero? ). The subset-sum problem is a NP-complete problem. Our
problem however looks like much harder to solve (since there are many more
operations possible than sum only and they could be nested and the accuracy
should reach to an infinite amount of digits) and hence this problem could be an
NP-hard problem. The only thing that makes it not similar to other NP problems is
the fact that the verification of a proposed solution might not be easy as we
should then check an infinite amount of digits.
Conjectures:
1. the number of ways to draw π up to an agreed number of digits is infinite
2. the listing of the set of ways and finding the best approximation is an NP
hard problem
Conclusion:
1. The drawing of π is theoretically possible to any desired number of digits
2. Squaring the circle can be fairly approximated using a simple technique
Acknowledgements
I would like to thank my wife for keeping the faith in me during all these nights
that I spend at my desk instead of in my bed.
References

10. 1. De zeven grootste raadsels van de wiskunde;Alex van den Brandhof,
Roland van der Veen, Jan van de Craats, Barry Koren; Uitgeverij Bert
Bakker
2. Unknown Quantity; John Derbyshire; Atlantic Books
3. Geometric Constructions; Lesley Lamphier; Iowa State University
4. Geometrical Constructions; Mathematics is not a Spectator Sport; Phillips
G., Springer
5. Fundamentals of Geometry; Oleg A. Belyaev
3. http://en.wikipedia.org/wiki/P_versus_NP_problem
4. http://en.wikipedia.org/wiki/Subset_sum_problem
5. http://en.wikipedia.org/wiki/Pentagon