CONVERGENCE OF THE RATIO OF PERIMETER OF A REGULAR POLYGON TO THE LENGTH OF ITS LONGEST DIAGONAL AS THE NUMBER OF SIDES OF POLYGON INCREASES

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1 CONVERGENCE OF THE RATIO OF PERIMETER OF A REGULAR POLYGON TO THE LENGTH OF ITS LONGEST DIAGONAL AS THE NUMBER OF SIDES OF POLYGON INCREASES Pawa Kumar Bishwakarma Idepedet Researcher Correspodig Author: bishowkarmapawa2020@gmail.com ABSTRACT A regular polygo is a plaar geometrical structure with all sides of equal legth ad all agles of equal magitude. The ratio of perimeter of ay regular polygo to the legth of its logest diagoal is a costat term ad the ratio coverges to the value of as the umber of sides of the polygo icreases. The result has bee show to be valid by actually calculatig the ratio for each polygo by usig correspodig formula ad geometrical reasoig. A computatioal calculatio of the ratio has also bee preseted to validate the covergece. The values have bee calculated up to 30 sigificat digits. Keywords: Regular Polygo, covergece INTRODUCTION A regular polygo is a plaar geometrical structure with equal sides ad equal agles. For a regular polygo of sides there are diagoals. Each agle of a regular polygo of sides is give by while the sum of iterior agles is. The agle made at the ceter of ay polygo by lies from ay two cosecutive vertices (ceter agle) of a polygo of sides is give by. It is evidet that each exterior agle of a regular polygo is always equal to its ceter agle. The ratio of the perimeter to the logest diagoal or diameter is characteristic feature of ay regular polygo. For a regular polygo of eve umber of sides the ratio is give by [ ], while for a regular polygo of odd umber of sides the ratio is give by [ ]. As the umber of sides of a regular polygo becomes ifiitesimally large, i.e. the resultig polygo is called regular apeirogo which resembles a circle. The regular polygo at this state has coutably ifiite umber of equal edges. The ratio of the perimeter to the logest diagoal reduces to, where is circumferece of the resemblig circle, is agle opposite to the perpedicular arm ad diameter of the circle. is perpedicular arm of the right agled triagle whose hypoteuse is

2 For iclusio of all regular polygos equilateral triagle poses a sigificat hurdle. It has o recogizable diagoal or diameter. The ratio of the perimeter of the equilateral triagle to the legth of oe of its sides is take i this case which is equal to The formulas for the ratio of perimeter to the logest diagoal ca be summarized as I. BK = 3, for =3 II. BK = f() =, -, for is a eve umber III. BK = f() = 2, -, for is a odd umber IV. BK =, for The ratio for idividual regular polygo has bee represeted by a sigle term, BK; BK is short form of Bishwakarma. BK ratio for coveiece. CALCULATION OF BK RATIOS FOR SOME REGULAR POLYGONS 1. For =3, represetig a equilateral triagle 2. For a regular polygo with eve umber of sides, i.e. is a eve umber ad Square, Hexago, Octago are the basic examples of this types of polygos. The legth of the logest diagoal of these regular polygos is give as by [ref. MM], where is the legth of the side of the polygo. If a regular polygo has eve umber of edges, the legth of the logest diagoal of that polygo will be hypoteuse of a right agled triagle for which oe of the sides of the polygo is the perpedicular side as illustrated i figure

3 Figure 1: Illustratio of the logest diagoal as the perpedicular of a right agled triagle ILLUSTRATION I The values of legth of the logest diagoal for some regular polygos of have bee geometrically preseted below. The calculatio of the ratio of the perimeter of the correspodig polygo to the legth of the logest diagoal has bee preseted alogside. I. Square (=4) = The ratio of the perimeter of the square to the legth of its logest diagoal is

4 From geometry, BK = = = 2 2 From formula, BK =, - = 4, - = 4 = 2 2 II. Regular Hexago (=6) = The ratio of the perimeter of the square to the legth of its logest diagoal is From geometry, BK = = = 3 From formula, BK =, -

5 = 6, - = 6 = 3 Regular Decago (=10) The ratio of the perimeter of the square to the legth of its logest diagoal is From geometry, BK = = = From formula, BK =, - = 10, -

6 = 10 = Similarly we ca use BK=, edges. Therefore BK=, - formula for all regular polygos havig eve umber of - is a valid formula for give domai. 3. For a regular polygo with odd umber of sides, i.e. is a odd umber ad Petago, Heptago, Noago are the basic examples of this types of polygos. The legth of the logest diagoal MM], where is the legth of the side of the polygo. of these regular polygos is give as by [ref. ILLUSTRATION 2 The value of legth of the logest diagoal for some regular polygos of have bee geometrically preseted below. The calculatio of the ratio of the perimeter of the correspodig polygo to the legth of the logest diagoal has bee preseted alogside. I. Regular Petago ( = 5) Diagoals from D ad E are draw to B such that it subteds a agle ad creates equal agles o either side of the agle. Similarly diagoals AC ad AD subted a agle at A creatig equal agles o either side of the agle. Side AB is exteded upto F ad AN CD such that Triagle ACD ad BDE are isosceles Tiragle CAN ad DAN are right agled triagles From ABD, (α+ β) + (α+ β) + α = 180 (sum of iterior agles of a triagle) 3α + 2β = 180.(i)

7 O straight lie ABF, ABE + EBD + DBC + CBF = 180 (agle add up o a straight lie) Β + α+ β + = β = (α + ) (ii) From equatio (i) ad (ii) (α + ) = 180-3α α = (iii) I triagle ACD,lie AN divides triagle ACD ito equal two parts CAN= DAN= & CN = ND = /2 From right agled triagle CAN ad DAN, AC=AD, both AC ad AD are logest diagoal(d) for petago ad hypoteuse for right agled triagles CAN ad DAN respectively. = AC = d = AC = (iv) From equatio (iii) ad (iv) AC = = For petago BK= = =

8 From formula, BK = 2, - = 10 [ ] = II. Regular Heptago(=7) From both vertex agle A ad B of Heptago a agle α take from the middle so that by symmetry all agles marked β are equal i diagram. Lie AB exteded upto H Triagle ADE ad BEF are isosceles Tiragle DAN ad EAN are right agled triagle AEB DAE EBF α (isosceles triagles) From ABE triagle, (α+ β) + (α+ β) + α = 180 * um of ter or agle of a tr agle+ 3α + 2β = 180.(i) O straight lie ABH, ABF + FBE + EBC + CBH = 180 Β + α+ β + = β = (α + ) (ii)

9 From equatio (i) ad (ii) (α + ) = 180-3α α = (iii) I triagle ADE,lie AN divides triagle ADE ito equal two parts DAN EAN= ad DN = NE = From right agled triagle DAN ad EAN, AD=AE, both AD ad AE are logest diagoal(d) for heptago ad hypoteuse for right agled triagles DAN ad EAN respectively. Here i right agled triagle DAN = AD= d = AD = ( v) From equatio (iii) ad (iv) AD = = = This beautiful rule holds all the regular polygos havig odd umber of sides Sice =7 for heptago BK= per meter of regular petago = =7 =14 = From formula, BK = 2, - BK = (2 7), - = 14, - =

10 Similarly this formula gives a accurate ratio of perimeter to the diagoal of all polygos havig odd umber of sides. 4. For represetig a apeirogo For a give perimeter, if become very large (or coutably ifiite) the a regular polygo becomes regular apeirogo [ref]. Regular apeirogo is ultimate form of regular polygo ad it has coutably ifiite umber of equal edges. The ratio of the perimeter of this polygo to the legth of its largest diagoal is calculated by the limitig value as The limitig value is calculated to be Proof For polygos of eve umber of sides = l m [ ] For polygos of odd umber of sides = l m, - ANALOGY TO CIRCLE By aalogy to regular polygos with large umber of edges, regular apeirogo resembles a circle. I the figure below we ca see that is circumferece of circle, is agle opposite to the perpedicular arm ad is legth of the perpedicular arm of the right agled triagle whose hypoteuse is diameter of the circle.

11 If two lies from diametrically opposite poits of the circle itersect at ay poit o the circumferece the agle formed o the circumferece by these two lies is always 90. ILLUSTRATION From or, The BK ratio is the give by COMPUTATIONAL ILLUSTRATION Figure 2: Covergece of the BK ratio as the umber of sides of a regular polygo icreases

12 Figure 3: Fluctuatios of the BK ratio for the umber of sides of a regular polygo from 3 to 250 Figure 2 shows the covergece of the BK ratios as the polygo umber icreases. However, at lower polygo umber there are sigificat fluctuatios i the value of the ratio. As the polygo umber exceeds 250 the fluctuatio is approximately egligible ad approaches the value of APPENDIX The umerical value of the BK Ratios have bee preseted below: Serial umber of regular polygo Name of regular polygo BK Ratio 1 Equilateral triagle BK = 3 2 Square 3 Petago 4 Hexago BK = 6 BK =, -=, - = 4si45 BK= 2, -=2, - = 10si18 5 Heptago BK = 14 6 Octago BK = 8 7 Noago BK = 18 BK ratio with precisio up to 30 decimal digits

13 8 Decago BK =10 9 Hedecago/Udeca go BK = Dodecago BK = Tridecago BK = Tetradecago BK = 14si 13 Petadecago BK = Hexadecago BK = Heptadecago BK = Octadecago BK = Eeadecago BK = Icosago BK = Icosiheago BK = Icosikaidigo BK = Icosikaitrigo BK = Icosikaitetrago BK = Icosikaipetago BK = Icosikaihexago BK = Icosikaiheptago BK = Icosikaioctago BK = Icosikaieeago BK = Tricotago BK = Tricotakaiheago BK = Tricotakaidigo BK = Tricotakaitrigo BK = Tricotakaitetrago BK =

14 33 Tricotakaipetago BK = Tricotakaihexago BK = Tricotakaiheptago BK = Tricotakaioctago BK = Tricotakaieeago BK = Tetracotago BK = Tetracotakaiheago BK = Tetracotakaidigo BK = Tetracotakaitrigo BK = Tetracotakaitetrago BK = Tetracotakaipetag BK = 90 o Tetracotakaihexago BK = Tetracotakaiheptag BK = 94 o Tetracotakaioctago BK = Tetracotakaieeag BK = 98 o Petacotago BK = Petacotakaiheag BK = 102 o Petacotakaidigo BK = hectoheaga BK = Hectokaidigo BK = Dihectogo BK = Dihectoheago BK = Dihectokaidigom BK =

15 498 Petahectogo BK = Petahectoheago BK = Petahectokaidigo BK = Chilliago BK = Chilliaheago BK = Chilliaidigo BK = Petakischilliago BK = Petakischilliaheag BK = o Petakischilliadigo BK = Myriago BK = Myriaheago BK = Myriakaidigo BK = Cetachilliago BK = Cetachilliaheago BK = Cetachilliakaidigo BK = Megago BK = Megaheago BK = Megakaidigo BK = Terago(10 12 sides) BK = ( )-go BK = 2( )si(90 /( ))

16 10 12 ( )-go BK = ( )si(180 /( )) Petago(10 15 sides) BK = ( )-go BK = 2( )si(90 /( )) ( )-go BK = ( )si(180 /( )) Regular Apeirogo (Precise Circle) BK= = π It is ultimate regular polygo REFERENCES [1] "Murderous Maths, The LOgest Diagoal Formula!," [Olie]. Available: [2] [Olie]. Available: [3] "Polygos ad Ratios," [Olie]. Available:

GEOMETRY AREA UNIT. Day I Area Review. Triangle Parallelogram Rectangle. A b h. Square Trapezoid Kite. b 2. b 1. 1 Atrapezoid h b1 b2

GEOMETRY AREA UNIT. Day I Area Review. Triangle Parallelogram Rectangle. A b h. Square Trapezoid Kite. b 2. b 1. 1 Atrapezoid h b1 b2 GEOMETRY RE UNIT Day I Review Objectives: SWBT fid the area of various shapes Triagle Parallelogram Rectagle triagle base height b h parallelog ram base height b h recta gle base height b h Square Trapezoid