Individual Events I1 Given
- 107 See the remark
- 14 See the remark
30
bc a
c = 3600
c
1 2
30
3600 (2) + (3) – (1):
a = 121
a
11 2
30
3600 (1) + (3) – (2):
b = 361
b
19 2
(1) + (2) – (3):
2
3
3
5
1
1
1
2
1
1 c a b c a b
1
1
1
2
1
1
=
3600 361
2
2
2 2014 2011 2014 2015 2014 2015 2014 2013 2014 2013 2011 2014
2
1 x x x x x x
=
2
=
2
2
2
2
4
2
a b b c c a
121 3600 = 121
2
19 I2 Given that a = 2014x + 2011, b = 2014x + 2013 and c = 2014x + 2015.
Find the value of a
2
2
2 – ab – bc – ca.
a
2
1
2
2
2
2
2
1 = 12 As shown in Figure 1, a point T lies in an equilateral
9
13-14 Group
49
5
4
2
4 15 –
144 65
3
23
2
5
1
15
10
8
4
4
1
121
19
2
12
3 120
193
8
5
90
13-14 Individual
6
7
±1
6
29
. Find the value of
bc a
.
5
2
1
3
1
2
1
ca a c bc c b ab b a
a c ca c b bc b a ab
3
(=7.25) 7 –1 8 1584
2
9
3
16
3
10
5 3
5
Individual Events
I1 Given that a , b, c > 0 and
- b
- c
- b
- c
- – ab – bc – ca =
I3
triangle PQR such that TP = 3, TQ =
3 3 and TR = 6. Find PTR. Rotate PTR anticlockwise by 60 to QSR. Then PTR QSR, SR = 6 and SRT = 60 Consider
TRS,
SR = 6 = TR TRS is isosceles.
SRT = 60 RTS = RST = 60 (s sum of isos. )
P
TRS is an equilateral triangle
TS = 6
3
Consider TQS,2
2
2
2
2
2 T QS + QT = 3 + (
3 3 ) = 9 + 27 = 36 = 6 = TS TQS = 90 (converse, Pythagoras’ theorem)
3
3
6
tan TSQ = = 3
3 3
3 TSQ = 60 60
QSR = TSQ + RST = 60 + 60 = 120
R Q
PTR = QSR = 120 (corr. s, PTR QSR)
3
6 S Reference: http://www2.hkedcity.net/citizen_files/aa/gi/fh7878/public_html/Geometry/construction/others/345.pdf
2 I4 Let and be the roots of the quadratic equation x – 14x + 1 = 0. 2 2 Find the value of . 2 2
1
1 2 ; 2 ;
+ 1 = 14 + 1 = 14 + = 14 and = 1
2
2
2
= ( + ) – 2 = 196 – 2 = 194 + 2 2 2 2 2 2 2
14
1 14 1 196
14 196 14 196
14
= = = = 193 2 2
1
1
14 14 196 196 As shown in Figure 2, ABCD is a cyclic quadrilateral,
I5 where AD = 5, DC = 14, BC = 10 and AB = 11.
Find the area of quadrilateral ABCD.
Reference: 2002 HI6
2
2
2 AC = 10 + 11 – 2 1110 cos B ...............(1)
2
2
2 AC = 5 + 14 – 2
514 cos D ...................(2) (1) = (2): 221 – 220 cos B = 221 – 140 cosD (3) B + D = 180 (opp. s, cyclic quad.) cos D = –cos B (3): (220 + 140) cos B = 0 B = 90 = D Area of the cyclic quadrilateral = area of ABC + area of ACD
1
1 = 11 10 5 14 = 90
2
2
2014
2
I6 Let be a positive integer and n < 1000. If (n – 1) is divisible by (n – 1) , n find the maximum value of n.Let p = 2014. p p 1 p 2 p 1 p 2
n
1 n 1 n n n 1 n n n
1
2 = = 2 n
1
n
1 n
1
p 1 p 2 n
1 n 1 n 1 p
= p 1 p 2
n
1
n
1 n 1 p
= 1
n
1 n 1 n
1
p –1 p –2 Clearly n – 1 are factors of n – 1, n – 1, , n – 1. p 1 p 2 n
1 n
1 1 is an integer.
n
1 n
1
p 2014
2 19
53 = = is an integer
n
1 n 1 n
1 The largest value of n – 1 is 2 53 = 106.
i.e. The maximum value of n = 107.
2014 Remark: The original question is Let n be a positive number and n < 1000. If (n – 1) is
2
divisible by (n – 1) , find the maximum value of n. 設 n 為正數，且 n < 1000。 Note that n must be an integer for divisibility question.
3
2 I7 If x + x + x + 1 = 0,
- –2014 –2013 –2012 –1 2 2013 2014 + find the value of x + x + x + x + + 1 + x + x + x + x .
Reference: 1997 FG4.2
2 The given equation can be factorised as (1 + x)(1 + x ) = 0 x = –1 or ±i
- –2014 –2013 –2012 –1 2 2013 2014
- x + x + 1 + x + x + x +
x + x + x
- –2014
2 3 –6
2 3 –2 –1
2
3
2
3
= x (1 + x + x + x ) + + x (1 + x + x + x ) + x + x + 1 + x + x + x (1 + x + x + x )
2011
2
3
+ x (1 + x + x + x ) +
- –2 –1 2 –2
2
3
2
2
= x + x + 1 + x + x = x (1 + x + x + x ) + x = x
2
2 When x = –1, x = 1; when x = ±i, x = –1
I8 Let xy = 10x + y. If xy + yx is a square number, how many numbers of this kind exist? xy + yx = 10x + y + 10y + x = 10(x + y) + x + y = 11(x + y) Clearly x and y are integers ranging from 1 to 9.
2 x + y 18. In order that xy + yx = 11(x + y) is a square number, x + y = 11 (x, y) = (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3) or (9, 2).
There are 8 possible numbers.
I9 Given that x , y and z are positive real numbers such that xyz = 64.
2
2 If S = x + y + z, find the value of S when 4x + 2xy + y + 6z is a minimum.
2
2
2
2
4x + 2xy + y + 6z = 4x – 4xy + y + 6xy + 6z
2
= (2x – y) + 6(xy + z)
6 2 xyz = 96 (A.M. G.M.)
2
2 When 4x + 2xy + y + 6z is a minimum, 2x – y = 0 and xy = z
2
y = 2x, z = 2x xyz = 64
2
4
) = 64 = 16 x(2x)(2x x
x = 2, y = 4, z = 8 S = 2 + 4 + 8 = 14 Remark: The original question is: Given that x, y and z are real numbers such that xyz =64 Note that the steps in inequality fails if xy < 0 and z < 0.
I10 Given that ABC is an acute triangle, where A > B > C.
If x is the minimum of A – B, B – C and 90 – A, find the maximum value of x. In order to attain the maximum value of x, the values of A – B, B – C and 90 – A must be equal.
A – B = B – C = 90 – A 2 B = A + C (1) 2 A = 90 + B (2) A + B + C = 180 (3) (s sum of ) Sub. (1) into (3), 3 B = 180 B = 60 (4) Sub. (4) into (2): 2 A = 90 + 60 A = 75 (5) Sub. (4) and (5) into (1): 2(60 ) = 75 + C C = 45 The maximum value of x = 75 – 60 = 15
Method 2
90 – A x 90 – A + A + B + C 180 + x (s sum of ) B + C 90 + x (1) B – C x (2) ((1) + (2)) 2: B 45 + x (3) A – B x (4) (3) + (4): A 45 + 2x (5) 90 – A x 90 – x A 90 – x A 45 + 2x by (5) 90 – x 45 + 2x 45 3x 15 x The maximum value of x is 15.
Group Events 2 2 2 2 G1 Given that 2014 x 2004 x 2 , find the value of 2014 x 2004 x .
Reference: 1992 FI5.4 2 2 2 2
2014 2004 2014 2004 x x x x
2 2
2 2 2014 x 2004 x 2 2014 2004 x x
2 2
2 2014 x 2004 x 2 2 10 = 2( 2014 x 2004 x ) 2 2
2014 x 2004 x = 5 G2 Figure 1 shows a
ABC, AB = 32, AC = 15 and
BC = x, where x is a positive integer. If there are
points D and E lying on AB and AC respectively such that AD = DE = EC = y, where y is a positive integer. Find the value of x. Let BAC = , AE = 15 – y, y = 1, 2, , 14. Apply triangle inequality on ADE, y + y > 15 – y y > 5 (1) AED = (base s, isos. ) By drawing a perpendicular bisector of AE, 15 y cos =
(2) 2 y Apply cosine formula on ABC,
2
2
2 x = 15 + 32 – 2(15)(32)cos
15 y
2 x = 1249 – 480 by (2) y
2 x = 1729 –
7200 (3)
y
x is a positive integer
2
x is a positive integer 7200 is a positive integer.
y
y is a positive factor of 7200 and y = 6, 7, 8, , 14 by (1) and (3) y = 6, 8, 9, 10 or 12.
2 When y = 6, x = 1729 – 1200 = 529 x = 23, accepted.
2 When y = 8, x = 1729 – 900 = 829, which is not a perfect square, rejected.
2 When y = 9, x = 1729 – 800 = 929, which is not a perfect square, rejected.
2 When y = 10, x = 1729 – 720 = 1009, which is not a perfect square, rejected.
2 When y = 12, x = 1729 – 600 = 1129, which is not a perfect square, rejected.
Conclusion, x = 23
7
3
5 G3 If 180 and cos + sin = , find the value of cos + cos + cos + .
13 Reference: 1992 HI20, 1993 HG10, 1995 HI5, 2007 HI7, 2007 FI1.4
7 cos + sin = (1)
13
49
2
(cos + sin ) = 169
49
2
2
cos + 2 sin cos + sin = 169
49 1 + 2 sin cos = 169
120 2 sin cos = (*) 169
120
- –2 sin cos =
169 289 1 – 2 sin cos = 169
289
2
2
cos – 2 sin cos + sin = 169
289
2
(cos – sin ) = 169
17
17 cos – sin = or
13
13 From (1), sin cos < 0 and 0 180 cos < 0 and sin > 0
17 cos – sin = (2)
13
10 (1) + (2): 2 cos =
13
5 cos =
13
5 cos
65
3
5
13 cos = = + cos + cos + = 2
25 1 cos 144 1
169
G4 As shown in Figure 2, ABCD is a square. P is a point lies in A D
ABCD such that AP = 2 cm, BP = 1 cm and APB = 105.
2
2
2 If CP + DP = x cm , find the value of x.
Reference: 1999 HG10
2 cm
Let CP = c cm, DP = d cm Rotate APB about A anticlockwise through 90 to AQD.
105 P Rotate APB about B clockwise through 90 to CRB.
Join PQ, PR.
1 cm
AQ = AP = 2cm, PAQ = 90, BR = BP = 1cm, PBR = 90B C APQ and BPR are right angled isosceles triangles.
AQP = 45, BRP = 45
Q PQ =
2 2 cm, PR = 2 cm (Pythagoras’ theorem)
DQP = 105 – 45 = 60, CRP = 105 – 45 = 60
A D Apply cosine formula on DQP and CRP. 2 2
2 cm
2
2 DP =
2 2 1
2
1
2 2 cos 60 =
9
2 2 cm
2 2 2
2 CP =
2
2
2
2 2 cos 60 =
6
2 2 cm
2 2 + 9 –
2 2 = 15 –
4
2
x = 6 –
P
1 cm
B C
R
2
2
2
2 G5 If x , y are real numbers and x + 3y = 6x + 7, find the maximum value of x + y .
1
2
2
2
2
2 2 x + 3y = 6x + 7 (x – 3) + 3y = 16 (1) and y = x
6 x 7 (2)
3
2
2 Sub. (2) into x + y :
1
1
1
2
2 2 2 2 2
- y =
3
6 7 =
2
6 7 =
2
3
7
x x x x x x x x
3
3
3
1 2 2 2
1 2
2 2
5 = = = 2 x x
3 1 . 5 2 1 . 5
7 2 x 1 . 5 2 . 5 x 1 . 5
3
3
3
6
2
2 From (1), 3y = 16 – (x – 3)
≥ 0 –4 ≤ x – 3 ≤ 4 –1
≤ x ≤ 7
0.5 ≤ x + 1.5 ≤ 8.5
2
0.25 ≤ (x + 1.5) ≤ 72.25 2 2 289
x 1 .
5
1 ≤ ≤
6
3
6
2 2
5
5
2
2
1 x 1 .
5 + y is 49.
289 = 49 The maximum value of x
≤ ≤
3
6
6
6
G6 As shown in Figure 3, X, Y and Z are points on BC , CA and AB of ABC respectively such that
AZY = BZX, BXZ = CXY and CYX = AYZ. If AB = 10, BC = 6 and CA = 9, find the length of AZ.
Let AZY = , BXZ = and CYX = . ZXY = 180 – 2 (adj. s on st. line) XYZ = 180 – 2 (adj. s on st. line)
C b-tc
YZX = 180 – 2 (adj. s on st. line)
Y
X
ZXY + XYZ + YZX = 180 (s sum of )
180 – 2 + 180 – 2 + 180 – 2 = 180
tc
+ + = 180 (1) In CXY, C + + = 180 (s sum of )
C = 180 – ( + ) = by (1)
A c-tb B tb Z
Similarly, B = , A = AYZ ~ ABC, BXZ ~ BAC, CXY ~ CAB (equiangular) Let BC = a, CA = b, AB = c.
AZ AY
t (corr. sides, ~ ’s), where t is the proportional constant
AC AB AZ AY
t AZ = bt, AY = ct
b c BZ = AB – AZ = c – tb; CY = AC – AY = b – tc BX BZ
(corr. sides, ~ ’s)
BC AB
2 c tb BX c bct
BX = (1)
a c a CX CY
(corr. sides, ~ ’s)
BC AC
2 b tc CX b bct
CX = (2)
a b a BX + CX = BC 2 2 c bct b bct
- = a by (1) and (2)
a a
2
2
2 b + c – 2bct = a 2 2 2 2 2 2 b c a
9
10 6 145
29
AZ = tb = = = = (= 7.25)
2 c 2
10
20
4
Method 2 Join AX, BY, CZ.
C
Let AZY = , BXZ = and CYX = .Y
X
ZXY = 180 – 2 (adj. s on st. line)
XYZ = 180 – 2 (adj. s on st. line)
qp
YZX = 180 – 2 (adj. s on st. line) ZXY + XYZ + YZX = 180 (s sum of )
A B
Z
180 – 2 + 180 – 2 + 180 – 2 = 180 + + = 180 (1) In CXY, C + + = 180 (s sum of ) C = 180 – ( + ) = by (1) Similarly, B = , A = ABXY, BCYZ, CAZX are cyclic quadrilaterals (ext. = int. opp. ) Let XZC = p, YZC = q.
Then XBY = CBY = CZY = q (s in the same segment) XAY = XAC = XZC = p (s in the same segment) But XAY = XBY (s in the same segment) p = q On the straight line AZB, + q + p + = 180 (adj. s on st. line) AZC = BZC = 90 i.e. CZ is an altitude of ABC. 2 2 2 2 2 2
b c a
9
10 6 145
29
By cosine formula, cos A = = = = 2 bc 2 9 10 180
36
29
29 AZ = AC cos A = 9 =
36
4 G7 Given that a , b, c and d are four distinct numbers, where (a +c)(a +d)= 1 and (b +c)(b +d)= 1. Find the value of (a + c)(b + c). (Reference: 2002 HI7, 2006 HG6, 2009 FI3.3)
2
a ac ad cd
1
1
2 b bc bd cd
1
2
2
2
(1) – (2): a – b + (a – b)c + (a – b)d = 0 (a – b)(a + b + c + d) = 0 a – b 0 a + b + c + d = 0 b + c = –(a + d) (a + c)(b + c) = –(a + c)(a + d) = –1
G8 Let a 1 = 215, a 2 = 2014 and a n +2 = 3a n +1 – 2a n , where n is a positive integer.
Find the value of a 2014 – 2a 2013 .
a n +2 = 3a n +1 – 2a n a n +2 – 2a n +1 = a n +1 – 2a n
a 2014 – 2a 2013 = a 2013 – 2a 2012 = a 2012 – 2a 2011 = 2 – 2a
1 = 2014 – 2(215) = 1584
= a
8
2 G9 Given that the minimum value of the function y = sin x – 4 sin x + m is .
3
y Find the minimum value of m .
2
2 y = sin x – 4 sin x + m = (sin x – 2) + m – 4
2 m – 3 (sin x – 2) + m – 4 m + 5
8
m – 3 =
3
1
m =
3
8
16 y
3
3
16
8
8
3
3
1 y
3 3 =
m 1
3
3
16
16
3 y
3
is = 3 . The minimum value of m
1
3
90
G10 Given that tan tan 90 tan 1 and 1 < tan x < 3. Find the value of tan x.
x tan x
90
90
90
90 tan 90 or 90 tan 270 or 90 tan
90
2 1 , m x x x m Z tan x tan x tan x
1
1
1 tan x 1 or tan x 3 or tan x 2 m
1 tan x tan x tan x
2
2
2
tan x – tan x + 1 = 0 or tan x – 3tan x + 1 = 0 or tan x – (2m + 1)tan x + 1 = 0
2
3
5 2 m 1 2 m 1
4
= –3 < 0, no solution or tan x = or
2
2 3 5 3
5 2.6, 0.4
1 < tan x < 3 and
2
2 3 5 only
tan x =
2
Geometrical Construction
1. Figure 1 shows a ABC. Construct a circle with centre O inside the triangle such that the three sides of the triangle are tangents to the circle.
The steps are as follows: (The question is the same as 2009 construction sample paper Q1) (1) Draw the bisector of BAC. B (2) Draw the bisector of ACB. 3 S
T 4 4 O is the intersection of the two angle 1
2
bisectors.O 4 (3) Join BO .
A R
(4) Let R , S, T be the feet of perpendiculars
C from O onto AC, BC and AB respectively. B
AOT AOR (A.A.S.)
S
TCOS COR (A.A.S.)
OT = OR = OS (Corr. sides, 's) 5 O
BOT BOS (R.H.S.)
A
OBT = OBS (Corr. s, 's)
R C BO is the angle bisector of ABC.
The three angle bisectors are concurrent at one point. (5) Using O as centre, OR as radius to draw a circle. This circle touches ABC internally at R, S, and T. It is called the inscribed circle .
2. Figure 2 shows a rectangle PQRS. Construct a square of area equal to that of a rectangle.
Reference: http://www2.hkedcity.net/citizen_files/aa/gi/fh7878/public_html/Geometry/construction/others/rectangle_into_rectangle.pdf
作圖方法如下： 假設該長方形為 PQRS，其中 PQ = a，QR = b。
5 c
(1) 以 R 為圓心，RS 為半徑作一弧，交 QR 的
E P
3 D
延長線於 G。
S
1
(2) 作 QG 的垂直平分線，O 為 QG 的中點。
a
2
(3) 以 O 為圓心，OQ 為半徑作一半圓，交 RS 的延長線於 D，連接 QD、DG。
(4) 以 R 為圓心，RD 為半徑作一弧，交 QR 的
4 F Q
G b O R
延長線於 F。 (5) 以 F 為圓心，FR 為半徑作一弧，以 D 為圓
心，DR 為半徑作一弧，兩弧相交於 E。 (6) 連接 DE、FE。 作圖完畢，證明如下： GDQ = 90 (半圓上的圓周角)
RG = RS = a
DRG ~ QRD (等角)
DR RG (相似三角形三邊成比例)
DR QR
2 DR = ab
(1)
RF = DR = DE = EF (半徑相等)
DRF = 90 (直線上的鄰角) DEFR 便是該正方形，其面積與長方形 PQRS 相等。(由(1)式得知) 證明完畢。
3. Figure 3 shows two line segments AB and AC intersecting at the point A. Construct two
circles of different sizes between them such that (i) They touch each other at a point; and (ii) the lines AB and AC are tangents to both circles.
圖三所示為兩相交於 A 點的綫段 AB 及 AC。試在它們之間構作兩個大小不同的圓使得 (i)
該兩圓相 切 於一點；及 (ii) 綫段 AB 及 AC 均為該圓的切綫。
6 5 1 4 3 2 F H G E B A C P
Steps (Assume that BAC < 180, otherwise we cannot construct the circles touching BAC.) (1) Draw the angle bisector AH of BAC. (2) Choose any point P on AH. Construct a line through P and perpendicular to AH, intersecting
AB and AC at E and F respectively.
(3) Draw the angle bisector EG of AEF, intersecting AH at G. (4) Draw the angle bisector EH of BEF, intersecting AH at H. (5) Use G as radius, GP as radius to draw a circle. (6) Use
H as radius, HP as radius to draw another circle.
The two circles in steps (5) and (6) are the required circles satisfying the conditions. Proof:
G is the incentre of AEF and H is the excentre of AEF The two circles in steps (5) and (6) are the incircle and the excircle satisfying the conditions.
Remark: The question Chinese version and English version have different meaning, so I have
changed it. The original question is: 圖三所示為兩相交於 A 點的綫段 AB 及 AC。試在它們之間構作兩個大小不同的圓使得 (i)
該兩圓相
交 於一點；及
(ii) 綫段 AB 及 AC 均為該圓的切綫。 A suggested solution to the Chinese version is given as follows:
作圖方法如下： (1) 作 AOB 的角平分綫 OU。 (2) 找一點 P 不在角平分綫上，連接 OP。
在角平分綫上任取一點 D。 分別作過 D 且垂直於 OA 及 OB 之綫段，E 和 G 分別為兩垂 (3)
足。 (4) 以 D 為圓心，DE 為半徑作一圓，交 OP 於 C 及 F，其中 OC < OF。
B
V 6 H 2 1 U H 1 S 6 4 5 Q 5 G 3 7 D 5 2 P 5 3 6 F 7 6 C A E R O T
連接 DF，過 P 作一綫段與 DF 平行，交角平分綫於 Q。 (5)
連接 CD，過 P 作一綫段與 CD 平行，交角平分綫於 U。 (6) 分別作過 Q 且垂直於 OA 及 OB 之綫段，R 和 S 分別為兩垂足。
分別作過 U 且垂直於 OA 及 OB 之綫段，T 和 V 分別為兩垂足。 (7) 以 Q 為圓心，QR 為半徑作一圓 H
1 。以 U 為圓心，UT 為半徑作另一圓 H 2 。
作圖完畢。 證明如下： 一如上文分析，步驟 4 的圓分別切 OA 及 OB 於 E 及 G。
(角平分綫) QOR = QOS
OQ = OQ (公共邊)
(由作圖所得) QRO = 90 = QSO QOR QOS (A.A.S.)
QR = QS (全等三角形的對應邊)
圓 H 分別切 OA 及 OB 於 R 及 S。 (切綫
1
半徑的逆定理) (等角)
ODG ~ OQS 及ODF ~ OQP
OQ QP QS 及 OQ (相似三角形的對應邊) DG OD OD DF QP
QS
DG DF
DG = DF QS = QP
經過 P。 圓 H
1
利用相同的方法，可證明圓 H
2 分別切 OA 及 OB 於 T 及 V，及經過 P。
證明完畢。