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