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

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

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

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 ( 180 4 ) = 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 = 2 2 4 si ( 180 6 ) = d i e d = 2 From geometry, perimeter of the hexgo legth of the logest digol = 6 2 = 3

Usig formul, we get f() = si ( 180 ) = 6 si (180 6 ) = 3 III Regulr Decgo (=10) si ( 180 10 ) = d i e d = si18 From geometry, perimeter of the decgo legth of the logest digol = 10 si18 = 309016 Usig formul, we get f() = si ( 180 ) = 10si (180 ) = 309016 10 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

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 + 360 = 180 2 β = 180 - (α + 360 ) (ii) From equtio (i) d (ii), we hve 180 - (α + 360 ) = 180-3α α = 360 2 (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 = 2 1 2 ) 2si( 90 )

From Geometry, perimeter of the petgo logestdigollegth = 5 2si( 90 5 ) = 3090169 Usig formul, we get f() = 2si ( 90 ) = 10 si (90 5 ) = 3090169 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 + 360 = 180 2 β = 180 - (α + 360 ) (ii) From equtio (i) d (ii), we get 180 (α + 360 ) = 180 3α α = 360 2 (iii)

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( 360 2 1 2 ) = 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 ) = 3115293 Usig formul, we get f() = 2si ( 90 ) = (2 7)si (90 7 ) = 3115293 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

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

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

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 3 30000000000000000000000000 0000 2 Squre f() = 4 si( 180 4 ) = 4si45 28284271247461900976033774 4842 3 Petgo f() = 2 5si( 90 ) = 10si18 30901699437494742410229341 5 7183 30000000000000000000000000 4 Hexgo f()= 6 si 30 5 Heptgo f()= 14si(90 /7) 0000 31152930753884016600446359 0295

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 15 11 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 10 17 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) 30614674589207181738276798 7224 31256671980047462793308992 8185 30901699437494742410229341 7183 31309264420122730897634387 0956 31058285412302491481867860 5149 31339536866383993870757598 7377 31152930753884016600446359 0295 31358538980296041419950246 4407 31214451522580522855725578 9563 31371242217522678381481376 4325 31256671980047462793308992 8185 31380151279486283348130695 0102 31286893008046173802021063 8934 31386639306298186802194693 2086 31309264420122730897634387 0956 31391510147748648923746700 3503 31326286132812381971617494 6949 31395259764656688038089112 2828 31339536866383993870757598 7377 31398207611656947410923929 0974 31350053308926200371237661 8762 31400566979542165166946829 0652 31358538980296041419950246 4407 31402484680001881612826151 6408

30 Tricotkidigo f()= 32si(180 /32) 31365484905459392638142580 4444 31 Tricotkitrigo f()= 66si(90 /33) 31404064443669916316859581 0608 32 Tricotkitetrgo f()= si(180 /34) 31371242217522678381481376 4325 33 Tricotkipetgo f()= si(90 /70) 31405381245360447820663235 7613 34 Tricotkihexgo f()= 36si 5 31376067389156942480903137 5015 35 Tricotkiheptgo f()= 74si(90 /37) 31406490365149746349771198 7755 36 Tricotkioctgo f()= 38si(180 /38) 31380151279486283348130695 0102 37 Tricotkieego f()= 78si(90 /39) 31407433285343811822324491 1686 38 Tetrcotgo f()= 40si(180 /40) 31383638291137978013184098 3974 39 Tetrcotkihego f()= 82si(90 /41) 31408241625828986018504475 8480 40 Tetrcotkidigo f()= 42si(180 /42) 31386639306298186802194693 2086 41 Tetrcotkitrigo f()= 86si(90 /43) 31408939829586598072905648 4846 42 Tetrcotkitetrgo f()= 44si(180 /44) 31389240607662229744028611 1695 43 Tetrcotkipetgo f()= 90si 2 31409547032250874481395663 4628 44 Tetrcotkihexgo f()= 46si(180 /46) 31391510147748648923746700 3503 45 Tetrcotkiheptgo f()= 94si(90 /47) 31410078387214096622748496 1792 46 Tetrcotkioctgo f()= 48si(180 /48) 31393502030468672071351468 2121 47 Tetrcotkieego f()= 98si(90 /49) 31410546020222070748879514 8387 48 Petcotgo f()= 50si(180 /50) 31395259764656688038089112 2828 49 Petcotkihego f()= 102si(90 /51) 31410959727293760952056192 8758 50 Petcotkidigo f()= 52si(180 /52) 31396818659588747787770416 4220 99 Hectoheg f()= 202si(90 /101) 31414660079108765500514002 4123 100 Hectokidigo f()= 102si(180 /102) 31410959727293760952056192 8757

198 Dihectogo f() = 200si(180 /200) 31414634623641351506590706 6198 199 Dihectohego f() = 402si(90 /201) 31415606760582980684791262 4424 200 Dihectokidigom f() = 202si(180 /202) 31414660079108765500514002 4123 498 Pethectogo f() = 500si(180 /500) 31415719827794756248676550 7897 499 Pethectohego f() = 1002si(90 /501) 31415875064885465734491186 7315 500 Pethectokidigo f() = 502si(180 /502) 31415721471587027737755298 9356 998 Chilligo f() = 1000si(180 /1000) 31415874858795633519332270 3549 999 Chillihego f() = 2002si(90 /1001) 31415913642417427419638104 5095 1000 Chilliidigo f() = 1002si(180 /1002) 31415875064885465734491186 7315 4998 Petkischilligo f() = 5000si(180 /5000) 31415924468812861167262676 6426 4999 Petkischillihego f() = 10002si(90 /5001) 31415926019333303442935539 1453 5000 Petkischillidigo f() = 5002si(180 /5002) 31415924470465537519715259 6898 9998 Myrigo f() = 10000si(180 /10000) 31415926019126656929793464 7928 9999 Myrihego f() = 20002si(90 /10001) 31415926406730947731331068 8367 10000 Myrikidigo f() = 10002si(180 /10002) 31415926019333303442935539 1453 99998 Cetchilligo f() = 100000si(180 /100000) 31415926530730219604831480 2075 99999 Cetchillihego f() = 200002si(90 /100001) 31415926534606030027806206 1722 100000 Cetchillikidigo f() = 100002si(180 /100002) 31415926530730426307141571 8326 999998 Meggo f() = 1000000si(180 / 31415926535846255256825959 1000000) 6341

999999 Meghego f() = 2000002si(90 /1000001) 1000000 Megkidigo f() = 1000002si(180 / 1000002) 10 12-2 Tergo(10 12 sides) f() = 10 12 si(180 /10 12 ) 10 12-1 (10 12 +1)-go f() = 2(10 12 +1)si(90 /(10 12 +1)) 31415926535885013128514835 6440 31415926535846255463533850 7120 31415926535897932384626382 1556 31415926535897932384626420 9135 10 12 (10 12 +2)-go f() = (10 12 +2)si(180 /(10 12 +2)) 31415926535897932384626382 15566722854897991273463 10 15-2 Petgo(10 15 sides) f() = 10 15 si(180 /10 15 ) 10 15-1 (10 15 +1)-go f() = 2(10 15 +1)si(90 /(10 15 +1)) 31415926535897932384626433 832743351714171194293458 31415926535897932384626433 832782109560021569094516 10 15 (10 15 +2)-go f() = (10 15 +2)si(180 /(10 15 +2)) 31415926535897932384626433 832743351714171194500168 Regulr Apeirogo (Precise Circle) It is ultimte regulr polygo lim f() = lim si (180 ) = π 31415926535897932384626433 lim f() = lim 2si (90 ) = π 832743351714171194500168 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: 101007/978-1-4612-1412-0_11

[3]TÓth LF (1964), Regulr Figures, Elsevier, Netherlds, doi: 101016/C2013-0-01735-5 [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