Answers: (1998-99 HKMO Heat Events) Created by: Mr.

Francis Hung Last updated: 7 January 2016

98-99 1 1 2 8 3 56 4 405 5 100000

Individual 6 2401 7 9 8 36 9 11 10 9

1 1
98-99 1 3 2 –24 3 4 5 6
2 2
Group
6 12 7 4 8 7 9 12 10 135
Individual Events
I1 The circumference of a circle is 14π cm. Let X cm be the length of an arc of the circle, which
1
subtends an angle of radian at the centre. Find the value of X.
7

Let r be the radius of the circle.

2πr = 14π
⇒r=7
1
X = rθ = 7× = 1
7

I2 In Figure 1, ABCDEF is a regular hexagon with area equal to

3 3 cm2. Let X cm 2 be the area of the square PQRS, find the value
of X.
Area of the hexagon = 6×areas of ∆AOB
1 3 3
3 3 = 6 ⋅ ⋅ OB 2 sin 60o = ⋅ OB 2
2 2
2
OB = 2
Area of the square = (2OB)2 = 4×2 = 8
I3 8 points are given and no three of them are collinear. Find the number of triangles formed by
using any 3 of the given points as vertices.
The number of triangles formed
8× 7 × 6
= 8C3 = = 56
1× 2 × 3
I4 In Figure 2, there is a 3 × 3 square.
Let ∠a + ∠b + …+ ∠i = X°, find the value of X.
Reference: 廣州、
廣州、武漢、武漢、福州、 福州、重慶、 重慶、洛陽 初中數學聯賽
∠c = ∠e = ∠g = 45°
∠a + ∠i = 90°, ∠b + ∠f = 90°, ∠d + ∠h = 90°
∠a + ∠b + …+ ∠i = 45°×3 + 90°×3 = 405°
X = 405
I5 How many integers n are there between 0 and 106, such that the unit digit of n3 is 1?
13 = 1, the unit digit of n must be 1
There are 106 ÷ 10 = 100000 possible integers.

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Answers: (1998-99 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 7 January 2016

I6 Given that a, b, c are positive integers and a < b < c = 100, find the number of triangles
formed with sides equal a cm, b cm and c cm.
By triangle inequality: a + b > c = 100
Possible pairs of (a, b): (2, 99), (3, 98), (3, 99), (4, 97), (4, 98), (4, 99), … ,
(50, 51), (50, 52), … , (50, 99), ....................................
(98, 99)
Total number of triangles = 1 + 2 + … + 48 + 49 + 48 + … + 2 + 1
1 + 49
= × 49 × 2 − 49 = 2401
2
I7 A group of youngsters went for a picnic. They agreed to share all expenses. The total amount
used was $288. One youngster had no money to pay his share, and each of the others had to
pay $4 more to cover the expenses. How many youngsters were there in the group?
Let the number of youngsters be n.
288 288
− =4
n −1 n
72 = n2 – n
n=9
I8 A two-digit number is equal to 4 times the sum of the digits, and the number formed by
reversing the digits exceeds 5 times the sum of the digits by 18. What is the number?
Let the unit digits of the original number be x and the tens digit by y.
10y + x = 4(x + y) ..............(1)
10x + y – 5(x + y) = 18 ......(2)
From (1), 6y = 3x ⇒ x = 2y ..........(3)
Sub. (3) into (2): 20y + y – 5(2y + y) = 18
⇒ y = 3, x = 6
The number is 36.
I9 Given that the denominator of the 1001th term of the following sequence is 46, find the
1 1 2 1 2 3 1 2 3 4
numerator of this term. , , , , , , , , , , ⋅⋅⋅
2 3 3 4 4 4 5 5 5 5
Suppose the numerator of the 1001th term is n.
1 + 2 + 3 + … + 44 + n = 1001, n ≤ 45
1
(45)(44) + n = 1001
2
n = 1001 – 990 = 11
I10 In the following addition, if the letter ‘S’ represents 4, what digit does the letter ‘A’ SEE
represent? SEE
3E + 4 = 10a + Y .........(1), where a is the carry digit in the tens digit. 4EE SEE
4E + a = 10b + 4 ......(2), where b is the carry digit in the hundreds digit. 4EE + YES
4×3 + Y + b = 10E + A ..........(3) 4EE EASY
From (3), E = 1 or 2 + YE4
When E = 1, (1) ⇒ Y = 7, a = 0, (2) ⇒ b = 0, (3) ⇒ A = 9 EA4Y
When E = 2, (2) ⇒ a = 1, Y = 0 reject because YE4 is a 3-digit number.
∴A=9

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Answers: (1998-99 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 7 January 2016

Group Events
G1 If a is a prime number and a2 – 2a – 15 < 0, find the greatest value of a.
(a + 3)(a – 5) < 0
⇒a<5
The greatest prime number is 3.
G2 If a : b : c = 3 : 4 : 5 and a + b + c = 48, find the value of a – b – c.
a = 3k, b = 4k, c = 5k; sub. into a + b + c = 48
⇒ 3k + 4k + 5k = 48
⇒k=4
a = 12, b = 16, c = 20
a – b – c = 12 – 16 – 20 = –24
G3 Find the value of log 3 + 5 + 3 − 5  .
 
Reference: 1993 FI1.4, 2001 FG2.1, 2011 HI7, 2015 FI4.2, 2015 FG3.1

 
log 3 + 5 + 3 − 5  = log
 6+2 5
 +
6 − 2 5 


= log
(1 + 5 )2
+ ( 5 − 1)
2 


   2 2 
   2 
 
1+ 5 + 5 −1
= log
 
( )
 = log 2 5  = log 2 5 = log 10 = 1
   2
 2   2 
G4 Find the area enclosed by the straight line x + 4y – 2 = 0 and the two coordinate axes.
1 1 1 1
x-intercept = 2, y-intercept = ; the area = × × 2 =
2 2 2 2

G5 Natural numbers are written in order starting from 1 until 198th digit as shown
1234567891
1444 4201112
444 L . If the number obtained is divided by 9, find the remainder.
L4
3
198 digits

123456789 has 9 digits

10111213…9899 has 90×2 = 180 digits
∴ 1234567891011…9899100101102 has 198 digits.
1 + 2 + 3 + … +9 = 45, 11+12+…+19 is also divisible by 9,...,
91+92+…+99 is divisible by 9.
10 + 20 + … + 90 is divisible by 9
∴the remainder is the same as 100101102 divided by 9.
1 + 1 + 1 + 1 + 2 = 6, the remainder is 6.
G6 The average of 2, a, 5, b, 8 is 6. If n is the average of a, 2a+1, 11, b, 2b+3, find the value of n.
2 + a + 5 + b + 8 = 30 .........(1), a + 2a + 1 + 11 + b + 2b + 3 = 5n .............(2)
From (1): a + b = 15
(2) 5n = 3a + 3b + 15 = 3(a + b) + 15 = 3×15 + 15 = 60
⇒ n = 12
G7 If p = 2x2 – 4xy + 5y2 – 12y + 16, where x and y are real numbers, find the least value of p.
p = 2x2 – 4xy + 2y2 + 3y2 – 12y + 16 = 2(x – y)2 + 3(y2 – 4y + 4) + 4 = 2(x – y)2 + 3(y – 2)2 + 4
p ≥ 4, the least value of p is 4.

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Answers: (1998-99 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 7 January 2016

G8 Find the unit digit of 333335.

31 = 3, 32 = 9, 33 = 27, 34 = 81, the unit digit of 34m is 1, where m is any positive integer.
333335 = 3334×83+3 = (3334)83×3333 = (...1)83×(...33) = ...7, the unit digit is 7.
G9 In Figure 1, ∠MON = 20°, A is a point on OM, OA = 4 3 ,
D is a point on ON, OD = 8 3 , C is any point on AM, B is
any point OD. If l = AB + BC + CD, find the least value of
l.
Reflect the figure along the line OM, then reflect the figure
between ∠MON1 along the line ON1.
∠NOM2 = 3×20° = 60° M2
l = AB + BC + CD = AB1 + B1C + CD
l = A2B1 + B1C + CD
N1
l is the shortest when A2, B1, C, D are collinear. C2
By cosine formula on ∆OA2D,
D1
Shortest l =A2D = (4 3 ) + (8 3 ) − 2(4 3 )(8 3 )cos 60
2 2 o A2
B1 C
M

= 48 + 192 − 96 = 12 A
60°

O N
B D

G10 In figure 2, P is a point inside the square ABCD, PA = a,

A D
PB = 2a, PC=3a (a > 0). If ∠APB=x°, find the value of x
Reference: 2014 HG4 a
Rotate ∆APB by 90° in anti-clockwise direction about B.
Let P rotate to Q, A rotate to E. P
∆APB ≅ ∆EQB (by construction)
EQ = a, BQ = 2a = PB. Join AQ.
∠PBQ = 90° (Rotation) 2a 3a
∠ABQ = 90° – ∠ABP = ∠PBC
AB = BC (sides of a square)
∆ABQ ≅ ∆CBP (SAS)
B C
AQ = CP = 3a (corr. sides ≅ ∆’s)
Q ∠PBQ = 90° (Rotation) A D
∴ PQ2 = PB2 + QB2 (Pyth. Theorem)
a
= (2a)2 + (2a)2 = 8a2
AP2 + PQ2 = a2 + 8a2 = 9a2 P
AQ2 = (3a)2
∴ AP2 + PQ2 = AQ2
2a 3a
∠APQ = 90°
Q
Q∠PBQ = 90° and PB = QB
a 2a
∴∠BPQ = 45°
∠APB = 45° + 90° = 135° E B C

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