Nongeneral preliminary method
Theminute handsp makes a full revolutionperhour
If atthebeginning of anhour thehourhandisnot at 12 theminute hand will
passthehour hand onceandonlyonce in the hour claim provenbelow
Withthis in mind weknowthat the minute hour hands will neveragainmeetin
the 12 Hr Thus starting at 1300 we seek the time when theminute El hourunique
hands meetwlinthe 1 o'clockhour
Inthe 1 o'clockhr we've afullrevolutionis za thiscanactuallybedispensedw I canbeeacharcevery minutes thehrhand madeourunitofrotation butthatwouldmakegetsthrough ofitsrevolution peoplethinktheclock'sshapedoesn'tmatter itdoesnt n mhtt Estgo.az za mlt Fa2x where t is time inminutes
halt FaIna w
at1300theminutehandis zththrough eachminute theminutehandgetsitsrevolution tooththrough its revolutionWeseek a solution where two
men ha 21ItF IT 2 FaleIz
s IstGaeeminutes t
5 minutes t l ewenI3o5I8and1305IIIamin Gas 1817th
1min4.5460
Claim a Everyhour theminute hand meetsthehour hand onceand only once
Proof Let his a II 1921 bethepositionat hour hand during the K oclockhour
and let m 0,11 La2aIbetheposition oftheminute hand duringanyhour
Facts hide L 2 0 II 2x 0
hi Litho f E 12 2 2x 2KEqclassesofthistypecannotbeused asis bc 2Aand0 are consideredequivalent
mla D However inequalities h.coza h e2xremain validgiventheconstraintsoftheproblem
MH 2x
hi and m are differentiableanTutu dn his m is differentiable on 1911anymacroscopicphysicalabjectoughtto have anatleast 2times
He've Inca hulasmcazGand delk hide m 1 I G differentiable displacementfunction
The intermediate value theorem tells us 7te4,11 st dutt D hit mltUnlesstheclock is broken theminute hand is alwaysmoving faster than the hourhandu i.e
andbothhandsneverm an za die40 movebackwards
inanhour
suppose now thatthe hands meetmore than once say at teand ta wt tee12Then ducts distr D distle did14 0
integral
Themean value 4hm tells us It cHe12 gt duke duct dict teta a dictydomain
Thus thehands meetonce andonly onceperhour D
I l
GeneralSolution Theequations above are actually correctfor allhrs however theproblem w
identifying solutions lies in thefact that a position is equivalent to x 2am forany nThus weneed a scheme for reducingvourposition sothat its always in 10,2
abuseoflanguage ifthereareuncountablymanyeqclasses
Beginning at 000000 we trackpositions
GH a a 2x htt htt 2x where istime inminutes after0000
ma a21T MH mct talkie