hongkong olympiad

6. 對任意實數 a、b、c 及 d ，定義運算  ：(a, b)*(c, d) = (ad + bc, bd)。

1, 5   8 
 若x,y 3 及 a = x ，求 a 的值。
7  5, 3 y

For any real number a, b, c and d, we define the operation : (a, b)*(c, d) = (ad + bc, bd).

1, 5   8 
x, Ify 3 and a = x , find the value of a.
7  5, 3 y

7. 已知 sin  – cos  = 1 及 0 <  < 180。若 tan  = B，求 B 的值。
5

Given that sin  – cos  = 1 and 0 <  < 180. If tan  = B, find the value of B.
5

8. 如圖三，PAC、QBA、RCB 及 ABC 皆是等邊三角形。點 X、Y 及 Z 分別為 PAC、
QBA 及 RCB 的內心。若 PA 的長度是 10 cm 及 XYZ 的周界是 w cm，求 w 的
值。(註：三角形的內心為該三角形三條內角平分綫的交點。)
In Figure 3, PAC, QBA, RCB and ABC are equilateral triangles. The points X, Y and Z
are the incentres of PAC, QBA and RCB respectively. If the length of PA is 10 cm and the
perimeter of XYZ is w cm, find the value of w. (Remark: the incentre of a triangle is the
point of intersection of the three interior angle bisectors of the triangle.)

圖三
Figure 3

 9. 設 f x  1 4x2  60x  9  4x2  60x  9 ，
2
若 k  f 1  f 2  f 3    f 15  f 16 ，求 k 的值。

 Let f x  1 4x2  60x  9  4x2  60x  9 .
2

If k  f 1  f 2  f 3    f 15  f 16 , find the value of k.

10. 在平面上點 P 的坐標是 (3, 4)。以 (0, 0) 為中心，點 P 順時針方向旋轉 45 後，再
沿 y-軸反射到達點 Q = (x, y)。若 z = x + y，求 z 的值。
The coordinates of point P on the plane is (3, 4). After rotating 45 clockwise about the
centre (0, 0) and reflecting along the y-axis, the point P reaches the point Q = (x, y).
If z = x + y, find the value of z.

*** 試卷完 End of Paper ***

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Hong Kong Mathematics Olympiad 2006-2007

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. 若由 1 至 50 內與 50 互質的整數有 N 個，求 N 的值。 (註：正整數 a 與 b 稱為互

質若 a 與 b 的最大公因數是 1。)

If there are N integers from 1 to 50 that are relatively prime to 50, find the value of N.

(Remark: positive integers a and b are said to be relatively prime if their greatest common

divisor is 1.)

2. 如圖一，在梯形 ABCD 中， AB // CD ，BCE  ECD ，CE  AD 及 DE  2AE 。若
DEC 的面積是 2007 cm2 及四邊形 ABCE 的面積是 T cm2，求 T 的值。
In Figure 2, ABCD is a trapezium, AB // CD , BCE  ECD , CE  AD and DE  2AE .
If the area of DEC is 2007 cm2 and the area of quadrilateral ABCE is T cm2, find the value

of T.

AB

ET 圖一
Figure 1

DC
3. 已知 a2 – 3a + 1 = 0。若 A  2a 5  5a 4  2a 3  8a 2  7a ，求 A 的值。

3a 2  3
Given that a2 – 3a + 1 = 0. If A  2a 5  5a 4  2a 3  8a 2  7a , find the value of A.

3a 2  3

4. 已知點 A、B 及 C 的坐標分別為 (3, 4)、(6, 4) 及 (8, 10)。 M 及 N 分別為 AB 及 BC
的中點。X 為 AN 上一點使得 AX：XN = 2：1。若 r  CX ，求 r 的值。
XM
Given that the coordinates of the points A, B and C are (3, 4), (6, 4) and (8, 10) respectively.
M and N are the midpoints of AB and BC respectively. X is a point on AN such that
AX：XN = 2：1. If r  CX , find the value of r.
XM

http://www.hkedcity.net/ihouse/fh7878 P.3 HKMO 2007 Heat Event (Group)

5. 如圖二，兩個邊長為 1 cm 的正方體組成一個 1 cm  1 cm  2 cm 的長方體。一隻螞蟻
沿著長方體爬行，其爬行路綫須為正方體的棱。牠從頂點 A 出發，以每分鐘爬行 1 cm
的速度，於 4 分鐘後到達頂點 B。若螞蟻可行路綫數目共有 S 個，求 S 的值。

In Figure 2, a 1 cm  1 cm  2 cm rectangular box is made up by two cubes with side length
1 cm. An ant is climbing along the box in a way that it must stay on the edges of the cubes
through out the climbing. Starting from vertex A and climbing with a speed of 1 cm per
minute, it reaches vertex B after 4 minutes. If the total number of possible paths taken by the
ant is S, find the value of S.

圖二
Figure 2

6. 若以 5 除 72007 所得的餘數是 R，求 R 的值。
If the remainder of 72007 when dividing by 5 is R, find the value of R.

7. 設 k  sin 30  cos 60  sin 90  cos120    sin1890  cos1920 ，求 k 的值。
Let k  sin 30  cos 60  sin 90  cos120    sin1890  cos1920 , find the value of k.

8. 如圖三，已知半圓的直徑為 10 cm。 A、B 和 C 是半圓上任意的三點使 B 在 AC 上。
設 x 為綫段 AB 及 BC 的長度之和，求 x 可取的最大值。
In figure 3, given that the diameter of the semicircle is 10 cm. A, B and C are three arbitrary
points on the semicircle where B is on AC. If x is the sum of the length of the line segments
AB and BC, find the greatest possible value of x.

圖三
Figure 3

9. 在坐標平面上，點 A  (6, 2) 、 B  (3, 3) 、 C  (0, n) 及 D  (m, 0) 組成一個四邊形
ABCD。求 n 的值使得該四邊形 ABCD 的周界為最短。
In the coordinate plane, the points A  (6, 2) , B  (3, 3) , C  (0, n) and D  (m, 0) form

a quadrilateral ABCD. Find the value of n so that the perimeter of the quadrilateral ABCD is
the least.

10. 已知整數 x 及 y 滿足 3x + 5y = 1。若 S  x  y 及 S > 2007，求 S 可取的最小值。
Given that integers x and y satisfying the equation 3x + 5y = 1. If S  x  y and S > 2007,
find the least possible value of S.

*** 試卷完 End of Paper ***

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

1 157.5 2 30 3 57 4 2006 5 500
06-07 9
4
Individual 6 5 7 3 8 30 82 10 3 2

1 20 14049 4 4 2 5 12
06-07 283 9 10 2011

Group 3 7 3 8 10 2 9 4
6 2 3

Individual Events
I1 In Figure 1, a clock indicates the time 3:45. If the angle between the

hour-hand and the minute-hand is , find the value of .

Reference 1984 FG7.1, 1985 FI3.1, 1987 FG7.1, 1989 FI1.1, 1990 FG6.3

At 3:00 pm, the minute-hand lags the hour-hand by 360 1 = 90.
4

At 3:45 pm, the minute-hand has moved 360 3 = 270;
4

the hour-hand has moved 360 1  3 = 22.5.
12 4

The angle between the hour-hand and the minute-hand is

270 – 90 – 22.5 = 157.5   = 157.5
I2 In Figure 2, there is a paper net that can be wrapped into a regular

polyhedron. If this polyhedron has y edges, find the value of y.
There are altogether 12 pentagons. Each pentagon has 5 edges. The

total number of edges is 125 = 60.
When the paper net is wrapped to form a polyhedron, every 2 edges
are stuck together to form 1 edge.

 The number of edges in the polyhedron = 60  2 = 30

I3 Among 4 English books, 6 Chinese books and 9 Japanese books, two books are selected. It is
found that they are of the same language. If there are X such choices, find the value of X.
X = 4C2 + 6C2 + 9C2 = 6 + 15 + 36 = 57

I4 Let r1 and r2 be the two real roots of the equation (x – 2006)(x – 2007) = 2007
If r is the smaller real root of the equation (x – r1)(x – r2) = –2007, find the value of r.
(x – 2006)(x – 2007) = 2007  x2 – 4013x + 20052007 = 0

 r1 + r2 = 4013, r1r2 = 20052007...............(*)
(x – r1)(x – r2) = –2007  x2 – (r1 + r2)x + r1r2 + 2007 = 0
x2 – 4013x + 20052007 + 2007 = 0 by (*)
x2 – 4013x + 20062007 = 0
(x – 2006)(x – 2007) = 0
x = 2006 or x = 2007
r = the smaller real root = 2006

I5 Given that  and  are the roots of the equation x2 – 52007x+51000 = 0. If s = log25 2 +log25 2 ,
 

find the value of s.
  = 51000

s = log25 2 + log25 2 = log25( 2  2 )
   

= log25( ) = log 51000 = 1000 log 5 = 500
log 25 2 log 5

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

I6 For any real number a, b, c and d, we define the operation *:
(a, b)*(c, d) = (ad + bc, bd)

 If(x, y) = 1, 3  * and a = x , find the value of a.
 7  5  8 5,3 y

 (x, 5 ,9
y) =  3  38 5 7 5 
7

=  21 3 5  24  3 5, 9 5 
7 5 7

=  7 45 5 , 7 9 5 
   

a= x =5
y

I7 Given that sin  – cos  = 1 and 0 <  < 180. If tan  = B, find the value of B.
5

Reference: 1992 HI20, 1993 HG10, 1995 HI5, 2007 FI1.4, 2014 HG3

(sin  – cos )2 = 1
25

sin2  – 2 sin  cos  + cos2  = 1
25

1 – 2 sin  cos  = 1
25

sin  cos  = 12
25
25 sin  cos  = 12(sin2  + cos2 )
12sin2  – 25 sin  cos  + 12cos2  = 0

(3 sin  – 4 cos )(4 sin  – 3 cos ) = 0

tan  = 4 or 3
3 4

Check when tan  = 4 , then sin  = 4 , cos  = 3
3 5 5

LHS = sin  – cos  = 4 – 3 = 1 = RHS
5 5 5

When tan  = 3 , then sin  = 3 , cos  = 4
4 5 5

LHS = sin  – cos  = 3 – 4 = – 1  RHS
5 5 5

 B = tan  = 4
3

I8 In Figure 3, PAC, QBA, RCB and ABC are equilateral

triangles. The points X, Y and Z are the incentres of PAC, QBA,

RCB respectively. If the length of PA is 10 cm and the perimeter

of XYZ is w cm, find the value of w. (Remark: the incentre of a
triangle is the point of intersection of the three interior angle

bisectors of the triangle.)

It can be easily proved that XYZ is an equilateral triangle whose
side is half of PQ.

 w = 320  2 = 30

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

I9 Let f(x) = 1 (4x2 – 60x + 9 + |4x2 – 60x + 9|).
2

If k = f(1) + f(2) + f(3) ++ f(15) + f(16), find the value of k.

Similar question: 2004 FI3.2

Note that if a  0, 1 (a + |a|) = a; if a < 0, 1 (a + |a|) = 0
2 2

Now, 4x2 – 60x + 9 = (2x – 15)2 – 216 = (2x – 15 + 6 6 )(2x – 15– 6 6 )

= 4(x – 7.5 + 3 6 )(x – 7.5 – 3 6 )
If 7.5 – 3 6 < x < 7.5 + 3 6 , then 4x2 – 60x + 9 < 0
If x  7.5 – 3 6 or 7.5 + 3 6  x, then 4x2 – 60x + 9  0

7.5 – 3 6 = 56.25 – 54 > 0, 56.25 – 54 < 56.25 – 7 = 0.5

0 < 7.5 – 3 6 < 0.5

7.5 + 3 6 = 56.25 + 54 > 7.5 + 7 = 14.5, 56.25 + 54 < 7.5 + 7.5 = 15

14.5 < 7.5 + 3 6 < 15

Let g(x) = 4x2 – 60x + 9,  g(1) < 0, g(2) < 0, , g(14) < 0, g(15) > 0, g(16) > 0

f(1) = 0, f(2) = 0, , f(14) = 0, f(15) = g(15), f(16) = g(16)

k = f(1) + f(2) + f(3) ++ f(15) + f(16) = f(15) + f(16) = g(15) + g(16)
= 4152 – 6015 + 9 + 4162 – 6016 + 9

= 6015 – 6015 + 9 + 6416 – 6016 + 9

= 9 + 416 + 9

= 82

I10 The coordinates of point P on the plane is (–3, 4). After rotating 45 clockwise about the

centre (0, 0) and reflecting along the y-axis, the point P reaches the point Q = (x, y). If z = x +y,

find the value of z. Reference: 2015 HG3

Let P(–3, 4) makes an angle  with the positive y-axis. Q(x, y) R(a, b)

Then sin  = 3 , cos  = 4 , OP = 5. P(-3, 4)
5 5
4
Let R(a, b) be the point after rotating P clockwise about O.

Then OR = OP = 5 5
a = 5 sin(45 – ) = 5 sin 45 cos  – 5 cos 45 sin  52

= 5 2  4 – 5 2  3 = 2 
2 5 2 5 2 45

b = 5 cos(45 – ) = 5 cos 45 cos  + 5 sin 45 sin 

= 5 2  4 + 5 2  3 = 7 2 O
2 5 2 5 2

Q = (–a, b) = (– 2 ,7 2 )
22

z=x+y=– 2 + 7 2 =3 2
2 2

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

Group Events
G1 If there are N integers from 1 to 50 that are relatively prime to 50, find the value of N.
(Remark: positive integers a and b are said to be relatively prime if their greatest common divisor is 1.)

We first find the number of positive integers less than or equal to 50 that are not relatively
prime to 50.

They are 2, 4, 6, , 50 (There are 25 multiples of 2). Also, 5, 15, 25, 35, 45 are integers
which are not relatively prime to 50 (There are 5 of them).

 The number of integers which are not relatively prime to 50 is 30.
The number of integers which are relatively prime to 50 is 20.

G2 In Figure 2, ABCD is a trapezium, A B
E T
AB // CD, BCE = ECD, CE  AD and DE = 2AE.
If the area of DEC is 2007 cm2 and the area of
quadrilateral ABCE is T cm2, find the value of T.
Produce DA and CB to meet at F.

Then it is easy to prove that CDE  CFE (ASA) D C
Area of CDF = 22007 cm2 = 4014 cm2
F
Again, FAB ~ FDC (equiangular)
FA : FD = (FE – AE) : 2DE A B
E
= (DE – AE) : 2DE
= (2 AE – AE) : 4AE
=1:4
By the ratio of areas of similar triangles,

Area of ΔABF   1 2 2007cm 2
Area of DCF  4 

Area of ΔABF 1 DC
4014 cm2  16

Area of ABF = 2007 cm2
8

T = 2007 – 2007 = 14049 cm2 (= 1756.125 cm2)
8 8

G3 Given that a2 – 3a + 1 = 0. If A = 2a5  5a4  2a3  8a2  7a , find the value of A.
3a2  3

Reference 1993 HI9, 2000 HG1, 2001 FG2.1, 2009 HG2
By division algorithm, 2a5 – 5a4 + 2a3 – 8a2 + 7a = (a2 – 3a + 1)(2a3 + a2 + 3a) + 4a = 4a
3a2 + 3 = 3(a2 + 1) = 3(3a) = 9a

A = 4a = 4
9a 9

G4 Given that the coordinates of the points A, B and C are (3, 4), C(8, 10)
(6, –4) and (8,10) respectively. M and N are the midpoints of AB 10

and BC respectively. X is a point on AN such that AX : XN = 2 : 1. 8

If r = CX , find the value of r. 6 X
XM
A(3, 4)
M = (4.5, 0), N = (7, 3)
4

X =  2  7  3 , 2  3 4    17 , 10  N
 3 3   3 3 
2

slope of CM = 10  20 , slope of MX = 10  20 M5 1
8  4.5 7 3 7
-2
17  9
3 2

CXM are collinear -4 B(6, -4)

r = CX  y-coordinate of X = 10 = 10  r = 2
XM 3 1 r

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

G5 In Figure 2, a 1 cm  1 cm  2 cm rectangular box is made by two 3 B = 12
cubes with side length 1 cm. An ant is climbing along the box in a
way that it must stay on the edges of the cubes through out the 2
climbing. Starting from vertex A and climbing with a speed of 1 cm
per minutes, it reaches vertex B after 4 minutes. If the total number 1 6
of possible paths taken by the ant is S, find the value of S. 2
1 3
Reference: 1983 FI4.1, 1998 HG6, 2000 HI4
1
From A the ant can only climb upwards, or to the right, or towards
B. 2
Add the numbers on the corners of the box. The numbers shows the
number of possible ways for the ant to climb from A to reach there.
 The number of possible ways to reach B is 12.

A1

G6 If the remainder of 72007 when dividing by 5 is R, find the value of R.
7  5 ....... 2, 72  5 ....... 4, 73  5 ........ 3, 74  5 ........ 1

The remainder repeats as the exponent increases.

2007 = 4501 + 3, the remainder is 3.

Method 2
72 = 49 = 50 – 1
72007 = 772006 = 7(72)1003 = 7(50 – 1)2003 = 7[50m + (–1)2003], where m is an integer.

= 5(70m) – 7 = 5(70m – 2) + 3, the remainder is 3.

G7 Let k = sin 30 + cos 60 + sin 90 +cos 120 +  +sin 1890 +cos 1920, find the value of k.

sin 30 + cos 60 + sin 90 +cos 120 + sin 150 + cos 180 + sin 210 + cos 240 + sin

270 + cos 300 + sin 330 + cos 360

= 1  1  1 1  1 1  1  1 1  1  1 1 = 0
2 2 2 2 2 2 2 2

The cycle repeats for every multiples of 360, and 1800 = 3605

k = sin 1830 + cos 1860 + sin 1890 + cos 1920 = 1  1  1  1 = 3
2 2 2 2

G8 In figure 3, given that the diameter of the semicircle is 10 cm. A, B
and C are three arbitrary points on the semi-circle where B is on the

arc AC. If x is the sum of the length of the line segments AB and
BC, find the greatest possible value of x. (Reference: 2011 HG6)
Let O be the centre, the radius is 5 cm.
Let OD, OE be the respective perpendicular bisectors.

It is easy to prove that COD = BOD, BOE = AOE.

Let COD = , AOE = 

 0 < 2 + 2 < 180

0 <  +  < 90

BD = 5 sin , BE = 5 sin  5 cm C
D
AB + BC = 10 sin  + 10 sin  O 
= 10(sin  + sin ) 10 cm 



= 20 sin    cos   
2 2
5 cm B
sin     sin 45, cos     1, equality holds when  =  = 45 A E
2 2

x  20 sin 45 = 10 2

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Answers: (2006-07 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 11 March 2017

G9 In the coordinate plane, the points A = (–6, 2), B = (–3, 3), C = (0, n) and D = (m, 0) form a

quadrilateral ABCD. Find the value of n so that the perimeter of the quadrilateral ABCD is the

least.

In order that the perimeter is the least, n > 0, m < 0. B(-3, 3) 4
Reflect B(–3, 3) along y-axis to P(3, 3).
Reflect A(–6, 2) along x-axis to Q(–6, –2). A(-6,2) P(3, 3)
2 C(0, n)

By the property of reflection, the perimeter is equal to -5 D(m, 0) O
QD + CD + CP + AB
It is the least when P, C, D, Q are collinear. Q(-6, -2) -2

In this case, slope of CP = slope of PQ.

3  n  3  2  n = 4 .
3 3  6 3

G10 Given that integers x and y satisfying the equation 3x + 5y = 1. If S = x – y and S > 2007, find

the least possible value of S.

32 + 5(–1) = 1, one possible pair solution is (x, y) = (2, –1)

The slope of 3x + 5y = 1 is  3 .
5

 The parametric equation of 3x + 5y = 1 is  x  2  5k , where k is an integer.
 y  1 3k


S = x – y = 2 + 5k – (–1 – 3k) = 3 + 8k

S > 2007  3 + 8k > 2007

k > 250.5

The least possible k = 251

The least possible S = 3 + 8(251) = 2011

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Hong Kong Mathematics Olympiad 2007-2008

Heat Event (Individual)

除非特別聲明，答案須用數字表達，並化至最簡。 時限：40 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: 40 minutes

1. 如圖一，ABC 為一個三角形且 AB  13 cm、BC 14 cm 及 A

AC  15 cm。D 為 AC 上的一點使得 BD  AC。若 CD 比 AD

長 X cm，求 X 的值。 D

In Figure 1, ABC is a triangle, AB = 13 cm, BC = 14 cm and

AC = 15 cm. D is a point on AC such that BD  AC. If CD is

longer than AD by X cm, find the value of X. B

C
圖一 Figure 1

2. 已知梯形 PQRS 的邊長分別為 PQ  6 cm、QR  15 cm、RS  8 cm 及 SP  25 cm，並有
QR // PS。若 PQRS 的面積為 Y cm2，求 Y 的值。

Given that a trapezium PQRS with dimensions PQ = 6 cm, QR = 15 cm, RS = 8 cm and
SP = 25 cm, also QR // PS. If the area of PQRS is Y cm2, find the value of Y.

3. 已知 x0 及 y0 為正整數且滿足方程 11 1 若 35 < y0 < 50 及 x0 + y0 = z0 求 z0 的值。
x y 15

Given that x0 and y0 are positive integers satisfying the equation 1  1  1 . If 35 < y0 < 50
x y 15

and x0 + y0 = z0, find the value of z0.

4. 設 a、b、c 和 d 為實數。若 a 1，b  3，c 4 及 ac = e，求 e 的值。
b2 c2 d5 b2  d 2

Let a, b, c and d be real numbers. If a1, b  3, c 4 and ac = e, find the value
b2 c2 d5 b2  d 2

of e.

5. 如圖二，利用 8 個相同的小長方形能拼出一個大的長 40 cm
圖二 Figure 2
方形。已知在大長方形中較短的邊長為40 cm。若小長
方形的面積是 A cm2，求 A 的值。

In Figure 2, the large rectangle is formed by eight
identical small rectangles. Given that the length of the
shorter side of the large rectangle is 40 cm and the area
of the small rectangle is A cm2, find the value of A.

6. 在圖三中，ABC為等邊三角形。它由多個相同的等邊三角形組 A
成。若圖中共有N個等邊三角形，求N的值。

In Figure 3, ABC is an equilateral triangle. It is formed by several
identical equilateral triangles. If there are altogether N equilateral
triangles in the figure, find the value of N.

BC
圖三 Figure 3

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7. 設 r 為方程 4  5   3 的較大實根。求 r 的值。
y 1 y 5 2

Let r be the larger real root of the equation 4  5   3 . Find the value of r.
y 1 y 5 2

8. 設 x 為有理數及 w  x  2007  x  2007 。求 w 的最小可能值。
2008 2008

Let x be a rational number and w = x  2007  x  2007 . Find the smallest possible value of w.
2008 2008

      9. 設 m 和 n 為正整數。已知表達式  22 2  2   44 4  4 含有 m 個 2 及 n 個 4。

若 k  m ，求 k 的值。
n

Let m and n be a positive integers. Given that the number 2 appears m times and the number 4

     appears n times in the expansion  22 2  2   44 4  4 . If k = m , find the value of k.
   n

10. 求 log2(sin2 45°) + log2(cos2 60°) + log2(tan2 45° 的值。
Find the value of log2(sin2 45°) + log2(cos2 60°) + log2(tan2 45°

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Hong Kong Mathematics Olympiad 2007-2008 時限：20 分鐘

Heat Event (Group)

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. 已知 5  11 的小數部分為 A 及 5  11 的小數部分為 B。設 C = A + B，求 C 的值。

Given that the decimal part of 5 + 11 is A and the decimal part of 5 – 11 is B.
Let C = A + B, find the value of C.

2. 有一批糖共 x 粒，x 為正整數，這批糖能分別為 851 人及 943 人所均分。求 x 的最小可
能值。

A total number of x candies, x is a positive integer, can be evenly distributed to 851 people as
well as 943 people. Find the least possible value of x.

3. 如圖一，正四面體 ABCD 的邊長為 2 cm。若該正四面體 D

的體積是 R cm3，求 R 的值。 C

In Figure 1, ABCD is a tetrahedron with side length of 2 cm.
If the volume of the tetrahedron is R cm3, find the value of
R. A

B
圖一 Figure 1

4. 已知 x 為正整數及 x < 60。若 x 恰有 10 個正因子，求 x 的值。
Given that x is a positive integer and x < 60. If x has exactly 10 positive factors, find the value
of x.

5. 已知 90 <  < 180 及 sin  = 3 。若 A = cos(180 – )，求 A 的值。
2

Given that 90 <  < 180 and sin  = 3 . If A = cos(180 – ), find the value of A.
2

6. 設 x 為正實數，求 x2008  x1004  1 的最小值。
x1004

Let x be a positive real number. Find the minimum value of x 2008  x1004  1 .
x1004

7. 設 x 及 y 為實數，且滿足 yx741333  2008 x  1   5 ， z = x + y，求 z 的值。
 2008 y  3   若
7 
4 5。

 x  1 3  2008 x  1   5
 3   3
Let x and y be real numbers satisfying   2008 y  7   5 . If z = x + y, find the
 3
7   4
y  4 

value of z.

http://www.hkedcity.net/ihouse/fh7878 P.3 HKMO 2008 Heat Event (Group)

8. 設 R 為 135791113151719 被 4 除後的餘數，求 R 的值。
Let R be the remainder of 135791113151719 divided by 4. Find the value of R.

9. 已知 k、 x1 及 x2 為正整數且 x1 < x2。設 A、B 及 C 為曲綫 y = kx2 上的三點，其 x

坐標分別為 x1、x1 及 x2。若 ABC 的面積是 15 平方單位，求所有可能 k 值的總和。

Given that k, x1 and x2 are positive integers with x1 < x2. Let A, B and C be three points on the
curve y = kx2, with x-coordinates being –x1, x1 and x2 respectively. If the area of ABC is 15
square units, find the sum of all possible values of k.

10. 如圖二，ABCD 是長方形紙張並有 AB  4 cm 及 BC  5 cm。 A ED
將該紙張對摺，使 C 點與 A 點重合，得摺痕 EF。若 EF  r cm， C
求 r 的值。
In Figure 2, ABCD is rectangular piece of paper with AB = 4 cm
and BC = 5 cm. Fold the paper by putting point C onto A to create
a crease EF. If EF = r cm, find the value of r.

BF

圖二 Figure 2

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