hongkong olympiad

Hong Kong Mathematics Olympiad (1988 – 89) 時限：40 分鐘

Heat Event (Individual)
除非特別聲明，答案須用數字表達，並化至最簡。

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.
每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 40 minutes

1. Given that x  1  3 , find x2  1 .
x x2

已知 x  1  3 ，求 x2  1 。
x x2

2. If x # y  xy  2x , find the value of 2 # 3.

設 x # y  xy  2x ，求 2 # 3 的值。

3. Find the number of sides of a regular polygon if an interior angle exceeds an exterior angle by
150.
若一正多邊形的某內角較其外角大的 150，求該正多邊形邊的數目。

4. Find the value of b such that 10log10 9  8b  5 .
已知 10log10 9  8b  5 ，求 b 的值。

5. 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.
某人以 15 km/h 速率乘單車由 P 至 Q，然後以 10 km/h 速率由 Q 返回 P。求該人來回全
程的平均速率。

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

     If 51  5 2    5 n  n 14 , find n.
     [x]是小於或等於 x 的最大整數。例如，[3]  3，[5.7]  5。若 51  5 2    5 n  n 14，

求 n。

7. 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.
某小孩以平行四邊形的兩條相鄰邊長的乘積當作該圖形的面積，他計算的答案是正確面

積的 2 倍。若該平行四邊形的銳角是 x，求 x。

8. If the points A (–8, 6), B (–2, 1) and C (4, c) are collinear, find c .
已知三點 A (–8, 6)、B (–2, 1) 及 C (4, c) 共綫，求 c。

9. The graphs of x2 + y = 8 and x + y = 8 meet at two points. If the distance between these two

points is d , find d.

曲綫 x2 + y = 8 與直綫 x + y = 8 相交於兩點。若該兩點的距離是 d ，求 d。

10. The sines of the three angles of a triangle are in the ratio 3 : 4 : 5. If A is the smallest interior
angle of the triangle and tan A  x , find x.
16
在某三角形中，各內角正弦的比是 3 : 4 : 5。若 A 是這個三角形的最小內角，且

tan A  x ，求 x。
16

11. Two dice are thrown. Find the probability that the sum of the two numbers shown is greater
than 7.
兩骰同擲，求兩數的和大於 7 的概率。

12. F is a function defined by F ( x)  2x 1 , if x  3 F F (3) .
3x2 , . Find
if x  3

函數 F 定義為 F ( x)  2x 1 , if x  3 。求 F F (3) 。
3x2 ,
if x  3

http://www.hkedcity.net/ihouse/fh7878 P.1 HKMO 1989 Heat Event (Individual)

13. If a b c x   ax  by  cz and 1 2 31y4   26 , find y.
y  

 z   2 

設 a b c x   ax  by  cz ，且 1 2 31y4  26 ，求 y。
y 

 z   2 

14. If 1  sin 37sin 45cos 60sin 60 , find B.
B cos30cos 45cos53

設 1  sin 37sin 45cos 60sin 60 ，求 B。
B cos30cos 45cos53

15. If x  y  4 , y  z  5 and z  x  7 , find the value of xyz.

已知 x  y  4 、 y  z  5 及 z  x  7 ，求 xyz 的值。

16. ,  are the roots of the equation x2 10x  c  0 . If   11 and  >  , find the value of

 – .
已知 、 為 x2 10x  c  0 的兩根，且   11 及  > ，求  –  的值。

17. In figure 1, FE // BC and ED // AB. If AF : FB  3 : 2, find A
the ratio area of DEF : area of ABC. FE
如圖一，FE // BC 及 ED // AB。若 AF : FB  3 : 2，求

DEF 的面積 : ABC 的面積。

(Figure 1) (圖一) B D F C

18. In figure 2, a regular hexagon ABCDEF is inscribed in a A E

circle centred at O. If the distance of O from AB is 2 3 23
and p is the perimeter of the hexagon, find p.
如圖二，ABCDEF 為一正六邊形內接於圓形上，O 為圓 O
心。若 O 至 AB 的距離為 2 3 ，且 p 為該正六邊形的周 B
界，求 p。

(Figure 2) (圖二) C D

19. In figure 3, ABCD and ACDE are cyclic quadrilaterals. A

Find the value of x + y.

在圖三，ABCD 及 ACDE 是圓內接四邊形，求 x + y 的 B x° 50°

值。

y° E

(Figure 3) (圖三) C D

20. Find the value of a in figure 4. A
如圖四，求 a。
42 C

4O a
6
B

(Figure 4) (圖四)

D

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.2 HKMO 1989 Heat Event (Individual)

Hong Kong Mathematics Olympiad (1988 – 89)

Heat Event (Group) 時限：20 分鐘
除非特別聲明，答案須用數字表達，並化至最簡。

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.
每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 20 minutes

1. Given a and b are distinct real numbers satisfying a2  5a 10 and b2  5b 10 . Find the

value of 1  1 .
a2 b2

a、b 為兩相異實數，且 a2  5a 10 及 b2  5b 10 ，求 1  1 的值。
a2 b2

2. 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.
一凸 n 邊形的一個內角是 x，其他內角的和是 800，求 n 的值。

3. It is known that 12  22    n2  n(n 1)(2n 1) for all positive integers n.
6

Find the value of 212  222    302 .

已知對所有正整數 n，12  22    n2  n(n  1)(2n  1) ，求 212  222    302 的值。
6

4. One of the positive integral solutions of the equation 19x + 88y = 1988 is given by (100, 1).

Find another positive integral solution.
方程 19x + 88y = 1988 的其中一組正整數解是 (100, 1)，求另一組正整數解。

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

A(2, 3)與 B(17, 23)的連綫交 2x  y  7 於 P，求 AP 的值。
PB

6. Find the remainder when 72047 is divided by 100.

求 72047 被 100 除所得的餘數。

7. If log2log3log7 x  log3log7log2 y  log7log2log3 z  0 , find the value of x + y + z.
若 log2log3log7 x  log3log7log2 y  log7log2log3 z  0 ，求 x + y + z 的值。

8. In figure 1, AB // MN // CD. If AB  4, CD  6 and MN  x , D
6
find the value of x . B C

在圖一中，AB // MN // CD。若 AB  4、CD  6 及 MN  x， 4 N
求 x 的值。 x

(Figure 1)(圖一) A M

9. In figure 2, B = 90, BC  3 and the radius of the inscribed A

circle of ABC is 1. Find the length of AC.
在圖二中，B = 90、BC  3，且 ABC 的內切圓半徑

長 1 單位，求 AC 的長度。

1

1

(Figure 2)(圖二) B C

3

10. In the attached division (see figure 3), the dividend in (a) is *8*

divisible by the divisor in line (b). Find the dividend in line (b) ..... * * * ) * * * * * * ..... (a)

(a). * * * *

(Each asterisk * is an integer from 0 to 9.) ***
在所附除法算式中(見圖三)，(a)列的被除數可被(b) 列的 ***

除數整除。求(a)列的被除數。 ****

(每一星號*為由 0 至 9 的整數。) (Figure 3)(圖三) ****

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.3 HKMO 1989 Heat Event (Group)

Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

172 2 3 24 4 1
45 8 –4 9 5 12 km/h
88-89 6 45 7 147 13 3 14
6 : 25 18 24 19 2
Individual 5 2 10 12
11 12 2 15 24
12 230 20 12

16 12 17

88-89 1 9 2 7 3 6585 4 (12, 20) 5 3
20 2

Group 6 43 7 480 8 12 9 5 10 110768

5

Individual Events

I1 Given that x + 1 = 3 , find x2 + 1 .
x x2

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

x2 + 1 =  x + 1 2 − 2
x2  x 

= 32 – 2 = 7

I2 If x # y = xy – 2x, 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.

x = 180o(n − 2) , y = 360o

nn

x = y + 150°

180o(n − 2) = 360o + 150°

nn

180(n – 2) = 360 + 150n

18n – 36 = 36 + 15n
⇒ n = 24

I4 Find the value of b such that 10log10 9 = 8b + 5 .

9 = 8b + 5

⇒b=1
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.

Total distance travelled = 2x km. Total time = x + x hour.
15 10

Average speed = 2x km/h
x+x
15 10

= 2 = 12 km/h
2+3

30

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

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

[ ] [ ] [ ]If 51 + 5 2 + L + 5 n = n +14 , find n.

Reference 1991 HI13

[ ] [ ] [ ] [ ] [ ] [ ]51 = 1, 5 2 = 1, … , 5 31 = 1; 5 32 = 2, … , 5 242 = 2; 5 243 = 3
[ ] [ ] [ ]If n ≤ 31, 51 + 5 2 + L + 5 n = n
[ ] [ ] [ ]If 32 ≤ n ≤ 242, 51 + 5 2 + L + 5 n = 31 + 2(n – 31) = 2n – 31

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°.
ab = 2absin xo
sin x° = 1

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

mCB = mBA
c −1 = 1−6
4+2 −2+8
c = –4
I9 The graphs of x2 + y = 8 and x + y = 8 meet at two points. If the distance between these two

points is d , find d.
From (1), y = 8 – x2 …… (3)
From (2), y = 8 – x …… (4)
(3) = (4): 8 – x = 8 – x2

x = 0 or 1

When x = 0, y = 8; when x = 1, y = 7

Distance between the points (0, 8) and (1, 7) = 12 + (7 − 8)2 = 2

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

angle of the triangle and tan A = x , 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.
a2 + b2 = (3k)2 + (4k)2 = (5k)2 = c2

∴ ∠C = 90° (converse, Pythagoras’ theorem)
tan A = a = 3 = 12

b 4 16
⇒ x = 12

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

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

Method 2 P(7) = 6 = 1
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

2P(8, 9, 10, 11, 12) = 1 – P(7) = 5
6

P(8, 9, 10, 11, 12) = 5
12

I12 F is a function defined by F (x) = 32xx2+,1 , if x ≤ 3 F(F(3)) .
. Find
if x > 3

F(3) = 2×3 + 1 = 7

F(F(3)) = F(7) = 3×72 = 147

I13 If (a b  x  and (1 2 14  = 26 , find y. (Reference: 1986 FI3.4)

c) y  = ax + by + cz 3) y
 z   2 

(1 2 3)1y4  = 14 + 2y + 6 = 26 ⇒ y = 3

 2 

I14 If 1 = sin 37°sin 45°cos 60°sin 60° , find B.
B cos30°cos 45°cos53°

Reference: 1990 HI14

1 = sin 37°sin 45°cos 60°sin 60°
B cos30°cos 45°cos53°

= sin 37°sin 45°cos 60°sin 60° = cos 60° = 1
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

I16 α, β are the roots of the equation x2 – 10x + c = 0. If αβ = –11 and α > β, find the value of

α – β.

α + β = 10

α – β = (α + β)2 − 4αβ

= 102 − 4(−11) = 12

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

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

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

BDEF is a parallelogram formed by 2 pairs of parallel lines

∆DEF ≅ ∆FBD (A.S.A.) FE

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

∆AEF ~ ∆ACB (Q FE // BC, equiangular)

S ∆AEF  3  2 9 B C
S ∆ACB  +  25
= 2 3 = …… (1) D
(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)

S∆CDE =  2 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
⇒ area of ∆DEF : area of ∆ABC = 6 : 25

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. 23

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

∠AOB = 360° ÷ 6 = 60° (∠s at a point) B CD E
(Figure 2)
∠AOH = 30° A

AH = OH tan 30° = 2 3 × 1 = 2 x° 50°
3 y° E

⇒ AB = 4

Perimeter = 6×4 = 24

I19 In figure 3, ABCD and ACDE are cyclic quadrilaterals. Find the

value of x + y.

Reference: 1992 FI2.3 B

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

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

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

x + y = 230

(Figure 3) C D

I20 Find the value of a in figure 4. A C
∆AOB ~ ∆DOC (equiangular) a
a=6 42
42
a = 12 4O
B

6

(Figure 4) D

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

Group Events
G1 Given a and b are distinct real numbers satisfying a2 = 5a + 10 and b2 = 5b + 10. Find the

value of 1 + 1 .
a2 b2

Reference: 1991 HI14
a and b are the roots of x2 = 5x + 10; i.e. x2 – 5x – 10 = 0

a + b = 5; ab = –10

1 + 1 = a2 + b2 = (a + b)2 − 2ab = 52 − 2(−10) = 9
a2 b2 a 2b 2 (ab)2 (−10)2 20

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

Q 0 < x < 180

∴ 800 + x = 180×5 = 180(n – 2)

n=7

G3 It is known that 12 + 22 + … + n2 = n(n +1)(2n +1) for all positive integers n.
6

Find the value of 212 + 222 + … + 302.

Reference: 1993 HI6
212 + 222 + … + 302 = 12 + 22 + … + 302 – (12 + 22 + … + 202)

= 1 ⋅30⋅31⋅61 – 1 ⋅20⋅21⋅41 = 9455 - 2870 = 6585
66

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)

The line has a slope of − 19 = y2 − y1
88 x2 − x1

Given that (100, 1) is a solution.

− 19 = y2 −1
88 x2 −100

Let y2 – 1 = –19t; x2 – 100 = 88t, where t is an integer.
y2 = 1 – 19t, x2 = 100 + 88t
For positive integral solution of (x2, y2), 1 – 19t > 0 and 100 + 88t > 0
− 25 < t < 1

22 19

Q t is an integer ∴ t = 0 or –1
When t = –1, x2 = 12, y2 = 20 ⇒ another positive integral solution is (12, 20).

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

Reference: 1990 HG3

Equation of AB: y − 3 = 23 − 3 ⇒ 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

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

G6 Find the remainder when 72047 is divided by 100.

Reference: 2002 HG4
The question is equivalent to find the last 2 digits of 72047.
71 = 7, 72 = 49, 73 = 343, 74 = 2401

The last 2 digits repeats for every multiples of 4.
72047 = 74×511+3

The last 2 digits is 43.

G7 If log2[log3(log7 x)] =log3[log7(log2 y)] =log7[log2(log3 z)] = 0, find the value of x + y + z.

log2[log3(log7 x)] = 0
⇒ log3(log7 x) = 1
⇒ log7 x = 3
⇒ x = 73 = 343

log3[log7(log2 y)] = 0
⇒ log7(log2 y) = 1
⇒ log2 y = 7
⇒ y = 27 = 128

log7[log2(log3 z)] = 0
⇒ log2(log3 z) = 1
⇒ log3 z = 2
⇒ z = 32 = 9

x + y + z = 343 + 128 + 9 = 480

G8 In figure 1, AB // MN // CD. If AB = 4, CD = 6 and MN = x, D

find the value of x. B

Reference: 1985 FI2.4, 1990 FG6.4 N 6
∆AMN ~ ∆ACD (equiangular) 4

x = AM … (1) (ratio of sides, ~∆’s) x
6 AC
∆CMN ~ ∆CAB (equiangular) AM C

(Figure 1)

x = MC … (2) (ratio of sides, ~∆’s)
4 AC

(1) + (2): x + x = AM + MC = 1⇒ x = 12
64 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) R 1O
OPBR is a rectangle ⇒ OPBR is a square. 1

BP = BR = 1 (opp. sides of rectangle)

CP = 3 – 1 = 2 B P3 C
CQ = CP = 2 (tangent from ext. point)

Let AR = AQ = t (tangent from ext. point)
32 + (1 + t)2 = (2 + t)2 (Pythagoras’ theorem)
9 + 1 + 2t + t2 = 4 + 4t + t2

6 = 2t

t = 3 ⇒ AC = 2 + 3 = 5

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Answers: (1988-89 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 13 September 2015

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 = (a18a3)x, ***
b = (b1b2b3)x ****
c = (c1c2c3c4c5c6)x ****
d = (d1d2d3d4)x

e = (e1e2e3)x a1 8 a3

f = (f1f2f3)x (b) ......... b1 b2 b3 ) c1 c2 c3 c4 c5 c6 ......... (a)
g = (g1g2g3g4)x
d1 d2 d3 d4
Q 8b = e and a1×b = d, a3×b = g 4-digit numbers

∴ a1 = 9 and a3 = 9 and d = g e1 e2 e3

d = 9b > 1000 and f = 8b < 999 f1 f2 f3
g1 g2 g3 g4
112 ≤ b ≤ 124 …… (1) g1 g2 g3 g4
b1 = 1, d1 = 1
b2 = 1 or 2, f1 = 8 or 9, d2 = 0 or 1

c1 = 1 989

e1 – f1 = 1 (b) ......... 1 b2 b3 ) 1 c2 c3 c4 c5 c6 ......... (a)
∴ f1 = 8 and e1 = 9 1 d2 d3 d4
8b = (8f2f3)x < 900

b < 112.5 …… (2) e1 e2 e3
Combine (1) and (2) f1 f2 f3
b = 112 1 d2 d3 d4
c = 989×112 = 110768

1 d2 d3 d4

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Hong Kong Mathematics Olympiad (1989 – 90)

Heat Event (Individual) 時限：40 分鐘
除非特別聲明，答案須用數字表達，並化至最簡。

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.
每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 40 minutes

1. Find the value of 1  1  1  1  1 .
3 8 8  7 7  6 6  5 5 2

求下式的值： 1  1  1  1  1 。
3 8 8  7 7  6 6  5 5 2

2. If b < 0 and 22b4  20  2b  4  0 , find b.

若 b < 0 及 22b4  20  2b  4  0 ，求 b。

3. If f (a)  a  2 and F (a, b)  a  b2 , find F 3, f (4) .

若 f (a)  a  2 ，且 F (a, b)  a  b2 ，求 F 3, f (4) 。

4. For positive integers a and b, define a # b = ab + ba. If 2 # w = 100, find the value of w.
對正整數 a 及 b，定義 a # b = ab + ba，若 2 # w = 100，求 w 的值。

5. a and b are constants. The straight line 2ax + 3by = 4a + 12b passes through a fixed point P

whose coordinates do not depend on a and b. Find the coordinates of P.
a 及 b 為常數。直綫 2ax + 3by = 4a + 12b 恆過一定點 P(其座標與 a 和 b 無關)。求 P
點的座標。

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

of the triangle and cos A = x , find the value of x.
5

某三角形各內角正弦的比為 3 : 4 : 5。若 A 為該三角形的最小內角，且 cos A = x ，求 x
5

的值。

7. If x + y = 9, y + z = 11 and z + x = 10, find the value of xyz.

若 x + y = 9、y + z = 11 及 z + x = 10，求 xyz 的值。

8. If ,  are the roots of the equation 2x2 + 4x – 3 = 0 and 2, 2 are the roots of the equation
x2 + px + q = 0, find the value of p.
若、是方程 2x2 + 4x – 3 = 0 的根，且 2、2 是方程 x2 + px + q = 0 的根，求 p 的值。

9. If xlog10 x  x3 and x > 10, find the value of x.
100

若 xlog10 x  x3 ，且 x > 10，求 x 的值。
100

10. Given that a0 = 1, a1 = 3 and an2  an1an1  (1)n for positive integers n. Find a4.
已知 a0 = 1，a1 = 3 及 an2  an1an1  (1)n ，其中 n 為正整數。求 a4。

11. Find the unit digit of 2137754.
求 2137754 的個位數。

12. If  r  1  2  3 , find r3  1 .
 r r3

若  r  1  2  3 ，求 r3  1 。
 r r3

http://www.hkedcity.net/ihouse/fh7878 P.1 HKMO 1990 Heat Event (Individual)

13. A positive integer N, when divided by 10, 9, 8, 7, 6, 5, 4, 3 and 2, leaves remainders 9, 8, 7, 6,

5, 4, 3, 2 and 1 respectively. Find the least value of N.
正整數 N 被 10、9、8、7、6、5、4、3 及 2 除所得的餘數依次是 9、8、7、6、5、4、
3、2 及 1，求 N 的最小值。
14. If 1  cos 45sin 70cos 60 tan 40 , find the value of A.

A cos340sin135 tan 220
若 1  cos 45sin 70cos 60 tan 40 ，求 A 的值。

A cos340sin135 tan 220
15. If 10 men can make 20 tables in 5 days, how many days are required to make 60 tables by 15

men?

若 10 人需要 5 天製成 20 張檯，請問 15 人需要多少天製成 60 張檯?

16. In figure 1, the exterior angles of the triangle are in the ratio z'
x’ : y’ : z’  4 : 5 : 6 and the interior angles are in the ratio z
x : y : z  a : b : 3. Find the value of b.
圖一的三角形的三個外角的比是 x’ : y’ : z’  4 : 5 : 6，而

三個內角的比是 x : y : z  a : b : 3，求 b 的值。 y y'

x

17. In ABC, C = 90 and D, E are the mid-points of BC and x'
CA respectively. If AD  7 and BE  4, find the length of (Figure 1)(圖一)
AB. (See figure 2.)
在 ABC 中，C = 90 及 D、E 分別為 BC 及 CA 的 A

中點。若 AD  7 及 BE  4，求 AB 的長度。(參考圖二) E

BD C

(Figure 2)(圖二)

18. Figure 3 shows 3 semi-circles of diameters a, 2a and 3a
respectively. Find the ratio of the area of the shaded part to
that of the unshaded part.
如圖三，三個半圓的直徑分別為 a、2a 及 3a。求陰影

部分的面積與沒有陰影部分的面積的比值。 aa a
(Figure 3)(圖三)

19. Find the value of 2 1 3  1 4  4 1 5    1 20 .
 3  19 

求 2 1 3  1 4  4 1 5    19 1 20 的值。
 3  

20. In figure 4, C = 90, AD  DB and DE is C

perpendicular to AB. If AB  20 and AC  12, find

the area of the quadrilateral ADEC. E
D
在圖四中，C = 90、AD  DB 及 DE 垂直於

AB。若 AB  20 及 AC  12，求四邊形 ADEC 的

面積。 A B

(Figure 4)(圖四)

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http://www.hkedcity.net/ihouse/fh7878 P.2 HKMO 1990 Heat Event (Individual)

Hong Kong Mathematics Olympiad (1989 – 90)

Heat Event (Group)

除非特別聲明，答案須用數字表達，並化至最簡。 時限：20 分鐘

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.

每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 20 minutes

1. If 11 5 and 1  1  13 , find the value of 1  1 .
ab a2 b2 a5 b5

若 11 5 及 1  1  13 ，求 1  1 的值。
ab a2 b2 a5 b5

2. There are N pupils in a class.
When they are divided into groups of 4, 1 pupil is left behind.
When they are divided into groups of 5, 3 pupils are left behind.
When they are divided into groups of 7, 3 pupils are left behind.
Find the least value of N.
某班有學生 N 人。
若將學生分為每 4 人一組，有 1 人餘下，
若將學生分為每 5 人一組，有 3 人餘下，
若將學生分為每 7 人一組，有 3 人餘下。
求 N 的最小值。

3. The coordinates of A, B, C and D are (10, 1), (1, 7), (–2, 1) and (1, 3) respectively. AB and CD
meet at P. Find the value of AP .
PB
A、B、C 及 D 的座標依次是(10, 1)、(1, 7)、(–2, 1)及(1, 3)。AB 與 CD 相交於 P。求 AP
PB
的值。

4. Find the remainder when 21989 + 1 is divided by 3.
求 21989 + 1 被 3 除所得的餘數。

5. Euler was born and died between 1700 A.D. and 1800 A.D. He was n + 9 years old in n3 A.D.
and died at the age of 76. Find the year in which Euler died.
歐拉在 1700 A.D.和 1800 A.D.之間出生和去世。在 n3 A.D.時，他剛好 n + 9 歲，而他在
76 歲時去世。求歐拉去世的年份。

6. Let N! denote the product of the first N natural numbers, i.e. N! = 123N.
If k is a positive integer such that 30! = 2k an odd integer, find k.

設 N! 為首 N 個自然數的乘積，即 N! = 123N。
若 k 是正整數使得 30! = 2k一奇數，求 k。

http://www.hkedcity.net/ihouse/fh7878 P.3 HKMO 1990 Heat Event (Group)

7. The graph of the parabola y  x2  4x  9 cuts the y
4 y  x2  4x  9
4
x-axis at A and B (figure 1). If C is the vertex of the
parabola, find the area of ABC. A x
拋物綫 y  x2  4x  9 的圖像交 x-軸於 A 及 B(圖一)。
BO
4
若 C 是拋物綫的頂點，求ABC 的面積。 C

8. In figure 2, FE // BC and ED // AB. If AF : FB  1 : 4, (Figure 1)(圖一)
find the ratio of area of EDC : area of DEF.
在圖二中，FE // BC 及 ED // AB。若 AF : FB  1 : 4， A
求EDC 的面積: DEF 的面積。 FE

BD C

(Figure 2)(圖二)

9. In the attached multiplication (figure 3), the letters O , L , Y , M , P , I , O L Y M P I A D

A and D represent different integers ranging from 1 to 9. Find the  D

integer represented by A. OOOOO OOOO

(Figure 3)(圖三)

在所附乘法算式中(圖三)，字母 O、L、Y、M、P、I、A 及 D 代

表由 1 至 9 的不同整數，求 A 所代表的整數。

10. Three circles, with centres A, B and C respectively, touch one P
another as shown in figure 4. If A, B and C are collinear and PQ
is a common tangent to the two smaller circles, where PQ  4, BA C
find the area of the shaded part in terms of .
以 A、B 及 C 為圓心的三個圓兩兩相切如圖四。若 A、B

及 C 共綫，且 PQ 是兩個較小圓的公切綫，其中 PQ  4，

試以  表陰影面積。

Q
(Figure 4)(圖四)

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http://www.hkedcity.net/ihouse/fh7878 P.4 HKMO 1990 Heat Event (Group)

Answers: (1989-90 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 24 April 2017

152 –2 3 7 4 6 5 (2, 4)
647 9 100 10 109
89-90 11 9 12 120 8 –7 14 2 15 10 days
Individual 9
16 5 17 0 13 2519 19
20 58.5
2 13 18 1 : 2 20

1 275 2 73 3 2 4 0 5 1783
89-90
Group 6 26 7 125 =15 5 8 4 : 1 9 7 10 2
88

Individual Events
I1 Find the value of 1  1  1  1  1 .

3 8 8 7 7  6 6  5 5 2

1 1  1  1 1
3 8 8 7 7  6 6  5 5 2

=3  8  8  7   7  6  6  5  5  2

=5
I2 If b < 0 and 22b+4 – 202b + 4 = 0, find b.

Let y = 2b, then y2 = 22b, the equation becomes 16y2 – 20y + 4 = 0
4y2 – 5y + 1 = 0
(4y – 1)(y – 1) = 0

y = 2b = 1 or y = 1
4

b = –2 or 0
 b < 0  b = –2 only
I3 If f (a) = a – 2 and F(a, b) = a + b2, find F(3, f (4)).
Reference: 1985 FI3.3, 2013 FI3.2, 2015 FI4.3
f (4) = 4 – 2 = 2
F(3, f (4)) = F(3, 2) = 3 + 22 = 7
I4 For positive integers a and b, define a#b = ab + ba. If 2#w = 100, find the value of w.
Reference: 1999 FI3.1
2w + w2 = 100 for positive integer w.
By trail and error, 64 + 36 = 100
w = 6.
I5 a and b are constants. The straight line 2ax + 3by = 4a + 12b passes through a fixed point P
whose coordinates do not depend on a and b. Find the coordinates of P.
Reference: 1991 HI6, 1996 HI6
2ax + 3by = 4a + 12b  2a(x – 2) + 3b(y – 4) = 0
Put b = 0  x = 2,
Put a = 0  y = 4
P(2, 4)

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Answers: (1989-90 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 24 April 2017

I6 The sines of the angles of a triangle are in the ratio 3 : 4 : 5. If A is the smallest interior angle

of the triangle and cos A = x , find the value of x.
5

Reference: 1989 HI10

By Sine rule, a : b : c = sin A : sin B : sin C = 3 : 4 : 5

Let a = 3k, b = 4k, c = 5k.
a2 + b2 = (3k)2 + (4k)2 = (5k)2 = c2

 C = 90 (converse, Pythagoras’ theorem)

cos A = b  4
c5

x=4

I7 If x + y = 9, y + z = 11 and z + x = 10, find the value of xyz.

Reference: 1986 FG10.1, 1989 HI15

(1) + (2) – (3): 2y = 10  y = 5

(1) + (3) – (2): 2x = 8  x = 4

(2) + (3) – (1): 2z = 12  z = 6

 xyz = 120
I8 If ,  are the roots of the equation 2x2 + 4x – 3 = 0 and 2, 2 are the roots of the equation

x2 + px + q = 0, find the value of p.

 +  = –2

 =  3
2

p = –(2 + 2) = –( + )2 + 2
= –(–2)2 – 3 = –7

I9 If xlog10 x  x3 and x > 10, find the value of x.
100

Take log on both sides, log xlog x = 3 log x – log 100
(log x)2 – 3 log x + 2 = 0

(log x – 1)(log x – 2) = 0

log x = 1 or log x = 2

x = 10 or 100

 x > 10  x = 100 only

I10 Given that a0 = 1, a1 = 3 and a 2 – an–1an+1 = (–1)n for positive integers n. Find a4.
n

Put n = 1, a12  a0a2  (1)1  32 – a2 = –1  a2 = 10

Put n = 2, a22  a1a3  (1)2  102 – 3a3 = 1  a3 = 33

Put n = 3, a32  a2a4  (1)3  332 – 10a4 = –1  a4 = 109
I11 Find the unit digit of 2137754.

Reference 1991 HG1
71 = 7, 72 = 49, 73 = 343, 74 = 2401

The pattern of unit digit repeats for every multiples of 4.

2137754  (74)18872  9 mod 10

The unit digit is 9.

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Answers: (1989-90 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 24 April 2017

I12 If  r  1  2  3 , find r3  1 .
 r r3

Reference: 1985 FI1.2, 2017 FI1.4

r1 3
r

r2  1 =  r  1 2 2= 3 – 2 = 1
r2  r

r3  1 =  r  1  r 2 1 1 
r3  r  r2 

=  311= 0

I13 A positive integer N, when divided by 10, 9, 8, 7, 6, 5, 4, 3 and 2, leaves remainders 9, 8, 7, 6,
5, 4, 3, 2 and 1 respectively. Find the least value of N.

Reference: 1985 FG7.2, 2013FG4.3
N + 1 is divisible by 10, 9, 8, 7, 6, 5, 4, 3, 2.

The L.C.M. of 2, 3, 4, 5, 6, 7, 8, 9, 10 is 2520.
 N = 2520k – 1, where k is an integer.
The least positive integral of N = 2520 – 1 = 2519
I14 If 1  cos 45sin 70cos 60 tan 40 , find the value of A.

A cos340sin135 tan 220
Reference: 1989 HI14
1 = cos 45cos 20cos 60 tan 40
A cos 20cos 45 tan 40

= cos 60 = 1
2

A=2

I15 If 10 men can make 20 tables in 5 days, how many days are required to make 60 tables by 15

men?

1 man can make 20 = 2 table in 1 day.
10 5 5

15 men can make 2 15 = 6 tables in one day.
5

They can make 60 tables in 10 days

I16 In figure 1, the exterior angles of the triangle are in the ratio z'
x’ : y’ : z’  4 : 5 : 6 and the interior angles are in the ratio z
x : y : z  a : b : 3. Find the value of b.

Let x’ = 4k, y’ = 5k, z’ = 6k y'
4k + 5k + 6k = 360 (sum of ext.  of polygon) xy

15k = 360 x'

 k = 24 (Figure 1)

x’ = 96, y’ = 120, z’ = 144

x = 84, y = 60, z = 36 (adj. s on st. line)

x:y:z=7:5:3

b=5

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Answers: (1989-90 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 24 April 2017
A
I17 In ABC, C  90 and D, E are the mid-points of BC and E

CA respectively. If AD  7 and BE  4, find the length of BD C

AB. (See figure 2.) (Figure 2)

Let BD = x = DC, AE = y = EC aa a
x2 + (2y)2 = 72  (1) (Figure 3)
(2x)2 + y2 = 42  (2)
4(1) – (2): 15y2 = 180  y2 = 12
4(2) – (1): 15x2 = 15  x2 = 1
AB2 = (2x)2 + (2y)2 = 4 + 48

 AB = 52 = 2 13

I18 Figure 3 shows 3 semi-circles of diameters a, 2a and 3a
respectively. Find the ratio of the area of the shaded part to

that of the unshaded part.

Area of the shaded part =   a2    a 2 = 3 a2
2 2 2 8

Area of the unshaded part =   3a 2  3  a2 = 6  a2
2 2 8 8

The ratio = 3 : 6 = 1 : 2

I19 Find the value of 1  1  1    1 .
23 34 45 19 20

1  1  1  1
23 34 45 19 20

=  1  1    1  1    1  1      1  1 
2 3 3 4 4 5 19 20 

=1 1 = 9
2 20 20

I20 In figure 4, C = 90, AD  DB and DE is perpendicular C
E
to AB. If AB  20 and AC  12, find the area of the
D
quadrilateral ADEC.
(Figure 4)
BD = 10, BC = 16 (Pythagoras’ theorem) A B
BDE ~ BCA (equiangular)

BD : DE : BE = 16 : 12 : 20 (ratio of sides, ~’s)

DE =7.5, BE = 12.5

CE = 16 – 12.5 = 3.5

SADEC = 1 10  7.5  1 12  3.5 = 58.5
2 2

Method 2

BD = 10, BC = 16 (Pythagoras’ theorem)

BDE ~ BCA (equiangular)

SBDE =  BD 2  S ABC =  10 2  1 12 16 = 37.5
 BC   16  2

SADEC = 1 12 16 – 37.5 = 58.5
2

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