Answers: (1999-00 HKMO Heat Events) Created by: Mr.

Francis Hung Last updated: 26 December 2015

170
1 2 3 3 10 4 35 5 540
99-00 891
Individual 1
6 190 7 8 428571 9 24 10 0
3

99-00 1 –3 2 5 3 6 4 10 5 10
Group 6 60 7 0.93 8 421 9 12 10 0
Individual Events
     
I1 Let x  0.17 0.017 0.0017 ... , find the value of x . (Reference: 2009 HI1)
  17   17   17 17 1
0.17 = ; 0.017 = ; 0.0017 = ,  It is an infinite geometric series, a = , r =
99 990 9900 99 10
17 17 17
x=   
99 990 9900
17  1 1 
= 1    
99  10 100 
17 10 170
=  =
99 9 891
I2 Solve the following equation:
1 1 1 1 1 1 1
     
x  12 x  1x  2 x  2 x  3 x  3x  4 x  10x  11 x  11x  12 4
1  1 1   1 1   1 1   1 1   1 1  1
            
x  12  x  1 x  2   x  2 x  3   x  3 x  4   x  10 x  11   x  11 x  12  4
1 1

x 1 4

x=3
I3 Using digits 0, 1, 2, and 5, how many 3-digit numbers can be formed, which are divisible by 5?
(If no digit may be repeated.)
Possible numbers are: 105, 120, 125, 150, 205, 210, 215, 250, 510, 520.
Altogether 10 numbers.
I4 Figure 1 represents a 4  3 rectangular spiderweb. If a North D C
北
spider walks along the web from A to C and it always 

walks either due East or due North. Find the total number A  East 東 B
of possible paths. Figure 1 圖一
D(1) C(35)
Reference: 1983 FI4.1, 1998 HG6, 2007 HG5 4 10 20
The numbers at each of the vertices of in the following
1
3 6 10 15
figure show the number of possible ways.
So the total number of ways = 35 1
2 3 4 5

A(1) 1 1 1 B(1)

http://www.hkedcity.net/ihouse/fh7878 Page 1
Answers: (1999-00 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 26 December 2015

I5 In Figure 2, let A+B+C+D+E+F+G = x, find A

the value of x.
Reference: 1992 HI13, 2012 FG3.2 B P V
G
In the figure, let P, Q, R, S, T, U, V be as shown.
AVP + BPQ + CQR + DRS + EST + FTU+GUV=360 Q U
(sum of ext.  of polygon)
A = 180 – (AVP + BPQ) (s sum of ) C
F
B = 180 – (BPQ + CQR) (s sum of ) R
C = 180 – (CQR + DRS) (s sum of ) T
D = 180 – (DRS + EST) (s sum of ) S
E = 180 – (EST + FTU) (s sum of ) D E
F = 180 – (FTU + GUV) (s sum of )
G = 180 – (GUV + AVP) (s sum of )
A+B+C+D+E+F+G =1807 – 2360
x = 540
I6 Twenty straight lines were drawn on a white paper. Among them, no two or more straight
lines are parallel; also no three or more than three straight lines are concurrent. What is the
maximum number of intersections that these 20 lines can form?
2 lines give at most 1 intersection.
3 lines give at most 3 intersections.
4 lines give at most 6 intersections. (6 = 1 + 2 + 3)
.................................................................................
1  19
20 lines give at most 1 + 2 + 3 +  + 19 intersections =  19 = 190 intersections
2
I7 In a family of 2 children, given that one of them is a girl, what is the probability of having
another girl? (Assuming equal probabilities of boys and girls.)
Sample space = {(girl, boy), (girl, girl), (boy, girl)} and each outcome is equal probable.
1
 P(another child is also a girl) =
3
I8 A particular 6-digit number has a unit-digit “1”. Suppose this unit-digit “1” is moved to the
place of hundred thousands, while the original ten thousand-digit, thousand-digit,
hundred-digit, … are moved one digit place to the right. The value of the new 6-digit number
is one-third of the value of the original 6-digit number. Find the original 6-digit number.
(Reference: 1986 FG8) Let the original number be: abcde 1 , and the new number be: 1abcde .
3 1abcde = abcde1
3(100000 + 10000a + 1000b + 100c + 10d + e) = 100000a+10000b + 1000c + 100d + 10e + 1
Compare the unit digit: e = 7 with carry digit 2 to the tens digit
Compare the tens digit: d = 5 with carry digit 1 to the hundreds digit
Compare the hundreds digit: c = 8 with carry digit 2 to the thousands digit
Compare the thousands digit: b = 2 with no carry digit to the ten-thousands digit
Compare the ten-thousands digit: a = 4 with carry digit 1 to the hundred-thousands digit
The original number is 428571
12 sin 2 48  12 sin 2 42
I9 Find the value of .
sin 330 tan 135  sin 2 48 sin 2 42 tan 180
12 sin 2 48  12 sin 2 42 12 sin 2 48  12 cos 2 48 12
= = =24
sin 330 tan 135  sin 2 48 sin 2 42 tan 180  1  1
   1  sin 48 sin 42  0
2 2 2
 2
I10 Find the shortest distance between the line 3x – y – 4 = 0 and the point (2, 2).
Ax 0  By 0  C 3 2  2  4
d= = =0
A B
2 2
3 2   12
Method 2 Sub. (2, 2) into 3x – y – 4 = 0, LHS = 32 – 2 – 4 = 0 = RHS
 (2, 2) lies on the line, the shortest distance = 0

http://www.hkedcity.net/ihouse/fh7878 Page 2
Answers: (1999-00 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 26 December 2015

Group Events
a 5  3a 4  3a 3  a 2
G1 If a is a root of x2 + 2x + 3 = 0, find the value of .
a2  3
Reference: 1993 HI9, 2001 FG2.1, 2007 HG3, 2009 HG2
Divide (a5 + 3a4 + 3a3 – a2) by (a2 + 2a + 3), quotient = a3 + a2 – 2a, remainder = 6a
 
a 5  3a 4  3a 3  a 2 a 2  2a  3 a 3  a 2  2a  6a 
a2  3
=

a 2  2a  3  2 a 
6a
= = –3
 2a
G2 There are exactly n roots in the equation (cos2  – 1)(2 cos2  – 1) = 0, where 0 <  < 360.
Find the value of n.
1 1
cos  = 1, –1, or  .
2 2

 = 180, 45, 315, 135, 225

n=5
G3 Find the unit digit of 20042006.
41 = 4, 42 = 16, 43 = 64, 44 = 256, 
So the unit digit of 20042006 is 6.
G4 Let x = |y – m| + |y – 10| + |y – m – 10|, where 0 < m < 10 and m  y  10. Find the minimum
value of x.
x = y – m + 10 – y + 10 – y + m = 20 – y  20 – 10 = 10
The minimum = 10
G5 There are 5 balls with labels A, B, C, D, E respectively and there are 5 pockets with labels A,
B, C, D, E respectively. A ball is put into each pocket. Find the number of ways in which
exactly 3 balls have labels that match the labels on the pockets.
First choose any 3 bags out of five bags. Put the balls according to their numbers. The
remaining 2 balls must be put in the wrong order.
The number of ways is 5C3 = 10.
G6 In Figure 1, PQR is an equilateral triangle,
PT = RS; PS, QT meet at M; and QN is
perpendicular to PS at N. Let QMN = x, find
the value of x.
PT = RS (given)
QPT = 60 = PRS ( of an equilateral )
PQ = PR (side of an equilateral )
PQT  RPS (SAS)
PTQ = PSR (corr. s  )
R, S, M, T are concyclic (ext.  = int. opp. )
QMN= x =TRS =60 (ext. , cyclic quad.)
x = 60

http://www.hkedcity.net/ihouse/fh7878 Page 3
Answers: (1999-00 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 26 December 2015

G7 In Figure 2, three equal circles are tangent to P C Q

each other, and inscribed in rectangle PQRS,
QR
find the value of . (Use 3  1.7 and give
SR
the answer correct to 2 decimal places)
O1
Let the radii of the circles be r.
Suppose the 3 circles touch the rectangle at A,
B and C. Join O1O2, O2O3, O1O3, O1C, O2A,
O3B as shown. Then O1O2 = O2O3 = O1O3 = 2r
O1C = O2A = O3B = r
O1O2O3 is an equilateral  O2 O3
QR = O1C + O1O2 sin 60 + O2A
3
= r + 2r  + r = r(2 + 3 )
2 S B
QR   A R
r 2 3 2  1.7 37
SR = 4r, = = =  0.93
SR 4r 4 40
G8 The sum of two positive integers is 29, find the minimum value of the sum of their squares.
Let the two numbers be a and b.
a2 + b2 = a2 + (29 – a)2 = 2a2 – 58a + 841 = 2(a – 14.5)2 + 420.5
 a and b are integers, the minimum is attained when a = 15, b = 14
The minimum value of a2 + b2 = 152 + 142 = 225 + 196 = 421
G9 Let x = 3  3 and y = 3  3 , find the value of x2(1 + y2) + y2.
  
x2(1 + y2) + y2 = 3  3 1  3  3  3  3
= 3  3 4  3   3  3
= 12  4 3  3 3  3  3  3 = 12
Method 2
x2(1 + y2) + y2 = (x2 + 1)(y2 + 1) – 1
 
= 3  3 1 3  3 1 1 
= 16 – 3 – 1 = 12
G10 There are nine balls in a pocket, each one having an integer label from 1 to 9. A draws a ball
randomly from the pocket and puts it back, then B draws a ball randomly from the same
pocket. Let n be the unit digit of the sum of numbers on the two balls drawn by A and B, and
P(n) be the probability of the occurrence of n. Find the value of n such that P(n) is the
maximum.
P(1) = P((2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2))
P(2) = P((1,1), (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3))
P(3) = P((1,2), (2,1), (4,9), (5,8), (6,7), (7,6), (8,5), (9,4))
P(4) = P((1,3), (2,2), (3,1), (5,9), (6,8), (7,7), (8,6), (9,5))
P(5) = P((1,4), (2,3), (3,2), (4,1), (6,9), (7,8), (8,7), (9,6))
P(6) = P((1,5), (2,4), (3,3), (4,2), (5,1), (7,9), (8,8), (9,7))
P(7) = P((1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (8,9), (9,8))
P(8) = P((1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (9,9))
P(9) = P((1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1))
P(0) = P((1,9), (2,8), (3,7), (4,6), (5,5), (6,4), (7,3), (8,2), (9,1))
 When n = 0, P(n) is a maximum.

http://www.hkedcity.net/ihouse/fh7878 Page 4