UK SENIOR MATHEMATICAL CHALLENGE November 4th 2010 SOLUTIONS.

Transcription

1 UK SENIOR MTHEMTICL CHLLENGE Nvember th SOLUTIONS These slutins augment the printed slutins that we send t schls Fr cnvenience, the slutins sent t schls are cnfined t tw sides f paper and therefre in many cases are rather shrt The slutins given here have been etended In sme cases we give alternative slutins, and we have included sme Etensin Prblems fr further investigatins The Senir Mathematical Challenge (SMC) is a multiple chice cntest, in which yu are presented with five alternative answers, f which just ne is crrect It fllws that ften yu can find the crrect answers by wrking backwards frm the given alternatives, r by shwing that fur f them are nt crrect This can be a sensible thing t d in the cntet f the SMC, and we ften give first a slutin using this apprach Hwever, this des nt prvide a full mathematical eplanatin that wuld be acceptable if yu were just given the questin withut any alternative answers S fr each questin we have included a cmplete slutin which des nt use the fact that ne f the given alternatives is crrect Thus we have aimed t give full slutins with all steps eplained We therefre hpe that these slutins can be used as a mdel fr the type f written slutin that is epected in the British Mathematical Olympiad and similar cmpetitins We welcme cmments n these slutins, and, especially, crrectins r suggestins fr imprving them Please send yur cmments, either by t r by pst t enquiry@ukmtcuk SMC Slutins, UKMT Maths Challenges Office, Schl f Mathematics, University f Leeds, Leeds LS 9JT UKMT, These slutins may be used freely within yur schl r cllege Yu may, withut further permissin, pst these slutins n a website which is accessible nly t staff and students f the schl r cllege, print ut and distribute cpies within the schl r cllege, and use them within the classrm If yu wish t use them in any ther way, please cnsult us at the address given abve UKMT, Senir Mathematics Challenge, Slutins

2 What is the digit crss cube cube in this crssnumber? Dwn One less than a cube B C D 5 E Slutin: C The nly -digit cubes are 7 and S the first clumn is either,, r Of the pssibilities fr dwn, nly is less than a cube S acrss is (and acrss is 7) Hence, What is the smallest pssible value f p q r when p, q and r are different psitive integers? B C 5 D E Slutin: B It seems bvius that p, q and r shuld be as small as pssible, and, as the cefficient f p is larger than that f q, that we shuld take p q, and, similarly, q r This suggests that the smallest value f p q r, where p, q and r are different psitive integers, is btained by putting p, q, and r, when we btain the value We can prve that this is the smallest pssible value as fllws If p, then p, and hence, as q, p q r ( ) ( ) 5 S the smallest pssible value ccurs when p Nw, if q, q, and s p q r ( ) ( ) 5 S the smallest pssible value ccurs when p and q Finally, if r, r, and s p q r ( ) ( ) S the smallest pssible value is and ccurs when p, q and r The diagram shws an equilateral triangle tuching tw straight lines What is the sum f the fur marked angles? B 8 C D E Slutin: C w y z Let the angles be as marked Since the angles in an equilateral triangle are each the angles n a straight line is w y z 8, w y z 8, and hence w y z Therefre,, and the sum f UKMT, Senir Mathematics Challenge, Slutins

3 The year is ne in which the sum f the digits is a factr f the year itself Hw many mre years will it be befre this is net the case? B C 9 D E 5 Slutin: B In the cntet f the SMC, we need nly check the given alternatives until we find ne that is crrect is nt a divisr f B 9 9 is a divisr f T prve that this really is crrect, we als need t check the ther years frm t 5 We have that which is nt a divisr f, 5 which is nt a divisr f, 7 which is nt a divisr f, and 5 8 which is nt a divisr f 5 Hence is the first year after with the required prperty 5 ntice n Mrecambe prmenade reads It wuld take millin years t fill Mrecambe Bay frm a bath tap ssuming that the flw frm the bath tap is litres a minute, what des the ntice imply is the apprimate capacity f Mrecambe Bay in litres? B C D E Slutin: D We are nly asked fr an apprimate answer, s we can ignre leap years There are minutes in a day, and hence 5 minutes in a year, and hence ( 5) ( ) minutes in millin years Thus, accrding t the ntice, the capacity f Mrecambe Bay is apprimately ( 5) ( ) litres We nw need t find an efficient way t apprimate this number We use the symbl fr is apprimately equal t We have that ( 9) (5 ) Therefre ( 5) ( ) ( 9) ( ) ( 9 ) ( ) ( 5 ) ( ) Nte: This estimate culd be btained in ther ways and yu may be able t find a mre efficient methd The ntice mentined in the questin may be fund n the stretch f the prmenade between the centre f Mrecambe and Heysham We dn t knw hw the capacity f the Bay was measured Etensin Prblem: Check the plausibility f the estimated capacity by using a map t estimate the area f Mrecambe Bay, t see what the estimated capacity implies abut the average depth f the water Dean runs up a muntain rad at 8 km per hur It takes him ne hur t get t the tp He runs dwn the same rad at km per hur Hw many minutes des it take him t run dwn the muntain? B C 5 D 5 E 9 Slutin: B Since it takes hur at 8 km per hur t get t the tp, the length f the muntain rad is 8km Hence, running at km per hur it takes 8 th f an hur, that is, minutes, t run dwn the rad UKMT, Senir Mathematics Challenge, Slutins

4 Nte: It isn t really necessary t wrk ut that the length f the rad is 8 km We can argue directly that running at km per hur takes 8 th f the time taken t run the same distance at 8 km per hur S the time taken t run dwn the rad is 8 th f the time taken t run up it 7 There are arrangements f the five letters in the wrd NGLE If all are listed in alphabetical rder starting with EGLN and finishing with NLGE, which psitin in the list des NGLE ccupy? 8th B th C nd D th E th Slutin: C There are! arrangements f the letters NGLE In alphabetical rder these run frm EGLN t NLGE S the first arrangements f the letters in NGLE run frm EGLN t NLGE Since there are! arrangements f letters, the first f these arrangements begin E, the net with G, the net with L, and the final with N S the 9th t th arrangements are NEGL, NELG, NGEL, NGLE, NLEG, NLGE Thus NGLE is the nd arrangement in the list 8 Which f the fllwing is equivalent t ( y z)( y z)? y z B y z C y z z D ( y z) E ( y z) Slutin: D We shuld try t avid having t epand the prduct ( y z)( y z), and, instead lk fr smething better backgrund idea in prblems f this kind is that we might be able t eplit the standard factrizatin f the difference f tw squares, that is, a b ( a b)( a b) With this in mind we see that ( y z)( y z) ( ( y z))( ( y z)) ( y z) cmplete mathematical answer requires us t shw that nne f the ther epressins is equivalent t ( y z)( y z) Hw can we d this? Yu mustn t fall int the trap f thinking that just because epressins lk different, they are different Fr eample, we learn in trignmetry that the epressins cs and sin, which lk different, really are the same What d we mean that tw algebraic epressins are the same? We mean that they give the same values fr all the (relevant) values f the variables S we can shw that tw algebraic epressins are different by giving a single set f values f the variables fr which the epressins take different values There are lts f ways f ding this here If we put, y, z (always try a simple eample first!), we have that ( y z)( y z) 9, and y z 5 ; B y z ; C y z z ; D ( y z) 9 ; and E ( y z) This shws that nne f the epressins, B, C and E is equivalent t ( y z)( y z) (Indeed, since the epressins, B, C, D and E have different values when, y and z, we can deduce that nne f these epressins are equivalent) Etensin Prblem: Shw that y is equivalent t eactly ne f the fllwing epressins (i) ( y)( y y ), (ii) ( y)( y y ), (iii) ( y)( y y ), (iv) ( y)( y y ) UKMT, Senir Mathematics Challenge, Slutins

5 9 The symbl is defined by y y y What is the value f ( )? B C D E Slutin: D We have that 8 9 Hence ( ) () ( ) square is cut int 7 squares f which have area cm What is the length f the side f the riginal square? cm B 7 cm C 8 cm D 9 cm E cm Slutin: E If the 7th square tgether with the squares, makes up a square, bth the 7th square, and the riginal square must have side lengths which are an integer number f cms S, if the side f the riginal square is cm, we require that is als a square In the cntet f the SMC, it is easy t check the given alternatives in turn We have that, 7, 8 8, 9 5 and 8 S we see that the answer is cm n apprach which des nt wrk backwards frm the given alternatives is t seek a slutin f y, with and y psitive integers We have that y y ( y)( y) Since y, y are psitive integers with y y, the nly pssibilities are y, y ; y 8, y ; y, y and y 9, y Only in the case y 8, y, are and y integers, namely and y 8 Finally, we need t check that squares f area cm, can be fitted tgether with a square f side length 8, t make a square f side length The diagram alngside shws ne way this can be dne S is the slutin Nte: We can see that nly in the case y 8, y are and y integers, by directly slving each pair f equatins lternatively, we culd use a parity argument If and y are integers, then y, y must either bth be even numbers r bth be dd numbers (why?) This rules ut the cases y, y ; y, y and y 9, y What is the median f the fllwing numbers? 9 B 9 C D 5 7 E 5 Slutin: D T arrange these numbers in rder f magnitude it helps t calculate their squares We see that UKMT, Senir Mathematics Challenge, Slutins 5

6 (9 ( ) 5) 8 ; ( 9) 9 9 7; ( ) 7 ; (5 7) ; and 5 8 Using the fact that fr, y, if y then y, it fllws that Hence the median f these numbers is 5 7 The diagram, which is nt t scale, shws a square with side length, divided int fur rectangles whse areas are equal What is the length labelled? 7 B C 5 9 D E 5 Slutin: We have labelled the vertices f the square P, Q, R and S, P T Q and the ther pints as shwn Each f the fur rectangles int which the square is divided U V has the same area S each rectangle has area S WS Hence PW Hence, as the rectangle PTXW has area, W X Y WX Hence UV XY WX [There are several ther rutes t this cnclusin] S R Nte: In this case the rectangles TQVU and UVYX are cngruent The backgrund t this questin is the prblem f dividing a square int rectangles f equal area, n tw f which are cngruent The simplest slutin invlves dividing a square int 7 nn-cngruent rectangles with equal areas, as shwn n the right This result is due t Blanche Descartes, Eureka, 97 Blanche Descartes was a cllabrative pseudnym used by R Lenard Brks, rthur Stne, Cedric Smith, and W T Tutte, wh met in 95 as undergraduate students in Cambridge They prved a number f results abut tessellatins Mst ntably, they slved the prblem f squaring the square by shwing that a square may be divided int smaller squares, n tw f which are the same Etensin Prblems () Shw that it is nt pssible t divide a square int, r nn-cngruent rectangles with equal areas () Shw that if a square is divided int 5 rectangles, with equal areas, as shwn n the right, then at least tw f the rectangles will be cngruent () [Hard] Shw that it is nt pssible t divide a square int 5 r nn-cngruent rectangles with equal areas () [Hard] Calculate the dimensins f the rectangles in the divisin f a square int 7 rectangles as shwn abve (assuming that the square has side length ) UKMT, Senir Mathematics Challenge, Slutins

7 Hw many tw-digit numbers have remainder when divided by and remainder when divided by? 8 B 7 C D 5 E Slutin: The psitive integers that have remainder when divided by are,, 7,,, and thse that have remainder when divided by are,,,, The least psitive integer in bth these lists is Nw n is an integer which has the same remainder as when divided by and when divided by, if and nly if n is divisible by bth and That is, if and nly if n is divisible by S the integers that have remainder when divided by, and remainder when divided by, are thse f the frm k, where k is an integer The tw-digit psitive integers f this frm are,,,, 58, 7, 8 and 9 There are 8 numbers in this list The parallel sides f a trapezium have lengths and y respectively The diagnals are equal in length, and ne diagnal makes an angle with the parallel sides as shwn What is the length f each diagnal? y Slutin: y E B y sin C ( y) cs D ( y) tan E y cs We label the vertices f the trapezium P, Q, R and S as shwn, P Q and we let T, U be the pints where the perpendiculars frm P, Q, respectively, t the line RS meet this line Since the diagnals SQ and PR are equal and PT QU, we see, that the right-angled triangles QUS and PTR are cngruent S S T y U R SU TR and hence ST UR Nw ST UR SR TU SR PQ y Hence ST y and therefre SU ST TU ST PQ ( y ) y Frm the right-angled triangle QSU we have SU SQ cs Hence SU y SQ cs cs 5 What is the smallest prime number that is equal t the sum f tw prime numbers and is als equal t the sum f three different prime numbers? 7 B C D 7 E 9 Slutin: E The sum f tw dd prime numbers will be an even number greater than, and s cannt be a prime S the nly way fr a prime number t be the sum f tw prime numbers is fr it t be f the frm p, where p is a prime number Similarly, the sum f three prime numbers can nly be prime if it is the sum f three dd primes The three smallest dd prime numbers are, 5, 7, but 5 7 5, which is nt prime The net smallest sum f three dd primes is S 9 is the smallest prime number which is bth the sum f tw prime numbers and the sum f three different prime numbers UKMT, Senir Mathematics Challenge, Slutins 7

8 PQRS is a quadrilateral inscribed in a circle f which PR is a diameter The lengths f PQ, QR and RS are, 5 and 5 respectively What is the length f SP? Slutin: E B 8 C D E 9 Because PR is a diameter, the angles PQR and PSR are Q bth right angles Therefre, by Pythagras Therem PR PQ QR and PR PS RS, and hence 5 PQ QR RS SP P R Therefre, SP PQ QR RS 5 5 We nw need t find the value f SP withut using a calculatr 5 We have that 5 (5 ) (5 5) 5 ( 5 ) 5 ( 5) 5 (5 ) S Therefre SP It fllws that SP 9 Ntes: 5 5 (5 ) ( ) ( 5 ) ) Yu shuld already knw that the angle in a semi-circle is a right angle Etensin Prblem: Can yu prve this? ( ) ) We have that PR PQ QR S the side lengths f the right angled triangle PQR are 5, and 5 Nte that these are in the rati 5:: The numbers 5,, frm what is called a primitive Pythagrean triple, that is they are integer slutins f the equatin y z, where, y and z have n cmmn factrs ther than Etensin Prblem: Find the primitive Pythagrean triple which crrespnds in a similar way t the side lengths f triangle PRS 7 One f the fllwing is equal t B C 9 fr all values f Which ne? 8 D 9 E 8 9 Slutin: E We use the fact that a a when a, and the inde rule 9 ( ) 8 (9 ) 9 9 ( a b ) c a bc It fllws that T cmplete the slutin we need t shw that nne f the ther epressins is equal t 9 fr all values f s in the slutin t Questin 8, it is gd enugh t find a single value f fr which epressins, B, C and D, d nt have the same value as the values f, B, C and D are equal t, 9 fr all values f, 9 When, while 8 and 9, respectively This shws that nne f them is Etensin Prblem: Find a value f t shw that the epressins and B are nt equivalent, and a value f t shw that epressins C and D are nt equivalent UKMT, Senir Mathematics Challenge, Slutins 8

9 8 slid cube f side cm is cut int tw triangular prisms by a plane passing thrugh fur vertices as shwn What is the ttal surface area f these tw prisms? 8( ) B (8 ) C 8( ) Slutin: D ( ) E 8 We label the vertices f the cube as shwn The ttal surface area f P the tw prisms is the surface area f the cube plus twice the area f SQUW, as this is a face f bth prisms S Q Each f the si faces f the cube has side cm and s has area cm S the surface area f the cube is cm cm R By Pythagras Therem, SQ and WU each have length cm W U Hence the rectangle SQUW has area cm S the ttal surface area f the tw prisms is cm = V 8 cm 8( ) cm 9 The diagrams shw tw different shaded rhmbuses each inside a square with sides f length Each rhmbus is frmed by jining vertices f the square t midpints f the sides f the square What is the difference between the shaded areas? B C D E Slutin: B The square has side and hence has area By adding three vertical lines and ne hrizntal line, as shwn, we divide the square int cngruent triangles The shaded rhmbus is made up f f these triangles S the shaded area is ne quarter f the ttal area f the square S the shaded rhmbus has area 9 P Q R We add the diagnal frm the tp-left verte t the bttm-right verte f the square, and label the pints P, Q, R, S, T, U, V and W as shwn The triangles QRT, PQT, PTS have bases PQ, QR, PS T respectively f equal lengths, and they have equal heights S they have S the same area They make up the triangle PRS whse area is ne-quarter V f the area f the square S each f the triangles QRT, PQT and PTS has area ne-twelfth that f the square, namely S triangle PTR has area Triangle RVW is cngruent t triangle PTR and s als has area Nw U W triangle PRW has area 8, and hence the area f triangle RTV is 8 Similarly, triangle TUV has area Hence the area f the shaded rhmbus is It fllws that the difference in the areas f the shaded rhmbuses in the tw figures is 9 UKMT, Senir Mathematics Challenge, Slutins 9

10 There are girls in a mied class If tw pupils frm the class are selected at randm t represent the class n the Schl Cuncil, then the prbability that bth are girls is 5 Hw many bys are in the class? B C 5 D 8 E Slutin: C Suppse there are bys in the class Then there are pupils altgether, f whm are girls S the prbability that the first pupil chsen is a girl is If a girl is chsen, there remain 9 9 pupils, f whm 9 are girls, s that the prbability that the secnd pupil chsen is als a girl is 9 9 Hence the prbability that the tw chsen pupils are bth girls is 9 9 ( )(9 ) 9 Therefre, we need t slve the equatin 5 In the cntet f the SMC yu ( )( 9) culd just try the suggested alternatives in turn mathematically better methd is t slve this 9 equatin algebraically We have that 5 9 ( )(9 ) ( )(9 ) ( )( 5) This gives 5 r Since yu shuldn t have negative pupils (and certainly yu can t have a negative number f them), the slutin is 5 The diagram shws a regular heagn, with sides f length, inside a square Tw vertices f the heagn lie n a diagnal f the square and the ther fur lie n the edges What is the area f the square? B C D Slutin: We give three alternative slutins t this prblem 7 E () Suppse that the square has side length y We let J, K, L and M J N P K be the vertices f the square and let P, Q, R, S, T and U be the vertices f the heagn, as shwn We let O be the centre f the heagn, and N U be the pint where the perpendicular frm S t JK meets JK Q Since PQ is parallel t the diagnal JL, QPK 5 The triangle T O PQO is equilateral Hence QPO It fllws that R SPN SNJM is a rectangle and hence M S L SN MJ y ls PO OS PQ, and hence PS SN y Frm the right angled triangle PSN, we have that sin 75 Therefre PS y sin 75 sin( 5 ) [sin cs 5 cs sin 5 ] [ It fllws that y ( ) ( ) ] ( ) UKMT, Senir Mathematics Challenge, Slutins

11 () We apply the Sine Rule t the triangle JPU J P In this triangle, PJU 5 JPU and hence PUJ We als have that JP JK KP y and PU Therefre, applying the Sine Rule U PU JP t the triangle JPU, we btain sin PJU sin PUJ that is y sin 5 sin and hence sin sin y Therefre y ( ), and, as in sin 5 sin 5 the first methd, y () Our third methd avids the use f trignmetry We let K, M, O, P K P and Q be as in the diagram f slutin () We jin the cmmn centre, O, f the square and the heagn, t the pints K, M, P and X Q Let X be the pint where KM meets PQ Then OX is the height Q f the equilateral triangle OPQ and XK is the height f the issceles O right-angled triangle KPQ Nw OQ and QX S, applying Pythagras Therem t triangle OXQ, we have OX We als have XK XQ M Hence KM OK ( OX XK) It fllws that the side length f the square is KM ( ), and hence the area f the square is ( Nte: The first tw methds require us t knw (withut a calculatr) the sines and csines f the angles, 5 and These can be remembered using the right angled issceles triangle with angles 5, 5 and equilateral triangle C 9, and the triangle with degrees, and 9 btained by bisecting an If the sides, B, BC, adjacent t the right angle have length, then, by Pythagras Therem, the hyptenuse, C, has length B BC 5 9 B Therefre sin 5 and cs 5 C C P If PQ has length, then QR Therefre by Pythagras Therem applied t the right-angled triangle PQR, we have that Q R and Etensin Prblems: Find epressins fr Find epressins fr sin5 sin 75 PQ, and hence PQ We have frm triangle PQR that sin sin cs PR PQ cs5 in terms f surds and cs 75 in terms f surds and cs, giving, QR PQ PR UKMT, Senir Mathematics Challenge, Slutins

12 If p q, where p and q are psitive integers, which f the fllwing culd nt equal? B 8 5 C 8 7 D E 5 Slutin: B If p q, then p q Hence p q p( p q) q ( p q) pq the epressins given in the questin, we see that we need t cnsider the cases pq, pq 5 pq 7 Frm and Since p and q are psitive integers, if pq, then either p and q, r p and q S p q is either r S alternatives and D culd equal Similarly, if pq 5, then either p and q 5, r p 5 and q S p q is either r S alternative E culd equal but B culd nt Finally, if pq 7, then either p and q 7, r p 7 and q S p q either 8 r 5 S alternative C culd equal, is The diagram shws tw different semicircles inside a square with sides f length The cmmn centre f the semicircles lies n a diagnal f the square What is the ttal shaded area? B ( ) C D ( ) E 8 ( ) Slutin: B W P R X Z Q Y We let W, X, Y, Z be the vertices f the square, as shwn We let O be the cmmn centre f the tw semicircles, and we let P, Q be the pints where the tw semicircles tuch the edges WX and ZY, respectively Then as OP and OQ are radii f the semicircles, they are perpendicular t WX and ZY S POQ is a straight line which is parallel t WZ and XY We let R be the pint where the larger semicircle meets WX, as shwn Suppse that the smaller semicircle has radius r Frm the rightangled issceles triangle POR we see that the larger semicircle has radius r Then PQ r r Hence, as the square has side length, r r, and hence ( ) r ( ) ( )( ) The shaded area is the sum f the areas f the tw semicircles S the shaded area is r r) r (( )) ( ) ( ) ( UKMT, Senir Mathematics Challenge, Slutins

13 Three spheres f radius are placed n a hrizntal table and inside a vertical hllw cylinder f height units which is just large enugh t surrund them What fractin f the internal vlume f the cylinder is ccupied by the spheres? 7 Slutin: E B C D E 7 The diagram represents a hrizntal crss sectin thrugh the centres, say P, Q, R, f the spheres T We let O be the pint where this crss sectin meets the central ais f the cylinder We let S be the pint where the line QO meets PR and we let T be the pint where the P line OP meets the cylinder We let be the length f OP S PQR is an equilateral triangle f side length S SPO is O a triangle with angles 9, and degrees, in which R Q PS, OP and SO By Pythagras Therem applied t this triangle (), and hence, s Hence the radius f the cylinder, which is equal t the length f OT is OP PT 7 It fllws that the vlume f the cylinder is (7 ) The ttal vlume f the three spheres f radius is Hence the rati f these vlumes is (Nte that O is the centrid f the equilateral triangle (7 ) 7 PQR It is a standard result that the centrid divides the medians f a triangle in the rati :) 5 ll the digits f a number are different, the first digit is nt zer, and the sum f the digits is There are N 7! such numbers What is the value f N? 7 B 97 C D 7 E 8 Slutin: D 9 5, s we btain a set f digits with sum by mitting digits with sum 9 There are eight cmbinatins f nn-zer different digits with sum 9, namely 9, 8, 7,, 5,, 5 and Nw k digits can be arranged in rder in k! different ways S, deleting when it is the first digit, these k digits give rise t k! different numbers, all with the same sum f digits The numbers btained fr any set f digits will be different t thse btained frm a different set, s we get the ttal number f such numbers, by adding up the number f numbers in each f the separate cases If we mit the digit 9, there remain 9 digits which frm 9! different numbers whse digits have sum Similarly, mitting 8, r 7, r, r 5,, in each case we can frm 8! different numbers whse digits have sum, and mitting,, r 5,, r,,, in each case we can frm 7! such numbers S the ttal number f such numbers is 9! ( 8!) ( 7!) ( ) 7! (7 ) 7! 7 7! UKMT, Senir Mathematics Challenge, Slutins