Individual Event 1 I1.1 Determine the area of the shaded region, , in the figure. (Reference: 2011 FG4.4)
Individual Events
4 I1
5 I2
I3
11 I4
6
3 55 24 45
5
6
3
45
7.5
33
16 7 81
64 Group Events
1
11 G1 area
48 G2 Product G3 Product G4 PZ
1.6
80
20 3 2 2 3 S +S +S x y +2x y +xy minimum
6 17 33 50
1 Sum
1
5 remainder
0 Day
5 15 d
2
1 a
30 5 product
2
100 10100
Individual Event 1 I1.1 Determine the area of the shaded region, , in the figure. (Reference: 2011 FG4.4) x Label the unmarked regions by x and y respectively.
1 3 + + y = area of //-gram = 4 + + x
2
y
y = x + 1 (1) 1+x+3+3+
+y+4 +1 = area of //-gram = 2(4 + + x) 12 + x + y + = 8 + 2 + 2x (2) Sub. (1) into (2): 12 + x + x + 1 + = 8 + 2 + 2x = 5
I1.2 If the average of 10 distinct positive integers is 2 , what is the largest possible value of the
largest integer, , of the ten integers? Let the 10 distinct positive integers be 0 < x
1 < x 2 < < x 10 , in ascending order. x x x 1 2 10
2 5
10
10 1 + x 2 + x
- x +
9 = 100
If is the largest possible, then x
1 , x 2 , , x 9 must be as small as possible.
The least possible x
1 , x 2 , , x 9 are 1, 2, 3, , 9.
The largest possible = 100 – (1 + 2 + + 9) = 100 – 45 = 55
I1.3 Given that 1, 3, 5, 7, , and 1, 6, 11, 16, , + 1 are two finite sequences of positive integers. Determine , the numbers of positive integers common to both sequences.
The two finite sequences are: 1, 3, 5, 7, , 55 and 1, 6, 11, 16, , 56. The terms common to both sequences are 1, 11, 21, 31, 41, 51. = 6
I1.4 If log 2 a + log 2 b , determine the smallest positive value for a + b.
log a + log b 6
2
2
6 ab 2 = 64 2 a + b = a b
2 ab 2 64 = 16
The smallest positive value of = 16
Individual Event 2
I2.1 Determine the positive real root, , of x x x x x . 2 x x x x x
2 x x 2 2 x x x x x x 2 x x
2
2 x = 4(x – x)
2
3x = 4x
4
x = 0 (rejected) or
3
4 Check: When x = ,
3
4
4
4
4 4
2
3 4
2
3 3
1 3
1
4 L.H.S. = = = = = R.H.S.
3
3
3
3
3
3
3
3
3
4 =
3 I2.2 In the figure, two circles of radii 4 with their centres
4 placed apart by . Determine the area , of the shaded region.
Let the centres of circles be A and B as shown. AB = 3 Suppose the two circles touches the two given line segments at E, F, G, H as shown. Then EFGH is a rectangle with FE = AB = GH = 3, EH = FG = 8
E F
= Area of semi-circle FIG + area of rectangle EFGH
- – area of semi-circle EDH = Area of rectangle EFGH
=
3 8 = 24
D
I B C A G H
I2.3 Determine the smallest positive integer such that the equation x
4 2 has an integer solution in x. x 24
4
2 x
2 6
4
2 The smallest positive integer = 3
I2.4 Determine the unit digit, , of .
3 3 3 9 3
27 3 =
3 = 3
2
3
4 The unit digit of 3, 3 , 3 , 3 are 3, 9, 7, 1 respectively.
This pattern repeats for every multiples of 4. 27 = 6 4 + 3 = 7
Individual Event 3 11 1 11 2 11 3 11
, , ,
I3.1 If the product of numbers in the sequence
10
10 10 , 10 is 1 000 000, determine the value of the positive integer . 1 2 3 11 11 11 11
6 10 10 10 10 = 10
1
2 3
6
11
11
11
11
1
1
66
2
2
– 132 = 0 + (
– 11)( + 12) = 0 = 11
.
I3.2 Determine the value of if
1 2
3 2 3
4 8 9
10 Reference: 2003 HG1
1 1 r 2 r
1
2
r r
1 r 1 r 2 r r 1 r 2 r r 1 r 2
1
1
1 Put r = 1,
2 1
2 2
3 1 2
3
1
1
1 Put r = 2,
2 2
3 3
4 2 3
4 .........................................................
1
1
1 Put r = 8, 2 8 9 9
10 8 9
10 Add these equations together and multiply both sides by and divide by 2:
1
1
= = 11
2
2 9
10
1
2
3
2
3
4
8
9
10
= 45
I3.3 In the figure, triangle ABC has
ABC = 2, AB = AD and CB = CE. If = DBE, determine the value of . Let ABE = x ABC = 90 CBE = 90 – x ADB = x + (base s isos. ) CEB = CBE = 90 – x (base s isos. ) In BDE, + x + + 90 – x = 180 (s sum of ) = 45
I3.4 For the sequence 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, , determine the sum of the first terms.
= 1 + 2 + 1 + 2 + 2 + 1 + 2 + 2 + 2 + 1 + 2 + 2 + 2 + 2 + 1 + 2 ++ 2 + (5 terms) + 1 + 2 + + 2 (6 terms) + 1 + 2 ++ 2 (7 terms) + 1 + 2 + + 2 (8 terms) + 1 = 9 + 2 36 = 81
Individual Event 4
6
3 I4.1 If 3 6 , determine the value of .
3 2
2
3 Reference: 1989 FG10.1
6
3
3 2
2
3 3
6
3 2
2
3
3 2
2
3
3 2
2
3
6 3 3
6 18
12
3 6 6 3
6 = 6
n
I4.2 Consider fractions of the form , where n is a positive integer. If 1 is subtracted from n
1 both the numerator and the denominator, and the resultant fraction remains positive and is
7
n
1
6
n
7 7n – 7 < 6n and n > 1 1 < n < 7 Possible n = 2, 3, 4, 5, 6 = 5
I4.3 The perimeters of an equilateral triangle and a regular hexagon are equal. If the area of the triangle is square units, determine the area, , of the hexagon in square units.
Reference: 1996 FI1.1, 2016 FI2.1
Let the length of the equilateral triangle be x, and that of the regular hexagon be y. Since they have equal perimeter, 3x = 6y x = 2y The hexagon can be divided into 6 identical
y equilateral triangles. x
1
1
2
2
Ratio of areas = x sin 60 : 6 y sin 60 = 2 : 3
2
2 = 5
3 = 5 = 7.5
2
3
5
9
17
33
65 I4.4 Determine the value of = .
2
4
8
16
32
64
3
5
9
17
33
65
1 =
7
2
4
8
16
32
64
2
1
1
1
1
1
1
1 =
1
2
4
8
16
32
64
2
33 =
64
Group Event 1 G1.1 If an isosceles triangle has height 8 from the base, not the legs, and perimeters 32, determine the area of the triangle. A
Let the isosceles triangle be ABC with AB = AC = y
D is the midpoint of the base BC, AD BC, AD = 8
Let BD = DC = x
y
8 yPerimeter = 2x + 2y = 32 y = 16 – x (1) B C
2
2
2 D x x x + 8 = y
(2)
2
2 Sub. (1) into (2): x + 64 = 256 – 32x + x x = 6, y = 10
Area of the triangle = 48 6
1
1 6
x x
6
x x
G1.2 If f(x) = where x is a positive real number, determine the minimum 3
1
1 3
x x
3
x x
value of f(x).
Reference: 1979 American High School Mathematics Examination Q29 3 2 2
1
1 3 x x
3 x x
f(x) = 3
1 3
1 x x 3 x x
3
1
1 3
= x x
3
x x
1
= 3 x
x
2 1 f(x) =
3 x 6 6
x
G1.3 Determine the remainder of the 81-digit integer 111
1 divided by 81.
80
79
- 11 1 = 10 + 10
+ 10 + 1 81 digits
80
79
= (10 – 1) + (10 – 1) + + (10 – 1) + 81
=
99
9
99
9
+ + + 9 + 81 80 digits 79 digits
=
9 11 1 11 1
1
81 80 digits 79 digits
Let x =
11 1 11 1
1 80 + 79 + + 1 (mod 9) 80 digits 79 digits
81
x 940 0 mod 9
80
2
x = 9m for some integer m
11 1 = 9 9m + 81 0 (mod 81) 81 digits
The remainder = 0
G1.4 Given a sequence of real numbers a 1 , a
2 , a 3 , that satisfy
1 1) a
1 = , and
2
2 1 + a 2 + a k = k a k , for k 2.
- 2) a Determine the value of a 100 .
Reference: 2013 HG10
1
2
1
1
- a
2 = 2 a 2 a 2 =
2
2
3
6
1
1
2
1
1
- a +
3 = 3 a 3 a 3 =
2
6 3
4
12
1 Claim: a n = for n 1
1
n n Pf: By M.I., n = 1, 2, 3, proved already.
1 Suppose a k = for some k 1
k k
1
1
1
1
2
+ = (k + 1) a + + a + k +1 k +1
2 6 k k
1
1
1
1
1
1 2
1 k 2 k a
k 1
2
2 3 k k
1
1
1
1
a k +1 =
1 The statement is true for all n 1
k k
2 k 1 k 1 k
2
1
1
a 100 = =
100 101 10100
Group Event 2
1
1
1
1
1
1 G2.1 By removing certain terms from the sum, , we can get 1. What is the
2
4
6
8
10
12 product of the removed term(s)?
1
1
1
1
3
1
1
2
4
6
12
4
4
1
1 The removed terms are ; .
8
10
1 Product =
80
n –1 G2.2 If S = 1 – 2 + 3 – 4 + n, where n is a positive integer, n + (–1)
determine the value of S
17 + S 33 + S 50 .
If n = 2m, where m is a positive integer,
S 2 m = (1 – 2) + (3 – 4) + + (2m – 1 – 2m) = –m S 2 m +1 = –m + 2m + 1 = m + 1
S + S + S = 9 + 17 – 25 = 1
17
33
50 G2.3 Six persons A, B, C, D, E and F are to rotate for night shifts in alphabetical order with A
serving on the first Sunday, B on the first Monday and so on. In the fiftieth week, which day does A serve on? (Represent Sunday by 0, Monday by 1, , Saturday by 6 in your answer.) 50 7 = 350 = 658 + 2
B serves on Saturday in the fiftieth week. A serves on Friday in the fiftieth week.
Answer 5.
G2.4 In the figure, vertices of equilateral triangle ABC are connected to D in straight line segments with AB = AD.
If BDC = , determine the value of .
Reference: 2003 HG8, 2011 HG9
Use A as centre, AB as radius to draw a circle to pass through B, C, D.
ce
BAC = 2BDC ( at centre twice at ☉ ) 60 = 2 = 30
Group Event 3
1
1
1
G3.1 Determine the value of the product
1
1 1 . 2 2 2
2
3
10
Reference: 1986 FG10.4, 1999 FIS.4
1
1
1
1
1
1
1
1
1
1
1 1 =
1
1
1
1
1
1 2 2 2
2
3
10
2
3
10
2
3
10
1
2
3
9
3
4
11
11
11
= = 1 =
2
3
4
10
2
3
10
10
2
20
1
1
1
1 G3.2 Determine the value of the sum ,
log 100 ! log 100 ! log 100 ! log 100 ! 2 3 4 100 where 100! = 100 9998321.
1
1
1
1
log 2 100 ! log 3 100 ! log 4 100 ! log 100 100 ! log 2 log 3 log 4 log 100
=
log 100 ! log 100 ! log 100 ! log 100 !
log 100 !
= = 1
log 100 ! G3.3 In the figure, ABCD is a square, ADE is an
equilateral triangle and E is a point outside of the square ABCD. If AEB = , determine the value of . (Reference: 1991 FI1.1)
180 90
60 (
= s sum of isos. )
2 = 15
G3.4 Fill the white squares in the figure with distinct non-zero
digits so that the arithmetical expressions, read both horizontally and vertically, are correct. What is ? = 5
6 2 = 3
- 1 + 4 = 5
= =
7
8
Group Event 4 G4.1 In the figure below, ABCD is a square of side length 2. A
circular arc with centre at A is drawn from B to D. A semicircle with centre at M, the midpoint of CD, is drawn from C to D and sits inside the square. Determine the shortest distance from P, the intersection of the two arcs, to side AD, that is, the length of PZ. Join AP, DP, MP. Let F be the foot of perpendicular from P to CD. Let AZ = x. Then DZ = 2 – x = PF, DM = MP = 1
2
2
2 In AZP, AZ + ZP = AP (Pythagoras’ theorem)
2
2 ZP = 4 – x (1)
2
2
2 A B
In + PF = PM (Pythagoras’ theorem) PMF, MF
2
2
2 MF = 1 – (2 – x) = 4x – x – 3 (2)
PZ = DF = 1 + MF
2 x
2 2
2
4 – x = (1 +
4 x x 3 )
2 2
2
4 – x = 1 + 2
4 x x 3 + 4x – x – 3 Z 2 P
3 – 2x =
4 x x
3
2
2 2 - x
9 – 12x + 4x = 4x – x – 3
2
1
5x – 16x + 12 = 0 (5x – 6)(x – 2) = 0
D C
1
M F6 x = or 2 (rejected)
5 2 2 PZ = = = = 1.6 4 x 4 1 .
2 2 .
56 Method 2 Let D be the origin, DC be the x-axis, DA be the y-axis.
2
2
2
2 Equation of circle DPC: (x – 1) + y = 1 x – 2x + y = 0 (1)
2
2
2
2
2 Equation of circle BPD: x + (y – 2) = 2 x + y – 4y = 0 (2) x
(1) – (2) y = (3)
2 2 2 x
Sub. (3) into (1): x = 0 (rejected) or 1.6 PZ = 1.6
x 2 x
4
5
1 5
1
3
2
2
3 G4.2 If x = and y = ，determine the value of x y + 2x y + xy .
2
2 5
1
xy = = 1
4
1
3
2
2
3
2
2
2
2 x y + 2x y + xy = xy(x + 2xy + y ) = x + y + 2 =
5 1 5 1 2 = 5
4
a a b c d G4.3 If a, b, c and d are distinct digits and d a a b c , determine the value of d.
2
1 4 d Consider the unit digit subtraction, c = 0 and there is no borrow digit in the tens digit.
Consider the tens digit, 10 + 0 – b = 4 b = 6 and there is a borrow digit in the hundreds. Consider the hundreds digit, 5 – a = 1 a = 4 and there is no borrow digit in the thousands.
44602 Consider the ten thousands digit, 4 – d = 2
d = 2; Check: 24460 2 0142
4
4 G4.4 Determine the product of all real roots of the equation x + (x – 4) = 32.
4
4 Let t = x – 2, then the equation becomes (t + 2) + (t – 2) = 32
4
2
2(t + 24t + 16) = 32
4
2 t + 24t = 0
2 t = 0 or –24 (rejected) t = 0
x = 2 Product of all real roots = 2