Answers: (2003-04 HKMO Heat Events) Created by: Mr.

Francis Hung Last updated: 16 November 2016

1 –2009010 2 7 3 45 4 700 5 6
03-04 3 19
Individual 6 12.5 7 6 8  9 12 10
16 4

1 2475 2 1 3 6 4 32 5 5
03-04 1 5
Group 6 500 7 34.56 8 9 10 10
6 3
Individual Events
I1 Let A = 12 – 22 + 32 – 42 + .... + 20032 – 20042, find the value of A.
Reference: 1997 HI5, 2002 FG2.3, 2015 FI3.2, 2015 FG4.1
A = (12 – 22) + (32 – 42) + .... + (20032 – 20042)
= –3–7–11 – .....– 4007, this is an arithmetic series, a= –3, l = –4007=a+(n–1)(–4), n =1002
3  4007
=  1002 = –2009010
2
I2 If 2003 B  2003 , C is the unit digit of B, find the value of C.
B = 20032003; 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243; the unit digit repeats for every
multiples of 4. 20032003 = 20034500+3; the unit digit is 7; C = 7.
I3 If x + y + z = 10, x2 + y2 + z2 = 10 and xy + yz + zx = m, find the value of m.
(x + y + z)2 = 102  x2 + y2 + z2 + 2(xy + yz + zx) = 100  10 + 2m = 100  m = 45
I4 Arrange the natural numbers in the following order. In this arrangement, 9 is in the row 3 and
the column 2. If the number 2003 is in the row x and the column y, find the value of xy.
1 2 4 7 11 16 ...
3 5 8 12 17 ...
6 9 13 18 ...
10 14 19 ...
15 20 ...
21 ...
Reference: 2003 FI1.4
Consider the integers in the first column of each row: 1, 3, 6, 10, ...
They are equivalent to 1, 1 + 2, 1 + 2 + 3, 1 + 2 + 3 + 4, ...
nn  1
The first integer in the nth row = 1+ 2 + 3 + ... + n =
2
nn  1
< 2003  n(n + 1) < 4006
2
 6263 = 3906, 6364 = 4032
 The greatest possible n = 62
3906  2 = 1953
The 63rd element of the first row = 1954
The 62nd element of the second row = 1955, ......... and so on.
2003 = 1953 + 50; 63 – 50 + 1 = 14
The 14th element of the 50th row is 2003; x = 50, y = 14
xy = 5014 = 700
I5 Let E = 12  6 3  12  6 3 , find the value of E.
Reference: 1993 FI1.4, 1999 HG3, 2001 FG2.1, 2011 HI7, 2015 FI4.2, 2015 FG3.1
12  6 3 = 9  3  2 9  3 = a  b  2 ab = a  b = 3  3
12  6 3 = 9  3  2 9  3 = a  b  2 ab = a  b = 3  3
12  6 3  12  6 3 = 3  3  3  3 = 6

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Answers: (2003-04 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 16 November 2016

I6 In the figure, O is the centre of the bigger semicircle

with radius 10 cm, OB is the diameter of the smaller
semicircle and C is the midpoint of arc OB and it lies
on the segment OA. Let the area of the shaded region
be K cm2, find the value of K. (Take  = 3)
Shaded area = area of sector OAB – area of OCB
1  1
= 10 2   10  5 = 12.5
2 4 2

I7 In the figure, let the shaded area formed by the three straight lines
y = –x + 3, y = x + 1 and y = –5x + 19 be R, find the value of R.
Intersection points are A(1, 2), B(3,4), C(4, –1).
CAB = 90
1
Area = 8 18  6 sq.unit
2

 2
I8 If t = sin 4  cos 2 , find the value of t.
6 6
 2
t = sin 4  cos 2
6 6
4 2
1 1
=   
2 2
1 1
= 
16 4
3
=
16
I9 In the figure, C lies on AE, ABC and CDE are
equilateral triangles, F and G are the mid-points of
BC and DE respectively. If the area of ABC is
24 cm2, the area of CDE is 60 cm2, and the area of
AFG is Q cm2, find the value of Q.
FAC = GCE = 30
AF // CG (corr. s eq.)
Area of AFG = Area of ACF = 12 cm2
(They have the same bases AF and the same height)
I10 If  and  are the roots of the quadratic equation 4x2 – 10x + 3 = 0 and k = 2 + 2,
find the value of k.
k = 2 + 2
= ( + )2 – 2
2
5 3
=   2 
2 4
19
=
4

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Answers: (2003-04 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 16 November 2016

Group Events
1 1 2 1 2 3 1 2 3 4  1 2 99 
G1 If x                       ,
2 3 3  4 4 4 5 5 5 5  100 100 100 
find the value of x.
1 2 3 4 99 100
x =     =  99 = 2475
2 2 2 2 2 4
G2 If z is the positive root of the equation 64x – 136x + 69x = 0, find the value of z.
(32x – 23x)(22x – 33x) = 0
32x = 23x or 22x = 33x
1
2x 2 2x 3  2 
 or   
3x 3 3x 2  3 
x = 1 or –1 (rejected)
z = positive root = 1
G3 If there are at most k mutually non-congruent isosceles triangles whose perimeter is 25cm and
the lengths of the three sides are positive integers when expressed in cm, find the value of k.
Possible triangles are {7,7,11}, {8,8,9}, {9,9,7}, {10,10,5}, {11,11,3}, {12,12,1}
k=6
G4 Given that a, b are positive real numbers satisfying a3 = 2004 and b2 = 2004. If the number of
integers x that satisfy the inequality a < x < b is h, find the value of h.
123 = 1728, 442 = 1976
a3 = 2004  12 < a < 13; b2 = 2004  44 < b < 45
a < x < b  12 < x < 45  number of integral values of x = 32
G5 If the sum of R consecutive positive integers is 1000 (where R > 1), find the least value of R.
Let the smallest positive integer be x. (Reference: 2006 HG5)
x + (x + 1) + ... + (x + R – 1) = 1000
 2 x  R  1  1000
R
2
2000
R(2x + R – 1) = 2000  2x + R – 1 = , which is an integer.
R
Possible R are: 1,2,4,5,8,10,16,20,25,40,50,80,100,125,250,400,500,1000,2000.
When R = 4m + 2, where m is an integer.
(4m + 2)(2x + 4m + 1) = 2000  (2m + 1)(2x + 4m + 1) = 1000
L.H.S. is odd, R.H.S. is even  reject 2, 10, 50, 250.
When R = 4m, where m is an integer.
4m(2x + 4m – 1) = 2000  m(2x + 4m – 1) = 500 = 4125  m is a multiple of 4
 R = multiple of 16  reject 4, 8, 20, 40, 100, 500, 1000
2000
2x + R – 1 = > R – 1  2000 > R(R – 1)  2000 > R – 1  45 > R
R
The possible values of R are 1, 5, 16, 25.
When R = 1, 1(2x) = 2000  x = 1000
When R = 5, 5(2x + 4) = 2000  x = 198
When R = 16, 16(2x + 15) = 2000  x = 55
When R = 25, 25(2x + 24) = 2000  x = 28
The least value of R > 1 is 5, x = 198.
198 + 199 + 200 + 201 + 202 = 1000

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Answers: (2003-04 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 16 November 2016

G6 If a, b and c are positive integers such that abc + ab + bc + ac + a + b + c = 2003, find the
least value of abc.
(a + 1)(b + 1)(c + 1) = 2004 = 22  3  167
abc is the least when the difference between a, b and c are the greatest.
a + 1 = 2, b + 1 = 2, c + 1 = 501
a = 1, b = 1, c = 500
abc = 500

G7 In the figure, ABCD is a trapezium, the segments AB and CD are both perpendicular to BC
and the diagonals AC and BD intersect at X. If AB = 9 cm, BC = 12 cm and CD = 16 cm, and
the area of BXC is W cm2, find the value of W.
Reference: 1993 HI2, 1997 HG3, 2000 FI2.2, 2002 FI1.3, 2010HG4, 2013 HG2
ABX ~ CDX
AX : CX = AB : CD = 9 : 16
SABX : SCDX = 92 : 162 = 81 : 256
Let SABX = 81y, SCDX = 256y
Let AX = 9t, CX = 16t ( ABX ~ CDX)
ABX and BCX have the same height.
16t 16
SBCX = SABX  = 81 y   144 y
9t 9
SABC = SABX + SBCX
9  12
= 81y + 144y
2
6
y
25
6
 SBCX = 144y = 144 = 34.56
25

G8 Let y = log1400 2 + log1400 3 5 + log1400 6 7 , find the value of y.

log 2  3 5  6 7 12 log 2  13 log 5  16 log 7 3 log 2  2 log 5  log 7
y = =
log1400 log1400 6 log1400
log 8  log 25  log 7 log2  4  25  7  log1400 1
y= = = =
6 log1400 6 log1400 6 log1400 6

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Answers: (2003-04 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 16 November 2016

G9 In the figure, ABC is an isosceles triangle with AB = AC and ABC = 80. If P is a point on
the AB such that AP = BC, ACP = k, find the value of k.
Reference:《數學教育》第八期（一九九九年六月）, 2010 HG10 A
ACB = 80 = ACB (base s isos. )
BAC = 20 (s sum of )
BPC = (20 + k) (ext.  of APC) P
AP CP
  (1) (sine rule on ACP)
sin k  sin 20
BC CP
  (2) (sine rule on BCP)
sin 20  k  sin 80

80
sin 20  k 

sin 80 cos10 1
(1)  (2):  =  
=
sin k 
sin 20 2 sin 10 cos10 2 sin 10

B C
2 sin(20 + k) sin 10 = sin k
cos(10 + k) – cos(30 + k) = sin k
cos(10 + k) = sin(60 – k) + sin k
cos(10 + k) = 2 sin 30 cos(30 – k)
cos(10 + k) = cos(30 – k)
10 + k = 30 – k
k = 10
Method 2
Rotate A 60 in anti-clockwise direction about P as shown. A
APQ is an equilateral triangle. Join QC.
ACB = 80 = ACB (base s isos. ) Q
BAC = 20 (s sum of )
QAP = 60 = AQP (s of an equilateral ) P
QAC = 60 + 20 = 80 = ACB
QA = AP = BC (given)
AC = AC (common)
 ACB  CAQ (S.A.S.)
AQC = ABC = 80 (corr. s  ’s)
CQP = 80 – 60 = 20 = CAP
CP = CP (common)
80
AP = QP (by construction)
APC  QPC (S.S.S.)
 ACQ = BAC = 20 (corr. s ACB  CAQ) B C
ACP = QCP = 10 (corr. s APC  QPC)

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Answers: (2003-04 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 16 November 2016

G10 Suppose P(a, b) is a point on the straight line x – y + 1 = 0 such that the sum of the distance
between P and the point A(1,0) and the distance between P and the point B(3,0) is the least,
find the value of a + b.
Regard x – y + 1 = 0 as mirror. 4.0
y

C(–1,2) is the mirror image of A(1,0). x-y+1=0

Sum of distance is the least 3.0

 P(a, b) lies on BC. C(-1,2)

2.0
P(a, b) lies on x – y + 1 = 0
b=a+1
P(a,b)
1.0

mPB  mBC
x
a 1 2 -4.0 -3.0 -2.0 -1.0 1.0
A(1,0) 2.0 3.0
B(3,0) 4.0 5.

a 3 4 -1.0

–2a – 2 = a – 3
-2.0
1 4
a , b
3 3 -3.0

5
a+b= -4.0
3

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