hongkong olympiad

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

Individual Event 3 (2x + 10)
I3.1 In the figure, find x.

x + 20 + 2x + 10 + 2x = 180 (s sum of )
x = 30

(x + 20) 2x

I3.2 The coordinates of the points P and Q are (a, 2) and (x, 6) respectively.
If the coordinates of the mid-point of PQ is (18, b), find a.

1 (a + 30) = 18
2

a=6

I3.3 A man travels from X to Y at a uniform speed of a km/h and returns at a uniform speed of

2a km/h. If his average speed is c km/h, find c.

Let the distance between X and Y be s km.

c = 2s = 2 = 8

s  s 1  1
a 2a 6 12

I3.4 If f (y) = 2y2 + cy – 1, find f (4).

f (4) = 2(4)2 + 8(4) – 1 = 63

Individual Event 4 A bB
I4.1 If the curve y = 2x2 – 8x + a touches the x-axis, find a. a

 = (–8)2 – 4(2)a = 0 15
a=8
I4.2 In the figure, AB is a diameter of the circle.

If AC  15, BC  a and AB  b, find b.
ACB = 90 ( in semi-circle)
b2 = 152 + 82 (Pythagoras’ theorem)
b = 17

C

I4.3 The line 5x + by + 2 = d passes through (40, 5). Find d.

Reference: 1984 FI2.3
d = 5(40) + 17(5) + 2 = 287
I4.4 X sold an article to Y for $d at a profit of 2.5%.
If the cost price of the article for X is $K, find K.
K = 287(1 + 2.5%) = 280

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

Individual Event 5

I5.1 Let x =19.87 . If 19.8 7 = a , find a. (Hint: 99x = 100 x – x)
99

99x = 100 x – x = 1987 + 0.87 – 19. 87 = 1968

x = 1968
99

 a = 1968

I5.2 If f (y) = 4 sin y and f (a – 18) = b, find b.

b = f (a – 18) = f (1950)

= 4 sin 1950

= 4 sin(3605 + 150)

= 4 sin 150 = 2

I5.3 If b 3 =2 21  3 , find c.
7 3 c

2 3 3  2 7 3=2 21  3
7 2 7 3 c

c = 4(7) – 3 = 25

I5.4 In the figure, ST is a tangent to the circle at P. M
S
If MQP = 70, QPT = c and MRQ = d, find d. R
d
MRP = 70 (s in the same segment)

PRQ = c = 25 ( in alt. segment) 70 Q
c
d = 70 + 25 = 95 T
P

Sample Group Event
SG.1 If 100A = 352 – 152, find A.

Reference: 1984 FI1.1

100A = (35 – 15)(35 + 15) = 1000

A = 10
SG.2 If (A – 1)6 = 27B, find B.

(10 – 1)6 = 27B
312 = 33B

B=4
SG.3 In the given diagram, the sum of the three marked angles is C. Find C.

Reference: 1984 FSI.1, 1989 FSI.1
Sum of interior angles of a triangle = 180
angle sum of three vertices = 3360 = 1080
C = 1080 – 180 = 900
SG.4 If the lines x + 2y + 1 = 0 and 9x + Dy + 1 = 0 are parallel, find D.

Reference: 1986 FI4.2, 1989 FSG.2

 1   9
2 D

 D = 18

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