UK SENIOR MATHEMATICAL CHALLENGE November 6th 2012 EXTENDED SOLUTIONS

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1 UK SNIOR MTHMTIL HLLNG Novembe 6th 0 XTN SOLUTIONS These solutions augment the pinted solutions that we send to schools. Fo convenience, the solutions sent to schools ae confined to two sides of pape and theefoe in many cases ae athe shot. The solutions given hee have been extended. In some cases we give altenative solutions, and we have included some xtension Poblems fo futhe investigations. (The shot solutions have been appended and appea at the end of this document.) The Senio Mathematical hallenge (SM) is a multiple choice contest, in which you ae pesented with five options, of which just one is coect. It follows that often you can find the coect answes by woking backwads fom the given altenatives, o by showing that fou of them ae not coect. This can be a sensible thing to do in the context of the SM. Howeve, this does not povide a full mathematical explanation that would be acceptable if you wee just given the question without any altenative answes. So fo each question we have included a complete solution which does not use the fact that one of the given altenatives is coect. Thus we have aimed to give full solutions with all steps explained. We theefoe hope that these solutions can be used as a model fo the type of witten solution that is expected when pesenting a complete solution to a mathematical poblem (fo example, in the itish Mathematical Olympiad and simila competitions). We welcome comments on these solutions, and, especially, coections o suggestions fo impoving them. Please send you comments, eithe by to: enquiy@ukmt.co.uk o by post to: SM Solutions, UKMT Maths hallenges Office, School of Mathematics, Univesity of Leeds, Leeds LS 9JT. Quick Making Guide suppoted by UKMT, 0. These solutions may be used feely within you school o college. You may, without futhe pemission, post these solutions on a website which is accessible only to staff and students of the school o college, pint out and distibute copies within the school o college, and use them within the classoom. If you wish to use them in any othe way, please consult us at the addess given above.

2 . Which of the following cannot be witten as the sum of two pime numbes? Solution: If the odd numbe is the sum of two pime numbes, it must be the sum of the only even pime,, and an odd pime. ut = + 9 and 9 is not pime. So is not the sum of two pime numbes. On the othe hand, 5 = +, 7 = + 5, 9 = + 7 and 0 = + 7. Theefoe, these ae all the sum of two pime numbes. xtension Poblem. In 7 the Geman Mathematician histian Goldbach (690-76) made the conjectue that evey even numbe geate than is the sum of two pimes. The Goldbach onjectue, as it is now called, has, so fa, not been poved, but no counteexample to it has been found. Investigate this conjectue by finding the even numbe in the ange fom to 00 which can be expessed as the sum of two pime numbes in the lagest numbe of ways.. The diagam shows an equilateal tiangle, a squae and a egula pentagon which all shae a common vetex. What is the value of θ? θ Solution: The inteio angles of an equilateal tiangle, a squae and a egula pentagon ae 08, espectively. Hence θ = 60 ( ) = = 0. 60, 90 and. The pice of my favouite soft dink has gone up in leaps and bounds ove the past ten yeas. In fou of those yeas it has leapt up by 5p each yea, whilst in the othe six yeas it has bounded up by p each yea. The dink cost 70p in 00. How much does it cost now? Solution: The cost has isen by 5p fou times, and by p six times. So the total pice ise in pence has been ( ) =. Theefoe the pice now is ( 70 + )p = 0p =. 0.. ccoding to one astonome, thee ae one hunded thousand million galaxies in the univese, each containing one hunded thousand million stas. How many stas is that altogethe? Solution: 5 6 One hunded thousand is 0 and one million is0. So one hunded thousand million is0 0 = 0 = 0 +. Theefoe the total numbe of stas is0 0 = 0 = 0.

3 5. ll six digits of thee -digit numbes ae diffeent. What is the lagest possible sum of thee such numbes? Solution: Suppose that the six digits that ae used ae a, b, c, d, e and f and that we use them to make the thee -digit numbes ad, be and cf. The sum of these numbes is ( 0a + d) + (0b + e) + (0c + f ), that is 0( a + b + c) + ( d + e + f ). To make this as lage as possible, we need to choose a, b and c so that a + b + c is as lage as possible, and then d, e and f fom the emaining digits so that d + e + f is lage as possible. lealy, this is achieved by choosing a, b, c to be 7, 8, 9, in some ode, and then d, e, f to be, 5, 6 in some ode. So the lagest possible sum is0 ( ) + ( ) = 55. [Fo example, the thee -digit numbes could be 7, 85 and 96.] 6. What is the sum of the digits of the lagest -digit palindomic numbe which is divisible by 5? [Palindomic numbes ead the same backwads and fowads, e.g. 77.] Solution: digit palindomic numbe has the fom deed whee d and e ae digits. This numbe is divisible by 5 if and only if it is divisible by both and 5. It will be divisible by 5 if and if only if d is 0 o 5, but the fist digit of a numbe cannot be 0, so d is 5, and the numbe has the fom 5ee5. numbe is divisible by if and only if the sum of its digits is divisible by. So 5ee5 is divisible by if and only if 0 + e is divisible by, that is, when e is 7, o. So the lagest -digit palindomic numbe which is divisible by 5 is The sum of the digits of 5775 is =. xtension poblems 6. Find the lagest -digit palindomic numbe which is a multiple of Find the lagest 5-digit and 6-digit palindomic numbes which ae multiples of The agument above uses the fact that a positive intege is divisible by if and only if the sum of its digits is divisible by. xplain why this test woks. 6. Find a simila test fo divisibility by Find a test, in tems of its digits, to decide when a positive intege is divisible by. 7. Given that x + y + z =, x + y z = and x y z =, what is the value of xyz? 0 Solution: dding the equations, x + y + z = and x y z =, gives x =, fom which it follows that x =. dding the equations x + y + z = and x + y z =, gives ( x + y) =. Hence x + y =. Theefoe, as x =, it follows that y =. Since x + y + z =, it now follows that z = ( x + y) = =. Theefoe, xyz = =.

4 8. The diagams below show fou types of tile, each of which is made up of one o moe equilateal tiangles. Fo how many of these types of tile can we place thee identical copies of the tile togethe, without gaps o ovelaps, to make an equilateal tiangle? Solution: 0 n equilateal tiangle may be divided into,, 9, 6,, that is, a squae numbe, of equilateal tiangles of the same size. Thee tiles, each made up of k smalle equilateal tiangles of the same size, contain altogethe k of the smalle tiangles. So, if it is possible to place thee copies of the tile to make an equilateal tiangle, then k must be a squae numbe. Fo the above tiles k takes the values,, and, espectively. Of these, only fo k = and k = do we have that k is a squae. So the only possible cases whee thee of the above tiles can be eaanged to make an equilateal tiangle ae the two tiles on the ight. We see fom the diagams above that in both cases we can use thee of the tiles to make an equilateal tiangle. So thee ae such cases. xtension poblem 8. In the agument above we have stated, but not poved, that if an equilateal tiangle is divided into smalle equilateal tiangles of the same size, then the numbe of the smalle tiangles is always a pefect squae. Pove that this is coect. 8. etemine a necessay and sufficient condition fo a positive intege, k, to be such that k is a squae. [sk you teache if you ae not sue what is meant by a necessay and sufficient condition.] 8. Is it possible to find a tile which is made up of k equilateal tiangles of the same size, whee k is a squae, but thee copies of the tile cannot be placed togethe, without gaps o ovelaps, to make an equilateal tiangle? 9. Piee said, Just one of us is telling the tuth. Qad said, What Piee says is not tue. Ratna said, What Qad says is not tue. Sven said, What Ratna says is not tue. Tanya said, What Sven says is not tue. Solution: How many of them wee telling the tuth? 0 If Piee is telling the tuth, eveyone else is not telling the tuth. ut, also in this case, what Qad said is not tue, and hence Ratna is telling the tuth. So we have a contadiction. So we deduce that Piee is not telling the tuth. So Qad is telling the tuth. Hence Ratna is not telling the tuth. So Sven is also telling the tuth, and hence Tanya is not telling the tuth. So Qad and Sven ae telling the tuth and the othe thee ae not telling the tuth.

5 0. Let N be the smallest positive intege whose digits add up to 0. What is the fist digit of N +? 5 6 Solution: If N is the smallest positive intege whose digits add up to 0, it will have the smallest possible numbe of digits among numbes whose digits add up to 0. This means that, as fa as possible, each digit should be a 9. So N will have the fom d , whee k is a positive intege and d is a non-zeo digit. Now, 0 = So k = and d = 5. Hence N = Theefoe N + = So the fist digit of N + is a 6.. oco is making clown hats fom a cicula piece of cadboad. The cicumfeence of the base of each hat equals its slant height, which in tun is equal to the adius of the piece of cadboad. What is the maximum numbe of hats that oco can make fom the piece of cadboad? Solution: k Let the adius of the cicle be. This must also be the slant height of each hat. If we flatten out a hat we get a secto of a cicle in which the cicula pat of the bounday has length, as shown in the diagam on the left. The cicumfeence of a cicle of adius has length π. Since < π <.5, it follows that 6 < π < 7. So oco can cut out 6 sectos fom the cicula piece of cad as shown, each of which can be made into a hat. The aea of cad needed fo each hat is and the aea of the cicle is π. s π < 7( ), oco cannot make 7 hats. So the maximum numbe of hats that oco can make is 6. xtension Poblems In this solution we have used the fact that < π <. 5. How do we know that this is tue? Hee we ask you to show this, stating with the definition of π as the atio of the cicumfeence of a cicle to its diamete (and making one geometical assumption).. Suppose we have cicle with diamete. Its adius is theefoe π. We conside a egula hexagon with its vetices on the cicle, and a egula hexagon which touches the cicle, as shown in the diagam. Use the assumption that the cicumfeence of the cicle lies between the peimetes of the two hexagons to show that < π <. educe that < π <. 5.. The method of appoximating the cicumfeence of a cicle by a egula polygon was discoveed independently in moe than one cultue. The Geek mathematician chimedes who lived fom 87 to used egula polygons with 96 sides to obtain the appoximation 0 < π <. The hinese mathematician Liu Hui, who lived aound 60, obtained the appoximation < π <. Use the appoximate value π. 59 to estimate the pecentage eos in these appoximations. [Today, using iteative methods that convege vey apidly, and poweful computes, π has been calculated to billions of decimal places.]

6 . The numbe can be expessed as the sum of one o moe positive integes in fou diffeent ways: ; + ; + ; + +. Solution: In how many ways can the numbe 5 be so expessed? y systematically listing all the possibilities, we see that the numbe 5 may be witten as a sum of one o moe positive integes in the following 6 ways: 5; + ; + ; + ; + ; + + ; + + ; + + ; + + ; + + ; + + ; ; ; ; ; xtension Poblems. In how many ways can the numbe be expessed as the sum of one o moe positive integes?. To genealize these esults we need a bette method than listing all the possibilities. Fom the cases n =, and 5, it seems eason to guess that a positive intege n can be expessed as the sum of one o moe positive integes in n ways. an you pove that this conjectue is coect?. We can also conside the numbe of diffeent ways of expessing a positive intege, n, as the sum of positive integes when the ode does not matte. Fo example we count + + as being the same as + +. These aangements, whee the ode does not matte, ae called patitions. You can see that thee ae just 7 diffeent patitions of 5, namely, 5, +, +, + +, + +, and ount the numbe of diffeent patitions of n, fo n 8. [Thee is no simple fomula fo the numbe of diffeent patitions of n.]. cube is placed with one face on squae in the maze shown, so that it completely coves the squae with no ovelap. The uppe face of the cube is coveed in wet paint. The cube is olled aound the maze, otating about an edge each time, until it eaches squae 5. It leaves paint on all of the squaes on which the painted face lands, but on no othes. The cube is emoved on eaching the squae 5. What is the sum of the numbes on the squaes which ae now maked with paint? Solution: T S L T We tack the position of the side of the cube which is coveed in wet paint. We imagine that the maze is hoizontal, and that we ae looking at it fom the side with the squaes which ae maked with the numbes, 6, 5, and. We use T,, F, S, L and R fo the Top, ottom, Font, Sten (back), Left and Right sides of the cube, as we look at it, espectively. Initially, the wet paint is on Top. We see fom the diagam that the wet paint is on the bottom of the cube when the cube is on the squaes labelled, 7,, 5, 0 and. The sum of these numbes is = 80. F S L F S F L S S L S T S L T

7 . Six students who shae a house all speak exactly two languages. Helga speaks only nglish and Geman; Ina speaks only Geman and Spanish; Jean-Piee speaks only Fench and Spanish; Kaim speaks only Geman and Fench; Lionel speaks only Fench and nglish whilst May speaks only Spanish and nglish. If two of the students ae chosen at andom, what is the pobability that they speak a common language? Solution: We can select two of the six students in 5 ways, as shown in the table. We have put a tick ( ) in each box coesponding to a pai of students who speak a common language, and a coss ( ) in each box coesponding to a pai of students who do not speak a common language. We see that thee ae ticks in the 5 boxes. Theefoe, if two students ae chosen at andom, the pobability that they speak a common language is = Ina G & S ltenative method Note that each of the students has a common language with out of 5 of the othe students. It theefoe follows that if two students ae chosen at andom, the pobability that they speak a common language is 5. Helga & G Ina G & S J-P F & S Kaim G & F Note: We may epesent the fou languages by lines Helga and the students by points which ae on the lines coesponding Geman Spanish to the languages they speak, as shown. This makes it easy to see that fo each student, thee is just one othe student with whom they have no common language. Lionel F & J-P F & S Kaim G & F Lionel F & May S & nglish May Lionel Ina Fench Kaim Jean-Piee 5. Pofesso Rossefop uns to wok evey day. On Thusday he an 0% faste than his usual aveage speed. s a esult, his jouney time was educed by x minutes. How many minutes did the jouney take on Wednesday? Solution: x 0x 9x 8x 5x Suppose that the distance that Pofesso Rossefop uns is d and his usual aveage speed (in tems of the units used fo distance and minutes) is s. So, on Wednesday his jouney takes him d minutes. On Thusday his aveage speed was 0 s = 00 0 s and so the jouney takes him s d 0d d 0d d d = minutes. We theefoe have = x. This gives = x and hence = x. So s s s s s s 0 on Wednesday his jouney took him x minutes. xtension Poblem 5. Now conside the geneal case whee Pofesso Rossefop uns the same distance p% slowe on Monday than on Tuesday, and q% faste on Wednesday than on Tuesday. Suppose that his un on Wednesday took him x minutes less than his un on Monday. How many minutes did his un take on Tuesday?

8 6. The diagam shows the ellipse whose equation is x + y xy + x y =. The cuve cuts the y-axis at points and and cuts the x-axis at points and. y What is the aea of the inscibed quadilateal? Solution: The gaph cuts the y-axis at points whee x = 0, and hence whee y y =, that is, whee y y = 0, that is, ( y + )( y 6) = 0, giving y = and y = 6. Hence is the point (0,6), is the point ( 0, ) and so has length 8. Similaly, and ae points whee y = 0 and hence x + x = 0, that is, ( x + )( x ) = 0, giving x = and x =. So is the point (,0), is the point (,0), and so has length 7. The aea of the inscibed quadilateal is theefoe. = (8 7) = 8. xtension Poblems 6. The solution above takes it fo ganted that the aea of the quadilateal is half the poduct of the lengths of its diagonals. Give an example to show that this is not tue fo all quadilateals. 6. What is the special popety of that makes it coect that in this case the aea is half the poduct of the length of the diagonals? Give a poof to show that you answe is coect. x 7. The diagam shows a patten found on a floo tile in the cathedal in Spoleto, Umbia. cicle of adius suounds fou quate cicles, also of adius, which enclose a squae. The patten has fou axes of symmety. What is the side length of the squae? Solution: Let O be the cente of the cicle, let P be one of the points whee two of the quate cicles meet, let Q be the cente of one of these quate cicles, and let R be the vetex of the squae that lies on OQ, as shown. Then in the tiangle OPQ thee is a ight angle at P and OP = QP =, and theefoe, by Pythagoas Theoem, OQ =. Since QR =, it follows that OR =. It follows that the diagonal of the squae has length ( ). Using, Pythagoas Theoem again, it follows that the side length of the squae is ( ( ) ) =. xtension Poblem 7. In the poof above we claim, by Pythagoas Theoem, that if the diagonal of a squae has length x, then its side length is x. Show that this is indeed a consequence of Pythagoas Theoem. Q P R O

9 8. The diagam shows two squaes, with sides of length, Solution: inclined at an angle α to one anothe. What is the value of x? cos α sin α tan α cosα sinα onside the tiangle at the bottom of the diagam. We have labelled the points P, Q and R, as shown. We let N be the point whee the pependicula fom P to QR meets QR.. Now, QPR = α 90 = 80 α. In the tiangle PQR we have PQ = PR = as they ae sides of the squaes. It follows that the tiangle PQR is isosceles and hence PQN = PRN = α. So the tiangles PQN and PRN ae conguent. Hence QN x QN = NR = x. Theefoe, fom the tiangle PQN we have cosα = = = x.that is, x = cosα. QP 9. The numbes,,, 5, 6, 7, 8 ae to be placed, one pe squae, in the diagam shown so that the sum of the fou numbes in the hoizontal ow equals and the sum of the fou numbes in the vetical column also equals. In how many diffeent ways can this be done? α Solution: Let the numbes in the squaes be t, u, v, w, x, y and z, as shown. We have ( w + x + y + z) + ( t + u + x + v) = + =, that is t ( t + u + v + w + x + y + z) + x =. ut t + u + v + w + x + y + z = u w x y z = 5 and hence x = 7. So t, u, v ae thee numbes chosen fom,,, 5, 6, 8 which add up to. It can be checked that thee v ae just two choices fo these numbes,, 8 and, 5, 6. If the numbes in the hoizontal ow ae,, 8 and 7, thee ae choices fo w and then choices fo y and then choice fo z, making choices altogethe, and then similaly 6 choices fo t, u, v, making 6 6 = 6 combinations. Likewise thee ae 6 combinations whee the numbes in the hoizontal ow ae, 5, 6 and 7. This makes = 7 diffeent ways altogethe. P x 80 α α Q x N x R 0. In tapezium PQRS, SR = PQ = 5 cm and SP is paallel to RQ. ll fou sides of PQRS ae tangent to a cicle with cente. The aea of the tapezium is 600 cm. What is the adius of the cicle? 7.5cm 8cm 9cm 0cm cm R S P Q Solution: Let the adius of cicle be cm, and let the points whee the tapezium touches the cicle be W, X, Y and Z as shown. s the adii W, X, Y S Z P and Z ae pependicula to the edges of the tapezium PQ, QR, RS and SP, Y W espectively, the aea of the tapezium, which is the sum of the aeas of the tiangles PQ, QR, RS and SP, is PQ + QR + RS + SP = ( PQ + QR + RS + SP ). R X Q Now using the popety that the two tangents to a cicle fom a given point have equal length, it follows that SP + QR = SZ + PZ + QX + RX = SY + PW + QW + RY = ( PW + QW ) + ( RY + SY ) = PQ + RS = = 50 cm. Hence PQ + QR + RS + SP = = 00 cm. Theefoe, as the aea of the tapezium of 600 cm, we have that (00) = 600 and hence =.

10 . Which of the following numbes does not have a squae oot in the fom x + y, whee x and y ae positive integes? Solution: Suppose a + = ( x + y ), whee a, x and y ae positive integes. It follows that a + = x + xy + y = ( x + y ) + xy. Theefoe xy = 6 and x + y = a. Since x and y ae positive integes, the only possibilities fo x and y ae x =, y = 6 ; x =, y = ; x =, y = and x = 6, y =. Theefoe the only possibilities fo a ae + (6 ) = 7; + ( ) = ; + ( ) = 7 and 6 + ( ) = 8. Theefoe options,, and ae numbes which have squae oots of the equied fom, but not option. xtension Poblems. Which numbes of the fom a + fom x + y, whee x and y ae positive integes?, whee a is a positive intege, have squae oots of the. Which numbes of the fom a +, whee a is a positive intege, have squae oots of the fom x + y, whee x and y ae positive integes.. Give an example of a numbe of the fom a + b 5, whee a and b ae positive integes, which does not have a squae oot of the fom x + y 5, whee x and y ae positive integes.. semicicle of adius is dawn with cente V and Y diamete UW. The line UW is then extended to the Z point X, such that the UW and WX ae of equal length. n ac of the cicle with cente X and adius is then dawn so that the line XY is a tangent to the semicicle U V W X at Z, as shown. What, in tems of, is the aea of tiangle YVW? 9 Solution: Since UW = WX, we have WX = UW =. s V is Y the cente of the cicle with UW as diamete, VW = VZ =. Z Hence VX = VW + WX =. The tiangles YVX and VVW have the same height and VW = VX. Theefoe aea( YVW) = aea( YVX ). s VZ is a adius of the U V W X semicicle, and XY is a tangent to the semicicle at Z, the lines VZ and XY ae pependicula. Theefoe aea( YVX ) = VZ. XY = (.) =. Hence, aea( YVW ) = aea( YVX ) = ( ) =.

11 . Tom and Gei have a competition. Initially, each playe has one attempt at hitting a taget. If one playe hits the taget and the othe does not then the successful playe wins. If both playes hit the taget, o if both playes miss the taget, then each has anothe attempt, with the same ules applying. If the pobability of Tom hitting the taget is always 5 and the pobability of Gei hitting the taget is always, what is the pobability that Tom wins the competition? Solution: Let the pobability that Tom wins the competition be p. The pobability that initially Tom hits and Gei misses is =. The pobability that initially they both hit is 8 = and that they both miss is =. So the pobability that eithe they both hit o both miss is 8 + = 9 =. If they both hit o both miss the competition is in the same position as it was initially. So Tom s pobability of winning is then p. Theefoe, p = + p. So = p and hence p = The top diagam on the ight shows a shape that tiles the plane, as shown in the lowe diagam. The tile has nine sides, six of which have length. It may be divided into thee conguent quadilateals as shown. What is the aea of the tile? Solution: The tile is made up of thee conguent quadilateals one of which, PQRS, is P Q shown in the diagam on the ight. We let T be the foot of the pependicula 0 o fom Q to SR and we let U be the point shown. Then SR = UR =. U lso PQ=QU. Let thei common length be x. Hence QR = + x and TR = x. ecause the thee conguent quadilateals fit togethe at Q, we 0 have PQR =0. s PQ and SR ae both pependicula to PS, they ae 60 o 0 paallel and hence QRT = 60. Hence, fom the ight-angled S T R TR 0 x tiangle QRT we have = sin 60, that is, =, and theefoe x = + x. Hence x = and QR + x so x =. Theefoe TR = and QR =. Now 0 60 = ( ) 0 QT QR = cos 60 and hence 0 QT = QR cos 60 = cos =. [We could also use Pythagoas Theoem applied to the tiangle QRT to deduce this length.] We can now wok out the aea of the quadilateal PQRS in two diffeent ways. Most staightfowadly, aea(pqrs) = aea(pqts) + aea(qrt) = + ( ) =. ltenatively, 9 as PQ is paallel to SR, PQRS is a tapezium. We now use the fact that the aea of a tapezium is its height multiplied by the aveage length of its paallel sides. Hence, PQRS has aea. ( + ) = =. The tile is made up of thee copies of PQRS. So its aea is =. 9 9

12 5. How many distinct pais ( x, y) of eal numbes satisfy the equation ( x + y) = ( x + )( y )? 0 Solution: Method. The most staightfowad method is to ewite the given equation as a quadatic in x, whose coefficients involve y, and then use the b ac 0 condition fo a quadatic, ax + bx + c = 0, to have eal numbe solutions. Now, ( x + y) = ( x + )( y ) x + xy + y = xy x + y 6 x + xy + x + y y + 6 = 0 x + ( y + ) x + ( y y + 6) = 0. So hee a =, b = y + and c = y y + 6. We see that b ac = ( y + ) ( y y + 6) = y + y 8 = ( y 8y + 6) = ( y ). So thee is a eal numbe solution fo x if and only if ( y ) 0. Now, as fo all eal numbes y, ( y ) 0, it follows that ( y ) 0 y = 0 y =. When y = the quadatic equation becomes x + 8x + 6 = 0, that is ( x + ) = 0, which has just the one solution x =. So (,) is the only pai of eal numbes, ( x, y), fo which ( x + y) = ( x + )( y ). Method. We can simplify the algeba by making the substitution w = x + and z = y. Then x + y = ( w ) + ( z + ) = w + z, and the equation becomes ( w + z) = wz. Now, ( w + z) = wz w + wz + z = wz w + wz + z = 0 ( w + z) + ( z) = 0. The sum of the squaes of two eal numbes is zeo if and only if each eal numbe is 0. So the only eal numbe solution of ( w + z) + ( z) = 0 is w + z = z = 0. This is equivalent to w = z = 0 and hence to x = and y =. So again we deduce that thee is just this one solution. xtension Poblems 5. It is possible to use the b ac citeion to show that ( 0,0) is the only pai of eal numbes that satisfy the equation w + wz + z = 0. heck this. 5. Show that the b ac citeion is coect by poving that fo all eal numbes, a, b, c, with a 0, the quadatic equation ax + bx + c = 0 has a eal numbe solution if and only if b ac. 5. [Fo those who know about complex numbes.] If we allow the possibility that x and y ae complex numbes, then the equation ( x + y) = ( x + )( y ) has moe than one solution. heck that x = 6, y = + 5 i is one solution of this equation. How many moe can you find?

13 0. The two tangents dawn fom a point outside a cicle to that cicle ae a b S P equal in length. This theoem has been used to mak fou pais of X a equal line segments on the diagam. In the cicle the diamete, XY, b has been maked. It is also a pependicula height of the tapezium. We ae given that SR = PQ = 5 cm so we can deduce that d c Y (a + d) + (b + c) = = 50. The aea of tapezium R c Q PQRS = (SP + QR) XY = 600 cm d. Theefoe (a + b + c + d) = 600. So 50 = 600, i.e. =.. (x + y ) = x + xy + y. Note that all of the altenatives given ae of the fom a + so we need xy = 6. The only odeed pais (x, y) of positive integes which satisfy this ae (, 6), (, ), (, ), (6, ). Fo these, the values of x + y ae 7,, 7, 8 espectively. So the equied numbe is UKMT UKMT UKMT UK SNIOR MTHMTIL HLLNG Oganised by the United Kingdom Mathematics Tust suppoted by. Let the pependicula fom Y meet UV at T and let ZXV = α. Note that VZX = 90 as a tangent to a cicle is pependicula to the adius at the point of contact. Theefoe sin α = =. onside tiangle YTX: sin α = YT YX. So YT = YX sin α =. So the aea of tiangle YVW = VW YT =. = Y Z a U T V W X SOLUTIONS. Tom wins afte one attempt each if he hits the taget and Gei misses. The pobability of this happening is 5 = 5. Similaly the pobability that Gei wins afte one attempt is 5 = 5. So the pobability that both competitos will have at least one moe attempt is 5 5 = 5. Theefoe the pobability that Tom wins afte two attempts each is 5 5. The pobability that neithe Tom no Gei wins afte two attempts each is 5 5. So the pobability that Tom wins afte thee attempts each is ( 5) 5 and, moe geneally, the pobability that he wins afte n attempts each is ( 5) n 5. Theefoe the pobability that Tom wins is 5 + ( 5) 5 + ( 5) 5 + ( 5) 5 +. This is the sum to infinity of a geometic seies with fist tem 5 and common atio 5. Its value is 5 ( 5) =.. The diagam shows one of the thee quadilateals making up the x tile, labelled and with a line inseted. Note that it is a tapezium. x s thee quadilateals fit togethe, it may be deduced that = 60 = 0, so = 60. It may also be deduced that the length of is + x, whee x is the length of. Now cos = cos60 = = x +x. So + x = x, i.e. x =. The aea of is ( + ) = ( + ) sin60 = = 9. So the aea of the tile is 9 =. x x 5. Stating with (x + y) = (x + ) (y ) and expanding both sides gives x + xy + y = xy x + y 6, i.e. x + (y + ) x + y y + 6 = 0. To eliminate the xy tem we let z = x + y and then eplace x by z y. The equation above becomes z + (z y) + y y + 6 = 0. Howeve, z + (z y) + y y + 6 = (z + ) + y 6y + = (z + ) + (y 8y + 6) = (z + ) + (y + ). So the only eal solution is when z = and y = ; i.e. x = and y = Keep these solutions secue until afte the test on TUSY 6 NOVMR 0 This solutions pamphlet outlines a solution fo each poblem on this yea's pape. We have tied to give the most staightfowad appoach, but the solutions pesented hee ae not the only possible solutions. Occasionally we have added a Note (in italics). Please shae these solutions with you students. Much of the potential benefit of gappling with challenging mathematical poblems depends on teaches making time fo some kind of eview, o follow-up, duing which students may begin to see what they should have done, and how many poblems they could have solved. We hope that you and they agee that the fist 5 poblems could, in pinciple, have been solved by most candidates; if not, please let us know. 5. The UKMT is a egisteed chaity.

14 . If an odd numbe is witten as the sum of two pime numbes then one of those pimes is, since is the only even pime. Howeve, 9 is not pime so cannot be witten as the sum of two pimes. Note that 5 = + ; 7 = + 5; 9 = + 7; 0 = + 7, so is the only altenative which is not the sum of two pimes.. The inteio angles of an equilateal tiangle, squae, egula pentagon ae 60, 90, 08 espectively. The sum of the angles at a point is 60. So θ = 60 ( ) = 0.. The cost now is ( ) p =.0.. One hunded thousand million is = 0. So the numbe of stas is 0 0 = Let the equied addition be ab + cd + ef, whee a, b, c, d, e, f ae single, distinct digits. To make this sum as lage as possible, we need a, c, e (the tens digits) as lage as possible; so they must be 7, 8, 9 in some ode. Then we need b, d, f as lage as possible, so, 5, 6 in some ode. Hence the lagest sum is 0( ) + ( ) = = In ode to be a multiple of 5, a numbe must be a multiple both of and of 5. So its units digit must be 0 o 5. Howeve, the units digit must also equal the thousands digit and this cannot be 0, so the equied numbe is of the fom 5aa5. The lagest such foudigit numbes ae 5995, 5885, Thei digit sums ae 8, 6, espectively. In ode to be a multiple of, the digit sum of a numbe must also be a multiple of, so 5775 is the equied numbe. The sum of its digits is. 7. dd the fist and thid equations: x =, so x =. dd the fist two equations: x + y =, so y =. Substitute fo x and y in the fist equation: + ( ) + z = so z =. Theefoe xyz = ( ) ( ) =. 8. If an equilateal tiangle is split into a numbe of smalle identical equilateal tiangles then thee must be one small tiangle in the top ow, thee small tiangles in the ow below, five small tiangles in the ow below that and so on. So the total numbe of small tiangles is o 9 o 6 etc. These ae all squaes and it is left to the eade to pove that the sum of the fist n odd numbes is n. So, fo thee copies of a given tile to fom an equilateal tiangle, the numbe of tiangles which compise the tile must be one thid of a squae numbe. Only the tiles made up of thee equilateal tiangles and twelve equilateal tiangles satisfy this condition. Howeve, it is still necessay to show that thee copies of these tiles can indeed make equilateal tiangles. The diagams above show how they can do this. 9. If Piee is telling the tuth then Qad is not telling the tuth. Howeve, this means that Ratna is telling the tuth, so this leads to a contadiction as Piee stated that just one peson is telling the tuth. So Piee is not telling the tuth, which means that Qad is telling the tuth, but Ratna is not telling the tuth. This in tun means that Sven is telling the tuth, but Tanya is not. So only Qad and Sven ae telling the tuth. 0. It can be deduced that N must consist of at least digits since the lagest -digit positive intege consists of nines and has a digit sum of 007. It is possible to find -digit positive integes which have a digit sum of 0. The lagest of these is and the smallest is So N = and N + = ( zeos).. Let the adius of the cicula piece of cadboad be. The diagam shows a secto of the cicle which would make one hat, with the mino ac shown becoming the cicumfeence of the base of the hat. The cicumfeence of the cicle is π. Now 6 < π < 7. This shows that we can cut out 6 hats in this fashion and also shows that the aea of cadboad unused in cutting out any 6 hats is less than the aea of a single hat. Hence thee is no possibility that moe than 6 hats could be cut out.. Two diffeent ways of expessing 5 ae + and +. In the following list these ae denoted as {, : two ways}. The list of all possible ways is {5: one way}, {, : two ways}, {, : two ways}, {,, : thee ways}, {,, : thee ways}, {,,, : fou ways}, {,,,, : one way}. So in total thee ae 6 ways. {iffeent expessions of a positive intege in the above fom ae known as patitions. It may be shown that the numbe of distinct compositions of a positive intege n is n.}. The table below shows the position of the face maked with paint when the base of the cube is on the 5 squaes. ode: T - top, - base; F - font; H - hidden (ea); L - left; R - ight T H F T R L T F H T L R R R R L L L F So the equied sum is = 80.. Note that each student has a language in common with exactly fou of the othe five students. Fo instance, Jean-Piee has a language in common with each of Ina, Kaim, Lionel and May. Only Helga does not have a language in common with Jean-Piee. So whicheve two students ae chosen, the pobability that they have a language in common is /5. 5. Let Pofesso Rossefop's usual jouney take t minutes at an aveage speed of v metes/ minute. Then the distance to wok is vt metes. On Thusday his speed inceased by 0%, i.e. it was v / 0 metes/minute. The time taken was (t x) minutes. Theefoe v / 0 (t x) = vt. So (t x) = 0t, i.e. t = x. 6. t points and, x = 0. So y y =, i.e. (y 6)(y + ) = 0, i.e. y = 6 o y =. So is (0, ) and is (0, 6). t points and, y = 0. So x + x =, i.e. (x )(x + ) = 0, i.e. x = o x =. So is (, 0) and is (, 0). Theefoe the aeas of tiangles and ae 7 6 = and 7 = 7. So has aea 8. {It is left to the eade to pove that aea =.} 7. In the diagam, is the cente of the quate-cicle ac ; is the point whee the cental squae touches ac ; F is the point whee the cental squae touches ac ; O is the cente of the cicle. s both the cicle and ac have adius, O is a squae of side. y Pythagoas' Theoem: O = +. So O =. Theefoe O = O =. y a simila agument, OF =. Now F = O + OF = O since O = OF. So the side of the squae is O = ( ) =. 8. In the diagam, is the midpoint of. Tiangle is isosceles since = =. Theefoe, bisects and α is pependicula to. The angles at a point total 60, so = α = 80 α. Theefoe = = 90 α. So = = α. x Theefoe x = = = cos α = cos α = cos α. 9. Note that the numbe epesented by x appeas in both the hoizontal ow and the vetical column. Note also that = 5. Since the x numbes in the ow and those in the column have sum, we deduce that x = 5 = 7. We now need two disjoint sets of thee numbes chosen fom,,, 5, 6, 8 so that the numbes in both sets total. The only possibilities ae {,, 8} and {, 5, 6}. We have six choices of which numbe to put in the top space in the vetical line, then two fo the next space down and one fo the bottom space. That leaves thee choices fo the fist space in the hoizontal line, two fo the next space and one fo the final space. So the total numbe of ways is 6 = 7. O F