hongkong olympiad

Hong Kong Mathematics Olympiad 2011 – 2012

Heat Event (Geometric Construction)

香港數學競賽 2011 – 2012

初賽(幾何作圖)

每隊必須列出詳細所有步驟(包括作圖步驟)。 時限：20 分鐘

All working (including geometric drawing) must be clearly shown.

此部份滿分為十分。The full marks of this part is 10 marks. Time allowed: 20 minutes

School Code: _________________________________________________________

School Name: ________________________________________________________

Question No. 3 第三題

Figure 2 shows a triangle PQR. Construct a line MN parallel to QR so that

(i) M and N lie on PQ and PR respectively; and

(ii) the area of PMN = 1  the area of PQR.
2

圖二所示為一三角形 PQR。試構作一綫段 MN 平行於 QR 使得

(i) M 及 N 分別位於 PQ 及 PR 上;且

(ii) PMN 的面積= 1 PQR 的面積。
2

Figure 2
圖二

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.5 HKMO 2012 Heat Event (Geometric Construction)

Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

1 9 2 504510 3 8 4 23 5 8
11-12
Individual 6 293 (= 8 21 ) 7 68 4 9 x = 13, y = 2 10 2041 (= 811265 = 81.64)
34 34 25

11-12 1 6 2 2037 3 2 + 21006 4 2 503 1 5 2012
Group 6 16 7 10 *1248 9see the remark 180 10 5

Individual Events

I1 Find the value of the unit digit of 22 + 32 + 42 +  + 201220122. (Reference: 1996 HG10)
12 + 22 + 32 +  + 102  1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0 (mod 10)

 5 (mod 10)

22 + 32 + 42 +  + 201220122

 (12 ++ 102) +  + (201220012 ++ 201220102) + 201220112 + 201220122 –12 (mod 10)

 52012201 + 1 + 4 – 1 (mod 10)

 9 (mod 10)

I2 Given that a, b and c are positive even integers which satisfy the equation a + b + c = 2012.

How many solutions does the equation have?

Reference: 2001 HG2, 2006 HI6, 2010 HI1
Let a = 2p, b = 2q, c = 2r, where p, q, r are positive integers.
a + b + c = 2012  2(p + q + r) = 2012  p + q + r = 1006
The question is equivalent to find the number of ways to put 1006 identical balls into 3
different boxes, and each box must contain at least one ball.
Align the 1006 balls in a row. There are 1005 gaps between these balls. Put 2 sticks into three
of these gaps, so as to divide the balls into 3 groups.
The following diagrams show one possible division.

  |           | 
The three boxes contain 2 balls, 1003 balls and 1 ball. p = 2, q = 1003, r = 1.
The number of ways is equivalent to the number of choosing 2 gaps as sticks from 1005 gaps.

The number of ways is C1005 = 1005 1004 = 504510.
2 2

I3 In Figure 1, ABCD is a square. The coordinates of B and

D are (5, –1) and (–3, 3) respectively. If A(a, b) lies in the

first quadrant, find the value of a + b.

Mid-point of BD = M(1, 1)
MB2 = MA2  (a – 1)2 + (b – 1)2 = (5 – 1)2 + (1+ 1)2 = 20
a2 + b2 – 2a – 2b – 18 = 0  (1)

MA  MB  b 1  11  1
a 1 5 1

b = 2a – 1  (2)
Sub. (2) into (1): a2 + (2a – 1)2 – 2a – 2(2a – 1) – 18 = 0
5a2 – 10a – 15 = 0  a2 – 2a – 3 = 0  (a –3)(a + 1) = 0

a = 3 or –1(rejected)
b = 2(3) – 1 = 5
a+b=3+5=8

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

Method 2 mBD = 3  1 =  1 ; mAD = b  3
2 a  3
35

 BDA = 45  tan 45 = 1

= b 3 = mBD  tan 45 =  1 1 = 1
a 3 1  mBD tan 45  2 3
 mAD
1  1 1
2

3b – 9 = a + 3  3b – a = 12  (1)

Mid-point of BD = M(1, 1)  b = 2a – 1  (2) (similar to method 1)

Solving (1) and (2) gives a = 3, b = 5  a + b = 8

Method 3 by Mr. Jimmy Pang from Lee Shing Pik College y y'
Mid-point of BD = M(1, 1) A = A'

Translate the coordinate system by x’ = x – 1, y’ = y – 1 D(-3, 3) = D'
The new coordinate of M is M’ = (1 – 1,1 – 1) = (0, 0)
The new coordinate of B is B’ = (5 – 1, –1 – 1) = (4, –2) M(1, 1)=M'(0,0) x'
Rotate B’ about M’ in anticlockwise direction through 90 O x
The new coordinate of A’ = (2, 4)
B(5, -1) = B'

Translate the coordinate system by x = x’ + 1, y = y’ + 1 C = C'

The old coordinate of A = (2 + 1, 4 + 1) = (3, 5) = (a, b)

a+b=8
I4 Find the number of places of the number 2202512. (Reference: 1982 FG10.1, 1992 HI17)

2202512 = 220524 = 102054 = 6251020

The number of places = 23

I5 Given that log4 N =1  1  1  1  , find the value of N. (Reference: 1994 HI1)
3 9 27

log4 N =1 1  1  1  = 1 1 = 3 (sum to infinity of a geometric series, a = 1, r = 1 .)
3 9 27 1 3 2 3

N = 432 = 8
I6 Given that a and b are distinct prime numbers, a2 – 19a + m = 0 and b2 – 19b + m = 0. Find

the value of a  b . (Reference: 1996 HG8, 1996FG7.1, 2001 FG4.4, 2005 FG1.2)
ba

a and b are prime distinct roots of x2 – 19x + m = 0

a + b = sum of roots = 19 (odd)

 a and b are prime number and all prime number except 2, are odd.

 a = 2, b = 17 (or a = 17, b = 2)

a  b = 17  2 = 293 (= 8 21 )
b a 2 17 34 34

I7 Given that a, b and c are positive numbers, and a + b + c = 9. Suppose the maximum value

among a + b, a + c and b + c is P, find the minimum value of P.

WLOG assume that a + b = P, a + c  P, c + a  P.

18 = 2(a + b + c) = (a + b) + (b + c) + (c + a)  3P

6P

The minimum value of P is 6.

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

I8 If the quadratic equation (k2 – 4)x2 – (14k + 4)x + 48 = 0 has two distinct positive integral

roots, find the value(s) of k.
Clearly k2 – 4  0; otherwise, the equation cannot have two real roots.

Let the roots be , .
 = (14k + 4)2 – 4(48) (k2 – 4) = 22[(7k + 2)2 – 48k2 + 192] = 22(k2 + 28k + 196) = [2(k + 14)]2

= 14k  4 2k  142 = 7k 2 k 14 = 8k  16 = 8 , = 6k 12 = 6 .
2k k2 4 k2 4  k2 4 
2 4
  k 2 k 2

For positive integral roots, k – 2 is a positive factor of 8 and k + 2 is a positive factor of 6.

k – 2 = 1, 2, 4, 8 and k + 2 = 1, 2, 3, 6

k = 3, 4, 6, 10 and k = –1, 0, 1, 4

 k = 4 only

Method 2 provided by Mr. Jimmy Pang from Po Leung Kuk Lee Shing Pik College

The quadratic equation can be factorised as: [(k – 2)x – 8][(k + 2)x – 6] = 0

 k  2 and k  –2  x = k 8 2 or 6
 k2

By similar argument as before, for positive integral root, k = 4 only.
I9 Given that x, y are positive integers and x > y, solve x3 = 2189 + y3.

x3 – y3 = (x – y)(x2 + xy + y2) = 2189 = 11199 and both 11 and 199 are primes.
x2 + xy + y2 = (x – y)2 + 3xy

x–y (x – y)2 + 3xy xy xy

1 2189 = 1 + 3xy 729.33 (rejected)

11 199 = 121 + 3xy 26 13 2

199 11 = 1992 + 3xy – (rejected)
2189 1 = 21892 + 3xy – (rejected)

 x = 13, y = 2

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

I10 In figure 2, AE = 14, EB = 7, AC = 29 and BD = DC = 10.
Find the value of BF2.

Reference: 2005 HI5, 2009 HG8
AB = 14 + 7 = 21, BC = 10 + 10 = 20
AB2 + BC2 = 212 + 202 = 841 = 292 = AC2

 ABC = 90 (converse, Pythagoras’ theorem)

Let BF = a, CBF = , ABF = 90 – 

Area of BEF + area of BCF = area of BCE

1  20 a sin   1 a  7 cos   20  7
2 2 2

20a sin  + 7a cos  = 140  (1)

Area of BDF + area of ABF = area of ABD

1  21 a cos   1  a 10 sin   10  21
2 2 2

21a cos  + 10a sin  = 210  (2)

2(2) – (1): 35 a cos  = 280

a cos  = 8  (3)

3(1) – (2): 50 a sin  = 210

a sin  = 21  (4)
5

(3)2 + (4)2: BF2 = a2 = 82 +  21 2 = 2041 (= 811265 = 81.64)
 5  25

Method 2 ABC = 90 (similar to method 1) A
Regard B as the origin, BC as the x-axis, BA as the 20

y-axis, then d = 10i, c = 20i, e = 7j, a = 21j 15
Suppose F divides AD in the ratio p and 1 – p.
Also, F divides EC in the ratio t : 1 – t.

f = p d + (1 – p) a = 10pi + 21(1 – p)j  (1)

f = t c + (1 – t) e = 20ti + 7(1 – t)j  (2) p 29

Compare coefficients: 10 F
1-p1-t
10p = 20t and 21(1 – p) = 7(1 – t) E D10

 p = 2t  (3) and 3(1 – p) = 1 – t  (4) 5 t

3(3) + (4): 3 = 5t + 1  t = 2 B
5

BF2 = | f |2 = |20( 2 )i + 7(1 – 2 )j|2 C
5 5
20

= |8i + 21 )j|2 = 82 +  21 2 = 2041
5  5  25

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

Method 3 ABC = 90 (converse, Pythagoras’ theorem, similar to method 1)
Regard B as the origin, BC as the x-axis, BA as the y-axis, then

Equation of AD: x  y  1  (1); equation of EC: x  y  1  (2)
10 21 20 7

2(2) – (1): 5y  1  y = 21 ; 3(1) – (2): x  2  x = 8
21 5 4

BF2 = x2 + y2 = 82 +  21 2 = 2041 (= 811265 = 81.64)
 5  25

Method 4 ABC = 90 (similar to method 1) A
Join BF. Let AF = x, FD = y, BF = z, ADB = 

x + y = AD = 212 102 = 541 (Pythagoras’ theorem)

Apply Menelau’s theorem on ABD with EFC. 14

AE  BC  DF  1 x 29
EB CD FA

14  20  y  1  y  1 E
7  10 x x 4
t
x  y 541 7z F
5 5
 y = =  (1) 10  y
D
Apply cosine formula on BDF B 10 C

z2 = 102 + y2 – 20y cos 

= 100 +  541 2 – 20 541  10 = 100 + 541 – 40 = 2041
5 5 541 25 25

Method 5 (Provided by Mr. Lee Chun Yu, James from St. Paul’s Co-educational College)

Area of ABC = 1 2120 = 210
2

Area of ABD : area of ADC = 1 : 1 = Area of BDF : area of DCF

 Area of ABF : area of ACF = 1 : 1

Area of CEB : area of CEA = 1 : 2 = Area of FEB : area of FEA

 Area of CFB : area of CFA = 1 : 2

 Area of ABF : area of ACF : area of BCF = 2 : 2 : 1

Area of BCF = 210  2  1  1 = 42
2

Distance from F to BC = 42  2 = 4.2
20

Area of ABF = 210  2  2  1 = 84
2

Distance from F to AB = 84  2 = 8
14  7
BF2 = 82 + 4.22 = 81.64 (Pythagoras’ theorem)

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

Group Events

G1 Given that x, y and z are three consecutive positive integers, and y  z  x  z  x  y is an
xxyyzz

integer. Find the value of x + y + z.

x = y – 1, z = y + 1

y  z  x  z  x  y = y y 1  y 1 y 1  y 1 y
x x y y z z y 1 y y 1

= 2 y  1y  y  1 2 y y2  1 2 y  1y y  1

y 1yy  1

= 2y3  3y2  y  2y3  2y  2y3  3y2  y = 6y3 = 6y2
y2 1
y 1yy 1 y  1 y y 1
Clearly y2 – 1 does not divide y2, so y + 1 and y – 1 are factors of 6.

y – 1 = 1  y = 2, y + 1 = 3  x + y + z = 6

y – 1 = 2  y = 3, but y + 1 = 4 which is not a factor of 6, rejected.

y – 1 = 3 or 6 are similarly rejected.

Method 2 x = y – 1, z = y + 1

y  z  x  z  x  y = y y 1  y 1 y 1  y 1 y =6 3  3
x x y y z z y 1 y y 1  y 1
y 1

3 is an integer  y > 1  (1); 3 is an integer  y  2  (2)
y 1 y 1

Solving (1) and (2) gives y = 2, x = 1, z = 3  x + y + z = 6

G2 Given that x is a real number and x  2012  5  x2  x . Find the value of x.

If 5  x, the equation is equivalent to x  2012  5  x  x

x  2012  2x  5
x – 2012 = 4x2 – 20x + 25
4x2 – 21x + 2037 = 0
 = 212 – 4(4)(2037) < 0  no real solution, rejected

If 5 < x, then the equation becomes x  2012  x  5  x

x – 2012 = 25
x = 2037

G3 Evaluate 22  21008  22012 . (Answer can be expressed in index form.)

22  21008  22012 = 2  1  21006  22010

 = 2  1 2  21005  21005 2

 = 2  1 21005 2

= 2 + 21006.

G4 Evaluate 1  1 2010  1  1 .
2012  2011 2011  3 2 2 1

(Answer can be expressed in surd form.)

1 2011  1 2010  1 2 1 1
2012  2011  3 2

= 2012  2011  2011  2010    3  2  2 1
2012  2011 2011  2010 3  2 1
2

= 2012 1

= 2 503 1

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

G5 Find the minimum value of x2 + y2 – 10x – 6y + 2046.
x2 + y2 – 10x – 6y + 2046

= (x – 5)2 + (y – 3)2 + 2012  2012

G6 In Figure 3, ABC is an isosceles triangle. Suppose AB = AC = 12.

If D is a point on BC produced such that DAB = 90 and CD = 2,

find the length of BC.

Let ABC =  = ACB (base , isos. )

ACD = 180 –  (adj. s on st. line)

BD = 12 sec 

BC = 212 cos  = BD – 2 = 12 sec  – 2
12 cos2  + cos  – 6 = 0

(3 cos  – 2)(4 cos  + 3) = 0

cos  = 2 or  3 (rejected)
3 4

BC = 212 cos  = 16

Method 2 Draw AE  BD. A
ABE  ACE (R.H.S.) Let BE = x = EC.

cos B = x = 12 2  x  x 6 12 12
12 2x  12 1
x2 + x – 72 = 0 B x E x C2 D

 (x – 8)(x + 9) = 0

x=8

 BC = 2x = 16

G7 Given that ax = by = cz = 30w and 1  1  1  1 , where a, b, c are positive integers (a ≤ b ≤ c)
xyzw

and x, y, z, w are real numbers, find the value of a + b + c.

log ax = log by = log cz = log 30w

x log a = y log b = z log c = w log 30

1  1  1  1 = log a  log b  log c = log abc
w x y z w log 30 w log 30 w log 30 w log 30

abc = 30

 a  1 and b  1 (otherwise x log a = y log b = z log c = w log 30  0 = w log 30  w = 0)
 a = 2, b = 3, c = 5

a + b + c = 10
G8 Given that the roots of the equation x2 + px + q = 0 are integers and q > 0.

If p + q = 60, find the value of q. Remark the original question is: Given that the
Let the roots be  and . roots of the equation x2 + px + q = 0 are integers
 +  = –p  (1) and p, q > 0. If p + q = 60, find the value of q.
 = q > 0  (2)
p + q = 60  +  = –p < 0  (1),  = q > 0  (2)
 –( + ) +  = 60   < 0 and  < 0 and ( – 1)( – 1) = 61
1 –  – (1 – ) = 61   – 1 = –1,  – 1 = –61   = 0,  = –60
( – 1)( – 1) = 61, which is a prime   = q = 0 (rejected), no solution

 – 1 = –1,  – 1 = –61 or  – 1 = 1,  – 1 = 61

When  – 1 = –1,  – 1 = –61

  = 0,  = –60

  = q = 0 (contradicting to the given condition q > 0,  rejected)

 – 1 = 1,  – 1 = 61

  = 2,  = 62, q =  = 124

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

G9 Evaluate sin2 1 + sin2 2 + sin2 3 +  + sin2 359 + sin2 360.

Reference 2010 FG1.1
sin2 1 + sin2 89 = sin2 1 + cos2 1 = 1
sin2 2 + sin2 88 = sin2 2 + cos2 2 = 1

.................................................................
sin2 44 + sin2 46 = sin2 44 + cos2 44 = 1
sin2 1 + sin2 2 +  + sin2 89 = (sin2 1 + sin2 89) +  + (sin2 44 + sin2 46) + sin2 45

= 44.5
sin2 1 + sin2 2 + sin2 3 +  + sin2 359 + sin2 360
= 44.5 + sin2 90 + 44.5 + 0 + 44.5 + sin2 270+ 44.5 + 0 = 180

G10 In a gathering, originally each guest will shake hands with every other guest, but Steven only

shakes hands with people whom he knows. If the total number of handshakes in the gathering

is 60, how many people in the gathering does Steven know? (Note: when two persons shake

hands with each other, the total number of handshakes will be one (not two).)

Suppose there are n persons and Steven knows m persons (where n > m).

If everyone shakes hands with each other, then the total number of hand-shaking = C2n
In this case, Steven shakes hands with n – 1 persons. However, he had made only m

hand-shaking.

 C2n  n 1  m  60

nn 1  n  m  59

2

n2 – n – 2n + 2m = 118

n2 – 3n  118 = n2 – 3n + 2m < n2 – 3n + 2n = n2 – n

n2 – 3n – 118  0 and n2 – n – 118 > 0

–9.5  n  12.5 and (n < –10.4 or n > 11.4)

 11.4 < n  12.5

For integral value, n = 12

12 11  12  1  m  60
2

m=5

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Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

Geometrical Construction
1. In the space provided, construct an equilateral triangle ABC with sides equal to the length of

MN below.

C

(1) 作綫段 AB，使得 AB = MN。

(2) 以 A 為圓心，AB 為半徑作一弧；以 B 為圓心，BA 為半徑作一弧；

兩弧相交於 C。

(3) 連接 AC 及 BC。 AB
ABC 為等邊三角形。

2. As shown in Figure 1, construct a circle inside the triangle ABC, so that AB, BC and CA are

tangents to the circle.

(1) 作ABC 的角平分綫。 A

(2) 作ACB 的角平分綫。P 為兩條角平分綫的交點。

(3) 連接 AP。 T S
(4) 分別過 P 作垂直綫至 BC、AC 及 AB，R、S、T 為 3
4
對應的垂足。 4

5

P2

BPR  BPT (A.A.S.) B 1 C

CPR  CPS (A.A.S.) R
PT = PR = PS (全等三角形的對應邊)

(5) 若 P 至 BC 的垂足為 R，以 P 為圓心，PR 為半徑作一圓，此圓內切於三角形的三

邊，稱為內切圓(inscribed circle)。(切綫半徑的逆定理)

3. Figure 2 shows a triangle PQR. Construct a line MN parallel to QR so that

(i) M and N lie on PQ and PR respectively; and

(ii) the area of PMN = 1  the area of PQR.
2

首先，我們計算 MN 和 QR 的關係： P

QPR = MPN (公共角)

PQR = PMN (QR // MN, 對應角)

PRQ = PNM (QR // MN, 對應角)

 PQR ~ PMN (等角)

PMN 的面積  1   PM 2 MN
PQR 的面積 2 PQ

PM = 1  PM = PQ QR
PQ 2 2

作圖步驟: 2P
(1) 利用垂直平分綫，找 PQ 之中點 O。

(2) 以 O 為圓心 OP = OQ 為半徑，向外作一半圓，與 F

剛才的垂直平分綫相交於 F。 1

PFQ = 90 (半圓上的圓周角) 3O N
M
PFQ 為一個直角等腰三角形

QPF = 45 QR
PF = PQ sin 45 = PQ

2

(3) 以 P 為圓心，PF 為半徑，作一圓弧，交 PQ 於 M。PM = PQ 。
2

(4) 自 M 作一綫段平行於 QR，交 PR 於 N，則PMN 平分PQR 的面積。

http://www.hkedcity.net/ihouse/fh7878 Page 9

Answers: (2011-12 HKMO Heat Events) Created by: Mr. Francis Hung Last updated: 23 November 2016

Percentage of correct questions

1 34.48% 2 12.58% 3 32.66% 4 35.40% 5 30.73%

6 30.83% 7 42.19% 8 13.08% 9 25.96% 10 4.26%

1 57.49% 2 35.22% 3 38.06% 4 48.99% 5 53.44%

6 18.62% 7 16.60% 8 6.07% 9 30.77% 10 42.51%

http://www.hkedcity.net/ihouse/fh7878 Page 10

Hong Kong Mathematics Olympiad 2012-2013

Heat Event (Individual)

除非特別聲明，答案須用數字表達，並化至最簡。 時限：40 分鐘

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.

每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 40 minutes

1. 化簡 94  2 2013 。

Simplify 94  2 2013 .

2. 一個平行四邊形可被分成 178 個邊長為 1 單位的等邊三角形，若該平行四邊形的周界

為 P 單位，求 P 的最大值。

A parallelogram is cut into 178 pieces of equilateral triangles with sides 1 unit. If the

perimeter of the parallelogram is P units, find the maximum value of P.
3. 圖一所示為一直角三角形 ACD，其中 B 是 AC 上的

點且 BC = 2AB。已知 AB = a 及 ACD = 30，求  的

值。

Figure 1 shows a right-angled triangle ACD where B is a

point on AC and BC = 2AB. Given that AB = a and

ACD = 30, find the value of .

圖一 Figure 1

4. 已知 x2 + 399 = 2y，其中 x、y 為正整數。求 x 的值。
Given that x2 + 399 = 2y, where x, y are positive integers. Find the value of x.

5. 已知 y = (x + 1)(x + 2)(x + 3)(x + 4) + 2013，求 y 的最小值。

Given that y = (x + 1)(x + 2)(x + 3)(x + 4) + 2013, find the minimum value of y.
6. 從一個有 n 條邊的凸多邊形中，選取其中一隻內角。若餘下的 n – 1 隻內角之和是

2013，求 n 的值。

In a convex polygon with n sides, one interior angle is selected. If the sum of the remaining

n – 1 interior angle is 2013, find the value of n.
7. 圖二所示為一通過 B 點及 C 點的圓，而 A 點則

在圓之外。已知 BC 是圓的直徑，AB 及 AC 分

別與圓相交於 D 點及 E 點，且 BAC = 45，

求 ADE 的面積 。
BCED 的面積

Figure 2 shows a circle passes through two points B

and C, and a point A is lying outside the circle.

Given that BC is a diameter of the circle, AB and AC

intersect the circle at D and E respectively and

BAC = 45, find area of ADE . 圖二 Figure 2
area of BCED

8. 解 31 31 x = x。

Solve 31 31 x = x.
9. 圖三所示為五邊形 ABCDE。AB = BC = DE = AE + CD = 3，且A =

C = 90，求該五邊形的面積。
Figure 3 shows a pentagon ABCDE. AB = BC = DE = AE + CD = 3
and A = C = 90, find the area of the pentagon.

圖三 Figure 3

10. 若 a 及 b 為實數，且 a2 + b2 = a + b。求 a + b 的最大值。
If a and b are real numbers, and a2 + b2 = a + b. Find the maximum value of a + b.

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.1 HKMO 2013 Heat Event (Individual)

Hong Kong Mathematics Olympiad 2012-2013

Heat Event (Group)

除非特別聲明，答案須用數字表達，並化至最簡。 時限：20 分鐘

Unless otherwise stated, all answers should be expressed in numerals in their simplest form.

每題正確答案得一分。Each correct answer will be awarded 1 mark. Time allowed: 20 minutes

1. 已 知 一 個 直 角 三 角 形 三 邊 的 長 度 皆 為 整 數 ， 且 其 中 兩 邊 的 長 度 為 方 程
x2 – (m + 2)x + 4m = 0 的根。求第三邊長度的最大值。

Given that the length of the sides of a right-angled triangle are integers, and two of them are
the roots of the equation x2 – (m + 2)x + 4m = 0. Find the maximum length of the third side of

the triangle.

2. 圖一所示為一梯形 ABCD，其中 AB = 3、CD = 5 及 圖一 Figure 1
AC、BD 相交於點 O。若 AOB 的面積是 27，求
梯形 ABCD 的面積。

Figure 1 shows a trapezium ABCD, where AB = 3,
CD = 5 and the diagonals AC and BD meet at O. If
the area of AOB is 27, find the area of the trapezium
ABCD.

3. 設 x 及 y 為實數使得 x2 + xy + y2 = 2013。求 x2 – xy + y2 的最大值。
Let x and y be real numbers such that x2 + xy + y2 = 2013.
Find the maximum value of x2 – xy + y2.

4. 若 、 是方程 x2 + 2013x + 5 = 0 的根，求 (2 + 2011 + 3)(2 + 2015 + 7) 的值。
If ,  are roots of x2 + 2013x + 5 = 0, find the value of (2 + 2011 + 3)(2 + 2015 + 7).

5. 如圖二所示，ABCD 為一個邊長為 10 單位的正方形，E 及
F 分別為 CD 及 AD 的中點，BE 及 FC 相交於 G。
求 AG 的長度。
As shown in Figure 2, ABCD is a square of side 10 units, E
and F are the mid-points of CD and AD respectively, BE
and FC intersect at G. Find the length of AG.

圖二 Figure 2

http://www.hkedcity.net/ihouse/fh7878 P.2 HKMO 2013 Heat Event (Group)

6. 若 a 及 b 為正實數，且方程 x2 + ax + 2b = 0 及 x2 + 2bx + a = 0 都有實數根。求 a + b
的最小值。
Let a and b are positive real numbers, and the equations x2 + ax + 2b = 0 and x2 + 2bx + a = 0
have real roots. Find the minimum value of a + b.

7. 已知 ABC 的三邊的長度組成一個等差數列，且為方程 x3 – 12x2 + 47x – 60 = 0 的根，
求 ABC 的面積。

Given that the length of the three sides of ABC form an arithmetic sequence, and are the
roots of the equation x3 – 12x2 + 47x – 60 = 0, find the area of ABC.

8. 圖三中，ABC 為一等腰三角形，其中 圖三 Figure 3
AB = AC，BC = 240。已知 ABC 的內接
圓的半徑是 24，求 AB 的長度。
In Figure 3, ABC is an isosceles triangle
with AB = AC, BC = 240. The radius of the
inscribed circle of ABC is 24. Find the
length of AB.

9. 從 1、2、3、、2012、2013 中最多可取出多少個數，使得在取出的數中任意兩數之和
都不是這兩個數之差的倍數？
At most how many numbers can be taken from the set of integers: 1, 2, 3,  , 2012, 2013
such that the sum of any two numbers taken out from the set is not a multiple of the difference
between the two numbers?

10. 對所有正整數 n，定義函數 f 為

(i) f (1) = 2012,
(ii) f (1) + f (2) +  + f (n – 1) + f (n) = n2 f (n)，n > 1
求 f (2012) 的值。

For all positive integers n, define a function f as
(i) f (1) = 2012,
(ii) f (1) + f (2) +  + f (n – 1) + f (n) = n2 f (n)，n > 1.
Find the value of f (2012).

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.3 HKMO 2013 Heat Event (Group)

Hong Kong Mathematics Olympiad 2012 – 2013

Heat Event (Geometric Construction)
香港數學競賽 2012 – 2013

初賽(幾何作圖)

每隊必須列出詳細所有步驟(包括作圖步驟)。 時限：20 分鐘

All working (including geometric drawing) must be clearly shown. Time allowed: 20 minutes
此部份滿分為十分。The full marks of this part is 10 marks.

School Code: _________________________________________________________
School Name: ________________________________________________________

Question No. 1 第一題

Line segment PQ and an angle of size  are given below. Construct the isosceles triangle PQR with
PQ = PR and QPR = .
下圖所示為綫段 PQ 及角 。試構作一個等腰三角形 PQR，其中 PQ = PR 及 QPR = 。

http://www.hkedcity.net/ihouse/fh7878 P.4 HKMO 2013 Heat Event (Geometric Construction)

Hong Kong Mathematics Olympiad 2012 – 2013

Heat Event (Geometric Construction)
香港數學競賽 2012 – 2013

初賽(幾何作圖)

每隊必須列出詳細所有步驟(包括作圖步驟)。 時限：20 分鐘

All working (including geometric drawing) must be clearly shown. Time allowed: 20 minutes
此部份滿分為十分。The full marks of this part is 10 marks.

School Code: _________________________________________________________

School Name: ________________________________________________________

Question No. 2 第二題

Construct a rectangle with AB as one of its sides and with area equal to that of ABC below.
試構作一個面積與下圖所示的 ABC 面積相等的長方形，長方形其中一邊為 AB。

http://www.hkedcity.net/ihouse/fh7878 P.5 HKMO 2013 Heat Event (Geometric Construction)

Hong Kong Mathematics Olympiad 2012 – 2013

Heat Event (Geometric Construction)
香港數學競賽 2012 – 2013

初賽(幾何作圖)

每隊必須列出詳細所有步驟(包括作圖步驟)。 時限：20 分鐘

All working (including geometric drawing) must be clearly shown. Time allowed: 20 minutes
此部份滿分為十分。The full marks of this part is 10 marks.

School Code: _________________________________________________________

School Name: ________________________________________________________

Question No. 3 第三題

The figure below shows two straight lines AB and AC intersecting at the point A. Construct a circle
with radius equal to the line segment MN so that AB and AC are tangents to the circle.
下圖所示為兩相交於 A 點的綫段 AB 及 AC。試構作一半徑等於綫段 MN 的圓使得 AB 及
AC 均為該圓的切綫。

*** 試卷完 End of Paper ***

http://www.hkedcity.net/ihouse/fh7878 P.6 HKMO 2013 Heat Event (Geometric Construction)