1. Extend QR downwards to meet the x-axis at U(6, 0). y

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1 In the digrm, two stright lines re to be drwn through so tht the lines divide the figure OPQRST into pieces of equl re Find the sum of the slopes of the lines R(6, ) S(, ) T(, 0) Determine ll liner functions f () b such tht if g () f () for ll vlues of, then f() g() for ll vlues of (Note : f is the inverse function of f ) () Determine ll pirs (, b) of positive integers for which b 0 (b) Determine ll rel vlues of for which log ( ) log ( ) Integers cn be written in bses other thn the usul bse 0 For emple, the nottion (5) 7 stnds for the bse 7 representtion of the integer (which equls in bse 0) In generl, if, nd z re integers between 0 nd b, inclusive, then (z) b b b z Find ll triples (,, z) for which (z) 0 (z) 7, where ech of, nd z comes from the list,,,, 5, 6 5 Vernon strts with first number n with 0 < n < Vernon enters the first number into mchine to produce second number He then enters the second number bck into the mchine to produce third number Vernon continull enters the result bck into the mchine giving chin of numbers When Vernon enters number into the mchine, z if, the mchine outputs, nd z if >, the mchine outputs ( ) If the mchine ever produces the number, Vernon stops the process () A chin strts with This gives 6 0 Wht re the net four numbers in this chin? (b) Vernon enters number with 0 < < into the mchine nd the mchine produces the number Determine the vlue of (c) The fourth number in chin is Determine ll of the possible vlues of the first number in this chin Solutions Etend QR downwrds to meet the -is t U(6, 0) R(6, ) S(, ) U T(, 0) The re of figure OPQRST equls the sum of the res of squre OPQU (which hs side length 6, so re 6) nd rectngle RSTU (which hs height nd width 6, so re ) Thus, the re of figure OPQRST is If we re to divide the figure into three pieces of equl re, then ech piece hs re 6 Let V be the first point on the perimeter Mthemtics TODAY NOVEMBER 7 Pge 7

2 (mesured clockwise from P) so tht the line through O nd V cuts off n re of 6 Note tht the re of DOPQ is hlf of the re of squre OPQU, or, so V is to the left of Q on PQ Thus, V hs coordintes (v, 6) for some number v 7 Mthemtics TODAY NOVEMBER V R(6, ) S(, ) U T(, 0) Consider DOPV hving bse OP of length 6 nd height PV of length v Since the re of DOPV is 6, then (6)(v) 6 or v 6 or v Therefore, the slope of OV is 6 Let W be the second desired point Since the re of DOTS is (OT)(TS) () () (less thn of the totl re) nd the re of trpezoid ORST is (RS OT) (ST) (6 ) () (more thn of the totl re), then W lies on RS V R W S(, ) U T(, 0) Suppose tht W hs coordintes (w, ) for some number w We wnt the re of trpezoid WSTO to be 6 Therefore, (WS OT) (ST) 6 ( w ) () 6 w 6 w Thus, the coordintes of W re (, ), nd so the slope of OW is Thus, the sum of the two required slopes is 9 Since f () b, we cn determine n epression for g() f () b letting f () nd to obtin b We then interchnge nd to obtin b which we solve for to obtin b or b Therefore, f b () Note tht 0 (This mkes sense since the function f() b hs grph which is horizontl line, nd so cnnot be invertible) Therefore, the eqution f() g() b becomes ( b ) or b b 0, nd this eqution is true for ll Since the eqution b b 0 is true for ll, then the coefficients of the liner epression on the left side must mtch the coefficients of the liner epression on the right side Therefore, 0 nd b b From the first of these equtions, we obtin FOR MORE DETAILED SOLUTION REFER NOVEMBER ISSUE OF MATHEMATICS TODAY or, which gives or If, the eqution b b becomes b b, which gives b If, the eqution b b becomes b b, which is not possible Therefore, we must hve nd b, nd so f() Pge 7

3 () First, we fctor the left side of the given eqution to obtin ( b) 0 Net, we fctor the integer 0 s Note tht ech of, nd 6 is prime, so we cn fctor 0 no further (We cn find the fctors of nd using tests for divisibilit b nd, or b sstemtic tril nd error) Since 0 6, then the positive divisors of 0 re,,,, 6,, 67, 0 Since nd b re positive integers, then nd b re both positive integers Since nd b re positive integers, then nd b > 0, so b > Since ( b) 0, then nd b must be divisor pir of 0 (tht is, pir of positive integers whose product is 0) with < b We mke tble of the possibilities: b b b N/A Note tht the lst cse is not possible, since b must be positive Therefore, the three pirs of positive integers tht stisf the eqution re (, 006), (, ), (, ) (We cn verif b substitution tht ech is solution of the originl eqution) (b) We successivel mnipulte the given eqution to produce equivlent equtions : log ( ) log ( ) log ( ) log ( ) log (( ) ) (using log A log B log AB) ( ) (eponentiting both sides) 6 (multipling b ) 6 (dividing both sides b 0) 6 6 Net, we mke the substitution t, noting tht t Thus, we obtin the equivlent equtions t 6t 6t t 0 (t ) (t ) 0 Therefore, t or t Since t > 0, then we must hve t Thus, log( / ) log log log / ( / ) log( / ) log log log log log log log log Since we re told tht (z) b b b z, then (z) z 00 0 z nd (z) z 9 7 z From the given informtion (z) 0 (z) z (9 7 z) 00 0 z 9 z z Mthemtics TODAY NOVEMBER 75 Pge 75

4 Since the left side of this eqution is n even integer (), then the right side must lso be n even integer Since is n even integer, then for z to be n even integer, it must be the cse tht z is n even integer This gives us three possibilities: z, z nd z 6 Cse : z Here, or We tr the possible vlues for : z gives () ; this gives the triple (,, z) (,, ) z gives () 5; this gives the triple (,, z) (5,, ) z If is t lest, then is t lest 7, which is impossible Therefore, there re two triples tht work when z Cse : z Here, or We tr the possible vlues for : z gives () ; this gives the triple (,, z) (,, ) z gives () 6; this gives the triple (,, z) (6,, ) z If is t lest, then is t lest, which is impossible Therefore, there re two triples tht work when z Cse : z 6 Here, 6 or We tr the possible vlues for : z gives () 5; this gives the triple (,, z) (5,, 6) z If is t lest, then is t lest 7, which is impossible Therefore, there is one triple tht works when z 6 Finll, the triples tht stisf the eqution re (,, z) (,, ), (5,, ), (,, ), (6,, ), (5,, 6) 5 () Since the fourth number in the chin is nd is less thn, then the fifth number is Since the fifth number in the chin is nd is less thn, then the sith number is Since the sith number in the chin is nd is lrger thn, then the seventh number is 6 Since the seventh number in the chin is 6 nd 6 is lrger thn, then the eighth 6 number is 5 0 Therefore, the net four numbers in the chin re, 6 0,, (b) Therefore re two possibilities : nd > If nd is entered into the mchine, then the mchine outputs Since the input is to be the sme s the output, then or 0 This is impossible since > 0 If > nd is entered into the mchine, then the mchine outputs ( ) Since the input is to be the sme s the output, then ( ) or Thus, nd so, which stisfies the restrictions Therefore, if is entered into the mchine nd is produced, then (c) Suppose tht the chin is, b, c, In this prt, we hve to produce the chin 76 Mthemtics TODAY NOVEMBER Pge 76

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6 bckwrds (tht is, we hve to determine c, b nd ) We mke the following generl observtion: Suppose tht is entered into the mchine nd is produced If, then Solving for in terms of, we obtin (This gives us the input in terms of the output) If >, then ( ) Solving for in terms of, we obtin or or ( ) Since the third number in the chin is c nd the fourth is, then from bove, c () or c ( ) In either cse, the chin is, b,, Since the second number in the chin is b nd the third is, then from bove, b or b Therefore, the chin is either,,, or,,, If the chin is,,,, then the first number is nd the second number is Thus, either or 7 7 If the chin is,,,, then the first number is nd the second number is Thus, either or 5 5 Therefore, the possible first numbers in the chin re 5 7,,, We cn double check tht ech of these gives fourth number of : Mthemtics TODAY NOVEMBER Pge 7