CHAPTER1
1. Theanswersaregivenintheanswersectionofthetext.FortheEgyptianhieroglyphics,375isthreehundreds,seventens,andfiveones,while4856isfour thousands,eighthundreds,fivetens,andsixones.ForBabyloniancuneiform,notethat 375=6 × 60+15,while4856=1 × 3600+20 × 60+56.
(invertthirdline)
Wemultiply10by 3 30:
Thetotalofthetwomarkedlinesisthen7,asdesired.
Notethatthesumofthethreelasttermsinthesecondcolumnis99 2 4.Wetherefore needtofigureoutbywhattomultiply7 2 4 8togive 4sothatwegetatotalof100. Butsinceweknowfromthefourthlinethatmultiplyingthatvalueby8gives63,we alsoknowthatmultiplyingitby 63gives 8.Thustherequirednumberisdouble 63, whichis 42 126.Thusthefinalresultofourdivisionis12 3 42 126.
9. x + 1 7 x =19.Choose x =7;then7+ 1 7 · 7=8.Since19 ÷ 8=2 3 8 ,thecorrectanswer is2 3 8 × 7=16 5 8 . 10. (x + 2 3 x) 1 3 (x + 2 3 x) =10.Inthiscase,the“obvious”choicefor x is x =9.Then9 addedto2/3ofitselfis15,while1/3of15is5.Whenyousubtract5from15,youget 10.Sointhiscaseour“guess”iscorrect.
11. Theequationhereis (1+ 1 3 + 1 4 )x =2.Therefore.wecanfindthesolutionbydividing 2by1+ 1 3 + 1 4 .Wesetupthatproblem:
Thesumofthenumbersintheright-handcolumnbeneaththeinitiallineis1 141 144 .So weneedtofindmultipliersgivingus 3 144 = 144 72.But1 3 4times144is228.It followsthatmultiplying1 3 4by 228gives 144andmultiplyingby 114gives 72. Thus,theansweris1 6 12 114 228.
12. Sincethereare10hekatstobedividedamong10men,theaverageforeachis1hekat. Sotogetthelargestshare,weshouldaddtotheaverage1/8timeshalfthenumber ofdifferences.Sincethereare9differences,weadd1/84 1 2 times,or9/16.Therefore, thelargestshareis1 9 16 hekats,whichthescribewritesas1 2 16.Wethensubtract1/8 fromthisvalue9timestogettheshareofeachman.Theanswersare1 9 16 ,1 7 16 ,1 5 16 , 1 3 16 ,1 1 16 , 15 16 , 13 16 , 11 16 , 9 16 ,and 7 16
13. Since x mustsatisfy100:10= x :45,wewouldgetthat x = 45×100 10 ;thescribebreaks thisupintoasumoftwoparts, 35×100 10 and 10×100 10
14. Theratioofthecrosssectionareaofalogof5handbreadthsindiametertooneof4 handbreadthsdiameteris52 :42 =25:16=1 9 16 .Thus,100logsof5handbreadths diameterareequivalentto1 9 16 × 100=156 1 4 logsof4handbreadthsdiameter.
16. Themodernformulaforthesurfaceareaofahalf-cylinderofdiameter d andheight h is A = 1 2 πdh.Similarly,themodernformulaforthesurfaceareaofahemisphereof diameter d is A = 1 2 πd2.Theseformulasareidenticalif h = d
17. 7/5=1;2413/15=0;5211/24=0;27,3033/50=0;39,36
18. 0;22,30=3/80;08,06=27/2000;04,10=5/720;05,33,20=5/54
19. Since60is (2 1 2 ) × 24,thereciprocalof24is2;30.Since60is (1 7 8 ) × 32,and 7 8 can beexpressedas0;52,30thereciprocalof32is1;52,30.Since60is (1 1 3 ) × 45,the reciprocalof45is1:20.Since60=1 1 9 × 54,and 1 9 canbeexpressedas 1 10 + 1 90 = 6 60 + 40 3600 =0;06,40,thereciprocalof54is1;06,40.Iftheonlyprimedivisorsof n are 2,3,5,then n isaregularsexagesimal.
20. 25 × 1,04=1,40+25,00=26,40.18 × 1,21=6,18+18,00=24,18.50 ÷ 18= 50 × 0;3,20=2;30+0;16,40=2;46,40.1,21 ÷ 32=1,21 × 0;01,52,30= 1;21+1;10,12+0;00,40,30=2;31,52,30.
21. 1,08,16 × 3,45=4,16,0,0.4,16 × 3,45=16,0,0.16 × 3,45=1,0,0.
22. Sincethelengthofthecircumference C isgivenby C =4a,andbecause C =6r,it followsthat r = 2 3 a.Thelength T ofthelongtransversalisthen T = r√2= ( 2 3 a)( 17 12 ) = 17 18 a.Thelength t oftheshorttransversalis t =2(r t 2 ) =2a( 2 3 17 36 ) = 7 18 a.Thearea A ofthebargeistwicethedifferencebetweentheareaofaquartercircleandthearea oftherighttriangleformedbythelongtransversalandtwoperpendicularradiidrawn fromthetwoendsofthatline.Thus
23. Sincethelengthofthecircumference C isgivenby C =3a,andbecause C =6r,it followsthat r = a 2 .Thelength T ofthelongtransversalisthen T = r√3= ( a 2 )( 7 4 ) = 7 8 a. Thelength t oftheshorttransversalistwicethedistancefromthemidpointofthearc tothecenterofthelongtransversal.Ifwesetupourcirclesothatitiscenteredonthe origin,themidpointofthearchascoordinates ( r 2 , √3r 2 ) whilethemidpointofthelong transversalhascoordinates ( r 4 , √3r 4 ).Thusthelengthofhalfoftheshorttransversalis r 2 andthen t = r = a 2 .Thearea A ofthebull’seyeistwicethedifferencebetweenthe areaofathirdofacircleandtheareaofthetriangleformedbythelongtransversal andradiidrawnfromthetwoendsofthatline.Thus
24. If a isthelengthofoneofthequarter-circlearcsdefiningtheconcavesquare,thenthe diagonalisequaltothediameterofthatcircle.Sincethecircumferenceisequalto4a, thediameterisone-thirdofthatcircumference,or1 1 3 a.Thetransversalisequaltothe diagonalofthecircumscribingsquarelessthediameterofthecircle(whichisequal tothesideofthesquare).Sincethediagonalofasquareisapproximatedby17/12of theside,thetransversalisthereforeequalto5/12ofthediameter,or 5 12 4 3 a = 5 9 a.
25. √3= √22 1 ≈ 2 1 2 1 1 2 =2 0;15=1;45.Sinceanapproximatereciprocalof1;45is0;34,17,09,wegetfurtherthat √3= (1;45)2 0;03,45= 1;45 (0;30)(0;03,45)(0;34,17,09) =1;45 0;01,04,17,09=1;43,55,42,51,which wetruncateto1;43,55,42becauseweknowthisvalueisaslightover-approximation.
26. v + u =1;48=1 4 5 and v u =0;33,20= 5 9 .So2v =2;21,20and v =1;10,40= 106 90 . Similarly,2u =1;14,40and u =0;37,20= 56 90 .Multiplyingby90gives x =56, d =106.Inthesecondpart, v + u =2;05=2 1 12 and v u =0;28,48= 12 25 .So 2v =2;33,48and v =1;16,54= 769 600 .Similarly,2u =1;36,12and u =0;48,06= 481 600 Multiplyingby600gives x =481, d =769.Next,if v = 481 360 and u = 319 360 ,then v + u =2 2 9 =2;13,20.Finally,if v = 289 240 and u = 161 240 ,then v + u =1 7 8 =1;52,30.
27. Theequationsfor u and v canbesolvedtogive v =1:22,08,27= 295707 216000 = 98569 72000 and u =0;56,05,57= 201957 216000 = 67319 72000 .ThustheassociatedPythagoreantripleis67319, 72000,98569.
28. Thetwoequationsare x 2 + y 2 =1525; y = 2 3 x +5.Ifwesubstitutethesecondequation intothefirstandsimplify,weget13x 2 +60x =13500.Thesolutionisthen x =30, y =25.
29. Ifweguessthatthelengthoftherectangleis60,thenthewidthis45andthediagonal is √602 +452 =75.Sincethisvalueis1 7 8 timesthegivenvalueof40,thecorrect lengthoftherectangleshouldbe60 ÷ 1 7 8 =32.Thenthewidthis24.
30. Onewaytosolvethisistolet x and x 600betheareasofthetwofields.Thenthe equationis 2 3 x + 1 2 (x 600) =1100.Thisreducesto 7 6 x =1400,so x =1200.The secondfieldthenhasarea600.
31. Let x betheweightofthestone.Theequationtosolveisthen x 1 7 x 1 13 (x 1 7 x) =60. Wedothisusingfalsepositiontwice.First,set y = x 1 7 x.Theequationin y isthen y 1 13 y =60.Weguess y =13.Since13 1 13 13=12,insteadof60,wemultiply ourguessby5toget y =65.Wethensolve x 1 7 x =65.Hereweguess x =7and calculatethevalueoftheleftsideas6.Toget65,weneedtomultiplyourguessby 65 6 =10 1 6 .Soouransweris x =7 × 65 6 =75 5 6 gin,or1 mina 15 5 6 gin.
32. Wedothisinthreesteps,eachusingfalseposition.First,set z = x 1 7 x + 1 11 (x 1 7 x) Theequationfor z isthen z 1 13 z =60.Weguess13for z andcalculatethevalueof theleftsidetobe12,insteadof60.Thuswemustmultiplyouroriginalguessby5 andput z =65.Thenset y = x 1 7 x.Theequationfor y is y + 1 11 y =65.Ifwenow guess y =11,theresultontheleftsideis12,insteadof65.Sowemustmultiplyour guessby 65 12 toget y = 715 12 =59 7 12 .Wenowsolve x 1 7 x =59 7 12 .Ifweguess x =7, theleftsidebecomes6insteadof59 7 12 .Sotogetthecorrectvalue,wemustmultiply 7by 715 12 /6= 715 72 .Therefore, x =7 × 715 72 = 5005 72 =69 37 72 gin =1 mina 9 37 72 gin
33. Startwithasquareofside x andcutoffastripofwidth a fromtherightside.The remainingrectanglethenhasarea x 2 ax,or b.Thisrectanglecanthenbethoughtof asasquareofside x a/2thatismissingasmallsquareofside a/2.Ifoneaddsback thatsmallsquare,thenthesquareofside x a/2hasarea b + (a/2)2,sowecanfind x
34. Theequation x 60 x =7isequivalentto x 2 60=7x orto x 2 7x =60.Thesolution isthen x = ( 7 2 )2 +60+ 7 2 = 17 2 + 7 2 =12.Thusthetwonumbersare12and5.
35. Giventheappropriatecoefficients,theequationbecomes 4 9 a 2 + a + 4 3 a = 23 18 ,where a isthelengthofthearc.Ifwescaleupby 4 9 ,wegettheequation
Thealgorithmforthistypeofequationgives
. Thus a = 1 2
36. Theequationis 2 3 x 2 + 1 3 x = 1 3 .Tosolve,wescaleby 2 3 : ( 2 3 x)2 + 1 3 ( 2 3 x) =
.The solutionis
37. AllthetrianglesinFigure1.15aresimilartooneanother,andthereforetheirsides areallintheratio3:4:5.Therefore, AE = 3 5 AD =0;36 × 0;36=0;21,36.Also, DE = 4 5 AD =0;48 × 0;36=0;28,48.Also, EF = 4 5 DE =0;48 × 0;28,48=0;23,02,24. Finally, DF = 3 5 DE =0;36 × 0;28,48=0;17,16,48.
38. Ifthecircumferenceis60,thentheradiusis10.Thus,ifthedistanceofthechord fromthecircumferenceis x,thenwehavearighttriangleofsides6and10 x,with hypotenuse10.ThePythagoreantheoremleadstotheequation62 + (10 x)2 =100, or x 2 +36=20x,forwhichtheonlyvalidsolutionis x =2.
Solutions7
39. Thetwoequationsare ℓ + w =7, ℓw + 1 2 ℓ + 1 3 w =15.ToputthisintoastandardBabylonianform,wecanrewritethesecondequationintheform (ℓ + 1 3 )(w + 1 2 ) =15 1 6 Wecanthenrewritethefirstequationas (ℓ + 1 3 ) + (w + 1 2 ) =7 5 6 .ThentheBabylonian algorithmyields ℓ + 1 3 =
=
3 .Therefore, ℓ =4andso w =3.
CHAPTER2
1. 125= ρκϵ,62= ξβ,4821= ′δωκα,23,855= Mβ ′ γωνϵ
2. 8 9 = ∠ ´ γ ´ ι ´ η (8/9=1/2+1/3+1/18)
3. Theanswerisinthebackofthetext.Thebasicideaisthat200/9=22 2 9 = 22+1/6+1/18.
4. Theaverageof a and c is1/4+1/16+1/64.Theaverageof b and d is1/2+1/4+1/8+1/16.Theproductofthetwoaveragesis 1/8+1/16+1/32+1/64+1/32+1/64+1/128+1/256+1/128+1/256+1/512+1/1024, or1/4+59/1024.Thisisslightlylessthanthegivenanswerof1/4+1/16.
5. Since AB = BC;sincethetwoanglesat B areequal;andsincetheanglesat A and C are bothrightangles,itfollowsbytheangle-side-angletheoremthat △EBC iscongruent to △SBA andthereforethat SA = EC
6. Becausebothanglesat E arerightangles;because AE iscommontothetwotriangles; andbecausethetwoangles CAE areequaltooneanother,itfollowsbytheangle-sideangletheoremthat △AET iscongruentto △AES.Therefore SE = ET
7. Thedistancefromthecenterofthepyramidtothetipoftheshadowis378+342=720 feet.Thereforetheheightofthepyramidis6/9=2/3ofthisvalue,or480feet.
8. Tn =1+2+ + n = n(n +1) 2 .Thereforetheoblongnumber n(n +1) isdoublethe triangularnumber Tn.
9. n 2 = (n 1)n 2 + n(n +1) 2 ,andthesummandsarethetriangularnumbers Tn 1 and Tn
11. Suppose a 2 + b2 = c 2.Suppose a isodd.Then a 2 isodd.If b isodd,then b2 isoddand c 2 iseven,so c iseven.If b iseven,then b2 isevenand c 2 isodd,so c isodd.Asimilar resultholdsif c isodd.
12. Examplesusingthefirstformulaare(3,4,5),(5,12,13),(7,24,25),(9,40,41), (11,60,61).Examplesusingthesecondformulaare(8,15,17),(12,35,37),(16,63,65), (20,99,101),(24,143,145).
13. Letusassumethatthesecondlegiscommensurabletothefirstandlet b, a benumbers representingthetwolegs(intermsofsomeunit).Wemayaswellassumethat b and a arerelativelyprime.Sincethehypotenuseisdoublethefirstleg,wehave b2 + a 2 = (2a)2 =4a 2,or b2 =3a 2.Since b2 isamultipleof3,itmustalsobeamultipleof9,so b2 =9c 2 and b =3c.Then9c 2 =3a 2,or a 2 =3c 2.Thisimpliesthat a 2 isamultipleof 9,sothat a isamultipleof3.Butthenboth a and b aremultiplesof3,contradicting thefactthattheyarerelativelyprime.
14. Sincesimilarsegmentsaretotheircorrespondingcirclesinthesameratio,theareasof similarsegmentsaretooneanotherasthesquaresonthediametersofthecircles.Thus, theareasofsimilarsegmentsarealsotooneanotherasthesquaresontheradiiofthe circles.Butinsimilarsegments,thetrianglesformedbythetworadiiandchordsare similartriangles.Thusthechordofonesegmentistothechordinthesimilarsegment astheradiusofthefirstcircletotheradiusofthesecond.Thatis,thesquaresonthe radiiaretooneanotherasthesquaresonthechords.Therefore,theareasofsimilar segmentsaretooneanotherasthesquaresontheirchords.
15. ByExercise14,theareaofsegment BD istheareaofsegment AB asthesquareon BD isthesquareon AB.Butthisratioisequalto3.Thus,theareaofsegment BD is threetimestheareaofsegment AB,orisequaltothesumoftheareasofsegments AB, AC,and CD.Therefore,theareaofluneisequaltothedifferencebetweenthearea ofthelargesegmentandtheareaofsegment BD.Butthisisequaltothedifference betweentheareaofthelargesegmentandtheareasofthethreesmallsegments,which isinturnequaltotheareaofthetrapezoid.Toconstructthetrapezoid,notethatone cancertainlyconstructalinesegmentequalto √3timesthelengthofagivenline segment.Toplacethislinesegmentbothparalleltotheoriginaloneandsuchthatthe linesconnectingtheendpointsofthetwosegmentsareeachequaltotheoriginalline segment,wesimplyneedtofindthedistancebetweenthetwosegments.Andthat canbeconstructedbyusingthePythagoreanTheoremappliedtothetrianglewhose hypotenuseisequaltotheoriginalsegmentandonelegofwhichisequaltohalfthe differencebetweenthenewlinesegmentandtheoriginalone.Tocircumscribeacircle aroundthistrapezoid,notethatonecanconstructacirclethroughthreepoints,say B, A,and C.Bythesymmetryofthetrapezoid,thiscirclewillalsogothroughpoint D
21. Ifoneequatesthetimesofthetworunners,where d isthedistancetraveledbyAchilles, theequationis d/10= (d 500)/(1/5).Thisisequivalentto49d =25,000,so d = 510.2yards.SinceAchillesistravelingat10yardspersecond,thiswilltakehim51.02 seconds.
CHAPTER3
1. OnewaytodothisistouseI-4.Namely,considertheisoscelestriangle ABC with equalsides AB and BC alsoasatriangle CBA.Thenthetriangles ABC and CBA have twosidesequaltotwosidesandtheincludedanglesalsoequal.Thus,byI-4,theyare congruent.Therefore,angle BAC isequaltoangle BCA,andthetheoremisproved.
2. Putthepointofthecompassonthevertex V oftheangleandswingequalarcsintersectingthetwolegsat A and B.Thenplacethecompassat A and B,respectively,and swingequalarcs,intersectingat C.Thelinesegmentconnecting V to C thenbisects theangle.Toshowthatthisiscorrect,notethattriangles VAC and VBC arecongruent by SSS.Therefore,thetwoangles AVC and BVC areequal.
3. Letthelines AB and CD intersectat E.Thenangles AEB and CED arebothstraight angles,anglesequaltotworightangles.Ifonesubtractsthecommonangle CEB from eachofthese,theremainingangles AEC and BED areequal,andthesearethevertical anglesofthetheorem.
4. Supposethethreelineshavelength a, b,and c,with a ≥ b ≥ c.Onthestraightline DH oflength a + b + c,letthelengthof DF be a,thelengthof FG be b,andthelength of GH be c.Thendrawacirclecenteredon F withradius a andanothercirclecentered at G withradius c.Let K beanintersectionpointofthetwocircles.Thenconnect FK and GK.Triangle FKG willthenbethedesiredtriangle.For FK = FD,andthishas length a.Also FG haslength b,while GK = GH andthishaslength c.Notealso thatwemusthave b + c > a,forotherwisethetwocircleswouldnotintersect.That a + b > c and a + c > b areobviousfromhowwehavelabeledthethreelengths.
5. Supposeangle DCE isgiven,andwewanttoconstructanangleequaltoangle DCE atpoint A ofline AB.Drawtheline DE sothatwenowhaveatriangle DCE.(Here D and E arearbitrarypointsalongthetwoarmsofthegivenangle.)Then,bytheresult ofExercise4,constructatriangle AGF,with AG alongline AB,where AG = CE, AF = CD,and FG = DE.Bytheside-side-sidecongruencetheorem,triangle AGF is congruenttotriangle CED.Therefore,angle FAG isequaltoangle DCE,asdesired.
6. Let ABC bethegiventriangle.Extend BC to D anddraw CE parallelto AB.ByI–29, angles BAC and ACE areequal,asareangles ABC and ECD.Thereforeangle ACD equalsthesumoftheangles ABC and BAC.Ifweaddangle ACB toeachofthese,we getthatthesumofthethreeinterioranglesofthetriangleisequaltothestraightangle BCD.Becausethislatterangleequalstworightangles,thetheoremisproved.
7. Placethegivenrectangle BEFG sothat BE isinastraightlinewith AB.Extend FG to H sothat AH isparallelto BG.Connect HB andextendituntilitmeetstheextensionof
FE at D.Through D draw DL parallelto FH andextend GB and HA sotheymeet DL in M and L,respectively.Then HD isthediagonaloftherectangle FDLH andsodivides itintotwoequaltriangles HFD and HLD.Becausetriangle BED isequaltotriangle BMD andalsotriangle BGH isequaltotriangle BAH,itfollowsthattheremainders, namelyrectangles BEFG and ABML,areequal.Thus ABML hasbeenappliedto AB andisequaltothegivenrectangle BEFG
8. Becausetriangles ABN, ABC,and ANC aresimilar,wehave BN : AB = AB : BC,so AB2 = BN · BC,and NC : AC = AC : BC,so AC2 = NC · BC.Therefore AB2 + AC2 = BN · BC + NC · BC = (BN + NC) · BC = BC2,andthetheoremisproved.
9. Inthisproof,weshallrefertocertainpropositionsinEuclid’sBookI,allofwhich areprovedbeforeEuclidfirstusespostulate5.(Thatoccursinproposition29.)First, assumePlayfair’sAxiom.Supposeline t crosseslines m and l andthatthesumofthe twointeriorangles(angles1and2inthediagram)islessthantworightangles.We knowthatthesumofangles1and3isequaltotworightangles.Therefore ∠2< ∠3. Nowonline BB′ andpoint B′ constructline B′C′ suchthat ∠C′B′B = ∠3(Proposition 23).Therefore,line B′C′ isparalleltoline l (Proposition27).Therefore,byPlayfair’s Axiom,line m isnotparalleltoline l.Itthereforemeets l.Wemustshowthatthetwo linesmeetonthesamesideas C′.Ifthemeetingpoint A isontheoppositeside,then ∠2isanexteriorangletotriangle ABB′,yetitissmallerthan ∠3,oneoftheinterior angles,contradictingProposition16.WehavethereforederivedEuclid’spostulate5.
Second,assumeEuclid’spostulate5.Let l beagivenlineand P apointoutsidetheline. Constructtheline t perpendicularto l through P (Proposition12).Next,constructthe line m perpendiculartoline t at P (Proposition11).Sincethealternateinteriorangles formedbyline t crossinglines m and l arebothrightandthereforeareequal,itfollows fromProposition27that m isparallelto l.Nowsuppose n isanyotherlinethrough P.Wewillshowthat n meets l andisthereforenotparallelto l.Let ∠1betheacute anglethat n makeswith t.Thenthesumofangle1andangle PQR islessthantwo rightangles.Bypostulate5,thelinesmeet.
Notethatinthisproof,wehaveactuallyprovedtheequivalenceofEuclid’spostulate 5tothestatementthatgivenaline l andapoint P noton l,thereisatmostoneline through P whichisparallelto l.TheotherpartofPlayfair’sAxiomwasproved(inthe secondpartabove)withoutuseofpostulate5andwasnotusedatallinthefirstpart.
10. Onepossibilityforanalgebraictranslation:Ifthelinehaslength a andiscutata pointwithcoordinate x,then4ax + (a x)2 = (a + x)2.Thisisavalididentity.Hereis ageometricdiagram,with AB = ME = a and CB = BD = BK = KR = x:
11. If ABC isthegivenacute-angledtriangleand AD isperpendicularto BC,thenthe theoremstatesthatthesquareon AC islessthanthesquareson CB and BA bytwice therectanglecontainedby CB and BD.Ifwelabel AC as b, BA as c,and CB as a, then BD = c cos B.Thusthetheoremcanbetranslatedalgebraicallyintotheform b2 = a 2 + c 2 2ac cos B,exactlythelawofcosinesinthiscase.
12. Supposethediameter CD ofacirclewithcenter E bisectsthechord AB at F.Then join EA and EB,formingtriangle EAB.Triangles AEF and BEF arecongruentby side-side-side(since AE = BE arebothradiiofthecircleand F bisects AB).Therefore angles EFA and EFB areequal.Butthesumofthosetwoanglesisequaltotworight angles.Henceeachisarightangle,asdesired.Toprovetheconverse,usethesame constructionandnotethattriangle AEB isisosceles,soangle EAF isequaltoangle EBF,whilebothangles EFA and EFB arerightbyhypothesis.Itfollowsthattriangles AEF and BEF areagaincongruent,thistimebyangle-angle-side.So AF = BF,and thediameterbisectsthechord.
13. Inthecircle ABC,lettheangle BEC beanangleatthecenterandtheangle BAC bean angleatthecircumferencewhichcutsoffthesamearc BC.Connect ∠EAB.Similarly, ∠FEC isdouble ∠EAC.Thereforetheentire ∠BEC isdoubletheentire ∠BAC.Note thatthisargumentholdsaslongasline EF iswithin ∠BEC.Ifitisnot,ananalogous argumentbysubtractionholds.
14. Let ∠BAC beananglecuttingoffthediameter BC ofthecircle.Connect A tothe center E ofthecircle.Since EB = EA,itfollowsthat ∠EBA = ∠EAB.Similarly, ∠ECA = ∠EAC.Thereforethesumof ∠EBA and ∠ECA isequalto ∠BAC.Butthe sumofallthreeanglesequalstworightangles.Therefore,twice ∠BAC isequalto tworightangles,andangle BAC isitselfarightangle.
15. Lettriangle ABC begiven.Let D bethemidpointof AB and E themidpointof AC Drawaperpendicularat D to AB andaperpendicularat E to AC andletthemmeetat point F (whichmaybeinsideoroutsidethetriangle,oronside BC).Assumefirstthat
Solutions13
F isinsidethetriangle,andconnect FB, FA,and FC.Since BD = BA,triangles FDB and FDA arecongruentbyside-angle-side.Therefore FB = FA.Similarly,triangles FEA and FEC arecongruent.So FC = FA.Thereforeallthreelines FA, FB,and FC areequal,andacirclecanbedrawnwithcenter F andradiusequalto FA.Thiscircle willcircumscribethegiventriangle.Finally,notethattheidenticalconstructionworks if F isonline BC orif F isoutsidethetriangle.
16. Let G bethecenterofthegivencircleand AGD adiameter.Withcenterat D andradius DG,constructanothercircle.Let C and E bethetwointersectionsofthetwo(equal) circles,andconnect DC and DE.Then DE and DC aretwosidesofthedesiredregular hexagon.Tofindtheotherfoursides,drawthediameters CGF and EGB.Then CB, BA, AF,and FE aretheothersides.Todemonstratethatwehaveinfactconstructeda regularhexagon,notethatallthetriangleswhosebasesaresidesofthehexagonand whoseothersidesareradiiareequilateral;thusallthesidesofthehexagonareequal andalltheanglesofthehexagonarealsoequal.
17. Inthecircle,inscribeaside AC ofanequilateraltriangleandaside AB ofanequilateralpentagon.Thenarc BC isthedifferencebetweenone-thirdandone-fifthofthe circumferenceofthecircle.Thatis,arc BC = 2 15 ofthecircumference.Thus,ifwe bisectthatarcat E,thenlines BE and EC willeachbeasideofaregular15-gon.
18. Let a = s1b + r1, b = s2r1 + r2, , rk 1 = sk + 1rk.Then rk divides rk 1 andtherefore also rk 2, , b, a.Iftherewereagreatercommondivisorof a and b,itwoulddivide r1, r2, , rk.Sinceitisimpossibleforagreaternumbertodivideasmaller,wehave shownthat rk isinfactthegreatestcommondivisorof a and b
19.
20.
21.
963=1 657+306
657=2 · 306+45
306=6 · 45+36
45=1 · 36+9
36=4 · 9+0
Therefore,thegreatestcommondivisorof963and657is9.
4001=1 2689+1312 2689=2 1312+65 1312=20 65+12
65=5 12+5
12=2 5+2 5=2 2+1
Therefore,thegreatestcommondivisorof4001and2689is1.
46=7 6+423=7 3+2 6=1 4+23=1 2+1 4=2 · 2 2=2 · 1
Notethatthemultiples7,1,2inthefirstexampleequalthemultiples7,1,2inthe second.
33=2 · 12+911=2 · 4+3 12=1 · 9+34=1 · 3+1 9=3 · 3 3=3 · 1
Itfollowsthatbothratioscanberepresentedbythesequence (2,1,3). 22. Since1 x = x 2,wehave
Thus1: x canbeexpressedintheform (1,1,1, )
Solutions15
23. If d isthediagonalofasquareofside s,thenthefirstdivisiongives d =1s + r.To understandthenextsteps,itisprobablyeasiesttoset s =1anddealwiththenumerical values.Therefore, d = √2,and r = √2 1.Ournextdivisiongives s =2r + t,where t =3 2√2.Geometrically,ifwelayoff s alongthediagonal,then r istheremainder d s.Thendrawasquareofside r withpartofthesideoftheoriginalsquarebeingits diagonal.Notethatifwenowlayoff r alongthediagonalofthatsquare,theremainder is t.Inotherwords, r isthedifferencebetweenthediagonalofasquareofside s and s,while t isthedifferencebetweenthediagonalofasquareofside r and r.Itfollows thatifoneperformsthenextdivisionintheprocess,wewillgetthesamerelationship. Thatis, r =2t + u,wherenow u isthedifferencebetweenthediagonalofasquare ofside t and t.Thus,thisprocesswillcontinueindefinitelyandtheratio d : s canbe expressedas (1,2,2,2, ...).
24. Since a > b,thereisanintegralmultiple m of a b with m(a b) > c.Let q bethe firstmultipleof c thatexceeds mb.Then qc > mb ≥ (q 1)c,or qc c ≤ mb < qc. Since c < ma mb,itfollowsthat qc ≤ mb + c < ma.Butalso qc > mb.Thuswehave amultiple(q)of c thatisgreaterthanamultiple(m)of b,whilethesamemultiple(q) of c isnotgreaterthanthesamemultiple(m)of a.Thusbydefinition7ofBookV, c : b > c : a.
25. Let A : B = C : D = E : F.Wewanttoshowthat A : B = (A + C + E) : (B + D + F).Takeanyequimultiples mA and m(A + C + E) ofthefirstandthirdand anyequimultiples nB and n(B + D + F) ofthesecondandfourth.Since m(A + C + E) = mA + mC + mE,andsince n(B + D + F) = nB + nD + nF,andsincewhenever mA > nB, wehave mC > nD and mE > nF,itfollowsthat mA > nB impliesthat m(A + C + E) > n(B + D + F).Sinceasimilarstatementholdsforequalityandfor“lessthan,”theresult followsfromEudoxus’sdefinition.Amodernproofwouldusethefactthat a1bi = b1ai forevery i andthenconcludethat a1(b1 + b2 + + bn) = b1(a1 + a2 + + an)
26. Giventhat a : b = c : d,wewanttoshowthat a : c = b : d.Sotakeanyequimultiples ma, mb of a and b andalsoequimultiples nc, nd of c and d.Now ma : mb = a : b = c : d = nc : nd.Thusif ma > nc,then mb > nd;if ma = nc,then mb = nd;andif ma < nc, then mb < nd.Thus,bythedefinitionofequalratio,wehave a : c = b : d.
27. Let AB =9and BC =5.Drawacirclewith AC asdiameteranderectaperpendicular to AC at B,meetingthecircleat E.Then BE isthedesiredlength x
28. Supposethefirstoftheequalandequiangularparallelogramshassidesoflength a and b whilethesecondhassidesoflength c and d,eachpairsurroundinganangle equalto α.Sincetheareaofaparallelogramistheproductofthetwosideswiththe sineoftheincludedangle,weknowthat ab sin α = cd sin α.Itfollowsthat ab = cd or that a : c = d : b,asdesired.Conversely,if a : c = d : b andtheparallelogramsare equiangularwithangle α betweeneachpairofgivensides,then ab = cd and ab sin α = cd sin α,sotheparallelogramsareequal.Euclid’sproofis,ofcourse,differentfrom thismodernone.Namely,ifthetwoparallelogramsare P1 = ADBF and P2 = BGCE, withequalanglesat B,Euclidplacesthemsothat FB and BG areinastraightline asare EB and BD.Hethencompletesthethirdparallelogram P3 = FBEK.Since P1 = P2,wehave P1 : P3 = P2 : P3.But P1 : P3 = DB : BE,since BF iscommon, and P2 : P3 = BG : BF,since BE iscommon.Thus, DB : BE = BG : BF,thedesired conclusion.Theconverseisprovedbyreversingthesteps.
29 Wewanttoprovethatnumberstowhichthesamenumberhasthesameratioareequal. Sosuppose a : b = a : c.Therefore, a isthesamemultipleorthesamepartorthe samepartsofboth b and c.Thatis, a = mb and a = mc forsomerationalnumber m Butthismeansthat b = 1 m a and c = 1 m a,thus b = c
30. Suppose a : b = f : g andsupposethenumbers c, d, , e arethenumbersincontinued proportionbetween a and b.Let r, s, t, , u, v bethesmallestnumbersinthesame ratioas a, c, d, , e, b.Then r, v arerelativelyprimeand r : v = a : b = f : g.Itfollows that f = mr, g = mv forsomeinteger m andthatthenumbers ms, mt, , mu areinthe sameratioastheoriginalsetofnumbers.Thusthereareatleastasmanynumbersin continuedproportionbetween f and g astherewerebetween a and b.Sincethesame argumentworksstartingwith f and g,itfollowsthatthereareexactlyasmanynumbers incontinuedproportionbetween f and g asbetween a and b.Sincethereisnointeger between n and n +1,itfollowsthattherecannotbeameanproportionalbetweenany pairofnumbersintheratio (n +1) : n.
31. Thenumber ab isthemeanproportionalbetween a 2 and b2 .
32. Thenumbers a 2b and ab2 arethetwomeanproportionalsbetween a3 and b3 .
33. If a 2 measures b2,then b2 = ma 2 forsomeinteger m.Sinceeveryprimenumber whichdivides b2 mustdivide ma 2 andthereforemustdivideeither m or a 2,itfollows bycountingprimesthat m mustitselfbeasquare.Thus m = n 2 and b = na,so a measures b.Conversely,if a measures b,then b = na and b2 = n 2 a 2,so a 2 measures b2 .
34. Suppose m factorstwodifferentwaysasaproductofprimes: m = pqr s = p ′ q ′ r ′ s ′.Since p divides pqr s,itmustalsodivide p ′ q ′ r ′ s ′.ByVII–30, p mustdivideoneoftheprimefactors,say p ′.Butsinceboth p and p ′ areprime,we musthave p = p ′.Aftercancelingthesetwofactorsfromtheirrespectiveproducts,we canthenrepeattheargumenttoshowthateachprimefactorontheleftisequaltoa primefactorontherightandconversely.
35. Onestandardmodernproofisasfollows.Assumethereareonlyfinitelymanyprime numbers p1, p2, p3, , pn.Let N = p1p2p3 ··· pn +1.Therearethentwopossibilities. Either N isprimeor N isdivisiblebyaprimeotherthanthegivenones,sincedivision byanyofthoseleavesremainder1.Bothcasescontradicttheoriginalhypothesis, whichthereforecannotbetrue.
36. Wekeepaddingpowersof2untilwegetaprime.After1+2+4+8+16+32+64= 127,thenextsumsare255,511,1023,2047,4095,and8191.Thefirstfiveofthese arenotprime(notethat2047=23 × 89).But8191isprime(checkbydividingbyall primeslessthan √8191).Sothenextperfectnumberis8191 × 4096=33,550,336.
Solutions17
37. Since BC isthesideofadecagon,triangle EBC isa36-72-72triangle.Thus ∠ECD = 108◦.Since CD,thesideofahexagon,isequaltotheradius CE,itfollowsthattriangle ECD isanisoscelestrianglewithbaseanglesequalto36◦.Thustriangle EBD isa3672-72triangleandissimilartotriangle EBC.Therefore BD : EC = EB : BC or BD : CD = CD : BC andthepoint C dividesthelinesegment BD inextremeand meanratio.
38. ByExercise37,ifweset d tobethelengthofthesideofadecagon,wehave (1+ d) : 1=1: d or d2 + d 1=0.Itfollowsthat d = √5 1 2 .Thelength p oftheside ofapentagonis(p.91) p = 1 2 10 2√5.Itisthenstraightforwardtoshowthat p 2 =12 + d2 asasserted.
39. Theedgelengthoftheicosahedroninasphereofdiameter1is ei = 1 10 50 10√5= 0.52573111.Thecircumradiusoftheequilateraltrianglewiththisedgelengthis r = ei 2sin60◦ =0.303531.Theedgelengthofthedodecahedroninthesame sphereis ed = 1 6 (√15 √3) =0.35682209.Thecircumradiusofthepentagonwiththisedgelengthis r = ed 2sin36◦ =0.303531,thesamevalueasinthe icosahedron.
40. Webeginwitharectangleofsides x and y.Wethenlayoff y along x,withtheremainder being x y =7.Dividetherectanglewithsides7and y inhalfandmoveonehalfto thebottom.Wethenaddthesquareofside 7 2 togetasquareofside y + 7 2 = x 7 2 with area18+ ( 7 2 )2 = 121 4 .Thus y = 11 2 7 2 =2and x = 11 2 + 7 2 =9.
41. First,tosolvethetwoequations,weset x = a/y andsubstituteinto y 2 αx 2 = b.After multiplyingby y 2,wegetthefourthdegreeequation y 4 αa 2 = by2.Thisisquadratic in y 2,sowecansolvetoget
(Notethatweonlyusetheplussign,since y 2 mustbepositive.)Then
Wecanthensolvefor x:
(Ofcourse,since a isassumedpositive,wetakethepositivesolutionfor x whenwe havethepositiveonefor y,aswellasthenegativesolutionfor x withthenegativeone for y.)Theasymptotesofthehyperbola xy = a arethelines x =0and y =0,andthese aretheaxesofthehyperbola y 2 αx 2 = b.Thustheasymptotesofonehyperbolaare theaxesoftheother.
CHAPTER4
1. If x isthedistancetothe14kgend,then10(10 x) =14x,100=24x,and x =4 1 6 m fromthe14kgend.
2. Since8 10=80and12 8=96,theleverinclinestowardthe12kgweight.
3. Sinceaweight W ofgolddisplacesavolume V1 offluid,aweight w1 ofgoldwill displace w1 W V1 offluid.Similarly,aweight w2 ofsilverdisplaces w2 W V2 offluid.Thus thewreath,oftotalweight W,displacesthesumofthesetwoamountsoffluid.That is, V = w1 W V1 + w2 W V2.Since W = w1 + w2,itfollowsthat (V V1)w1 = (V2 V)w2, andthedesiredresultfollows.
4. Lemma1: DA/DC = OA/OC by Elements VI–3.Therefore DA/OA = DC/OC = (DC + DA)/(OC + OA) = AC/(CO + OA).Also, DO2 = OA2 + DA2 bythe PythagoreanTheorem.
Lemma2: AD/DB = BD/DE = AC/CE = AB/BE = (AB + AC)/(CE + BE) = (AB + AC)/BC.Therefore, AD2/BD2 = (AB + AC)2/BC2.But AD2 = AB2 BD2 . So (AB2 BD2)/BD2 = (AB + AC)2/BC2 and AB2/BD2 =1+ (AB + AC)2/BC2 .
5. Set r =1, ti and ui asinthetext,and Pi theperimeterofthe ithcircumscribedpolygon. Thenthefirstteniterationsofthealgorithmgivethefollowing:
t1 =0.577350269 u1 =1.154700538 P1 =3.464101615
t2 =0.267949192 u2 =1.03527618 P2 =3.21539031
t3 =0.131652497 u3 =1.008628961 P3 =3.159659943
t4 =0.065543462 u4 =1.002145671 P4 =3.146086215
t5 =0.03273661 u5 =1.0005357 P5 =3.1427146
t6 =0.016363922 u6 =1.00013388 P6 =3.141873049
t7 =0.0081814134 u7 =1.000033467 P7 =3.141662746
t8 =0.004090638249 u8 =1.000008367 P8 =3.141610175
t9 =0.002045310568 u9 =1.000002092 P9 =3.141597032
t10 =0.001022654214 u10 =1.000000523 P10 =3.141593746
6. Let d bethediameterofthecircle, ti thelengthofonesideoftheregularinscribed polygonof3 2i sides,and ui thelengthoftheotherlegoftherighttriangleformed fromthediameterandthesideofthepolygon.Then