Cycloidal Areas without Calculus

Transcription

1 TOM M APOSTOL ad MAMIKON A MNATSAKANIAN Cycloidal Areas without Calculus Itroductio For ceturies atheaticias have bee iterested i curves that ca be costructed by siple echaical istruets Aog these curves are various cycloids used by Apolloius aroud 00 BC ad by Ptoley aroud 00 AD to describe the apparet otios of plaets The siplest cycloid is the curve traced out by a poit o the circuferece of a circular dis that rolls without slippig alog a horizotal lie; it fors a sequece of arches restig o the lie, as show i Figure Let S deote the area of the regio above the lie ad below oe of these arches (show shaded i Figure ) A routie use of itegral calculus reveals that S is three ties the area of the rollig circular dis, which we express sybolically as follows: () S J The derivatio of this forula usig itegral calculus requires paraetric or Cartesia equatios for the cycloid This paper solves the ore geeral proble i which the rollig circle is replaced by ay regular polygo The result is obtaied by a geoetrical ethod, ad the area forula for the cycloid is obtaied as a liitig case We use the forula for the area of a circular sector, but there is o eed to ow the equatios represetig the cycloid Cyclogos Whe a regular polygo rolls without slippig alog a straight lie, a give vertex o its circuferece traces out a curve we call a cyclogo Lie the cycloid, a cyclogo cosists of a sequece of arches restig o the lie, as show by the exaple of a rollig petago i Figure Each arch, i tur, is coposed of circular arcs, equal i uber to oe fewer tha the uber of vertices of the polygo The arcs eed ot have the sae radius Figure Cycloid traced out by a poit o the circuferece of a rollig circle If S deotes the area of the regio above the lie ad below oe of these arches, we will show that, i place of (), we have the elegat ad surprisig result J () S +, where deotes the area of the rollig polygo ad J is the area of the dis that circuscribes the polygo The circle ca be regarded as the liitig case obtaied by lettig the uber of edges icrease without boud i a regular polygo Siilarly, the cycloid is the liitig case of a cyclogo Equatio () for the area of the regio uder oe arch of a cycloid is ow revealed as a liitig case of Equatio () We begi with two siple exaples, a rollig triagle, ad a rollig square ollig equilateral triagle Figure shows oe arch of a cyclogo traced out by a rollig equilateral triagle whose edges have legth a The regio uder this arch ad above the lie cosists of two equal circular sectors of radius a ad oe equilateral triagle Each circular sector has area (π/)a which is also the area of the circular dis that circuscribes a equilateral triagle of edge-legth TOM M APOSTOL, Eeritus Professor of Matheatics at Caltech, is Director of Project Matheatics! MAMIKON MNATSAKANIAN has devised rearable geoetric iovatios for teachig atheatics Figure Cyclogo traced out by a vertex o the boudary of a rollig regular petago Math Horizos Septeber 999

2 Figure Oe arch of a cyclogo traced out by a rollig equilateral triagle a Therefore π J S area of + a area of +, which proves () i this case ollig square Figure shows a cyclogo traced out by a rollig square The regio uder this arch cosists of two right triagles plus three circular quadrats, two of radius a (the edge-legth of the square), ad oe of radius a (the diagoal of the square) The two right triagles have total area a, the area of the rollig square, ad the total area of the three circular quadrats is π π J a ca h π a + F H G I K J Therefore we have S a + J, which proves () i this case as well Figure Oe arch of a cyclogo traced out by a rollig square 5 ollig -go I the geeral case of a regular polygo with vertices, the regio uder oe arch of the cyclogo cosists of triagles ad circular sectors, each subtedig a agle of π/ radias Figure 6 Triagular dissectio of a regular hexago ad distributio of its footprits The triagles ca be regarded as footprits left by the triagular pieces obtaied by dissectig the origial polygo with diagoals fro a give vertex to each of the oadjacet vertices, as illustrated i Figure 5 for 6 The su of the areas of these triagles is equal to the area of the regio eclosed by the regular polygo This is illustrated for the regular hexago i Figure 6 The radii of the circular sectors are the legths of the segets fro oe vertex to each of the reaiig vertices A sector of radius r subtedig a agle of π/ radias has area πr /, so the su of the areas of the sectors is equal to π r I the ext sectio we will show that the su of the squares of these radii is equal to, where is the radius of the circle that circuscribes the polygo Therefore the su of the areas of the sectors is equal to π, which is twice the area of the circuscribig dis I other words, () is a cosequece of the relatio () r 6 A extesio of the Pythagorea Theore for regular polygos The result i (), which is eeded to calculate the su of the areas of the circular sectors i the foregoig sectio, is of Figure 5 Footprits left by triagular pieces of a rollig hexago Math Horizos Septeber 999

3 idepedet iterest because it reduces to the Pythagorea theore whe the regular polygo is a square The authors have ot bee able to locate this surprisig theore i ay published wor, so it ay be ew Theore The su of the squares of the segets draw fro oe vertex of a regular -go to the reaiig vertices is equal to, where is the radius of the circuscribig circle 7 Proof for regular polygos with a eve uber of sides The proof for eve aes repeated use of the Pythagorea Theore It is illustrated for the case 0 i Figure 7, which shows ie segets draw fro oe vertex of a regular decago to the other ie vertices I Figure 7 there are four segets, labeled as r, r,, r, ad four irror iages, plus the diaeter of legth, so the su i questio is () r + r r r r r r Figure 7 Nie segets draw fro oe vertex of a regular decago to the other ie Figure 8 shows two segets fro the opposite extreity of the diaeter to cosecutive vertices By syetry with respect to a horizotal diaeter, these segets have legths r ad r The ew seget r eets the old seget r o the circle ad, together with the diaeter, fors a right triagle with hypoteuse (Here we use the fact that ay triagle iscribed i a seicircle is a right triagle with the diaeter as hypoteuse) Applyig the Pythagorea Theore to this right triagle we fid (5) r + r Siilarly, the ew seget r itersects the old seget ad fors aother right triagle with hypoteuse Applyig the Pythagorea Theore oce ore we fid (6) r + r so the su i () is equal to r which proves the Theore for 0 I the geeral case of eve, oe of the segets is the diaeter of the circuscribig circle, ad the other segets for ( )/ pairs syetrically located with respect to the diaeter The sae arguet just give for the case 0 shows that ( ) / r + + c h which proves the theore for every eve This proof does ot wor if is odd 8 Proof for regular polygos with a odd uber of sides A differet ethod that applies to all regular polygos with a odd uber of sides is illustrated for a regular heptago i Figure 9 The three segets ad their irror iages i the diaeter are the 6 segets draw fro oe vertex of a regular heptago to the other 6 vertices We wish to prove that c r + r + h, or that (7) r + r + r 7 We apply the law of cosies to each of three isosceles triagles i Figure 9 havig a vertex at the ceter of the heptago, two edges of legth ad base of legth r, where,, The correspodig vertex agles are θ, where θ π/7, θ π/7, θ 6π/7 The law of cosies for the isosceles triagle with vertex agle θ states that Figure 8 earrageet of segets r ad r i Figure 7 r r r Math Horizos Septeber 999

4 r (8) r cos θ so the su of these equatios gives us (9) r 6 cos θ But, by a trigooetric idetity described below i (), the su of cosies is equal to ½, so (9) iplies (7) I the geeral case of a regular polygo with + sides we wish to prove that (0) r ( + ) I this case we apply the law of cosies to isosceles triagles, usig (8) for each of these triagles, where ow θ π/( + ) Istead of (9) we have the equatio () r ( ) cos θ I this case we have the followig idetity (which we prove below i Sectio 9), () cos θ, so () reduces to (0) 9 Origi of the trigooetric idetity () The trigooetric idetity i () ca be writte as () cos θ + 0 r r Figure 9 egular heptago with edge r iscribed i a circle of radius where θ π/( + ) This is a cosequece of a ore geeral trigooetric idetity si θcos( + ) θ () cos( θ), si θ which holds for ay positive iteger ad ay θ that is ot a iteger ultiple of π (See Exercise, p 06, of Apostol s Calculus, Vol I, d ed, Joh Wiley & Sos, Ic, 967) If we tae θ π/ the right eber vaishes ad () becoes (5) cos π F 0 H G I K J Whe is odd, say + the last ter i the su is equal to The reaiig ters ca be arraged i pairs, by couplig the ters with ad, which have the sae cosie Cosequetly (5) ca be writte as π I cos 0 HG + K J +, which is the sae as () F Note The foregoig ethod, usig the law of cosies, also wors if the polygo has a eve uber of sides, say + sides, but oe ior chage is eeded There are ow + segets fro a give vertex to the reaiig vertices Oe of these is a diaeter, ad the other ca be arraged i pairs by couplig each seget with its irror iage i that diaeter However, we do ot give further details because the proof preseted i Sectio 7 is ore eleetary 0 Alterate proof for a regular petago A alterate proof for a regular petago ca be give by a ethod that is of iterest because it aes use of Ptoley s rearable theore o cyclic quadrilaterals (quadrilaterals iscribed i a circle) Ptoley s theore states that, for ay cyclic quadrilateral, the product of the legths of the diagoals is equal to the su of the products of the legths of opposite sides (A siple proof of Ptoley s theore ca be foud i the Worboo that accopaies the videotape Sies & Cosies, Part III, produced by Project MATHEMATICS!, Caltech, 99 The videotape also gives a coputer aiated versio of this proof) Figure 0 shows a regular petago ibedded i a regular decago with edges of legth r The segets r ad r ad their irror iages i a diaeter are the four segets draw fro oe vertex of the regular petago to the other four vertices We are to prove that r + r + r + r 0 or the equivalet stateet (6) r + r 5 Apply Ptoley s theore to the cyclic quadrilateral i Figure 0 with two itersectig diagoals of legth r to obtai (7) r rr + r Math Horizos Septeber 999 5

5 r r r r r Figure 0 A cyclic quadrilateral with three edges r, oe edge, ad two diagoals r Next, refer to the two siilar isosceles triagles show i Figure, ad equate ratios of correspodig sides to get /r /, or (8) rr Figure The isosceles triagle with equal edges ad base is siilar to that with equal edges ad base r Substitute (8) i (7) ad the use the Pythagorea relatio r + r ( ) to obtai r + r + r 5 r, which iplies (6) 6 Math Horizos Septeber 999