2024年1月11日发(作者：秦安中考数学试卷真题)

SolutionstotheFifty-EighthWilliamLowellPutnamMathematicalCompetitionSaturday,December7,1996ManjulBhargavaandKiranKedlayaA-1Ifandarea1,thouttheshorteroverlapsideofgen-eralitythatmustofaberectangleatleast,findthismaximum,hisvalueisthede-siredvalueof.A-2Lettively.(Weandareassumingbethecentersofand,respec-radius3.)Thenthedesiredlocushasisradiusanannulus1andcenteredhasatthemidpointofradius2.,withinnerradius1andouterForafixedpointthesegmentsunderahomothetyforoncenteredlying,es,ofradius,whichthecenterofthissmallercircletracesoutacircleofradius(againbyhomothety).Byconsideringthetwopositionsofonthelineofcentersofthecircles,oneseesthatcenteredatthemidpointofisclearlythespecifiedannulus.,rewaystochoose3ofthe6courses;chpairofcoursesischosenby4students(correspondingtothefourwaystocompletethispairtoasetof3courses)andisnotchosenby4students(correspondingtothe3-elementsubsetsoftheremaining4courses).Note:Assumingthatnotwostudentschoosethesamecourses,theabovecounterexampleisunique(uptoper-mutingstudents).Thismaybeseenasfollows:Givenagroupofstudents,supposethatforanypairofcourses(amongthesix)thereareatmost4studentstakingboth,erearewhereatmostpairs,isastudent,therhand,ifastudentistakingcourses,/sheoccursoccursisminimizedininatleastfor,tyHencethereonlycanifeachbeatstudentmoststu-dents,takes3courses,andforeachsetoftwocourses,hereareonly4waystocompleteagivenpairofcoursestoasetof3,andonly4waystochoose3coursesnotcontainingthegivenpair,theonlywayfortheretobe20students(underourhypotheses)r,RobinChapmanhaspointedoutthattheso-lutionisnotuniqueintheproblemasstated,ernatesolutionistoidentifythe6courseswithpairsofantipodalverticesofanicosahe-dron,andhaveeachstuex-ample,eachof10selectionsismadebyapairofstu-dents.A-4Infact,dIfbyinductionandonthenumberofel-withwith.,otherwisechoose,claimlabeledthatsuchtherethatex-,existencefienosuchforall,wehaveexists;general,holdsbynotingpropertythat1forandbyinductiononApplyingthiswhen,weget,

e;thenforany,y,weputinifaboveshowsthatotherwise;rto,andthatA-5(duetoLennyNg)For,dividesandwherethecongruenceisarationalnumberwhosenumerator,meansinreducedthatform,tsuffictionon,forall,,soforisallcourseconstantandthismeanson,.Ifandisdefinedbedefinedbecause.;Notethelatterthatasbefore,thenbyin-duction,followsthesequencebynotingcanthatthepolynomialispositiveatandhasitsminimumat,ase,constantFinally,,wedefiwehadwededuceforall,bythesameargumentasinthefirstcaseActually,thisdoesn’thappen;,ludethatbywhatwefunction.(ThankstoMarshallBuckforcatchingisaconstantanin-accuracyinapreviousversionofthissolution.)esequencedefiwedefineasanycontinuousfunctionwithequalvaluesonontheendpoints,andextendthedefinitionfromtobytherelationextendthedefinition,er,,theresultingfurthertoanyfunctionfunctionwithhasbytherelationthatthepropertydesiredclearlyhasthisform.B-1Letdenotetheset,andletdenotethenumberofminimalselfienumberofminimalselfishsubsetsofnotcontaining

therhand,foranymini-malselfishsubsetofcontaining,bysubtracting1fromeachelement,andthentakingawaytheelementfromtheset,weobtainaminimalselfishsubsetof(sinceandcannotbothoccurinaselfishset).Conversely,anyminimalselfishsubsetofgivesrisetoaminimalselfialselfi,wehave,wheredenotesthethtermoftheFibonaccisequence.B-2Byestimatingtheareaunderthegraphofperandlowerrectanglesofwidth2,wegetusingup-Since,wehave,uponexpo-nentiatingandtakingsquareroots,

er,,bytheCayley-Hamiltontheorem).Thisisacontradiction.B-5Consideracheckerboard,inwhichwewritean-letterstring,tringisbalanced,wecancovereachpairofadjacentsquarescontainingthesameletterwithadomino,andthesewillnotoverlap(becausenothreeinarowcanbethesame).Moreover,anydominoisseparatedfromthenextbyanevennumberofsquares,sincetheymustcoveroppositeletters,sely,anyarrangementofdominoeswheread-jacentdominoesareseparatedbyanevennumberofsquarescorrespondstoauniquebalancedstring,wechoosewhetherthestringstartswithotherwords,tthesearrangementsbynumberingthesqisisdecided,havearrangementsinthefirstcaseandinthesecond,henumberofbalancedstringsisB-6Wewillprovetheclaimassumingonlythattheconvexhullofthepointsterior.(ThankstoMarshallcontainsBuckforthepointingorigininoutitsthatin-thelastthreewordsarenecessaryintheprevioussen-tence!)Letsideofthegivenequationissothattheleft-hand(1)4Nownotethat(1)isthegradientofthefunctionandsoitsuffiywehaveNotethatthismaximumispositiveforifwehadforall:ofthe-planewould,thenbeathehalf-planesubsetcontainingallofthepointswouldthennotcontaintheorigin,a,onvexhullThefunctiontheunitcircle,uousHenceonithasaglobalminimum,andsoforall,.Since,theinfimumofisthesameovertheentire-planeasoverthisdisk,alueatsomepointinthedisk,whichattainsistheitsinfikieshassuggestedanalternatesolutionasfol-lows:for,drawthelooptracedby(1),thisofcoursehaswindingnumber0aboutanypoint,butforwindingnumber1aboutlarge,theoneorigin,canshowsosomewherethisloophasinbetweentheloopmustpassthroughtheorigin.(Prov-ingthislatterfactisalittletricky.)


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