<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">APM</journal-id><journal-title-group><journal-title>Advances in Pure Mathematics</journal-title></journal-title-group><issn pub-type="epub">2160-0368</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/apm.2022.129041</article-id><article-id pub-id-type="publisher-id">APM-120068</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  A Method for the Squaring of a Circle
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Lyndon</surname><given-names>O. Barton</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>Delaware State University, Dover, USA</addr-line></aff><pub-date pub-type="epub"><day>07</day><month>09</month><year>2022</year></pub-date><volume>12</volume><issue>09</issue><fpage>535</fpage><lpage>540</lpage><history><date date-type="received"><day>19,</day>	<month>July</month>	<year>2022</year></date><date date-type="rev-recd"><day>24,</day>	<month>September</month>	<year>2022</year>	</date><date date-type="accepted"><day>27,</day>	<month>September</month>	<year>2022</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  This paper presents a Method for the squaring of a circle (
  <em>i.e.</em>, constructing a square having an area equal to that of a given circle). The construction, when applied to a given circle having an area of 12.7 cm
  <sup>2</sup>, it produced a square having an area of 12.7 cm
  <sup>2</sup>, 
  using only an unmarked ruler and a compass. This result was a clear demonstration that not only is the construction valid for the squaring of a circle but also for achieving absolute results (independent of the number pi (π) and in a finite number of steps) when carried out with precision.
 
</p></abstract><kwd-group><kwd>Famous Problems in Mathematics</kwd><kwd> Archimedes</kwd><kwd> College Mathematics</kwd><kwd> Cycloidal Construction</kwd><kwd> Mean Proportional Principle</kwd><kwd> Squaring the Circle</kwd><kwd> Quadrature</kwd><kwd> Geometer’s Sketch Pad</kwd><kwd> College Geometry</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>The quadrature of a circle (i.e., the squaring of a circle or finding the square whose area is exactly equal to that of a given circle) is one of the three famous geometric problems that have intrigued mathematicians for centuries, dating back to the days of the ancient Greeks such as Hippocrates, Plato, and the creator of the number pi (π), Archimedes. The other two famous problems are the trisection of an angle and the doubling of a cube [<xref ref-type="bibr" rid="scirp.120068-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.120068-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120068-ref3">3</xref>].</p><p>Despite tireless efforts on the part of these mathematicians, this problem not only has remained unsolved until now, but proofs have been offered notably by Ferdinand Von Lindermann (1882) and others, who applied algebraic methods to geometry to show that the problem cannot be solved using a straightedge (or unmarked ruler) and a compass alone. The basis for that conclusion is the fact that the number pi (π), a function of the area of a circle, is not an algebraic number, nor is it constructible by a straightedge and a compass.</p><p>Despite the formidability of the quadrature problem, the construction presented in this paper will demonstrate how a construction, using only an unmarked straightedge and a compass, can be developed to produce, in a finite number of steps, a square whose area is equivalent to that of a given circle.</p></sec><sec id="s2"><title>2. Example Problem</title><p>Given a circle having an area of 12.7 cm<sup>2</sup>, construct a square whose area is equivalent to that of the circle, using an unmarked straightedge and compass only.</p><p>Procedure</p><p>Since the procedure being presented is based on the mechanics of the rolling wheel, it is, therefore, instructive to proceed with the following steps:</p><p>1) Develop a typical cycloid construction for a circle in <xref ref-type="fig" rid="fig1">Figure 1</xref>.</p><p>This construction illustrates the periodic locations of a point on the circumference of the given circle as it completes one revolution while rolling without slipping on a flat surface [<xref ref-type="bibr" rid="scirp.120068-ref4">4</xref>]. For example, if one were to imagine a bug that gets attached to the rim of a wheel at point A, which is coincident with station L0; then, as the wheel rolls on towards the right, the points 0, 1, 2, 3, 4, 5, and 6 represent the positions of the bug on the wheel at stations L0, L1, L2, L3, L4, L5, and L6 (or point B), where the bug is reunited with the surface. Therefore, the distance covered by the bug between contacts with the flat surface is equivalent to the circumference of the circle, which is denoted by the straight line AB.</p><p>It should be noted that while points 1, 2, 3, 4, 5, and 6 are necessary for the layout of a cycloidal path, in this construction, they are required here only for determining the distance covered, AB, between the points of contact after one revolution, Hence, there is no need for any additional tool besides the straightedge and compass for this construction.</p><p>For further details on cycloid construction, see reference [<xref ref-type="bibr" rid="scirp.120068-ref4">4</xref>].</p><p>2) Convert the circumference of the circle to a rectangle. See <xref ref-type="fig" rid="fig1">Figure 1</xref>.</p><p>Knowing that the circumference is 2πr and the area of a circle is πr<sup>2</sup>, then we can think of this area as being represented by the rectangle ABCD where the side AB is known to be 2πr, and the other side BC is unknown but can be determined since this distance is related to the circumference of the circle.</p><p>3) Determine BC</p><p>BC is found by equating the area of the rectangle to the area of the circle.</p><p>AB &#215; BC = π r 2 Equation (1)</p><p>which yields</p><p>BC = π r 2 2 π r</p><p>BC = r/2 … half radius.</p><p>Therefore, from Equation (1), the rectangle ABCD can be written as</p><p>AB &#215; r/2 = Area of given circle Equation (2)</p><p>(2πr) &#215; r/2 = Area of given circle Equation (3)</p><p>4) Determine the equivalent rectangular area to that of the circle</p><p>Applying Equation (3) to the problem at hand, given that the area of the circle is equal to 12.7 cm<sup>2</sup>, this area can be represented as a rectangle ABCD, Figure2, where side AB = 12.7 cm (See FigureA1 for construction), and the other side BC = 1 cm.</p><p>5) Convert the rectangle ABCD to a square, using the Mean Proportional Principle [<xref ref-type="bibr" rid="scirp.120068-ref5">5</xref>] [<xref ref-type="bibr" rid="scirp.120068-ref6">6</xref>]. See <xref ref-type="fig" rid="fig3">Figure 3</xref>.</p><p>a) Referring to rectangle ABCD, with center at C and radius CB, describe an arc cutting DC (ext’d) at K.</p><p>b) With KD as the base, construct a semicircle cutting BC (extended) at point F.</p><p>This point will define on line BC (extended) at F, there defining the segment CF as one side of the required square.</p><p>c) Complete the required square C I J F.</p></sec><sec id="s3"><title>3. Proof</title><p>The proof of this construction is in the final measurements of the rectangle ABCD and square C I J F.</p><p>Final Construction Results</p><p>Area of Circle = 12.7 cm<sup>2</sup> (Given);</p><p>Area of Rectangle = 12.7 cm<sup>2</sup>;</p><p>Area of square = 12.7 cm<sup>2</sup>.</p></sec><sec id="s4"><title>4. To Summarize</title><p>Given the area of a circle, the quickest way to convert this area to that of a square is to do the following:</p><p>1) Form a rectangle ABCD, as in the present example, with one side AB = 12.7 cm and the other side BC = 1 cm, so that the area of the rectangle formed agrees with the given area, which is 12.7 cm<sup>2</sup>.</p><p>2) Then, using the Mean Proportional Principle [<xref ref-type="bibr" rid="scirp.120068-ref6">6</xref>] [<xref ref-type="bibr" rid="scirp.120068-ref7">7</xref>], convert the rectangle to a square.</p><p>In the example problem presented here where the segment AB = 12.7 cm is required, this construction is done graphically as shown in FigureA1. As follows:</p><p>1) Construct two segments—one AM that is 12 cm long and the other MO that is 1 cm long</p><p>2) Divide segment MO into 10 equal parts (See reference [<xref ref-type="bibr" rid="scirp.120068-ref5">5</xref>] on “Dividing A Line Segment Into Equal Parts,”)</p><p>3) Select segment MN that is 7/10 th of MO, and add it to the 12 cm segment AM.</p></sec><sec id="s5"><title>5. Practical Benefits</title><p>Apart from the mere satisfaction of an academic interest, which the quadrature of a circle problem has generated for centuries, there are several practical benefits which the construction presented here offers.</p><p>As with all other graphical solutions, it offers an alternative to the analytical approach normally used to solve the problem and, in so doing, it provides additional insight into a problem solution that otherwise may be too theoretical or abstract.</p><p>In the field of engineering mechanics, particularly that of kinematic analysis and synthesis, many graphical solutions rely on geometric constructions. In this context, the quadrature construction could be a part of a more complex graphical procedure, without which such a procedure cannot be considered totally graphical. Also, in the field of engineering mechanics, the construction can provide a means to make useful comparisons between area-dependent relationships such as those involving stress, flow and moment of inertia.</p><p>In addition, the construction provides a means whereby the number pi (π) can be generated as a precision scale for making or checking measurements that contain pi (π) as a factor.</p><p>Finally, what is significant to note is that it is difficult to foresee at this time the various ramifications that the solution of this problem could bring. But already, one could see that the area of a circle does not need to be expressed solely as a function of pi (π). That is, once the equivalent square for the circle is found, one can quickly determine the other side of an area equivalent rectangle once one side is specified.</p></sec><sec id="s6"><title>6. Summary</title><p>A graphical method for constructing a square that is equivalent in area to that of a given circle, using only an unmarked straightedge and a compass, has been presented. The construction, when applied to a circle with a given area of 12.7 cm<sup>2</sup>, it produced a square with an area of 12.7 cm<sup>2</sup>, which is equivalent to that of the given circle. This equivalency clearly validates the logic of the method, which likes any mathematical formula; it guarantees that the required quadrature is achievable provided the construction is carried out with precision.</p><p>To be specific, despite the formidability of this age-old challenge, the construction presented has produced a square that is equivalent in area to that of a given circle, using an unmarked straightedge and compass alone. Also, by circumventing the use of the number pi (π), the construction has made it possible to achieve a complete solution in a finite number of steps (actually 5 steps), unlike other attempted solutions that this author has encountered in the literature. Therefore, for these reasons, one can only conclude that, finally, the age-old challenge of squaring the circle has been met. [<xref ref-type="bibr" rid="scirp.120068-ref1">1</xref>].</p><p>NOTE:</p><p>The use of the Geometer’s Sketch Pad [<xref ref-type="bibr" rid="scirp.120068-ref7">7</xref>] was solely for the layout of arcs and lines involved in building a presentable construction and not for measurements except for determining final results. Hence, its use is not a violation of the unmarked straightedge and compass rule.</p></sec><sec id="s7"><title>Conflicts of Interest</title><p>The author declares no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s8"><title>Cite this paper</title><p>Barton, L.O. (2022) A Method for the Squaring of a Circle. Advances in Pure Mathematics, 12, 535-540. https://doi.org/10.4236/apm.2022.129041</p></sec><sec id="s9"><title>Appendix</title></sec></body><back><ref-list><title>References</title><ref id="scirp.120068-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Bennet, D. (2002) Exploring Geometry with Geometer’s Sketch Pad. Key Curriculum Press, Emeryville.</mixed-citation></ref><ref id="scirp.120068-ref2"><label>2</label><mixed-citation publication-type="journal" xlink:type="simple"><name name-style="western"><surname>Barton</surname><given-names> L.O. </given-names></name>,<etal>et al</etal>. 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