UK I NTERMEDIATE MATHEMATICAL OLYMPIAD

Transcription

1 UK I NTERMEDIATE MATHEMATICAL OLYMPIAD Maclaurin Question Papers and Solutions 008 to 00 Organised by the United Kingdom Mathematics Trust

2 i UKMT UKMT UKMT UK Intermediate Mathematical Olympiad 008 to 00 Maclaurin Question Papers and Solutions Organised by the United Kingdom Mathematics Trust Contents ackground ii Rules and Guidelines 008 paper paper 4 00 paper solutions solutions 3 00 solutions 0 UKMT 0

3 ii ackground The Intermediate Mathematical Olympiad and Kangaroo (IMOK) are the follow-up competitions for pupils who do extremely well in the UKMT Intermediate Mathematical Challenge (about in 00 are invited to take part). The IMOK was established in 003. There are three written papers (Cayley, Hamilton, Maclaurin) and two multiple-choice papers (the Pink and Grey Kangaroo). The written papers each take two hours and contain six questions. oth Kangaroo papers are one hour long and contain 5 questions. The Maclaurin paper is for pupils in: Y or below (England and Wales); S4 or below (Scotland); School Year or below (Northern Ireland).

4 The United Kingdom Mathematics Trust Intermediate Mathematical Olympiad and Kangaroo (IMOK) Olympiad Maclaurin Paper All candidates must be in School Year (England and Wales), S4 (Scotland), or School Year (Northern Ireland). READ THESE INSTRUCTIONS CAREFULLY EFORE STARTING. Time allowed: hours.. The use of calculators, protractors and squared paper is forbidden. Rulers and compasses may be used. 3. Solutions must be written neatly on A4 paper. Sheets must be STAPLED together in the top left corner with the Cover Sheet on top. 4. Start each question on a fresh A4 sheet. You may wish to work in rough first, then set out your final solution with clear explanations and proofs. Do not hand in rough work. 5. Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. 6. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks; also, incomplete or poorly presented solutions will not receive full marks. 7. These problems are meant to be challenging! The earlier questions tend to be easier; the last two questions are the most demanding. Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many: you will have done well if you hand in full solutions to two or more questions. DO NOT OPEN THE PAPER UNTIL INSTRUCTED Y THE INVIGILATOR TO DO SO! The United Kingdom Mathematics Trust is a Registered Charity. Enquiries should be sent to: Maths Challenges Office, School of Mathematics, University of Leeds, Leeds, LS 9JT. (Tel )

5 Advice to candidates Do not hurry, but spend time working carefully on one question before attempting another. Try to finish whole questions even if you cannot do many. You will have done well if you hand in full solutions to two or more questions. Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols such as π, fractions, or square roots, if appropriate, but NOT decimal approximations. Give full written solutions, including mathematical reasons as to why your method is correct. Just stating an answer, even a correct one, will earn you very few marks. Incomplete or poorly presented solutions will not receive full marks. Do not hand in rough work.

6 All the digits of a certain positive three-digit number are non-zero. When the digits are taken in reverse order a different number is formed. The difference between the two numbers is divisible by eight. Given that the original number is a square number, find its possible values.. The diagram shows a regular pentagon ACDE. A circle is drawn such that A is a tangent to the circle at A and CD is a tangent to the circle at D. The side AE of the pentagon is extended to meet the circumference of the circle at F. Prove that DE = DF. F E A D C 3. Show that the equation has no solutions for positive integers x, y. x + y = 5 4. A circle is inscribed in a right-angled triangle, as shown. The point of contact of the circle and the hypotenuse divides the hypotenuse into lengths x and y. Prove that the area of the triangle is equal to xy. x y 5. An ant lives on the surface of a cuboid which has points X, Y and Z on three adjacent faces. The ant travels between X, Y and Z along the shortest possible path between each pair of points. The angles x, y and z are the angles between the parts of the ant's path, as shown. Prove that x + y + z = 70. X x Y y z Z 6. An artist is preparing to draw on a sheet of A4 paper (a rectangle with sides in the ratio : ). The artist wishes to place a rectangular grid of squares in the centre of the paper, leaving a margin of equal width on all four sides. Show that such an arrangement is possible for a or a 3 grid but impossible for any other g (g + ) grid.

7 Five numbers are arranged in increasing order. As they get larger the difference between adjacent numbers doubles. The average of the five numbers is more than the middle number. The sum of the second and fourth numbers is equal to the largest number. What is the largest number?. Miko always remembers his four-digit PIN (personal identification number) because (a) it is a perfect square, and (b) it has the property that, when it is divided by, or 3, or 4, or 5, or 6, or 7, or 8, or 9, there is always a remainder of. What is Miko's PIN? 3. Solve the simultaneous equations 5xy x + y = 6 4xz x + z = 3 3yz y + z =. 4. In a trapezium ACD the sides A and DC are parallel and AD = AC < 90. Point P lies on A with CPD = AD. Prove that PC + PD = A DC. 5. A lottery involves five balls being selected from a drum. Each ball has a different positive integer printed on it. Show that, whichever five balls are selected, it is always possible to choose three of them so that the sum of the numbers on these three balls is a multiple of In the figure, p, q, r and s are the lengths of four arcs which together form the circumference of the circle. Find, in simplified form, an expression for s in terms of p, q and r. q p r s

8 00 5. How many different ways are there to express 5 in the form a + b, where a and b are positive integers with a b?. The diagram shows a regular heptagon, a regular decagon and a regular 5-gon with an edge in common. Find the size of angle XYZ. X Z Y 3. Solve the equations x + xy + x = 9 y + xy + y = The diameter AD of a circle has length 4. The points and C lie on the circle, as shown, so that A = C =. Find the length of CD. C A 4 D 5. The diagram shows a rectangle divided into eight regions by four straight lines. Three of the regions have areas, and 3, as shown. What is the area of the shaded quadrilateral? 3 6. Every day for the next eleven days I shall eat exactly one sandwich for lunch, either a ham sandwich or a cheese sandwich. However, during that period I shall never eat a ham sandwich on two consecutive days. In how many ways can I plan my sandwiches for the next eleven days?

9 6 008 Solutions. All the digits of a certain positive three-digit number are non-zero. When the digits are taken in reverse order a different number is formed. The difference between the two numbers is divisible by eight. Given that the original number is a square number, find its possible values. Solution Consider the three-digit number abc = 00a + 0b + c. When the digits are taken in reverse order we obtain the number cba = 00c + 0b + a. If these numbers are different, then a c. Case I: a > c If a > c, then the difference between abc and cba is (00a + 0b + c) (00c + 0b + a) = 99a 99c = 99 (a c). We are told that this difference is divisible by 8. ut 99 is not divisible by, so we conclude that a c is divisible by 8. Since a and c are different single-digit numbers, the only possible value of a c is 8, which occurs when a = 9 and c =. It now remains to check the nine possible values of b. Of these, only b = 6 yields a square number for abc, namely 96 = 3. Case II: a < c Using a similar argument, we obtain a = and c = 9. Again, checking the nine possible values of b, we find only one square number, namely 69 = 3. Hence, the only numbers satisfying the given conditions are 69 and 96.. The diagram shows a regular pentagon A ACDE. A circle is drawn such that A is a tangent to the circle at A and CD is a tangent to the circle at D. The side AE of E the pentagon is extended to meet the circumference of the circle at F. Prove that DE = DF. F First solution Since ACDE is a regular pentagon DEF = D = 7. () We also deduce that each interior angle of the regular pentagon is 08, since angles on a straight line are supplementary. C

10 A 7 F E O D Let O be the centre of the circle. The angle between a tangent and the radius drawn to the point of contact is 90. Hence OA = ODC = 90. The sum of the interior angles of the pentagon OACD is 540. Hence AOD = = 44. The angle subtended at the centre of a circle is twice the angle subtended at the circumference, so EFD = AOD C = 7. () From equations () and () we deduce that EFD = DEF, that is, triangle DEF has equal base angles; therefore the sides opposite these angles are equal. Hence DE = DF, as required. Second solution A E F D C Another solution makes use of the alternate segment theorem, from which AD = AFD, as shown. Can you see how to complete the argument?

11 8 3. Show that the equation has no solutions for positive integers x, y. x + y = 5 Solutions The first two solutions make use of the following: We are given x + y = 5, where x and y are positive integers. Multiplying the given equation throughout by xy we obtain First solution Equation ( ) may be rearranged to give y + x = 5xy. 5xy x y = 0, ( ) or 5xy 55x 55y = 0. Therefore (5x ) (5y ) = 0, so that (5x ) (5y ) =. () Now we know that x and y are integers and so 5x and 5y are also integers. Therefore equation () tells us that 5x is a factor of, that is, 5x = ±, ±, or ±. Of these six cases, only 5x = gives a positive integer for x but in this case 5y = leading to y =. So we conclude that the original equation has no solutions for positive integers x and y. Second solution Use equation ( ) to find y in terms of x: y + x = 5xy so x = 5xy y = (5x ) y, x and thus = y, () 5x noting that 5x is non-zero since x is an integer. Now y > 0, so 5x > 0 and hence x 3 since x is an integer. We may also assume, without loss of generality, that y x, so that 5x and hence x 4. Therefore x = 3 or x = 4. However, from equation (), neither of these values of x gives an integer value for y and hence the original equation has no solutions for positive integers x and y.

12 Third solution Assume, without loss of generality, that x y. Then so that and hence Therefore x + y x 5 x 5x. x 4 since x is an integer. Can you see how to proceed? 9 Note A unit fraction has numerator. An Egyptian fraction is constructed by adding different unit fractions together, for example, 5 = In ancient Egypt fractions were written like this. It is possible to write any given proper fraction as an Egyptian fraction, but finding the smallest number of unit fractions needed is not easy the problem above shows that requires more than two. 4. A circle is inscribed in a right-angled triangle, as shown. The point of contact of the circle and the hypotenuse divides the hypotenuse into lengths x and y. Prove that the area of the triangle is equal to xy. Solution Use the notation in the diagram below. x x y 5 Q r O r y C R A The two tangents to a circle from an external point are of equal length. Hence AR = y and Q = x. Now OR and OQ are radii of the circle and AC and C are tangents to the circle, so that ORC = 90 and OQC = 90. It follows that OQCR is a square, and hence CR = CQ = r. From Pythagoras' theorem applied to triangle AC we get A = C + AC and so (x + y) = (x + r) + (y + r) which may be expanded and simplified to give xy = xr + yr + r. ()

13 0 Now the area of the triangle is base height = AC C = (y + r) (x + r) = (xy + xr + yr + r ), = (xy + xy) from equation () = xy, as required. 5. An ant lives on the surface of a cuboid which has points X, Y and Z on three adjacent faces. The ant travels between X, Y and Z along the shortest possible path between each pair of points. The angles x, y and z are the angles between the parts of the ant's path, as shown. Prove that x + y + z = 70. X x Y y z Z First solution Consider any two adjacent faces along which the ant moves and unfold them as shown in the diagram. Since the ant takes the shortest possible path between any two points, on the unfolded diagram the path will be a straight line. b a It follows that the path, when it crosses an edge, always forms a pair of supplementary angles, such as a and b in the diagram above. Y y V X x z Z Now consider the three quadrilaterals, shown shaded in the diagram above, which are bounded by the path and which have one vertex V of the cuboid in common. The angle sum of each quadrilateral is 360 and so the sum of the angles in all three quadrilaterals is = 080.

14 This sum comprises: the three angles x, y and z ; three pairs of supplementary angles, as described above; and three right angles at the vertex V. We thus obtain the equation x + y + z = 080 and hence as required. x + y + z = 70, Second solution Another solution considers a net of three faces of the cuboid, drawn on a sheet of paper, as shown in the diagram below. Z Y T X On the net, one section of the ant's path is disconnected. The two broken parts of the path may be extended, as shown by the dashed lines, to meet at a point T. Can you see how to show that the angle at T is a right angle, and so complete the proof? 6. An artist is preparing to draw on a sheet of A4 paper (a rectangle with sides in the ratio : ). The artist wishes to place a rectangular grid of squares in the centre of the paper, leaving a margin of equal width on all four sides. Show that such an arrangement is possible for a or a 3 grid but impossible for any other g (g + ) grid. Solutions Without any loss of generality, we may suppose that the sheet of paper measures units by unit. Let the width of the margin be m units and let the g (g + ) grid be composed of squares with sides of length d units, as shown in the diagram.

15 d m The length of the sheet is and the width of the sheet is d (g + ) + m = dg + m =. () Eliminate d from this pair of equations to yield the relationship We now use the fact that m has to be positive to obtain and hence so that m = g + g. () ( )g < 0 < g + g g < = + + = +. Restricting our solutions to positive integers, we deduce that there are only two possibilities: g = or g =. If g =, then from equations () and () we find that and m = > 0 d = ( ) = > 0, so that the arrangement is indeed possible for a grid. Similarly, if and g =, then m = 3 > 0 d = (3 ) so that the arrangement is also possible for a = > 0, 3 grid. Hence the arrangement is possible for grids of size or 3 but impossible for any other grids of the form g (g + ).

16 009 Solutions 3. Five numbers are arranged in increasing order. As they get larger the difference between adjacent numbers doubles. The average of the five numbers is more than the middle number. The sum of the second and fourth numbers is equal to the largest number. What is the largest number? Solution Let a be the smallest number and d be the difference between the first and second numbers. Then the other differences are d, 4d and 8d, so the five numbers are a, a + d, a + 3d, a + 7d and a + 5d. The condition that the average of the five numbers is more than the middle number gives 5a + 6d = a + 3d +, 5 which means that d = 5. The condition that the sum of the second and fourth numbers is equal to the largest number now gives a + 8d = a + 5d and hence a = 35. So the five numbers are 35, 40, 50, 70 and 0. Hence the largest number is 0. Remark Slightly neater algebra is obtained if instead we define a and d so that the numbers are represented as a 3d, a d, a, a + 4d and a + d.. Miko always remembers his four-digit PIN (personal identification number) because (a) it is a perfect square, and (b) it has the property that, when it is divided by, or 3, or 4, or 5, or 6, or 7, or 8, or 9, there is always a remainder of. What is Miko's PIN? Solution Since the required number has four digits, it is between 000 and Let the number be N. We know that N is a perfect square, and also that N is divisible by, 3, 4, 5, 6, 7, 8 and 9. Hence N is divisible by the lowest common multiple of these numbers, which is = 50. There are therefore only three possible values for N in the range, namely 5, 504 and 756, and of these only the middle one is a perfect square, being 7. Hence Miko's PIN is 504. Remark It would be possible to start by listing all the four-digit perfect squares, and then checking them one by one for the condition (b) concerning remainders. This is, however, not a very efficient method, since there are 68 numbers to check. If a method like this is

17 4 used, it must be made very clear that the check is done properly in every case. For example, one might exclude 599 by showing that it is a multiple of 7 and thus has remainder 0. It would not be enough simply to claim that this check had been made without giving details in each case. The task can be reduced by, for instance, excluding the even squares, leaving only 34 numbers to check, or realising that the number has to end in a to satisfy the conditions for divisibility by and 5. However, by the time you are applying this logic, you might as well go the whole way and use the method given above. 3. Solve the simultaneous equations 5xy x + y = 6 4xz x + z = 3 3yz y + z =. Solutions It is worth beginning with the observation that none of x, y or z can be zero, since that would immediately invalidate the equations. This allows us, in subsequent work, to cancel a factor of x, y or z from both sides of an equation. Note also that we should not assume that x, y or z are whole numbers. First solution Label the equations as follows: From (), and from (), so 5xy x + y = 6 () 4xz x + z = 3 () 3yz =. y + z (3) 5xy = 6x + 6y 8xz = 6x + 6z, We can, however, also deduce from () and () that so, cancelling x, 5xy 8xz = 6y 6z. (4) 5xy x + y = 8xz x + z, 5y (x + z) = 8z (x + y)

18 5 and therefore 5xy 8xz = 3yz. (5) ut, from (3), Putting (4), (5) and (6) together, we have and so 3yz = y + z. (6) 6y 6z = y + z y = z. Now, substituting y = z in (3) and cancelling z, we have 6z 3 =, so z = and y =. Then x = 3, by using (5), for example. Note that it is now necessary to check that this triple of values works for all the equations. All we have shown so far is that, if the equations are true, then the only possible values of x, y and z are 3, and ; this does not mean that, if x, y and z are 3, and, then the equations are true for example, the equations may have no solutions. Second solution Since none of x, y or z is zero, we can define a = x, b = y and c = z. y taking the reciprocals of each side of the three equations, we obtain b + a = 5 6 a + c = 4 3 Hence, by adding, we deduce that c + b = 3. a + b + c = 6, so c =, b = and a = 3. Therefore x = 3, y = and z =. Again we should check these values satisfy the original equations, or show that our logic is reversible.

19 6 4. In a trapezium ACD the sides A and DC are parallel and AD = AC < 90. Point P lies on A with CPD = AD. Prove that PC + PD = A DC. Solutions We begin with a diagram showing the information given in the question. The three marked angles are equal. D C A First solution Since A is parallel to DC, the alternate angles DCP and CP are equal. Hence the two triangles CDP and PC are similar. Exactly the same argument shows that DPA is also similar to CDP. We therefore have three similar triangles. Notice that we have been careful to describe the triangles in the correct order of vertices, so that the ratios of sides can now be read off conveniently. Now corresponding sides are in the same ratio, so we have and it follows that P PC = PC DC and P AP PD = PD DC, PC = P DC and PD = AP DC. Adding these two equations, we obtain PC + PD = (AP + P) DC = A DC. Second solution This proof starts by using the converse of the alternate segment theorem. ecause DA = CPD it follows that CP is a tangent at P to the circumcircle of DAP. Similarly, DP is a tangent at P to the circumcircle of CP. Let DC (extended) meet the two circles at X and Y, as shown. D X C Y A P

20 7 Now CXP = DAP, by the external angle of a cyclic quadrilateral, and also CXP = XPA, by alternate angles. Hence XPA = DAP = CP and so XP is parallel to C and PCX is a parallelogram. Similarly, APY D is a parallelogram. In particular, CX = P and DY = AP. Now we use the tangent-secant theorem for each circle. This gives CP = CX CD = P CD and DP = DC DY = DC AP. The last stage of the proof proceeds in the same way as the first solution. Remark When the three marked angles are right angles the result reduces to Pythagoras' theorem. 5. A lottery involves five balls being selected from a drum. Each ball has a different positive integer printed on it. Show that, whichever five balls are selected, it is always possible to choose three of them so that the sum of the numbers on these three balls is a multiple of 3. Solution It is worth observing that, for the purposes of this question, we can replace the numbers on the balls by their remainders when they are divided by 3. This is because, for any two numbers a and b, their sum a + b is a multiple of 3 if, and only if, the sum of their remainders is divisible by 3. To prove this, suppose that, when divided by 3, a has remainder r and b has remainder s. Then we can write a = 3m + r and b = 3n + s, for some integers m and n. Therefore a + b = 3m + 3n + r + s = 3 (m + n) + r + s and this is divisible by 3 if, and only if, r + s is divisible by 3. Using this idea, any selection of five balls will result in a set of five numbers, each of which is 0 or or. If three of the balls have the same number, then their sum will be divisible by 3, and we are done. If not, then there is one pair of numbers equal to x, another pair equal to y and a fifth number equal to z, where x, y and z are 0, and, in some order. This means that there are three balls numbered 0, and, and the sum of these numbers is divisible by 3. Hence, in either case, we have a choice of three of the five balls whose sum is a multiple of 3.

21 8 6. In the figure, p, q, r and s are the lengths of four arcs which together form the circumference of the circle. Find, in simplified form, an expression for s in terms of p, q and r. q p r First solution First note that the length of an arc of a circle is proportional to the angle subtended at the centre. Hence, by appropriate choice of radius, we can arrange that the arc lengths are actually equal to these angles. (If this is not done, the calculation proceeds in the same way, but there is some cancelling to do at the end.) The diagram shows these four angles at the centre of the circle. s q r p s Now there are two relationships between p, q, r and s. First there is the obvious one p + q + r + s = 360. () We also have the angle sum of a quadrilateral; one of the angles is 90, one is (p + q + r), and the others can be calculated in terms of p and r using the fact that there are two isosceles triangles. This gives and so Hence, from (), and so (p + q + r) + (90 p) (90 r) = 360 p + q + r = 80. p + q + r + s = p + 4q + r s = p + 3q + r.

22 9 Second solution We construct the two radii perpendicular to the two given chords, as shown in the diagram. p r q Since a radius perpendicular to a chord bisects the chord, it also bisects the corresponding arc. So we have arcs of length p and r, as shown. Now the diagram contains a quadrilateral with three right angles, so the fourth angle is also a right angle and therefore the two radii are perpendicular. Hence the radii determine a quarter of the circle, and so which gives p + q + r + s = 4 ( p + q + r), s = p + 3q + r. s

23 0 00 Solutions How many different ways are there to express in the form 5 a + b, where a and b are positive integers with a b? Solution There are infinitely many integers and hence the number of potential values of a and b is infinite, so any satisfactory method first needs to reduce the problem to a finite number of cases. We demonstrate two methods of doing this. Method Since 0 < a b we have b and therefore a 5 = a + b a. Hence a 5. Also, since b > 0, we have b > 0 and so 5 = a + b > a. Hence a > 5. Therefore 8 a 5 and the possible values of a are 8, 9,, 5. In order to see which of these correspond to integer values of b, it is helpful to find b in terms of a: b = 5 a a 5 = 5a and so 5a b = a 5. We now see that we require a 5 to be a positive divisor of 5a. So we can determine which values of a will give integer values of b from the table: a a 5 5a Divisor? b yes yes yes no 9 5 yes no no yes 5 Hence there are five ways to express 5 in the required form. Method We may multiply every term in the equation 5 = a + b by 5ab in order to clear the fractions. We obtain ab = 5b + 5a. We now rearrange this equation, first writing it in the form 4ab 30a 30b = 0,

24 then adding 5 to both sides to give 4ab 30a 30b + 5 = 5, that is, (a 5)(b 5) = 5. Therefore a 5 is a divisor of 5 = 3 5, so that a 5 =, 3, 5, 9, 5. Larger values are not possible since a b and so a 5 b 5. Also, negative values are not possible since then a 5 would be at most 5, but a > 0 so that a 5 > 5. Each of the corresponding values of a and b is an integer: (a, b) = (8, 0), (9, 45), (0, 30), (, 0) or (5, 5). Hence there are five ways to express in the required form. 5 The diagram shows a regular heptagon, a regular decagon and a regular 5-gon with an edge in common. Find the size of angle XYZ. X Z Y Solution We shall give two methods, each of which involves finding the exterior angle of a regular polygon. If a regular polygon has n sides, then each exterior angle is given by exterior angle = 360 n. (.) Method Extend the common edge PO to point T as shown below. The angles labelled x, y and z are exterior angles of the regular decagon, regular heptagon and regular 5-gon respectively. P Hence, using the result (.), X O x y T z Z Y x = = 36, y = = and z = = 4. It follows that XOY = x + y = and ZOY = y z = Now the sides of the three polygons are all equal, so the triangles XOY and ZOY are isosceles. We can therefore find their base angles: XYO = = 46 7

25 and ZYO = = Thus XYZ = ZYO XYO = = 30. Method Let PO be the common edge, as shown below. Since OX = OY = OZ = OP the points X, Y, Z and P lie on a circle centre O. P Y X Z Now the angle at the circumference is half the angle at the centre, so ut from (.) O x z XYZ = XOZ = (x + z). x = = 36 and z = = 4, so that XYZ = (36 + 4) = Solve the equations x + xy + x = 9 y + xy + y = 3. Solution Substitution is one of the standard methods of solving simultaneous equations: use one equation to find an expression for one unknown, then substitute this expression into the other equation, thereby forming a single equation in just one of the unknowns. Though it is possible to use a substitution method straight away here, the algebra is rather unpleasant, so we demonstrate two other approaches. In passing, we note that the question places no restrictions on x and y. In particular, we cannot assume that they are integers.

26 3 Method Adding the two given equations, we get x + y + x + xy + y = 6 so that (x + y) + (x + y) 6 = 0, which factorises to give Hence (x + y ) (x + y + 3) = 0. x + y = or x + y = 3. (3.) Also, subtracting the two given equations, we get x y + x y = which factorises to give (x y) + (x y)(x + y) = so that Hence, using (3.), (x y)( + x + y) =. x y = 4 or x y = 6, (3.) which occur when x + y = and x + y = 3 respectively. We can now solve equations (3.) and (3.) by, for example, adding and subtracting, to obtain (x, y) = (3, ) or ( 9, 3 ). We now need to check whether these two pairs of values really do satisfy the equations given in the question. Each of them does, so the required solutions are x = 3, y = and x = 9, y = 3. Method Factorise each of the given equations to give x ( + y + x) = 9 (3.3) and y( + x + y) = 3. (3.4) Since no side of either equation is zero, we may divide equation (3.3) by (3.4) to obtain x y = 3, so that x = 3y. Now substitute this expression for x into equation (3.4) to get y ( y) = 3,

27 4 which may be rearranged to y y 3 = 0, or (y + ) (y 3) = 0. Therefore y = or y = 3 and, since x = 3y, we have solutions (x, y) = (3, ) or ( 9, 3 ). We now need to check whether these two pairs of values really do satisfy the equations given in the question. Each of them does, so the required solutions are x = 3, y = and x = 9, y = 3. 4 The diameter AD of a circle has length 4. The points and C lie on the circle, as shown, so that A = C =. Find the length of CD. C A 4 Solution Let O be the centre of the circle, so that OA = O = OC =, and let chord AC and radius O meet at X. C X D A O D Triangle OAC is isosceles and O bisects angle AOC (because the chords A and C are equal and so subtend equal angles at the centre). Hence O is the perpendicular bisector of the base AC of the isosceles triangle AOC. In other words, AX = XC and AXO = 90, as shown in the diagram above. Since angle ACD is 90 (angle in a semicircle) triangle ACD is a right-angled triangle with CD, whose length we have to find, as one side. We know AD = 4, so can find CD, using Pythagoras' theorem, from CD = AD AC, (4.) provided we can find the length of AC. We shall do this by using areas, but there are other methods. Now consider isosceles triangle OA and let N be the midpoint of A, so that triangle ANO is right-angled. A N O

28 5 Then, from Pythagoras' theorem, NO = ( ) = 5 4, so that NO = 5. Hence the area of triangle AO is 5 A NO = 4. ut the area of triangle OA is also O AX. Therefore AX = 5 4 and so AC = AX = 5. Using this value in equation (4.). we get and hence CD = 7. Solution CD = AD AC = = 49 4 Let O be the centre of the circle. Reflecting the sector OD, shown shaded, about the diameter perpendicular to D, so that and D are interchanged, gives the right-hand diagram. If C is the reflection of C, then C = CD. Also, since reflection about a diameter reflects a circle to itself, we know that C lies on the circle. C C A O D A O D Now angles AC and C AD are both angles at the circumference subtended by chords of length. These angles are therefore equal and so C and AD are parallel (alternate angles). Draw perpendiculars from and C to AD to create a rectangle XYC, as shown below, with C = XY. C A X O Y D Then AX = A cos θ = cos θ and, by constructing the perpendicular bisector of A in isosceles triangle OA, shown below, we see that cos θ = 4. Therefore AX = 4. A O Similarly DY = 4 and so CD = C = XY = = 3. N

29 6 Solution 3 Let P be the intersection of A produced and DC produced. P C A 4 D Now angles AD and DC are both angles at the circumference subtended by chords of length. These angles are therefore equal. Also, AD = 90 (angle in a semicircle). Therefore triangles AD and PD are congruent (ASA). Hence P = and PD = 4. Further, in triangles CP and DAP, angle P is common and CP = DAP (exterior angle of cyclic quadrilateral). These triangles are therefore similar and hence PC : = : 4. So PC = and CD = PD PC = 4 = 3. 5 The diagram shows a rectangle divided into eight regions by four straight lines. Three of the regions have areas, and 3, as shown. What is the area of the shaded quadrilateral? Solution Notice that the shaded area is the intersection of the two triangles shown shaded in the following diagrams. 3 The area of each of these triangles is half the area of the rectangle, since each has the same base and height as the rectangle. Therefore the total unshaded area within the rectangle in the right-hand figure is also half the area of the rectangle. In other words, the shaded area in the left-hand figure is equal to the total unshaded area in the righthand figure. 3 s b a Using the notation indicated above, we therefore have a + s + b = a b + 3 and hence s = = 6. So the area of the shaded quadrilateral is 6. Remark The result also applies if the rectangle is replaced by a parallelogram. Moreover, there is clearly nothing special about the values, and 3.

30 7 6 Every day for the next eleven days I shall eat exactly one sandwich for lunch, either a ham sandwich or a cheese sandwich. However, during that period I shall never eat a ham sandwich on two consecutive days. In how many ways can I plan my sandwiches for the next eleven days? Solution n ( ) We shall use the notation for the number of ways of choosing r objects from n. This r is a binomial coefficient, sometimes written. We have n ( ) = n! r r! (n r)! n C r n (n ) (n r + ) =. r We give three different methods, all of which create a plan for the next eleven days by constructing a line of Cs and Hs. Method Let h (n) be the number of ways of creating a line of length n starting with H, and c (n) be the number starting with C. For any line of length n starting with C, we may construct a line of length n + starting with H by adding H at the start. Also, for any line of length n + starting with H, we may construct a line of length n starting with C by deleting the initial H. oth of these statements depend on the rule that there are no consecutive Hs. Hence h (n + ) = c (n). Similarly, adding C to the front of a line starting with H or one starting with C creates a longer line starting with C (and vice versa), so that c (n + ) = c (n) + h (n). Therefore h (n + ) = c (n + ) = c (n) + h (n) = h (n + ) + h (n) and c (n + ) = c (n + ) + h (n + ) = c (n + ) + c (n). We thus have two sequences generated like the Fibonacci sequence: each term is the sum of the previous two terms. Moreover, letting t (n) be the total number of ways to construct a line for n days, then t (n) = h (n) + c (n) and therefore t (n + ) = t (n + ) + t (n). Hence t (n) is also a Fibonacci-like sequence. Now t () =, corresponding to the lines C and H, and t () = 3, corresponding to the lines CC, CH and HC. So the first eleven terms of t (n) are, 3, 5, 8, 3,, 34, 55, 89, 44 and 33. Hence the number required is 33.

31 8 Method We create a plan for the next eleven days by constructing a line of Cs and Hs from two types of tile, C or HC, which ensures that two Hs are never placed together. There are various possibilities, determined by the number of C tiles which are used, which can be, 9, 7, 5, 3 or. Eleven tiles can be placed in just one way, or ways. 0 ( ) C ( ) C HC 0 HC ( ) C HC 9 Nine tiles and one tile can be placed in ways, the number of ways of choosing the position of one tile in a row of 0 tiles. Seven tiles and two tiles can be placed in ways, the number of ways of choosing the position of two HC tiles in a row of 9 tiles. Continuing in this way, we see that the total number of ways is ( ) ( ) ( ) + 0 ( ) ( ) ( ) = = 44. However, placing tiles in this way always ends the line with a C, so does not allow for the possibility of ending with an H. ut if the line ends with an H, then the remainder of the line has ten letters and ends with a C. We may count the number of ways for this in a similar way to the above, but with a total of 0 letters instead of : 0 ( ) ( ) ( ) + 9 ( ) ( ) ( ) = = 89. Therefore the total number of ways altogether is = 33. Method 3 We create a plan for the next eleven days by starting with k Cs in a line, where k, and adding some Hs (possibly none) to construct a line of Cs and Hs. There are k + slots into which a single H may be placed at either end or between two Cs. We need to place ( k) Hs so that each H is in a different slot, so we have to choose k of k + the slots; there are ways in which this can be done. k k + ( ) Now we cannot add more Hs than the number of available slots, so k k +, that is, k 5. k + Therefore the total number of ways is the sum of for 5 k, in other words k ( ) 6 ( ) ( ) ( ) + 7 ( ) ( ) + 8 ( ) ( ) = = 33.