Answers: (1997-98 HKMO Heat Events) Created by: Mr.
Francis Hung Last updated: 30 December 2015
1
97-98 1 2 2 40 3 –12 4 3 5 −
Individual 2
6 466 7 19 8 3 9 2 10 744
1 2 2 12 3 27 4 64 5 14
97-98
1
Group 6 14 7 − 8 1 9 20 10 19
2
Individual Events
I1 Given that x3 – 5x2 + 2x + 8 is divisible by (x – a) and (x – 2a), where a is an integer, find the
value of a.
Let f (x) = x3 – 5x2 + 2x + 8
f (–1) = –1 – 5 – 2 + 8 = 0 ⇒ x + 1 is a factor
f (2) = 8 – 20 + 4 + 8 = 0 ⇒ x – 2 is a factor
f (x) = (x + 1)(x – 2)(x – 4)
a=2
I2 Given that 8, a, b form an A.P. and a, b, 36 form a G.P. If a and b are both positive numbers,
find the sum of a and b.
8+b
a= … (1); b2 = 36a … (2)
2
Sub. (1) into (2): b2 = 18(8 + b)
b2 – 18b – 144 = 0
(b + 6)(b – 24) = 0
b = –6 (rejected) or b = 24
8 + 24
a= = 16
2
a + b = 40
x x
I3 Find the smallest real root of the following equation: = .
(x − 4)(x + 3) (x + 4)(x − 6)
Reference: 1995 FI2.1
x(x + 4)(x – 6) = x(x – 4)(x + 3)
x(x2 – 2x – 24) = x(x2 – x – 12)
0 = x(x + 12)
x = 0 or –12
The smallest root = –12
I4 In figure 1, ABCD is a square. E is a point on AB such that BE = 1 and A E
B
CE = 2. Find the area of the square ABCD.
BC2 = 22 – 12 (Pythagoras’ theorem on ∆BCE)
Area of the square = BC2 = 3
D C
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Answers: (1997-98 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 30 December 2015
I5 If 2x + 3 = 2 + 2 + 2 + L , find the value of x.
(2 x + 3)2 = 2 + 2 + 2 + 2 + L = 2 + (2 x + 3)
2
4x + 12x + 9 = 2x + 5
4x2 + 10x + 4 = 0
2x2 + 5x + 2 = 0
(2x + 1)(x + 2) = 0
1
x =− or –2
2
Q 2x + 3= 2 + 2 + 2 + L > 0
1
∴ x ≠ –2, x = − only
2
I6 Given that n is a positive integer which is less than 1000. If n is divisible by 3 or 5, find the
number of possible values of n. (Reference: 1993 FG8.3-4, 1994 FG8.1-2, 2015 FI3.1)
Number of multiples of 3 = 333
Number of multiples of 5 = 199
Number of multiples of 15 = 66
Number of possible n = 333 + 199 – 66 = 466
I7 In figure 2, ABCD is a rectangle with CD = 12. E is a point on
CD such that DE = 5. M is the mid-point of AE and P, Q are
points on AD and BC respectively such that PMQ is a straight
line. If PM : MQ = 5 : k , find the value of k.
Draw a straight line HMG // CD (H lies on AD, G lies on BC)
AH = HD (Intercept theorem) A B
∆PHM ~ ∆QGM (equiangular)
∆AHM ~ ∆ADE (equiangular)
P M
PM : MQ = HM : MG (ratio of sides, ~∆'s) H G
1
= DE : (HG – HM) (ratio of sides, ~∆'s) Q
2
5 7 C
= 2.5 : (12 – 2.5) = 5 : 19 (opp. sides, rectangle) D E
k = 19
I8 Find the last digit of the value of 620 – 512 – 8.
61 = 6, 62 = 36, …, the last digit of 620 is 6, the last digit of 512 is 5.
The last digit if the number is 6 – 5 – 8 (mod 10) = –7 = 3 (mod 10)
x+2 x −1 5
I9 Let a be the positive root of the equation + = , find the value of a.
x −1 x+2 2
Cross multiplying: 2(x + 2 + x – 1) = 5 ( x − 1)( x + 2)
4(4x2 + 4x + 1) = 25(x2 + x – 2)
⇒ 9x2 + 9x – 54 = 0
⇒ 9(x – 2)(x + 3) = 0
⇒a=x=2
I10 Find the sum of all positive factors of 240.
Reference 1993 HI8, 1994 FI3.2, 1997 HI3, 1998 FI1.4, 2002 FG4.1, 2005 FI4.4
240 = 24×3×5
Positive factors are in the form 2a3b5c, 0 ≤ a ≤ 4, 0 ≤ b, c ≤ 1, a, b, c are integers.
Sum of positive factors = (1 + 2 + 22 + 23 + 24)(1 + 3)(1 + 5) = 31×4×6 = 744
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Answers: (1997-98 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 30 December 2015
Group Events
1 1
G1 If x + = 2 , find the value of x 3 + 3 .
x x
Reference: 1984 FG10.2
2
1 1
x + 2 = x + − 2= 2
2
x x
1 1 1
x3 + 3
= x + x 2 − 1 + 2 = 2×(2 – 1) = 2
x x x
G2 In Figure 1, ABC is a triangle. AD and BE are the bisectors of the E
exterior angles A and B respectively meeting CB and AC C
produced at D and E. Let AD = BE = AB and ∠BAC = a°. Find
the value of a. Reference: 1986 上海市初中數學競賽
A B
180o − a o ao
∠BAD = = 90o − (adj. ∠s on st. line, ∠ bisector)
2 2
180o − ∠BAD ao
∠ABD = = 45o + (∠s sum of ∆ABD) D
2 4
∠ABD ao
∠CBE = = 22.5o + (vert. opp. ∠s, bisector)
2 8
∠ABE = ∠ABC + ∠CBE = 180° – ∠ABD + ∠CBE
ao ao ao
= 135o − + 22.5o + = 157.5o −
4 8 8
∠AEB = a° (base ∠s isosceles ∆)
ao
a° + a° + 157.5o − = 180° (∠s sum of ∆ABE)
8
a = 12
G3 If –6 ≤ a ≤ 4 and 3 ≤ b ≤ 6, find the greatest value of a2 – b2.
0 ≤ a2 ≤ 36 and 9 ≤ b2 ≤ 36
–36 ≤ a2 – b2 ≤ 27
⇒ The greatest value = 27.
G4 Let a, b, c be integers such that a2 = b3 = c. If c > 1, find the smallest value of c.
Reference: 1999 FG3.1
Let a = k3, b = k2, c = k6
c>1⇒k>1
The smallest k = 2
⇒ The smallest c = 26 = 64
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Answers: (1997-98 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 30 December 2015
G5 In figure 2, the area of the parallelogram ABCD A D
is 120. M and N are the mid-points of AB and
BC respectively. AN intersects MD and BD at P
points P and Q respectively. Find the area of M
BQPM.
Produce DM and CB to meet at R. Q
Let BC = 2a. Then BN = NC = a (mid-point)
B C
∆AQD ~ ∆BQN (equiangular) N
A D
BQ BN
= (ratio of sides, ~∆’s) P
QD AD M 2t
Q
1
=
t
(N = mid-point, opp. sides of //-gram) a a
2 R 2a B N C
1
Area of ∆ABD = × 120 = 60
2
2
Area of ∆AQD = × ∆ABD = 40
3
2
Area of ∆BQN BN 1
= =
Area of ∆AQD AD 4
1
∴Area of ∆BQN = × 40 = 10 ……..(1)
4
As M is the mid-point, ∆AMD ≅ ∆BMR (ASA)
⇒ RM = MD (corr. sides ≅ ∆’s) ……………(2)
Also ∆APD ~ ∆NPR (equiangular)
DP AD
= (ratio of sides, ~∆’s)
PR NR
2a 2
= = (opp. sides of //-gram, corr. sides ≅ ∆’s) ………..(3)
3a 3
Combine (2) and (3)
2 1
PD = RD ; MD = RD
5 2
1 2 1
MP = MD − PD = RD − RD = RD
2 5 10
1
RD
MP 10 1
⇒ = = ……………..(4)
PD 2 4
RD
5
1
Area of ∆AMD = × 120 = 30
4
1 1
By (4): Area of ∆AMP = × Area of ∆AMD = × 30 = 6 …………(5)
5 5
1
Area of ∆ABN = × 120 = 30
4
∴ Area of BQPM = Area of ∆ABN –Area of ∆AMP –Area of ∆BQN
= 30 – 6 – 10 = 14 (by (1) and (5))
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Answers: (1997-98 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 30 December 2015
G6 In figure 3, find the number of possible paths from point A to
point B following the direction of arrow heads.
Reference 1983 FI4.1, 2000 HI4, 2007 HG5 A 1 1 1 1
The numbers at each of the vertices of in the
4 4
following figure show the number of 1 2 3
possible ways. 1
3 6 10
So the total number of ways = 14 B 14 Figure 3
G7 Find the smallest real root of the equation (x – 2)(2x – 1) = 5.
2x2 – 5x – 3 = 0
(2x + 1)(x – 3) = 0
1
x =− or 3
2
1
The smallest real root is − .
2
G8 In figure 4, four circles with radius 1 touch each other inside a
square. Find the shaded area. (Correct your answer to the nearest
integer.)
The line segments joining the four centres form a square of sides = 2
Shaded area = 22 – π⋅12 ≈ 1
G9 In figure 5, ABCD is a square and points E, F, G, H are the
H
mid-points of sides AB, BC, CD, DA respectively, find the number of A D
right-angled triangles in the figure. (Reference: 1995 HG8)
Let the shortest side of the smallest right-angled triangle be 1.
Then AE = 2 , EH = 2, AB = 2 2 , AC = 4 E G
We count the number of right-angled triangles with different
hypotenuses.
Hypotenuse Number of triangles B C
F
2 8
2 4
2 2 4
4 4
Total number of triangles = 20
G10 A test is composed of 25 multiple-choice questions. 4 marks will be awarded for each correct
answer and 1 mark will be deducted for each incorrect answer. A pupil answered all questions
and got 70 marks. How many questions did the pupil answer correctly?
Reference: 1994 FI1.2
Suppose he answer x questions correctly and 25 – x question wrongly.
4⋅x – (25 – x) = 70
x = 19
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