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

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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 APPROACHES TO Pw Kumr BK Kthmdu, Nepl Correspodig to: Pw Kumr BK, emil: bishwkrm2020@gmilcom ABSTRACT Regulr polygos re plr geometric structures tht re used to gret extet i mthemtics, egieerig d physics For ll size of regulr polygo, the rtio of perimeter to the logest digol legth is lwys costt d coverges to the vlue of π s the umber of sides of the polygo pproches to The purpose of this pper is to itroduce Bishwkrm Rtio Formuls through mthemticl expltios The Bishwkrm Rtio Formule clculte the rtio of perimeter of regulr polygo to the logest digol legth for ll possible regulr polygos These rtios re clled Bishwkrm Rtios- ofte deoted by short term BK rtios- s they hve bee obtied vi Bishwkrm Rtio Formule The result hs bee show to be vlid by ctully clcultig the rtio for ech polygo by usig correspodig formul d geometricl resoig Computtiol clcultios of the rtios hve lso bee preseted upto 30 d 50 sigifict figures to vlidte the covergece Keywords: Regulr Polygo, limit, π, covergece INTRODUCTION A regulr polygo is plr geometricl structure with equl sides d equl gles For regulr polygo of sides there re ( 3) digols Ech gle of regulr polygo of sides 2 is give by ( 2 ) 180 while the sum of iterior gles is ( 2)180 The gle mde t the ceter of y polygo by lies from y two cosecutive vertices (ceter gle) of polygo of sides is give by 360 It is evidet tht ech exterior gle of regulr polygo is lwys equl to its ceter gle The rtio of the perimeter to the logest digol or dimeter is chrcteristic feture of y regulr polygo For regulr polygo of eve umber of sides, the rtio is give by si ( 180 ), while for regulr polygo of odd umber of sides the rtio is give by 2si ( 90 ) As the umber of sides of regulr polygo becomes ifiitesimlly lrge, ie, the resultig polygo is clled regulr peirogo which resembles circle The regulr polygo t this stte

2 hs coutbly ifiite umber of equl edges The rtio of the perimeter to the logest digol reduces to C siθ, where C is circumferece of the resemblig circle, θ is gle opposite to the p perpediculr rm d p is perpediculr rm of the right gled trigle whose hypoteuse is dimeter of the circle The expressio C siθ provides exctly π Agles re used both i degrees p d i rdis s required For iclusio of ll regulr polygos equilterl trigle poses sigifict hurdle It hs o recogizble digol or dimeter The rtio of the perimeter of the equilterl trigle to the legth of oe of its sides is tke i this cse which is equl to 3 The Bishwkrm Rtio Formule for the rtio of perimeter to the logest digol legth c be summrized s follows: I f() = 3, for = 3 II f() = si ( 180 ), for is eve umber, 4 III f() = 2si ( 90 ), for is odd umber, 5 IV f() = C siθ p, for CALCULATION OF BK RATIOS FOR SOME REGULAR POLYGONS 1 For =3, represetig equilterl trigle totl perimeter of equilterl trigle side legth = 3 = 3 2 For regulr polygo with eve umber of sides Squre ( = 4), Hexgo ( = 6), Octgo ( = 8) re the bsic exmples of this types of polygos The legth of the logest digol (d) of these regulr polygos is give by, si( 180 where is the legth of the side of the polygo If regulr polygo hs eve ) umber of edges, the legth of the logest digol of tht polygo will be hypoteuse of

3 right gled trigle for which oe of the sides of thepolygo is the perpediculr side s illustrted i figure From figure 1, we get si ( 180 ) = d i e d = si ( 180 ) Figure 1: Illustrtio of the logest digol s the perpediculr of right gled trigle ILLUSTRATION I The vlues of legth of the logest digol for some regulr polygos of hve bee geometriclly preseted below The clcultio of the rtio of the perimeter of the correspodig polygo to the legth of the logest digol hs bee preseted logside I Squre (=4) si ( ) = d i e d = = ( 2 ) si45 From geometry, the rtio of the perimeter of the squre to the legth of its logest digol is perimeter of the squre = 4 = legth of the logest digol ( 2) 2 2 Usig formul, we get f() = si ( 180 II Regulr Hexgo (=6) ) = 4si (180 ) = 4 si 45 = si ( ) = d i e d = 2 From geometry, perimeter of the hexgo legth of the logest digol = 6 2 = 3

4 Usig formul, we get f() = si ( 180 ) = 6 si (180 6 ) = 3 III Regulr Decgo (=10) si ( ) = d i e d = si18 From geometry, perimeter of the decgo legth of the logest digol = 10 si18 = Usig formul, we get f() = si ( 180 ) = 10si (180 ) = Similrly, we c usef() = si ( 180 ) formul for ll regulr polygos hvig eve umber of edges Therefore, f() = si ( 180 ) is vlid formul for give domi 3 For regulr polygo with odd umber of sides Petgo ( = 5), Heptgo ( = 7), Nogo ( = 9) re the bsic exmples of this types of polygos The legth of the logest digol d of these regulr polygos is give by, 2 si( 90 ) where is the legth of the side of the polygo ILLUSTRATION 2 The vlues of legth of the logest digol for some regulr polygos hve bee geometriclly preseted below The clcultio of the rtio of the perimeter of the correspodig polygo to the legth of the logest digol hs bee preseted logside I Regulr Petgo ( = 5) Digols from D d E re drw to B such tht it subteds gle α, d cretes equl gles β o either side of the gle αsimilrly,digols AC d AD

5 subted gle α t A cretig equl gles β o either side of the gle α Side AB is exteded upto F d AN CD such tht ACD d BDE re iscosceles d CAN d DAN re right gled trigles From ABD, we get (α+ β) + (α+ β) + α = 180 (sum of iterior gles of trigle) 3α + 2β = 180 (i) O stright lie ABF, we get ABE + EBD + DBC + CBF = 180 (gle dd up o stright lie) ie β + α+ β = β = (α ) (ii) From equtio (i) d (ii), we hve (α ) = 180-3α α = (iii) I ACD,lie AN divides ACD ito equl two prts CAN= DAN= 2 d CN = ND = 2 From right gled trigles CAN d DAN, AC=AD, both AC d AD re logest digol (d) for petgo d hypoteuse for right gled trigles CAN d DAN respectively si ( )= 2 ie AC = d = 2 2 AC AC = 2si( 2 si( 2 ) ) (iv) From equtio (iii) d (iv), we get AC = 2si( 360 = ) 2si( 90 )

6 From Geometry, perimeter of the petgo logestdigollegth = 5 2si( 90 5 ) = Usig formul, we get f() = 2si ( 90 ) = 10 si (90 5 ) = II Regulr Heptgo(=7) From both vertex gle A d B of Heptgo gle α tke from the middle so tht by symmetry ll gles mrked β re equl i digrm Lie AB is exteded upto H, ADE d BEF re isosceles, d DAN d EAN re right gled trigle Let AEB= DAE = EBF = α(isosceles trigles), the From ABE, we get (α+ β) + (α+ β) + α = 180 {sum of iterior gles of trigle} 3α + 2β = 180 (i) O stright lie ABH, we get ABF + FBE + EBC + CBH = 180 ie β + α+ β = β = (α ) (ii) From equtio (i) d (ii), we get 180 (α ) = 180 3α α = (iii)

7 I ADE,lie AN divides ADE ito equl two prts DAN= EAN= 2 ddn = NE = 2 From right gled trigles DAN d EAN, AD=AE, both AD d AE re logest digol (d) for heptgo d hypoteuse for right gled trigles DAN d EAN respectively From right gled trigle DAN, we get si ( 2 ) = 2 AD i e AD = d = 2 si ( ) 2 AD= 2si( (iv) ) 2 From equtio (iii) d (iv), we get AD= 2si( ) = 2si( 90 ) Thus, this beutiful rule holds ll the regulr polygos hvig odd umber of sides Sice =7 for heptgo, we get perimeter of the heptgo logest digol legth = 7 2si( 90 ) = (2 7)si ( 90 7 ) = Usig formul, we get f() = 2si ( 90 ) = (2 7)si (90 7 ) = Similrly, this formul gives ccurte rtio of perimeter to the digol of ll polygos hvig odd umber of sides 4 For represetig peirogo For give perimeter, if become very lrge (or coutbly ifiite) the regulr polygo becomes regulr peirogo Regulr peirogo is ultimte form of regulr polygo d it hs coutbly ifiite umber of equl edges Fct: The rtio of the perimeter of this polygo to the legth of its lrgest digol is clculted by the limitig vlue s d it is clculted to be π

8 ANALYTICAL PROOF We hve two fcts: I 180 = π rdis six II limf(x) = lim = 1 x 0 x For is eve umber, we get lim f() = lim si (180 ) lim f() = lim si (π ) π π si ( π lim f() = lim ) π π = 1 π = π For is odd umber, we get lim f() = lim = 2 si (90 ) lim f() = lim si ( π 2 ) 2 π π si ( π lim f() = lim ) π 2 π 2 = 1 π = π ANALOGY TO CIRCLE By logy to regulr polygos with lrge umber of edges, regulr peirogo resembles circle I the figure right, we c see tht C is circumferece of circle,θ is gle opposite to the perpediculr rm d p is legth of the perpediculr rm of the right gled trigle whose hypoteuse is dimeter of the circle If two lies from dimetriclly opposite poits of the circle itersect t y poit o the circumferece the gle formed o the - circumferece by these two lies is lwys 90

9 ILLUSTRATION From ABC, we get siθ = p d p ie d = d the rtio is the give by siθ lim f() = c d = C p siθ lim f() = C si θ = π p = C si θ p COMPUTATIONAL ILLUSTRATION Figure2: covergece of the BK rtio s the umber of sides of regulr polygo icreses

10 Figure 2: Fluctutios of the BK rtio for the umber of sides of regulr polygo from 3 to 250 Figure 2 shows the covergece of the rtios s the polygo umber icreses However, t lower polygo umber there re sigifict fluctutios i the vlue of the rtio As the polygo umber exceeds 250 the fluctutio is pproximtely egligible d pproches the vlue of π APPENDIX The umericl vlues of the rtios hve bee preseted below: Seril umber Nme of regulr polygo Bishwkrm(BK) Rtio BK rtio with precisio upto 30 d 50 sigifict figures 1 Equilterl trigle Squre f() = 4 si( ) = 4si Petgo f() = 2 5si( 90 ) = 10si Hexgo f()= 6 si 30 5 Heptgo f()= 14si(90 /7)

11 6 Octgo f()= 8si(180 /8) 7 Nogo f()= 18si 10 8 Decgo f()= 10si 18 9 Udecgo f()= 22si(90 /11) 10 Dodecgo f()= 12si Tridecgo f()= 26si(90 /13) 12 Tetrdecgo f()= 14si(180 /14) 13 Petdecgo f()= 30si 6 14 Hexdecgo f() = 16si(180 /16) 15 Heptdecgo f() = 34si(90 /17) 16 Octdecgo f() = 18si Eedecgo f() = 38si(90 /19) 18 Icosgo f() = 20si(180 /20) 19 Icosihego f()= 42si(90 /21) 20 Icosikidigo f()= 22si(180 /22) 21 Icosikitrigo f() = 46si(90 /23) 22 Icosikitetrgo f()= 24si(180 /24) 23 Icosikipetgo f()= 50si(90 /25) 24 Icosikihexgo f()= 26si(180 /26) 25 Icosikiheptgo f()= 54si(90 /27) 26 Icosikioctgo f()= 28si(180 /28) 27 Icosikieego f()= 58 si(90 /29) 28 Tricotgo f()= 30si 6 29 Tricotkihego f()= 62si(90 /31)

12 30 Tricotkidigo f()= 32si(180 /32) Tricotkitrigo f()= 66si(90 /33) Tricotkitetrgo f()= si(180 /34) Tricotkipetgo f()= si(90 /70) Tricotkihexgo f()= 36si Tricotkiheptgo f()= 74si(90 /37) Tricotkioctgo f()= 38si(180 /38) Tricotkieego f()= 78si(90 /39) Tetrcotgo f()= 40si(180 /40) Tetrcotkihego f()= 82si(90 /41) Tetrcotkidigo f()= 42si(180 /42) Tetrcotkitrigo f()= 86si(90 /43) Tetrcotkitetrgo f()= 44si(180 /44) Tetrcotkipetgo f()= 90si Tetrcotkihexgo f()= 46si(180 /46) Tetrcotkiheptgo f()= 94si(90 /47) Tetrcotkioctgo f()= 48si(180 /48) Tetrcotkieego f()= 98si(90 /49) Petcotgo f()= 50si(180 /50) Petcotkihego f()= 102si(90 /51) Petcotkidigo f()= 52si(180 /52) Hectoheg f()= 202si(90 /101) Hectokidigo f()= 102si(180 /102)

13 198 Dihectogo f() = 200si(180 /200) Dihectohego f() = 402si(90 /201) Dihectokidigom f() = 202si(180 /202) Pethectogo f() = 500si(180 /500) Pethectohego f() = 1002si(90 /501) Pethectokidigo f() = 502si(180 /502) Chilligo f() = 1000si(180 /1000) Chillihego f() = 2002si(90 /1001) Chilliidigo f() = 1002si(180 /1002) Petkischilligo f() = 5000si(180 /5000) Petkischillihego f() = 10002si(90 /5001) Petkischillidigo f() = 5002si(180 /5002) Myrigo f() = 10000si(180 /10000) Myrihego f() = 20002si(90 /10001) Myrikidigo f() = 10002si(180 /10002) Cetchilligo f() = si(180 /100000) Cetchillihego f() = si(90 /100001) Cetchillikidigo f() = si(180 /100002) Meggo f() = si(180 / ) 6341

14 Meghego f() = si(90 / ) Megkidigo f() = si(180 / ) Tergo(10 12 sides) f() = si(180 /10 12 ) ( )-go f() = 2( )si(90 /( )) ( )-go f() = ( )si(180 /( )) Petgo(10 15 sides) f() = si(180 /10 15 ) ( )-go f() = 2( )si(90 /( )) ( )-go f() = ( )si(180 /( )) Regulr Apeirogo (Precise Circle) It is ultimte regulr polygo lim f() = lim si (180 ) = π lim f() = lim 2si (90 ) = π Refereces [1] Poskitt K (1997), Murderous Mths: The Logest Digol Formul, Scholstic, Uited Kigdom [2] Artm B (1999), Euclid Books IV: Regulr Polygos I: Euclid-The cretor of Mthemtics, Spriger, New York, doi: / _11

15 [3]TÓth LF (1964), Regulr Figures, Elsevier, Netherlds, doi: /C [4] Che F (2001), Regulr Polygo, Volume I: Applied New Theory of Trisectio to Costruct Regulr Heptgo for Ceturies i the History of Mthemtics, Itertiol Sciece Mth & Scieces Istitut [5] Ftuer R (2017) Regulr Polygo Tesselltios Activity Book, Tesselltios, Uited Sttes

LEVEL I. ,... if it is known that a 1

LEVEL I. ,... if it is known that a 1 LEVEL I Fid the sum of first terms of the AP, if it is kow tht + 5 + 0 + 5 + 0 + = 5 The iterior gles of polygo re i rithmetic progressio The smllest gle is 0 d the commo differece is 5 Fid the umber of