Numbers Answers 1. The smallest amount that cannot be paid with ≤ 4 coins is 38p.[1p, 2p, 5p, 10p, 20p can all be paid with just one coin.3p, 4p, 6p, 7p, 11p, 12p, 15p, 21p, 22p, 25p, 30p require two coins.8p, 9p, 13p, 14p, 16p, 17p, 23p, 24p, 26p, 27p, 31p, 32p, 35p all require three coins.18p (= 10 + 5 + 2 + 1), 19p (= 10 + 5 + 2 + 2), 28p (= 20 + 5 + 2 + 1), 29p (= 20 + 5 + 2 + 2), 33p (= 20 + 10 + 2 + 1), 34p (= 20 + 10 + 2 + 2), 36p (= 20 + 10 + 5 + 1),37p (= 20 + 10+ 5 + 2), all require four coins. 38p cannot be paid with four coins.]

2. The surface of the initial 2 by 2 by 2 arrangement consists of 6 × 22 = 24 unit squares. The surface of each unit cube consists of just 6 unit squares. Hence we have to leave at least 4 unit cubes to ensure a surface area of 24 unit squares.

In fact, if one removes the unit cubes, with shaded faces, at the four corners, the resulting “shape” has a surface area of exactly 24 unit squares.

3. The first twelve students scored a total of 12 × 6.5 = 78 marks.

The remaining eight students might have scored any total between 8 × 0 = 0 and 8 × 10 = 80.Thus the twenty students could have a total score as low as 78 + 0 = 78 marks, or as high as78 + 80 = 158 marks. Hence all we can say about the average M for the whole group is that it must be ≥ 78/20 = 3.9 and ≤ 158/20 = 7.9

4. 0.1 = 1 / 10, so 0.n = n / 10. 0.1 = 1 / 9, so 0. n = n / 9.

(0.n) / (0. n ) = (n / 10) / (n / 9) = 9 / 10

5. 1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2))))))))))

= 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211.

Now (xn − 1 + xn − 2 + … + x2 + x + 1) (x − 1) = (xn − 1)

211 + 210 + 29 + … + 22 + 2 + 1) = (212 − 1)/ (2 − 1) = 212 − 1

6. Suppose there are h pieces along each horizontal edge and v pieces along each vertical edge. Then h.v = 1000 = 23.53.Thus the only possibilities for the pair {h, v} are {1, 1000}, {2, 500}, {4, 250}, {5, 200},{8, 125}, {10, 100}, {20, 50}, {25, 40}.The total number of edge pieces is 2h + 2v − 4 = 2 (h + v − 2) (since 2h + 2v counts eachof the four corners twice).126 (= 2 (25 + 40 − 2)), 136 (= 2 (20 + 50 − 2)), 216 (= 2 (10 + 100 − 2)) are all possible; but 316 is not.

7. The question really asks how many of the first 15 positive integers can be expressed as the sum of 4 or fewer perfect squares. These can be listed as follows:−1 = 12; 2 = 12 + 12; 3 = 12 + 12 + 12; 4 = 22. Expressions for 5, 6, 7 and 8 can be obtained by adding 22 to each of the ones already given. 9 = 32 and expressions for 10, 11, 12, 13, 14, and 15 can be obtained by adding to the ones for 1, 2, 3, 4, 5 and 6.

Note: In fact, the French mathematician Lagrange proved in 1770 that every positive integer can be

expressed as the sum of four squares.

8. Using their initial to denote a candidate’s mark gives:−

P + N = 32; P + F = 26; N + F = 36. Adding these three equations gives 2P + 2N + 2F = 94, hence P + N + F = 47. Thus 47 + G = 4 × 16 giving G = 17.

Biographical note: although the lives of Pascal and Fermat overlapped with the lives of the other two mathematicians, Galileo died in 1642, the year in which Newton was born (on Christmas Day).

9. Let Mary’s height at age 5 be h. So at age 10 her height is 1.3h and by age 15 this will be1.2 × 1.3h = 1.56h.

10. Number of years = 52

144

245

346

447

548

649

= 49 × 1 × 47 × 46 × 3 × 44 ÷ 52

3750003503 .

Note: 375000 is clearly an overestimate since 49 × 1 × 47 × 46 is less than 503 and 3 × 44 ÷ 52 is less than 3.

11. The smallest number of possible prime divisors of 457 that Damien needs to check is the number of prime numbers less than or equal to the square root of 457. Since 212 < 457 < 222, he needs to check only primes less than 22. These primes are 2, 3, 5, 7, 11, 13, 17 and 19.

12. Let Jon’s and Jan’s ages be 3x and x respectively. Then 3x − 3 = 4(x − 3) which gives x = 9. Therefore Jon is 27 and Jan is 9. If the ratio will be 2 : 1 in y years time then 27 + y = 2(9 + y) which gives y = 9. In 9 years time, Jon will be 36 and Jan will be 18.

13. Let N be the two-digit number ‘ab’, that is N = 10a + b. So the sum of N and its ‘reverse’ is10a + b + 10b + a = 11a + 11b = 11 (a + b). As 11 is prime and a and b are both single digits, 11 (a + b) is a square if, and only if, a + b = 11. So the possible values of N are 29, 38, 47, 56, 65, 74, 83, 92.

14. Let the original cost price and original selling price of the dress be C and S respectively. Then 0.8 × S =

1.04 × C. So CCS 3.1

8.004.1

.Therefore the shopkeeper would have made a profit of 30% by selling the dress at its original price.

15. The volume of water which fell at Sprinkling Tarn in 1954 is approximately equal to (25 000 × 6) m3, that is 150 000 m3. Now 1 m3 = 106 cm3 = 106 ml = 1000 litres.So approximately 150 million litres of water fell on Sprinkling Tarn in 1954.

16. Let A, B, C, D, E be five vertices of the star, as shown. Then AB = BC = CD = DE = 1. Each exterior angle of a regular octagon is 360o ÷ 8, that is 45o, so CBD = CDB = 45o. Hence BCD is a right angle and we deduce from the symmetry of the figure that each interior angle of the star is either 90o or 225o. The length of BD is 2, so the area of the star is the area of a square of side 2 + 2 plus the area of four congruent triangles with sides 1, 1, 2.

The required area, therefore, is (2 +2)2 + 4( 21

× 1 × 1), that is 6 + 42 + 2, that is 8 + 4 2.Possible alternative ending: Dissect the star into 8 congruent kites such as OBCD. As for a rhombus, the area of a kite is half the product of its diagonals. In this case that is

).22(212)21(

21

21

BDOC

Required area is 4 )22( .

A B

C

D E

O

17. First note that 1, 11, 111, 1111, 11111 are not divisible by 7 but that 111111 = 15873 × 7. So any number in which all the digits are the same is divisible by 7 if the number of digits is a multiple of 6. Therefore the 2004-digit number 888…888 is a multiple of 7. Similarly, the 2004-digit number 222…222 is a multiple of 7, as is the 2004-digit number 222…229 since it differs from 222…222 by 7. Furthermore, the 2004-digit number 222…22n is not a multiple of 7 if n is any digit other than 2 or 9. Now N = 222…22n × 102004 + 888…888, so if N is divisible by 7 then 222…22n is divisible by 7 and we deduce that n = 2 or 9.

18. Note that n2 − 1 is divisible by n − 1. Thus:

).1(1–

8–11–

8–1–1–

1–9– 22

nn

nnn

nn

n

So, if n is an integer, then 1–9–2

nn

is an integer if and only if n − 1 divides exactly into 8. The possible values of n − 1 are −8, −4, −2, −1, 1, 2, 4, 8, so n is −7, −3, −1, 0, 2, 3, 5, 9. The sum of these values is 8.(Note that the sum of the 8 values of n – 1 is clearly 0, so the sum of the 8 values of n is 8.)

19. When n! is written in full, the number of zeros at the end of the number is equal to the power of 5 when n! is written as the product of prime factors, because there is at least that high a power of 2 available. For example, 12! = 1 × 2 × 3 × … × 12 = 210 × 35 × 52 × 7 × 11.This may be written as 28 × 35 × 7 × 11 × 102, so 12! ends in 2 zeros, as 28 × 35 × 7 × 11 is not a multiple of 10.We see that 24! ends in 4 zeros as 5, 10, 15 and 20 all contribute one 5 when 24! is written as the product of prime factors, but 25! ends in 6 zeros because 25 = 5 × 5 and hence contributes two 5s. So there is no value of n for which n! ends in 5 zeros. Similarly, there is no value of n for which n! ends in 11 zeros since 49! ends in 10 zeros and 50! ends in 12 zeros. The full set of values of k less than 50 for which it is impossible to find a value of n such that n! ends in k zeros is 5, 11, 17, 23, 29, 30 (since 124! ends in 28 zeros and 125! ends in 31 zeros), 36, 42, 48.

20. Let N have x digits, so that x ≤ 2002. When the digit 1 is placed at its end, N becomes 10N + 1. When 1 is placed in front of it, N becomes 10x + N. Therefore: 10N + 1 = 3 (10x + N) i.e.7N = 3 × 10x − 1. So we need to find which of the numbers 2, 29, 299, 2999, 29999, …. are divisible by 7. The first such number is 299999 (corresponding to x = 5), giving N = 42857 and we check that 428571 = 3 × 142857.The next such numbers correspond to x = 11, x = 17, x = 23, x = 29 and the largest number in the given range corresponds to x = 1997.The number of different values of N, therefore, is 1 + (1997 − 5) ÷ 6 = 333.

Algebra Answers 1. The first twelve students scored a total of 12 × 6.5 = 78 marks.

The remaining eight students might have scored any total between 8 × 0 = 0 and 8 × 10 = 80.Thus the twenty students could have a total score as low as 78 + 0 = 78 marks, or as high as78 + 80 = 158 marks. Hence all we can say about the average M for the whole group is that it must be ≥

78/20 = 3.9 and ≤ 158/20 = 7.9.

2. Using their initial to denote a candidate’s mark gives:−

P + N = 32; P + F = 26; N + F = 36. Adding these three equations gives 2P + 2N + 2F = 94, hence P + N + F = 47. Thus 47 + G = 4 × 16 giving G = 17.

Biographical note: although the lives of Pascal and Fermat overlapped with the lives of the other two mathematicians, Galileo died in 1642, the year in which Newton was born (on Christmas Day).

3.2 d d

V m s - 1 U m s- 1

t 1 t 2

Ud

tVd

t =;2

= 21

TUd

Vd

2

TUV

d

12

d (2U + V) = TUV

)2/(33 VUTUVd

4. B cannot be true since a + b > a but a ÷ b ≤ a; C cannot be true since a − b < a but a × b ≥ a; similarly, D

cannot be true. Furthermore, E cannot be true since 2ba a +b. However, A is true if a = 4 and b = 2.

5. Let the smaller number be x and y the larger number y.

Then ;5=3;+=4–4);+(

41

=– xyxyxyxyxy hence x : y = 3 : 5

6. The sequences have common differences of 7 and 9 respectively. The lowest common multiple of 7 and 9 is 63, so the next term after 2005 to appear in both sequences is 2005 + 63, that is 2068.

7. Let the smallest of the three even numbers be n. Then the other two numbers are n + 2 andn + 4. So 4n + 2(n + 4) = 3(n + 2) + 2006, that is 6n + 8 = 3n + 2012, that is n = 668.

8. Let Rachel and Heather have x and x2 pennies respectively. So x + x2 = 100n, where x and n are positive integers. We require, therefore, that x(x + 1) = 100n = 22 × 52 × n. Now x andx + 1 cannot both be multiples of 5, so their product will be a multiple of 25 if and only if x orx + 1 is a multiple of 25. The smallest value of x which satisfies this condition is 24 which is a multiple of 4 so 24 × 25 is a multiple of 100. Therefore Rachel has 24 pennies, Heather has 576 pennies and, in total, they have £6.

9. It is not possible for all five statements to be true at the same time since if a < b, a < 0, b < 0, are all true

then a1

b1

is not true since abba

ab–1–1

which is negative. However, when these three statements

are true, a2 > b2 is also true, so it is possible for four of the statements to be true at the same time.

10. Since yx

1=

, it follows that xy 1

. Hence .––1–1 22 xyxyyx

yy

xx

11. For the equation to have integer solutions, it must be possible to write x2 + nx − 16 in the form(x − α) (x − β), where α and β are integers. Therefore x2 + nx − 16 = x2 − (α + β)x + αβ and we require

that αβ = −16. The possible integer values of α, β are 1,−16; −1, 16; 2, −8; −2, 8; 4, –4 (we do not count −16, 1 as being distinct from 1, −16, for instance).As n = −(α + β), the possible values of n are 15, −15, 6, −6 and 0.

12. Re-ordering (x − 1)(x4 + 1)(x2 + 1)(x + 1) = (x − 1)(x + 1)(x2 + 1)(x4 + 1)

= (x2 − 1) (x2 + 1) (x4 + 1) = (x4 − 1) (x4 + 1) = x8 − 1

Note :(x − 1) (x4 + 1) (x2 + 1) (x + 1) has value 0 when x = 1. This enables B, C and D to be eliminated.

13.

P

Q

R

S

T

U

r cm

r cm

Let the radius of the arc with centre R be r cm.Then QT = (17 − r) cm and PU = (15 − r) cm.Now QS = QT (radii of arc ST) and PS = PU(radii of arc SU), so QP = (17 − r + 15 − r)cm= (32 − 2r)cm.But, by Pythagoras’ Theorem:

QP = 22 15– 17 cm = )15– 17)(15 + 17( cm = 8 cm

So 32 − 2r = 8, that is r = 12.

14. (a + b + c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3b2c + 3bc2 + 3c2a + 3ca2 + 6abc = a3 + b3 + c3 + 3(a + b + c)(ab + bc + ca) − 9abc + 6abc

Hence: a3 + b3 + c3 = (a + b + c)3 − 3(a + b + c)(ab + bc + ca) + 3abc.

15. Let Jon’s and Jan’s ages be 3x and x respectively. Then 3x − 3 = 4(x − 3) which gives x = 9. Therefore Jon is 27 and Jan is 9. If the ratio will be 2 : 1 in y years time then 27 + y = 2(9 + y) which gives y = 9. In 9 years time, Jon will be 36 and Jan will be 18.

16. Let the costs in pence of a peach, an orange and a melon be x, y, z respectively. We need x − y. We are given that 5x + 3y + 2z = 318 and 4x + 8y + 3z = 449. Multiplying the first by 3 and the second by 2 and subtracting gives 7x − 7y = 3 × 318 − 2 × 449 = 954 − 898 = 56, so x − y = 8.(Note that as we have three unknowns, but only two equations, it is impossible to determine unique values of x, y and z. However, as has been shown, in this case it is possible to calculate the difference between the values of x and y.)

17. The terms on the left-hand side of the equation form an arithmetic progression which has

n3 − 5 terms. So the sum of these terms is 25–3–3

25– 3

3

3

3

3 nn

nn

n

. Hence n3 − 5 = 120, so n = 5.

18.

)0≠1++,0≠+,0≠(=+)+()+(

=

++

= yxyxxxyxx

yxx

yxx

x

xy

1+++

=yx

yx

i.e. y(x + y + 1) = x + y

i.e. y2 + xy − x = 0.

For y to be real, this quadratic equation must have real roots so x2 + 4x ≥ 0, i.e. x (x + 4) ≥ 0.This condition is satisfied when x ≤ −4 or when x ≥ 0.However, x 0 so y is real when x ≤ −4 or when x > 0.

19.

1)2002(1–)2002()1)2002((–1–)2002(

11)2002(1–)2002(

1–1)2002(1–)2002(

1)2005(1–)2005()2008(

ffff

ffff

fff

= )2002(1–

)2002(22–

ff

Hence f (2002) × f (2008) = −1 provided that f (2002) ≠ 0.

20. Let X consist of x digits, each of which is 1. So 91–10x

X . Let pX2 + qX + r consist of y digits, each of

which is 1. So pX2 + qX + r = 91–10 y

. Then p

2

91–10

x

+ q 91–10

91–10 y

rx

, that is p(102x − 2 × 10x + 1) + 9q(10x − 1) + 81r = 9(10y − 1), that is (on dividing throughout by 102x), p + (9q − 2p)10−x + (p − 9q + 81r)10−2x = 9 × 10y−2x − 9 × 10−2x.

We now let x tend to infinity (through integer values). The LHS of the above equation tends to p, and the second term on the right goes to 0. By continuity of the function f (u) = 10u = eu log10, we can deduce that y − 2x must tend to a limit. Let this limit be L. Since y − 2x is always an integer, it must actually equal L for all x sufficiently large. Passing to the limit, therefore, we obtainp = 9 × 10L. Since p is to be an integer, we must have that L (also an integer) is a non-negative integer. Substituting for p in the previous equation and simplifying leads to 9q − 18 × 10L + (9 × 10L − 9q + 81r)10−x = −9 × 10−x.

Passing to the limit again leads to q = 2 × 10 L and the previous line then also gives 9 × 10L − 18 × 10L + 81r =

−9. So r = 91–10L

. Possible values of (p, q, r) therefore are (9, 2, 0), (90, 20, 1),(900, 200, 11), etc. So of the values given in the question for q, only q = 2 is possible.(Observe that the three triples above correspond to L = 0, L = 1, L = 2 respectively and we note that increasing L by1 corresponds to multiplying pX2 +qX + r by 10 and adding 1. As pX2 +qX + r consists only of 1s, 10(pX2 + qX + r) + 1 will also consist only of 1s, explaining why there is an infinite family of quadratics which satisfy the required condition.)

Shapes Answers 1. Let the cube have side length s. Using Pythagoras on the right angled triangle ABC (with

AB = BC = s), we find AC = s 2. Using Pythagoras on the right angled triangle ACG, wefind AG = s 3. Hence cos CAG = AC / AG = (2 / 3)

2. The perimeter of the rectangle is 12 cm. So each side of the square is 3 cm.

3.

AB

CP

ΔAPBis equilateral hence ABP = 60°

But BC = AB = PB so ΔPBC is isosceles with 7560–135PBC .

)75–180(21BPC

and 605.52APC .

4.C

A B

AB is a diameter so ACB = 90° 222 – ACABCB

= 400 – 144 = 256CB = 16

5. Let the exterior angle be x°. Then x + 4x = 180 and therefore x = 36. As is the case in all convex polygons, the sum of the exterior angles = 360° and therefore the number ofsides = 360/36 = 10.

6.

6

6

2

2

T h e b l a c k s q u a r e i s8 c m b y 8 c m .

The dotted line shows the locus of P as thewhite square slides around the black square.The locus is a square of side 16 cm and hencemoves through a distance of 64 cm.(Note: It may be shown that the total distancemoved by P is 64 cm, irrespective of the position ofP on the white square.)

7. New area %99)(

21

10099

1011

109

109

21

hbahba

of previous area of trapezium.

8. Joining A to C and each of A, C and E to the centre of the hexagon divides it into six congruent triangles

four of which make up ABCE. So the area of ABCE is 32

of 60.

9. The shaded area = 2222

35)–4(

31

32 rrrr

.

The unshaded area = 222

37

35–4 rrr

. Hence the required ratio is 5 : 7.

10. The shaded area is a right-angled isosceles triangle with height 1. The altitude divides it into two other

right-angled isosceles triangles so the base of the shaded area is 2. Therefore the area is 112

21

.

11. The diagram shows the original rectangle with the corner cut from it to form a pentagon. It may be deduced that the length of the original rectangle is 20 and that a, b, c, d are 8, 10, 13, 15 in some order. By Pythagoras’ Theorem c2 = (20 – b) 2 + (a – d)2. So c cannot be 8 as there is no right-angled triangle having integer sides and hypotenuse 8. If c = 10, then (20 – b) and (a – d) are 6 and 8 in some order, but this is not possible using values of 8, 13 and 15. Similarly, ifc = 15, then (20 – b) and (a – d) are 9 and 12 in some order, but this is not possible using values of 8, 10 and 13. However, if c = 13, then (20 – b) and (a – d) are 5 and 12 in some order, which is true if and only if a = 15, b = 8, d = 10.

So the area of the pentagon is 20 × 15 − 270125

21

.

a

bc

d

2 0

12. The shaded area is 2436015)1–4(

36022 xxx

. So 24x

= 64× 2

; thus x = 64.

13. In the given diagram, there are four hexagons congruent to the hexagon in Figure (i), four hexagons congruent to the hexagon in Figure (ii) and eight hexagons congruent to the hexagon in Figure (iii).

Figure (i) Figure (ii) Figure (iii)

14. The pyramid has 2n edges and n + 1 faces, so the required difference is 2n − (n + 1), that isn − 1.

15.P

R

Q

O

S V

U T

Let O be the centre of square PQRS. The medians of triangle PSR

intersect at T so OT = OS

31

.Hence the area of triangle PTR is one third of the area of trianglePSR, that is one sixth of the area of square PQRS. So the required

fraction = 32

21

61

.

16.

P

Q

RS

TU

Let U be the point of intersection of QS and RT. As QS andRT are medians of the triangle, they intersect at a point

which divides each in the ratio 2:1, so QU = 3168

32

.

Therefore the area of triangle QTR = QURT

21

323

161221

.

As a median of a triangle divides it into two triangles of equalarea, the area of triangle PTR is equal to the area of triangle QTR,so the area of triangle PQR is 64.

17. Let the lengths of the sides of the cuboid, in cm, be a, b and c. So 4 (a + b + c) = x. Also, by Pythagoras’ Theorem a2 + b2 + c2 = y2. Now the total surface area of the cuboid is

2ab + 2bc + 2ca = (a + b + c)2 – (a2 + b2 + c2) = 1616––

4

222

2 yxyx

.

18.

P Q

R

S

TU

The diameter of the sphere is l – 2h where l is thelength of a space diagonal of the cube and S is theperpendicular height of one of the tetrahedral cornerswhen its base is an equilateral triangle.The diagram shows such a tetrahedron: S is a corner ofthe cube; the base of the tetrahedron, which is consideredto lie in a horizontal plane, is an equilateral triangle,PQR, of side 2 units; T is the midpoint of PQ. AlsoU is the centroid of triangle PQR, so RU : UT = 2 : 1. AsU is vertically below S, the perpendicular height of thetetrahedron is SU.

As RTP is a right angle, RT2 = RP2 – TP2 = (2)2 – 23

22

2

.

Also, RU = RT

32

, so 32

23

94

94 22 RTRU

.

So SU2 = SR2 – RU2 = 1 – 31

32

. Therefore h =

331

31

.

Now l2 = 22 + 22 + 22 = 12, so l = 12 = 2 3 . Therefore the diameter

of the sphere is 334

332–32

.

19. Let O be the centre of the cube. Consider triangle ABO: from Pythagoras’ Theorem,

OA = AB = BO =

22

21

21

cm = 21

cm. So triangle OAB is equilateral. A similar argument may

be applied to triangles OBC, OCD, etc. The area of each of these equilateral triangles is 21

21

21

×

sin 60° cm2, that is 3

81

cm2. So the area of hexagon ABCDEF is 6 × 83

cm2. However, the total red

area exposed by the cut is twice the area of this hexagon, that is 233

cm2.20. Each arc is x/360 times the full circumference.

arc AA′ = (x / 360) 2πa, and arc BB′ = (x / 360) 2πb.

Now equate the lengths of the two routes (arc AA’ direct and via arc BB′) (x / 360) 2π = (a − b) + (x / 360) 2πb + (a − b)(x/360)2π(a − b) = 2(a − b) (x / 360)π = 1 x = 360/π ≈ 360/ (22/7) = 1260/11 ≈ 114.5.

Circles Answers 1. In returning to its original position, the centre of the disc moves around the circumference of a circle of

diameter 2d, i.e. a distance 2πd. As the disc does not slip, its centre moves a distance πd when the disc makes one complete turn about its centre, so two complete turns are made.

2. The average speed of the balloon, in km/h, ≈

12040241000

242113

12750

3. Let the small circles have radius r. Then the large circle has radius 3r. The unshaded area is 7πr2, while the shaded area is π(3r)2 − 7πr2 = 2πr2. So the required ratio is 7 : 2.

4. Each spoke of the London Eye is about 201

mile long. As 1 mile is approximately 1600 m, this means that the radius of the giant wheel is roughly 80 m. So the circumference is approximately 500 m.

5. The cross-section of the 20 m tape has area π (42 − 32) cm2 = 7π cm2.Therefore, the 80 m tape should have a cross-section of area 28π cm2.

Hence, the outer radius of the 80 m roll will be approximately 37 cm.

6.

21

ACB(angle subtended by an arc at the centre of a circle is twice the angle subtended at the

circumference) and, similarly,

21

CAD. Therefore

21–

21

AXB(the exterior angle of a

triangle is equal to the sum of the two interior opposite angles).

7.

2 - 1

2 - 1

1

145°

45°

The triangle whose vertices are the centres of the

three circles has sides of length 2,2 and 2 andis, therefore, a right-angled isosceles triangle.The perimeter of the shaded region is

222)1–2(2

412

812

.

8.x

x

O

B

T '

TU

V

x

y

V 'U 'T '

2

Let the point on the ground vertically below T be T′,let O be the point where line DB meets the walland let OT′ = x.Then, since T′OB = T′BO = 45 o T′B = x.

As TB = 2, TT′ = 2–4 x .

Hence, with respect to axes shown in the diagram,the equation of the curve on which T, U, V, … lie is

0,0,4.i.e,–4 222 yxyxxywhich is the equation of part of a circle of radius 2

9.

U

V

W

Y

X

Z

8 8 °

1 58 °

The opposite angles of a cyclic quadrilateral add up to 180° so XWZ = 180° − 158° = 22°;VWZ = 180° − 88° = 92° hence; VWX = 92° + 22° = 114°.

10. The core of the trunk occupies 81% of the volume of the trunk. Assuming that the trunk is cylindrical, this means that 81% of the cross-sectional area of the trunk is occupied by the core. Now 0.81 = 0.9, so the diameter of the core is 90% of the diameter of the trunk, that is 36 cm. Hence the thickness of the bark is 4 cm ÷ 2, that is 2 cm.

11. The diameter of the largest cylinder is 24cm, so the sum of the areas of the horizontal parts of the sculpture, excluding its base, is that of a circle of diameter 24cm, that is 144π cm2. The sum of the areas of the vertical parts of the sculpture is (2π × 1 × 2 + 2π × 2 × 2 + 2π × 3 × 2 + … + 2π × 12 × 2)cm2, that

is 312π cm2. So, excluding the base, the total surface area of the sculpture is 456π cm2.

12. ΔAOM is a 30-60-90 triangle, so OA = 2OM.

ΔAOM and ΔAO′M′ are similar, and

OO′ = OM + O′M′ (sum of two radii)

A

O

O 'MM '

OAOO

OAOOOA

OAAO

OMMO '''''

–1–

= OMOM

OMMOOM

.2–

21

.2)(–1

'''

O’M’ : OM = 1 : 3

(O’M’)2 : (OM)2 = 1 : 9

13. Join AB, BC and CA.

Then is ABC is equilateral, with sides length 3cm. The three medians meet at O, whereAO = BO = CO = (2/3) × (33/2) = 3.

OX = OY = OZ = 3 − 3.

Hence the circle centre O through X, Y, Z has area π (3 − 3)2 = π (12 − 63) = 6π (2 − 3).A

BCO

Y Z

X

14.

OA

B

The small circle can be enlarged to the size of the othercircle by using a scale factor of R / r. So if A and B and arethe centres of the circles

22r

RrrOAOB

rR

2R = r2 + r + R( )

( )( ) 22+3=1+21–2

1+2=

1–21+2

=2

rR

.

15.

r r

r

Let the radius of each large semicircle be R. Then πR2 = 4. The circle in the left-hand diagram has radius R / 2 and therefore its area is π(R/2)2 = (πR2)/4 = 1. Let the radii of the circles in the right-hand diagram be r. Then R = 2r + r.

Therefore r =

1–2

1–2121–2

12RRR

.

The grey shaded area =2πr2 = 2πR2 (2 − 1)2 = 2 × 4 × (2 − 22 + 1) = 24 − 162.The difference in areas is therefore 23 − 162.

16.C

O

D

AB

Tangents to a circle from an exterior point are equalin length.

Thus AD = AB = 1 and CD = 2 − 1In ΔOCD, ODC = 90° and OCD = 45°

Therefore ΔOCD is isosceles and hence radius OD = CD = 2 − 1

17.

D

A

E FC

B

s + t

r+

t

rs+

s

rr

t

The diagram shows points A, B and C, the centres of circles C1 C2, and C3 respectively. Let the radius of circle C3 be r. The positions of points D, E and F are as shown on the diagram.In ΔACE:AE2 = AC2 − EC2

= (r + s)2 − (r − s)2 = 4rs.In ΔBCF:BF2 = BC2 − CF2 = (r + t)2 − (r − t)2 = 4rt.

Hence DA = DE − AE = BF − AE = 2 rt − 2 rs .Also, DB = 2r − (s + t).Therefore, in ΔABD:

(s + t)2 = (2r − (s + t))2 + (2 rt − 2 rs )2

i.e. (s + t)2 = 4r2 − 4r (s + t) + (s + t)2 + 4rt − 8 rt rs + 4rs

i.e. 0 = 4r2 − 8r st

i.e. = 0 (impossible) or r = 2 st

18. Let the points of contacts of the tangents be P Q and R asshown and let xXOQ and yQOY . Then since OXis the axis of symmetry of the tangent kite OPXQ, it bisects

POQ so xXOP . A

B

C

O

X

Y

4 8 °

P

Q

R

xy

Similarly yROY . Thus, in the quadrilateral OPBRwe have2x + 2y + 2 × 90 o + 48 o = 360 o

i.e. 2 (x + y) = 132 o

i.e. x + y = 66 o

19. PQ2 = (2)2 + (2)2 = 4 PQ = 2 OP = 1

1

1

2

OA

B

P

Q2

Hence triangles APO and BPO are isosceles withBPQAPQ = 135°. The area of each of these triangles is

1121

sin 135° = 1 / (22). Also BOPAOPAOB = 45°

The shaded area APB = sector AOB − 2 × 221

= 21–

81 2OA

=

81

(12 +12 – 2 cos 135 ) – 21

=

81

(2 + 2) – 1 / 2.

Multiplying by 4 gives π (1 + 21

) 22– =22–

2)22(

.Note: as an alternative to using the cosine rule to calculate , it is

possible to apply Pythagoras’ Theorem giving O

A P

222

12

1122

2

OA

20. Let the radii of the outer and inner circles be R and r Rr

p

respectively. Then, by the Theorem of Pythagoras:R2 = r2 + p2 and therefore R2 − r2 = p2. The area of the shaded region = πR2 − πr2 = π(R2 − r2) = πp2

Triangles Answers 1. The longest side of any triangle is shorter than the sum of the lengths of the other two sides. This

condition means that the only possible triangles having different sides of integral unit length, and having perimeters less than 13 units, have sides of length 2, 3, 4 or 2, 4, 5 or 3, 4, 5.

2. Let the cube have side length s. Using Pythagoras on the right angled triangle ABC (withAB = BC = s), we find AC = s 2. Using Pythagoras on the right angled triangle ACG, wefind AG = s 3. Hence cos CAG = AC / AG = (2 / 3)

3. If (x, y) lies on the curve, so does (x, –y): this excludes A, C. Values of x which produce negative values of sin x do not feature: this excludes D. B is possible: this excludes E.

4. cos θ = 21

; sin2θ = 43

; sin θ = ±3

21

sin 2θ = 2 cos θ sin θ = 2 × 21

× 3

21

321

Note: sin 2θ is equal to 2 cos θ sin θ for all values of θ and since in this case cos θ = 21

, sin 2θ will also equal sin θ.

5. We note that the graph of y = sin (x2) passes through the origin and also is symmetrical about the y-axis,

since (−x)2 = x2. For x– , sin (x2) ≥ 0 and the only one of the graphs to satisfy all of these conditions is A.

[Note that the graph of y = sin (x2) has range −1 ≤ y ≤ 1 and crosses the x-axis when nx for all natural numbers n.]

6. By the Theorem of Pythagoras, QR = 8 cm = 2 2 cm, T

S

2 2

55

QS = 5 cm and RS = 5 cm. If T is the midpoint of QR,then TS2 = (5)2 − (2)2 = 5 − 2 = 3and therefore TS = 3 cm.

The area of triangle QRS = TSQR

21

= (2 × 3) cm2 = 6 cm2

7. Let the side of the cube be of length 2.

Then LM = MN = 211 22 ; LN = 222 211 = 6. So LMN is an isosceles triangle with sides

2, 2, 6. Thus cosNLM = 23

22/6

; hence

NLM = 30° = MNL. So LMN = 120°.(Alternatively, it may be shown that L, M and N, together with the midpoints of three other edges of the

cube, are the vertices of a regular hexagon. SoLMN may be shown to be 120°.)

8. The altitude of an equilateral triangle of side 1 is 3

21

. Hence XY = 212 1–3 .

9. Let the length in metres of the side of a pane be x. Then the area of one

pane = xx

21

sin 60° = 2

43

x.

x

x6 0 °

So 2

43

x 3300

6000

, that is x2 ≈ 3300360004

.

We conclude that 4

372 x

.

10. The five options all have h < h′ < h″. Label the triangle ABC so that h, h′, h″ are the altitudes from A, B, C respectively. ABC has area Δ = ah / 2 = bh′ / 2 = ch″ / 2.

a > b > c (since h < h < ′ h″)

Three positive lengths a, b, c with a > b > c form the sides of a triangle precisely whena < b + c (by the triangle inequality). This condition is equivalent to a/2Δ < b/2Δ + c/2Δ;that is,

1 / h < 1 / h′ + 1 / h″.

The ratios A, B, D, E all satisfy this condition, but C does not.

11. Note, from the information given, that PRQ = RPQ = 45o; RQS = 60o, PQS = 30o. Applying the

Sine Rule to: ΔSRQ : 60sin o

RS

= o

45sin

SQ

and to ΔSPQ:

oo

45sin=

30sin

SQSP

.

Hence oo 30sin

=60sin

SPRS

, so RS : SP = sin60o : sin30o = 3 : 1.

12. As α + β = 90o, sin α = cos β; cos α = sin β. So sin α sin β = sin α cos α; sin α cos β = sin2 α; cos α sin β = cos2 α; cos α cos β = cos α sinα. As α< β, α ≠ 45o. Sosinα ≠ cosα. Thus three of the four expressions have different values.

13. Let the equal sides have length k. The height of the triangle on the left is k cos x° and its base is 2k sin x°, so its area is k2 sin x° cos x°. The height of the triangle on the right is k sin x° and its base is k, so its area is

2

21 k

sin x°. Hence cos x° = 21

and so x = 60.

2 x

x

(Alternatively, the formula Δ = ab

21

sin C can be used to show that sin x° = sin 2x°; hence x + 2x = 180.)

14. The exterior angle at D of triangle ADC is equal to the sum of the two interior opposite angles:

)–( BCDACBDCADACBDC

= θ + ( ACB − θ) = ACB = DBC (given).

Hence CBD is isosceles with the same angles as ACB . AC / BC = BC / BD = BC / (AC / 2)

AC2 = 2.BC2

BC2 = AC2 + AB2 − 2.AC.AB. cos θ (by the cosine rule in ABC )

BC2 = (2.BC)2 + (2.BC)2 − 2. (2.BC) (2.BC) cos θ

cos θ = (3.BC2) / (4.BC2) = 3 / 4.

15. Each arc is x/360 times the full circumference. arc AA′ = (x / 360) 2πa, and arc BB′ = (x / 360) 2πb.

Now equate the lengths of the two routes (arc AA’ direct and via arc BB′) (x / 360) 2π = (a − b) + (x / 360) 2πb + (a − b)(x/360)2π(a − b) = 2(a − b) (x / 360)π = 1 x = 360/π ≈ 360/ (22/7) = 1260/11 ≈ 114.5.

16.

A

B CP3 0 °

3 0 °3 0 °

Points inside the triangle ABC which are closer to AB are to the left of AP. So the probability is the ratio of the area of ΔABP to the area of ΔABC which is the ratio of BP to BC since the triangles have the same height. BP to BC is equal to tan 30° : tan 60 i.e. 1 /3 : 3 = 1 : 3.[Alternatively, draw the perpendicular from to P to AC show three congruent triangles.]

17.Triangles BCQ, and CAP and ABR are congruent since each has sides of x and 3x and an included angle

of 60° (SAS). Consider triangle BCQ: its base, BQ, and height are 43

and 41

respectively of the base, QR, and height of triangle PQR.. So

area of triangle BCQ 163

× area of triangle PQR and the area of

triangle ABC = (1 − 3 × 163

) × area of triangle. PQR 1671

167

.

18. a3 + b3 = (a + b) (a2 − ab + b2)

A

B

C

P

Q R

3 x

3 x

3 x x

x

x

sin3 x + cos3 x = (sin x + cos x) (sin2 x − sin x cos x + cos2 x)

= (sin x + cos x) (1 − sin x cos x)

19. PQ2 = (2)2 + (2)2 = 4 PQ = 2 OP = 1

1

1

2

OA

B

P

Q2

Hence triangles APO and BPO are isosceles withBPQAPQ = 135°. The area of each of these triangles is

1121

sin 135° = 1 / (22). Also BOPAOPAOB = 45°

The shaded area APB = sector AOB − 2 × 221

= 21–

81 2OA

=

81

(12 +12 – 2 cos 135 ) – 21

=

81

(2 + 2) – 1 / 2.

Multiplying by 4 gives π (1 + 21

) 22– =22–

2)22(

.Note: as an alternative to using the cosine rule to calculate , it is

possible to apply Pythagoras’ Theorem giving O

A P

222

12

1122

2

OA

20. Let the original triangle be ABC and usea, b, c for the opposite sides as normal. Thenthe sides of the three inner squares are a, band c. Hence A1 = a2 + b2 + c2.We now move towards the three outersquares but first apply the Cosine Rule to theoriginal triangle in three ways.

a2 = b2 + c2 − 2bc cos A 2bc cos A = b2 + c2 − a2 (1)b2 = a2 + c2 − 2ac cos B 2ac cos B = a2 + c2 − b2 (2)c2 = a2 + b2 − 2ab cos C 2ab cos C = a2 + b2 − c2 (3)

From (1), the area of the lower left square is given byb2 + c2 − 2bc cos(180° − A) = b2 + c2 + 2bc cosA = 2(b2 + c2) − a2.

From (2), the area of the lower right square is given bya2 + c2 − 2ac cos(180° − B) = a2 + c2 + 2ac cosB = 2(a2 + c2) − b2.

C

A B

From (3), the area of the top square is given bya2 + b2 − 2ab cos(180° − C) = a2 + b2 + 2ab cosC = 2(a2 + b2) − c2.

Thus A2 = 4 (a2 + b2 + c2) − (a2 + b2 + c2) = 3A1.