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# Approximations in drawing π and squaring the circle

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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.
Afterwards, 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.

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### Approximations in drawing π and squaring the circle

1. 1. Elsevier Editorial System(tm) for Journal of Mathematical Analysis and ApplicationsManuscript DraftManuscript Number:Title: Approximations in drawing π and squaring the circleArticle Type: Regular ArticleSection/Category: MiscellaneousKeywords: Lindemann-Weierstrass, pi (π), squaring, circle, subset-sum problem, NP-complete, NP-hardCorresponding Author: Mr. Chris De Corte,Corresponding Authors Institution: KAIZY BVBAFirst Author: Chris De CorteOrder of Authors: Chris De CorteAbstract: In this document, I will explain how one can theoretically draw π (pi), up to any desiredaccuracy, using only a ruler and a compass.Afterward, I will show the easiest way to approximate the squaring of a circle.
2. 2. Approximations indrawing π and squaring the circleBy: Chris De CorteCover Letter
3. 3. Approximations in drawing π and squaring the circleAbstractIn this document, I will explain how one can theoretically draw π (pi), up to anydesired 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 transcendentalnumber, 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 πhence to square the circle.Key-wordsLindemann–Weierstrass, pi (π), squaring, circle, subset-sum problem, NP-complete, NP-hardIntroductionIn 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 compassonly.Methods & Techniques1. Approximating pi (π)Using simple geometry, it is easy to draw the following basic starting angles: 72°(pentagram), 60°, 45°. Lets call them αi.The above basic angles can be split n times in 2, each time using a simplebisection, 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 thex- or y-axis to obtain the cosine, sinus and even the tangent.*Manuscript
4. 4. So, I will list a limited list of all these possible angles in a spreadsheet softwareand also calculate the respective trigonometric values of them. Then I will sortthe values from high to low and manually start selecting the best fit values toconstruct 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 digitsaccuracy.Because of possible rounding inaccuracies in the spreadsheet software, I didntmake attempts of higher accuracy than 8 digits but this should not be a problemwith the right software.
5. 5. Out of the above spreadsheet, we can deduct following approximations for π:Approximation 1:Approximation 2:Approximation 3:
6. 6. 2. Squaring the circleUsing the above method to determine and draw π, it is clear that there will be aninfinite number of ways to square any circle up to a desired accuracy.However, during my attempts, I discovered a particularly easy way toapproximate 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 asecond 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 squarethat 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 asgood as unnoticeable by the naked eye. This is because to be correct, thesecond circle should have had radius sqrt(π/2) which is 1.2533141 while weapproximated it to be 1.25.
7. 7. ResultsDrawing π up to any agreed number of digits is theoretically possible using only acompass 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 limitedlist of basic angles followed by addition and subtraction of projections. Nomultiplication, division, squares, roots or nested roots are used of the projections.So, many more complicated scenarios are possible.I think that it is clear from the above spreadsheet and from the simple examplesthat:1. approximations to any more digits accuracy is theoretically possible. Foreach extra digit in accuracy that is wanted in π, we will always find acombination 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 thedesired accuracy in this way (as we can always increase one of thecoefficients of the last approximation and recalculate the other coefficientsso that the new approximation again achieves the desired accuracy).
8. 8. 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 Ferdinandvon Lindemann also include the use of (x=) transcendentals to come to anew transcendental (π)? I wouldnt think so because otherwise, thepolynomial π = 1 * π would proof the contrary. Suppose this reasoning iscorrect, how could one be sure that there does not exist a luckycombination that will work out true exactly?2. If the answer to the above question that there is definitively no solutionthat results in an exact value for π remains valid, then, which formula (withcombinations of addition, subtraction, division and multiplication of nestedroots and squares of basic angles and powers of bisected angles) willcome most close to π?I think the finding of the answer to either of the above 2 questions has similaritiesto the subset-sum problem (given a set of integers, is there a non-empty subsetwhose sum is zero? ). The subset-sum problem is a NP-complete problem. Ourproblem however looks like much harder to solve (since there are many moreoperations possible than sum only and they could be nested and the accuracyshould reach to an infinite amount of digits) and hence this problem could be anNP-hard problem. The only thing that makes it not similar to other NP problems isthe fact that the verification of a proposed solution might not be easy as weshould then check an infinite amount of digits.Conjectures:1. the number of ways to draw π up to an agreed number of digits is infinite2. the listing of the set of ways and finding the best approximation is an NPhard problemConclusion:1. The drawing of π is theoretically possible to any desired number of digits2. Squaring the circle can be fairly approximated using a simple techniqueAcknowledgementsI would like to thank my wife for keeping the faith in me during all these nightsthat I spend at my desk instead of in my bed.References
9. 9. 1. De zeven grootste raadsels van de wiskunde;Alex van den Brandhof,Roland van der Veen, Jan van de Craats, Barry Koren; Uitgeverij BertBakker2. Unknown Quantity; John Derbyshire; Atlantic Books3. Geometric Constructions; Lesley Lamphier; Iowa State University4. Geometrical Constructions; Mathematics is not a Spectator Sport; PhillipsG., Springer5. Fundamentals of Geometry; Oleg A. Belyaev3. http://en.wikipedia.org/wiki/P_versus_NP_problem4. http://en.wikipedia.org/wiki/Subset_sum_problem5. http://en.wikipedia.org/wiki/Pentagon