数学专业英语14Word文件下载.docx
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数学专业英语14Word文件下载.docx
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Somepuzzlesareoriginal.
Explaininghowananswerisarrivedatismoreimportantthantheansweritself.
Tothisend,hints,answers,andfullyworkedsolutionsareprovided,togetherwithlinkstorelatedmathematicaltopics.
Furtherreferencesareprovidedwithmanyofthesolutions.
Thepuzzlesareintendedtobefun,withaneducationalelement.
1.Triangulararea
IntriangleABC,producealinefromBtoAC,meetingatD,andfromCtoAB,meetingatE.
LetBDandCEmeetat
X.
LetBXEhaveareaa,BXChaveareab,andCXDhaveareac.
FindtheareaofquadrilateralAEXDintermsof
a,b,and
c.
Hint:
Considerpairsoftriangles,withcommonheightandcolinearbase.
Solution
BXEhasareaa,BXChasareab,andCXDhasareac.
Wewillusethefactthattheareaofatriangleisequalto½
×
base×
perpendicularheight.
Anysidecanserveasthebase,andthentheperpendicularheightextendsfromthevertexoppositethebasetomeetthebase(oranextensionofit)atrightangles.
ConsiderBXEandBXC,EX/XC=a/b.Similarly,consideringBXCandCXD,BX/XD=b/c.
NowdrawlineAX.
LetAXEhaveareaxandAXDhaveareay.SinceAXBandAXDhavecommonheight,wehave(a+x)/y=b/c.
Similarly,x/(y+c)=a/b.
Cross-multiplyingyields:
by=cx+ac,bx=ay+ac.
Solvingthesesimultaneousequations,weobtainx=ac(a+b)/(b2-ac),y
=
ac(b
+
c)/(b2
-
ac).
ThereforetheareaofquadrilateralAEXDis
.
2.Twologicians
Twoperfectlogicians,SandP,aretoldthatintegersxandyhavebeenchosensuchthat1
<
x
yandx+y
100.
Sisgiventhevaluex+yandPisgiventhevaluexy.
Theythenhavethefollowingconversation.
P:
Icannotdeterminethetwonumbers.
S:
Iknewthat.
NowIcandeterminethem.
SocanI.
Giventhattheabovestatementsaretrue,whatarethetwonumbers?
(Computerassistanceallowed.)
Firstofall,trivially,xycannotbeprime.
Italsocannotbethesquareofaprime,forthatwouldimplyx
y.
Wenowdeduceasmuchaspossiblefromeachofthelogicians'
statements.
Wehaveonlypublicinformation:
theproblemstatement,thelogicians'
statements,andtheknowledgethatthelogicians,beingperfect,willalwaysmakecorrectandcompletedeductions.
Eachlogicianhas,inaddition,onepieceofprivateinformation:
sumorproduct.
Solution
P'
sstatementimpliesthatxycannothaveexactlytwodistinctproperfactorswhosesumislessthan
Callsuchapairoffactorseligible.
Forexample,xycannotbetheproductoftwodistinctprimes,forthenPcoulddeducethenumbers.
Likewise,xycannotbethecubeofaprime,suchas33
27,forthen3×
9wouldbeauniquefactorization;
orthefourthpowerofaprime.
Othercombinationsareruledoutbythefactthatthesumofthetwofactorsmustbelessthan
Forexample,xycannotbe242
2×
112,since11×
22istheuniqueeligiblefactorization;
2×
121beingineligible.
Similarlyforxy
318
3×
53.
IfSwassurethatPcouldnotdeducethenumbers,thennoneofthepossiblesummandsofx+ycanbesuchthattheirproducthasexactlyonepairofeligiblefactors.
Forexample,x+ycouldnotbe
51,sincesummands
17and
34producexy
578,whichwouldpermitPtodeducethenumbers.
Wecangeneratealistofvaluesofx+ythatareneverthesumofpreciselytwoeligiblefactors.
ThefollowinglistisgeneratedbyJavaScript;
thefunctionmaybeinspectedbyviewingJavaScript:
functiongenSum(plaintext.)
Eligiblesums:
11,17,23,27,29,35,37,41,47,53.
(WecanuseGoldbach'
sConjecture,whichstatesthateveryevenintegergreaterthan2canbeexpressedasthesumoftwoprimes,todeducethattheabovelistcancontainonlyoddnumbers.
Althoughtheconjectureremainsunproven,ithasbeenconfirmedempiricallyupto4×
1014.)
Pnowknowsthatx+yisoneofthevalueslistedabove.
IfthisenablesPtodeducexandy,then,oftheeligiblefactorizationsofxy,theremustbepreciselyoneforwhichthesumofthefactorsisinthelist.
Thetablebelow,generatedbyJavaScript(viewplaintextJavaScript:
functiongenProd),showsallsuchxy,togetherwiththecorrespondingx,y,andx+y.
Thetableissortedbysumandthenproduct.
Notethataproductmaybeabsentfromthetableforoneoftworeasons.
Eithernoneofitseligiblefactorizationsappearsintheabovelistofeligiblesum
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