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CONVERGENCE of the RATIO of PERIMETER of a REGULAR POLYGON to the LENGTH of ITS LONGEST DIAGONAL AS the NUMBER of SIDES of POLYGON APPROACHES to ∞ Pawan Kumar B.K

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CONVERGENCE of the RATIO of PERIMETER of a REGULAR POLYGON to the LENGTH of ITS LONGEST DIAGONAL AS the NUMBER of SIDES of POLYGON APPROACHES to ∞ Pawan Kumar B.K CONVERGENCE OF THE RATIO OF PERIMETER OF A REGULAR POLYGON TO THE LENGTH OF ITS LONGEST DIAGONAL AS THE NUMBER OF SIDES OF POLYGON APPROACHES TO ∞ Pawan Kumar B.K. Kathmandu, Nepal Corresponding to: Pawan Kumar B.K., email: [email protected] ABSTRACT Regular polygons are planar geometric structures that are used to a great extent in mathematics, engineering and physics. For all size of a regular polygon, the ratio of perimeter to the longest diagonal length is always constant and converges to the value of 휋 as the number of sides of the polygon approaches to ∞. The purpose of this paper is to introduce Bishwakarma Ratio Formulas through mathematical explanations. The Bishwakarma Ratio Formulae calculate the ratio of perimeter of regular polygon to the longest diagonal length for all possible regular polygons. These ratios are called Bishwakarma Ratios- often denoted by short term BK ratios- as they have been obtained via Bishwakarma Ratio Formulae. The result has been shown to be valid by actually calculating the ratio for each polygon by using corresponding formula and geometrical reasoning. Computational calculations of the ratios have also been presented upto 30 and 50 significant figures to validate the convergence. Keywords: Regular Polygon, limit, 휋, convergence INTRODUCTION A regular polygon is a planar geometrical structure with equal sides and equal angles. For a 푛(푛−3) regular polygon of 푛 sides there are diagonals. Each angle of a regular polygon of n sides 2 푛−2 is given by ( ) 180° while the sum of interior angles is (푛 − 2)180°. The angle made at the 푛 center of any polygon by lines from any two consecutive vertices (center angle) of a polygon of 360° sides 푛 is given by . It is evident that each exterior angle of a regular polygon is always equal 푛 to its center angle. The ratio of the perimeter to the longest diagonal or diameter is characteristic feature of any 180° regular polygon. For a regular polygon of even number of sides, the ratio is given by 푛sin ( ), 푛 90° while for a regular polygon of odd number of sides the ratio is given by 2푛푠푛 ( ). As the 푛 number of sides of a regular polygon becomes infinitesimally large, i.e. 푛 → ∞, the resulting polygon is called regular apeirogon which resembles a circle. The regular polygon at this state has countably infinite number of equal edges. The ratio of the perimeter to the longest diagonal 푠푛휃 reduces to 퐶 , where 퐶 is circumference of the resembling circle, 휃 is angle opposite to the 푝 perpendicular arm and 푝 is perpendicular arm of the right angled triangle whose hypotenuse is 푠푛휃 diameter of the circle. The expression 퐶 provides exactly 휋. Angles are used both in degrees 푝 and in radians as required. For inclusion of all regular polygons equilateral triangle poses a significant hurdle. It has no recognizable diagonal or diameter. The ratio of the perimeter of the equilateral triangle to the length of one of its sides is taken in this case which is equal to 3. The Bishwakarma Ratio Formulae for the ratio of perimeter to the longest diagonal length can be summarized as follows: I. f(n) = 3, for n = 3 II. f(n) = nsin (180°), for n is an even number, n ≥ 4 n III. f(n) = 2nsin (90°), for n is an odd number, n ≥ 5 n sinθ IV. f(n) = C , for n → ∞ p CALCULATION OF BK RATIOS FOR SOME REGULAR POLYGONS 1. For n=3, representing an equilateral triangle 푡표푡푎푙 푝푒푟푚푒푡푒푟 표푓 푒푞푢푙푎푡푒푟푎푙 푡푟푎푛푔푙푒 3a = = 3. 푎 푠푑푒 푙푒푛푔푡ℎ a 2. For a regular polygon with even number of sides Square (푛 = 4), Hexagon (푛 = 6), Octagon (푛 = 8) are the basic examples of this types of polygons. The length of the longest diagonal (푑) of these regular polygons is given by a 180° , where ‘a’ is the length of the side of the polygon. If a regular polygon has even 푠푛( ) 푛 number of edges, then length of the longest diagonal of that polygon will be hypotenuse of a right angled triangle for which one of the sides of thepolygon is the perpendicular side as illustrated in figure. From figure 1, we get 180° a a 푠푛 ( ) = . 푒. 푑 = 180° 푛 푑 푠푛 ( ) 푛 Figure 1: Illustration of the longest diagonal as the perpendicular of a right angled triangle ILLUSTRATION I The values of length of the longest diagonal for some regular polygons of have been geometrically presented below. The calculation of the ratio of the perimeter of the corresponding polygon to the length of the longest diagonal has been presented alongside. I. Square (n=4) 180° a a 푠푛 ( ) = . 푒. 푑 = = (√2 )a 4 푑 푠푛45° From geometry, the ratio of the perimeter of the square to the length of its longest diagonal is 푝푒푟푚푒푡푒푟 표푓 푡ℎ푒 푠푞푢푎푟푒 4a = = 2√2 푙푒푛푔푡ℎ 표푓 푡ℎ푒 푙표푛푔푒푠푡 푑푎푔표푛푎푙 (√2)a Using formula, we get 180° 180° 푓(푛) = 푛 푠푛 ( ) = 4푠푛 ( ) = 4 푠푛 45° = 2√2 푛 4 II. Regular Hexagon (n=6) 180° a 푠푛 ( ) = . 푒. 푑 = 2a 6 푑 From geometry, 푝푒푟푚푒푡푒푟 표푓 푡ℎ푒 ℎ푒푥푎푔표푛 6a = = 3 푙푒푛푔푡ℎ 표푓 푡ℎ푒 푙표푛푔푒푠푡 푑푎푔표푛푎푙 2a Using formula, we get 180° 180° 푓(푛) = 푛 sin ( ) = 6 sin ( ) = 3 푛 6 III. Regular Decagon (n=10) 180° a a 푠푛 ( ) = . 푒. 푑 = 10 푑 푠푛18° From geometry, 푝푒푟푚푒푡푒푟 표푓 푡ℎ푒 푑푒푐푎푔표푛 10a = = 3.09016 푙푒푛푔푡ℎ 표푓 푡ℎ푒 푙표푛푔푒푠푡 푑푎푔표푛푎푙 a 푠푛18° Using formula, we get 180° 180° 푓(푛) = 푛푠푛 ( ) = 10푠푛 ( ) = 3.09016 푛 10 180° Similarly, we can use푓(푛) = 푛푠푛 ( ) formula for all regular polygons having even number 푛 180° of edges. Therefore, 푓(푛) = 푛푠푛 ( ) is a valid formula for given domain. 푛 3. For a regular polygon with odd number of sides Pentagon (푛 = 5), Heptagon (푛 = 7), Nonagon (푛 = 9) are the basic examples of this types of polygons. a The length of the longest diagonal 푑 of these regular polygons is given by 90° , ( ) 2 sin 푛 where 푎 is the length of the side of the polygon. ILLUSTRATION 2 The values of length of the longest diagonal for some regular polygons have been geometrically presented below. The calculation of the ratio of the perimeter of the corresponding polygon to the length of the longest diagonal has been presented alongside. I. Regular Pentagon (n = 5) Diagonals from D and E are drawn to B such that it subtends an angle 훼, and creates equal angles 훽 on either side of the angle 훼.Similarly,diagonals AC and AD subtend an angle 훼 at A creating equal angles 훽 on either side of the angle 훼. Side AB is extended upto F and AN ⊥ CD such that ∆ACD and ∆BDE are iscosceles and ∆CAN and ∆DAN are right angled triangles From △ABD, we get (α+ β) + (α+ β) + α = 180° (sum of interior angles of a triangle) 3α + 2β = 180°…………………….(i) On straight line ABF, we get ∠ABE + ∠EBD + ∠ DBC + ∠CBF = 180° (angle add up on a straight line) 360° i.e. β + α+ β + = 180° 푛 360° ∴ 2 β = 180°- (α + )…………………(ii) 푛 From equation (i) and (ii), we have 360° 180°- (α + ) = 180° - 3α 푛 360° ∴ α = ……………………(iii) 2푛 In ∆ACD,line AN divides ∆ACD into equal two parts a a ∠CAN= ∠DAN= and CN = ND = 2 2 From right angled triangles ∆CAN and ∆DAN, AC=AD, both AC and AD are longest diagonal (d) for pentagon and hypotenuse for right angled triangles ∆CAN and ∆DAN respectively. a a 푎 ⁄2 ⁄2 푠푛 ( )= i.e. AC = d = 푎 2 퐴퐶 푠푛( ) 2 a ∴ AC = 푎 ………………… (iv) 2푠푛( ) 2 From equation (iii) and (iv), we get a a AC = 360° 1 = 90° 2푠푛( × ) 2푠푛( ) 2푛 2 푛 From Geometry, 푝푒푟푚푒푡푒푟 표푓 푡ℎ푒 푝푒푛푡푎푔표푛 5a = = 3.090169 … 푙표푛푔푒푠푡푑푎푔표푛푎푙푙푒푛푔푡ℎ a 90° ( ) 2푠푛 5 Using formula, we get 90° 90° 푓(푛) = 2푛푠푛 ( ) = 10 sin ( ) = 3.090169 … 푛 5 II. Regular Heptagon(n=7) From both vertex angle A and B of Heptagon an angle ‘α’ taken from the middle so that by symmetry all angles marked ‘β’ are equal in diagram. Line AB is extended upto H, ∆ADE and ∆BEF are isosceles, and ∆DAN and ∆EAN are right angled triangle Let ∠ AEB= ∠DAE = ∠EBF = α(isosceles triangles), then From ∆ABE, we get (α+ β) + (α+ β) + α = 180° {sum of interior angles of a triangle} ∴ 3α + 2β = 180°…………………….(i) On straight line ABH, we get ∠ABF + ∠FBE + ∠ EBC + ∠CBH = 180° 360° i.e. β + α+ β + = 180° 푛 360° ∴ 2 β = 180°- (α + )…………………(ii) 푛 From equation (i) and (ii), we get 360° 180° − (훼 + ) = 180° − 3훼 푛 360° ∴ α = ……………………(iii) 2 푛 In ∆ADE,line AN divides ∆ADE into equal two parts a a ∠DAN= ∠EAN= andDN = NE = 2 2 From right angled triangles ∆DAN and ∆EAN, AD=AE, both AD and AE are longest diagonal (d) for heptagon and hypotenuse for right angled triangles ∆DAN and ∆EAN respectively. From right angled triangle ∆DAN, we get a a 푎 ⁄2 ⁄2 푠푛 ( ) = . 푒. 퐴퐷 = 푑 = 푎 2 퐴퐷 푠푛 ( ) 2 a ∴ AD= 푎 ………………… (iv) 2푠푛( ) 2 From equation (iii) and (iv), we get a a AD= 360° 1 = 90° 2푠푛( × ) 2푠푛( ) 2푛 2 푛 Thus, this beautiful rule holds all the regular polygons having odd number of sides. Since n=7 for heptagon, we get perimeter of the heptagon 7푎 90° = a = (2 × 7)푠푛 ( ) = 3.115293 … 7 푙표푛푔푒푠푡 푑푎푔표푛푎푙 푙푒푛푔푡ℎ 90° 2푠푛( ) 푛 Using formula, we get 90° 90° 푓(푛) = 2푛푠푛 ( ) = (2 × 7)푠푛 ( ) = 3.115293 … 푛 7 Similarly, this formula gives an accurate ratio of perimeter to the diagonal of all polygons having an odd number of sides.
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