Organised by the

United Kingdom Mathematics Trust

MATHEMATICAL OLYMPIAD

UK I NTERMEDIATE

2008 to 2010

Maclaurin Question Papers and Solutions

i

UKMT

UKM

TUKM

T

UK Intermediate Mathematical Olympiad2008 to 2010

Maclaurin Question Papers and Solutions

Organised by the United Kingdom Mathematics Trust

Contents

Background ii

Rules and Guidelines 1

2008 paper 3

2009 paper 4

2010 paper 5

2008 solutions 6

2009 solutions 13

2010 solutions 20

©UKMT 2011

ii

Background

The IntermediateMathematicalOlympiad and Kangaroo(IMOK)arethefollow-up competitionsfor pupilswho do extremelywell inthe UKMT IntermediateMathematicalChallenge(about1 in 200are invited to take part). The IMOK was established in 2003.

Thereare threewritten papers(Cayley,Hamilton, Maclaurin)andtwo multiple-choicepapers(the Pink and Grey Kangaroo). Thewritten paperseachtaketwo hoursandcontainsix questions.BothKangaroo papers are one hour long and contain 25 questions.

The Maclaurinpaperis for pupils in: Y11 or below (EnglandandWales); S4 or below (Scotland); School Year 12 or below(Northern Ireland).

1

The United Kingdom Mathematics Trust

Intermediate Mathematical Olympiad and Kangaroo(IMOK)

Olympiad Maclaurin Paper

READ THESE INSTRUCTIONS CAREFULLY BEFORE STARTING

1. Time allowed: 2 hours.

2. 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 topleft 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 explanationsand proofs.

Do not hand in rough work.

5. Answers must be FULLY SIMPLIFIED, and EXACT. They may contain symbols such as ,fractions, or square roots, if appropriate, but NOT decimal approximations.

π

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 orpoorly presented solutions will not receive full marks.

7. These problems are meant to be challenging! The earlier questions tend to be easier; the lasttwo 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 handin full solutions to two or more questions.

DO NOT OPEN THE PAPER UNTIL INSTRUCTED BY 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, LS2 9JT.

(Tel. 0113 343 2339)

http://www.ukmt.org.uk

All candidates must be in School Year 11 (England and Wales), S4 (Scotland), orSchool Year 12 (Northern Ireland).

2

Advice to candidates

• Do not hurry, but spend time working carefully on one question before attemptinganother.

• 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 suchas , fractions, or square roots, if appropriate, but NOT decimal approximations.π

• Give full written solutions, including mathematical reasons as to why your method iscorrect.

• 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.

3

2008

1. All the digits of a certain positive three-digit number are non-zero. When the digits aretaken in reverse order a different number is formed. The difference between the twonumbers is divisible by eight. Given that the original number is a square number, find its possible values.

2. The diagram shows a regular pentagon .A circle is drawn such that is a tangent to thecircle at and is a tangent to the circle at .The side of the pentagon is extended to meetthe circumference of the circle at .

ABCDEAB

A CD DAE

FProve that .DE = DF

A

B

CD

E

F

3. Show that the equation

1x

+1y

=511

has no solutions for positive integers .x, y

4. A circle is inscribed in a right-angled triangle, as shown.The point of contact of the circle and the hypotenusedivides the hypotenuse into lengths and .x yProve that the area of the triangle is equal to .xy

x

y

5. An ant lives on the surface of a cuboid which haspoints and on three adjacent faces.X, Y ZThe ant travels between , and along the shortestpossible path between each pair of points. The angles

, and are the angles between the parts of theant's path, as shown. Prove that .

X Y Z

x° y° z°

x + y + z = 270X

Y

Zx°

y°

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.1 : 2

Show that such an arrangement is possible for a or a grid but impossible forany other grid.

1 × 2 2 × 3g × (g + 1)

4

2009

1. Five numbers are arranged in increasing order. As they get larger the difference betweenadjacent numbers doubles.The average of the five numbers is 11 more than the middle number. The sum of thesecond and fourth numbers is equal to the largest number.What is the largest number?

2. 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 2, or 3, or 4, or 5, or 6, or 7, or 8, or 9,

there is always a remainder of 1.What is Miko's PIN?

3. Solve the simultaneous equations5xy

x + y= 6

4xz

x + z= 3

3yz

y + z= 2.

4. In a trapezium the sides and are parallel and .ABCD AB DC ∠BAD = ∠ABC < 90°Point lies on with .P AB ∠CPD = ∠BADProve that .PC2 + PD2 = AB × DC

5. A lottery involves five balls being selected from a drum. Each ball has a differentpositive integer printed on it.Show that, whichever five balls are selected, it is always possible to choose three of themso that the sum of the numbers on these three balls is a multiple of 3.

6. In the figure, , , and are the lengths of four arcswhich together form the circumference of the circle.

p q r s

Find, in simplified form, an expression for in termsof , and .

sp q r

p

q

r

s

5

2010

1. How many different ways are there to express in the form , where and arepositive integers with ?

215

1a + 1

b a ba ≤ b

2. The diagram shows a regular heptagon, aregular decagon and a regular 15-gon with anedge in common.

Find the size of angle .XYZ

Y

Z

X

3. Solve the equations

x + xy + x2 = 9

y + xy + y2 = −3.

4. The diameter of a circle has length 4. The points and lie on the circle, as shown, so that .

AD BC AB = BC = 1

Find the length of .CD

4

1

1

A

B

C

D

5. The diagram shows a rectangle divided into eight regionsby four straight lines. Three of the regions have areas 1,2 and 3, as shown.

What is the area of the shaded quadrilateral?

1

23

6. Every day for the next eleven days I shall eat exactly one sandwich for lunch, either aham sandwich or a cheese sandwich. However, during that period I shall never eat aham sandwich on two consecutive days.

In how many ways can I plan my sandwiches for the next eleven days?

6

2008 Solutions

1. All the digits of a certain positive three-digit number are non-zero. When the digits aretaken in reverse order a different number is formed. The difference between the twonumbers is divisible by eight. Given that the original number is a square number, find its possible values.

SolutionConsider the three-digit number . When the digits are takenin reverse order we obtain the number . If these numbers aredifferent, then .

‘abc’ = 100a + 10b + c‘cba’ = 100c + 10b + a

a ≠ c

Case I: If , then the difference between and isa > c a > c ‘ abc’ ’cba’

(100a + 10b + c) − (100c + 10b + a) = 99a − 99c

= 99 (a − c) .We are told that this difference is divisible by 8. But 99 is not divisible by 2, so weconclude that is divisible by 8. a − cSince and are different single-digit numbers, the only possible value of is 8,which occurs when and .

a c a − ca = 9 c = 1

It now remains to check the nine possible values of . Of these, only yields asquare number for , namely .

b b = 6‘abc’ 961 = 312

Case II: Using a similar argument, we obtain and . Again, checkingthe nine possible values of , we find only one square number, namely .

a < c a = 1 c = 9b 169 = 132

Hence, the only numbers satisfying the given conditions are 169 and 961.

2. The diagram shows a regular pentagon. A circle is drawn such that is

a tangent to the circle at and is atangent to the circle at . The side ofthe pentagon is extended to meet thecircumference of the circle at .

ABCDE ABA CD

D AE

FProve that .DE = DF

A

B

CD

E

F

First solutionSince is a regular pentagonABCDE

∠DEF =360°

5

= 72°. (1)We also deduce that each interior angle of the regular pentagon is , since angles on astraight line are supplementary.

108°

7A

B

CD

E

F

O

Let be the centre of the circle. The angle between a tangent and the radius drawn tothe point of contact is 90°. Hence

O

∠OAB = ∠ODC = 90°.The sum of the interior angles of the pentagon is . HenceOABCD 540°

∠AOD = 540° − 90° − 90° − 108° − 108°

= 144°.The angle subtended at the centre of a circle is twice the angle subtended at thecircumference, so

∠EFD = ∠AOD ÷ 2

= 72°. (2)From equations (1) and (2) we deduce that , that is, triangle hasequal base angles; therefore the sides opposite these angles are equal. Hence ,as required.

∠EFD = ∠DEF DEFDE = DF

Second solutionA

B

CD

E

F

Another solution makes use of the ‘alternate segment theorem’, from which, as shown.∠BAD = ∠AFD

Can you see how to complete the argument?

83. Show that the equation

1

x+

1

y=

5

11

has no solutions for positive integers .x, y

SolutionsThe first two solutions make use of the following:We are given

1x

+1y

=511

,

where and are positive integers. x yMultiplying the given equation throughout by we obtain11xy

11y + 11x = 5xy. (∗)First solutionEquation may be rearranged to give(∗)

= 0,5xy − 11x − 11y

or = 0.25xy − 55x − 55y

Therefore = 0,(5x − 11) (5y − 11) − 121

so that = 121. (1)(5x − 11) (5y − 11)Now we know that and are integers and so and are also integers. x y 5x − 11 5y − 11Therefore equation (1) tells us that is a factor of 121, that is,5x − 11

5x − 11 = ±1, ±11, ± 121.or

Of these six cases, only gives a positive integer for but in this case leading to . So we conclude that the original equation has no

solutions for positive integers and .

5x − 11 = −1 x5y − 11 = −121 y = −22

x y

Second solutionUse equation to find in terms of :(∗) y x

= 11y + 11x 5xy

so = 11x 5xy − 11y

= (5x − 11) y,

and thus = (2)11x

5x − 11y,

noting that is non-zero since is an integer.5x − 11 xNow , so and hence since is an integer. y > 0 5x − 11 > 0 x ≥ 3 xWe may also assume, without loss of generality, that , so that andhence .

y ≥ x 11 ≥ 5x − 11x ≤ 4

Therefore or . However, from equation (2), neither of these values of gives aninteger value for and hence the original equation has no solutions for positive integers and

x = 3 x = 4 xy x

y.

9Third solutionAssume, without loss of generality, that . Thenx ≤ y

1

x+

1

y≤

2

x

511

≤2x

so that

5x ≤ 22.and hence

Therefore since is an integer. Can you see how to proceed?x ≤ 4 x

NoteA unit fraction has numerator 1. An Egyptian fraction is constructed by adding differentunit fractions together, for example,

511

=13

+111

+133

.

In ancient Egypt fractions were written like this.

It is possible to write any given proper fraction as an Egyptian fraction, but finding thesmallest number of unit fractions needed is not easy—the problem above shows that requires more than two.

511

4. A circle is inscribed in a right-angled triangle, as shown.The point of contact of the circle and the hypotenusedivides the hypotenuse into lengths and .x yProve that the area of the triangle is equal to .xy

x

y

SolutionUse the notation in the diagram below.

x

y

r

r

A

B

C

Q

R

O

The two tangents to a circle from an external point are of equal length. Hence and .AR = y BQ = xNow and are radii of the circle and and are tangents to the circle, so that

and . It follows that is a square, and hence.

OR OQ AC BC∠ORC = 90° ∠OQC = 90° OQCRCR= CQ = r

From Pythagoras' theorem applied to triangle we getABC

AB2 = BC2 + AC2

(x + y)2 = (x + r)2 + (y + r)2and sowhich may be expanded and simplified to give

xy = xr + yr + r2. (1)

10

Now the area of the triangle is

‘ 12 base× height’ = 1

2 × AC × BC

= 12 (y + r) (x + r )

= 12 (xy + xr + yr + r2) ,

= 12 (xy + xy) from equation (1)

= xy,as required.

5. An ant lives on the surface of a cuboid which haspoints and on three adjacent faces.X, Y ZThe ant travels between , and along the shortestpossible path between each pair of points. The angles

, and are the angles between the parts of theant's path, as shown. Prove that .

X Y Z

x° y° z°

x + y + z = 270X

Y

Zx°

y°

z°

First solutionConsider any two adjacent faces along which the ant moves and unfold them as shown inthe diagram. Since the ant takes the shortest possible path between any two points, onthe unfolded diagram the path will be a straight line.

ab

It follows that the path, when it crosses an edge, always forms a pair of supplementaryangles, such as and in the diagram above.a b

X

V

Y

Z

x°

y°

z°

Now consider the three quadrilaterals, shown shaded in the diagram above, which arebounded by the path and which have one vertex of the cuboid in common. VThe angle sum of each quadrilateral is and so the sum of the angles in all threequadrilaterals is .

360°3 × 360° = 1080°

11

This sum comprises:the three angles , and ;x° y° z°three pairs of supplementary angles, as described above; andthree right angles at the vertex .V

We thus obtain the equation

x + y + z + 3 × 180 + 3 × 90 = 1080

x + y + z = 270,and hence

as required.

Second solutionAnother solution considers a net of three faces of the cuboid, drawn on a sheet of paper,as shown in the diagram below.

X

Y

Z

T

On the net, one section of the ant's path is disconnected. The two broken parts of the pathmay be extended, as shown by the dashed lines, to meet at a point .TCan you see how to show that the angle at is a right angle, and so complete the proof?T

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.1 : 2

Show that such an arrangement is possible for a or a grid but impossible forany other grid.

1 × 2 2 × 3g × (g + 1)

SolutionsWithout any loss of generality, we may suppose that the sheet of paper measures units by 1 unit. Let the width of the margin be units and let the grid becomposed of squares with sides of length units, as shown in the diagram.

2m g × (g + 1)

d

12

2

m

d

1

The length of the sheet is

d (g + 1) + 2m = 2

and the width of the sheet is

dg + 2m = 1. (1)Eliminate from this pair of equations to yield the relationshipd

2m = g + 1 − g 2. (2)We now use the fact that has to be positive to obtainm

0 < g + 1 − g 2

( 2 − 1)g < 1and hence

g <1

2 − 1so that

=1

2 − 1×

2 + 1

2 + 1

= 2 + 1.Restricting our solutions to positive integers, we deduce that there are only twopossibilities: or .g = 1 g = 2If , then from equations (2) and (1) we find that g = 1

2m = 2 − 2 > 0

d = 1 − (2 − 2)and

= 2 − 1 > 0,so that the arrangement is indeed possible for a grid.1 × 2

Similarly, if , theng = 2

2m = 3 − 2 2 > 0

2d = 1 − (3 − 2 2)and

= 2 2 − 2 > 0,so that the arrangement is also possible for a grid.2 × 3

Hence the arrangement is possible for grids of size or but impossible forany other grids of the form .

1 × 2 2 × 3g × (g + 1)

13

2009 Solutions

1. Five numbers are arranged in increasing order. As they get larger the difference betweenadjacent numbers doubles.The average of the five numbers is 11 more than the middle number. The sum of thesecond and fourth numbers is equal to the largest number.What is the largest number?

SolutionLet be the smallest number and be the difference between the first and secondnumbers. Then the other differences are , and , so the five numbers are , ,

, and .

a d2d 4d 8d a a + d

a + 3d a + 7d a + 15dThe condition that ‘the average of the five numbers is 11 more than the middle number’gives

5a + 26d5

= a + 3d + 11,

which means that .d = 5The condition that ‘the sum of the second and fourth numbers is equal to the largestnumber’ now gives

2a + 8d = a + 15d

and hence .a = 35So the five numbers are 35, 40, 50, 70 and 110. Hence the largest number is 110.

RemarkSlightly neater algebra is obtained if instead we define and so that the numbers arerepresented as , , , and .

a da − 3d a − 2d a a + 4d a + 12d

2. 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 2, or 3, or 4, or 5, or 6, or 7, or 8, or 9,

there is always a remainder of 1.What is Miko's PIN?

SolutionSince the required number has four digits, it is between 1000 and 9999.Let the number be . We know that is a perfect square, and also that isdivisible by 2, 3, 4, 5, 6, 7, 8 and 9. Hence is divisible by the lowest commonmultiple of these numbers, which is .

N N N − 1N − 1

8 × 9 × 5 × 7 = 2520There are therefore only three possible values for in the range, namely 2521, 5041 and 7561,and of these only the middle one is a perfect square, being . Hence Miko's PIN is 5041.

N712

RemarkIt would be possible to start by listing all the four-digit perfect squares, and thenchecking 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

14

used, it must be made very clear that the check is done properly in every case. Forexample, one might exclude 5929 by showing that it is a multiple of 7 and thus hasremainder 0. It would not be enough simply to claim that this check had been madewithout giving details in each case.The task can be reduced by, for instance, excluding the even squares, leaving only 34numbers to check, or realising that the number has to end in a 1 to satisfy the conditionsfor divisibility by 2 and 5. However, by the time you are applying this logic, you mightas well go the whole way and use the method given above.

3. Solve the simultaneous equations

5xyx + y

= 6

4xz

x + z= 3

3yz

y + z= 2.

SolutionsIt is worth beginning with the observation that none of , or can be zero, since thatwould immediately invalidate the equations. This allows us, in subsequent work, tocancel a factor of , or from both sides of an equation. Note also that we should notassume that , or are whole numbers.

x y z

x y zx y z

First solutionLabel the equations as follows:

5xyx + y

= 6 (1)

4xz

x + z= 3 (2)

3yz

y + z= 2. (3)

From (1),

5xy = 6x + 6y

and from (2),

8xz = 6x + 6z,

so

5xy − 8xz = 6y − 6z. (4)

We can, however, also deduce from (1) and (2) that

5xy

x + y=

8xzx + z

,

so, cancelling ,x

5y (x + z) = 8z(x + y)

15

and therefore

5xy − 8xz = 3yz. (5)

But, from (3),

3yz = 2y + 2z. (6)

Putting (4), (5) and (6) together, we have

6y − 6z = 2y + 2z

and so

y = 2z.

Now, substituting in (3) and cancelling , we havey = 2z z

6z3

= 2,

so and . Then , by using (5), for example.z = 1 y = 2 x = 3

Note that it is now necessary to check that this triple of values works for all theequations. All we have shown so far is that, if the equations are true, then the onlypossible values of , and are 3, 2 and 1; this does not mean that, if , and are 3, 2and 1, then the equations are true—for example, the equations may have no solutions.

x y z x y z

Second solutionSince none of , or is zero, we can definex y z

a =1x

, b =1y

and c =1z

.

By taking the reciprocals of each side of the three equations, we obtain

b + a =56

a + c =4

3

c + b =32

.

Hence, by adding, we deduce that

a + b + c =116

,

so , and . Therefore , and .c = 1 b = 12 a = 1

3 x = 3 y = 2 z = 1

Again we should check these values satisfy the original equations, or show that our logicis reversible.

16

4. In a trapezium the sides and are parallel and .ABCD AB DC ∠BAD = ∠ABC < 90°Point lies on with .P AB ∠CPD = ∠BADProve that .PC2 + PD2 = AB × DC

SolutionsWe begin with a diagram showing the information given in the question. The threemarked angles are equal.

A B

CD

P

First solutionSince is parallel to , the alternate angles and are equal. Hence thetwo triangles and are similar.

AB DC ∠DCP ∠CPB�CDP �PCB

Exactly the same argument shows that is also similar to . We thereforehave three similar triangles. Notice that we have been careful to describe the triangles inthe correct order of vertices, so that the ratios of sides can now be read off conveniently.

�DPA �CDP

Now corresponding sides are in the same ratio, so we have

PBPC

=PCDC

and APPD

=PDDC

,

and it follows that

PC2 = PB × DC and PD2 = AP × DC.Adding these two equations, we obtain

PC2 + PD2 = (AP + PB) × DC

= AB × DC.

Second solutionThis proof starts by using the converse of the alternate segment theorem. Because

it follows that is a tangent at to the circumcircle of .∠DAB = ∠CPD CP P �DAPSimilarly, is a tangent at to the circumcircle of .DP P �BCPLet (extended) meet the two circles at and , as shown.DC X Y

A B

CD

P

X Y

17

Now , by the external angle of a cyclic quadrilateral, and also, by alternate angles. Hence and so is

parallel to and is a parallelogram. Similarly, is a parallelogram. Inparticular, and .

∠CXP = ∠DAP∠CXP = ∠XPA ∠XPA = ∠DAP = ∠CBP XP

CB PBCX APYDCX = PB DY = AP

Now we use the tangent-secant theorem for each circle. This gives

CP2 = CX × CD

= PB × CD

and

DP2 = DC × DY

= DC × AP.The last stage of the proof proceeds in the same way as the first solution.

RemarkWhen 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 differentpositive integer printed on it.Show that, whichever five balls are selected, it is always possible to choose three of themso that the sum of the numbers on these three balls is a multiple of 3.

SolutionIt is worth observing that, for the purposes of this question, we can replace the numberson the balls by their remainders when they are divided by 3. This is because, for any twonumbers and , their sum is a multiple of 3 if, and only if, the sum of theirremainders is divisible by 3. To prove this, suppose that, when divided by 3, hasremainder and has remainder . Then we can write and ,for some integers and . Therefore

and this is divisible by 3 if, andonly if, is divisible by 3.

a b a + ba

r b s a = 3m + r b = 3n + sm n

a + b = 3m + 3n + r + s = 3(m + n) + r + sr + s

Using this idea, any selection of five balls will result in a set of five numbers, each ofwhich is 0 or 1 or 2. If three of the balls have the same number, then their sum will be divisible by 3, and weare done.If not, then there is one pair of numbers equal to , another pair equal to and a fifthnumber equal to , where , and are 0, 1 and 2, in some order. This means that thereare three balls numbered 0, 1 and 2, and the sum of these numbers is divisible by 3.

x yz x y z

Hence, in either case, we have a choice of three of the five balls whose sum is a multipleof 3.

18

6. In the figure, , , and are the lengths of four arcswhich together form the circumference of the circle.

p q r s

Find, in simplified form, an expression for in terms of, and .

sp q r

p

q

r

s

First solutionFirst note that the length of an arc of a circle is proportional to the angle subtended at thecentre. Hence, by appropriate choice of radius, we can arrange that the arc lengths areactually equal to these angles. (If this is not done, the calculation proceeds in the sameway, but there is some cancelling to do at the end.)The diagram shows these four angles at the centre of the circle.

p°q°

r°s°

Now there are two relationships between , , and . First there is the obvious onep q r s

p + q + r + s = 360. (1)We also have the angle sum of a quadrilateral; one of the angles is , one is

, and the others can be calculated in terms of and using the fact thatthere are two isosceles triangles. This gives

90°(p + q + r ) ° p r

(p + q + r) + (90 − 12p) + 90 + (90 − 1

2r) = 360

and so

p + 2q + r = 180.

Hence, from (1),

p + q + r + s = 2p + 4q + 2r

and so

s = p + 3q + r .

19

Second solutionWe construct the two radii perpendicular to the two given chords, as shown in thediagram.

q

s

12p

12r

Since a radius perpendicular to a chord bisects the chord, it also bisects thecorresponding arc. So we have arcs of length and , as shown.1

2p12r

Now the diagram contains a quadrilateral with three right angles, so the fourth angle isalso a right angle and therefore the two radii are perpendicular. Hence the radiidetermine a quarter of the circle, and so

p + q + r + s = 4 × ( 12p + q + 1

2r) ,

which gives

s = p + 3q + r.

20

2010 Solutions1 How many different ways are there to express in the form , where and are

positive integers with ?

215

1a + 1

b a ba ≤ b

Solution There are infinitely many integers and hence the number of potential values of and isinfinite, so any satisfactory method first needs to reduce the problem to a finite numberof cases. We demonstrate two methods of doing this.

a b

Method 1 Since we have and therefore . Hence .0 < a ≤ b 1

b ≤ 1a

215 = 1

a + 1b ≤ 2

a a ≤ 15

Also, since , we have and so . Hence .b > 0 1b > 0 2

15 = 1a + 1

b > 1a a > 15

2

Therefore and the possible values of are 8, 9, …, 15. In order to seewhich of these correspond to integer values of , it is helpful to find in terms of :

8 ≤ a ≤ 15 ab b a

1b

=215

−1a

=2a − 15

15a and so

b =15a

2a − 15.

We now see that we require to be a positive divisor of . So we candetermine which values of will give integer values of from the table:

2a − 15 15aa b

a 2a − 15 15a ?Divisor b

8 1 15 × 8 yes 120

9 3 15 × 9 yes 45

10 5 15 × 10 yes 30

11 7 15 × 11 no −

12 9 15 × 12 yes 20

13 11 15 × 13 no −

14 13 15 × 14 no −

15 15 15 × 15 yes 15

Hence there are five ways to express in the required form.215

Method 2 We may multiply every term in the equation by in order to clear thefractions. We obtain . We now rearrange this equation, first writingit in the form

215 = 1

a + 1b 15ab

2ab = 15b + 15a

4ab − 30a − 30b = 0,

21then adding 225 to both sides to give

4ab − 30a − 30b + 225 = 225,

that is,

(2a − 15) (2b − 15) = 225.

Therefore is a divisor of , so that .Larger values are not possible since and so . Also, negativevalues are not possible since then would be at most , but so that

.

2a − 15 225 = 32 × 52 2a − 15 = 1, 3, 5, 9, 15a ≤ b 2a − 15 ≤ 2b − 15

2a − 15 −15 a > 02a − 15 > −15

Each of the corresponding values of and is an integer:a b or .(a, b) = (8, 120), (9, 45), (10, 30), (12, 20) (15, 15)

Hence there are five ways to express in the required form.215

2 The diagram shows a regular heptagon, aregular decagon and a regular 15-gon withan edge in common.

Find the size of angle .XYZ

Y

Z

X

Solution We shall give two methods, each of which involves finding the exterior angle of aregular polygon. If a regular polygon has sides, then each exterior angle is given byn

=360°

n. (2.1)exterior angle

Method 1 Extend the common edge to point as shown below. The angles labelled , and

are exterior angles of the regular decagon, regular heptagon and regular 15-gonrespectively.

PO T x° y°z°

Y

ZX

P

T

Ox°

z°

y°

Hence, using the result (2.1),

x =36010

= 36,

y =3607

= 5137

and z =360

15= 24.

It follows that and . Now thesides of the three polygons are all equal, so the triangles and are isosceles.We can therefore find their base angles:

∠XOY = x° + y° = 8737° ∠ZOY = y° − z° = 273

7°XOY ZOY

∠XYO =180° − 873

7°2

= 4627°

22

and ∠ZYO =180° − 273

7°2

= 7627°.

Thus .∠XYZ = ∠ZYO − ∠XYO = 7627° − 462

7° = 30°

Method 2

Let be the common edge, as shown below. Since thepoints , , and lie on a circle centre .

PO OX = OY = OZ = OPX Y Z P O

Y

ZX

P

Ox°

z°

Now ‘the angle at the circumference is half the angle at the centre’, so

∠XYZ = 12∠XOZ

=12

(x + z) °.

But from (2.1)

x =360

10= 36

and z =36015

= 24,

so that .∠XYZ = 12 (36 + 24) ° = 30°

3 Solve the equations

x + xy + x2 = 9

y + xy + y2 = −3.

Solution

‘Substitution’ is one of the standard methods of solving simultaneous equations: use oneequation to find an expression for one unknown, then substitute this expression into theother equation, thereby forming a single equation in just one of the unknowns. Thoughit is possible to use a substitution method straight away here, the algebra is ratherunpleasant, so we demonstrate two other approaches.

In passing, we note that the question places no restrictions on and . In particular, wecannot assume that they are integers.

x y

23

Method 1

Adding the two given equations, we get

x + y + x2 + 2xy + y2 = 6

so that

(x + y)2 + (x + y) − 6 = 0,

which factorises to give

(x + y − 2) (x + y + 3) = 0.Hence

x + y = 2 or x + y = −3. (3.1)

Also, subtracting the two given equations, we get

x − y + x2 − y2 = 12

which factorises to give

(x − y) + (x − y) (x + y) = 12

so that

(x − y) (1 + x + y) = 12.Hence, using (3.1),

x − y = 4 or x − y = −6, (3.2)

which occur when and respectively.x + y = 2 x + y = −3

We can now solve equations (3.1) and (3.2) by, for example, adding and subtracting, toobtain

(x, y) = (3, −1) or (−92, 3

2) .

We now need to check whether these two pairs of values really do satisfy the equationsgiven in the question. Each of them does, so the required solutions are , and , .

x = 3 y = −1x = −9

2 y = 32

Method 2

Factorise each of the given equations to give

x (1 + y + x) = 9 (3.3)

and y(1 + x + y) = −3. (3.4)

Since no side of either equation is zero, we may divide equation (3.3) by (3.4) to obtain, so that . Now substitute this expression for into equation (3.4) to getx

y = −3 x = −3y x

y(1 − 2y) = −3,

24which may be rearranged to

2y2 − y − 3 = 0,

or

(y + 1) (2y − 3) = 0.

Therefore or and, since , we have solutionsy = −1 y = 32 x = −3y

(x, y) = (3, −1) or (−92, 3

2) .

We now need to check whether these two pairs of values really do satisfy the equationsgiven in the question. Each of them does, so the required solutions are , and , .

x = 3 y = −1x = −9

2 y = 32

4 The diameter of a circle has length 4. The points and lie on the circle, as shown, so that .

AD B CAB = BC = 1

Find the length of .CD

4

1

1

A

B

C

D

Solution 1Let be the centre of the circle, so that , and let chord andradius meet at .

O OA = OB = OC = 2 ACOB X

2

2

21

1

A

B

C

DO

X

Triangle is isosceles and bisects angle (because the chords and areequal and so subtend equal angles at the centre). Hence is the perpendicular bisectorof the base of the isosceles triangle . In other words, and

, as shown in the diagram above.

OAC OB AOC AB BCOB

AC AOC AX = XC∠AXO = 90°

Since angle is (angle in a semicircle) triangle is a right-angled trianglewith , whose length we have to find, as one side. We know , so can find ,using Pythagoras' theorem, from

ACD 90° ACDCD AD = 4 CD

CD2 = AD2 − AC2, (4.1)provided we can find the length of . We shall do this by using areas, but there areother methods.

AC

Now consider isosceles triangle and let be the midpointof , so that triangle is right-angled.

OAB NAB ANO

2

2

A

B

O

N12

12

25

Then, from Pythagoras' theorem, , so that . Hence thearea of triangle is .

NO2 = 22 − (12)2 = 15

4 NO = 152

AOB 12 × AB × NO = 15

4

But the area of triangle is also . Therefore and so. Using this value in equation (4.1). we get

OAB 12 × OB× AX AX = 15

4

AC = 2AX = 152

CD2 = AD2 − AC2 = 42 −15

4=

49

4and hence .CD = 7

2

Solution 2

Let be the centre of the circle. Reflecting the sector , shown shaded, about thediameter perpendicular to , so that and are interchanged, gives the right-handdiagram. If is the reflection of , then . Also, since reflection about adiameter reflects a circle to itself, we know that lies on the circle.

O OBDBD B D

C′ C BC′ = CDC′

1 1

A

B

DO

1

1

A

B

C

DO

C′

Now angles and are both angles at the circumference subtended by chords oflength 1. These angles are therefore equal and so and are parallel (alternateangles).

AC′B C′ADBC′ AD

Draw perpendiculars from and to to create a rectangle , as shown below,with .

B C′ AD BXYC′BC′ = XY

1 1

A

B

DX YO

C′

Then and, by constructing the perpendicular bisector of inisosceles triangle , shown below, we see that . Therefore .

AX = AB cosθ = cosθ ABOAB cosθ = 1

4 AX = 14

2

2

A

B

O

N12

12

Similarly and so .DY = 14 CD = BC′ = XY = 4 − 1

4 − 14 = 31

2

26Solution 3

Let be the intersection of produced and produced.P AB DC

4

1

1

A

B

C

P

D

Now angles and are both angles at the circumference subtended by chords oflength 1. These angles are therefore equal. Also, (angle in a semicircle).Therefore triangles and are congruent (ASA). Hence BP = 1 and PD = 4.

ADB BDC∠ABD = 90°

ABD PBD

Further, in triangles and , angle is common and (exteriorangle of cyclic quadrilateral). These triangles are therefore similar and hence

. So and .

BCP DAP P ∠BCP = ∠DAP

PC : 1 = 2 : 4 PC = 12 CD = PD − PC = 4 − 1

2 = 312

5 The diagram shows a rectangle divided into eight regionsby four straight lines. Three of the regions have areas 1,2 and 3, as shown.

What is the area of the shaded quadrilateral?

1

23

Solution

Notice that the shaded area is the intersection of the two triangles shown shaded in thefollowing diagrams.

The area of each of these triangles is half the area of the rectangle, since each has thesame base and height as the rectangle. Therefore the total unshaded area within therectangle 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 right-hand figure.

1

23

a

b

s

Using the notation indicated above, we therefore have and hence . So the area of the

shaded quadrilateral is 6.a + s + b = a + 1 + 2 + b + 3 s = 1 + 2 + 3 = 6

Remark The result also applies if the rectangle is replaced by a parallelogram. Moreover, there isclearly nothing special about the values 1, 2 and 3.

27

6 Every day for the next eleven days I shall eat exactly one sandwich for lunch, either aham sandwich or a cheese sandwich. However, during that period I shall never eat aham sandwich on two consecutive days.

In how many ways can I plan my sandwiches for the next eleven days?

Solution

We shall use the notation for the number of ways of choosing objects from . This

is a binomial coefficient, sometimes written . We have

( )nr

r n

nCr

( ) =n!

r! (n − r )!=

n (n − 1) … (n − r + 1)1 × 2 × … × r

.nr

We give three different methods, all of which create a plan for the next eleven days byconstructing a line of Cs and Hs.

Method 1

Let be the number of ways of creating a line of length starting with H, and be the number starting with C.

h (n) n c(n)

For any line of length starting with C, we may construct a line of length startingwith H by adding H at the start. Also, for any line of length starting with H, wemay construct a line of length starting with C by deleting the initial H. Both of thesestatements depend on the rule that there are no consecutive Hs. Hence

.

n n + 1n + 1

n

h (n + 1) = c (n)

Similarly, adding C to the front of a line starting with H or one starting with C creates alonger line starting with C (and vice versa), so that .c (n + 1) = c (n) + h(n)

Therefore

h (n + 2) = c (n + 1)

= c (n) + h (n)

= h (n + 1) + h (n)

and

c (n + 2) = c (n + 1) + h (n + 1)

= c (n + 1) + c(n) .

We thus have two sequences generated like the Fibonacci sequence: each term is the sumof the previous two terms. Moreover, letting be the total number of ways toconstruct a line for days, then and therefore

. Hence is also a Fibonacci-like sequence.

t (n)n t (n) = h (n) + c (n)

t (n + 2) = t (n + 1) + t (n) t (n)

Now , corresponding to the ‘lines’ C and H, and , corresponding tothe lines CC, CH and HC. So the first eleven terms of are 2, 3, 5, 8, 13, 21, 34, 55,89, 144 and 233. Hence the number required is 233.

t (1) = 2 t (2) = 3t (n)

28

Method 2

We create a plan for the next eleven days by constructing a line of Cs and Hs from twotypes of tile, or , which ensures that two Hs are never placed together. Thereare various possibilities, determined by the number of tiles which are used, whichcan be 11, 9, 7, 5, 3 or 1.

C HCC

Eleven tiles can be placed in just one way, or ways.C ( )110

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 10 tiles.

C HC ( )101

HC

Seven tiles and two tiles can be placed in ways, the number of ways of

choosing the position of two tiles in a row of 9 tiles.

C HC ( )92

HC

Continuing in this way, we see that the total number of ways is

( ) + ( ) + ( ) + ( ) + ( ) + ( ) = 1 + 10 + 36 + 56 + 35 + 6110

101

92

83

74

65

= 144.

However, placing tiles in this way always ends the line with a C, so does not allow forthe possibility of ending with an H. But if the line ends with an H, then the remainder ofthe line has ten letters and ends with a C. We may count the number of ways for this in asimilar way to the above, but with a total of 10 letters instead of 11:

( ) + ( ) + ( ) + ( ) + ( ) + ( ) = 1 + 9 + 28 + 35 + 15 + 1100

91

82

73

64

55

= 89.

Therefore the total number of ways altogether is .144 + 89 = 233

Method 3 We create a plan for the next eleven days by starting with Cs in a line, where ,and adding some Hs (possibly none) to construct a line of Cs and Hs. There are slots into which a single H may be placed—at either end or between two Cs. We need toplace Hs so that each H is in a different slot, so we have to choose of

the slots; there are ways in which this can be done.

k k ≤ 11k + 1

(11 − k) 11 − k

k + 1 ( )k + 111 − k

Now we cannot add more Hs than the number of available slots, so ,that is, .

11 − k ≤ k + 1k ≥ 5

Therefore the total number of ways is the sum of for , in other words( )k + 111 − k

5 ≤ k ≤ 11

( ) + ( ) + ( ) + ( ) + ( ) + ( ) + ( ) = 1 + 21 + 70 + 84 + 45+ 11 + 16

6

7

584

9

3102

111

120

= 233.