SI.3 Given that there are C integers that satisfy the equation |x – 2| + |x + 1| = B, find the value of C
Individual Events SI A
15 I1 R
30 I2 a
16 I3 m
3 I4 m
3
3
9 B
3 S 120 b n 9 n
2
4 C
4 T 11 c 36 p 2 p
9 D
8 U 72 d 42 q 1141 q
8 Group Events
17
3 SG z q G2 A G3 A G4 P 540 G1
3
5
13
8
1 R 6 k
1 B
13 R 4018 R
2
4
5
k w C Q S
5
25 46 320
5
3
xyz p D T Q
1 30 –1 5
2
3
2 Sample Individual Event (2008 Final Individual Event 1) SI.1 Let A = 15tan 44tan45tan46, find the value of A.
Similar question 2012 FG2.1
1 A 15 tan 44 1 = 15 tan 44 n 2008 ' s
SI.2
Let n be a positive integer and 20082008 2008 15 is divisible by A. If the least possible value of n is B, find the value of B.
The given number is divisible by 15. Therefore it is divisible by 3 and 5. The last 2 digits of the given number is 15, which is divisible by 15. n 2008 ' s
The necessary condition is: 20082008 2008 must be divisible by 3.
2 + 0 + 0 + 8 = 10 which is not divisible by 3. The least possible n is 3: 2+0+0+8+2+0+0+8+2+0+0+8 = 30 which is divisible by 3.
SI.3
Given that there are C integers that satisfy the equation |x – 2| + |x + 1| = B, find the value of C
Reference: 1994 HG1, 2001 HG9, 2004 FG4.2, 2008 HI8, 2008 FI1.3, 2010 HG6, 2012 FG2.3
|x – 2| + |x + 1| = 3 If x < –1, 2 – x – x – 1 = 3 x = –1 (rejected) If –1 x 2, 2 – x + x + 1 = 3 3 = 3, always true
- –1 x 2 If 2 < x, x – 2 + x + 1 = 3 x = 2 (reject)
- –1 x 2 only x is an integer, x = –1, 0, 1, 2; C = 4
SI.4 In the coordinate plane, the distance from the point (–C, 0) to the straight line y = x is D , find the value of D.
Ax By C The distance from P(x , y ) to the straight line Ax + By + C = 0 is . 2 2 A B
4
4 D The distance from (–4, 0) to x – y = 0 is = 2 2
2 2 8 ; D = 8
2 1
1
Individual Event 1
4
2 I1.1 Let a, b, c and d be the distinct roots of the equation x – 15x + 56 = 0.
2
2
2
2 If R = a + b + c + d , find the value of R.
4
2
2
2 x
- – 15x + 56 = 0 (x – 7)(x – 8) = 0
a = 7 , b =
7 , c = 8 , d =
8
2
2
2
2 R = a + b + c + d = 7 + 7 + 8 + 8 = 30
I1.2 In Figure 1, AD and BE are straight lines with AB = AC and AB // ED. If ABC = R and ADE = S, find the value of S.
ABC = 30 = ACB (base isos. ) BAC = 120 (s sum of ) ADE = 120 (alt. s AB // ED)
S
= 120 S log F 1
2
3 I1.3 Let F = 1 + 2 + 2 + 2 + + 2 and T = , find the value of T.
log
2
121
2
1
2 3 120 121 F = 1 + 2 + 2 + 2 + + 2 = = 2 – 1
2
1
121
log 1 F log
2 T = = 11 log 2 log
2 I1.4 Let f (x) be a function such that f (n) = (n – 1) f (n – 1) and f (n) 0 hold for all integers n 6.
f T If U = , find the value of U.
T
1 f T 3
Reference: 1999 FI5.4 f (n) = (n – 1) f (n – 1) = (n – 1)(n – 2)f (n – 2) = f
11 10 9 8 f
8 U = = = 89 = 72
11 f 1 11 3 10 f
8
Individual Event 2 2009
I2.1 Let [x] be the largest integer not greater than x. If a =
3 2 16 , find the value of a.
1 3 2
1 3
2
2009 2009
3 2 1 a = 3 2 16 = 0 + 16 = 16
I2.2 In the coordinate plane, if the area of the triangle formed by the x-axis, y-axis and the line 3x + ay = 12 is b square units, find the value of b.
3 3x + 16y = 12; x-intercept = 4, y-intercept =
4
1
3
3 Area = b = 4 =
2
4
2
1
3 I2.3 1
Given that x 2 b and x c , find the value of c.
3 x x
Reference: 1990 FI2.2
1
1
2
2 x 1
3 x 2 9 x
11
2
2 x x x
1
1
1
3 2 c x x x
1 3 11 1 = 36
3
2 x x x
I2.4 In Figure 1, = c, = 43, = 59 and = d, find the value of d.
A
BAC = (s in the same seg.) ACD = (s in the same seg.)
D
BAD + BCD = 180 (opp. s cyclic quad.)
c + d + 43 + 59 = 180 d = 180 – 43 – 59 – 36 = 42 ( c = 36) B C
Individual Event 3
4
1 I3.1 Given that a b . If m = a – b, find the value of m. 6
2 3
2
4 6
2
1 3
2
4
1 = 6
2 3
2 6
2 6
2 3
2 3
2
4
6
2
3
2
= 6 2 3
2 = 6 2 3
2
= 6
3
a = 6, b = 3; m = 6 – 3 = 3
I3.2 In figure 1, PQR is a right-angled triangle and RSTU is a
rectangle. Let A, B and C be the areas of the corresponding regions. If A : B = m : 2 and A : C = n : 1, find the value of n.
A : B = 3 : 2, A : C = n : 1 A : B : C = 3n : 2n : 3
Let TS = UR = x, QU = y PTS ~ TQU ~ PQR (equiangular)
Method 2 S PTS : S TQU : S PQR = A : C : (A + B + C) = 3n : 3 : (5n + 3)
2
2
2 x
Let SR = x, PS = z : y : (x + y) = 3n : 3 : (5n + 3)
Join TR which bisects the area
y
1 x y 5 n
3 (1), (2) of the rectangle.
x y n
3 A
3 A
3
B
x
5 n
3 B
2
1 From (2):
1 (3)
2
y3 S z
3
3
TPS
(4) 5 n
3 S x
1
1
TSR
Sub. (1) into (3): n 1
3 PTS ~ TQU (equiangular)
2
3 n 3 5 n
3
n z n
3 n
3 5 n
3 C 1 x
2 A
1
2
n = 3 = 9 (by (4))
3 n 6 n 3 5 n
3 6 n 1 2 n n = 9
I3.3 Let x , x , x , x be real numbers and x x . If (x + x )(x + x ) = (x + x )(x + x ) = n – 10
2
3
4
1
2
1
3
1
4
2
3
2
4
and p = (x
1 + x 3 ) (x 2 + x 3 ) + (x 1 + x 4 ) (x 2 + x 4 ), find the value of p.
Reference:
2002 HI7, 2006HG6, 2014 HG7
2
2 x x x x x x x x x x x x x x
1
1
1
3
1
4
3
4
2
2
3
2
4
3
4
2
2 x x x x x x x x x x
1
2
1
3
2
3
1
4
2
4
(x – x )(x + x + x + x ) = 0 x + x + x + x = 0 (1)
1
2
1
2
3
4
1
2
3
4 p = (x 1 + x 3 ) (x 2 + x 3 ) + (x 1 + x 4 ) (x 2 + x 4 )
= –(x
2 + x 3 )(x 2 + x 4 ) – (x 1 + x 3 )(x 1 + x 4 ) (by (1), x 1 + x 3 = –(x 2 + x 4 ), x 2 + x 4 = –(x 1 + x 3 ))
= 1 + 1 = 2 (given (x
1 + x 3 )(x 1 + x 4 ) = (x
2 + x
3 )(x 2 + x 4 ) = –1)I3.4 The total number of students in a school is a multiple of 7 and not less than 1000. Given that
the same remainder 1 will be obtained when the number of students is divided by p + 1, p + 2 and p + 3. Let q be the least of the possible numbers of students in the school, find q.
p + 1 = 3, p + 2 = 4, p + 3 = 5; HCF = 1, LCM = 60 q
= 60m + 1, where m is an integer. q 1000 and q = 7k, k is an integer. 60m + 1 = 7k 7k – 60m = 1
k
= 43, m = 5 satisfies the equation
k = 43 + 60t; 7k 1000 7(43 + 60t) 1000 t 2 Least q = 7(43 + 602) = 1141
Individual Event 4
2
3
2 I4.1 Given that x x 1 . If m = x x 2 2 , find the value of m.
3
2
3
2
2 m = x x 2 2 = x x x x x 1 3
3 I4.2 In Figure 1, BAC is a right-angled triangle, AB = AC = m cm.
Suppose that the circle with diameter AB intersects the line BC at
2 D
, and the total area of the shaded region is n cm . Find the value of n.
AB = AC = 3 cm; ADB = 90 ( in semi-circle)
1
1
1
2 Shaded area = area of ACD = area of ABC =
3 3 cm
2
2
2
9
n =
4
log
1
1
I4.3 Given that p = 4 n , find the value of p.
2009
2
1
9
p =
4 n 4 = 9
2009
4
2
2
2 I4.4 Let x and y be real numbers satisfying the equation x p y p 1 .
y
2 If k and q is the least possible values of k , find the value of q. x
3
2
2
(x – 3) + (y – 3) = 1 (1)
2
2 Sub. y = k(x – 3) into (1): (x – 3) + [k(x – 3) – 3] = 1
2
2
2
(x – 3) + k (x – 3) – 6k(x – 3) + 9 = 1
2
2
2
2
(1 + k )(x – 3) – 6k(x – 3) + 8 = 0 (1 + k )t – 6kt + 8 = 0; where t = x – 3 For real values of t: = 4[3 k – 8(1 + k )] 0
2 k 8
2 The least possible value of k = q = 8.
Method 2 4
2
2 The circle (x – 3) + (y – 3) = 1 intersects with
the variable line y = k(x – 3) which passes 3.5 through a fixed point (3, 0), where k is the 3 H slope of the line.
From the graph, the extreme points when the 2.5 C B variable line touches the circle at B and C.
H (3, 3) is the centre of the circle. 2 In ABH, ABH = 90 ( in semi-circle) 1.5 Let HAB = , AH = 3, HB = 1, AB = 8
2 2 (Pythagoras' theorem) 1
1 tan = 0.5
8 Slope of AB = tan (90 – ) = 8 O 1 2 3 A(3, 0) 4
2 Least possible k = q = 8
Sample Group Event (2008 Final Group Event 2) GS.1 In Figure 1, BD, FC, GC and FE are straight lines.
If z = a + b + c + d + e + f + g, find the value of z.
a + b + g + BHG = 360 ( s sum of polygon ABHG ) c + f = CJE (ext. of CFJ) c + f + e + d +JHD = 360 ( s sum of polygon JHDE )
H a + b+ g + BHG + c+f+e+d + JHD = 720
J a
+ b + c + d + e + f + g + 180 = 720
z = 540
6
6
6
6
6
6 GS.2 If R is the remainder of 1 + 2 + 3 + 4 + 5 + 6 divided by 7, find the value of R.
6
6
5
4
3
2
2
3
4
5
6 x + y = (x + y)(x – x y + x y – x y + xy – y ) + 2y
6
6
6
6
6
6
6
6
6 + 1 = 7Q
1 + 2; 5 + 2 = 7Q 2 + 22 ; 4 + 3 = 7Q 3 + 23
6
6
2 + 22 + 23 = 2(1 + 64 + 729) = 1588 = 7226 + 6; R = 6
6
6
6
6
6
6
6
6
6
6
6
6 Method 2
1 + 2 + 3 + 4 + 5 + 6 1 + 2 + 3 +(–3) + (–2) + (–1) mod 7
6
6
6
2(1 + 2 + 3 ) 2(1 + 64 + 729) mod 7 k 2(1 + 1 + 1) mod 7 6 mod 7 GS.3 If 14! is divisible by 6 , where k is an integer, find the largest possible value of k. We count the number of factors of 3 in 14!. They are 3, 6, 9, 12. So there are 5 factors of 3.
k = 5
1
1
1
7 GS.4 Let x, y and z be real numbers that satisfy x + = 4, y + = 1 and z + = . y z x
3 Find the value of xyz. (Reference 2010 FG2.2) Method 1 Method 2
1 4 y
1
1 From (1), x = 4 – = x
4
1
y y y
1 y
1
(4)
y
1
2 x 4 y
1 z
y
7
1
7
4 y
1
3 x
3 7 y x
1 z = (5) xy
(1)(2):
1
4
3 4 y
1 z yz
1
1
1
From (2): = 1 – y
x y
3 z z yz
1 x z = (6)
x
Sub. (2) into the eqt.:
3 1 y xyz
1 7 y x
(5) = (6): =
x
Let a = xyz, then
3
4 1 y 3 4 y
1 a
1 28 y 7 3 y 7 y
4
7
1
=
4
y y
(2)(3):
5 1 y
3 4 y
1
3 a
3 3 a
3
3(4y – 1) = (1 – y)(25y – 7)
z
25
1
25
2 z z
(1)(3):
4 4
6
12y – 3 = –25y – 7 + 32y
a 3 a
3
2
25y – 20y + 4 = 0
1
7
2
a
1
1 100 (4)(5)(6):
1 4
(5y – 2) = 0
a a a
3 3
2 y = a
1 7 a
3 4 a 1 100
5 2 a
3
3
2
1
5
3
2 Sub. y = into (6): z = = 2 which reduces to 28a – 53a + 22a + 3 = 0
5
1 5
3
2
(a – 1) (28a + 3) = 0
2
5
3 Sub. y = into (1): x + = 4 x =
a = 1
5
2
2
2
5
3 xyz = = 1
5
3
2 Method 3 (1)(2)(3) – (1) – (2) – (3):
1
1
1
1
1
1
1
28
22
xyz x y z x y z xyz 1
2
x y z xyz x y z
3 3 xyz
xyz = 1
Group Event 1 G1.1 Given some triangles with side lengths a cm, 2 cm and b cm, where a and b are integers and a 2 b. If there are q non-congruent classes of triangles satisfying the above conditions, find the value of q.
When a = 1, possible b = 2 When a = 2, possible b = 2 or 3 q = 3 3 x
4 G1.2 Given that the equation x has k distinct real root(s), find the value of k.
x x
2
2 When x > 0: x – 4 = 3x x – 3x – 4 = 0 (x + 1)(x – 4) = 0 x = 4
2
2 When x < 0: –x – 4 = –3x x – 3x + 4 = 0; D = 9 – 16 < 0 no real roots. k = 1 (There is only one real root.) y x
7 G1.3
Given that x and y are non-zero real numbers satisfying the equations and
12 y x x – y = 7. If w = x + y, find the value of w. x y
7
xy xy = 144 The first equation is equivalent to
12 xy
12
144
2 x 144 x
Sub. y = into x – y = 7: 7 – 7x – 144 = 0 (x + 9)(x – 16) = 0
x x x
= –9 or 16; when x = –9, y = –16 (rejected x is undefined); when x = 16; y = 9
w
= 16 + 9 = 25
x y
7 Method 2 The first equation is equivalent to xy 12 xy = 144 (1) xy
12
x – y = 7 and x + y = w
w w
7
7 x = , y =
2
2
w w
7 7 Sub. these equations into (1): = 144
2
2
2 w – 49 = 576 w =25 y x
7
From the given equation , we know that both x > 0 and y > 0
12 y x
w = x + y = 25 only
1
2 G1.4 Given that x and y are real numbers and x y 1 .
2 Let p = |x| + |y|, find the value of p.
Reference: 2005 FI4.1, 2006 FI4.2, 2011 FI4.3, 2013 FI1.4, 2015 HG4, 2015 FI1.1
1
2 x y
Both and 1 are non-negative numbers.
2 The sum of two non-negative numbers = 0 means each of them is zero
1
1
3
x = , y = 1; p = + 1 =
2
2
2
Group Event 2
5 G2.1 Given tan = , where 180 270. If A = cos + sin , find the value of A.
12
12
5 cos = , sin =
13
13
5
17 A = 12 =
13
13
13 G2.2 Let [x] be the largest integer not greater than x.
If B = 10 10 10 10 , find the value of B.
Reference:
2007 FG2.2 x
3 3 3 3 3 3
Let y =
10 10 10
2 y
= 10 + 10 10 10 = 10 + y
2 y
- – y – 10 = 0 1
41 1
41
y = or (rejected)
2
2
7 1
41
6 41 7
4
2
2 13 .
5 10 10 10 10 14 ; B = 13 G2.3 Let a b = ab + 10. If C = (12) 3, find the value of C.
12 = 2 + 10 = 12; C = 123 = 36 + 10 = 46
G2.4 In the coordinate plane, the area of the region bounded by the following lines is D square units, find the value of D.
L 1 : y – 2 = 0 L 2 : y + 2 = 0
L 3 : 4x + 7y – 10 = 0
L : 4x + 7y + 20 = 0
4 C(-1, 2)
D(-8.5, 2) y = 2 2 4x + 7y - 10 = 0 4x + 7y + 20 = 0
- -5 5 O -2 y = -2 A(-1.5, -2)
B(6, -2)
It is easy to show that the bounded region is a parallelogram ABCD with vertices A(–1.5, –2),
B (6, –2), C(–1, 2), C(–8.5, 2).
The area D = |6 – (–1.5)||2 – (–2)| = 7.54 = 30
Group Event 3 G3.1 Let [x] be the largest integer not greater than x.
2008 80 2009 130 2010 180 If A = , find the value of A.
2008 15 2009 25 2010
35 3 3
2008 4015 Reference: 2008 FGS.4 Calculate the value of . 3 3 2007 4015
Let a = 2009, b = 130, c = 25 2008 80 2009 130 2010 180 a 1 b 50 ab a 1 b 50
2008 15 2009 25 2010 35 a 1 c 10 ac a 1 c 10
ab b
50 a 50 ab ab b 50 a
50 =
ac c
10 a 10 ac ac c 10 a
10 3 ab 100 3 2009 130 100 783610
= 5 d 3 ac 20 3 2009 25 20 150695 where 0 < d < 1; A = 5
G3.2
There are R zeros at the end of
99
9
99
9
1
99 9 , find the value of R.
2009 of 9 ' s 2009 of 9 ' s 2009 of 9 ' s
99
9
99
9
1
99 9 =
1 1 1 1 2
1
s s s 2009 of ' 2009 of ' 2009 of ' s s s
2009 of 9 ' 2009 of 9 ' 2009 of 9 ' 2009 2009 2009
= (10 – 1) (10 – 1) + 210 – 1
4018 2009 2009 4018
= 10 – 210 + 1 + 210 – 1 = 10
R = 4018 G3.3 In Figure 1, a square of side length Q cm is inscribed in a semi-circle of radius 2 cm. Find the value of Q.
2
2 2 Q
Q
+ Q 4 (Pythagoras’ Theorem)
2
2
5Q = 16
4
5 Q = 4
5
5 G3.4 In Figure 2, the sector OAB has radius 4 cm and AOB is a right angle. Let the semi-circle with diameter OB be centred at I with IJ // OA, and IJ intersects the semi-circle at K. If
2 the area of the shaded region is T cm , find the value of T.
(Take = 3)
OI = 2 cm, OJ = 4 cm OI
1 cos IOJ =
OJ
2 IOJ = 60
S BIJ = S – S OIJ sector OBJ
1 1
2
2
= ( 4 2 4 sin 60 )cm
2
3
2
2
= ( 8
2 3 ) cm
3 Shaded area = S BIJ – S BIK 8
1 2
2
=
2 3 2 cm
3
4 5
2
=
2 3 cm
3
T = 5
2
3
Group Event 4 G4.1 Let P be a real number. If
3
2 P 1
2 P 2 , find the value of P.
2
2
3
2 P 2 1
2 P
3
2 P 4
4 1
2 P 1
2 P
4 1 P 2
2
4(1 – 2P) = 1
3 P =
8 G4.2 In Figure 1, let AB, AC and BC be the diameters of the corresponding three semi-circles. If AC = BC = 1 cm
2
and the area of the shaded region is R cm . Find the value of R.
Reference: 1994 HI9 AB =
2
2 Shaded area = R cm = S circle with diameter AC – 2 S segment AC
2
2
1
1
2
1
1
2
R =
1 =
2
2
2
2
2
G4.3
In Figure 2, AC, AD, BD, BE and CF are straight lines. If A +B +C +D = 140 and a+b + c = S, find the value of S. CFD = A + C (ext. of ) AEB = B + D (ext. of )
a = A + AEB = A + B + D (ext. of ) c = D + CFD = D + A + C (ext. of ) b
+ A + D = 180 (s sum of )
a + b + c = A + B + D + 180 – (A + D)
D + A + C + = A + B + C + D + 180
S = a + b + c = 140 +180 = 320
2 G4.4 Let Q = log
2 2 2 1 , find the value of Q.
2 2
1
log 2
3
Q =
log 2
3
log 2
3
=
2 3 log 2 3
2
3
log 2
3
=
1 log 2
3 log 2
3
= = –1 log 2
3
2
3
Method 2 Q = log
2 3
2
3
2
3
1 = log
2
3
2
3
1
= log
2 3 = –1
2
3