I3 Find the number of sides of a regular polygon if an interior angle exceeds an exterior angle by

  1

  7

  2 24 12 km/h

  1

  2

  

3

  4

  5

  2

  88-89

  45 45 –4

  2

  12

  6

  7

  

8

  9

  10 Individual

  5

  11 12 147

  

13

  3

  14

  2

  15

  24

  12

  16

  12

  17 6 : 25

  

18

  24

  19 230

  20

  12

  9 3 7 6585 (12, 20)

  1

  2

  

3

  4

  5 88-89

  20

  2 Group

  12

  6

  43

  7 480

  

8

  9

  5

  10 110768

  5 Individual Events

  1

  1

2 I1 Given that x 3 , find x .

  = + +

  2 x x

  

Reference: 1983 FG7.3, 1984 FG10.2, 1985 FI1.2, 1987 FG8.2, 1990 HI12, 1997 HG7

₂

   

  1

  1

  2 x =  x 

  2

• −

  2

   

  x x

  2

  = 3 – 2 = 7 I2 If x # y = xy – 2 x , find the value of 2 # 3. 2 # 3 = 2 3 – 2 2 = 2

  × ×

  

I3 Find the number of sides of a regular polygon if an interior angle exceeds an exterior angle by

  150 . (Reference 1997 HG6)

  °

  Let x be the size of each interior angle, y be the size of each exterior angle, n be the number of sides.

  o o

  180 n 2 360

  − ( ) x = , y = n n x = y + 150

  ° o o

  180 n 2 360

  − ( )

  = + 150

  ° n n

  180( n – 2) = 360 + 150 n 18 n – 36 = 36 + 15 n

  ⇒ n = 24

  10 8 b 5 . ^{log 10 9} = 9 = 8 b + 5

• I4 Find the value of b such that

  1

  ⇒ b =

  2 I5 A man cycles from P to Q with a uniform speed of 15 km/h and then back from Q to P with a uniform speed of 10 km/h. Find the average speed for the whole journey.

  Let the distance between P and Q be x km.

  x x

• Total distance travelled = 2 x km. Total time = hour.

  15

  10 2 x Average speed = km/h

  x x

• 15

  10

  2 = = 12 km/h

  3

• 2

  30

  I6 [x] denotes the greatest integer less than or equal to x. For example, [3] 3, [5.7] 5.

  = =

  5

  5

5 If

  1 2 n n 14 , find n.

  = + + + +

  L

  [ ] [ ] [ ] Reference 1991 HI13 _{5 5 5 5 5 5}

  1 = 1, 2 = 1, , 31 = 1; 32 = 2, , 242 = 2; 243 = 3

  … … [ ] [ ] [ ] [ ] [ ] [ ] _{5 5 5} If n 31,

  1 2 n = n

  ≤ L + + + [ ] [ ] [ ] ^{5 5 5}

  1 2 n = 31 + 2(n – 31) = 2n – 31

  ≤ ≤ L + + [ ] [ ] [ ]

• If 32 n 242,

  2n – 31 = n + 14

  ⇒

  n = 45

  

I7 A boy tries to find the area of a parallelogram by multiplying together the lengths of two

  adjacent sides. His answer is

  2 times the correct area. If the acute angle of the

  parallelogram is x , find x. (Reference: 1991 FSG.3-4)

  ° Let the lengths of two adjacent sides be a and b, where the angle between a and b is x .

  ° o ab 2 ab sin x

  =

  1 sin x =

  °

  2

  x = 45 I8 If the points A (–8, 6), B (–2, 1) and C (4, c) are collinear, find c.

  Reference: 1984 FSG.4, 1984 FG7.3, 1986 FG6.2, 1987 FG7.4 m = m

CB BA

  c

  1

  1

  6

  − − =

  4

  2

  2

  8

  − + + c = –4

  2 I9 The graphs of x + y = 8 and x + y = 8 meet at two points. If the distance between these two points is d , find d.

  2 From (1), y = 8 – x (3) ……

  From (2), y = 8 – x (4)

  ……

  2

  (3) = (4): 8 – x = 8 – x

  x = 0 or 1

  When x = 0, y = 8; when x = 1, y = 7 _{2 2}

  1

  7 8 =

  2 −

• Distance between the points (0, 8) and (1, 7) =

  ( ) d = 2

  

I10 The sines of the three angles of a triangle are in the ratio 3 : 4 : 5. If A is the smallest interior

x angle of the triangle and tan A , find x.

  =

  16 Reference: 1990 HI6 By Sine rule, a : b : c = sin A : sin B : sin C = 3 : 4 : 5 Let a = 3k, b = 4k, c = 5k.

  2

  2

  2

  2

  2

  2 a + b = (3k) + (4k) = (5k) = c C = 90 (converse, Pythagoras’ theorem) ∴ ∠ ° a

  3

  12 tan A =

  = = b

  4

  16

  ⇒

  x = 12

  

I11 Two dice are thrown. Find the probability that the sum of the two numbers shown is greater

than 7. (Reference: 2002 HG7) P ( sum > 7) = P(sum = 8 or 9 or 10 or 11 or 12)

  5

  4

  3

  2

  1

  =

  36

  36

  36

  36

  36

  15

  5 = =

  36

  12

  6

  1 Method 2 P (7) =

  =

  36

  6 P (2, 3, 4, 5, 6) = P (8, 9, 10, 11, 12)

  P (2, 3, 4, 5, 6) + P (7) + P (8, 9, 10, 11, 12) = 1

  5

  2 P (8, 9, 10, 11, 12) = 1 – P (7) =

  6

  5 P (8, 9, 10, 11, 12) =

  12

  2 x 1 , if x

  3

• 

  ≤ I12 F is a function defined by F ( x ) . Find F F ( 3 ) .

   =

  2 ( )

  3 x , if x

  3

   > F (3) = 2 3 + 1 = 7

  ×

2 F ( F (3)) = F (7) = 3

  7 = 147

  ×     x

  14     I13 If a b c  y  ax by cz and

  1

  2 3  y  26 , find y . (Reference: 1986 FI3.4) = = + +

  ( ) ( )     z

  2      

  14   ⇒

  1

  2 3  y  = 14 + 2 y + 6 = 26 y = 3 ( )

   

  2  

  1 sin 37 sin 45 cos 60 sin

  60

  ° ° ° ° I14 If , find B. =

  B cos

  30 cos 45 cos

  53

  ° ° ° Reference: 1990 HI14

  1 sin 37 sin 45 cos 60 sin

  60

  ° ° ° ° = B cos

  30 cos 45 cos

  53

  ° ° °

  sin 37 sin 45 cos 60 sin

  60

  1

  ° ° ° °

  = = cos 60 =

  °

  sin 60 sin 45 sin

  37

  2

  ° ° ° B = 2 I15 If x + y = –4, y + z = 5 and z + x = 7, find the value of xyz.

  Reference: 1989 HI15, 1990 HI7 ⇒

  (1) + (2) – (3): 2y = –6 y = –3

  ⇒

  (1) + (3) – (2): 2x = –2 x = –1

  ⇒

  (2) + (3) – (1): 2z = 16 z = 8

  xyz = 24

2 I16 , are the roots of the equation x – 10x + c = 0. If = –11 and > , find the value of

  α β αβ α β .

• – = 10
₂

  α β

  α β

• α β α β − αβ

  = ₂

  4

• –

  ( )

  =

  10

  4 11 = 12

  − − ( )

  A I17 In figure 1, FE // BC and ED // AB. If AF : FB 3 : 2, find the

  = ratio area of DEF : area of ABC . Reference: 1990 HG8

  ∆ ∆ BDEF is a parallelogram formed by 2 pairs of parallel lines

  (A.S.A.)

DEF FBD

  ∆ ≅ ∆ F E

  Let S DEF = x = S FBD (where S stands for the area) ∆ ∆

  AEF ~ ACB ∆ ∆ (Q FE // BC, equiangular) ₂ B C

    S

  3

  9 ∆ AEF

  D   (1)

  = = ……

  2 3 

  25 ∆ ACB

• S 

  (Figure 1) AE : EC = AF : FB = 3 : 2 (theorem of equal ratio)

  ∴ Q DE // AB AE : EC = BD : DC = 3 : 2 (theorem of equal ratio) ∴

  ~ CDE CBA (Q DE // BA, equiangular)

  ∆ ∆ ₂  

  S _CDE

  2

  4 ∆

    (2) = = ……

  S _CBA

  2

  3

  25 ∆

•  

  Compare (1) and (2) S AEF = 9k, S CDE = 4k, S ABC = 25k ∆ ∆ ∆

  9k + 4k + x + x = 25k x = 6k DEF : area of ABC = 6 : 25 ⇒ area of

  ∆ ∆ I18 In figure 2, a regular hexagon ABCDEF is inscribed in a circle

  A F centred at O. If the distance of O from AB is

  2 3 and p is the perimeter of the hexagon, find p.

  2

  3 Let H be the foot of perpendiculars drawn from O onto AB.

  O B E AOB = 360 6 = 60 ( s at a point)

  ∠ ° ÷ ° ∠ AOH = 30

  ∠ °

1 AH = OH tan 30 =

  2 3 = 2 ° ×

  3 D C ⇒ AB = 4 (Figure 2)

  Perimeter = 6 4 = 24 ×

  A I19 In figure 3, ABCD and ACDE are cyclic quadrilaterals. Find the value of x + y.

  Reference: 1992 FI2.3 B x °

  50°

  ADC = 180 – x (opp. s cyclic quad.) ∠ ° ° ∠

  ACD = 180 – y (opp. s cyclic quad.) ∠ ° ° ∠

  E y ° 180 – y + 180 – x + 50 = 180 ( s sum of )

  ° ° ° ° ° ° ∠ ∆ x + y = 230

  (Figure 3) C D A I20 Find the value of a in figure 4.

  C AOB ~ DOC (equiangular)

  ∆ ∆

  2

  4 a

  6

  =

  4

  2 O

  a

  4 a = 12

  B

  6 D (Figure 4)

  Group Events

  2

  2 G1 Given a and b are distinct real numbers satisfying a = 5a + 10 and b = 5b + 10. Find the

  1

  1

• value of .

  2

  2 a b

  Reference: 1991 HI14

  2

  2 a and b are the roots of x = 5x + 10; i.e. x – 5x – 10 = 0

• b = 5; ab = –10

  a

  1 1 a b a b 2 ab

  5

  2

  10

  9 ^{2 2 2} − − − ²

  ( ) ( )

• = = = =
_{2 2 2 2}

  2

  2 a b

  20

  a b ab

  10

  − ( ) ( )

  

G2 An interior angle of an n-sided convex polygon is x while the sum of other interior angles is

°

  800 . Find the value of n. (1990 FG10.3-4, 1992 HG3, 2002 FI3.4, 2013 HI6)

  °

  800 = 180 4 + 80

  ×

  800 + x = 180(n – 2) s sum of polygon

  ∠

  0 < x < 180

  Q

  800 + x = 180 5 = 180(n – 2)

  ∴ × n = 7 n ( n

  1 )( 2 n 1 )

  2

  2

  …

  2

+ +

G3 + It is known that 1 + 2 + n = for all positive integers n.

  6

  2

  2

  2 + Find the value of 21 + 22 + 30 .

  … Reference: 1993 HI6

  2

  2

  2

  2

  2

  2

  2

  2

  2

  21 + 22 + + 30 = 1 + 2 + 30 – (1 + 2 + 20 )

  … … …

  1

  1 =

  30 31 61 –

  20 21 41 = 9455 - 2870 = 6585

  ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

  6

  6 G4 One of the positive integral solutions of the equation 19x + 88y = 1988 is given by (100, 1). Find another positive integral solution. (Reference: 1991 HG8) 19 y y _{2 − 1} The line has a slope of =

  −

  88 x x _{2 − 1} Given that (100, 1) is a solution.

  19 y

  1 ₂ − =

  −

  88 x 100 ₂ − Let y

  2 – 1 = –19t; x 2 – 100 = 88t, where t is an integer. y 2 = 1 – 19t, x 2 = 100 + 88t

  For positive integral solution of (x , y ), 1 – 19t > 0 and 100 + 88t > 0

  2

  2

  25

  1

  t − < <

  22

  19 t is an integer t = 0 or –1

  Q ∴ ⇒

  When t = –1, x

  2 = 12, y 2 = 20 another positive integral solution is (12, 20).

  AP G5 The line joining A(2, 3) and B(17, 23) meets the line 2x – y = 7 at P. Find the value of . PB

  Reference: 1990 HG3 y

  3

  23

  3

  − − ⇒

  Equation of AB: 3y = 4x + 1 (2)

  = …… x

  2

  17

  2

  − −

  From (1): y = 2x – 7 (3)

  …… ⇒

  Sub. (3) into (2): 3(2x – 7) = 4x + 1 x = 11 Sub. x = 11 into (3): y = 2(11) – 7 = 15 The point of intersection is P(11, 15).

  AP

  11

  2

  3

  −

  = =

  PB

  17

  11

  2

  −

  2047 G6 Find the remainder when 7 is divided by 100.

  Reference: 2002 HG4 2047 The question is equivalent to find the last 2 digits of 7 .

  1

  2

  3

  4 7 = 7, 7 = 49, 7 = 343, 7 = 2401 The last 2 digits repeats for every multiples of 4.

  2047 4 × 511+3 7 = 7 The last 2 digits is 43.

  G7 If log 2 [log 3 (log 7 x)] =log 3 [log 7 (log 2 y)] =log 7 [log 2 (log 3 z)] = 0, find the value of x + y + z. log 2 [log 3 (log 7 x)] = 0 ⇒ log

  3 (log 7 x) = 1 ⇒ log x = 3

  7

  3 ⇒ x = 7 = 343 log

  3 [log 7 (log 2 y)] = 0 ⇒ log 7 (log 2 y) = 1 ⇒ log

  2 y = 7

  7 ⇒ y = 2 = 128 log

  7 [log 2 (log 3 z)] = 0 (log z) = 1 ⇒ log

  2

  3 ⇒ log 3 z = 2

  2 ⇒ z = 3 = 9 x + y + z = 343 + 128 + 9 = 480

  D 4, CD 6 and MN x,

  G8 In figure 1, AB // MN // CD. If AB = = = B find the value of x.

  Reference: 1985 FI2.4, 1990 FG6.4 N

  6

  4 AMN ~ ACD (equiangular) ∆ ∆ x x AM

  (1) (ratio of sides, ~ ’s)

  = … ∆ A

  6 AC

  C M CMN ~ CAB (equiangular)

  ∆ ∆ (Figure 1) x MC

  (2) (ratio of sides, ~ ’s)

  = … ∆

  4 AC

  12

• x x AM MC

  (1) + (2): 1 x =

• ⇒

  = =

  6

  4 AC

  5 G9 In figure 2, B = 90 , BC 3 and the radius of the inscribed

  ∠ ° = A circle of ABC is 1. Find the length of AC. ∆

  Reference: 1985 FG9.2

  Let the centre be O. Suppose the circle touches BC at P, AC at Q and AB at R respectively. Let AB = c and AC = b.

  Q OP BC, OQ AC, OR AB (tangent radius) ⊥ ⊥ ⊥ ⊥

  O

  1 ⇒ R OPBR is a rectangle OPBR is a square.

  1 BP = BR = 1 (opp. sides of rectangle) CP = 3 – 1 = 2

  C B

  3 CQ = CP = 2 (tangent from ext. point) P

  Let AR = AQ = t (tangent from ext. point)

  2

  2

  2

  3 + (1 + t) = (2 + t) (Pythagoras’ theorem)

  2

  2

  9 + 1 + 2t + t = 4 + 4t + t 6 = 2t

  ⇒ t = 3 AC = 2 + 3 = 5

  G10 In the attached division (see figure 3), the

• 8 * dividend in (a) is divisible by the divisor in line (b) ......... * * * ) * * * * * * ......... (a) (b). Find the dividend in line (a). (Each asterisk
• is an integer from 0 to 9.) Relabel the ‘*’ as shown.
• Let a = (a 8a ) ,

  1 3 x

  b = (b 1 b 2 b 3 ) x

  c = (c 1 c 2 c 3 c 4 c 5 c 6 ) x d = (d 1 d 2 d 3 d 4 ) x

  e = (e 1 e 2 e 3 ) x a 8 a

  1

  3 f = (f 1 f 2 f 3 ) x

  (b) ......... b 1 b 2 b 3 ) c 1 c 2 c 3 c 4 c 5 c 6 ......... (a) = (g ) g 1 g 2 g 3 g 4 x d 1 d 2 d 3 d

  4 1 b = d, a 3 b = g 4-digit numbers

  Q 8b = e and a × × a 1 = 9 and a

3 = 9 and d = g e

1 e 2 e

  3 ∴ d = 9b > 1000 and f = 8b < 999 f 1 f 2 f

  3 112 b 124 (1)

  ≤ ≤ …… g 1 g 2 g 3 g

  4 b 1 = 1, d 1 = 1 g 1 g 2 g 3 g

  4 b 2 = 1 or 2, f 1 = 8 or 9, d 2 = 0 or 1 c = 1

  1 9 8 9 e 1 – f 1 = 1

  (b) ......... 1 b b ) 1 c c c c c ......... (a)

  2

  3

  2

  3

  4

  5

  6 f 1 = 8 and e 1 = 9

  ∴ 1 d d d

  2

  3

  4 8b = (8f 2 f 3 ) x < 900 < 112.5 (2) b e e e

  ……

  1

  2

  3 Combine (1) and (2) f f f

  1

  2

  3 b = 112

  1 d d d

  2

  3

  4 c = 989 112 = 110768

  × 1 d d d

  2

  3

  4