CHAPTER2
1.125= ρκ ,62= ξβ,4821= δωκα,23, 855= M β γων
2. 8 9 = ´ γ ´ ι ´ η (8/9=1/2+1/3+1/18)
3.Theanswerisinthebackofthetext(exceptthelastcharactershouldbean η instead ofa β).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. Theproductofthetwoaveragesis1/8+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.Thisisslightlylessthanthe givenanswerof1/4+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.Thedistancefromthecenterofthepyramidtothetipofthe shadowis378+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. n2 = (n 1)n 2 + n(n+1) 2 ,andthesummandsarethetriangularnumbers Tn 1 and Tn.
10. 8n(n+1) 2 +1=4n2 +4n +1=(2n +1)2
11.Suppose a2 + b2 = c2 Suppose a isodd.Then a2 isodd.If b isodd,then b2 isoddand c2 iseven,so c iseven.If b iseven,then b2 isevenand c2 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 + a2 = (2a)2 =4a2,or b2 =3a2 .Since b2 isamultipleof3,itmustalsobeamultipleof9,so b2 =9c2 and b =3c.Then9c2 =3a2,or a2 =3c2 .Thisimpliesthat a2 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 isthree timestheareaofsegment AB,orisequaltothesumoftheareasofsegments AB, AC, and CD.Therefore,theareaofluneisequaltothedifferencebetweentheareaofthe largesegmentandtheareaofsegment BD.Butthisisequaltothedifferencebetween theareaofthelargesegmentandtheareasofthethreesmallsegments,whichisinturn equaltotheareaofthetrapezoid.Toconstructthetrapezoid,notethatonecancertainly constructalinesegmentequalto √3timesthelengthofagivenlinesegment.Toplace thislinesegmentbothparalleltotheoriginaloneandsuchthatthelinesconnectingthe endpointsofthetwosegmentsareeachequaltotheoriginallinesegment,wesimply needtofindthedistancebetweenthetwosegments.Andthatcanbeconstructedby usingthePythagoreanTheoremappliedtothetrianglewhose hypotenuseisequaltothe originalsegmentandonelegofwhichisequaltohalfthedifferencebetweenthenewline segmentandtheoriginalone.Tocircumscribeacirclearoundthistrapezoid,notethat onecanconstructacirclethroughthreepoints,Say B, A,and C.Bythesymmetryof thetrapezoid,thiscirclewillalsogothroughpoint D
21.Ifoneequatesthetimesofthetworunners,where d isthedistancetraveledbyAchilles, theequationis d/10=(d 500)/(1/5).Thisisequivalentto49d =25, 000,so d =510 2 yards.SinceAchillesistravelingat10yardspersecond,thiswilltakehim51.02seconds.