Prelim 1998 2002

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Forward

The present booklet contains test problems and solutions of five contests. Every

year during the month of May a contest (International Mathematical Olympiad Selection

Contest) is held and around five hundred secondary school students participate in the

contest. Since 1987 the contests have been organized by the International Mathematical

Olympiad Hong Kong Committee (IMOHKC). A purpose of the contests is to select

students for further training so that they may participate in national and international

mathematical competitions. A student may also learn something about problem solving

and consider participating in a contest as an interesting extracurricular activity. The

contests included in this booklet were conducted during May of the years 1997-2001.

Problems of the contests were set by members of IMOHKC.

Comments and suggestions concerning this booklet may be sent to:

Dr. Kin Li

Department of Mathematics

The Hong Kong University of Science and Technology

We would like to acknowledge the staff of the Mathematics Section, Education

Department who assisted in the production of this booklet. We would also like to

acknowledge the Quality Education Fund (QEF) for supporting the publication.

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Contest 97-98

1. Let a, b, c be real numbers such that

2c

1

b

1

a

1

=++ and 1c1

b

1

a

1222 =++

Find the value ofac

1

bc

1

ab

1 ++ . (1 mark)

2. On a Cartesian plane every point with integral coordinates is called a lattice point. Let Pn be

the lattice point (n, n + 3) and f(n) be the number of the lattice points (excluding the two

endpoints) on the line segment joining the origin O and the point Pn. Find f(1) + f(2) + f(3)

+ + f(1997). (1 mark)

3. A positive integer is said to be "good" if the digits of the integer can be divided into two

groups such that the sum of the digits in one group is equal to the sum in the other. Find the

smallest positive integer n such that n and n + 1 are "good". (1 mark)

4. Let x, y, z be real numbers such that x + y + z = 5 and xy + yz + zx = 3. Find the maximum

value of z. (1 mark)

5. Find the number of diagonals that can be drawn in a convex polygon of n (n 4) sides.(1 mark)

6. Find the time between 1:00 p.m. and 1:30 p.m., correct to the nearest minute, when the hour

and minute hand of a clock form an angle of 100o. (1 mark)

7. In ABC, AB = 41 cm, AC = 9 cm and BC = 40 cm. Find the radius of the inscribed circle ofABC. (1 mark)

8. Two perpendiculars chords of a circle are at distances a and b respectively from the center.

These two chords divide the circle into four pieces. Consider the sum of areas of the largest

and the smallest pieces, and the sum of the areas of the other two pieces. Find the difference

between these two sums. (1 mark)

9. How many integers from 1 to 1997 have the sum of their digits divisible by 5? (1 mark)

10. The faces of a cube are labeled with different positive integers such that the numbers on any

two adjacent faces differ by at least two. Find the minimum value of the sum of all six

numbers. (1 mark)

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11. A square whose sides are of integral lengths is cut into 25 smaller squares whose sides are

also of integral lengths.Exactly 24 of these smaller squares are unit squares. Find the area of

the original square. (1 mark)

12. Each of the numbers 1, 2, 3, , 25 is written into a square in a 55 table, such that thenumbers in each row are in increasing order. Find the maximum value of the sum of the

numbers in the third column. (1 mark)

13. Let S = 1 +80

1

3

1

2

1 +++ L . Find the integer n such that n 1 < S < n. (2 marks)

14. A positive integer N (in base 10) is composed of the digits 0 and 1 only, and is divisible by

2475. Find the smallest possible number of digits of N. (2 marks)

15. 20 football teams take part in a tournament. M matches have been played and it is found that

(a) between any two teams at most one match has been played, and

(b) among any three teams at least one match has been played between two of them.

What is the smallest possible value of M. (2 marks)

16. Let a n be the integer closest to n . Find1997321 a

1

a

1

a

1

a

1 ++++ L . (2 marks)

17. ABCDEF is a regular hexagon with AB = 1. P and S are the midpoints of AB and ED

respectively. The circle with PS as diameter cuts the lines PE and PD at Q and R respectively.

Find the area of the quadrilateral QRDE. (2 marks)

18. Points D and E are points inside an equilateral triangle ABC such that DE = 1, DA = EA

= 7 , DB = EC = 2. Find the length of AB. (2 marks)

19. In ABC, E, F, G are points on AB, BC, CA respectively such that AE/EB = BF/FC =CG/GA = 1/3. K, L, M are the intersecting points of the lines AF and CE, BG and AF, CEand BG, respectively. Suppose the area ofABC is 1, find the area ofKLM. (3 marks)

20. Three lines are drawn through a point in a triangle parallel to its sides. The segments

intercepted on these lines by the triangle turn out to have the same length. Given the

triangle's side lengths a, b and c, find the length of the segments. (3 marks)

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Contest 98-99

1. In a sequence {a1, a2, , an, }, a1 = 1, 4anan + 1 = (an + an + 1 1)2 , and an + 1 > an.Find a1998. (1 mark)

2. Given that Hk =2

)1k(kcos

2

)1k(k . Find H19 + H20 + + H98. (1 mark)

3. A book has 30 chapters. The lengths of the chapters are 1, 2, , 30 pages. Chapter One

starts from page 1 of the book, and each chapter starts from a new page. At most how many

chapters can start from an odd-numbered page? (1 mark)

4. On a square carpet of size 123123, each unit square is colored red or blue. Each red squarenot lying on the edge of the carpet has exactly five blue squares among its eight neighbors.

Each blue square not lying on the edge of the carpet has exactly four red squares among its

eight neighbors. Find the numbers of red squares on the carpet. (1 mark)

5. In ABC, C = 90o, A = 30o, and AB = 1. Let ABD, ACE, BCF be equilateraltriangle with D, E, and F lying outside ABC. Let DE intersects AB at G. Find the area ofDGF. (1 mark)

6. ABCDEF is a regular hexagon, M, N are points on the segments AC, CE respectively such

that rCE

CN

AC

AM == . If B, M, N are collinear, find r. (1 mark)

7. ABC is an equilateral triangle and P is a variable point on the same plane such that PAB,PBC, and PCA are isosceles triangles. In how many different positions can P lie?

(1 mark)

8. The lengths of the three medians AD, BE, and CF ofABC are 9, 12, and 15 respectively.Find the area ofABC. (1 mark)

9. Find the 4-digit number such that, when the order of its digits is reversed, the new value is 4

time the original one. (1 mark)

10. Determine the number of ordered pairs (x, y), where x and y are integers satisfying the

equation 2xy 5x + y = 55. (1 mark)

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11. Given that [x] represents the greatest integer not exceeding x, find the last three digits of the

integer

+ 310

1033

99

. (1 mark)

12. Let x0 = 5 andn

n1nx

1xx +=+ for n = 0, 1, 2, . Find the integer closest to x1998.

(2 marks)

13. At least how many of the + signs in the expression

+1 + 2 + 3 + + 100

must be replaced by signs so that its value is 1998. (2 marks)

14. Except for the first two terms, each term of the sequence 1000, x, 1000 x, is obtained bysubtracting the preceding term from the one before that. The last term is the first negative

term encountered. What positive integer x produces a sequence of maximum length?

(2 marks)

15. In ABC, tan A : tan B : tan C = 1 : 2 : 3. FindAB

AC. (2 marks)

16. Find the smallest positive integer n such that 1997n 1 is divisible by 21998. (2 marks)

17. In how many ways can 1998 be expressed as the sum of one or more consecutive integers.

(2 marks)

18. What is the largest integer k such thatk11

199810021001 Lis an integer? (2 marks)

19. Find the smallest multiple of 84 whose digits consist entirely of 6s and 7 s only.(2 marks)

20. An mnp rectangular box has half the volume of an (m + 2)(n + 2)(p + 2) rectangularbox, where m, n, and p are integers and m n p. What is the largest possible value of p?

(3 marks)

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11. A sequence of integers a1, a2, , an, is defined by a n = a n 1 a n 2 for n 3. The sum ofthe first 1588 terms of the sequence is 1997, and the sum of the first 1997 terms is 1588.

Find the sum of the first 1999 terms of the sequence. (2 marks)

12. Find the integer n satisfying 1999!10

n

!2

n

!1

n=

++

+

L . Here [x] denotes the greatest

integer less than or equal to x. (2 marks)

13. What is the largest positive integer n for which there is a unique integer k such that

13

7

kn

n

15

8 1, which shows that

m 3.Next rewrite this equation as (m 2)(n 2)(p 2) = 8(m + n + p).Let a = m 2, b = n 2, c = p 2. Then a 1. And so abc = 8(a + b + c + 6) can be rewritten

as8ab

151

8ab

158ab

8ab

61ab

8ab

6ba

8

c

+=

+=

++

++= because (a 1)(b 1) 0. This

shows that c cannot be larger than 816 = 128, and c attains this value if ab = 9 and a + b =ab + 1 = 10. Thus c = 128, when a = 1,b = 9, and m = 3, n = 11, and p = 130 are the

dimensions of a possible box.

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Contest 99-00 Solutions

1. Notice that Ak < Ak + 1 if and only if)!1k(

9919

!k

9919 1k1kkk

++

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8. Let the sides of such triangle be n 1, n, n + 1. Then (n 1) + n + (n + 1) 100 and(n 1) + n > n + 1, imply n > 2 and n 33, that is n = 3, 4, , 33. When n = 3, 22 + 32 < 42,the triangle is obtuse; when n = 4, 32 + 42 = 52, the triangle is right; when n 5, n2 + (n 1)2 (n + 1)2 = n2 4n = n(n 4) > 0, the triangle is acute. Thus the required number is 33 4= 29.

9. Draw CE // BD to meet AD produced at E. Since AC = 5, CF = 4, we have AF = 3. Using

similar triangles (ACF AEC), getCE

5

CE

AC

CF

AF

4

3 === , hence CE =3

20. Since area of

ABC = area ofCDE, we have finally area of ABCD = area ofACE =3

50

2

1

3

205 = .

10. Let EC = x, BE = y, ED = z. Since BC = CD, BAC = DAC, thus DCE ACD. Hence

DC

EC

CA

CD = , or4

x

x6

4 =+

, giving x = 2.

Now since ABCD is cyclic AE EC = BE ED, we get 62 = yz = 12.From BCD, y + z < 4 + 4 = 8, hence y = 3, z = 4 or y = 4, z = 3. In each case, y + z = 7.

11. Note an + 3 = an + 2 an + 1 = (an + 1 an) an + 1 = an, thus an + 6 = an + 3 = (an) = an, thus thesequence is periodic of period 6. Now Sn = an + an 1 + + a1 = (an 1 an 2) + (an 2 an 3) + + (a2 a1) + a2 + a1 = an 1 + a2. Thus S1588 = a1587 + a2 = a3 + a2 = 1997, S1997 = a1996

+ a2 = a4 + a2 = (a3 a2) + a2 = 1588, hence a2 = 1997 1588 = 409. Finally, S1999 = S1998 +a2 = a6 + a2 = a3 + a2 = 1588 + 409 = 1179.

12. Clearly n < 7!. Let n = a6! + b5! + c4! + d3! + e2! + f1! with a 6, b 5, c 4, d 3, e 2, f 1. Put into the original equation, gets 1237a + 206b + 41c + 10d + 3e + f = 1999.Solving get a = 1, b = 3, c = 3, d = 2, e = 0, f = 1, and n = 1165.

13. First,

8

15

n

k1

n

kn

7

31

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15. Let S be the sum of the remaining numbers. Then4n

S

= 51.5625 =

16

825. Hence 16S =

825(n 4) 16

4n is an integer n = 16k + 4.

Now 16

825

)4n(2

24n7n

4n

))6n()4n()2n(n()n21( 2

+

=++++++ L

8.2 < n < 106.

Also,16

825

)4n(2

40nn

4n

)8642()n21( 2 +=

++++++ L

n < 7.5 or n > 98.3.

For n = 16k + 4 lies between 98.3 and 106, n = 100.

Thus, S = 9616

825 = 825 6 = 4950. Note 1 + 2 + + 100 = 5050.

If x is the largest number removed, then x + (x 2) + (x 4) + (x 6) = 5050 4950 = 100,giving x = 28.

16. In base 3, the sequence is 1, 10, 11, 100, 101, 110, 111, . Since 200 in base 2 is

(11001000)2, hence the 200 th term in base 3 is (11001000)3 = 2943.

17. From the diagram, AE = AP = AD = 8, so that

APD = ADP, AEP = APE, and BAP =

2AEP. Now APD = APC CPD = B +BAP CPD = B + 2AEP CPD, ADP =C + CPD = B + CPD, as APD = ADP,we have AEP = CPD. Thus CPD BEP.

HenceBP

BE

CD

CP = or CPBP = CDBE = (12 8)(12 + 8) = 80.

18. Let = ABD, then AC + BD = 10(sin + cos) =

210 sin( + 45o). Now 45o >=+ . There are altogether 4 such squares. For the

squares whose distance is22

54 + from the origin, it is not covered, since

6415422 >=+ . There are 8 such squares. All other squares are covered. And hence

we get 144 44 4 8 = 88 squares covered by the circle.

20. Partition the integers into subset A1 = {1}, A2 = {2, 3}, A3 = {4, 5, 6}, A4 = {7, 8, 9, 10},

A5= {11, 12, , 16}, A6 = {17, 18, , 25}, A7 = {26, 27, , 39}, A8 = {40, 41, , 60}, A9

= {61, 62, , 91}. Choose one member from each group, then the ratio of any two of them

will exceed2

3or less than

3

2, thus k > 9. On the other hand, if 10 numbers are chosen, two

must be from the same group and will have a ratio bounded by2

3and

3.

2Thus the

minimum value of k is 10.

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1. Let a = 2x 4, b = 4x 2, then the equation becomesa3 + b3 = (a + b)3 = a3 + 3a2b+ 3ab2 + b3 0 = ab(a + b) a = 0 or b = 0 or a + b = 0.

Thus 2x 4 = 0 or 4x 2 = 0 or 4x + 2x 6 = (2x + 3)(2x 2) = 0, we have x = 2,21 or 1. The

required sum =2

7.

2. The numerator and the denominator are relatively prime, thus any prime factor must occur

entirely in the numerator or in the denominator (but not both). There are 10 prime factors of

30!, i.e. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. For each of these, one put it up or down, thus 210

choices, exactly half of them will have values between 0 and 1, i.e., 29 = 512 choices.

3. x17 can only be obtained by multiplying two x5 s and one x7. There are 20 ways to get x7 and

171C192 = ways to get two x5 s in the remaining 19 factors. So the answer is 20171 = 3420.

4. Note

=

=

2001

199911999

2001

19991999

2001

19992000.

Hence 1998119992001

19992000

2001

19991 ==

+

. Similarly, we have

1998

2001

19991999

2001

19992=

+

, , 1998

2001

19991001

2001

19991000=

+

.

Summing up, we get the sum = 1998000.

5. From x2 + xy + y2 = 0, get 1)yx)(yx(

xy =++

. Let t =yx

x

+, then

yx

y

t

1

+= and 1

t

1t =+ , t2

t + 1 = 0 = (t + 1)(t2 t + 1) = t3 + 1 t3 = 1. Thus t 2001 = t3(667) = 1, and hence

2t

1t

2001

2001 =+ .

6. Let m and n be integral roots of x2 + ax + 8a = 0, with m n.Then x2 + ax + 8a = (x m)(x n) = x2 (m + n)x + mn. We have a = (m + n), and 8a = mn.Thus a is an integer and 8(m + n) = mn, get mn + 8m + 8n = 0 or (m + 8)(n + 8) = 64.Consider the factorizations of 64 = 641 = 322 = 164 = 88 = 416 = 232 = 164 =(64)(1) = (32)(2) = (16)(4) = (8)(8) = (4)(16) = (2)(32) =(1)(64), we obtain 8 distinct pairs of m and n.

7. tan + tan = , tan tan = 2 12tantan1

tantan)tan(

=

+=+ .

Now sin2 ( + ) + sin ( + ) cos ( + ) + 2 cos2 ( + )= cos2 ( + ) [tan2 ( + ) + tan ( + ) + 2

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= ]2)tan()([tan)(tan1

1 22

++++++

= 2212)12()12(

)12( 2

2

2

22

2

=

+

+

+

8. After the 2000th operation, only those lamps with number which have an odd number of

factors will be on. This is equal to the number of perfect squares less than 2000. Since 44 2 =

1936, and 452 = 2025. Therefore the number of lamps which are on = 44.

9. Let m be the number of sides of the polygon determined by An, A1 and B. The degree

measures of the interior angles of the three polygons are 180 n

360, 60 and 180

m

360

respectively. If 6 < n, the polygon fit together at their common vertex A1, then

m

36018060

n

360180360 ++=

6m

366

6m

m6n

+=

= .

So m > 6, and n is a decreasing function of m. The largest value of n is 42, when m = 7.

10. The triangle is constructed by drawing APC = BPC= 60o on the lines, and constructing APBC as

circumcircle ofAPB. Then ABC = APC = 60o and

BAC = BPC = 60o. Now2

360sinAP

4 o == AP

=3

8, and

2

360sin

BP

3 o == AP =3

6. (AB)2

=3

148120cos

3

6

3

82

3

36

3

64 =+ o . Hence area ofABC =3

337

3

148

4

3 = .

11. Let DE = x, BC = a, then perimeter of ADE = p 2a

(tangent property). As ADE ABC, then p a2pax = ,

get x =

=

=

22

2

4

pa

4

p

p

2bap

2

1

p

2

p

)a2p(a.

When a =4

p, x attains its maximum. Thus maximum value

of DE =

8

p

4

p

p

22

=

.

L 1

L 2

L 3

A

B

CP

4

3

B

A

C

D E

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12. Area ofABC = 1252

1 = 30, sin A =13

5. Let AD = x,

AE = y, area ofADE = 15Asinxy2

1 = . Hence xy = 78.

By cosine formula, DE2 = x2 + y22xy cos A =

(x y)2 + 2xy(1 cos A) = (x y)2 + 278

13

121 = (x

y)2 + 12 12. Thus minimum value of DE = 12 .

13. Let area of ABC = S. Note that TUD ABC and their

areas are in the ratio UD2 : BC2. ThereforeBC

UD

S

8 = ,

Similarly, BC

PQ

S

128

= and BCDR

S

32

= . Furthermore, BC= BP + PQ + QC = UD + PQ + DR. Therefore,

1BC

DRPQUD

S

321288 =++=++ .

Hence 214S = or S = 392.

14. Observe that)1k(k

kxxx

1k

xx

k

xxxyy

1kk11k1k21

1kk +++

=+++

+++

= +++LLL

=)1k(k

kxkxx3x2x2xx 1kk33221

++++ +L

)1k(k

xxkxx2xx 1kk3221

++++

+L

Hence y1 y2 + y2 y3++ y1999 y2000

=

+

++++1999

1k

1kk3221

)1k(k

xxkxx2xx L=

= =

+

= =

+

+

=+

1999

1i

1999

ik

1ii1999

1k

k

1i

1ii

)1k(k

xxi

)1k(k

xxi

=

+

= =

+==

+i2000

i2000xxi

)1k(k

1xxi

1999

1i

1ii

1999

ik

1999

1i

1ii = =

+

1999

1i

1ii2000

i1|xx|

1999

1

1

11 | | 1999.2000

i i

i

x x += =

Take x1 = 2000, x2 = x3 = . = x2000 = 0, we observe that the estimate can be achieved.

15. Consider drawing the circles in the order C8, C7, , C1 and then locating the points in thecircles. Draw C8 and 8 points will be marked on the circle. Draw C7 so that it pass through 2

of the 8 points already exist, and 5 new points will have to be created. Now draw C6 so thatit passes through 2 existing points on C8 and 2 existing points on C7, and this leaves 2 morepoints to be created. Now the situation is: we have drawn C8, C7, C6 with 6 points fixed and

9 points (4 on C8 only, 3 on C7 only, 2 on C6 only) that can be fixed at later stage. Nowattempt to draw C5 and C4 by suitably selecting the positions of these 9 points. It may be

observed that there will be no difficulty to draw C3 and C2 by selecting three or two existingpoints not on the same circle. Finally, C1 may be drawn to passes through any one existingpoint. The minimum number of points is 8 + 5 + 2 = 15.

y

5x

12

CA

B

E

D

A

B

C

DQ

P

U

T

S R

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16. Denote the number of b,b pairs in f(n)(a) by Pn and the number of a, b pairs by Qn. The

number of b,b pairs in f(n)(a) is equal to the number of a, bpairs in f(n-1)(a). The numberof a, b pairs in f(n-1)(a) is equals to the number of a s plus the number of b, b pairs in f(n-2)(a). Moreover f (n-2)(a) consists of 2 n 2 letters, and half of them are a s. Therefore, Pn =

Qn 1= 2n 3 + Pn 2.

Inductively, we have

++

+++=

evennP2

oddnP222P

2

11

0

5n3n

n L .

Now P1 = 0, P2 = 1. It follows that if n is odd, Pn = 2n 2 + 2 n 5 + + 1 =

3

12 1n , if n is

even Pn = 2n 3 + 2 n 5 + + 2 + 1 =

3

121

3

)12(2 1n3n +=+

.

Combining the results, we obtain Pn =3

)1(2 n1n +.

17. Let T be on so that AT is tangent to as shown. Now O, E, F are collinear, EF is adiameter of . Let AE meets at K. Then EKF ACE (C = K = 90o, OE BC

OE//AC CAE = KEF). SoAE

EF

AC

EK =

22 CEAC

)OEOF(AC

AE

EFACEK

+

=

=

=5

224

28

)22)(22( =+

.

AK = AE + EK =5

234 + AT = 22246AKAE +=+= .

18. Let [XYZ] denote the area ofXYZ. Let AR : RB = BP : PC = CQ : QA = 1 : .

+=

1

1

]ABC[

]ACR[;

= 1

]ACZ[

]ABZ[;

+=

1

1

]ABZ[

]ARZ[

)1(

1

]ACZ[

]ARZ[

+= and

1

1

]ACR[

]ARZ[2 ++

= .

Thus )1)(1(

1

]ABC[

]ARZ[2 +++= ----- (1);

1]ACR[

]ACZ[2

2

+++= and

1]ABC[

]ACZ[2 ++

= .

1

12

1

31

]ABC[

]XYZ[2

2

2 +++=

++= ------ (2). Combining (1) and (2), we have

)1)(1(

1

1

1222

2

+++=

+++

( + 1)(2 2 + 1) = 1 2 1 = 0

= 251+

. As [ARZ] = 1 cm

2

, we have [ABC] = ( + 1)(2

+ + 1), and hence

A BO

C

E

FT

K

B C

A

P

Q

R

Y

Z

X

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[PCYX] =+1

1( + 1)(2 + + 1) 2 = 2 + 1 = 2 = (1 + 5 ) cm2.

19. Let AE intersects 2 at U, CE intersects 3 at V.Since AEC = AUB = BVC = 90o,EUBV is a rectangle. VUB = EBU =EAB which implies VU is tangent to2 at U.Similarly, UV is tangent to 3 at V, so U = U,V = V, and EUV ECA. Then

16

3

AC

BCAB

AC

EB

AC

UV

EACofarea

EUVofarea22

2

2

2

====

.

20. For the persons in the photographs we draw family tree. The 0th row corresponds to persons

where mothers do not appear in the photographs. Denote by rkthe numbers of persons in the

middle of the pictures that are in the kth row, tk numbers of other persons in the kth row. The

totality of mothers of women in row k is sk. We have sk + 12

1rk + 1 + tk + 1 (every middle

women has a sister in the pictures), and rk sk + 1 (every women has a daughter in the

pictures). Thus rk2

1rk + 1 + tk+ 1 , k = 0, 1, 2, . Also 1

2

1r0 + t0.

Sum up k from 0 to 1, 2, , we get (r0 + r1 + ) + 1 2

1(r0 + r1 + ) + (t0 + t1 + )

Hence (r0 + r1 + ) + (t0 + t1 + ) 2

3(r0 + r1 + ) + 1 = 12

2

3 + 1 = 19. We have such an

example:

r0 = 2, t0 = 0 1 2

r1 = 4, t1 = 0, s1 = 2 3 4 5 6

r2 = 6, t2 = 1, s2 = 4 7 8 9 10 11 12 15

r3 = 0, t3 = 6, s3 = 6 16 17 18 19 13 14

1, 2, 3, , 12 are middle women, 1 and 2 are sisters. The pictures are (3, 1, 2), (5, 2, 1), (7,

3, 4), (9, 4, 3), (11, 5, 6), (15, 6, 5), (16, 7, 8), (17, 8, 7), (18, 9, 10), (19, 10, 9), (13, 11, 12),

(14, 12, 11).

1 3AC

B

E

V'

U'

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1. For a number to be divisible by 11, the rule is the sum of the digits in the odd positions

minus the sum of the digits in the even positions should be a multiple of 11. So we need (2n

+ 7 + 9) (n + 2) = n + 14 divisible by 11. Hence the least n is 8. (Briefly, the rule is because10k = (11 1)k = 11m + (1)k by the binomial theorem and so N = am10m + + a110 + a0 =11M + am(1)m + + a1(1) + a0 = 11M + (a0 + a2 + ) (a1 + a3 + ).)

2. Suppose 1 armchair cost $ a, 1 bookcase cost $ b and 1 cabinet cost $ c. Then 8a + 11b + 2c

= 875 and 3a + 2b + 5c = 343. Taking the first equation and adding it to 3 times the second

equation, we get 17a + 17b + 17c = 1904. So a + b + c = 112.

3. T1 + T2 + T3 + + T2002

= F(12) F(1) + + F(20022) F(2002) = F(12) + + F(20022) F(1) F(2002)= 200(1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0) + 1 + 4 200(1 + 2 + + 9 + 0) 1 2= 2

4. 2 2002 (2 2003 1) = 416 500 (816 500 1). Now M = 16 500 1 is divisible by 16 1 = 15.Thus 2 2002 (2 2003 1) = 4(M + 1)(8M + 7) = 32M2 + 60M + 28 has last two digits equal 28because 32M2 and 60M are divisible by 455 = 100.

5. The equations can be put in the form

z2 = x2 + y2 2xy cos 30o and y2 = x2 + z2 2xz cos 45o,which remind us the cosine law. So we consider ABC with AB = z, BC = x, CA = y, B =45o and C = 30o. By sine law, y : z = sin B : sin C = 1:2 .

6. The equation can be rewritten as (x3 b)2 = 100, which implies x =3 10b . Thus 2= 33 10b10b + . Cubing both sides, we get

8 = b + 10 3 23 2 )10b)(10b(3)10b()10b(3 +++ (b 10)

8 = 20 ( )333 2 10b10b100b3 + = 20 3 2 100b6 .

Then 2100b3 2 = , which implies b = .36108 =

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7. Upon drawing an accurate figure and using a protractor,

GFC seems to be 20o. To confirm this, let H, I be the feetof perpendiculars from F and A to BC respectively. Now

BCA = 180oBAC ABC = 80o and BI = BA cos

60o = 2

1BA. Applying sine law to ABC, we get

oo 40sin

BC

80sin

BA = . So BA =o

o

40sin

80sinBC= 2 BC cos 40o

and BI = BC cos 40o. In BFC, BFC = 70o = BCF. SoBC = BF. Then BI = BF cos 40o = BH yielding H = I. Then

AF BC. So G = H = I and GFC = HFC = 20o.

8. Since the figure is the same after 90o, 180o or 270o

rotation about the centre of ABCD, the region must be a

square. Let this square has side length s. Now AF

=21

29BFAB 22 =+ . The area of AFCH is AHAB

=21

1and is also AFs = s

21

29. So s =

29

1and s2 =

841

1.

9. Let w = x

2

1, then w2 + y2 = 1. So w = cos and y = sin for some . Then x2 + 2xy + 4y2 +

x + 2y = 4 + 4 cos sin + 2 cos + 2 sin . Now let u = cos + sin = )45sin(2 o+ ,

then | u | 2 , u2 1 = 2 cos sin . Thusx2 + 2xy + 4y2 + x + 2y = 4 + 2(u2 1) + 2u = 2(u2 + u + 1) 2(2 + 2 + 1) = 6 + 2 2

with equality if u = 2 , = 45o, x = 2 , and y =2

2. So the maximum is 6 + 2 2 .

10. Let the length, width and height of the cuboid be L, W, H respectively. If L,W,H are at leasttwo, then x = (L 2)(W 2)(H 2), y = 2[(L 2)(W 2) + (W 2)(H 2) + (H 2)(L 2)] and z = 4[(L 2) + (W 2) + (H 2)]. If we let a = L 2, b = W 2, c = H 2, then1994 = x y + z 8 = abc2(ab + bc + ca)+4(a + b + c)8 = (a 2)(b 2)(c 2)=(L-4)(W-4)(H-4). Since no face is a square, taking all possible factorizations of 1994, the

numbers L,W,H are either 1001, 6, 5 or 1001, 3, 2. So LWH = 30030 or LWH=6006.

If one of L,W,H is 1, say H=1, then x=y=0 and z=2002=(L 2 )(W 2 ). Now.1311722002 = Then 2002 can be factored in 8 ways as product of two positive

integers. These give 8 more possible answers for the volume, which are 6012, 4012, 2592,

2392, 2340, 2320, 2232 and 2212.

B

A

C

D

E

F

G IH

s

D C

A B

H

E

F

G

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11. Attempting to draw a figure, it leads to the suspicion

that AB = AC. This is the case because the

circumradius of BMA beingBAMsin2

BM

would

equal the circumradius of CAN so that bothcircumcentres are symmetric with respect to the

perpendicular bisector of BC, which must then

contain A. Since AM, AN bisect BAN, CAM respectively, we getAC

AM

4

3

AB

AN == . If

AC = x, then AM = ANx4

3 = . By cosine law, 9 = MANcosx4

32x

4

32

22

and 16

= NACcosxx432xx

43 2

2

+

. Since MAN = NAC, solving the equations, we get

AC = x = 8.

12. Let a n be the number of n-digit positive integers with the two properties. First, notice that a1

= 0 and a2 = 1. For n 3, of the a n integers, there are a n 1 of them begin with 2, there are an 1 of them begin with 12 and there are 2

n 2 of them begin with 11.

So a n = an 1 + an 2 + 2n 2 for n 3. Then a3 = 3, a4 = 8, a5 = 19, a6 = 43, a7 = 94, a8 = 201,

a9 = 423, a10 = 880.

13. Let y = 2001 + n2. Without loss of generality, assume n 0. Since x > y, we have x = 2001 +(n + 1)2. Now, d divides x, y implies d divides x y = 2n + 1. Then d divides 2y n(2n + 1)= 4002 n and 2(4002 n) + (2n + 1) = 8005. Hence the maximum d can be is 8005. Thenn = 4002 + 8005k and x = 2001 + (4003 + 8005k)2. Since x < 5107, k = 0 and x = 2001 +40032 = 16026010.

14. Let {x} = x [x]. Note 0 {x} < 1. If x > 1, then the equation becomes x = 1x}x{

. This

leads to {x} = x(x 1) > x 1 x [x] = {x}, a contradiction. Also, if x < 1, then the

equation becomes x 2 =x1

}x{

. If x 3, then {x} = (x 2)(1 x) 14 > {x}, a

contradiction. If3 < x < 2 , then {x} = x + 3 and the equation becomes x 2 =x1

3x

+

,

which has the solution x = 5 . So the largest possible | x | is 5 .

4 3 4B C

A

NM

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15. Let [XY Z] denote the area of polygon XY Z.

Since1k

k

AB

AP

]SAB[

]SAP[

+== and

1k

1

DA

SA

]DAB[

]SAB[

+== ,

we get2)1k(

k

]DAB[

]SAP[

+= .

Similarly, .)1k(

k

]CDA[

]RDS[

]BCD[

]QCR[

]ABC[

]PBQ[2+

=== Now

[PQRS] = [ABCD] ([SAP] + [PBQ] + [QCR] + [RDS])

= [ABCD] )]CDA[]BCD[]ABC[]DAB[()1k(

k

]ABCD[2

2 444444 3444444 21=

++++

= ]ABCD[)1k(

1k2

2

++

.

So 52.0)1k(

1k2

2

=++

. Solving for k and noting k < 1, we get k =3

2.

16. There are 303C ways of choosing integers a, x, d such that 1 a < x < d 30. The equation a

+ d = x + x is possible if and only if a and d are both even or both odd, which account for

15

2

15

2 CC + cases. In the remaining 3850C2C15

2

30

3 = cases, we have a + d = x + x with x

x and a < x, x < d. So the answer is .19252

3850 =

17. Suppose 7n1nn21 aaaaa.0p

1++= &L&L . Let x and y be the integers with digits a1 a n and an +

1 an + 7 respectively. Then .110

yx

p

107

n

+= So 10n (107 1) = p[x(107 1) + y]. Then p

divides 10 or 107 1 = 322394649. Since 2, 3, 5 are not such prime, so if p 4649, then p

can only be 239. As 1041840.0239

1 &&= , the only other choice of p is 239.

A B

C

D

P

Q

R

S

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18. Consider ABC with a = BC =3

4, b =

AC = 2 , c = AB =3

10. Let D be the

midpoint of BC, E on BC such that AE

BC and F on AC such that FD BC.Consider folding along MN BC withM on BC and N on AB or AC. If N is on

AB, then the maximum overlapped area occurs when N = A. If N is on CF, then the

maximum overlapped area occurs when N = F. So for the problem, we may assume N is on

FA. By cosine law, cos C =2

1

ab2

cba 222 =+ and cos B =10

1

ac2

bca 222 =+ . Then C

= 45

o

and sin B = 10

3

.

Let C be on line BC such that M is the midpoint of CC and G be the intersection of AB andNC. In BCG, BGC = ABC NCC = ABC 45o. So sin BGC = sin B cos C

sin C cos B =5

1.

Let x = DM, then BC = 2x and CG =CBGsin

CGBsinCB

=

5

1

10

3x2

= x23 .

As usual, let [XY Z] denote the area of polygon XY Z. Then [CMN] =2

x3

2

2

1

+ ,

[BCG] =2

1BCCG sin 45o = 3x2 and the overlapped area [MBGN] =

2

x3

2

2

1

+ 3x2

= 15

4

15

2

x2

52

+

, which is maximum when DM = x =152

. (Since DE = CE DE = 1

3

1

3

2 = , M is on segment DE and N is on segment FA.) In that case [MBGN] =15

4.

x

C C'B

A

N

EM

G

D

F

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International Mathematical Olympiad (IMO)

Hong Kong Teams

29th IMO, July 1988, Canberra, Australia

Chan Shun Queens College

Choi Shu-Hung Kings College

Leung Wing-Hong Queen Elizabeth School

Tai Wai-Ling Diocesan Boys School

Yau Shuk-Han St. Paul s Co-educational College

Yip Nung-Kwan Tsuen Wan Government Secondary School

30th IMO, July 1989, Braunschweig, West Germany

Chan Kai-Pak St. Paul s Co-educational College

Chiu Shin-Yeung King s College

Choi Shu-Hung Kings College

Man Lai-Chee St. Paul s Co-educational College

Tam Ting-Kin Kwun Tong Government Secondary School

Yau Shuk-Han St. Paul s Co-educational College

31st IMO, July 1990, Beijing, China

Chan Kin-Wah TWGHs Mrs. Wu York Yu Memorial College

Cheng Wing-Leung Ha Kwai Chung Government Secondary Technical School

Ho Yuk King s College

Lau Chi-Hin Pentecostal School

Leung Wing-Kai Queen s College

Ma Jim-Lok Kings College

32nd IMO, July 1991, Sigtuna, Sweden

Chan Kin-Wah TWGHs Mrs. Wu York Yu Memorial College

Cheung Cheuk-Wah, Trevor Queens College

Ko Chi-Kin. Thomas Queens College

Ngai Chi-Ho Diocesan Boys School

To Kar-Keung Kings College

Wong Wilkie Diocesan Boys School

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33rd IMO, July 1992, Moscow, Russia

Chung Wai-Yin Kings College

Lam Chi-Wai Chuen Yuen College

Lam Pei-Fung Queens College

Lee Wai-Fun St. Mark s School

Suen Yun-Leung Kings College

To Kar-Keung Kings College

34th IMO, July 1993, Istanbul, Turkey

Chan Tsz-Ho Ying Wa College

Chu Hoi-Pun Chong Gene Hang College

Lam Chi-Wai Chuen Yuen College

Lin Kwong-Shing Tsuen Wan Government Secondary School

Tsui Ka-Hing Queen Elizabeth School

Yung Fai Chuen Yuen College

35th IMO, July 1994, Hong Kong

Chu Hoi-Pun Chong Gene Hang College

Ho Wing-Yip Clementi Secondary School

Poon Wai-Hoi, Bobby St. Pauls College

Suen Yun-Leung Kings College

Tsui Ka-Hing Queen Elizabeth School

Wong Him-Ting Salesian English School

36th IMO, July 1995, Toronto, Canada

Cheung Kwok-Koon SKH Bishop Mok Sau Tseng Secondary School


Mok Tze-Tao, Edmond Queens College


Wong Him-Ting Salesian English School

Yu Chun-Ling Ying Wa College

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37th IMO, July 1996, Mumbai, India


Law Siu-Lung Diocesan Boys School



Tse Shan-Shan Tuen Mun Government Secondary School

Yu Chun-Ling Ying Wa College

38th IMO, July 1997, Mar del Plata, Argentina

Chan Chung-Lam Bishop Hall Jubilee School

Cheung Pok-Man STFA Leung Kau Kui College

Lau Lap-Ming, Alvin St. Paul s College

Leung Wing-Chung Queen s Elizabeth School


Yu Ka-Chun Queens College

39th IMO, July 1998, Taipei, Taiwan

Chan Kin-Hang Bishop Hall Jubilee School

Cheung Pok-Man STFA Leung Kau Kui College

Choi Ming-Cheung Kings College

Lau Lap-Ming, Alvin St. Paul s College

Law Ka-Ho Queens Elizabeth School

Leung Wing-Chung Queen s Elizabeth School

40th IMO, July 1999, Bucharest, Romania

Chan Ho-Leung Diocesan Boys School


Chan Tsz-Hong Diocesan Boys School


Ng Ka-Wing STFA Leung Kau Kui College

Wong Chun-Wai Choi Hung Estate Catholic Secondary School

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41st IMO, July 2000, Taejon, Korea


Fan Wai-Tong St. Mark s School


Ng Ka-Wing STFA Leung Kau Kui College

Wong Chun-Wai Choi Hung Estate Catholic Secondary School

Yu Hok-Pun SKH Bishop Baker Secondary School

42nd IMO, July 2001, Washington D. C., U.S.A.


Chao Khek-Lun Harold St. Pauls College

Cheng Kei-Tsi, Daniel La Salle College

Ko Man-Ho Wah Yan College, Kowloon

Leung Wai-Ying Queen Elizabeth School

Yu Hok-Pun SKH Bishop Baker Secondary School

Prelim 1998 2002

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