hongkong olympiad

Answers: (1986-87 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 15 August 2016

Group Event 6
G6.1 If ,  are the roots of x2 – 10x + 20 = 0, and p = 2 + 2, find p.

 +  = 10,  = 20
p = ( + )2 – 2

= 102 – 2(20) = 60

G6.2 The perimeter of an equilateral triangle is p. If its area is k 3 , find k.

Reference: 1984FI4.4, 1985 FSI.4, 1986 FSG.3, 1988 FG9.1

Length of one side = 20

1  202 sin 60  k 3
2

k = 100

G6.3 Each interior angle of an N-sided regular polygon is 140. Find N.

Reference: 1997 FI4.1

Each exterior angle = 40 (adj. s on st. line)

360 = 40 (sum of ext. s of polygon)
N

N=9
G6.4 If M = (102 + 101 + 12)(102 – 12)(102 – 101 + 12), find M.

M = (102 + 101 + 12)(10 – 1)(10 + 1)(102 – 101 + 12)
= (103 – 1)(103 + 1)
= 106 – 1 = 999999

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Answers: (1986-87 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 15 August 2016

Group Event 7
G7.1 The acute angle formed by the hands of a clock at 3:30 p.m. is A. Find A.

Reference 1984 FG7.1, 1985 FI3.1, 1989 FI1.1, 1990 FG6.3, 2007 HI1

At 3:00 p.m., the angle between the arms of the clock = 90

From 3:00 p.m. to 3:30 p.m., the hour-hand had moved 360 1  1 = 15.
12 2

The minute hand had moved 180.
A = 180 – 90 – 15 = 75

G7.2 If tan(3A + 15) = B , find B.
tan(225 + 15) = B
B=3

G7.3 If log10AB = C log1015, find C.
log10 (753) = C log10 15
log10 225 = C log10 15
C=2

G7.4 The points (1, 3), (4, 9) and (2, D) are collinear. Find D.

Reference: 1984 FSG.4, 1984 FG7.3, 1986 FG6.2, 1989 HI8

D9  9  3
24 4  1

D – 9 = –4
D=5

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Answers: (1986-87 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 15 August 2016

Group Event 8

G8.1 If A = 5sin   4cos and tan  = 2, find A.
3sin   cos

Reference: 1986 FG10.3, 1989 FSG.4, 1989 FG10.3, 1990 FG7.2

A = 5sin   4cos  cos
3sin   cos  cos

= 5 tan   4
3tan   1

= 52  4 = 2
32  1

G8.2 If x  1 = 2A, and x3  1 = B, find B.
x x3

Reference: 1983 FG7.3, 1984 FG10.2, 1985 FI1.2, 1989 HI1, 1990 HI12, 2002 FG2.2
x1 =4

x

 x2  1 = 42 – 2 = 14
x2

B = x3  1
x3

=  x  1  x2 1 1 
 x  x2 

= 4(14 – 1) = 52

G8.3 There are exactly N values of  satisfying the equation cos3  – cos  = 0, where 0≤≤ 360.

Find N.

cos  (cos  + 1)(cos  – 1) = 0

cos  = 0, –1 or 1

 = 90, 270, 180, 0, 360

N=5
G8.4If the Nth day of May in a year is Thursday and the Kth day of May in the same year is Monday,

where 10 < K < 20, find K.

Reference: 1984 FG6.3, 1985 FG9.3, 1988 FG10.2
5th May is Thursday
9th May is Monday
16th May is Monday

 K = 16

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Answers: (1986-87 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 15 August 2016

Group Event 9 ABCD
In the given multiplication, different letters represent different integers ranging 9

from 0 to 9.  DCBA
 1 BC 9
G9.1 Find A.
G9.2 Find B. 9
G9.3 Find C. 9 CB 1
G9.4 Find D.

Reference: 1994 HI6

As there is no carry digit in the thousands digit multiplication, A = 1, D = 9

Consider the tens digit: 9C + 8  B (mod 10)  (1)

As there is no carry digit in the thousands digit, let the carry digit in the

hundreds digit be x.

9B + x = C and B, C are distinct integers different from 1 and 9

 B = 0, C = x

Sub. B = 0 into (1): 9C + 8  0 (mod 10)

 9C  2 (mod 10)

C=8

 A = 1, B = 0, C = 8, D = 9

Group Event 10
G10.1 The average of p, q, r and s is 5. The average of p, q, r, s and A is 8. Find A.

Reference: 1985 FG6.1, 1986 FG6.4, 1988 FG9.2
p + q + r + s = 20
p + q + r + s + A = 40
A = 20
G10.2 If the lines 3x – 2y + 1 = 0 and Ax + By + 1 = 0 are perpendicular, find B.

Reference: 1983 FG9.3, 1984 FSG.3, 1985 FI4.1, 1986 FSG.2, 1988 FG8.2

3    20  = –1  B = 30
2  B 

G10.3 When Cx3 – 3x2 + x – 1 is divided by x + 1, the remainder is 7. Find C.

C(–1) – 3 – 1 – 1 = –7

C=2

G10.4 If P, Q are positive integers such that P + Q + PQ = 90 and D = P + Q, find D.

(Hint: Factorise 1 + P + Q + PQ)

Reference: 2002 HG9, 2012 FI4.2
WLOG assume P  Q, 1 + P + Q + PQ = 91
(1 + P)(1 + Q) = 191 = 713
1 + P = 1  P = 0 (rejected)
or 1 + P = 7  P = 6
1 + Q = 13  Q = 12
D = 6 + 12 = 18

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Individual Events

SI a 900 I1 P 100 I2 k 4 I3 h 3 I4 a 18 I5 a 495
r 3b 2
b7 Q8 m 58 k6 9 99
4x
p2 R 50 a2 m4 M 109
1Y
q 9 S3 b3 15 w 18
p 16 60 G10 n 22
SG p 10 k 96
q Group Events 43 t 168
a 9h 25
75 G6 x 125 G7 M 5 G8 S 27 G9 p

0.5 n 10 N 6 T 135 t
A9 K
9 y 1000 a8

m 14 K 1003 k4 B0 C

Sample Individual Event (1984 Sample Individual Event)
SI.1 In the given diagram, the sum of the three marked angles is a.

Find a.
Sum of interior angles of a triangle = 180
angle sum of three vertices = 3360 = 1080
a = 1080 – 180 = 900

SI.2 The sum of the interior angles of a regular b-sided polygon is a. Find b.
a = 900 = 180(b – 2)
b=7

SI.3 If 8b = p21, find p.
87 = p21
221 = p21
p=2

SI.4 If p = logq 81, find q.
2 = p = logq 81 and q > 0
q2 = 81
q=9

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Individual Event 1
I1.1 If N(t) = 10018t and P = N(0), find P.

P = 100180 = 100
I1.2 A fox ate P grapes in 5 days, each day eating 6 more than on the previous day.

If he ate Q grapes on the first day, find Q.
Q + (Q + 6) + (Q + 12) + (Q + 18) + (Q + 24) = P = 100
5Q + 60 = 100
Q=8
I1.3 If Q% of 25 is 1 % of R, find R.

32 Q

25  8 = R 1
32 100 1008
 R = 50
I1.4 If one root of the equation 3x2 – ax + R = 0 is 50 and the other root is S, find S.

9

50 S = product of roots = R = 50
9 33
S=3

Individual Event 2

I2.1 If ab = ad – bc and 34 = k, find k.

cd 2k

3k – 8 = k

k=4
I2.2 If 50m = 542 – k2, find m.

Reference: 1984 FI1.1, 1987 FSG.1
50m = 542 – 42 = (54 + 4)(54 – 4) = 5850

 m = 58
I2.3 If (m + 6)a = 212, find a.

(58 + 6)a = 212
 64a = 212
 26a = 212

a=2

I2.4 A, B and C are the points (a, 5), (2, 3) and (4, b) respectively. If AB  BC, find b.

A(2, 5), B(2, 3), C(4, b).

AB is parallel to y-axis

 BC is parallel to x-axis

b=3

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Individual Event 3

I3.1 If 3  2 21  h , find h.
2 7 3 25

3  2 7  3  2 21  h
2 7 3 2 7 3 25

2 21  3  2 21  h O B
h=3 P
I3.2 The given figure shows a circle of radius 2h cm, centre O. A
If AOB =  , and the area of sector AOBP is k cm2, find k.

3

1  2  32    k

23
k=6

I3.3 A can do a job in k days, B can do the same job in (k + 6) days.

If they work together, they can finish the job in m days. Find m.

1 1 1
m k k6
1 1 1

m 6 12
m=4
I3.4 m coins are tossed. If the probability of obtaining at least one head is p, find p.
P(at least one head) = 1 – P(all tail)

=1  1 4 = 15
 2  16

Individual Event 4
I4.1 If f (t) = 2 – t , and f (a) = –4, find a.

3
f (a) = 2 – a = –4

3
 a = 18
I4.2 If a + 9 = 12Q + r, where Q, r are integers and 0 < r < 12, find r.
18 + 9 = 27 = 122 + 3 = 12Q + r
r=3
I4.3 x, y are real numbers. If x + y = r and M is the maximum value of xy, find M.
Reference: 1985 FI3.4
x+y=3
y=3–x
xy = x(3 – x) = 3x – x2 = –(x – 1.5)2 + 2.25
M = 2.25 = 9

4
I4.4 If w is a real number and 22w – 2w – 8 M = 0, find w.

9
22w – 2w – 8  9 = 0

94
 (2w)2 – 2w – 2 = 0
(2w + 1)(2w – 2) = 0
w=1

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Individual Event 5 B
I5.1 If 0.357  177 , find a. C

a A
0.357 = 3  0.057

10
= 3  57

10 990
= 297  57

990
= 354 = 59 = 177

990 165 495
= 177

a
a = 495
I5.2 If tan2 a + 1 = b, find b.
b = tan2 495 + 1

= tan2(1803 – 45) + 1
=1+1=2
I5.3 In the figure, AB  AD, BAC = 26 + b, BCD = 106.
If ABC = x, find x.
BCA = DCA = 1 BCD = 53 (eq. chords eq. s)

2
BAC = 28
x = ABC = 180 – 28 – 53 = 99 (s sum of )
x = 99

D

I5.4 If h k  m p   hm  kn hp  kq and 1 2 3 x   11 Y , find Y.
n q 4 5

1 3  2 4 x  2 5  11 Y 

 Y = 109

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Sample Group Event (1984 Group Event 7)
SG.1 The acute angle between the 2 hands of a clock at 3:30 p.m. is p. Find p.

At 3:00 p.m., the angle between the arms of the clock = 90
From 3:00 p.m. to 3:30 p.m., the hour-hand had moved 360 1  1 = 15.

12 2
The minute hand had moved 180.
p = 180 – 90 – 15 = 75
SG.2 In ABC, B  C  p. If q  sin A, find q.
B  C  75, A  180 – 75 – 75 = 30
q = sin 30 = 1

2
SG.3 The 3 points (1, 3), (2, 5), (4, a) are collinear. Find a.

95  a3= 2
4 2 41
a=9
SG.4 The average of 7, 9, x, y,17 is 10. If the average of x +3, x +5, y + 2, 8 and y + 18 is m, find m.
7  9  x  y 17 = 10

5
 x + y = 17
m = x  3  x  5  y  2  8  y 18

5

= 2x  y 36

5
= 14

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Group Event 6

G6.1 In the figure, the bisectors of B and C meet at I. A
If A = 70 and BIC = x, find x.

Let ABI = b = CBI, ACI = c = BCI I
2b + 2c + 70 = 180 (s sum of )

b + c = 55 B C
In BCI, b + c + x = 180 (s sum of )

x = 180 – 55 = 125

G6.2 A convex n-sided polygon has 35 diagonals. Find n.

Reference: 1984 FG10.3, 1985 FG8.3, 1989 FG6.1, 1991 FI2.3, 2001 FI4.2, 2005 FI1.4
C2n  n  35

 nn  3  35

2

n2 – 3n – 70 = 0

 (n – 10)(n + 7) = 0

n = 10

G6.3 If y = ab – a + b – 1 and a  49, b  21, find y.

Reference: 1985 FG8.4, 1986 FG9.3, 1990 FG9.1

y = (a + 1)(b – 1) = (49 + 1)(21 – 1) = 5020 = 1000

G6.4 If K = 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 +  + 1001 + 1002, find K.

Reference: 1985 FG7.4, 1990 FG10.1, 1991 FSI.1, 1992 FI1.4

K = 1 + (2 – 3 – 4 + 5) + (6 – 7 – 8 + 9) +  + (998 – 999 – 1000 + 1001) + 1002 = 1003

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Group Event 7 (Similar Questions 1985 FG8.1-2, 1990 FG7.3-4)
M, N are positive integers less than 10 and 8M420852  9  N9889788  11.
G7.1 Find M.

11 and 9 are relatively prime
 8M420852 is divisible by 11
 8 + 4 + 0 + 5 – (M + 2 + 8 + 2) is divisible by 11
 5 – M = 11k
M=5
G7.2 Find N.
N9889788 is divisible by 9
 N + 9 + 8 + 8 + 9 + 7 + 8 + 8 = 9t
N=6
G7.3 The equation of the line through (4, 3) and (12, 3) is x  y  1. Find a.

ab

y  3  3   3

x  4 4 12
3x – 12 + 4y – 12 = 0
 3x + 4y = 24
x  y 1
86
a=8
G7.4 If x + k is a factor of 3x2 + 14x + a, find k. (k is an integer.)
3(–k)2 + 14(–k) + 8 = 0
 3k2 – 14k + 8 = 0
(3k – 2)(k – 4) = 0

 k = 4 (reject 2 )
3

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Group Event 8

G8.1 If log9 S = 3 , find S.
2

S = 93 = 27
2

G8.2 If the lines x + 5y = 0 and Tx – Sy = 0 are perpendicular to each other, find T.

Reference: 1983 FG9.3, 1984 FSG.3, 1985 FI4.1, 1986 FSG.2, 1987 FG10.2

 1  T  1
5 27

T = 135

The 3-digit number AAA, where A  0, and the 6-digit number AAABBB satisfy the following
equality: AAA  AAA + AAA = AAABBB.
G8.3 Find A.

A(111)  A(111) + A(111) = A(111000) + B(111)
111A2 + A = 1000A + B
Consider the thousands digit: 9 < A2  81
 A = 4, 5, 6, 7, 8, 9
When A = 4: 11116 + 4 = 4000 + B (rejected)
When A = 5: 11125 + 5 = 5000 + B (rejected)
When A = 6: 11136 + 6 = 6000 + B (rejected)
When A = 7: 11149 + 7 = 7000 + B (rejected)
When A = 8: 11164 + 8 = 8000 + B (rejected)
When A = 9: 11181 + 9 = 9000 + B
A=9
G8.4 Find B.
B=0

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Group Event 9
G9.1 The area of an equilateral triangle is 50 12 . If its perimeter is p, find p.

Reference: 1984FI4.4, 1985 FSI.4, 1986 FSG.3, 1987 FG6.2
Each side = p

3

1  p 2 sin 60  50 12  100 3
2 3

p = 60

G9.2 The average of q, y, z is 14. The average of q, y, z, t is 13. Find t.

Reference: 1985 FG6.1, 1986 FG6.4, 1987 FG10.1

q  y  z = 14
3

 q + y + z = 42

q  y  z  t  13
4

 42  t = 13
4

t = 10
G9.3 If 7 – 24x – 4x2  K + A(x + B)2, where K, A, B are constants, find K.

Reference: 1984 FI2.4, 1985 FG10.2, 1986 FG7.3, 1987 FSI.1
7 – 24x – 4x2  –4(x2 + 6x) + 7  –4(x + 3)2 + 43

K = 43

G9.4 If C = 34n  9n4 , find C.
27 2 n  2

C = 34n  32n8 = 9
36n6

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Answers: (1987-88 HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December 2015

Group Event 10
G10.1 Each interior angle of an n-sided regular polygon is 160. Find n.

Each exterior angle = 20 (adj. s on st. line)
360  20

n
 n = 18
G10.2 The nth day of May in a year is Friday. The kth day of May in the same year is Tuesday, where

20 < k < 26. Find k.

Reference: 1984 FG6.3, 1985 FG9.3, 1987 FG8.4
18th May is Friday
22nd May is Tuesday
 k = 22

In the figure, AD  BC, BA  CA, AB  7, BC  25, AD  h A
and CAD = . 

7h

BD C
25

G10.3 If 100 sin  = t, find t.
AC2 + 72 = 252 (Pythagoras’ theorem)
AC = 24
ACD = 90 –  (s sum of )
ABC =  (s sum of )
t = 100 sin  =100 24 = 96
25

G10.4 Find h.
Area of ABC = 1  7  24  1  25h
22

h = 168
25

Method 2
In ABD,
h = AB sin 

= 7  24 = 168
25 25

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Individual Events

SI a 900 I1 a 35 I2 a 7 I3  8 I4 t 13 I5 a 30
s 4 b 150
b7 b7 b3 b 16 a 3 n 12
c 12 k 24
c3 c 10 c9 A 128
G9 x
d5 d2 d5 d7 y
k
Group Events r

SG a 2 G6 n 8 G7 G 1 G8 y 7 40 G10 a 6
3x 3
b9 k5 D8 k –96 8k 2
5y 4
p 23 u 35 L2 a1

k3 a1 E5 m2

Sample Individual Event
SI.1 In the given diagram, the sum of the three marked angles is a.

Find a.
Reference: 1984 FSI.1 1987 FSG.3
Sum of interior angles of a triangle = 180
angle sum of three vertices = 3360 = 1080
a = 1080 – 180 = 900

SI.2 The sum of the interior angles of a convex b-sided polygon is a. Find b.
Reference 1984 FSI.2
a = 900 = 180(b – 2)
b=7

SI.3 If 27b–1 = c18, find c.
33(7–1) = c18
c=3

SI.4 If c = logd 125, find d.
3 = c = logd 125
d3 = 125
d=5

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Individual Event 1
I1.1 The obtuse angle formed by the hands of a clock at 10:30 is (100 + a). Find a.

Reference 1984 FG7.1, 1985 FI3.1, 1987 FG7.1, 1990 FG6.3, 2007 HI1
At 10:00, the angle between the arms of the clock = 60

From 10:00 to 10:30, the hour-hand had moved 360 1  1 = 15.
12 2

The minute hand had moved 180.
100 + a = 180 – 60 + 15 = 135  a = 35
I1.2 The lines ax + by = 0 and x – 5y + 1 = 0 are perpendicular to each other. Find b.

 35  1  1
b5

b=7
I1.3 If (b + 1)4 = 2c+2, find c.

84 = 2c+2
23(4) = 2c+2
 c = 10
I1.4 If c – 9 = logc (6d – 2), find d.
10 – 9 = 1 = log10 (6d – 2)
 6d – 2 = 10
d=2
Individual Event 2
I2.1 If 1000a = 852 – 152, find a.
1000a = (85 + 15)(85 – 15) = 10070
a=7
I2.2 The point (a, b) lies on the line 5x + 2y = 41. Find b.
5(7) + 2b = 41
b=3
I2.3 x + b is a factor of x2 + 6x + c. Find c.
Put x = –3 into x2 + 6x + c = 0
(–3)2 + 6(–3) + c = 0
c=9
I2.4 If d is the distance between the points (c, 1) and (5, 4), find d.
d2 = (9 – 5)2 + (1 – 4)2 = 25
d=5

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Individual Event 3

I3.1 If  +  = 11,  = 24 and  > , find .

 and  are the roots of the equation x2 – 11x + 24 = 0

(x – 3)(x – 8) = 0

>=8
I3.2 If tan  =   , 90 <  < 180 and sin  = b , find b.

15 34

In the figure, P = (8, –15) P(8, -15)
r2 = 82 + (–15)2 (Pythagoras’ theorem)
2
r = 17 r
sin  = 8 = 16 8 
N -15
17 34 O
b = 16

I3.3 If A is the area of a square inscribed in a circle of diameter b, find A.

Reference: 1984 FG10.1, 1985 FSG.4 E D
Let the square be BCDE.

BC = 16 cos 45 = 8 2 16

 2

A = 8 2 = 128

45 C
B

I3.4 If x2 + 22x + A  (x + k)2 + d, where k, d are constants, find d.
x2 + 22x + 128  (x + 11)2 + 7
d=7

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Individual Event 4

I4.1 The average of p, q, r is 12. The average of p, q, r, t, 2t is 15. Find t.

p + q + r = 36

p + q + r + t + 2t = 75

3t = 75 – 36 = 39

t = 13

I4.2 k is a real number such that k4 + 1 = t + 1, and s = k2 + 1 . Find s.
k4 k2

k4 + 1 = 14
k4

k4 + 2 + 1 = 16
k4

(k2 + 1 )2 = 16
k2

 s = k2 + 1 = 4
k2

I4.3 M and N are the points (1, 2) and (11, 7) respectively. P(a, b) is a point on MN such that

MP : PN  1 : s. Find a.

MP : PN  1 : 4
a = 4 11 = 3

1 4
I4.4 If the curve y = ax2 + 12x + c touches the x-axis, find c.

y = 3x2 + 12x + c

 = 122 – 4(3)c = 0

 c = 12

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Individual Event 5

I5.1 In the figure, find a. T P
Reference: 1997 FG1.1, 2005 FI2.3 30 a
Label the vertices A, B, C, D, E, P, Q, R, S, T as shown. E
AEP = 40 + 45 = 85 (ext.  of SQE) D
EAP = 30 + 35 = 65 (ext.  of TRA) A
In AEP, 85 + 65 + a = 180 (s sum of ) 40 Q
a = 30 S
45
I5.2 If sin(a + 210) = cos b, and 90 < b < 180, find b.
B
C 35

R

sin 240 =  3 = cos b
2

b = 150

I5.3 Each interior angle of an n-sided regular polygon is b. Find n.
Each exterior angle = 30 (adj. s on st. line)

360 = 30 (sum of exterior angles of polygon)
n

 n = 12
I5.4 If the nth day of March in a year is Friday. The kth day of March in the same year is Wednesday,

where 20 < k < 25, find k.
12th March is Friday
17th March is Wednesday
24th March is Wednesday

 k = 24

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Sample Group Event
SG.1 If 2at2 + 12t + 9 = 0 has equal roots, find a.

(12)2 – 4(2a)(9) = 0

a=2

SG.2 If ax + by = 1 and 4x + 18y = 3 are parallel, find b.

Reference: 1986 FI4.2, 1987 FSG.4

2 4
b 18

b=9
SG.3 The b th prime number is p. Find p.

Reference: 1985 FSG.2, 1990 FI5.4

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 

p = 23

SG.4 If k = 4sin   3cos  and tan  = 3, find k.
2sin   cos 

Reference: 1986 FG10.3, 1987 FG8.1, 1989 FG10.3, 1990 FG7.2

k = 4sin   3cos  cos 
2sin   cos  cos 

= 4 tan   3
2 tan  1

= 43 3 = 3
23  1

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Answers: (1988-89 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 23 August 2016

Group Event 6

G6.1 An n-sided convex polygon has 20 diagonals. Find n.

Reference: 1984 FG10.3, 1985 FG8.3, 1988 FG6.2, 1991 FI2.3, 2001 FI4.2, 2005 FI1.4

Number of diagonals = C2n  n = nn 1  n  20

2

n2 – 3n – 40 = 0

(n – 8)(n + 5) = 0

n=8

G6.2 Two dice are thrown. The probability of getting a total of n is k . Find k.
36

Total = 8

Favourable outcomes = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

P(total = 8) = 5
36

k=5

G6.3 A man drives at 25 km/h for 3 hours and then at 50 km/h for 2 hours.

His average speed for the whole journey is u km/h. Find u.

u = 25 3  50 2 = 35
32

G6.4 If ab = ab + 1 and (2a)3 = 10, find a.

2a = 2a + 1

(2a)3 = (2a + 1)3 = 3(2a + 1) + 1 = 10

6a + 4 = 10

a=1

Group Event 7

In the attached calculation, different letters represent different integers GO L D E N
J
ranging from 1 to 9. 
If the letters O and J represent 4 and 6 respectively, find D E NGO L

G7.1 G.

G7.2 D. 1 4 L8EN

G7.3 L. 6
G7.4 E. 8EN1 4 L

Carry digit in the 100000 digit is 2

G = 1, D = 8

Carry digit in the hundreds digit is 3 142857
E=5 6
Carry digit in the tens digit is 4
N = 7, L = 2 857142

 G = 1, D = 8, L = 2, E = 5

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