Singapore Mathematical Olympiad (2005-2013)

Mathematics Olympiad

2005-2013

SINGAPORE

MATHEMATICAL OLYMPIAD

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Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)

2012

Junior Section (First Round)

Tuesday,

29

May

2012

0930-1200

hrs

Instructions to contestants

1. Answer ALL 35 questions. 2. Enter your answers on the answer sheet provided. 3. For the multiple choice questions, enter your answer on the answer sheet by shading the

bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.

4. For the other short questions, write your answer on the answer sheet and shade the ap propriate bubble below your answer. 5. No steps are needed to justify your answers. 6. Each question carries 1 mark. 7. No calculators are allowed. 8. Throughout this paper, let

example,

=

=

lxJ

denote the greatest integer less than or equal to x. For

l2.lj 2, l3. 9J 3.

PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO.

1

Multiple Choice Questions 1. Let

a

and (3 be the roots of the quadratic equation x2

value of

(a

- f3? is

(B) 1;

(A) 0;

(C)

2;

+

2bx + b

=

1. The smallest possible

(E) 4.

(D) 3;

2. It is known that n2012 + n2010 is divisible by 10 for some positive integer n. Which of the following numbers is not a possible value for n? (A)

2;

(B) 13;

•

(E) 59.

(D) 47;

(C) 35;

3. Using the vertices of a cube as vertices, how many triangular pyramid can you form?

(B) 58;

(A) 54;

4.

AB

is a chord of a circle with centre

with

AB

closer to

C

than to

(E) 70.

(D) 64;

(C) 60;

D.

0. CD

Given that

is the diameter perpendicular to the chord AB,

LAOB =

90° , then the quotient

area of .6.ABC area of .6.AOD (A) J2 -1;

(B)

v;;

2 - J2;

5. The diagram below shows that lines. Let

F and

BG = EF,

G

ABCD

DE AE.

2

is a parallelogram and both

be the intersections of

find the quotient

(E) �. -

(C)

BE

E

2

with

CD

and

AC

AE

and

BE

are straight

respectively. Given that

6. Four circles each of radius x and a square are arranged within a circle of radius 8 as shown in the following figure.

What is the range of x?

(B) 0

(A) 0
<

x < 8 ( J2 + 1);

(C)

(E) 4- J2
4- 2J2 < x < 8(J2 -1);

7 . Adam has a triangular field

4- 2J2
ABC

with

AB =

5,

BC =

8 and

CA =

separate the field into two parts by building a straight fence from BC

such that

(A)

4 J2I . 11 '

AD

(B)

bisects

�;

L.BAC.

4

(C)

A

(D)

11 '

5�;

11. H e intends to

to a point

Find the area of the part of the field

5J2I.

+ 1).

ABD.

a

and b be real numbers with b =/= 0 such that

Which of the following statements is incorrect? (A) If b is an integer then

a

is an integer;

(B) If a is a non-zero integer then b is an integer; (C) If b is a rational number then

a

is a rational number;

(D) If a is a non-zero rational number then b is a rational number;

(E) If b is an even number then a is an even number. 3

on side

(E) None of the above.

8. For any real number x, let lxJ be the largest integer less than or equal to x and x Let

D

=

x

- lxJ.

9. Given that

--;x;---;x;

y=

X

is an integer. Which of the following is incorrect?

x can admit the value of any non-zero integer; (B) x can be any positive number; (C) x can be any negative number; (D) can take the value 2; (E) can take the value -2.

(A)

y y

10. Suppose that

A, B, C

are three teachers working in three different schools X , Y, Z and spe

cializing in three different subjects: Mathematics, Latin and Music. It is known that A

(i)

does not teach Mathematics and

B

does not work in school Z;

(ii) The teacher in school Z teaches Music; (iii) The teacher in school X does not teach Latin; B

(iv)

does not teach Mathematics.

Which of the following statement is correct? (A)

B

works in school X and

(B)

A

teaches Latin and works in school Z;

(C)

B

teaches Latin and works in school

(D)

A

teaches Music and

C

C

works in school

Y;

Y;

teaches Latin;

(E) None of the above. Short Questions a

=

a >

=

b be real numbers such that b, 2a + 2b 75 and 2-a + 2-b 12-1 . Find the value of 2a-b . x3 - x + 120 . an mteger. . . . . 12 . Fm. d the sum of all positive mtegers x sueh t h at (x - 1 )(x + 1) 11. Let

and

IS

4

13. Consider the equation

vh

x2 - 8x + 1 + J 9x2 - 24x - 8

=

3.

It is known that the largest root of the equation is - k times the smallest root . Find k.

14. Find the four-digit number abed satisfying 2(abed) + 1000 deba. (For example, if a 1, b 2, e 3 and d 4, then abed =

=

=

=

=

=

1234.) =

15. Suppose x andy are real numbers satisfying x2 + y 2 - 22x - 20y + 221 0. Find xy. 16. Let m and n be positive integers satisfying mn2 + 876 4mn + 217n. Find the sum of all possible values of m. =

17. For any real number x, let lxJ denote the largest integer less than or equal to x. Find the value of lxJ of the smallest x satisfying lx 2 J - lx J 2 100. =

18. Suppose XI, x2 , . . . , X 49 are real numbers such that Find the maximum value of

XI + 2x2 + + 49x49 · ·

·

·

19. Find the minimum value of Jx 2 + ( 20 -y ) 2 + Jy 2 + ( 21 - z ) 2 + Jz 2 + ( 20 -w ) 2 + Jw 2 + ( 21 - x ) 2 . 20. Let A be a 4-digit integer. increased by

number is

n,

When both the first digit (left-most) and the third digit are

and the second digit and the fourth digit are decreased by

n times

A.

Find the value of A.

21. Find the remainder when 1021I 022 is divided by 1023. 5

n, the new

22. Consider a list of six numbers. When the largest number is removed from the list, the average

is decreased by 1. When the smallest number is removed, the average is increased by 1. When both the largest and the smallest numbers are removed, the average of the remaining four numbers is

20. Find the product of the largest and the smallest numbers.

23. For each positive integer Suppose that a1

a2012.

=

n 2:

1, we define the recursive relation given by an+l

=1 1

+an

.

Find the sum of the squares of all possible values of a1.

24. A positive integer is called

friendly if it is divisible by the sum of its digits. For example,

111 is friendly but 123 is not. Find the number of all two-digit friendly numbers.

25. In the diagram below, D and E lie on the side AB, and Flies on the side AC such that DA=DF=DE, BE= EF and BF= BC. It is given that L_ABC= 2 L_ACB. Find x, where L_BFD= X0•

B

F

26. In the diagram below, A and B(20,0) lie on the x-axis and C(O, 30) lies on they-axis such that L_ACB= 90°. A rectangle DEFG is inscribed in triangle ABC. Given that the area of triangle CGF is 351, calculate the area of the rectangle DEFG. y

c

A

D

0

6

E

B

X

27. Let

ABCDEF

be a regular hexagon. Let

G

be a point on

area of AGEF is 100, find the area of the hexagon G

E

ED

such that

ABCDEF.

EG =

3GD . If the

D

F

c B

A

28. Given a package containing 200 red marbles, 300 blue marbles and 400 green marbles. At each occasion, you are allowed to withdraw at most one red marble, at most two blue marbles and a total of at most five marbles out of the package. Find the minimal number of withdrawals required to withdraw all the marbles from the package.

29. 3 red marbles, 4 blue marbles and

5 green marbles are distributed to 12 students.

Each

student gets one and only one marble. In how many ways can the marbles be distributed so that Jamy and Jaren get the same colour and Jason gets a green marble?

30 . A round cake is cut into

n

pieces with 3 cuts. Find the product of all possible values of n.

31 . How many triples of non-negative integers ( x, y, z) satisfying the equation

xyz + x y + yz + zx + x + y + z = 2012? 32. There are 2012 students in a secondary school. Every student writes a new year card. The cards are mixed up and randomly distributed to students. Suppose each student gets one and only one card. Find the expected number of students who get back their own cards.

33. Two players

A

and B play rock-paper-scissors continuously until player

A

wins 2 consecutive

games. Suppose each player is equally likely to use each hand-sign in every game. What is the expected number of games they will play? 7

34. There are 2012 students standing in a circle; they are numbered 1, 2, . . . , 2012 clockwise. The counting starts from the first student ( number 1) and proceeds around the circle clockwise. Alternate students will be eliminated from the circle in the following way: The first student

stays in the circle while the second student leaves the circle. The third student stays while the fourth student leaves and so on. When the counting reaches number number

2012, it goes back to

1 and the elimination continues until the last student remains. What is the number

of the last student?

n

< n.

35. There are k people and chairs in a row, where 2 ::; k There is a couple among the k people. The number of ways in which all k people can be seated such that the couple is seated together is equal to the number of ways in which the (k - 2) people, without the couple present, can be seated. Find the smallest value of n.

8

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012 Junior Section (First Round)

Multiple Choice Questions

1. Answer: (D) . Since x 2 + 2bx + (b - 1) 0, we have a/3 b - 1 and a + f3 -2b. Then (a- /3) 2 (a + /3) 2 - 4af3 ( -2b) 2 - 4(b - 1) 4(b2 - b + 1) 4[(b - 1 2) 2 + 3 4] 3 The equality holds if and only if b 1 2. 2. Answer: (E) . Note that n 2o 12 n 201o n 201o (n 2 + 1). If n 2, then 5 (n 2 + 1) and 2 n2010 . If n 13 or 47, then 10 (n 2 + 1). If n 35, then 2 (n 2 + 1) and 5 n 2010 . If n = 59, then 5 f (59 2 + 1) and 5f n 2010 . 3. Answer: (B). There are 8 vertices in a cube. Any 4 vertices form a triangular pyramid unless they lie on the same plane. (!)- 6 -6 58 4. Answer: (B). Let the radius of the circle be 1. Then AE CE 2 CE 2(1 - 1 J2) 2 _ J2 sl::.ABc AE cE 1 OD � AE OD Sl::.AOD Sl::.AOD =

=

=

=

=

=

=

�

=

+

=

=

=

=

=

x

x

x

=

x

9

x

=

=

c

D

5. Answer: (A) . DE . Then = EF . Note that Let AE EB DF + FC EF + FG EF + EB - 2EF + 2 1 - DC AB - AB AB - EB GB - EB EF - (I__ ) 3 - V5 (0 1). Then + - 3 = 0, and thus = -;1 2 6. Answer: (D) . x

x

=

_

_

_

_

x

< x<

x

x

_

Consider the extreme cases:

Then Xmin

= 8 - 24� = 4 - 2� and

7. Answer: (D)

By Heron's formula,

Then

Xmax

= �28+ 1 = 8(� - 1).

s6.ABC s6.ABD s6.ACD

=J12(12 - 5)(12 - 8)(12 - 11) 4v'2I. 5 � X AB X AD X sin a AB � X AC X AD X sin a AC 11 5v'2I 5 S6.ABD = 5 + 11 X 451 =-4 =

10

x

B

c

A 8. Answer: (B) .

� = b ( l�J + { �} ) = b l�J + b { �} · It follows from a = b l�J + b { �} that b l�J = a. Hence, � = l�J is an integer. Note that a = b ·

Obviously, b is not necessary an integer even if a is an integer.

9. Answer: (C) .

0, then = -= ( -x) = �x -= --2x = -2.

x can take any nonzero real number. If x

If x

<

0, then

y

=

-=

-fE- -fE

X

-fE-

X

-

-fE- -$

y

>

-fE

X

- xX

-8-

=- = X

-

X

X

10. Answer: (C) .

The assignment is as follows:

A: in Z, teaches Music; B: in

Y,

teaches Latin;

C:

in X , teaches Mathematics.

Short Questions

11. Answer: 4. 75 (2a + 2b )(2-a 2-b ) = 2 2a-b + 2b-a . Then 2a-b 2b-a = 417 = 4 4"1 12 Since a b, we obtain 2a-b = 4. 12. Answer: 25. x 3 - x + 120 120 . It 1s. an .mteger "f and on y . f ( x2 - 1) -120 . =x Note t h at (x 1) (x 1) x2 1 Then x2 - 1 = ±1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120. Hence, x = 0 2 3 4 5 11; and 2 + 3 + 4 + 5 + 11 = 25. 13. Answer: 9. +

=

+

+

+

>

_

+

+

1

_

11

1

1

0.

Let y =v'3x 2 - 8x +

1 . Then the equation becomes y y'3 2 - 11 = 3 . Theny'3y 2 - 11 = 3 - y. Squaring both sides, we have 3y 2 - 11 = 9 - 6y y 2 + 3y - 10 = 0 . Then y = 2 or y = -5 (rej ected because y 0) . Solve 3x 2 - 8x 1 = 2 2 . Then x = 3 and x = -3- 1 . Hence, k = 3/3- 1 = 9 . 14. Answer: 2996 . y

+

2

+

+

y 2 ; that is,

Rewrite the equation as the following:

b c d b c d 0 0 + c b a Since a is the last digit of 2d, a is even; since 2a 1::::; d::::; 9, a::::; 4 . So a = 2 or a = 4 . If a = 4, then d 2a 1 = 9 and thus d = 9; but then the last digit of 2d would be 8 1- a, a a a 1 d

2

+

2

+

0

+

contradiction.

a = 2, then d 2a 1 = 5 and the last digit of 2d is 2; so d = 6 . The equation reduces to b c b c 1 1 c b There are 2 cases : either 2c 1 = b and 2b = 10 c, which has no integer solution; or 2c + 1 = 10 b and 2b 1 = 10 + c, which gives b = c 9 . 15 . Answer: 110 . Complete the square: (x - 11) 2 + (y - 10) 2 = 0 . Then x = 11 and y = 10; thus xy = 110 . 16 . Answer: 93 . If

+

+

+

+

+

=

Rearranging, we have

- 876 8 mn2 - 217n = 4mn - 876 n = 4mn mn - 21 7 = 4 - mn - 217 Then ( mn - 217) 8 . It follows that mn - 217 = ±1, ±2, ±4, ±8 . So mn = 218, 216, 219, 215, 221, 213, 225, 209, and n = -4, 12, 0, 8, 2, 6, 3, 5 respectively. mn 216 225 N ot e th at m = - 1s a positiVe mteger. T en m = - = 18 or m = - = 75 . n 12 3 =?-

.

. .

.

h

12

17. Answer: 50. Write x = lxJ + --fl: � Then 100:::; (lxJ + --fl:-) 2 - lxJ 2 = 2lxj--fl: -+ --fl:...J, 2lxJ + 1. S o lxJ 50 and x 2 lx2J = 100 + 502 = 2600. On the other hand, x = v'26QO is a solution. 18. Answer: 35. (X! + 2X 2 + + 49X 49 ) 2 = ( 1 X! + J2 J2X 2 + " " " + .j49 " .j49X 49 r :::; ( 1 + 2 + + 49) (xi + 2x� + + 49x�9) = 49 X2 50 X 1 = 352 . = X 49 = ± 351 The equality holds if and only if x1 = 19. Answer: 58. �

<

�

·

·

·

·

·

·

·

·

·

·

·

·

·

·

·

As shown below,

J x2 + (20 -y) 2 + Jy 2 + (21- z) 2 + J z2 + (20 -w) 2 + Jw 2 + (21 - x) 2 = AB + BC + CD +DA = A'B + BCJ + CD + DA." :::; A' A" = 422 + 402 = 58. 21-x A" A' A x • ::: ------- -.----"'7""o;::,...-----.,.. ----------:::,.. ', ;::,:, ,.. .... ....... """" ,' ;3 , .... ..I .... ',

,..,

o C'l

;3 I

B

'

21- z

C\

\ \

' \

\ jj

..

...

........

,,

...........

.......

,, A.. -" '

20. Answer: 1818.

A = abed. Then 1000(a + n) + 100(b - n) + 10(c + n) + (d - n) = nA. It gives A + 909n = nA; or equivalently, ( n - 1 )A = 909n. Note that (n - 1) and n are relatively prime and 101 is a prime number. We must have (n - 1) So n = 2 or n = 4. If n = 4, then A = 1212, which is impossible since b n. So n = 2 and A = 909 2 = 1818. Let the 4-digit number be

-9.

<

13

x

21. Answer: 4. Note that 1024 = 21 0 1 (mod 1023). Then 10211022 ( _ 2)1022 21022 21ox102+2 1024102 22 1102 22 4 22. Answer: 375. Let m and M be the smallest and the largest numbers. Then m + 80 + 1 = M + 80 1 = m + M + 80 5 5 6 Solving the system, m = 15 and M = 25. Then mM = 375. =

=

=

=

X

=

=

X

=

(mod

1023)

_

23. Answer: 3 .

+

1 a3 =- 1-,a a4 =--2 +- a, as =-3--, + 2a . . . . In general, a1 = a. Then a2 =-1 -, + a 2 + a 3 +2 a 5 + 3a an =FFnn+l++FnF-na1 a where F1 = 0, F2 = 1 and Fn +l = Fn Fn-1 for all 1. F + F2o11 a = a, then (a2 + a - 1)F2o12 = 0. If a 201 2 = 2o1 2 + L' 2012 a 2 L' 1 20 3 Since F 201 2 0, we have a + a - 1 = 0. Let and (3 be the roots of a 2 + a - 1 = 0. Then Let

-=-----'-==----

+

v

n�

v

a

>

23. ga + 1, we have (a + b) ga. . lOa + b = -Smce a+b a+b Case 1: If 3 f (a + b), then (a + b) a, and thus b = 0. We have 10 20 40 50 70 80. 3a ::::; 3. If 3a = 1, then Case 2: If 3 (a + b) but gf(a + b), then (a + b) 3a and 1::::; a+b a3a+ b 3a 2a = b, we have 12 24 48. If a + b = 2, then a = 2b, we have 21 42 84. If -= 3, then a + b b = 0, we have 30 60. Case 3 : If g (a+ b), then a+b = g or 18. If a+b = 18, then a = b = g, which is impossible. If a + b = g, then we have g friendly numbers 18 27 36 45 54 63 72 81 go.

24. Answer:

--

Therefore, in total there are 23 2-digit friendly numbers.

108. Since DA = DE = DF, LEFA = goo. Let LEBF = LEFB = x0• LBFC =goo - x0 and LCBF = 2x0, LBAC =goo - 2x0•

25. Answer:

14

Then

LBCF =

B

F

It is given that 3x = 2 ( 9 0 - x ) . Then x = 36. So x = 180 - ( 90 - x ) - ( 90 - 2x ) = 3x = 108. 26. Answer: 468.

AO = 30202 = 45. Then the area of !::,. ABC IS. ( 20 + 452 ) Let the height of !::,. CGF be Then 3 ( ) 2 = 351 = ( 3 ) 2 =

Note that

X

30

= 975 .

h.

h

30

Note that the rectangle

5

975

:::?

30

h

- h

DEFG has the same base as !::,. CGF. Then its area is 351

X

2 3

X

2 = 468

27. Answer: 240.

DE = 1 and denote A = EG = 3 4. EGAX = 1 +A AGEF 1 +A 1 1 + 2A . - -8 - Imp Ies 4 4 DEXY DEXY 8 ABCDEF 6 AGEF 1 + 2A Note that DEXY = 8" Then ABCDEF 6 12 If AGEF = 100, then ABCDEF = 100 = 240. S E A G D We may assume that -

2

1.

-

x

X/

I

I

I I

I

Fl

I

I

I

I

.'-------------.lL.----->L

A

B

28. Answer: 200. 15

5 12

Since at most one red marble can be withdrawn each time, it requires at least

200 withdrawals.

200 is possible. For example, 150 . (1, 2, 2) + 20 . (1, 0, 4) + 10 . (1, 0, 2) + 20 . (1, 0, 0) (200, 300, 400). 29. Answer: 3150. Case 1: Jamy and Jaren both take red marbles. So 1 red, 4 blue and 4 green marbles are distributed to 9 students: (�) (�) 630. Case 2: Jamy and Jaren both take blue marbles. So 3 red, 2 blue and 4 green marbles are distributed to 9 students: G) (�) 1260. Case 3: Jamy and Jaren both take green marbles. So 3 red, 4 blue and 2 green marbles are distributed to 9 students. The number is the same as case 2. 630 + 1260 + 1260 3150. 30. Answer: 840. With three cuts, a round cake can be cut into at least 1 + 3 4 pieces, and at most 1 + 1 + 2 + 3 7 pieces. Moreover, 4, 5, 6, 7 are all possible. 4 5 6 7 840. On the other hand,

=

=

X

=

X

=

=

=

n=

x

x

x

=

31. Answer: 27. ( + 1)(y + 1)( z + 1) 2013 3 11 61. I f all y, z are positive, there are 3 ! 6 solutions. If exactly one of y, z is 0, there are 3 6 18 solutions. If exactly two of y, z are 0, there are 3 solutions. 6 + 18 + 3 27. 32. Answer: 1. 1 . Then the expectation For each student , the probability that he gets back his card is 20 12 1 . of the whole class 2012 2012 1. x

=

x,

=

x

=

x,

=

x

x,

IS

x

x

=

=

16

33. Answer: 12. Let E be the expectation. If

A

does not win, the probability is

If A wins and then does not win, the probability is

2/3 and the game restarts.

(1/3)(2/3) and the game restarts. The probability that wins two consecutive games is (1/3)(1/3). Then E = 32 (E + 1) + 92 (E + 2) + 91 2. Solving the equation, we get E = 12. 34. Answer: 1976. If there are 1024 = 21 0 students, then the 1024th student is the last one leaving the circle. Suppose 2012 - 1024 = 988 students have left. Among the remaining 1024 students, the last student is (2 988 - 1) + 1 = 1976. 35. Answer: 12. (n - 1) 2 (� = �) ( k - 2)! = (k : 2) ( k - 2)!. Then 2 = (n k + � n k + 2 ) That is, 2n 2 - (4k - 6)n + (2k 2 - 6k + 4 - n) = 0. We can solve 7 - 5 (n k). v8k --n = k +-4 Note that the square of any odd number has the form 8k-7. Choose k so that v8k - 7-5 = 4, i.e . , k 11. Then n 12. A

X

X

x

x

x

x

x

x

_

>

=

=

17

1

_

"

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012

(Junior Section, Round 2) 0930-1230

Saturday, 23 June 2012

0 be the centre of a parallelogram ABCD and P be any point in the plane. Let M, N be the midpoints of AP, BP, respectively and Q be the intersection of MC and ND. Prove that 0, P and Q are collinear.

1. Let

A such that each of the ten digits 0, 1, exactly once as a digit in exactly one of the numbers A, A2, A3.

2. Does there exist an integer

. . . , 9 appears

L.ABC, the external bisectors of LA and LB meet at a point D. Prove that the circumcentre of L.ABD and the points C, D lie on the same straight line.

3. In 4.

Determine the values of the positive integer n for which the following system of equations has a solution in positive integers x1, x2, , Xn. Find all solutions for each such n. •

.

Xl+ X2 + · · · + Xn = 16 1 1 1 + + . . ·+ - = 1 Xn X1 X2 5.

.

(1) (2)

Suppose S = a1, a2, , a15 is a set of 1 5 distinct positive integers chosen from 2 , 3, ... , 2012 such that every two of them are coprime. Prove that S contains a prime number. (Note" Two positive integers m, n are coprime if their only common factor is 1.) •

.

.

18


(Junior Section, Round 2 solutions)

MN II AB II CD, we have �MQN"' �CDQ. Hence MN = AB/2 = CD/2. Thus QM = CQ/2. In �ACP, CM is a median and Q divides CM in the ratio 1:2. Thus Q is the centroid. Hence the median PO passes through Q.

1. Since

p

Since the total number of digits in A, A2 and A3 is 10, the total number digits in 32, 322, 32 3 is 11 and the total number of digits in 20, 202, 203 is any solution A must satisfy 21 :::; A :::; 31. Since the unit digits of A, A2, A3 are distinct, the unit digit of A can only be 2, 3, 7, 8. Thus the only possible values of A are 22, 23, 27, 28. None of them has the desired property. Thus no such number exists.

2.

9,

CD bisects LC. If CA = CB, then CD is the perpendicular bisector of AB. Thus the circumcentre of �ABD is on CD. 3. Note that

If C A =/:. CB, we may assume that CA > CB. Let E be a point on CB extended and F be the point on CA so that C F = CB. Then, since CD is the perpendicular bisector of

19

BF, we have LAFD = LDBE = LDBA. Thus AFBD is a cyclic quadrilateral, i.e . , F is on the circumcircle of 6.ABD. The circumcentre lies on the perpendicular bisector of BF which is CD. 4.

then Without loss of generality, we may assume that x1 S x2 S S Xn· If x 1 = from n = and cannot be satisfied. Thus x1 � If x2 = then n = 2 and again cannot be satisfied. Thus x2 � Similarly, X3 � Thus x4+ + Xn S with x4 � Thus n s ·

(2), (1)

1

(1) 4.

4.

·

2.

3.

1,

·

4.

2,

·

·

7

·

n =

(i)

1: No solution. (ii) 2: The only solution of + 1 is x1 x2 2 which doesn't satisfy (1) . Thus there is no solution. (iii) 3: The only solutions of; + +;3 1 are (x1 , x2 , x3) (2, 3, 6) , (2, 4, 4) and ( 3, 3, 3) . They all do not satisfy (1) . (iv) 4: According to the discussion in the first paragraph, the solutions of x 1 + + X4 16 are (x1 , x2 , x3 , x4) (2, 3, 4, 7), (2, 3,5, 6) ,(2, 4,4, 6) ,(2, 4,5,5), (3, 3, 4, 6) , (3, 3,5,5),(3, 4,4,5),(4, 4, 4,4) . Only the last one satisfy (2) . Thus the system of equations has a solution only when 4 and for this the only solution is x1 x2 X3 X4 4. __!_

n =

Xl

n =

n =

·

·

·

1

__!_ 2 X

x12

=

=

=

=

=

=

=

=

=

=

n =

=

n,

5.

Suppose, on the contrary, that S contains no primes. For each i, let Pi be the smallest prime divisor of ai . Then Pl, P2 , . . . , P1 5 are distinct since the numbers in S are pairwise coprime. The first primes are 5, If = = > and aj � Pj is the largest among Pl, P2 , . . . , p1 5 , then Pj � a contradiction. Thus S must contain a prime number.

15

2, 3, 7, 11, 13, 17, 19 , 23, 29 , 31, 37, 41, 43, 47. 47 472 472 2309 2012,

20


2012

Senior Section (First Round)

Tuesday,

29

May

2012

0930-1200

hrs

Instructions to contestants 1.

2.

3.

Answer ALL

35

questions.

Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.

4- For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. 5. 6. 7.

8.

No steps are needed to justify your answers. Each question carries

1

mark.

No calculators are allowed. Throughout this paper, let LxJ denote the greatest integer less than or equal to x. For example, L2.1J = 2, L3.9J = 3.

PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO

21


1. Suppose

a

and (3 are real numbers that satisfy the equation

) V J2+ 1

(V

x2 + 2 J2+ 1 x+

�3 +;3 .

=

0

Find the value of

(A) 3 J J2+ 1(J2 - 1) - 8

(B) 8 -6v\12 + 1(J2 - 1)

(C) 3 JJ2+ 1(J2 - 1)+ 8

(D) 6 J J2+ 1(J2 - 1) - 8

(E) None of the above

2. Find the value of 20112 X 2012 - 2013 20132 X 2014 - 2015 -- -------+ -------,-2012! 2014! 1 + 1 - 1 - 1 2 0 1 0! 2 011! 2 01 2 ! 2 009! 1 + 13 + 1 (D) 1 + 2 010! 2 01 ! 2 01 4 ! 2 009!

1 + 1 + 1 + 1 2 010! 2 011! 2 01 2 ! 2 009! 1 + 1 - 1 - 13 (C) 2 010! 2 01 2 ! 2 01 ! 2 009! 1 + 1 - 13 - 1 (E) 2 0 1 0! 2 01 ! 2 01 4! 2 009! (A)

(B)

3. The increasing sequence T = 2 3 5 6 7 8 10 11 consists of all positive integers which are not perfect squares. What is the 2012th term of T? (A) 2055

(B) 2056

(C) 2057

(D) 2058

(E) 2059

4. Let 0 be the center of the inscribed circle of triangle �ABC and D be the point on AC with OD _LAC. If AB = 10 AC = 9 BC = 11, find CD.

(A) 4

(B) 4.5

(C) 5

(D) 5.5 22

(E) 6

5. Find the value of

(A)

1

cos4 75° + sin4 75° + 3 sin2 75° cos2 75° cos6 75° + sin6 75° + 4 sin2 75° cos2 75° ·

(B) �

(C)

�

(D) 1

(E) cos 75° + sin 75°

6. If the roots of the equation x2 + 3x - 1 = 0 are also the roots of the equation x4 + ax2 + bx+ c = 0, find the value of a+ b+ 4c. (A) -13

(B) -7

(C) 5

(D) 7

(E) 11

7. Find the sum of the digits of all numbers in the sequence 1, 2, 3, 4, . . . , 1000. (A) 4501

(B) 12195

(C) 13501

(D) 499500


8. Find the number of real solutions to the equation X

100

=

. smx,

where x is measured in radians. (A) 30

(B) 32

(C) 62

(D) 63

(E) 64

9. In the triangle 6ABC, AB = AC, LABC = 40° and the point D is on AC such that BD is the angle bisector of LABC. If BD is extended to the point E such that DE= AD , find LECA. A

(A) 20°

(B) 30°

(C) 40°

(E) 50°

10. Let m and n be positive integers such that m > n. If the last three digits of 2012m and 2012n are identical, find the smallest possible value of m+ n. (A) 98

(B) 100

(C) 102

(D) 104

23


Short Questions

11. Let a b c and d be four distinct positive real numbers that satisfy the equations (a2m2 c2 o12)(a2012 d2012) 2011 and (b2o12 c2 o12) (b2o12 d2o12) 2011 Find the value of (cd ) 2 01 2 -(ab ) 2 012 . 12. Determine the total number of pairs of integers x and y that satisfy the equation 1 1 1 +2 3' 2X _

_

_

=

_

Y

=

Y

13. Given a set S 2 , a collectionF of subsets of S is said to be intersecting if for any two subsets A and B in F, we have A n B -=1 0. What is the maximum size ofF? =

1

10

14. The set M contains all the integral values of m such that the polynomial 2 ( m -1) x2 - ( m2 - m +12)x + 6m has either one repeated or two distinct integral roots. Find the number of elements of M.

15. Find the minimum value of

I

sin x+ cos x

+ coscosx -2xsin x I ----

16. Find the number of ways to arrange the letters A, A, B, B, C, C, D and E in a line, such that there are no consecutive identical letters. x

17. Suppose x 3V2+Iog3 is an integer. Determine the value of x. 18. Let f(x) be the polynomial (x-a1 )(x-a2)(x-a3)(x-a4)(x-as) where a1 , a2, a3, a4 and a5 are distinct integers. Given that f (104) 2012, evaluate a1+a2+a3+a4+a5. 19. Suppose x, y, z and are positive real numbers such that yz 6-Ax xz 6-Ay xy 6-Az x2+y2+z2 1 Find the value of (xyz.A) -1 . =

=

A

=

=

=

=

24

20. Find the least value of the expression real numbers satisfying the equation

(x +y )(y +z) , given that x, y, z are positive xyz(x+y+z) 1. =

21. For each real number let be the minimum of the numbers 2012. 4. Determine the maximum value of -2

x+

x, f(x)

6f (x) +

4x+1, x + 2 and

22. Find the number of pairs (A, B) of distinct subsets of 1, 2, 3, 4, 5, 6 such that A is a proper subset of B. Note that A can be an empty set. 23. Find the sum of all the integral values of that satisfy

x

JX+3- 4 �+ JX+ 8- 6� 24. Given that for real values of

=

1.

I x2+ 4x +5- v'x2+ 2x+5 1 ,

S = v'

x, find the maximum value of 84.

25. Three integers are selected from the set S = 1, 2, 3, ... , 19, 20 . Find the number of selections where the sum of the three integers is divisible by 3. 26. In the diagram below, ABCD is a cyclic quadrilateral with AB = AC. The line FG is tangent to the circle at the point C, and is parallel to BD. If AB = and BC = 4, find the value of 3AE.

6

A

25

27. Two Wei Qi teams, A and B, each comprising of 7 members, take on each other in a competition. The players on each team are fielded in a fixed sequence. The first game is played by the first player _of each team. The losing player is eliminated while the winning player stays on to play with the next player of the opposing team. This continues until one team is completely eliminated and the surviving team emerges as the final winner - thus, yielding a possible gaming outcome. Find the total number of possible gaming outcomes. 28. Given that m ( cos + ( sin and n ( .J2- sin ( cos where and are the usual unit vectors along the x-axis and the y-axis respectively, and E ( 1r 2n ) . If the length or magnitude of the vector m n is given by m n 8'f, find the value of 5 cos (� + i) 5.

= B) i

B) j

=

+

B) i + B ) j , i j B + =

+

29. Given that the real numbers x, y and the maximum value of j ( x y v2 x

z satisfies the condition x +y + z = 3, find +13 +�3y +5+V'8z +12. 30. Let P ( x ) be a polynomial of degree 34 such that P(k) = k (k + 1 ) for all integers z) =

k=0 1 2

34. Evaluate 42840 x P ( 35) .

31. Given that a is an acute angle satisfying

v369 - 360 cosa +v544 - 480 sina - 25 = 0 find the value of 40 tana. 32. Given that and

a b c d e are real numbers such that a+b +c+d+e = 8 a2+b2+ c2+d2+e2 = 16

Determine the maximum value of L eJ. 33. Let L denote the minimum value of the quotient of a 3-digit number formed by three distinct digits divided by the sum of its digits. Determine L10 £J . 34. Find the last 2 digits of

35. Let

f (n) be the integer nearest to yfii. Find the value of

26

Singapore Mathematical Society

Singapore Mathematical Olympiad (SMO) 2012 Senior Section (First Round Solutions)


1. Answer:

(D)

+

Note that a (3 =

-

(2V J2 +1) and a(3 J J2 +1. __!_ __!_ (a+ (3) 3- 3a(3(a + (3) + =

a3

(33

=

=

( af3) 3 s..)J2

+1 6vv'2+1 6(vv'2 +1)- s(J2 +1) v'2+1 6v,---.J'i- +-1 ( .J2 1) s -

2. Answer: (E)

27

-

3. Answer: ( C ) Note that 442 = 1936, 452 = 2025 and 462 = 2116. So 2, 3, . . . , 2012 has at most 2012 - 44 terms. For the 2012th term, we need to add the 44 numbers from 2013 to 2056. But in doing so, we are counting 452 = 2025, so the 2012th term should be 2012 + 44 + 1= 2057. 4. Answer: ( C )

B

is tangent to the circle at D, by constructing E and F as shown, we have CD= CF, AD= AE and BE= BF. Solving for the unknowns give CD= 5. AC

5. Answer: (D) cos6 x + sin6 x+ 4 sin2 x cos 2 x = ( cos 2 x + sin 2 x ) ( sin4 x + cos 4 x - sin 2 x cos 2 x ) + 4 sin 2 x cos 2 x = ( sin4 x+ cos4 x + 3 sin2 x cos 2 x ) . 6. Answer: ( B ) We have the factorization ( x2 + 3x - 1 ) ( x2 + mx - c) = x4 + ax 2 + bx + c.

Comparing coefficients give 3 + m= 0, -1 - c + 3m= a and -3c - m= b . We can solve these equations to obtain a+ b + 4c= -7. 7. Answer: ( C ) Among all numbers with 3 or less digits, each i, i= 0, 1, 2, , 9, appears exactly 300 times. Thus the sum of the digits of all the numbers in the sequence 1, 2, 3, 4, , 999 is 300 ( 1 + 2 + . . . + 9 ) = 13500, and so the answer is 13501. ·

·

·

·

28

·

·

8. Answer: (D) Since -1:::;; sin x:::;; 1, we have - 100:::;; x:::;; 100. We also observe that 317r < 100 < 327r. For each integer k with 1 :::;; k :::;; 16, 1 �0 = sin x has exactly two solutions in [2(k-1)7r (2k - 1)7r], but it has no solutions in ((2k - 1)7r 2k7r). Thus this equation has exactly 32 non-negative real solutions, i.e. x = 0 and exactly 31 positive real solutions. Then it also has exactly 31 negative real solutions, giving a total of 63. 9. Answer: ( C ) Construct a point F on EC such that EF= EA. Since LAED LFED, 6.AED is congruent to 6.FED. Thus DF = DA = DE and LFDE = LADE = 60°. We also have LEDC = LADE= 60°, which implies that LFDC= 60° and 6.CFD is congruent to 6.CED. In conclusion, LEGD= LFCD= 40°. =

A

E

10. Answer: (D) We want to solve 2012m

_

2012n (mod 1000) which is equivalent to

Since (12m-n - 1) is odd, we must have 8 12n so n 2:: 2. It remains to check that 125 12m-n - 1 i.e. 12m-n 1 ( mod 125). Let

29

Short Questions

11. Answer: 2011 Let A = a2 01 2 , B = b2 01 2 , C = c2 0 1 2 and D = d2 01 2 . Then A and B are distinct roots of the equation (x -C ) (x - D) = 2011. Thus the product of roots, AB = CD - 2011 and CD - AB = 2011. 12. Answer: 6 Note that x and y must satisfy 2x+ 1 · 3 = y(y + 2). We first assume x :;::: 0, which means both y and y + 2 are even integers. Either 3 y or 3 y + 2. In the first case, assuming y > 0, we have y = 3 · 2k and y + 2 = 2(3 . 2k-l + 1) = 2x+1 -k. The only way for this equation to hold is k = 1 and x = 3. So (x, y, y + 2) = (3, 6, 8). In the case 3 y+ 2, assuming y > 0, we have y+ 2 = 3· 2k and y = 2(3· 2k-1 - 1) = 2x+ 1 -k. Now the only possibility is k = 1 and x = 2, so (x, y, y + 2) = (2, 4, 6) . In the two previous cases, we could also have both y and y+ 2 to be negative, giving (x, y, y + = (3, -8, -6) or (2, -6, -4). Finally, we consider x < 0 so 3 = 2-x-ly(y + 2). In this case we can only have x = -1 and (x, y, y + 2) = (-1, 1, 3) or (-1, -3, -1). Hence possible (x, y) pairs are (3, 6) , (3, -8) , (2, 4) , (2, -6) , ( -1, 1) and ( -1, -3).

2)

13. Answer: 512 If A E F then the complement can belong to F, that is

S nA

fj. F. So at most half of all the subsets of

F:::;

S

210

2 = 512.

Equality holds because we can take F to be all subsets of S containing 1. 14. Answer: 4 If m = 1, the polynomial reduces to -12x + 6 = 0 which has no integral roots. For m-=/:- 1, the polynomial factorizes as ((2x - m) ( (m - 1)x - 6) , with roots x = r; and x = � . For integral roots, m must be even and m - 1 must divide 6. The m l values are m = -2, 0, 2 and 4. So M has 4 elements. only possible 15. Answer: 2 Note that cos 2x = cos2 x - sin2 x. So 1 cos x - sin x = sin x + cos x + . sin x + cos x + smx + cos x cos 2x Set w = sin x + cos x and minimize lw + �I· By AM-GM inequality, if w is positive then the minimum of w + � is 2; if w is negative, then the maximum of w + � is -2. Therefore, the minimum of lw + �I is 2.

I

I I

30

I·

16. Answer: 2220 We shall use the Principle of Inclusion and Exclusion. There are ways to arrange the letters without restriction. There are ways to arrange the letters such that both the As occur consecutively. (Note that this is the same number if we use B or C instead of A.) There are � ways to arrange the letters such that both As and Bs are consecutive. (Again this number is the same for other possible pairs.) Finally there are 5! ways to arrange the letters such that As, Bs and Cs occur consecutively. For there to be no consecutive identical letters, total number of ways is

21�i21

2i�1

� - 3 X _2!_ + 3 X 212121

2121

17. Answer: 9 J2 log3 Taking logarithm, we get log3 J2 solution for is 2. Therefore

y = +y

'

6! - 5. - 2220

21

-

x = + 2x Let y = log3 x. x = 3 = 9.

The only possible

18. Answer: 17 The prime factorization of 2012 is 2 503. Let b 104. If ai are distinct, so are b- ai, i.e. (b- a1 ) , (b- a3 ) , (b- a4 ) and (b- a5 ) must be exactly the integers 1 -1 2 -2 503 . Summing up, we have

= b( -a2) , 2 5(104) - ( a1 + a2+ a3 + a4 + a5 ) = 1 - 1 + 2 - 2 + 503 ·

19. Answer: 54 Multiplying the first three equations by

x, y and z respectively, we have xyz = 6-\x2 = 6-\y2 = 6-\z2 Since,\# 0 and x2+y2+z2 = 1, we deduce that x2 = y2 = z2 = !,sox = y = z = }s and x2 1 2 Hence \

_

/\-

xyz 6 x2

_

-

x3

- 6V3" _

20. Answer: 2 Observe that 1 (x+y)(y +z) = xy +xz+y2+yz = y (x+y +z) +xz = -+ xz xz 2: 2 where the equality holds if and only if xz = 1. Let x = z = 1 and y = .J2- 1, then we have the minimum value 2 for (x + y )(y + z ) . 31

21. Answer: 2028 Let L1, L2, L3 represent the three lines y + 1, y + 2 and y + respectively. Observe that L1 and L2 intersects at G, D, L1 and L3 intersects at (�, 3), and L2 and L3 intersects at Thus

= 4x

=x

=-2x 4

(�, �) .

1.

X< x=r3 ' �f (x) = x 2, 3 Thus the maximum value of f (x ) is � and the maximum value of 6f (x ) 4x+ 1, 7

3' 8

3'

1

+

3·

+

+

2028.

2012 is

22. Answer: Since B cannot be empty, the number of elements in B is between 1 to After picking B with k elements, there are 2k - 1 possible subsets of B which qualifies for A, as A and B must be distinct. Thus the total number of possibilities is

665

6.

t m (2.

_

1)

=t, m (2. 1) = t, m 2· - t, m =(2 1) 6 - (1 1) 6 =36 - 26 =665. _

+

+

23. Answer: The equation can be rewritten as (-,fx� - 2) 2 + Cv'x=-I. - 3)2 1. If -yX=l 2: 3, it reduces to -yX=l - + -yX=l - 3 1 i.e. -yX=l 3 giving 10. If -vX=1 ::;; 2, it reduces to 2 - -yX=l + 3 - -yX=l 1 i.e. -yX=l 2 giving

45

J

J

= 2 = = X= = = X=5. If 2 <-yX=l<3, i.e. 5
+

S

+

+

+

+

+

32

+

+

+

Let P = 0) , A = and B = then S represents the difference between the lengths PA and PB. S is maximum when the points P, A and B are collinear and that occurs when P = 0). So

(x,

(-2, 1)

(-1, 2) ,

(-3,

Thus the maximum value of S 4 =

4.

25. Answer: 384 Partition S into three subsets according to their residues modulo 3: S0 = 3, 6, . . . , 18 , = 1, 4, . . . , 19 and = 2, 5, . . . , 20 . In order for the sum of three integers to be divisible by 3, either all three must belong to exactly one or all three must belong to different Hence total number of such choices is (�) + 2 G) + 6 7 7 = 384. 26. Answer: 10 82

81

Si

si.

x

x

Since BD II FG and FG is tangent to the circle at C, we have LBCF = LOBE = LDCG = LBDC = LBAC. Furthermore LBEC = LBAC+ LABE = LOBE+ LABE = LABC = LACE. We can then conclude that BE = BC = DC = t:.DCE. If we let AE =

4. Also, f:.ABE is similar to DE AE 2 ===> DE = 3x. = DC AB By the Intersecting Chord Theorem, AE · EC = BE· ED, i.e. x(6- x ) = 4(�x ) , which gives x = so 3AE = 3x = 10. 27. Answer: 3432 x,

1 3°,

We use a1, a2, , a7 and b1, b2, , b7 to denote the players of Team A and Team B, respectively. A possible gaming outcome can be represented by a linear sequence of the above terms. For instance, we may have a1a2b1b2a3b3b4b5a4b5b7a5a6a7 which indicates player followed by player from Team A were eliminated first, and the third player eliminated was player from team B. However Team A emerged the final winner as all seven players of Team B gets eliminated with a5, a6 and a7 remaining uneliminated. ( Note a6 and a7 never actually played. ) Thus, a gaming outcome can be formed by choosing out of possible positions for Team A, with the remaining filled by Team B players. Therefore, the total number of gaming outcomes is given by C 74) = c�D'2 = •

14

•

•

•

1

•

•

1

2

7 14 3432.

33

28. Answer: 1

8� =

m

=V4 + 2 J2 cosB - 2 J2 sinB = j2 1+ cos ( �) = 2 J2 I cos (� + i) I

+n

() +

Since 1r < () < 27r, we have �7f < � + i < �7f. Thus, cos (� + i) < that cos (� + i) -� and hence

=

29. Answer: Using the

f(x

0. This implies

5 cos (� + i) + 5 = 1

8

AM-GM

inequality, we have

y z) =v2x + 13+{!3y + 5+�8z + 12 =v 2x ; 13. V4+�3y: 5 . {12. {12+«8z; 12 . � . � . � 2x+13 + 4 3y+5 + 2 + 2 8z+12 + 2 + 2 + 2 < + + ----8--"' 2 4 3 1 29 = 4 (x+ y+ z)+ 4 4

4

=8

The equality is achieved at x

_____

= �, y = 1 and z = �.

30. Answer: Let n 34 and Q(x) (x+ 1)P(x)-x Then Q(x) is a polynomial of degree n + 1 and Q(k) Thus there is a constant C such that

40460

=

=

= 0 for all k = 0 1 2

Q(x) (x+ 1)P(x)-x Cx(x- 1)(x- 2 ) (x- n ) Letting x -1 gives 1 C(-1)( -2 ) ( -n - 1) C(-1)n+l (n+ 1)! Thus C (-1)n+l ( n+ 1)! and 1_ (x+Q(x))__1_ x+ (-1)n+lx(x- 1)(x- 2 ) (x- n ) P(x)__ - x+ 1 x+ 1 ( n+ 1)1.

=

=

·

·

·

·

·

·

=

=

=

So

·

·

=

·

(

·

·

·

)

1 n (-1)n+1(n+1)1 = 1 n+1+(-1t+l = -P(n+1) = -( n+ 1+ ) ) --( " n+ 2 (n+ l ) ! n+ 2 n+ 2 34

n.

since n

= 34 is even. Hence 42840 X P(35) = 34 X 35 X 36 X 3436=40460

31. Answer: 30 v'369 - 360 cos a and Let

a.

Y = v'544- 480 sin Observe that X2 = 122+152-2(12)(15) cos Y2 = 122+ 202-2(12)(20) cos (90°- ) and 152+ 202 = 252 = (X+ Y) 2, so we can construct a right-angled triangle ABD X=

a

a

as shown.

A

X

c y

15 2

B"---D ---=--____, In particular LABC a and LCBD fact similar to 6ABD. So LAD B a and

=

a.

= 90°- We can check that 6ACB is in = 40tana = 40 1520 = 30 x

32. Answer: We shall apply the following inequality:

3

4(a2+b2+c2+d2) � (a+b+c+d) 2 Since a+b+ c +d = 8-e and a2+ b2+ c2+ d2 = 16-e2, we have 4(16-e2) � (8-e) 2 i.e. e(5e -16) :::; 0. Thus 0 :::; e :::; 16 5. Notethatifa = b = c = d = 6 5,wehavee = 16 5. Hence LeJ = 3. 33. Answer: 105 A three-digit number can be expressed as lOOa+lOb+c, and so we are minimizing +lOb+ c F(a b c) = lOOaa+b+c 35

Observe that with distinct digits Thus we assume that Note that

a b c, F(a b c) has the minimum value when a< b
_

where the last congruence can be calculated by the extended Euclidean algorithm. Thus by repeated squaring, we have

19a 1933 59 _

(mod

100)

35. Answer: Note that (n+ � ) 2 n2 � ' so f(n2 n but f(n2 So each of the sequences (n - f(n) )�=1 ) and (n f(n))�=1 ) increases by for every increment of n by except when n If n 2 and (n we have n - f(n) so the former sequence has f(n every perfect square repeated once. On the other hand, if n we have n f(n) so the latter sequence but (n f(n omits every perfect square. Thus

5

= +n+

+n+1 = n+1. +n = = (0 1 1 2 ) + 2 ) = (2 3 5 26 1 1, = m +m. = m +m, =m + 1- + 1)) = m2, 2+ m, = m = m2+2m + 1+ + 1)) = m2+2m+ 2,

+

oo

�

G) f(n) + (�) -f(n) (�r

( �) + (�) � 3 � ( 3) () () -I: 3 + I: 3 + I: 3 - I: ( 3) oo

n-f(n)

oo

n+ f(n)

_

-

oo

_

2 n

00

m=1

n=O

=5

36

2 2 m

oo

n=1

2 n

oo

m=1

2 2 m


(Senior Section, Round 2) 0900-1300

Saturday, 23 June 2012

1. A circle

w

through the incentre I of a triangle ABC and tangent to AB at A, intersects the segment BC at and the extension of BC at Prove that the line IC intersects w at a point M such that

D

E.

MD=ME . 2. Determine all positive integers n such that n equals the square of the sum of the digits of n. 3. If 46 squares are colored red in a 9 9 board, show that there is a 2 2 block on x

x

the board in which at least 3 of the squares are colored red.

an an+l be a finite sequence of real numbers satisfying ao an+l and ak-1- 2ak ak+l

= =0 + Prove that fork = 0 1 n+1, ak 5.

Prove that for any real numbers a

<

=:;

1

for

k=1 2

n

c

=

k(n + 1-k) 2

c

b d 2:: 0 with a + b= +d

37

2,


(Senior Section, Round 2 solutions)

Join AD, ID, IA and AE. Let IE intersect AC at N. We have LIAN=LIAB= LIEA so that the triangles NIA and AlE are similar. Thus LAN!=LEAl=LIDB. Also LDCI = LNG!. Therefore, the triangles DC! and NCI are congruent. Hence LDIC=LNIC implying MD=ME.

1.

M

Let s ( n ) denote the sum of all the digits of n. Suppose n is a positive integer such that s ( n ) 2 = n. Let s ( n) =k so that n =k2 . Then s (k2 ) = s ( n ) =k. Let 10r-1 � k < lOr, where r is a positive integer. That is k has exactly r digits. From 10r-1 � k, we have r � logk + 1. From k < 10r, we have k2 < 102r so that k2 has at most 2r digits. Therefore, s (k2 ) � x 2r =18r � 18 logk + 18 which is less than k if k 2 50. Thus the equation s (k2 ) =k has no solution ink ifk 2 50. Let k < 50 and s (k2 ) =k. Taking mod we get k2 k ( mod Thus k- 0, 1 ( mod That is k =1, 10, 18, 19, 27, 28, 36, 37, 45, 46. Only when k =1 and k = we have s (k2 ) =k. The corresponding solutions for n are n=1 or 81. 2.

9

9) .

9,

9,

3. Suppose that at most 2 squares are colored red in any 2

9).

x

9,

2 square. Then in any x 2 block, there are at most 10 red squares. Moreover, if there are 10 red squares, then there must be 5 in each row.

9

38

Now let the number of red squares in row i of the 9 x 9 board be ri. Then ri+ri+1 ::; 10, 1:::; i:::; 8. Suppose that some ri :::; 5 with i odd. Then (r1 +r2)+· · ·+(ri-2+ri-1)+ri+· · · +(rs+rg):::; 4 x + 5 = 45. On the other hand, suppose that r1, r3, r5, r7, rg 2: 6. Then the sum of any 2 consecutive ri's is :::; 9. So (r1+r2)+· · ·+(r7+rs)+rg :::; 4 x 9+9 = 45.

10

=

k n 1- k Let bk = ( -+;, ). Then bo bn+l = and bk-1 - 2bk+ bk+l = -1 for k = 1, 2 . .. , n. Suppose there exists an index i such that ai > bi, then the sequence a0 b0, ..., an+1-bn+1 has a positive term. Let j be the index such that aj-1-bj-1 < aj-bj and aj- bj has the largest value. Then (aj-1- bj-1)+(aj+l- bj+1) < 2(aj- bj)· Using ak-1- 2ak+ak+1 2: -1 and bk-1- 2bk+bk+1 = -1 for all k we obtain (aj-1- bj-1)+(aj+1- bj+1) 2: 2(aj- bj) a contradiction. Thus ak:::; bk for all k. Similarly, we can show that ak 2: -bk for all k and therefore ak :::; bk as required. 4.

0

5.

First note that (ae + bd)(ad +be) 2: (abed)2. To see this, we may assume a 2: e 2: d 2: b since a+b = e+d. Then ed- ab 2: Also we have the two obvious inequalities ae+bd 2: ed- ab and ad+be 2: ed-ab. Multiplying them together we get (ae + bd)(ad+be) 2: (abcd)2. Next (a2+e2)(a2+d2)(b2+e2)(b2+d2) 2 = ((ae+ bd)(ad+be)- (abed)2) +(abed)2 + ((a+b)2(e+d)2- 1) (abed)2 2 = ((ae+ bd)(ad+be)- (abed)2) +16(abed)2

0.

<; =

cac+bd :ad+be)' - (oh- cd)2r +16(abcd)2 ca b)'�c+d)' - (abcd)2r+16(abcd)'

by AM-GM

+

(4 - (abed)2)2+16(abed)2. This final expression is an increasing function of (abcd)2. The largest value of (abed)2 is 1 when (a, b, e, d) = (1, 1, 2), (1, 1, 2, 2, 1, 1), (2, 1, 1). Consequently, 2 (4 - (abcd)2) +16(abed)2:::; 25, proving the inequality. =

0,

0 ), (0,

39

0,

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Open Section, First Round)

Wednesday, 30 May 2012

2012

0930-1200 hrs

Instructions to contestants

1. Answer ALL 25 questions.

2. Write your answers in the answer sheet provided and shade the appropriate bubbles below

your answers.

3. No steps are needed to justify your answers.

4. Each question carries 1 mark. 5. No calculators are allowed.


40

Throughout this paper, let LxJ denote the greatest integer less than or equal to x. For example, ) 2 . 1 = 2 , 3.9 = 3 This notation is used in Questions 2 , 10, 16, 1 7, 18 and 22 .

L J

L J (

1 . The sum of the squares of 50 consecutive odd integers is 300850. Find the largest odd integer whose square is the last term of this sum. 1000 2. Find the value of

L Llog2 kJ . k=3

. 2 ) 3 . Given that f x is a polynomial of degree 201 2 , and that f k = k for k = 1 , 2 , 3, ·

(

x

find the value of 2014

()

( )

f 2014 .

(

· ·,

2013,

)

4. Find the total number of sets of positive integers x , y, z , where x , y and z are positive integers, with x < y < z such that x + y + z = 203. 5. There are a few integers n such that n 2 + n + 1 divides n 2 01 3 + 6 1 . Find the sum of the squares of these integers.

( ) 6. It is given that the sequence an �= l ' with a 1 = a2 = 2 , is given by the recurrence relation

for all n = 2 , 3, 4,

· · ·.

Find the integer that i s closest t o the value of

201 1 a + l k L2 a k . k=

7. Determine the largest even positive integer which cannot be expressed as the sum of two composite odd positive integers. 8 . The lengths of the sides of a triangle are successive terms of a geometric progression. Let A and C be the smallest and largest interior angles of the triangle respectively. If the shortest side has length 16 em and sin A - 2 sin B + 3 sin C sin G - 2 sin B + 3 sin A find the perimeter of the triangle in centimetres.

19 9'

9 . Find the least positive integral value of n for which the equation x +X +

y �

has integer solutions x 1 , x 2 , x 3 , · ·

(

···

+ X = 200 2 2 00 2

· , Xn ·

)

41

�

an be a real root of the cubic equation nx 3 + 2x - n

= 0, where n is a positive integer. 01 3 2 1 If f3n= l( n + 1)anJ for n= 2 3 4 · · · , find the value of f3k 1006 L · k=

10. Let

2

1 1 . In the diagram below, the point D lies inside the triangle such that and 90° . Given that 5 and 6, and the point M is the midpoint of find the value of 8 x DM 2 .

AC,

LBDC=

AB=

BC=

ABC

LEAD= LBCD

1 2 . Suppose the real numbers x and y satisfy the equations x 3 - 3x 2 + 5x

=1

Find x + y .

and y 3 - 3y 2 + 5y

=5

13. The product of two of the four roots of the quartic equation x4 - 18x 3 +kx 2 +200x- 1984 is -32. Determine the value of k.

=0

14. Determine the smallest integer n with n ?: 2 such that

is an integer. 1 5 . Given that f is a real-valued function on the set of all real numbers such that for any real numbers a and b, f af b ab

( )

Find the value of f 201 1 .

( ( ))=

16. The solutions to the equation x 3 - 4 x

l J= 5, wherek x is a real number, are denoted by XI X 2 X 3 . . . X k for some positive integer k. Find L x r . i=l

1 7. Determine the maximum integer solution of the equation

= 1001 + l !E_1 ! J + l !E_J l !E_3 ! J + . . . + l _!!_ 10! _ J 2! 42

18. Let A, B , C be the three angles of a triangle. Let L be the maximum value of sin 3A + sin 3B + sin 3C . Determine l10LJ. ( 19. Determine thex2number of sets of solutions x , y, z) , where x , y and z are integers, of the 2 2 2 2. + the equation

y +z

=

x y

20. We can find sets of 13 distinct positive integers that add up to 2142. Find the largest possible greatest common divisor of these 13 distinct positive integers. 21. Determine the maximum number of different sets consisting of three terms which2 form an arithmetic progressions that can be chosen from a sequence of real numbers a1 , a , · · · , a1 01 , where

22. Find the value of the series

� l 20121 + 2 k L...J

k=O

2 k+ 1

j.

23. The sequence (xn )�=1 is defined recursively by X n+ l with x1

=

=

(2 - J3) 1 (2 - J3) Xn + - Xn

'

1. Determine the value of X1001 - X401 ·

24. Determine the maximum value of the following expression

- ' ' -$ 1 - X 2 - X 3 - X4 - ' ' - X 2 0 1 4 -

where x1 , x 2 , · · · , x 2 014 are distinct numbers in the set

1 00 6 . -1 ). 25. Evaluate 22011 L ( -1) k 3k (2012 2k k=O

43

-1, 2, 3, 4, · · · , 2014---:

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012 (Open Section, First Round Solutions)

1. Answer: 121 Let the integers be X + 2 X + 4

·· ·

100. Then ( X + 2) 2 + ( X + 4 )2 + + ( X + 100) 2 = 300850 Let y = X + 51 and regrouping the terms, we obtain [ (y - 49 )2 + (y + 49 )2 ] + [ (y - 47) 2 + (y + 47) 2 ] + . . . + [ (y - 1) 2 + (y + 1) 2 ] = 300850 which simplifies to 50y2 + 2 ( 1 2 + 32 + 52 + 72 + . . . + 49 2 ) = 300850 . the fact that 1 2 + 32 + 52 + · · · + (2n - 1) 2 = 34 3 - 31 n, we obtam. y = 72. Hence Usmg X = 21, so that the required number is 121.

Solution.

X+

···

n

D

2. Answer: 7986 Note that 2k+1 - 2k = 2k , and that 2k t 2k+ 1 if and only if llog2 tJ = k. Then the requires sum (denoted by S) can be obtained by 1 023 = 2: 2: llog2 tj + llog2 3j - L llog 2 tj k(=2 2k 9<2)k+l 10 {; k2k + 1 - �23 9 8192 + 1 - 23(9) = 7986 :::;

Solution.

<

9

s

t=lOOl

9

t

l

D

3. Answer: 4 Let g(x) = xf(x) - 2, hence g(x) is a polynomial of degree 2013. Since g(l) = g(2) = g(3) = = g(2013) = 0, we must have g(x) = >.(x - l)(x - 2)(x - 3) (x - 2012)(x - 2013) 2 Hence, for some >.. Also, g(O) = -2 = ->. 2013 ! , we thus have >. = 20131 g(2014) =20�3! (2013! ) = 2014 . f(2014) - 2 concluding that 2014 f(2014) = 4 Solution.

···

···

·

·

D

44

4.

Answer: 3333 . . mteger . sets = 20301 positive First note that there are (2022 ) = 202(201) 2 ( x, y, z) which satisfy the given equation. These solution sets include those where two of the three values are equal. If x = then 2x+z = 203. By enumerating, z = 1, 3, 5, , 201. There are thus 101 solutions of the form (x, x, z). Similarly, there are 101 solutions of the form (x, y, x) and ( x, y, y). Since x y z, the required answer is

Solution.

y,

·

<

·

·

<

D

5. Answer: 62 Since n 3 - 1 = (n7 - 1)(n2 + n + 1), we know that n2 + n + 1 divides n3 - 1. 2 0 13 Also, since n - 1 = (n3 ) 6 1 - 1, we also know that n2 + n + 1 divides n20 13 - 1. As 2 2 n 0 1 3 + 61 = n 0 1 3 1 + 62, we must have that n2 + n + 1 divides n20 13 + 61 if and only if n2 + n + 1 divides 62. Case (i): If n2 + n + 1 = 1, then n = 0, -1. Case (ii): If n2 + n + 1 = 2, there is no integer solution for n. Case (iii): If n2 + n + 1 = 31, then n = 6, -5. Case (iv): If n2 + n + 1 = 62, there is no integer solution for n. Thus, all the integer values of n are 0, -1, 6, -5. Hence the sum of squares is 1 + 36 + 25 = 62. Solution.

_

D

6. Answer: 3015 The recurrence relation can be written as l a:: - a:: l = ( n � 1 - �) - ( � - n : 1 ) Summing for n = 2 to N, we obtain Solution.

showing that

·

3 1- 1 . -;;:;,;- - 2 - ( N N + 1 ) Summing this up for N = 2 to N = 2011, we obtain 2 0 11 -'1-1 � (2010) - ( � - -1- ) = 3014. 5 + -1ak S = '"""' = � ak 2 2 2012 2012 showing that the integer closest to S is 3015. aN + l

_

45

D

7. Answer: 38 Let n be an even positive integer. Then each of the following expresses n as the sum of two odd integers: n = (n - 15) + 15 (n - 25) + 25 or (n - 35) + 35. Note that at least one of n - 15 n - 25 n - 35 is divisible by 3 so that it is composite, and hence n can be expressed as the sum of two composite odd numbers if n 38. Indeed, it can be verified that 38 cannot be expressed as the sum of two composite odd positive integers. Answer: 76 2 Let the lengths of the sides of the triangle in centimetres be 16, 16r and 16r 2 (where r 1). Then 1r2--2r2r++3r3 -199 so that r = -32 . Hence, the perimeter of the D triangle = 16 ( 1 + � + � ) = 76cm 9. Answer: 4 Since 2002 4(mod9) 43 1(mod9) and 2002 = 667 3 + 1, it follows that 20022002 4667x3+ 1 4(mod9). Observe that for positive integers x, the possible residues modulo 9 for x3 are 0 ±1 . Therefore, none of the following X� X� + � � + X� + � can have a residue of 4 modulo 9. However, since 2002 = 103 + 103 + 13 + 13, it follows that 20022002 2002 . (2002667 ) 3 (103 + 103 + 1 3 + 1 3 ) (2002667) 3 (10 . 2002667 ) 3 + (10 . 2002667 ) 3 + (2002667 ) 3 + (2002667 ) 3 This shows that x� + x� + x� + x! = 20022002 is indeed solvable. Hence the least integral value of n is 4. 10. Answer: 2015 Let f(x) = nx3 + 2x - n. It is easy to see that f is a strictly increasing function for n = 2 3 4 Further, 1 ( n : 1 ) = n ( n : 1 ) 3 + 2 ( n : 1 ) - n =(n ; 1)3 ( -n2 + n + 1 ) < O for all n 2. Also, f(1) = 2 Thus, the only real root of the equation nx3 + 2x-n = 0 for n 2 is located in the interval n(�1 1 ) . We must have n < an < 1 n < (n + 1 )an n + 1 -n+1 so that f3n = l(n + 1)anJ = n for all n 2. Consequently, -1- "201(33 k =-1- "2013 =-1- . 2012 (2 + 2013) = 2015 1006 k=2 1006 k=2 1006 2 Solution.

>

D

8.

Solution.

>

Solution.

=

=

x

=

=

X

X

X

D

·

2:

·

· .

2:

>

0.

==?

<

2:

�

�

k

D

46

1 1 . Answer: 22 Solution.

Extend CD to E such that CD = DE . ' ' ', ' '', ' ' ' ', ' ' ' ' ' ' ' ' ' ' \ ' \ ' ' \

,,

,

' ' ' ' \ '

�-------=

c

6

It is clear that D.CDB and D.EDB are congruent . Hence EB = CB = and L_BED L_BCD . Thus, L_BED = L_BCD = L_BAD implies that the points B , D , A are E are concyclic. Given that L_BDC = goo , hence L_EDB = goo . BDAE is a cyclic quadrilateral with EB as a diameter. Thus, L_EAB = goo . In the right-angled triangle EAB , we have

Since D and M are the midpoints of EC and AC respectively, DM = 8 x DM2 = 22 .

1 2 . Answer: 2 Solution.

2 From x 3 - 3

2 and from y 3 - 3 y

x + 5x

+ 5y

=

=

(

=

� AE = �· Thus, D

1 , we have

x - 1 ) 3 + 2 (x - 1 ) = -2,

5 , we have

) y-1 3

+ 2 (y - 1 ) = 2 . Thus ) ( ) ) ) ( 0 ( x - 1 3 + 2 (x - 1 + y - 1 3 + 2 y - 1 )2) )( ) ( )2 = (x + y - 2 ) ( (x - 1 - (x - 1 y - 1 + y - 1 2 + (x + y - 2 ) )2) )( ) ( )2 = (x + y - 2 ) ( 2 + ( x - 1 - (x - 1 y - 1 + y - 1 . For any real numbers x and y, we always have ? )( ) ( )2 0 y-1 (x - 1 - (x - 1 y - 1 and thus x + y - 2 = 0, implying that x + y = 2 . (

=

+

47

�

D

13. Answer: 86 Let a, b, c, d be the four roots of x 4 - 18x 3 + kx 2 + 200x - 1984 = 0 such that ab = -32. Then a + b + c + d = 18, ab + ac + ad+ be + bd + cd = k, abc + abd + acd + bed = -200, abed = -1984. Since ab = -32 and abed = -1984, we have ed = 62. Then, from abc + abd + acd + bed = -200 we have -200 = -32c - 32d + 62a 62b = -32(e + d) + 62(a + b). Solving this equation together with the equation a + b + c + d = 18 gives that a + b = 4, c + d = 14. From ab + ac + ad + be + bd + ed = k, we have

{

Solution.

+

= 30 + (a + b)(c + d) = 86.

D

14. Answer: 337 Solution.

Assume that

and so

+

(n 1)(2n + 1) = 6m2 . Thus 6 (n + 1)(2n + 1), implying that n 1 or 5 ( mod 6). = 6k + 5 . Then m 2 = (k + 1)(12k + 11). Since (k + 1) and (12k + 11) are relatively prime, both must be squares . So there are positive integers a and b such that k + 1 = a2 and 12k + 11 = b2 . Thus 12a 2 = b2 1. But, as 12a 2 O ( mod 4) and b2 + 1 1 or 2 (mod 4), there are no integers a and b such that 12a2 = b2 + 1. Hence Case 1 cannot happen. n = 6k + 1. Then m 2 = (3k + 1)(4k + 1). Since 3k + 1 and 4k 1 are relatively prime, both must be squares. So there are positive integers a and b such that 3k + 1 = a 2 and 4k + 1 = b2 . Then 3b2 = (2a - 1)(2a + 1). Observe that in the left-hand side, every prime factor except 3 has an even power. So neither 2a - 1 nor 2a + 1 can be a prime other than 3. Now we consider positive integers a such that neither 2a - 1 nor 2a + 1 can b e a prime other than 3. If a = 1, then b = 1 and n = 1. So we consider a 2. The next smallest suitable value for a is 13. When a = 13, we have 3b2 = 25 27 and so b = 15, implying that k = 56 and so = 6k + 1 = 337. =

Case 1: n

+

=

=

Case 2 :

+

2:::

X

n

48

D

15. Answer: 2011

f(f(1)f(b))

f(b)

f(b)

From the recurrence relation, = f( 1 ) b and f(f(b)f( 1 ) ) = 1. Hence, = f( 1)b. By letting = / ( 1 ) , we obtain f(f( 1 ) ) = (f( 1 ) ) 2 . Also, from the given functional equation, we have f(f( 1 ) ) = 1, hence (! ( 1 ) ) 2 = 1 , following that / ( 1 ) is either 1 or - 1 . D Hence f(201 1 ) = 201 1 .

b

·

16. Answer: 19 <

�

2::

x - 1 LxJ x . Note that if x 3, there will be no solution as x3 - 4LxJ x3 - 4x = x(x2 - 4) 3(5) = 15. Also, if x - 2 , there will be no solution as x 3 -4 L xJ x3 -4(x - 1) = x(x 2 - 4) + 4 4 . Hence the solution must be in the interval ( - 2 , 3) . If LxJ = - 2 , then x3 = -3, giving x = N, which is a solution. If LxJ = - 1 , then x3 = 1, giving x = 1 which contradicts with LxJ = - 1 . LxJ = 0, then x3 = 5, hence there is no solution. If LxJ = 1 , then x3 = 9 . Since 2 W 3, there is no solution. If LxJ = 2, then x3 = 13. Since 2 m 3, X = m is a solution. D Thus, the required answer is -3 + 13 = 10. 17. Answer: 584 Solution.

Note that

2::

2::

�

<

If

<

<

<

Solution.

�

<

It is clear that

l � J + l;,J + l:,J + . . . + l 1�! J

x,

x

is a monotone increasing function of and when = 6 ! , the above expression has a value larger than 100 1 . Thus each solution of the equation

l �,J + l;,J + l:,J + . . . + l 1� ! J = 100 1

x is a solution, then l �,J + l;,J + l:,J + . . . + l 1� ! J = l �,J + l;,J + l :, J + l�J + l:,J .

is less than 6! and so if

As

x

<

6 ! , it has a unique expression of the form X

+ b 4! +

C X

X

C X

X

X

e,

+ d 2! + 1 . Note that where b, c, d, are non-negative integer with a 5, b 4, c 3, d 2 , if 3! + d 2 ! + X = a 5! + b 4! + then l�, J + l;,J + l:,J + l :,J + l � J = 206a + 41b + 10c + 3d + Since 41b + 10c + 3d+ 201 , we have 800 206a 1001 and so a = 4 . Thus 41b + lOc + 3d + = 1 77, X=a

a,

e

X

5!

3!

�

�

X

�

�

e

�

e,

e.

e

�

�

�

e

49

which implies that

c

b = 4 and so on, giving that = d = 1 and X

=4

X

5! + 4

X

4! + 1

X

3! + 1

X

e

= 0. Thus

2 ! + 0 = 584.

As 584 is the only integer solution, the answer is 584. D 18. Answer: 25 Solution.

We shall show that -2 ::::; sin 3A + sin 3B + sin 3C ::::; 3v'3/2.

Assume that A ::::; B ::::; C. Then A ::::; 60° . Thus sin 3A 2:: 0 . It is clear that sin 3B 2:: - 1 and sin 3C 2:: - 1 . Thus sin 3A + sin 3B + sin 3C 2:: -2. Let B = C. Then B = C = goo - A/2. If A is very small, B and C are close to goo , and thus sin 3A + sin 3B + sin 3C is close to -2. Now we show that sin 3A + sin 3B + sin 3C ::::; 3v'3/2. First the upper bound can be reached when A = B = 20° and C = 140° . Let X = 3A,

Y

= 3B and Z = 3(C - 120° ) . Then X + Y + Z = 180° and sin 3A + sin 3B + sin 3C = sin X + sin Y + sin Z.

Suppose that X, Y, Z satisfy the condition that X + Y + Z = 1 80° such that sin X + sin Y + sin Z has the maximum value. We can then show that X = Y = Z. Assume that

X ::::; Y ::::;

Z. If X < Z, then

X-Z . . . X+Z . X+Z sm X + sm Z = 2 sm cos < 2 sm , 2 2 2 implying that . . X+Z . X+Z . . y+� . y+� z < � +� �X + � 2 2

which contradicts the assumption that sin X + sin Y + sin Z has the maximum value. Hence X = Y = Z = 60° , implying that A = 20° , B = 20° and C = 140° and sin 3A + sin 3B + sin 3C = 3v'3/2. Since

y'3

�

1 . 732 , the answer is then obtained.

D

1 g . Answer: 1 Solution. Note that x = 0, y = 0 and z = 0 is a solution of this equation. We shall show that this is its only integer solution by proving that if x, y, z is a solution of this equation and whenever x, y, z are divisible by 2k , they are also divisible by 2k + 1 for any k 2:: 0.

Let

x

=

2k x' , y

= 2ky' and

z

= 2k z ' . Then

x2

+ y2 + z2 =

x2 y 2

is changed to

It is easy to verify that only when x' , y' , z ' are all even, x'2 + y '2 + z 12 and the same remainder when divided by 4. Thus x, y, z are divisible by 2 k + 1 .

50

2 2 k x12 y'2

have D

20. Answer: 21

Let d be the greatest common divisor ( gcd ) o f these 13 distinct positive integers. Then these integers can be represented as da1 , da2 , · · · , da13 , where gcd ( a1 , a2 , · · · , a13 ) = Let S denote a1 + a2 + · · · + a13 · Then Sd = In order for d to be the largest pos sible, S must be the smallest . Since S ;::: + + 3 + · · · + 3 = and that S divides and that = x 3 x 7 x the smallest possible value of S can be x = and the largest value of d is thus By choosing ( a1 , a2 , a3 , · · · , a12 , a13 ) = 2, 3, · · · , D we conclude that d = is possible. Solution.

13

1.

2142 2

1 2

51, 21.

21

2142.

1 91

2142, 2 51 102, (1, 12, 24),


First , for the following particular sequence, there are really three-term arithmetic progressions which can be chosen from this sequence: They are s , and t with

s

s,

1, 2, 3, . . . ' 101. i with 1 i 51 and 2i - i, �

s

<

�

2500 different

i, 2i - for all integers for all integers i 52 i 101. Now we show that for any given sequence of real numbers a1 a2 aw1 , there are at most 2500 different three-term arithmetic progressions which can be chosen from this sequence. Let as , ai , at represent a three-term arithmetic progression. It is clear that 2 i 100. If i 51, then the first term as has at most i - 1 choices, as must be an index in 1, 2, i - 1 . If 52 i 100, then the third term at has at most 101 - i choices, as must be an index in i + 1, i + 2, , 101 . 2 �

�

< t �

�

···,

t,

<

�

< ··· <

�

s

�

t

�

t

· ··

So the number of different three-term arithmetic progressions which can be chosen from this sequence is at most 1 00 51 + = + 2 + . . . + 5o + + 2 + . . . + = 25oo . 2 2 i=5 i= D

I)i - 1) L (101 - i) 1


Write

x

:= x

x

- L J . Then

1

49

if !!C2 < 12 otherwise

Thus, we have

Applying the above result for

x=

;,

f ( l ; J - l 2::1 J ) k=O l;J

In particular, when

n=

n.

20121, the infinite series converges to 20121. 51

D

23 . Answer: 0 Solution.

Let X n = tan O: n . Since 2 -

X n+I = tan a n + I =

-J3 = tan {;( ) , it follows that

(

tan an + tan ( � ) 1r = tan O:n + 1f 12 1 - tan an tan ( I 2 )

)

( So, X n+ l 2 = tan an + 1r ) = tan an X n , implying that this sequence has a period of 1 2 . Observe that 1 0 0 1 = 5 (mod 12) and 401 = 5(mod 12) . Consequently, =

Xl QO I - X40 1 = X5 - X5 = 0 D 24. Answer: 2013 Solution.

be 2014, as

First it is clear that the answer is an integer between 0 and 2014. But it cannot X I - X 2 - X 3 - X4 X 20 I4 and ·

·

·

·

X I + X2 +

·

·

·

·

·

-

+ X 2 0 I4 = 1

2+

+

·

·

·

+ 2014 = 1007

X

2015

have the same parity. Now we j ust need to show that 2013 can be achieved . For any integer k , ( 4k + 2) - ( 4k + 4) - ( 4k + 5) - ( 4k + 3) = 0 Thus 2 - 4 - 5 - 3 - . . . - ( 4k + 2) - ( 4k + 4) - ( 4k + 5) - ( 4k + 3) -2010 - 2012 - 2013 - 201 1 - 2014 - 1 =

0 - 2014 - 1

=

2013 D

25. Answer: 1 Solution. Consider the complex number binomial theorem one has

(

1f

1f

w

= cos

J + i sin J . On one hand, using the

) 20 I 2

Re cos 3 + i sin 3 20 I2 Re + i -;

( � V3)

(�)

c ) (�) ( ��::: ) [ ( ) ( )

2 0I 2

_

20 I O (; ) 2

o12 2

+...+

2012 1 1-3 2 2 I 0 2 2

_

20 12 + 4

( � ) ( �: ) 2 00 8

( )]

3 2 2012 + . . . + 3I 006 2012 + 201 2 4

52

On the other hand, using the De Moivre's theorem one has

( 20121f

Re ( w2 0 1 2 ) = Re cos Thus ,

+

i sin

20121f ) 3

_1_[1 - (2012) (2012) . . . 4 2 2012 (20124 ) . . . � [1 - ( 2 )

2 20 1 2 so that

3

2 20 11

3

+

3

32

+

32

- 3 1 004

+

+

53

= cos

20121f

(2012 2010)

- 3 1 00 4

3

+

2012) (2010

= cos

3 1 006

+

21f 3

=

2

-�

(2012 2012) ] - �2 2012) ] 1 (2012

3 1 006

=

=

D

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO ) 2012

(Open Section, Round 2) Saturday, 30 June 2012

1. The incircle with centre

0900-1300

of the triangle ABC touches the sides BC, CA and AB at and respectively. The line intersects the segment at K. Prove that A, K and M are collinear where M is the midpoint of BC.

D, E

F

2. Find all functions

f

:

IR

I

---+

ID

IR

EF

so that

+

+

=

(x y )(f(x ) -f (y )) (x-y ) f (x y ) for all

x, y

3. For each

i

E JR. =

1, 2, . . . , N, let ai, bi, ci be integers such that at least one of them is odd. Show that one can find integers z such that xai + ybi + zci is odd for at least 4N different values of i .

x, y,

/7

4.

Let p be an odd prime. Prove that ! P-2 + 2p -2 + 3p -2 + . . . +

5.

(� - 1 ) p -2

-

2 - 2P p

( mod p ) .

There are 2012 distinct points in the plane each of which is to be coloured using one of n colours so that the number of points of each colour are distinct. A set of n points is said to be multi-coloured if their colours are distinct. Determine n that maximizes the number of multi-coloured sets.

54


2012

(Open Section, Round 2 solutions)

1. A

c

B

Let the line AK intersect EC at M. We shall prove that M is the midpoint of EC. Since LFIK = LE and LEIK = LC, we have FK EK Also

FK sin LF AK

Therefore,

sin LFIK sin LEIK

AE sin LAKE

AF sin LAKF sin LFAK sin LKAE

sin E sin C '

FK EK

EK . sin LKAE

sin E sin C '

Consequently, EM EM . AM sin LFAK sin C = = = CM AM CM sin E sin LKAE so that EM = CM.

55

l,

2. Suppose that

f is a solution. Let a = 21 (! (1) - f (-1)), b = 21 (! (1) +! (-1)) and g (x ) = f (x ) - ax - bx2. Then (x +y )(g(x ) - g (y )) = (x - y)g (x +y ) and g (1) = g( -1) = 0. Letting y = 1 and y = -1 above give (x + 1) g (x ) = (x - 1) g (x + 1) xg (x +1) = (x +2)g(x) .

Thus

x(x +1)g (x) = x(x - 1) g(x +1) = (x - 1)(x + 2) g (x) for all x. So g (x ) = 0 for all x. Hence f(x) = ax+bx2. We can check directly that any function of this form (for some a, b E JR.) satisfies the given equation. Consider all the 7 triples where z are either 0 or but not all 0. For is odd. Thus among the 7 sums each at least one of the numbers 3 are even and 4 are odd. Hence there are altogether 4N odd sums. Thus there is choice of for which at least 4N 7 of the corresponding sums are odd. You can think of a table where the rows are numbered . . . , N and the columns correspond to the 7 choices of the triples The 7 entries in row are the 7 sums Thus there are 4 odd numbers in each row, making a total of 4N odd sums in the table. Since there are 7 columns, one of the columns must contain at least 4N 7 odd sums.

(x, y, z), x, y, ai, bi, Ci / 1, 2, i (x, y, z) .

3.

i, (x, y, z)

1

( xai +ybi +zci. / )

i 1, 2, . . P, ;1 , 2i ( p ) = (p -1) (p - 2) . . . (p - (2i -1)) (-1) (-2) (- (2i - 1)) p 2i (2i - 1) ! (2i - 1) !

4.

·

First, for each =

·

Hence

(p-1)/2

L i= 1

xai+Ybi+zci,

·

·

_

_

1 (mod p) .

. ( ) 2 (p-1)/2 ·p-1 ( ) 2 - L 2i. p 2i = p i=1 i= 1 (p- 1)/2 ( p ) 2 --p (mod p) (by Fermat 's Little Theorem. ) = i 1 2i

·p-2 = 't

(p-1)/2

_

-

·p- 2 't

't

P

_

-

-

� L....,;

't

P

� L....,;

=

The last summation counts the even-sized nonempty subsets of a p-element set, of which there are P - 1

2 - 1.

56

5.

Let m1 < m2 < < mn be the number of points of each colour. We call m1, m2, . . . , mn the colour distribution. Then m1 + + mn = and the num ber of multi-coloured sets is M = m1 m2 . . . mn. We have the following observations. ·

·

·

·

·

2012

·

(i) m1 > 1 . For if m1 = then m1m2 . . . mn < m2m3 . . . mn - 1(1 + mn ) This means if we use n- colours with colour distribution m2, m3, . . . , mn_ 1, + mn), we obtain a larger M.

1,

1

·

(1 (ii) mi+1- mi ::::; 2 for all i . For if there exists k with mk+1 - mk 2': 3, then the colour distribution with mk, mk+1 replaced by mk 1, mk+1- 1 yields a larger M. (iii) mi+1- mi 2 for at most one For if there exist i j with mi +1 - mi mj+1 - mj 2, the colour distribution with mi, mj+1 replaced by mi 1, mj+1 - 1 yields a larger M. (iv) mi+1 - mi 2 for exactly one For if mi +1 - mi 1 for all then m1 mn nm1 n(n2-1) 2012 4 503. Thus n(2m1 - 1 n) 8 503. Since 503 is prime, the parity of n and 2m1 -1 n are opposite and 2m1 - 1 n n, we have n 8 and m1 248. The colour distribution with m1 replaced by two numbers 2, 246 (using n 1 colours) yields a larger M. 2. If mn - mn 1 2, then from (iv) , we have m1 mn (v) m1 n(n-1) 1 2012. Thus -n(2m1 -1 ) 2 2011. Since 2011 is prime, we get nm1+ 2 n 2 and m1 1005 which will lead to a contradiction as in (iv) . Thus mn-mn_ 1 1. mi+1- mi 2 for some 1 ::::; i ::::; n - 2. Suppose m1 2': 3. Let m' mi+2 - 2. Then mi m' mi+1 with replacing mi+2 by 2, m' yields a larger M. Thus m1 2. From the above analysis, with n colours, we see that the colour distribution 2, 3, 2, . . ., n 1, n 2, with 3 ::::; i ::::;2 n, yields the maximum M. Now . . . , i- 1, i 1, we have 2: mi � (n 1) ( n 4) - i 2012. Thus n + 5n- 4020 2i, 3 :=::; i ::::; n, i.e. , n2 5n 2': 4026 and n2 3n ::::; 4020. Thus n 61 and i 3. Thus the maximum is achieved when n 61 with the colour distribution 2, 4, 5, 6, . . . , 63. =

·

·

·

+

=

+

=

=

<

+

<

=

+

+

i.

=

i.

=

=

=

=

=

+

+

=

=

<

·

=

+

=

=

+

+

n

=

+

=

i,

+

=

+

+

+

+

=

57

+

>

·

·

·

+

=

·

+

=

·

+

=

i +

=

=

=

=

=

Multiple Choice Questions 1. Calculate the following sum: 1 2 3 4 10 -+ -+ -+ -+ . . . + . 16 210 2 4 8

(A)

50 3 . 256'

507

505

509

(B) 5 ,. C ; (D) 5 ,. (E) None of the above. 2 6 ( ) 256 2 6

2. It is known that the roots of the equation x5 + 3x4- 40441 18x 3 - 12132362x 2- 12132363x- 201 12= 0

are all integers. How many distinct roots does the equation have? (A) 1;

(B) 2; ( C) 3; (D) 4; (E) 5.

3. A fair dice is thrown three times. The results of the first, second and third throw are recorded as x, x,

y

(A)

y

and

and 1 ; 12

z

z,

respectively. Suppose x + y=

z.

What is the probability that at least one of

is 2? 8

(B) � , ( C ) 5 ; (D) 1 8 ·

1 3;

(E)

!__. 13

4. Let

= 1 000 · · · 000 1 000 · · · 000 50 . '-....--" '-....--" 2012 times 2011 times

X

Which of the following is a perfect square? (A) x- 75;

(B) x- 25; (C) x ; (D) x + 25; (E) x + 75.

5. Suppose N 1, N2, ..., N2o11 are positive integers. Let

X= (N1 + N2 + · · · + N2o10) (N2 + N3 + · · · + N2ou ), Y= (N1 + N2 + · · · + N2ou) (N2 + N3 + · · · + N2o1o) . Which one of the following relationships always holds? (A) X=Y;

(B) X>Y; ( C ) X
. 6. Consider the following egg shaped curve. A BC D is a circle of radius 1 centred at 0. The arc AE is centred at C , CF is centred at A and EF is centred at D. 2

(A) 1680;

(B) 1712; (C) 3696; (D) 3760; (E) None of the above.

10. In the set {1, 6, 7, 9}, which of the numbers appear as the last digit of nn for infinitely many

positive integers n? (A) 1, 6, 7 only;

(B) 1, 6, 9 only; (C) 1, 7, 9 only; (D) 6, 7, 9 only; (E) 1, 6, 7, 9.

Short Questions z x y 11. Suppose -+ -+-= a b c

12. Suppose x=

a b c . ..J2 and-+-+-= 0. Fmd

13 v'19 + sv'3

x

y

z

. Find the exact value of x4- 6x 3- 2x 2 +18x + 23 x 2- 8x + 15

13. Let a1= 3, and define an+l=

1 a J3 n � for all positive integers n. Find a2011 · an + 3

14. Let a, b, c be positive real numbers such that a2 + ab + b 2= 25, b2 + be+ c2= 49, c2 + ca + a2= 64.

4

distinct ways of using two tiles of size 1

x

1, two tiles of size 1

x

2 and one tile of size 1

x

4.

It is not necessary to use all the three kinds of tiles. )

24. A 4 2

x

x

4 Sudoku grid is filled with digits so that each column, each row, and each of the four

2 sub-grids that composes the grid contains all of the digits from 1 to 4. For example,

Find the total number of possible 4

x

4

3

1

2

2

1

3

4

1

2

4

3

3

4

2

1

4 Sudoku grids.

25. If the 13th of any particular month falls on a Friday, we call it

Friday the 13th.

It is known

that Friday the 13th occurs at least once every calendar year. If the longest interval between two consecutive occurrences of Friday the 13th is

x

months, find

x.

26. How many ways are there to put 7 identical apples into 4 identical packages so that each package has at least one apple?

27. At a fun fair, coupons can be used to purchase food. Each coupon is worth $5, $8 or $12. For example, for a $15 purchase you can use three coupons of $5, or use one coupon of $5 and one coupon of $8 and pay $2 by cash. Suppose the prices in the fun fair are all whole dollars. What is the largest amount that you cannot purchase using only coupons?

28. Find the length of the spirangle in the following diagram, where the gap between adjacent parallel lines is 1 unit.

6

E

D

c

34. Consider an equilateral triangle A BC , where A B = BC = CA = 2011. Let P be a point

inside LA BC . Draw line segments passing through P such that DE

II BC , FG II CA and

HI II AB. Suppose DE : FG : HI= 8 : 7 : 10. Find DE +FG + HI . A

c

35. In the following diagram, A B

j_

BC . D and E are points on segments A B and BC respec

tively, such that BA + AE = E D + DC . It is known that A D = 2, BE = 3 and EC = 4. Find BA + AE .

8

Note that X= (10 1012 + 1) X 10 2014 +50= 104026 + 102014 +50 = (102013 )2 +2 X 102013 X 5 +50= (102013 +5) 2 +25. So x- 25 is a perfect square. 5. Answer: ( B ) .

Let K= N2 + · · · +N2010· Then X= (N1 +K) (K +N2on) andY= (N1 +K +N2on)K.

6. Answer: ( A ) .

7r (22 ) ,---... The area enclosed by A D , DE and AE is 8 -

The area of the wedge E DF So the area of the egg is:

. 7r (2- J2)2 = IS 4

(

-

1f - 1. 1= 2

)

3 J2 2-

1f

.

� + 1 +2 (�- 1) + (�- J2) 1r= (3- J2)1r- 1. x

7. Answer: ( B ) .

The left shows that 3 colours are not enough. The right is a painting using 4 colours.

8

.

Answer: (E). Since 5 I (24- 1), 71 (36- 1), 11 I (510- 1), 13 I (7 12- 1),

9. Answer:

n

is divisible by 5, 7, 11 and 13.

(C).

We consider the position of the Black Knight. The number of positions being attacked by the White Knight can be counted.

10

13. Answer: 3.

J33 x 01 . Then f ( (x))= Let f (x)= J X+ 3

vfsx-1 _J3 = J (J (J (x)))= x+vfs J3 vfsx 1 + 1 x+vfs

Since 2010= 6

x

J3 vs-1 _1

�vis 3x-1 +J3 x+vfs

3x-J3 - x-J3 J3x - 1 + V3x+ 3

-

x-J3 J3x+ (

3x - 1- v3x- 3 _.!. v/"il . So J (J (J (J (J (J (x))))))= x. = X 3x-J3 + x/"il+J3

335, a2o11= f (J (J··· f (J (3)) ·· ))= 3. '-----v--" 2010 copies ·

14. Answer: 129. Consider the following picture, where LAOB = L BOC= LCOA= 120°, OA= a, OB= b and OC= c. A

c

B

Then IBCI= 5, I CAl= 7 and lA B I= 8. The area of the triangle A BC is yl10 (10- 5) (10 - 7) (10 - 8)= 10.J3. 1V3 (ab +be+ ca)= 10J3. So ab +be+ ca= 40. Then 2 2 2 (a + b + c) 2= (a2+ ab + b2) + (b2 +be+ c2) + (c2 + ca + a2) + 3 (ab +be+ ca)= 258. Thus, (a+ b + c) 2= 129. 15. Answer: 0. Define Q (x) = (1 + x)P (x) - x. Then Q (x) is a polynomial of degree 2011. Since Q(O) = Q (1)= Q (2)= ···= Q (2010) = 0, we can write, for some constant A, Q (x)= Ax(x- 1) (x- 2)··· (x- 2010). 1= Q (-1)= A (-1) (-2) (-3)·· · (-2011)=-A· 2011!. Then Q (2012) = A· 2012!= -2012, Q (2012) + 2012 and P ( 2012)= = o. 2013 16. Answer: 9241. 12

Suppose x 1

=

x2

M

=

=

2:: >

2::

2::

=

=

Therefore, M

�

···

=

=

Xk

1

=

�

2

<

Xk+l

�

···

�

X2011 · Let M

=

x 1 · · · X20l1 · Then

Xk+1Xk+2 . . . X2QlOX20ll (Xk+l- 1)x k+2 · · · X2010X2011 +Xk+2 . . . X2010X2011 (Xk+l- 1) 2 +Xk+2 · · · X2010X2011 .

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

(x k+l- 1) 2 + · · · + (x 2009- 1)2 +X2010X2011 (x k+l- 1) 2 + · · · + (x 2009- 1)2 + (x 2o10- 1)2 + (x 2o11- 1) 2 2 (x k+l + · · · +x 2o11- (2011- k)) 2 (M- 2011) .

4022. On the other hand, (1, 1, . . . , 1, 2, 2011) is a solution to the equation.

So the maximum value is 4022. 2 1. Answer: 101. 2::

If n

M (91)

102, then M (n) =

M (M (102))

For each k

=

=

=

n- 10

M (92)

2::

92.

M (M ( 103))

=

1, . . . , 10, M (80 + k)

=

M (70 + k) M(k)

=

=

=

M (93)

M (M (91 + k))

=

M (M(81 + k))

···

M (91)

=

M (M(11 + k))

=

=

Hence, all integers from 1 to 101 are solutions to M (n)

=

=

M (91) M (91) =

=

=

M ( 101)

=

9 1.

9 1, and thus 9 1, 9 1.

91.

22. Answer: 19.

20 + 1 1 . (M) n 1 2on+l + 1 1n+l · n n - (n +1) (1 +(M) n) n +1 20 +1 1

An+l An

Then An+l

<

An if n

Note that 10 + If 10 n

�

�

n

�

>

�

10 +

11 ) n 1 + 20

18, then n

18 implies An

<

�

<

n ; and An+l 1 + 11 w)

10 +9

10 +8

An+l ·

�

<

=

19. Son

10 +

2::

An if n

>

<

19 implies An

9 . ; 1f n n 11 1 + ( 20 )

<

10 + >

1+

�!! w

)n

.

An+l ·

10, then n

<

10 +

9 n Hence, 1 + ( 11 20 ) .

23. Answer: 169.

Let an be the number of ways to pave a block of 14

1

X

n. Then an

=

an-1 +an-2 + an-4 with

Case 2:

Case 3:

The first year is a leap year. Jan

Feb

Mar

Apr

Mar

Jun

Jul

Aug

Sep

Oct

Nov

Dec

0

3

4

0

2

5

3

6

1

4

6

2

5

5

1

3

6

@] 1

4

@]

2

5

0

The second year is a leap year. Jan

Feb

Mar

Apr

Mar

Jun

Jul

Aug

Sep

Oct

Nov

Dec

0

3

3

6

1

4

2

5

5

4

5

1

3

[]]

4

@

@

3

1

[]]

5

0

1

2

From these tables we see that the answer is 14. The longest time period occurs when the Friday of 13th falls in July of the first year and in September of the second year, while the second year is not a leap year. 26. Answer: 350. By considering the numbers of apples in the packages, there are 3 cases: l) (4, l, l, l). 2) (3, 2, 1, 1 ) . 3) (2, 2, 2, l).

X 6 X 5 X 4 = 35. 4x3x2xl

(7) 7 (7) ( ) 7 � (7) ( ) ( ) 4

=

3

3! 2

4 2

=

5 2

6X5 = 3x2xl X

3 2

=

�

6

X

X

7 2

4 X3 = 35 2xl

X

6 = 210.

4 1

X

3 2

X X

6 1

X

5 2

X X

2 = l05. 1

X X

So the total number of ways is 35+ 210 + 105 = 350. 27. Answer: 19. Note that 8 + 12 = 20, 5 + 8+ 8 = 21, 5 + 5 + 12 = 22, 5 + 5 + 5 + 8 = 23, 8+ 8 + 8 = 24. If 2::

n

2::

25, write

n=

5k + m where 20

:::;

m

:::;

24 and k is a positive integer. So any amount

25 can be paid exactly using coupons.

However, 19 cannot be paid exactly using these three types of coupons. 28. Answer: 10301. . The broken line is constructed using

"

L ", with lengths 2, 4, 6, . . . , 200.

100 + 101 = 201. Then the total length is 2 (1 + 2 + 3 + 29. Answer: 15. 16

·

·

·

The last

+ 100) + 201 = 10301.

"

L " is

So c= 12 or c=-20 (ignored ) . Substitute y=

-x

+ 12 into the parabola:

2=

x

-x

+ 12

=*

= 3,-4. So A is (3, 9) and B is (-4, 16). Then

x

I A BI 2= (3- (-4)) 2+ (9- 16) 2= 98. 33. Answer: 30. Draw BF

..l

LE= 30°.

J2 CE , where F is on CE . If A B = 1, then BF = 2 and BE =

J2. Thus

E

D

34. Answer: 4022. Set D P= GP= a, I P= FP= b, E P= HP= c. Then DE +FG+HI= (a+ c)+ (a+ b) + (b + c)= 2 (a + b+ c)= 2 35. Answer: 10. By given, B D + 2 +AE= B D + DC. So 2 +AE= DC . Note that A B 2+ BE 2= AE 2 and B D 2+ BC 2 (2 + B D ) 2+ 32= AE 2, 4 (AE + B D )= 32. Then AE + BA

=

=

DC 2. Then

B D 2+ 72= (AE + 2) 2.

32 4 + 2= 10.

18

x

2011= 4022.


2011

(Junior Section, Round 2 solutions)

1. It's equivalent to prove x2y2 :;:::: (ac + bd)2 as all the numbers are nonnegative. This

is true s ince

x2 y 2

(a2 + b2)(c2 + d2) (ac)2 + (bd)2 + a2d2 + b2c2 :;:::: (ac)2 + (bd)2 + 2(ac)(bd) AM-GM (ac + bd)2.

= =

=

Let the tangent at P meet the tangent at R at the point S. Let 0 be the centre of r1. Then OR ST is a s quare. Hence LKPR L R P S 45°. Als o LNPS LNKP LPMS LMPK. Thus LMPR LRPN. 2.

=

=

=

=

=

=

K

3. Let ai 0,

maxSi, bi minSi and s uppos e that t1 then a1 :;:::: b j. Therefore a1 E Sj. =

=

=

min{ti}. For each j, if 81 n Sj i=

Note: Problem 4 in the Senior Section is the general vers ion. 4.

Replace 2011 by any pos itive odd integer n. We firs t s how by induction that am 3m2n-m - 1 for m 0. Suppos e 0, 1, ..., n- 1. This is certainly true for m + + nit's true for s ome m < n - 1. Then 3am 1 3m l2 m - 2. Since n- m > 1, the odd part is 3m+l2n-m-l - 1 which is am+l· Now an-1 3n-121 - 1. Thus =

=

=

=

=

20


Singapore Mathematical Olympiad (SMO)

2011

(Senior Section)

Tuesday,

31

May

2011

0930-1200

hrs

Instructions to contestants 1. Answer ALL 35 questions. 2. Enter your answers on the answer sheet provided. 3. For the multiple choice questions, enter your answer on the answer

sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.

4. For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. 5. No steps are needed to justify your answers. 6. Each question carries 1 mark. 7. No calculators are allowed.


22

6. Determine the value of

2

1 1 + 0+Vfs+2 0-V'8+2 (A) 4- 0 (E) 40-2 7. Let

x

=

is true?

(A)

(D)

1

(B) 2-20

1

log� 2

+

1

1

logi 4

1. 5 < X < 2

3
+

(C) 4 +0

1

1.

log� 8

(D)

20 + 4

Which of the following statements

(B) 2 < x < 2. 5 (E) 3.5
(C) 2.5

3

8. Determine the last two digits of 756•

(A) 01

(B) 0 7

9. It is given that

x

(C) 09

(D)

43

(E) 49

and y are two real numbers such that

y > 1. Find the smallest possible value of

x

>

1

and

=

c-1

logx2011 +logy2011 logxy 2011

(A) 4 10.

(B)

6

(C)

8

(D)

(E) 12

10

It is given that a, band care three real numbers such that a+ b

and ab

(A)

5

=

c2 - 7c +14. Find the largest possible value of a2 + b2.

(B)

6

(C)

8

(D)

9

(E) 10

Short Questions 11.

Find the value of

20112 +21112-2 X 2011 X 2111 25 24

18.

In the diagram below, the lengths of the 2three2 sides of the triangle are a +2 b = 2011. Find the value a em, b em and c em . It is given that of

c

cot C

cot A +cotE

b c a

19.

Su p pose there are a total of

petition, and at least

1011

partici pants, at

2011

partici pants in a mathematics com

1000 of them are female . Moreover, given any least 11 of them are male . How many male

partici pants are there in this com petition?

20.

Letf : Q \

{0, 1}

-t

Q be a function such that

2

xf ( x)+f for all rational numbers

2 (X -1) = 2x x

x #- 0, 1.

Here Q denotes the set of rational

�

numbers . Find the value off( ).

21.

In the diagram below , ABCD is a convex quadrilateral such that AC

intersects BD at the midpoint E of BD. The point His the foot of

the per pendicular from A onto DE, and Hlies in the interior of the segment DE. AH=

40

Su p pose LBCA =

em and CD=

x

90°, CE = 12

em . Find the value of A

c

26

em, EH=

x.

15

em,

28. It is given that

a,

b,

c

and d are four positive prime numbers such

that the product of these four prime numbers is e qual to the sum of

55 consecutive positive integers. Find the smallest possible value of

a+ b+ c +d. ( Remark: distinct. )

The four numbers

a,

b,

c,

d are not necessarily

29. In the diagram below, ABC is a triangle with AB= 39 em , BC = 45 em and CA= 42 em. The tangents at A and B to the circumcircle of

DABC meet at the point P. The point D lies on BC such that PD

is parallel to AC . It is given that the area of DABD is x cm 2. Find the value of x.

30. It is given that

a and

b are positive integers such that

a has

exactly

9 positive divisors and b has exactly 10 positive divisors. If the least

common multi ple

( LCM ) of

a and

b is 4400, find the value of

a

+ b.

31. In the diagram below, ABCD is a quadrilateral such that .!_ABC= 135° and L_BCD= 120°. Moreover, it is given that AB = 2 J3 em,

BC= 4 - 2 -/2 em, CD= 4 -/2 em and AD= of x2- 4x.

c

B

A

x

<-------"'

28

D

em. Find the value


Singapore Mathematical Olympiad (SMO) (Senior Section Solutions)

2011

1. Ans wer: (B)

a2 b2 < Since a, b and c are nonzero real numbers and b2+c 2 c 2 +a 2 that b 2 +c 2 c 2 +a 2 a 2 +b 2 > > c2 a2 b2 Adding 1 throughout, we obtain

c2 , we s ee a 2+b 2

--=--

--=---

Thus

<

1 1 1 > > 2 , which implies that a 2 < b2 < c 2. So we have I a I a b2 c

2

<

l bl

2. Ans wer: (A) Note that (s in()+cos ())2= sin2 ()+cos 2()+2 s in() cos ()= 1+s in 2()= 1+a. Since 0::; ()::;� ' we have sin()+cos ()> 0. So s in() +cos ()= v'f+(i". 3. Ans wer: (D) We have � [(a- b)2 + (b- e)2 + (c- a)2 ] 2 [(-1)2 + (-1)2 +22 ] = 3. ·

�.

4. Ans wer: (C) 1 1 X 2y -

1 2x+y

2x+y =1 2y X y X 1 2+-- -- -= 1 X y 2 1 y X 2 X y 2x+y

::::?-

::::?-

::::?-

Now we have

30

-

<

l ei ·

Moreover, 2x

2(log 3+log 5� +log 7i) log 2 log(9 x 5)+log(49i) log(45 x 27 i) > ---,-log 2 log 2 log(45 x 3) log(128) = 7' > �g 2 �g2

------

SO X >

-

-

3. 5.

8. Ans wer: (B) Note that 74-1 = 2400, s o that 74n-1 is divis ible by 100 for any n E z+. Now, 756 7(756-1-1 +1) 7(756-1-1)+7 7(74n-1)+7 ,

where

56-1 '77+ E !LJ . 4 Since 7(74n-1) is divis ible by 100, its las t two digits are 00. It follows that the last two digits of 756 are 07. =

n

9. Ans wer: (A) logx 2011+logy 2011 logxy 2011

log 2011 log 2011 . log xy + ) ( ) log x logy log 2011 1 1 (- +-) . (log x +logy) log x logy log x logy 2+--+-logy log x 4 (us ing A M 2:: G M), (

>

and the equality is attained when log x = logy, or equivalently, x= y. 10. Ans wer: (C) The roots of the equation x2- (c- 1)x+c2- 7c+ 14 = 0 are a and b, which are real. Thus the dis criminant of the equation is non-negative. In other words , (c-1)2- 4(c2- 7c+ 14)= -3c2+26c- 55= (-3c+11)(c- 5) 2:: 0. 11 < < So we have c - 5. Together with the equalities 3 (a+b?- 2ab (c- 1)2-2(c2- 7c+ 14) -c2+12c- 27= 9 - (c- 6)2,

32

15. Ans wer: 8001 Note that J5n- y'5n- 4 < 0. 01 if and only if 4 > 400. y'5; + v'5n- 4= J5n 5n- y'5n- 4 If n= 8000, then J5n+y'5n- 4= v' 40000+\1'39996 < 400. If n= 8001, then J5n+y'5n 4= v' 40005+v'40001 > 400. So the ans wer is 8001. -

16. Ans wer: 1006 The s eries can be paired as 1 1 1 1 1 1 ) + + ( ) + ( + )+( + 1 11 · · · 11 1 +11- 1+ 111 · 1 + 11-20 1+11-2oo9 1+112oo9 1 +1120 Each pair of terms is of the form 1 1 ----,-+--= 1. 1 1+ a 1+a There are 1006 pairs of s uch terms , and thus the s um of the s eries is 1006. 17. Ans wer: 54 X

7 77r . 4 (-7r ) +cos 4 (-7r ) +s in 4 (8) +cos 4 (87r) 8 8

Sill

sin 4 (i) +cos 4 (i) + s in 4 (i) +cos 4 (i) . 2 (47r) = 3 . 2 (87r) + COS 2 (87r) )2- 4 Sill . 2 (87r) COS 2 (87r) = 2- Sill 2(Sill 2 Thus 36x = 54. 18. Ans wer: 1005 By the laws of s ine and cos ine, we have s in A s inE sinG and c b a Then 1 cos C cote cosAsi B + �os B sinA � sinC cot A+cot B smAsm B s in A s in B cos C sin(A+B) s inC s in A s in B ( . 2 ) cos C Sill 0 (ab jc 2) s in2 C a 2 +b2- 2 ) )( ( 2ab s in2 C a 2 +b 2- c 2 2c 2 2011- 1 = 1005. 2 .

34

·

22. Ans wer: 3 Note that 1 1 -+X y

1 =? xy- 211x- 211y 211

= -

=

0 =? (x- 211)(y- 211)

=

2112.

Since 211 is a prime number, the factors of 2112 are 1, 211, 2112, -1, -211, -2112. Thus the pairs of integers (x, y) s atis fying the las t equation are given by: (x- 211, y- 211)

=

(1, 2112), (211, 211), (2112 , 1), (-1, -2112), (-211, -211), (-2112, -1).

Equivalently, (x, y) are given by (212, 211 +2112), (422, 422), (211+2112 , 212), (210, 211- 2112), (0, 0), (211- 2112 , 210). Note that (0, 0) does not s atis fy the firs t equation. Among the remaining 5 pairs which s atis fy the first equation, three of them satis fy the inequality x � y, and they are given by (x, y) (422, 422), (211+2112 , 212), (210, 211- 2112). =

23. Ans wer: 93 By long divis ion, we have x 4 +ax 2+bx+c

=

(x2+3x- 1) · (x 2- 3x+(a+10))+(b-3a- 33)x+(c+a+10).

Let m1 , m2 be the two roots of the equation x2 + 3x- 1 0. Note that m1 -=1m2 , since the dis criminant of the above quadratic equation is 32-4 · 1 · 1· (-1) 13 -=10. Since m1, m2 als o s atis fy the equation x4 +ax 2+bx+c 0, it follows that m1 and m2 als o satis fy the equation (b- 3a- 33)x+(c +a+ 10) 0. =

=

=

=

Thus we have (b- 3a- 33)m1 +(c +a+ 10)

=

0,

and (b- 3a- 33)m2 + (c +a+10) 0. Since m1 -=1m2 , it follows that b-3a-33 0 and c+a+lO 0. Hence we have b 3a+33 and c -a-10. Thus a+b+4c+100 a+(3a+33)+4(-a-10)+100 93. =

=

=

=

=

=

=

24. Ans wer: 1120 Let m and n be pos itive integers s atis fying the given equation. Then 3(n2- m) 2011n. Since 2011 is a prime, 3 divides n. By letting n 3k, we have (3k)2 m + 2011k. This implies that k divides m. Let m rk. Then 9k2 rk+2011k s o that 9k r+2011. The smalles t pos itive integer r s uch that r+2011 is divis ible by 9 is r 5. Thus k (5+ 2011)/9 224. The corres ponding values of m and n are m 1120 and n 672. =

=

=

=

=

=

=

=

=

36

=

=

� = c2�b2• Since GN is parallel to AD and G is the centroid of the triangle ABC, we have MD/MN= 3. It follows that c+b= aJ3. Thus , AB = aJ3-b = 15-6= 9.

28. Ans wer: 28 The s um of 55 pos itive cons ecutive integers is at leas t (55 x 56)/2 = 1540. Let the middle number of thes e cons ecutive pos itive integers be x. Then the product abed= 55x = 5 11 x. So we have 55x 2': 1540 and thus x 2': 28. The leas t value of a+b +c +d is attained when x= 5(7). Thus the ans wer is 5+11+ 5+ 7 = 28. ·

·

29. Ans wer: 168 Firs t L_BD P= L_BCA= L_BAP s o that P, B , D, A are concyclic. Thus L_ACD = L_ PBA= L_ PDA= L_DAC s o that DA= DC.

By cos ine rule, cos C = 3/5. Thus DC = � AC jcos C = 21 x 5/3 = 35. Hence ED = 10 and BC = 10 + 35= 45. Thus area(L,ABD ) = �� x area(L,ABC). By Heron's formula, area(L,ABC)= 756. Thus area(L, ABD)= �� x 756= 168. 30. Ans wer: 276 Since the number of pos itive divis ors of a is odd, a mus t be a perfect square. As a is a divis or of 4400 = 24 x 52 x 11 and a has exactly 9 pos itive divis ors , we s ee that a= 22 x 52. Now the leas t common multiple of a and b is 4400 implies that b mus t have 24 x 11 as a divis or. Since 24 x 11 has exactly 10 pos itive divis ors , we deduce that b = 24 x 11 = 176. Hence a+b = 276. 31. Ans wer: 20 Firs t we let /!, be the line which extends BC in both directions . Let E be the point on /!, s uch that AE is perpendicular t o f. Similarly, we let F be the point on /!, s uch that DF is perpendicular to f. Then, it is eas y to s ee that BE = AE = v'6, CF = 2J2 and DF = 2v'6. Thus EF = y'6 + 4- 2J2 + 2J2 = 4 +v'6. Now we let G be the point on DF s uch that AG is parallel to £. Then AG = EF = 4 +v'6 and

38

which is a contradiction to the given fact that each element of S is les s than or equal to 2011. Case 2. 1 is in A. We may let a16 = 1. Then a 1, a 2 , As in Cas e 1, we have

· · ·

, a 1 5 are compos ite numbers .

which is a contradiction. Thus we have shown that every 16-element s ubs et A of S s uch that all elements in A are pairwis e relatively prime must contain a prime number. Hence the smallest k is 16. 35. Ans wer: 12 Let the extens ions of A Q and BP meet at the point R. Then LP R Q= LPAB= LQPR s o that QP = QR. Since QA = QP, the point Q is the midpoint of A R. As A R is parallel to LP, the triangles A RB and LP B are similar s o that M is the midpoint of PL. Therefore, N is the centroid of the triangle P LB, and 3 MN= BM.

B

Let LABP= e. Thus tanB= A R/AB= 2A QjAB= 5/6. Then B L= PB cos B= AB cos 2 0. Als o BM/BL = BQ/BA s o that 3MN = EM = �� AB cos 2 0 cos 2B( QM + 3MN). Solving 'for MN, we have MN= 3 �:;_; e = 3x (�f6)2 = 12.

40


2011

(Senior Section, Round 2 solutions)

1. There is an error in this problem. The triangle is not neces s arily equilateral. In fact

we s hall prove that the altitude at A, the bis ector of LB and the median at C meet at a common point if and only if cos B= � where BC= a, CA = b and A B= c. a

c

Let D,E and F be the points on BC,CA and A B res pectively s uch that AD is the altitude at A , BE is the bis ector of LB and CF is the median at C. Suppos e that AD,BE,CF meet at a common point. The point of concurrence of AD,BE and CF mus t lie ins ide the triangle A BC. Since F is the midpoint of A B, by Ceva's theorem CE : EA = CD : DB. Us ing the angle bis ector theorem, CE : EA =a : c. Thus CD= a2/(a+c) and DB= acj(a+c). Thus cos B= �� = � a

c·

A

B

Convers ely, if cosB= � then LB is acute and BD= c cos B= acj(a+c)
' c

So given a and c, the acute angle B and hence the triangle A BC is determined. If a -=/- c, then the triangle A BC is not equilateral. 2. Yes , in fact, for any k E

N,

there is a s et 8k having k elements s atis fying the given condition. For k = 1, let 81 be any s ingleton s et. For k = 2, let 82 = {2 ,3}. Suppos e that 8k = {a1, ... , ak} s atis fies the given conditions. Let b1 = a1a2 ak bj = b1+aj -1, 2 :::; j :::; k+ 1. · · ·

Let 8k+1 = {b1, b2, ... , bk+1}. Then the fact that 8k+1 s atis fies the required property can be verified by obs erving that lm- nl = (m, n) if and only if (m- n) divides m. 42

Note that a1 a2 ··· an

=

1. It suffices to show that

since it is equivalent to 1 1 1 �n- 1. + +···+ 1+an 1 +a1 1+a2

--

We shall show that ( * ) is true for n 2:: 2. The case n 2 is obvious. We will now prove it by induction. Suppose ( * ) holds for n k. Now assume a1 ··· ak+l 1, ai > 0 for all i. To prove the inductive step, it suffices to show that =

=

which can be verified directly. Note: This is an extension of the problem :

44

=

Throughout this paper, let lxJ denote the greatest integer less than or equal to x. For example, l2.1J = 2, l3.9j = 3 ( This notation is used in Questions 7, 9, 19 and 20). 1.

A circular coin A is rolled, without sliding, along the circumference of another stationary circular coin B with radius twice the radius of coin A. Let x be the number of· degrees that the coin A makes around its centre until it first returns to its initial position. Find the value of x.

2. Three towns X,Y and Z lie on a plane with coordinates (0, 0), (200, 0) and (0, 300) respec tively. There are 100, 200 and 300 students in towns X,Y and Z respectively. A school is to be built on a grid point (x, y) , where x and y are both integers, such that the overall distance travelled by all the students is minimized. Find the value of x+y. 3. Find the last non-zero digit in 30!.

( For example, 5! = 120; the last non-zero digit is 2.) 4. The diagram below shows �A BC , which is isoceles with A B = AC and LA 20°. The point D lies on AC such that A D = BC. The segment E D is constructed as shown. Determine LA B D in degrees. A =

cos4 a sin4 a cos4 f3 sin 4 f3 . 5. G1ven that �(3 + -.-2- = 1, evaluate --2- + -.-2 - . cos sm f3 cos a sm a 6. The number 25 is expressed as the sum of positive integers XI, x 2, , Xk, where k What is the maximum value of the product of XI, x 2, X3, , and Xk ? ·

·

46

·

·

·

·

::;

25.

18.

A collection of 2011 circles divide the plane into N regions in such a way that any pair of

circles intersects at two points and no point lies on three circles. Find the last four digits of N.

19. If a positive integer N can be expressed as lxJ + l2xJ + l3xJ for some real numbers x, then we say that N is ''visible"; otherwise, we say that N is "invisible". For example, 8 is visible since 8= l1.5J + l2 (1.5 )J + l3 ( 1.5 )J , whereas 10 is invisible. If we arrange all the "invisible" positive integers in increasing order, find the 2011th "invisible" integer. 20. Let A be the sum of all non-negative integers

n

satisfying

Determine A. 21.

A triangle whose angles are A, B, C satisfies the following conditions sin A+ sinE + sinG cos A + cos B + cosC and

12 7'

12 . 25 Given that sin C takes on three possible values 81, 82 and 83 , find the value of 100818283 . sin A sin B sin C =

22. Let x

>

1,

y

>

1 and

z >

1 be positive integers for which the following equation 1! + 2! + 3! + . . . + x!=

y

z

is satisfied. Find the largest possible value of x + y+ z. 23. Let A BC be a non-isosceles acute-angled triangle with circumcentre 0, orthocentre H and LC = 41 o. Suppose the bisector of LA passes through the midpoint M of OH. Find LHAO in degrees. 24. The circle 'Yl centred at 01 intersects the circle 'Y2 centred at 02 at two points P and Q. The tangent to 'Y2 at P intersects 'Yl at the point A and the tangent to 'Yl at P intersects 'Y2at the point B where A and B are distinct from P. Suppose PQ · 0102= P01 P02 and LAP B is acute. Determine the size of LAP B in degrees. ·

25. Determine

n

1 . L (n). -+oo . n lim

i=O

( Note: Here

(�) 't

denotes

n�

.1 't. n

(

" ' 't .

)

�

fori= 0, 1, 2, 3, · · ·

48

,

n.

)

Thus, 30! 30! 107

1 226 . 3 4 . 5 7 . 74 . 112 . 132 . 17 . 19 . 23 . 29 1 1 2 9 . 3 4 . 7 4 . 112 . 132 . 17 . 19 . 23 . 29 1 6 4 . 25 . 74 . 112 . 132 . 17 . 19 . 23 . 29 6 (2) (1) (1) (9) (7) (9) (3) (9) (mod 10) 2 (-1) (-3) (-1) (3) (-1) (mod 10) 8(mod 10),

showing that the last non-zero digit is 8. D

4. Answer.

10

Solution.

Let E be the point inside f:,A BC such that f:,E BC is equilateral. Connect A and D to E respectively. It is clear that f:,AE B and f:,AEC are congruent, since AE = AE , A B = AC and BE= CE. It implies that L BAE = LCAE= lOa . Since A D = BC = BE , LE BA = L D A B = 20a and A B = BA, we have t:,A BE and f:, BA D are congruent, implying that LA B �= L BAE = lOa .

D

5.

Answer.

.

Solutwn.

1 cos2 a sin2 a cos4 a sin4 a . (3 --:--- . Then 13 and sine= ---Smce �(3 + . 2 (3 = 1, set cose= cos cos sm sm c

---

---

cos (e- a )= cose cos a + sine sin a= cos2 a + sin2 a= 1. 50

Now let x2 +ax+ b= (x- xl) (x- x2), where x1:::; x2. Then the set of integer solutions of x2+ ax+ b< 0 is {k : k is an integer, x1 < k< x2}. By the given condition, {k : k is an integer, x1 < k< x2}= {k : k is an integer, -11< k< 6} = {-10,-9, . .· , 5}. Thus -11 :::; x1 < -10 and 5 < x2 :::; 6. It implies that -6 < x1 + x2 < -4 and -66 :::; X1X2< -50. From x2+ax+b= (x-x1) (x-x2), we have a= - (x1+x2) and b= x1x2. Thus 4< a< 6 and -66:::; b<-50. It follows that 54< a - b< 72. Thus

max:Lia- biJ :::; 71.

It remains to show that it is possible that lla - biJ = 71 for some a and b. Let a= 5 and b= -66. Then x2+ab+b= (x+11) (x-6) and the inequality x2+ab+b< 0 has solutions {x: -11< x< 6}. So the set of integer solutions of x2 + ab + b< 0 is really the set of integers in A n B. Hence 10.

max:Lia- biJ = 71.

Answer. Solution.

D

8 We consider the polynomial P(t)= t3 + bt2 + ct+ d.

Suppose the root of the equation P (t)= 0 are x, y and z. Then

-b= X+y+ Z= 14,

and x3 +y3 + z3 +3d= (x+ y+ z) (x 2 +y2+ z2- xy- xz- yz). Solving for b, c and d, we get b= -14, c= 30 and d= -64. Finally, since t3 - 14t2+30tD 64= 0 implies t= 2 or t= 4 or t 8, we conclude that max{ a, ,B, 'Y}= 8. =

11.

Answer.

38

Solution. Let n be an even positive integer. Then each of the following expresses n as the sum of two odd integers: n= ( n- 15) + 15, ( n - 25) + 25 or ( n - 35) +35. Note that at least one of n - 15, n - 25, n - 35 is divisible by 3, hence n can be expressed as the sum of two composite odd numbers if n > 38. Indeed, it can be verified that 38 cannot be expressed as the sum of two composite odd positive integers. D

12.

Answer.

1936 52

We shall show that I AI bijection from A to B:

=

I BI

(63°)

=

=

34220 by showing that the mapping¢> below is a

First, since { a1, a2, a 3 } E A, we have a1+3 implying that {a1, a2- 2, a 3 - 5} E B.

�

a2 and a2+4

�

a 3 , and so a1

<

b3 , we have {b1, b2+2, b3+5}

<

a2-2

<

a 3-5,

It is clear that¢> is injective. It is also surjective, as for any {b1, b2, b3 } A and Hence¢> is a bijection and I AI 16.

Answer.

=

I BI

=

E

B with b1

<

b2

34220.

E

0

32

It is clear that 8(cos 40°) 3 -6 cos 40°+ 1 = 0, since cos 3A = 4 cos3 A-3 cos A . Observe that Solution.

1 3 + 64 sin2 20° 2 2 sin 20° cos 20° 2 6 + 32 (1- cos 40°) 1- cos 40° 1 + cos 40° 8 cos 40°+ 4 + 32 32 cos 400 1- (cos 40°) 2 8 COS 40° + 4- 32 COS 40° + 32 ( COS 40°) 3 + 32 1- (cos 40°) 2 1- 6 cos 40°+8(cos 40°) 3 4x +32 1- (cos 40°)2 32, ----

_

==

where the last step follows from 8(cos 40°) 3 - 6 cos 40° + 1 = 0. 17.

Answer. Solution.

0

6029 Given the original equation f(x2 + x) + 2f(x2-3x + 2) = 9x2- 15x,

we replace x by 1-x and obtain f(x2-3x+ 2) + 2f(x2+ x) = 9 (1-x) 2- 15 (1-x) = 9x2-3x-6. Eliminating f(x2-3x+ 2) from the two equations, we obtain 3f(x2 + x) = 9x2+ 9x- 12, thereby

f (x2+ x) = 3x2 + 3x-4 = 3 (x2 + x)- 4,

hence f (2011) = 3(2011)- 4 = 6029. 0

54

20.

Answer.

95004

Solution. We shall prove that for any positive integer nonnegative integer solutions to L�J= La�lJ, then

a, if f(a) denotes the sum of all

f(a)= 16a(a2-1)(a+ 2).

Thus

!(27)= 95004.

Let n be a solution to L�J = La�1J . Write n= aq + r, where 0::; r
A= LL(qa+r) q=O r=q a-1

a-1a-1

= L(a-q)qa+ LLr = = =

q=O r=q a-1

q=O a-1

a-1

q=O a-1

q=O a-1

r=O q=O a-1

q=O a-1

q=O a-1

r=O a-1

a-1

q=O

q=O

r=O

r=O

r

a2L:q-aL q2+ LLr a2L:q-aL q2+ L:r(r+1) a2L q-aL q2+ Lr2+ Lr

1) (a2+ 1 ) · �a( a-1 ) + (1- a)· �a(2a-1)(a2 6 1 = 6a(a2-1)(a+ 2). =

D

21.

Answer.

48

By using factor formulae and double angle formulae: sin A + sinE + sinG 4cos .:1 cos fl. cos Q 12 2 2 2 - --cosA + cos B + cosC 7' 1+4sin 4 sin � sin � and

A B C 12 . . . . A . B . C smA sm B smC = 8 sm 2 sm 2 sm 2 cos 2 cos 2 cos 2= .

25

Solving these equations, we obtain . A . B . C sm-sm-sm- - 0 .1

2

2

2

B C A cos 2 cos 2 cos 2 Furthermore, sin

C

2 = cos

(

A+ B

2

)

= cos 56

A

0. 6

B

A

B

2 cos 2 - sin 2 sin 2'

p

L

be the midpoint of BC . It is a known fact that AH = 20£. To see this, extend meeting the circumcircle of the triangle A BC at the point N. Then AN BH is a parallelogram. Thus AH = NB = 20£. Therefore in the right-angled triangle OLC, OC = OA = AH = 20£. This implies L.OCL = 30°. Since the triangle A BC is acute, the circumcentre 0 lies inside the triangle. In fact L.A = 60° and L.B = 79°. Then L.OAC = L.OCA = 41°-30° = 11°. Consequently, L.HAO = 2L.OAM = 2 X (30°- 11°) = 38°. D Let

CO

24.

Answer.

30

Let P01 = r1 and P02 = r2. First note that 0102 intersects PQ at the midpoint H (not shown in the figure) of PQ perpendicularly. Next observe that L.APQ = L.PB Q = L.P0201, and L.BPQ = L.PA Q = L.P0102. Therefore L.APB = L.APQ + L.BPQ = L.P0201 + L.P0102.

'/'1

'/'2

H and sin f3 = Let L.P0201 = a and L.P01 02 = f3. Then sin a = PrQ, cos a = 02 r 2 2 2 fr�, cos f3 = 0f1H. Thus sin L.APB = sin ( a + /3) = sin a cos f3+ cos a sin f3 = fr� · 0f1H +

25.

0r2H. E9.. 2 2r1

=

Answer.

2

Solution.

PQ(· 01 H+02H) = PQ0 · 102 2r1r2 = 21. Since L.APB is acute it is equal to 30°. 2r1r2 '

Let

Assume that

n

�

3. It is clear that

() � ; -1

an=

2+

n

�=1

58

-1 >

2.

·

D


2011

(Open Section, Round 2) Saturday, 2 July 2011

0900-1330

1. In the acute-angled non-isosceles triangle ABC, 0 is its circumcentre, His its orthocentre and AB > AC. Let Q be a point on AC such that the extension of HQ meets the extension of BC at the point P. Suppose BD= DP, where Dis the foot of the perpendicular from A onto BC. Prove that LODQ = 90°. 2. If 46 squares are colored red in a 9 x 9 board, show that there is a 2 x 2 block on the board in which at least 3 of the squares are colored red. 3. Let x, y, z > 0 such that

� +� + �

<

x!z·

Show that

2z 2x 2y + + yl + x2 y'l + y2 yl + z2 4.

Find all polynomials P(x) with real coefficients such that P(a)

5.

<

E Z

implies that

a E Z.

Find all pairs of positive integers (m, n) such that m+n-

3mn m+n

60

2011 3

3.

which leads to a contradiction. On the other hand, s uppos e that r1, r3, r5, r7, r9 2:: 6. Then the s um of any 2 cons ecutive ri's is :::; 9. Again we get a contradiction as (r1+r2)+

·

·

·

+(r7+rs)+rg

:::; 4

9+9=45.

x

Let r = 1/x, s= 1/y, t= 1/z. There exis ts a < 1 s uch that r + s + t=a2rst or a(r + s + t)=a3rst. Let a=ar, b=as, c=at. Write a= tan A, b tanB, c = tan C, then A+ B + C 1r. It is clear that 3.

=

=

1 2

- X

1 1 1 + ----= + -;= ::;;;: y1+r2 v1+s2 v1+t2 1 1 1 < + ----= + -;= ::;;;: v1+a2 v1 + b2 v1+c2 = cos A+ cos B+ cos B A+ B + C < 3 cos =�=� x RHS. 2 2 3

LHS=

(

)

2nd soln: Note that

1 1 1 -+ - + x y z

Hence

2x =;;;= y1+x2

<

1 xyz

:::::?

xy+yz+xz

2x = == Jx2+xy+xz+yz

----r=

< -r-::::::==

=

By AM-GM we have

==

<

1.

2x J(x+y)(x + z)

X 2x X -=;= < . + J(x+y)(x+z) - x+y x+z

----r-.;======:=;=

= --

Similarly, z z 2z =;= < -- + --. V( Z+X) (z + y) Z+X Z + y

2y Y + __ Y < __ J(y + z)(y+x) - y + z y + x'

----r-.;======:=;=

=

The des ired inequality then follows by adding up the three inequalities. 4.

Let P(x)=anxn + +a1x+a0. Define Q(x)=P(x + 1) - P(x). Then Q(x) is of degree n- 1. We'll prove by contradiction that IQ(x)l :::; 3 for all x. This will imply that n :::; 1. As s ume that IQ(a)i > 3 for s ome a E R Then IP(a + 1) - P(a)l > 3. Thus there are 3 integers between P(a) and P(a + 1). Hence there exis ts three values ·

·

·

62

Singapore Mathematical Society . Singapore Mathematical Olympiad (SMO)

2010

(Junior Section)

Tuesday, 1 June 20 10

0930-1200

Important:

Answer ALL 35 questions. Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer. For the other short questions, write your answer on the answer sheet and shade the appropriate bubbles below your answer. No steps are needed to justify your answers. Each question carries 1 mark. No calculators are allowed.

PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO.

1


1. Among the five real numbers below, which one is the smallest? (A)

v2010;

'JJYJ_J�

(B)

v2009;

(C) 2010;

20!.0MIV\n

(D)

2010 ; 2009

(E)

2009 . 2010

2. Among the five integers below, which one is the largest?

3. Among the four statements on real numbers below, how many of them are correct? "If a
0 then !b < ! a

"If a
"· '

"If a
(A) 0;

4.

(B)

(C) 2;

1;

(D)3;

(E) 4.

What is the largest integer less than or equal to (A) 2009;

(B)

2010;

(C) 2011;

(D) 2012;

.{/(2010)3 +3 x (2010)2+4 x 2010+1? (E) None of the above.

5. The conditions of the road between Town A and Town B can be classified as up slope, horizontal or down slope and total length of each type of road is the same. A cyclist travels from Town A to Town B with uniform speeds 8 km/h, 12 km/h and 24 km/h on the up slope, horizontal and down slope road respectively. What is the average speed of his journey?

12kmfh ----+

Town A

(A) 12 km/h;

(B)� km/h;

Town B

(C)

16 km/h; 2

(D) 17 km/h;

(E) 18 km/h.

6. In the diagram, MBC and f).CDE are equilateral triangles. Given that LEBD LAEB X0, what is the value of x?

=

62° and

=

A

D (A) 100;

(B) 118;

(C) 120;

(D) 122;

(E) 135.

7. A carpenter wishes to cut a wooden 3 x 3 x 3 cube into twenty seven 1 x 1 x 1 cubes. He can do this easily by making 6 cuts through the cube, keeping the pieces together in the cube shape as shown:

. . . . :·

.

.

:

·······.·······.·······

What is the minimum number of cuts needed if he is allowed to rearrange the pieces after each cut? (A) 2;

(B) 3;

(C) 4;

(D) 5;

(E) 6.

8. What is the last digit of 7(77)? (A) 1;

(B) 3;

(C) 5;

(D) 7;

(E) 9.

3

9.

Given that n is an odd integer less than many such integers are there ? (A) 3;

10.

(B)

4;

(C) 5;

(D) 6;

1000 and the product of all its digits is 252.

How

(E) 7.

What is the value of

(A) 451856;

(B)

691962;

(C) 903712;

(D) 1276392;

(E) 1576392.

Short Questions

11.

Let x and y be real numbers satisfying y=

2008x+2009 + 2010x- 2011

2008x+2009 +2010· 2011- 2010x

Find the value of y. For integers ah . . ., an

12.

E {1, 2, 3, . .., 9}, we use the notation a1a •••an to denote the number 2 10n-1a1+10n-2a2+· · · +10an-1+an. For example, when a = 2 and b = 0, ab denotes the number 20. Given that ab = b2 and ache = (ba)2• Find the value of abc.

13.

Given that (m- 2) is a positive integer and it is also a factor of 3m2- 2m+10. Find the sum of all such values of m.

14.

In triangle ABC, AB = 32 em, AC = find the length of AM in em.

36 em and BC = 44 em.

A

B

M

4

c

If M is the midpoint of BC,

15. Evaluate

678+690+702+ 714+...+ 1998+2010 3+9+15+21+... + 327+ 333

16. Esther and Frida are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1 to 160. Esther chose to do it row-wise so that the first row is numbered 1, 2, ..., 16 and the second row is 17, 18, . .., 32 and so on. Frida chose to do it column wise, so that her first column has 1, 2, . .., 10, and the second column has 1 1, 12, ..., 20 and so on. Comparing Esther's array with Frida's array, we notice that some numbers occupy the same position. Find the sum of the numbers in these positions.

... .. . ... ... ... ... ... ... 145 146 147 ... 2 18 ... .

1 17

.

3 19

.

... 16 . . . 32 . . . .. . . ... .. . 160 .

.

1 2 .. . .. .

11 12 ... . ..

10

20

.. ... . .. .. . 30 ...

21 22 ... ...

.

... ... .. . ... ...

151 152 .. .. .

.

160

Frida

Esther

17. The sum of two integers A and B is 2010. If the lowest common multiple of A and B is 14807, write down the larger of the two integers A or B.

18. A sequence of polynomials an(x) are defined recursively by a0(x) a1 (x) an(X)

=

=

=

1, x2+x+1, (� + 1)an-I(X)- an_z(X), for all n 2 2.

For example,

az(x) a3(x)

=

=

=

(�+1)(�+x+ 1)- 1 x4+x3 +2x2+x, (� +1)(x4+x3 + 2�+x) (� +x+1) x7 +x6+2x5 +2x4 + x3 +�- 1. =

-

Evaluate a 010(1). 2 19. A triangle ABC is inscribed in a semicircle of radius 5. If AB of s2 where s AC+BC. =

5

=

10, find the maximum value

102009>.

20.

Find the last two digits of 2011<20

21.

Your national football coach brought a squad of 18 players to the 2010 World Cup, consisting of 3 goalkeepers, 5 defenders, 5 midfielders and 5 strikers. Midfielders are versatile enough to play as both defenders and midfielders, while the other players can only play in their designated positions. How many possible teams of 1 goalkeeper, 4 defenders, 4 midfielders and 2 strikers can the coach field?

22. 23.

Given that value of x.

169(157-77x)2 +100(201-100x)2 = 26(77x-157)(1000x-2010),

Evaluate

find the

(20202-20100)(201002-1002)(20002+20100) 20106-106

24.

When 15 is added to a number x, it becomes a square number. When 74 is subtracted from x, the result is again a square number. Find the number x.

25.

Given that x and y are positive integers such that 56::::; x+y::::;

26.

the value of f-x2.

59 and 0.9

<

:: y

<

0.91, find

Let AA' and BB' be two line segments which are perpendicular to A' B'. The lengths of AA', BB' and A' B' are 680, 2000 and 2010 respectively. Find the minimal length of AX+XB where X is a point between A' and B'.

B

I

0 0 0 C'l

0 00 1,0

l

A'

27.

The product 1 x 2 x 3 x M= 1! x 2! x 3! x 4! x 5! ·

·

·

2010

n is denoted by n!. For example 4! = 1 x 2 x 3 x 4 = 24. Let 6! x 7! x 8! X 9!. How many factors of Mare perfect squares?

x

x

1

B'

6

28. Starting from any of the L's, the word LEVEL can be spelled by moving either up, down, left or right to an adjacent letter. If the same letter may be used twice in each spell, how many different ways are there to spell the word LEVEL?

L

L

L

E

L

E

v

E

L

E

L

L

'

L ...___

29. Let ABCD be a rectangle with AB 10. Draw circles C1 and C2 with diameters AB and CD respectively. Let P, Q be the intersection points of C1 and C2. If the circle with diameter PQ is tangent to AB and CD, then what is the area of the shaded region? =

30. Find the least prime factor of

10000 .. ·00 1. ""---v-"

2010-many

7

�

�

31. Consider the identity 1 + 2 +· · · + n = n(n+ 1). If we set P1(x) = x(x + 1), then it is the unique polynomial such that for all positive integer n, P1 (n)=1 + 2 +· · · + n. In general, for each positive integer k, there is a unique polynomial P (x) such that k P (n) =1k + 2k + 3 k + ··· + nk k Find the value of P2o10C

-

�)

for each n=1, 2, . . . .

.

32. Given that ABCD is a square. Points E and F lie on the side BC and CD respectively. such that BE=CF = AB. G is the intersection of BF and DE. If

�

Area of ABGD m ----- = Area of ABCD n is in its lowest term, find the value of m + n.

A

.-------�

B E

F

C

33. It is known that there is only one pair of positive integers a and b such that a :::; b and a2 + b2 + 8ab=2010. Find the value of a + b. 34. The digits of the number 123456789 can be rearranged to form a number that is divisible by 11. For example, 123475869, 459267831 and 987453126. How many such numbers are there? 35. Suppose the three sides of a triangular field are all integers, and its area equals the perimeter (in numbers). What is the largest possible area of the field?

8

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Junior Section Solutions) 1. Ans: (E) It is the only number less than 1. 2. Ans: (D) Other than (D), all numbers are less than 20102010. Now 37 9 3(3 ), the result follows.

>

2010. Thus 20102010

<

(37l

<

3. Ans: (B) Only the third statement is correct: a < b implies a+ c < b + c. For other statements, counterexamples can be taken as a= -1, b= 1; c= 0 and a= 0, b= -1 respectively.

4. Ans: (C) Since (2010+ 1)3= 20103 + 3 x 20102+ 3 x 2010+ 1. The result follows.

5. Ans: (A)

Let the distance between Town A and Town B be 3s. The total time taken for up slope, horizontal and down slope road are �, { and ;4 respectively. His average speed for the 2 3s . 1 . km/h. Wh 0 e JOUrney IS s s s = 12 8 + 2 + 24 1

6. Ans: (D)Observe that !:l.BCD is congruent to !:l.ACE (using SAS, BC= AC, CD= CE and LACE= 60° - LECB= LBCD). Thus LAEC= LBDC. We get x

o

= = = =

360° - LAEC- LBEC 360° - LBDC- LBEC

LEBD+ LECD 62° + 60°= 122°

Thus x= 122.

9

A

D

7.

Ans: (E)

There is no way to reduce the cuts to fewer than 6: Just consider the middle cube (the one which has no exposed surfaces in the beginning), each of the its sides requires at least one cut.

8. Ans: (B)

The last digit of 7k is 1, 7, 9, 3 respectively for k= 0, 1, 2, 3 (mod Since 77 = (-1? = 3 (m�d

9.

Ans: (C)

252 = 2 x 2 x 3

10.

x

3

x

4), the last digit of 7 <77) is 3.

4).

7. We can have: 667, 497, 479, 947, 749.

Ans: (C) We only need a rough estimate to rule out the wrong answers. -fiT� 3.3 and -Y5� 2.2, so the sum is� 5 . 58 + (1.1)8 � 5. 58 = 30.254 � 304 � 810000. Thus (C). Of course the exact answer can be obtained by calculations, for example,

t (�)< «t")l( �)8-1 t (�)< «t")'(- �)8-1 +

=2

.I (�)( m)i( vs)8-i

zeven

10

1 1. Ans: 20 10. Let

a=

2008x+ 2009 20 10x- 20 11·

Then a � 0 and- a � 0 since they are under the square root. Hence a= 0. Thus y= 20 10.

12. Ans: 369.

It is easy to see that b= 5, ab= 25 or b= 6, ab= 36. Upon checking, b = 6, a= 3 and so 2 (b a?= 63 = 3969. Therefore c = 9. Hence abc= 369.

13. Ans: 5 1. Since

2 3m - 2m+ 10

18 m- 2 m- 2 is an integer. Thus m- 2 is a factor of 18. m- 2= 1, 2, 3, 6, 9 , 18, thus m = 3, 4, 5, 8, 11, 20. The required sum is 5 1. -----

14. Ans: 26. Using the Median Formula,A_MZ

=

= 3m+ 4+

2 2 2 2AB + 2AC - BC ·

4

--

. ThusA M= 26 em.

15. Ans: 16. Note that both numerator and denominator are arithmetic progressions. Removing common factors gives

2X

113+ 115+ 117+ ... + 333 + 335 1+ 3+ 5 + ...+ 109+ 111

2 2 X 112 (335 + 1 13) = 5; (111+ 1)

:::::

16.

16. Ans: 322. The number in the r-th row and c-th column has value 16(r- 1) + c in Esther's array and value 10(c- 1) + r in Frida's array. So we need to solve

16(r- 1) + c= 10(c- 1)+ r

�

5r= 3c + 2.

There are exactly four solutions

(r, c)= {(1, 1), (4, 6), (7, 11), (10, 16)}.

11

17. Ans: 1139. A direct way to solve the problem is to factor 14807 directly. Alternatively, one may hope for A and B to have common factors to simplify the problem. This is a good strategy because of the following fact: "The greatest common divisor of A and B, equals the greatest common divisor of A + B and lcm(A, B). "

2010 is easily factored as 2 X 3 x 5 x 67. Checking that 67 is also a factor of 14807, we can conclude that 67 is also a factor of A and B. The problem is reduced to finding a and b such that 2010 1 07 =30 and ab= a + b= =221. 67 7 Since 221 can be factored easily, a andb must be 13 and 17. So the answer is 17x67=1139.

�

18. Ans: 4021. an(1) is the simple recurrence relation fn = 2fn-1 - fn- • fo = 1 and !1 = 3. Using standard 2 technique or simply guess and verify that fn=2n + 1. So a oio(1)=f2010 = 2(2010) + 1. 2 19. Ans: 200. ABC must be a right-angled triangle. Let x = AC and y = BC , by Pythagoras theorem r + r =102•

i =(x + y)2 =r

+ y2 + 2xy=100 + 2 x

area of ABC.

Maximum area occurs when x=y, i.e. LC AB=45°. So x=y= -{50.

20. Ans: 01. Note that 2011 = 11 (mod 100) and 112 = 21,11 3 (mod 100). Since 20102009 is divisible by 10,

20112 0102009

=

1110 x

·

·

·

x 1110

=

=

31 (mod 100) etc. So 1110

1

_

1

(mod 100).

21. Ans: 2250.

(i) x (;) x (�) choices for goalkeepers, strikers and midfielders respectively. The remaining

midfielder and defenders can all play as defenders, hence total number of possibilities are

3

X

lOX 5

12

X

(�)

=2250.

22. Ans: 31.

Let a= 1001x '- 2041 and b= 1000x- 2010.

Then the equation becomes a2+b2= 2ab. Thus (a- b?= 0. The result follows. 23. Ans: 100. Let x= 2010 and y= 10. The numerator becomes [(x+ y)2- xy] (�i ·

-

ii )

·

[(x - y)2+ xy]

= (� + xy + i ) iCx- y)(x+ y) (� - xy + i ) = Y2(� -l)(x3 +l) = y2(x6 -l). ·

·

Hence the answer is 100. 24. Ans: 2010. Let 15+x= m 2 and x-74= n2• We havem 2- n2= 89= 1x89. (m-n)(m+n)= 1x89. Let m-n= 1 and m+n= 89. Solving m = 45 and n= 44. Thus the number x is 452-15= 2010. 25. Ans: 177. From 0.9y < x < 0.91y, we get 0.9y + y < x + y < 0.91y + y. Thus 0.9y + y < 59 and 0.91y + y> 56. It follows that y < 31.05 and y> 29.3. Thus y= 30 or 31. If y= 30, then 27 < x < 27.3, no integer value of x. If y= 31, then 27.9 < x < 28.21, thus x= 28. Thus y2 - x2= (31 + 28)(31- 28)= 177. 26. Ans: 3350. Take the reflection with respect to A' B'. Let A" be the image of A. Then the minimal length is equal to the length of A" B = 20102 + (2000+ 680)2= 3350.

.V

27. Ans: 672. M

= l!x2!x3!x4!x5!x6!x7!x8!x9! = 28 X 37 X 46 X 55 X 64 X 73 X 82 X 9 = 230 X 313 X 55 X 73

A perfect square factor of M must be of the form 22xx32Yx52zx72w, where x, y, z and ware whole numbers such that 2x � 30, 2y � 13, 2z � 5, 2w � 3. Hence, the number of perfect square factors of M is 16x7x3x2= 672. 13

28. Ans: 144. There are total 12 ways starting from any of the L's to reach the middle V. Hence the total number of ways to spell the word LEVEL is 122 = 144. 29. Ans: 25. Let N be the midpoint of CD. Then LPNQ = 90°. So PQ = 5 Vi.

B

A

D

N

c

Then the area of the shaded region is

30. Ans: 11. Clearly 11 is a factor and 2, 3, 5 are not. We only need to rule out 7. 102011+1 = 4 (mod 7) because 103 = -1 (mod 7). 31. Ans: 0. Let k be a positive even numb�r.

14

k Define j(x) = Pk(x) - Pk(x- 1). Then j(n) = n for all integer n � 2. Note thatj is a . polynomial. We must have j(x) = xk. In particular, for integers n � 2, Pk(-n+ 1)- Pk(-n)

=

Pk(O)- Pk(-1)

=

j(-n+ 1)

=

(n- 1)\

Pk(-n+ 2)- Pk(-n+ 1) = j(-n+ 2) = (n- 2i,

Pk(1)- Pk(O)

k O , k j(1) = 1 . j(O)

=

=

k k k Summing these equalities, Pk(l)- Pk(-n) = 1 + O + 1 +

Define g(x) = Pk(-x) + Pk(x - 1). Then g(n) polynomial, g(x) = 0. In particular, Pk(

-�)+ Pk( -�)

32. Ans: 23. Join BD and CG and note that

=

0, i.e., Pk( -�)

��

=

A

=

=

·

·

·

+ (n- 1)k. That is,

0 for all integers n � 2. Since g is a

0.

2.

....-----.

B E

F

Assume the length of AB is 1. Let the area of ABGE and AFGC be x andy respectively.

Then the areas of t1EGC and ADGF are 2x arid 2y. Since the area of ABFC is

�

· we have

1 1 2 1 . . 3x+ y= 6. Smnlarly, 3y+ 2x=the area of AVEC= 3. Solve x = andy= . Thus 42 21 Area of ABGD Area of ABCD

=1

So m= 9 and n= 14. The result follows.

15

_

1 3( x + y)=l _ 5 42

=

�· 14

33. Ans: 42. Since a � 1,20 10 =a2+b2+8ab � 1+b2+8b. b2+8b-2009 � 0. However b2+8b-2009=0 has an integer solution 41. So a= 1 and b=41. The result follows. 34. Ans: 3 1680. Let X and Y be the sum of the digits at even and odd positions respectively. Note that 1 + 2 + 3 + + 9=45. We have X+ Y=45 and 11 divides IX- Yl. It's easy to see X= 17 and Y = 28; or X = 28 and Y = 17. Hence we split the digits into 2 sets whose sum is 17 and 28 respectively. ·

·

·

There are 9 ways for 4 digits to sum to 17: {9,5,2, 1}, {9,4,3, 1}, {8,6 ,2, 1}, {8,5,3, 1}, {8,4,3,2}, {7,6,3, 1}, {7,5,4, 1}, {7,5,3,2}, {6,5,4,2}. There are 2 ways for 4 digits to sum to 28: {9,8,7,4}, {9,8,6,5}. Thus the total number of ways is 11 x 4! X 5! =3 1680. 35. Ans. 60. Let the three sides of the triangle be a,b,c respectively. Then

where s=

�

a+ + c

�s(s - a)(s - b)(s - c)=a+ b + c=2s, . Note that s is an integer; otherwise s(s-a)(s-b)(s-a) is a non-integer.

Let x=s - a, y= s - b and z=s - c. Then x,y,z are positive integers satisfying xyz=4(x + y + z). Assume that x � y � z. Then xyz � 12x, i.e. , yz � 12, and thus z � 3. If z = 1, xy = 4(x + y + 1) implies (x - 4)(y - 4) = 20 = 20 (x,y)=(24,5),( 14,6),(9,8).

·

1

= 10 2 = 5 4. So ·

·

If z=2, 2xy=4(x+ y+ 2) implies (x- 2)(y- 2)=8=8· 1=4 2. So (x,y) =( 10 ,3 ),(6,4). ·

If z=3, 3xy=4(x + y + 3) implies (3x - 4)(3y - 4)=52, which has no solution x � y � 3. The area is 60, 42, 36, 30 , 24, respectively; and the largest possible value is 60.

16


2010 -'

(Junior Section, Round 2) Saturday, 25 June 2010

0930-1230

INSTRUCTIONS TO CONTESTANTS

1.

2. 3.

Answer ALL 5 questions. Show all the steps in your working. Each question carries

10

mark.

4. No calculators are allowed.

ABC D intersect at S and let P be the midpoint of AB. Let M be the intersection of AC and PD and N the intersection of BD and PC. A circle is incribed in the quadrilateral PMSN. Prove that the radius of the circle is MP- MS.

1. Let the diagonals of the square

2. F ind the sum of all the 5-digit integers which are not multiples of 1 1 and whose

digits are 1, 3,

4,

7, 9.

3. Let a1, a2, . . . , an be positive integers, not necessarily distinct but with at least five

si
distinct values. Suppose that for any 1 from

i

and

j such that ai + aj

�

ak +

a.e.

n,

there exist k, £, both different

What is the smallest possible value of

n? 4. A student divides an integer that

m

by a positive integer m ____: n

=

0

·

167ala2

·

·

·

n,

where

n

s 100, and claims

.

Show the student must be wrong. 5. The numbers

f, �, . .. , 20\ 0

are written on a blackboard.

A student chooses any

two of the numbers, say x, y, erases them and then writes down x

+y+

xy.

He

continues to do this until only one number is left on the blackboard. What is this

number?

17

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Junior Section, Round

2

2010

solutions)

Let 0 be the centre and r the radius of the circle. Let X, Y be its points of contact with the sides PM , MS, respectively. 1.

Since OY j_ MS and LYSO= LASP=45°, SY=YO=r. Also LOPX = LPDA ( since O P II DA) and LOXP= LPAD= 90°. Therefore 60X P c:::: 6 PAD. Hence OX/XP = PA/AD= 1/2. Hence PX = 2r. Therefore PM - MS = 2r + MX MY-r=r. c

D

p

A

B

First note that an integer is divisible by 11 if and only if the alternating sum of the digits is divisible by 11. In our case, these are the integers where 1,4 and 7 are at the odd positions. Let S be the sum of all the 5-digit integers formed by 1, 3, 4, 7, 9 and let T be the sum of those which are multiples of 11. Then 2.

s=4!(1 +

3 + 4 + 7 + 9)(1 + 10 + 100 + 1000 + 10000)

=6399936

T=2!2!(1 + 4 + 7)(1 + 100 + 10000) + 3!(3 + 9)(10 + 1000)=557568. Thus the sum is 6399936-557568=5842368. are the two smallest values. Then a1 = x and let s be the smallest index such that a8 = y. Now there are two other terms whose sum is x + y. Thus we have a2 = x and as+I = y. Since a1 + a2 = 2x, we must have a3 = a4 = x. Similarly, by considering the largest two values w < z, we have an =an-I =an-2 = an-3 = z and another two terms equal tow. Since there is one other value, there are at least 4 + 2 + 4 + 2 + 1=13 terms. The following 13 numbers 3.

a1 :::; a2 :::;

·

·

·

:::; an. Suppose

x

< y

18

satisfy the required property: 1, 1, 1, 1, 2, value of n is 13.

4. We have

m < 0 167 < 0 168 n ·

·

2, 3, 4, 4, 5, 5, 5, 5.

Thus the smallest possible

167n::; 1000m < 168n.

=>

Multiply by 6, we get

1002n::; 6000m < 1008n

6000m- 1000n < 8n::; 800.

=>

But 6000m -lOOOn 2': 2n > 0. Thus 6000m - 1000n 2': 1000 since it is a multiple of

1000. We thus get a contradiction.

We shall prove by induction that if the original numbers are

5.

the last number is

(1 + a1)

·

·

·

n 2': 2, then

(1 + an) - 1.

The assertion is certainly true for n

=

2, the base case. Now suppose it a1, ..., ak+l written on the board.

is true for

k + 1 numbers After one k numbers. Without loss of generality, we can assume that the student erases ak and ak+l and writes bk ak+ak+l +akak+l (1+ak)(1+ak+l)-l. After a further k operations, we are left with the number n

=

k

a1, ..., an,

2': 2.

Consider

operation, we are left with

=

=

This completes the proof of the inductive step. Thus the last number is

( 1 + �) ( 1 + �) ... ( 1 + 20 10 ) - 1 1

19

=

20 10

Singapore Mathematical Society Singapore Mathematical Oly mpiad (SMO) 2010 (Senior Section)

Tuesday, 1 June 2010

0930 - 1200 hrs

Important: Answer ALL 35 questions. Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer. For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. No steps are needed to justify your answers. Each question carries 1 mark. No calculators are allowed.


20


1.

. F mdthevalue of (A)

(B) (C)

2.

1

(1x2x3)+ (2 x4x6) + (3x6x9)+···+ (335 x670 x1005) (1x3x6)+ (2x 6xl2)+ (3x9x18) +··· + (335 x1005 x20 10)

-

3 2 3 1 -

6

(D)

1 2

(E)

-

4 9

If a, b, c anddare real

numbers suchthat

b+c+d= a+c+d= a +b +d= a +b+c = r, a d b c findthevalue of r. (A)

(B) (C) (D) (E)

3.

If

3 1 -1 3 or 1 3 or-1

1 a a nd tan x+- - =--,where a, b a nd c 2 4 tan x b-;rc are positiv e integers, find thevalue of a+ b +c. O< x<

(A)

(B) (C) . (D) (E)

;r

a nd sin x-cos x=

7r

8

32 34 48 50

21

(A) (B) (C) (D) (E)

5.

104 224 312

336 676

In the figure below, ABC is an isosceles triangle inscribed in a circle with centre 0 and diameter AD, with AB = AC. AD intersects BC atE , and F is the midpoint of OE. Given that BD is parallel to.FC and BC = in em. (A)

(B) (C)

(D) (E)

6.

D A

Find the number of ordered pairs (x, y), where x is an integer and y is a perfect square, such that y = (x-90)2 -4907. (A) (B) (C) (D) (E)

7.

3/5 2 J6 2J3 J7 2./6

2J5 em ,find the length of CD

0

1 2 3

4

LetS= {1, 2, 3, . , 9,10 }. A non-empty subset ofSis considered "Good" if the number of even integers in the subset is more than or equal to the number of odd integers in the same subset. For example, the subsets {4, 8}, {3, 4, 7, 8} and {1, 3, 6, 8,10} are "Good". How many subsets ofSare "Good"? . .

(A) (B) (C) (D) (E)

482 507 575

637 667

22

8.

If the graph of a quadratic function f (x) = ax

2 + bx + c (a i- 0) passes through

two distinct points (r, k) and (s, k), what is j(r + s)? (A)

(B) (C)

(D) (E)

9.

None of the above

Find the number of positive integers k < 100 such that 2(36n ) + k (23n +l) -1 is divisible by 7 for any positive integer n. (A)

10

(B)

12

(C)

13 14 16

(D) (E) 10.

2k c k-c 2k - c

Let

ABCD be

a trapezium with AD parallel to BC and LADC = 90° , as shown in

the figure below. Given that M is the midpoint of AB with CM = BC + CD+ DA

(A)

(B) (C) (D) (E)

=

13

2

em

17 em, fmd the area of the trapezium ABCD in cd.

26 28 30

A

,------,.-,

33 35

B

23

D

c

and

Short Questions

11.

12.

The area of a rectangle remains unchanged when either its length is increased by 6 units and width decreased by 2 units, or its length decreased by 12 units and its width increased by 6 units. If the perimeter of the original rectangle is x units, find the value of x. For r = 1, 2, 3, ..., let

2

1 -+

u,

=1 + 2+3 + ... + r. Find the value of 3 100 + ··· + + . + + + ··· +

(:J (: + J (: : :,) (: : .l

,

13.

If

14.

If

15.

2010!=Mx10k,

,

+

,

,

,

.�.J

.

where Mis an integer not divisible by 10, find the value of k.

1 1 a > b > 1 and -- + -- = -J1229 , find the value of log b 1ogb a a

rl

1 ·

logab b

1

logab a

For any real number X, let X denote the smallest integer that is greater than or

LxJ denote the largest integer that is less than or equal to x (for J1.23l = 2 and LL23J =1). Find the value of

equal to x and example,

I: r 2010 _ 2o10 o

k=ll

x2o1o

l Jl · k

k

.

Fmd the value of + (1_x)2010•

16.

Let f ( x)

17.

If a, b and c are positive real numbers such that

=

x2o1o

ab + a + b = be+b +c ca +c +a = 35, find the value of (a+ 1)(b + 1)(c + 1). =

24

18.

In the figure below, AB and CD are parallel chords of a circle with centre 0 and

radius rem. It is given that AB = 46 em, CD= 18 em and LAOB= 3 x LCOD. Find the value of r. A

B

19.

Find the number of ways that 2010 can be written as a sum of one or more positive integers in non.,.decreasing order such that the difference between the last term and the first term is at most 1.

20.

Find the largest possible value of n such that there exist n consecutive positive integers whose sum is equal to 2010. ·

21.

Determine the number of pairs of positive integers n and m such that 2 1! + 2! + 3! + . + n! = m • .

22.

·

.

The figure below shows a circle with diameter AB. C and D are points on the circle on the same side of AB such that BD bisects LCBA. The chords AC and BD intersect at E. It is given that AE = 169 em and EC = 119 em. If ED = x em, find the value of x.

25

23.

Find the number of ordered pairs (m, n) of positive integers in and n such that m + n = 190 and m and n are relatively prime.

24.

Find the least possible value of j(x) =

25.

Find the number of ways of arranging 13 identical blue balls and 5 identical red balls on a straight line such that between any 2 red balls there is at least 1 blue ball.

26.

LetS= {1, 2, 3, 4, . . . , 100000}. Find the least possible value of k such that any subset A ofSwith IAI = 2010 contains two distinct numbers a and b with Ia- bl::; k.

27.

Find the number of ways of traveling from A to B , as shown in the figure below, if you are only allowed to walk east or north along the grid, and avoiding all the 4 points marked x.

9

1+cos2x over all real numbers for which j(x) is defined.

+

25

1-cos2x

, where x ranges

North

L

East

A

28.

Two circles C1 and C2 of radii 10 em and 8 em respectively are tangent to each other internally at a point A. AD is the diameter of C1 and P and Mare points on C1 and Cz respectively such that PMis tangent to Cz, as shown in the figure below. If PM=

.fiO em and

LPAD

=

xo, find the value of x.

26

29.

30.

31.

Let a, b and cbe integers with a > b > c > 0. If b and c are relatively prime, b + c is a multiple of a, and a+ cis a multiple of b, determine the value of abc. Find the number of subsets {a, b, c} of {1, 2, 3, 4, ..., 20} such that a< b - 1 < c- 3. Let f(n) denote the number ofO's in the decimal representation ofthe positive

integer n. For example-, /(10001 123) = 3 and /(1234567) = 0. Let M = /(1)x2f(I)

+

j(2)x21<2>

+

/(3)x21<3>

+ . . · + f(99999)x21<9999� .

Find the value of M- 100000.

32.

33.

Determine the odd prime number p such that the sum of digits of the number p4 -5 p 2 + 13 is the smallest possible . The figure below shows a trapezium ABCD in which AD II BC and BC = 3AD. F is the midpoint of AB and E lies on BC extended so that BC = 3CE. The line segments EF and CD meet at the point G. It is given that the area of triangle GCE is 15 cm2 and the area of trapezium ABCD is k cm2• Find the value of k.

c

B

34.

Let P (x)= a0 + a1x + a x2 +

E

+

anxn be a polynomial in x where the coefficients a0, a1, a ,•••, an are non-negative integers . If P (1) 25 and P (27) 1771769 , 2 2

· · ·

find the value of ao + 2ar + 3� + ... + (n + 1)an.

27

=

=

35.

Let three circles rl' r2' r3 with centres At, Az, A3 and radii Tt, rz, r3 respectively be mutually tangent to each other externally. Suppose that the tangent to the circumcircle of the triangle A1AzA3 at A3 and the two external common tangents of r1 and r2 meet at a common point P, as shown in the figure below. Given that r1 = 18 em, r2 = 8 e m and r3 = k em," find the value of k.

p

28

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Senior Section Solutions)

1.

Answer: (A)

(1x2x3) + (2x 4x6) + (3x6x 9) +···+ (335 x 670x 1005) (1x3x 6) + (2x 6x12) + (3x 9x 18) + ···+ (335 x1005 x2010) =

=

2.

(1x2x3)[13 +23 +33 +···+3353] (1x3x6)[13 +23 +33 +···+3353]

�--��--��------�

1x2x3 1 = 1x3x6 3

Answer: (E) From the given equations, we obtain

a + b +c +d = a(r +1), a + b +c +d = b (r +1), a +b +c +d =c(r +1), a +b +c +d =d (r +1). Adding these four equations gives

4(a + b +c +d) = (a + b +c +d)(r +1), that is,

(3-r)(a +b +c +d) =0.

Thus r = 3, or a +b +c +d =0. If a +b +c +d =0, then we see from the original given equations that r = -1 Hence the value of r is either 3 or -1

.

.

3.

Answer: (E)

2

1[ . xcos x= . x-cosx)2 =-, We have (sm which .rmp1"1es that . sm 16

Therefore we obtain

tanx+

1 --

=

tan x

sin x --

cos x

cos x

1

sin x

sin xcos x

+--=

=

32 16-:r2

Hence a+ b+ c = 32+ 16+ 2 = 50 .

4.

Answer: (C) First, we note that 4n3

=

[n(n+ 1)]2 - [n(n- 1)]2 • Thus

29

.

16 -1[2 ----

32

Therefore 143+ 153 + . . . + 243 + 253 142 x152 132 x142 + = 4 4 152 x162 142 x152 + . . .. + 4 4 242 X 252 232 X 242 + 4 4 252 X 262 242 X 2s2 ---

��-

=

4 252 x26 2

4 132 x142

4 = (32 X 13)(18 Thus

5.

= (25 x 13+ 13 x7)(25 x 13-13 x7) 4 X 13) = 9 X 64 X 13 2 •

�143 +153 +163 +...+243 +253

= 3x8x 13 = .312.

Answer: (B) Since the diameter AD perpendicularly bisects the chord BC,

BE = EC =

.JS.

Also, given that BD II FC, we have LDBE = LFCE. Thus MDE is congruent to !1CFE, so DE = FE. As F is the midpoint of OE, we have OF = FE = ED. Let OF = x. Then AE = 5x. Using Intersection Chord Theorem, we have

AExED = BExEC, which leads to 5x2 = 5. Consequently we obtain x = 1. Now CD2 = C£2+ ED2 gives CD =

6.

.J5+1= ../6 .

Answer: (E) Let y = m2 and (x-90)2 = k?-, where m and k are positive integers. Then we obtain k 2 -m2 =4907 =7x701=1x4907, which gives (k -m)(k+m) =7x701 or (k-m)(k+m) =1x4907. It follows that k-m = 7 and k+ m = 701, or k-m = 1 and k+ m = 4907. Solving these two pairs of equations gives (k, m) = (354, 347) and (k, m) = (2454, 2453).

30

Therefore the ordered pairs (x, y) that satisfy the given equation are: (444, 3472 ), (-264, 3472 ), (2544, 24532 ), (-2364, 24532 ). Hence the answer is 4.

7.

Answer: (D) Let the number of even integers in a "Good" subset of S be i, where i = 1, 2, 3, 4, 5, and the number of odd integers in that subset be j, where j 0, 1, 2, ... , i. Then the number of "Good" subsets of S is =

t<(:J�(�) (:}(�) (:) +(�}(�) (:) (�) (�)<(�) (:) G) =

+

+

+

+

+

+. . . +

+. . .

=5(1 + 5) + 10(1 + 5 + 10) + ...+ (1 + 5 + 10 + 10 + 5 + 1) =30 + 160 + 260 + 155 + 32 =637.

8.

Answer: (B) Let g(x) = f (x) -k. Then g(r)

f (r)- k = k-k = 0. Similarly, g(s) = 0. Therefore r ands are roots of the quadratic equation g(x) =ax2 +hx+c-k =0, from which =

b

we deduce that r + s = - -. Hence a

b a

b a

b a

j(r + s ) = j(--) = a (-- )2 +b(--) +c = c.

9.

Answer: (D) We have 2(36n) +k (23n+l) -1= 2(272n) +2k(8n) -1 = 2( -1)2n +2k(ln) -1 (mod 7) = 2k +1 (mod 7). l Thus, for any positive integer n, 2(36n) + k(23n+ ) -1 is divisible by 7 if and only

if 2k +1 = 0 (mod 7 ). As k< 100, it is clear that the congruence holds for k =3, 10, 17, . .., 94. Thus the required number of positive integers k is 14. 10.

Answer: (C)

31

'

i

Extend DA and CM to meet atE as shown in the figure above. Since AM = MB, LAEM = LBCM and LAME = LBMC, we conclude that �M is congruent to tillCM. Therefore AE = BC and CM=EM. Thus CE = 2CM= 13.

Let the area of trapezium ABCD be S cm2 . Then S =_!_ (DE)(DC), and we have

2 (DE + DC)2 =DE2 + DC2 + 2(DE)(DC) =C£_2 + 4S =132 + 4S =169 + 4S. Now DE+DC=DA+AE+DC=DA+BC+DC=17. Hence 172 =169 + 4S, and it follows that S = 30. 11.

Answer: 132 Let the length and width of the original rectangle beL and W respectively. Then

LW= (L + 6)(W -2)

and

LW = (L-12)(W + 6).

Simplifying the above equations, we obtain

L-2W=12

and

3W-L=6 .

Solving the simultaneous equations, we get L=48 and W=18. Hence the perimeter of the rectangle is 132 units.

12.

Answer: As ur

2575

r r + 1) , we have =1 + 2 + 3 + ... + r = ( 2 2 =t �-� =2-__2_=�. =t t_!_ +1 +1 r�l r ( r + 1) r�l r r + 1 r�l ur

(

) i i i:-/ =i:( : J=i:(i +12 )=_!_ ( (n2 1) + n)=�(4 n + 3).

Hence, Sn : =

i i�l i+l

i�l "" __L i�l L...J u, r�l In particular, slOO =2575.

13.

2

n

+

Answer: 501 The number k is the number of the factor 10 that occurs in 2010!. This number is given by the number of pairs of prime factors 2 and 5 in 2010!. Now between 1 and 2010, there are: • • • •

402 integers with 5 as a factor; 80 integers with 25 as a factor; 16 integers with 125 as a factor; 3 integers with 625 as a factor.

32

Therefore the total number of prime factor 5 in 2010! is 402 + 80 + 16 + 3 =501. As there are clearly more than 501 prime factor 2 in 2010! , we obtain k =501.

14.

Answer: 35 First note that since a> b > 1,

1

--

logabb

-

1 logab a

1

1

logabb

logab a

·

>

0. Then

= Iogb ab -loga ab = (Iogb a+ 1) (loga b +1) = logb a-loga b

1

1

loga b

Iogb a

·

=-----

[�-�)' = [�+ : J { : X : J =

-

lo , a

lo . b

lo , a

= ..)1229-4

= .J1225

= 35.

15.

Answer: 1994

l� J

If

= 2010 - 2 10 =0, so Xl= .If k 12010, then X: r O k 2 10 - 2 10 y k ' 2010, then 0 < y:= < 1 so r l =1.

Consider k

=

1, 2, ..., 2010.

� l� J

'

Since the prime factorization of 2010 is 2 x 3 x 5 x 67, we see that 2010 has 16 distinct divisors . Hence

o l2010

Ik=l I 16.

k

l Jl = klL20!0JXl+kl2:20!01 l y

- 2010 k

=number of non-divisor of 2010 among k =2010 -16 =1994.

Answer: 1005 _

Observe that j(x) + j(l -x)-

It follows that

(l- xyolO x2o1o -1 . + x2oJo+ (1-x )2010 (1-x )2010+x20JO _

33

=

17.

1005.

Answer: 216 Adding 1 to both sides of the given equation ab + a + b

=

35, we obtain

(a+ 1)(b+ 1) = 36. Likewise, adding 1 to the other two given equations gives (b+ 1)(c+ 1) = 36 and (c+ 1)(a+ 1) = 36. Now multiplying the three resulting equations above leads to 3 [(a+1)(b+1)(c+1)] 2 = 36 = 66• 3 It follows that (a+1)(b+1)(c+1)=6 =216.

18.

Answer:

27

A

B

Let M and N be the midpoints of AB and CD respectively and let L CON = x. Then LAON=3x and 3 23 AM r s 3x 3sin x � 4sin x = =3_4sin 2 x . = = sm x r sm x 9 CN 23 1 1 1 . Thus sm 2 x = "4 3 g = g , and so sm x= "3 . Hence r=

--

(- )

�

.

CN =27. sin x

34

19.

Answer: 2010 Consider any integer k where 1 :::; k:::; 2010. By division algorithm, there exists unique pair of integers (q, r) such that 2010 = kq+ r with 0:::; r:::; k- 1. We rewrite this as 2010 = (k- r)q+ r(q+ 1). That is, k- r copies of q and r copies of q+ 1 add up to 2010. Thus there is one desired expression for each value of k, which is clearly unique . Hence there are 2010 such expressions in all .

20.

Answer: 60 Let a be a positive integer such that the sum of n consecutive integers a, a+ 1, ... , a+ (n- 1) is 2010, that is,

a+(a+1)+···+(a+n -1) 2010. =

n(2a+n -1) . =2010, or Th.I S gives

2 n(2a+n -1) =4020=22 x3x 5x67.

(1)

Since n < 2a+ n- 1, we have

n<.J22x3xSx67 =2.JlOOS <2x 32=64.

(2)

Now (1) and (2) imply that n E{1, 2,3,4,S,6,10,12,1S,20,30,60}. Sincenand 2a+ n- 1 have different parities, it follows that nand we have If n

21.

=

4020 n

have differ�nt parities. Consequently,

n E{1,3,4,S,12,1S,20,60}.

60, then 2a+n -1 =

4020 =67, so a 4. Thus the largest possible value of n is 60. 60 =

Answer: 2 First note that if n 2: 4, then

1!+2!+3!+···+n! =1!+2!+3!+4! (modS) =1+2+1+4=3 (modS).

Since the square of any integer is congruent to either 1 or 4 modulo S, it follows that 1 ! + 2! + 3! + + n! i= m2 for any integer m in this case. So we consider n< 4. Now we have · · ·

1!

=

12,

1! + 2!

=

3, 1! + 2! + 3!

=

32•

Hence we conclude that there are two pairs of positive integers (n, m ) , namely, (1, 1) and (3, 3 ), such that 1! + 2! + 3! + + n! = m2• · · ·

35

22.

Answer: 65 EC 119 = = . Thus we can let BC = 119y BA EA 169 and BA = 169y for some real number y. Since LBCA = 90°, we have Since BE bisects LCBA, we have

BC

AB2 = AC2+ B C2

(169yy =(169+119) 2 +(119y) 2 y 2 (169 -119X169+119)= (169+119) 2 y2 =

169+119

169-119 12 y=5

=

144 25

Hence, from triangle BCE, we have BE =

�119 2 + (119y) 2 =119 x 13 . 5

Finally, note that f).A[)E and MCE are similar, so we have AE xCE 169x119 =65 cm. = ED= 119x 153 BE

23.

Answer: 72 First we find the number of ordered pairs (m, n) of positive integers m and n such that m+ n = 190 and m and n are not relatively prime. To this end, write m = ka and n = kb, where k, a and b are positive integers with k > 1. Since m+ n = 190, we see that k is a factor of 190 = 2x5x19 with k t 190. We consider six cases: (i) (ii) (iii) (iv) (v) (vi)

k = 2. Then a+ b = 95, and there are 94 such pairs (a, b) of a and b such that the equation holds. k = 5. Then a + b = 38, and there are 37 such pairs (a, b) of a and b such that the equation holds. k = 19. Then a + b = 10, and there are 9 such pairs (a, b) of a and b such that the equation holds. k = 10. Then a+ b = 19, and there are 18 such pairs (a, b) of a and b such that the equation holds. k = 38. Then a+ b = 5, and there are 4 such pairs (a, b) of a and b such that the equation holds. k = 95. Then a+ b = 2, and there is 1 such pair (a, b) of a and b such that the equation holds.

(

It follows from the above cases that the number of ordered pairs m, n) of positive integers m and n such that m+ n = 190 and m and n are not relatively prime is 94+37+9-18-4-1 = 117.

36

Since the total number of ordered pairs (m, n) such that m + n = 190 is 189, we conclude that the required number of ordered pairs (m, n) where m and n are relatively prime is 189 - 117 = 72.

24.

Answer: 32 For all real values of x for which] (x) is defined, we have

9

j(x) =

l+cos 2x 1-cos 2x

= _!_

2

= 17 + _!_

� t an2

� 17 +

2

=

9

=

� t an2 x+9) + _!_ (25 cot

= 17

+

32.

2

� ( ��

2

X

+ 25 coe

2

X

x+ 25

)

) ))

X

t an2 x 25cot2 x

(AM-GM Inequality)

least possible value of j(x) is 32. fi) = 32. Thus the .

�3

Answer: 2002 First we place the 5 red balls on a straight line and then place 1 blue ball between 2 adjacent red balls. With this arr angement fixed, the condition of the question is satisfied. We are now left with 9 blue balls. We can place the remaining 9 blue balls into the spaces before, after or in between the 5 red balls. The number of ways that this can be done is the answer to the question. Including the two ends, there are 4 + 2 = 6 spaces into which these 9 blue balls can be placed. The number of ways of distributing the 9 blue balls into the 6 spaces is 4

( : J ( � J c :J 9

26.

25

+ 2cos2 x 2sin2 x

_!_(2J9x 25) 2

Note that j(t an-1

25.

25

+

+

-!

�

�

�

2002.

Answer: 49 Consider the following subsets of S: S1

= {1, 2, 3, ..., 49, 50},

{51, 52, 53, ..., 99, 100}, s3 = {1 01, 102, 1 03, ..., 149, 150}, S2 =

Szooo =

{99951, 99952, 99953, ..., 99999, 100000}.

In other words, Si = {50i- 49, 50i- 48, ..., 50i} fori= 1, 2, ..., 2000. Note that Sis partitioned into these subsets St, Sz, S 3 , ... , Szooo.

37

By Pigeonhole Principle, for any subset A of S with where 1:::; i:::; 2000, such that I AnSi � I 2. Let

a,b EAnsi. It

is clear that

IAI = 2010,

there exists i,

Ia- bl:::; 49.

To show that 49 is the least possible value of k, we find a subset A � S with lAI = that Ia- bl2: 49 for any distinct a,b EA. Let

2010 such

A={ 49j+1: j=0,1, 2, ... , 2009}={1,50, 99,1 48, ... , 98442}.

Then A is a subset of S with

27.

Answer:

IAI = 2010

and

Ia- bl2: 49

for any distinct

a,b EA.

11 2

E �--+---+---4----r--� n

Figure 1 A

c

We observe that to avoid the four points marked x, the path must cross either C, D orE as shown in Figure 1 above. Further, the paths that cross C, D orE are exclusive, that is, no path can cross both C and D or D and E, or C and E. There is only 1 way to get from A to C and from A toE. It is easy to see that there are 4 ways to get from A to D. 1

1 1

1

1

c

6

5

..

4 .. 3

2 1

21

• • •

B

15

10 6

Figure 2

3

1

To count the number of ways to get from either C, D orE to B, we note that the number of ways to get to a certain junction is the sum of the numbers of ways to get to the two junctions immediately preceding it from the left or from below (as shown in Figure 2). Therefore there are 21 ways to get from C to B. Similarly, there are 21 ways to get from D to Band 7 ways to get from E to B. Hence the number of ways to get from A to B that • pass through C: number of ways from A to C x number of ways from C to B= 1 X 21 = 21; • pass through D: number of ways from A to Dx number of ways from D to B= 4 X 21 = 84; • pass through E: number of ways from A to E x number of ways from E to B=1x7=7. It follows that the total number of ways from A to B is 21 + 84+7= 1 12. 38

28 .

Answer: 60

Let 0 be the centre of

C2

and let PA intersect

C2 atB .The

homothety centred atA

It mapsB toP. Thus AB = !__ (This 10 AP 10 can also be seen by connecting P to the centre 0' of C1 so that the triangles ABO and APO' are similar.) The power of P with respect to C2 is PM2 = 20. Thus mapping

Cz

to

C1

has similitude ratio

!__

PB·PA = 20,or equivalently (PA - AB)PA =20. Together with AB = !_ ,we AP 10 obtain AB =8 and AP = 10 . Consequently,the triangle ABO is equilateral,and hence LPAD = LBA0 = 60°.

29.

Answer: 6 We shall show that a =3,b =2 and c =1. Note that 2a > b + c.As b + c is a multiple of a ,it follows that a =b + c. Let a + c =kb . Then kb =a + c = b + c + c,so 2c = (k- 1)b . Since c < b,we must have k =2 and therefore b =2c. Since b and c are relatively prime,this implies that c = 1 and b =2 . Thus a =3 . Hence abc = 6.

30 .

Answer: 680 For any 3-element subset {a ,b,c},define a mapping f by f ( {a ,b,c}) = {a ,b - 1,c - 3}. Now observe that {a ,b,c} is a subset of {1, 2, 3, 4, . . . , 20} with a < b - 1< c - 3 if and only if j( {a ,b, c} ) is a 3-element subset of { 1, 2, 3, . . . , 17}. Hence the answer is

C:)

= 680.

39

31.

Answer: 2780 Note that 0 ::;; f(n)::;; 4 for 1 ::::;n::::; 99999. Fork= 0, 1, 2, 3, 4, let ak denote the number of integersn, where 1 ::::;n::::; 99999, such thatj (n) = k. Then 4 4 k k M= kak 2 = kak 2 . k =O k =l.

L:

L:

By considering the number of 2-digit, 3-digit, 4-digit and 5-digit positive integers with exactly one 0 in their decimal representation, we obtain � = 9+ 9 X 9 X 2+ 9 X 9 X 9 X 3+ 9 X 9 X 9 X 9 X 4=28602 .

Similarly, we have

a =9 + 9 X 9 X 2

( �)

+9 X 9 X 9 X

(�)

= 4626,

a3=9+9 2 x4=333, a =9. 4 Hence 3

M =1x28602x2+2x4626x22 +3x333x2 +4x9x24 =102780,

and it follows that M- 100000 = 2780. 32.

Answer: 5 Let n = p4 -5 p 2 +13. When p = 3, we haven= 49 and so the sum of digits is 13. When p= 5, we haven= 513 and the resulting sum of digits is 9. Now let p > 5 be a prime. We have n= p4 -5p 2 +13= ( p-2)( p-1)( p+1)( p+2)+9. Since p- 2, p- 1, p, p + 1 and p + 2 are five consecutive integers, at least one of them is divisible by 5. Since pi= 5, we must have 5 divides one of the numbers p- 2, p- 1, p + 1 and p + 2, so 5 divides the product (p- 2)(p- 1)(p + 1)(p + 2). Observe that at least one of the numbers p + 1 and p + 2 is even. Therefore we see that (p- 2)(p- 1)(p + 1)(p + 2) is divisible by 10. It follows that for any prime p > 5, the number n has at least two digits and the units digit is 9, so the sum of the digits is greater than 9. Consequently the smallest possible sum of digits is 9, and this value is attained when p= 5.

40

33.

Answer: 360 X

ExtendBA andCD to meet at X .LetH be the point onCD such thatFH II Let AD

=

CE

=

a. Then

BC

=

ECG,

3a , andFH

( ) CE

=

1

2

(AD +

BC)

By the similarity of trianglesFHG and we have 2 FH x area of (i) area of FHG = (ii)

Ll CG CE

HG

FH

Ll ECG

=

It follows from (i) and (ii) that the area of triangleFDH

(iii)

XA

XF

AD

XAD LlXAD ( )

60 cm2;

=

3

2

� x 60

2

HG

=

.

90 cm2 .

bey cm2 . By the similarity of triangles XAD and

1

- =- =

FH 2 area of ilXDF

(iv) area of LlXFH

, so that XA =AF and hence =

2x area of AD

FH

=

2

y

=

=

2y cm2;

4y.

It follows from (iii) and (iv) that the area of triangleFDH 2 Since the area of triangleFDH is 90 cm , we get y = 45. Finally, by the similarity of triangles andABC, area of

2a.

2 , so thatHG = 2CG andDH =HC =

-=-=

Now, let area of triangle XFH , we have

=

BC.

LlXBC (BC)2 =

AD

Hence the area of trapeziumABCD

XAD

y =9y . =

8y = 360 cm2•

41

=

4y-2y

=

2y cm2 .

34.

Answer: 75 First we note that every positive integer m can be written uniquely (in base 27) as m=b0 +b1 x27+b2x27 2 +···+b7 x277, where r and b0, bp b2,

• • •

, br (depending on m) are non-negative integers with

bi < 27 for i =

0, 1, 2, ..., r. Since the coefficients of P(x) are non-negative integers and P(1) = 25, we see that ak ::;; 25 < 27 for 0:::; k:::; n. Thus by the above remark, the polynomial P (x) is

uniquely determined by the value P(27) = 1771769. Writing 1771769 in base 27, 3 we obtain 1771769 = 2+11x27+9x27 +3x274• Therefore 3 P (x) =2+11x+9x +3x4• Hence a0 +2a1 +3a +···+(n+1)an =2+2x11+4x9+5x 3=75. 2 35.

Answer: 12

p

N,. Let P be the point of concurrence of the

First we shall show that r3 =

tangent to the circumcircle of the triangle ��As at As and the two external common tangents of f1 and f • Note that the line joining � and � also passes 2 through P. First we have PA; =P� · PA1 , so

( ) PAs

p�

2

=

PA1

PAs =..2. That is, = p� r2 p�

On the other hand, PAs sinLAg� P sin� r1 +r3 = -= =-sin�As P sin� r +r3 p� 2 1+ 3 Thus = ' ' • Solving for r3, we obtain r3 = + 2 ' ,3 r2 =8 , we get r3=12.

--

N,.

3 v;;

42

3. v;;

Substituting r1 =18 ,

',

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (

. Senior Section, Round 2) Saturday, 25 June 2010

0930-1230


1.

2.

3.


10

mark.

4. No calculators are allowed. 1. In the triangle ABC with AC > AB,

D is the foot of the perpendicular from A onto BC and E is the foot of the perpendicular from D onto AC. Let F be the point on the line DE such that EF DC = BD DE Prove that AF is perpendicular ·

·

.

to BF. 2. The numbers

t, �, . . . , 2 0\ 0

are written on a blackboard.

A student chooses any

two of the numbers, say x, y, erases them and then writes down x

+

y

+

xy. He

continues to do this until only one number is left on the blackboard. What is this number? 3. Given

a1 2': 1 and ak+l ;::: ak + 1 for all k 1, 2, , n, show that af + a� + + a � 2': (a1 + a2 + + an) 2 . ·

·

·

·

·

·

4. An infinite sequence of integers, a0 , = for any n 2': 0, where

an, a1

·

=

an+l an - bn,

·

·

a1, a3 , with a0 > 0, has the property that bn is the number having the same sign as .

.

•

but having the digits written in the reverse order. For example if =

1089 and

all n;:::l.

a2

=

-8712, etc. Find the smallest value of

a0

5. Let p be a prime number and let

so that

a0

=

1210,

an i= 0 for

a1, a2 , ... , ak be distinct integers chosen from n 1, 2, ..., p 1. For 1 ::; i ::; k, let r } ) denote the remainder of the integer nai upon n division by p, so 0 ::; r} ) < p. Define n S { n : 1 :S n :S p - 1, rin) < < r )} -

·

=

Show that

S has less than k2_f1

elements.

43

·

·

k

.

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Senior Section, Round

2

2010

solutions)

Since we are supposed to prove LAFB=goo, it means that the 4 points A, B, D, F are concyclic. Note that AC > AB implies that LB > LC. If TD is the tangent to the circumcircle w of the triangle ABD with B and T lying opposite sides of the line AD, then LADT = LB > LC = LADE so that w intersects the interior of DE at F. Therefore F can only be in the interior of DE. Now observe that the triangles ADE and DCE are similar so that AD/AE =DC/DE. By the given condition, this can be written as AD/AE = BD/ EF. This means the triangles ABD and AFE are similar. Thus LABD = LAFE. This shows that A, B, D, F are concyclic. Therefore LAFB=LADE=goo. 1.

A

c

B

2.

See Junior Section Question 5.

We will prove it by induction. First, it is clear that suppose it is true for n terms. Then 3.

ar 2:: af since a1 2:: 1.

n+l n n La� 2:: a�+l +La� � a�+l + (Lak) k=l k=l k=l n n+l (Lak) +a�+1-a�+1-2an+1Lak. k=l k=l 2

2

=

44

Next,

a�+l-a;+l-2an+l.L.:�=lak :?: 0. Since ak+l- ak :?: 1, we have a�+l- a� :?: ak+l + ak. Summing up over k 1, ... , n, and using ar - al :?: 0, we have n a�+l- at:?: an+l+ 2 L ak - a1 k=l To complete the induction, we'll now show that

=

4. If

ao has a single digit, then a1 0. Thus a0 has at least 2 digits. If a0 ab 10a+b, then a 1 9(a-b) which is divisible by 9. it follows that all subsequent terms are divisible by 9. Checking all 2-digit multiples of 9 shows that eventually 9 appears ( Note that ab and ba give rise to the same sequence, but with opposite signs ) : 81 63 27 45 9. =

=

=

=

---+

---+

---+

---+

ao abc, then a1 99(a - c). Thus if suffices to investigate 3-digit multiples of 99, 198, ..., 990. Here we find that 99 will eventually appear: 990 891 693 297---+ -495 ---+ 99. If ao =abed, then a1 999(a-d)+90(b-c). If b, care both 0, then a1 and all subsequent terms are multiples of 999. However, if such numbers appear in the sequence, eventually 999 will appear: 9990 8991 6993 2997 -4995 999. For 1010, we get 909 and for 1011 we get -90. For 1012, we get 1012 -1089 -8712 6534 2178 -6534 and the sequence becomes periodic thereafter. Thus the smallest a0 1012. If

=

=

i. e. ,

---+

---+

---+

=

---+

---+

---+

---+

---+

---+

---+

---+

---+

---+

=

5. Let

rbn) 0 and Tk�1 =

S' Note that

=

=

p.

Set

k {n: 1:::; n:::; p - 1, .2.: lr��- r�n)l p}. i=O =

k

L lr��- r�n) I i=O

=

P

iff

Tbn)

:S;

rin)

:S;

·

·

·

:S;

rk�1.

n E S', l r��)1 - r�n) I r��- dn) n(ai+l- ai) (mod p) and p f (ai+l- ai), the numbers r��)1- r�n), 1 :::; n:::; p- 1, are all distinct. Therefore Thus

lSI

Pl8'1

=

=

IS' I ·

Since for

=

lri�i- ri"ll t I.; lri�i- rl"ll t�j t nEB' i=O i=O nEB' i=O j=l I.;

Therefore

lSI

�

<

�

k2_J1.

45

�

�

(k + l)IS (IS'I + l).


( SMO ) 2010 1)

Singapore Mathematical Olympiad

( Open Section,

Round

0930-1200 hrs

Wednesday, 2 June 2010

Instructions to contestants 1. Answer ALL 25 questions. 2. Write your answers in the answer sheet provided and shade the appropriate bubbles below your answers.

3. No steps are needed to justify your answers. 4. Each question carries 1 mark.

5. No calculators are allowed.


46

8n3- 96n2 +360n- 400 . is an integer . Find 1. Let S be the set of all mtegers n such that 2n - 7 the value of lnl.

L

nES

2. Deter mine the largest value of x for which

�

lx2 - 4x - 396011 lx2 +4x - 3960 11.

3. Given that find the value of L

�

X=

1 1 1 L1 /3j + L2 /3j + L3 /3j

+ ... + L7999

1 /3j,

J, where LYJ denotes the greatest integer less than or equal to 1 0 (For example, L2.1J = 2, L30J = 30, L- 10.5J = - 1 1.)

4. Determine the smallest positive integer C such that 6n.�

:::;

y.

C for all positive· integers n.

.

5. Let CD be a chord of a circle r1 and AB a diameter of r1 per pendicular to CD at N with AN> NB. A circle r2 centr ed at C with radius CN intersects r1 at points P and Q, and the segments PQ and CD intersect at M . Given that the radii of r1 and r2 ar e 6 1 and 60 r espectively, find the length of AM. 6.

50

Deter mine the minimum value of

L Xk, k=l

positive number s x1, . . . , X5o satisfying

where the summation is done over all possible

50

L __!_ = 1. X k=l

k

4

7. Find the sum of all positive integer s p such that the expression (x- p)(x- 13 ) + can be expressed in the for m ( +q) ( +r ) for distinct integers q and r.

x

x

� - :2 - : 2

8. Let Pk = 1 + , wherek is a positive integer . Find the least positive integer n 3 such that the product P P Pn exceeds 20 10. 3 ·

·

·

9. Let B be a point on the circle centred at 0 with diameter AC and let D and E be the circumcentres of the tr iangles OAB and OBC respectively. Given that sin L.BOC = g and AC = 24, find the area of the tr iangle BDE. 10. Let f be a real-valued function with the rule f(x) = x3 +3x +6x+1 defined for all real value of It is given that a and bare two r eal numbers such that f(a ) = 1 and f(b) = 19. Find the value of ( a +b

x.

1 1. If cot a +cot ,6 +cot'/'

2

4

)2.

=

- 5,4 tan

cot a cot ,6 +cot ,6 cot')' +cot')' cot a

=

=

17 and B 1 -5, find the value of tan(a +,6 +')').

a +tan ,6 + tan')'

7

47

12.

The figure below shows a road map connecting two shopping malls A and B in a certain city. Each side of the smallest square in the figure represents a road of distance 1km. Regions C and D represent two large residential estates in the town. Find the number of shortest routes to travel from A to B along the roads shown in the figure. 8 km

.B

c 10 km

D A

1 3.

Let

a1

=

+ an> 2

14.

1, a2

=

2

and for all

·�� for all

n 2m,

n 2 2, an+l

where

m is

=

n-1 2n an + + 1 an-1· n n 1

It is known that

a positive integer. Find the least value of m.

It is known that

V9 - 8sin 50° =a + b sinc0 for exactly one set of positive integers

(a, b,c

+ ), where 0 < c < 90. Find the value of b c.

15. If a is a real root of the equation x5 -x3 + x -2 the least positive integer not exceeding x.

a

=

0, find the value of

La6 J, where Lx J is

16. If a positive integer cannot be written as the difference of two square numbers, then the integer is called a "cute" integer. For example, 1, 2 and 4 are the first three "cute" integers. Find the 2010th "cute" integer.

( Note: A

the square of a positive integer. As an illustration, 16 are the first four square numbers.) square number is

1, 4,

9 and

17. Let f (x ) be a polynomial in x of degree 5. When f (x ) is divided by x - 1, x- 2, x - 3, x -4 and x2 -x -1, f (x ) leaves a remainder of 3, 1, 7, 36 and x- 1 respectively. Find the square of the remainder when f (x ) is divided by x + 1.

48

18. Determine the number of or dered pairs of positive integer s (a, b) satisfying the equation 100( a +b)= ab- 100.

( Note:

As an illustr ation, ( 1, 2) and ( 2, 1) are considered as two distinct ordered pairs. )

19. Let p= ab +ba If a, b and p are all prime, what is the value of p? . 20. Determine the value of the following expression:

l J l

J l

J l

l

J

J

11 X 3 11 11 X 2 11 X 4 11 X 2009 +. . . + + + + 2010 2010 2010 2010 2010 '

where

LYJ

denotes the gr eatest integer less than or equal to y.

21. Numbers 1, 2, . . . , 2010 are placed on the circumference of a cir cle in some order. The numbers i and j, where i =F j and i,j E { 1, 2, . . . , 2010} form a friendly pair if

( i) i and j are not neighbour s to each other, and ( ii) on one or both of the arcs connecting i and j

along the circle, all numbers in between

them are greater than both i and j.

Determine the minimal number of friendly pairs. 2 2. LetS be the set of all non-zer o r eal-valued functions f defined on the set of all r eal numbers such that f(x2 +yf(z)) = xf ( x ) +zf(y) for all r eal number s x, y and z Find the maximum value of£( 1 2345) , where f E S. . 23. All possible 6-digit number s, in each of which the digits occur in non-increasing order from left to r ight ( e. g. , 966541) , are written as a sequence in increasing order ( the first thr ee 6-digit number s in this sequence are 100000, 110000, 111000 and so on) . If the 2010th number in this sequence is denoted by p, find the value of L J, where LxJ denotes the greatest integer less than or equal to x.

:0

24. Find the number of permutations a1 a2a3a4a5a6 of the six integers from 1 to 6 such that for all i from 1 to 5, ai+l does not exceed ai by 1. 25. Let A=

( ) c�� )) ( ) C )) ( ( ) c )) ( ( ( (��) - G��)) ( ) () ( (;) ) 2010 0

o

2

+

O10 o

2010 1

2

+

010 1

20 10 2

2

2

+. . . + Determine the minimum integer s such that sA �

( Note:

For a given positive integer n,

other values of r, define

n r

=

= 0.

49

40 20 2010

·

n! _ r ).' for r = 0, 1, 2, 3,

' r. n

·

·

·

,

n; and for all

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Open Section, Round

1)

2010

1. Answer: 50 27 8n3- 96n2 + 360n- 400 . 4n2- 34n + 61 + _ . Smce Note that 4n2 - 34n + 61 is 2n _7 2n 7 an integer for all integers n, we must have that 27 divisible by 2n- 7. Hence, 2n - 7 -1,1,-3,3,-9,9,-27,27, so that n 3,4,2,5,-1,8,-10,17. Hence the required sum equals 50. =

=

=

2. Answer: 199 By direct computation,

lx2- 4x- 396011 � lx2 + 4x- 396011 -¢:::=:?- (x2- 4x- 39601)2- (x2 + 4x- 39601)2 � 0 -¢:::=:?- 2(x2- 39601)(-8x) � 0 -¢:::=:?- x(x + 199)(x - 199):::; 0, we conclude that the largest possible value of

x is 199.

3. Answer: 1159 Note that X

-

L11/3j + L21/3j + L31/3j + ... + L79991/3j L Lkl/3j + L Lkl/3j + L Lkl/3j + ... +

L:

L:

1+

2+

L:

3 + ... +

L:

L

Lkl/3j

19

(23- 13) + 2 (33- 23) + 3 (43- 33) + ... + 18 (193- 183) + 19 (203- 193) 19 19(8ooo) - L: k3 k=l 19 20 2 19(8000) -

(

X

:. LlOOJ

)

;

115900 =

1159.

50

4. Answer: 65 6n Define f( n) =I for n=1, 2, 3, ···. It is clear that f( l ) =6, f( 2) =18, f(3) =36, f( 4 ) =54 n.

and f(5) =64.8. For all. n ;:::: 6, f( n) =

6n n!

�

6 6 6 6 6 � l X 2 X J X Ll X S X 6 =6 X 3 X 2 X 1.8=64.8

5. Answer: 78 Extend DC meeting r2 at H. Note that DN =NC =CH = 60. Since M is of equal power with respect to r1 and r2 Thus MN ·MH = MC ·MD . That is MN(MC+60) = . MC(MN +60) giving MN =MG. Thus MN =30. A

The power of N with respect to r 1 is DN · NC = 602, and is also equal to N A ·N B = NA·(AB -NA) =NA·(1 2 2 -NA). Thus NA·(1 2 2 -NA) =602. Solving this quadratic equation, we get NA = 7 2 or 50. Since NA > NB, we have NA = 7 2. Consequently AM =vfNA2+MN 2=vf7 22+302=78. 6. Answer: 2500 By Cauchy-Schwarz inequality,

and equality holds if and only if Xk =50 fork=1, . . . , 50. Therefore the required value is 2500. 7. Answer: 26 Let (x- p)(x- 13)+4=(x+q)(x+r ) . Substituting x=-q into the above identity yields (-q - p)(-q - 13) =-4, which becomes (q+p)(q+ 13) =-4. Since p and q are integers, we must have the following cases:

( a) ( b)

q+p=-4, q+13=1;

( d)

q+p=- 2, q+13= 2

q+p=4, q+13=-1;

(c) q+p= 2,q+13=- 2;or

51

() q = -14,p= 8 and hence (x- p)(x- 13)+4 = (x - 14)(x- 17); (b ), we obtain q = -12,p = 8 and hence (x- p)(x- 13)+4 = (x- 9)(x- 12); (c ), we obtain q =-15,p= 17 and hence (x- p)(x- 13) +4= (x- 15)2; which

For case a , we obtain -For case

For case is NOT what we want;

()

For case d , we obtain q = - l l,p = 9 and hence (x- p)(x- 13)+4 = (x- 11)2; which is also NOT what we want. Hence the two possible values of pare 8 and 18, the sum of which is 26.

8. Answer: 8038 First, note that

( ) (1+.!.k ) 2= (k - 1)(k+1)2 k3

! ! ]__ ]__ Pk = 1+ k _ k 2 _ k3= 1 - k Therefore,

1· 32 2· 42 3· 52 (n- 1)(n+1)2 (n+1) 2 = P 2P3··· Pn=� ·�·43··· n3 4n (n+1)2

> 2010 is equivalent to n2-8038n+1 > 0, which is equivalent Next, observe that 4n to n(n- 8038) > -1. Since n is a positive integer, the last inequality is satisfied if and only if n 2:: 8038. Consequently, the least n required is 8038.

9. Answer: 45 Let d = AC = 24. First, it is not difficult to see that LDEB =LACE and LEDB = LCAB, so that the triangles DEE and ABC are similar.

Let M and N be the feet of the perpendiculars from B onto DE and AC respectively. As M is the midpoint of OB, we have BM = �· Also BN = BO sinLBOC = � x � = 2r Therefore DE = AC x �lfr = d x �� = 5 . Thus the area of the triangle BDE is ! x BM x DE=! x � x 58d = s;;. Substituting d = 24, the area of the triangle BDE is

l

45.

52

10. Answer : 4

We note that f(x) = (x + 1)3 +3(x+ 1 ) + 10. Let g(y) = y3 + 3y, which is a str ictly increasing odd function. Hence f(a) = 1 implies g(a + 1) =-9 and f(b ) = 19 implies g(b +1) =9. Thus, a+1=-(b + 1), implying a+b =- 2, so that (a+b)2 =4.

11. Answer : 11 Let x=tan a, y =tan/3 and z=tan'Y. Then xy+yz+zx xyz x+y+z x+y+z xyz

-

-

4 5 17

6

-

-

17 5

-� and xy+yz +zx=�· It follows that

From the above three equations, xyz= tan(a +pa +'Y) =

x +y +z- xyz 17/6- (-5/6) = =11. 1- 2/3 1- (xy+yz+zx)

1 2. Answer: 2 20 23

Include the points E, F, G, H, I, J, K, L, M, Nand Pin the diagr am as shown and consider all possible routes: E

8km F

G H

10 km

B

c J

I

L

K

M

D A

For the route A

N

p

---+ E ---+ B,

there is 1 way.

For the route A ---+

F ---+

B, there are

For the route A ---+

G---+

B, there ar e

C2) · (D c ) (�) ° 2

53

·

=80 ways. =1260 ways.

For the route

A� H �I� B,

For the route

A� J �I� B,

For the route

A� J

�

K � B,

A� L �I� B,

For the route

A� L � K � B, A� L � M

�

=

=

·

=

=

there are

·

=

·

For the route

A� P � B,

there are

=

=

ways.

ways.

ways.

11 ways.

22023 ways.

Hence, by adding up, there are altogether

4021 an+l - an=

Rearranging the recurrence relation yields

n � 3, we have an- an-1 =

ways.

·

B, there are

there are

ways.

1176 ways.

there are

A� N � B,

for

=

there are

For the route

1 3. Answer:

ways.

there are

For the route

For the route

(�) · (�) 980 (�) · (D · (�) 5880 (�) (D (!) · (�) · (�) 7350 (!) (D (D 4410 (!) (�) 490 (�) · c21) 385 ell)

there are

-ll

n (an- an-1) for n � 2. Thus, n+ .

. n- 2 n- 2 n- 3 (an-1- an-2) = an-2- an-3 ) ( · n n n- 1. n- 2 n- 3 2 1 a1) ··· = n · n -1 ··· 4 · 3( a22 2 2 -n(n- 1 ) n -1 n --

--

--

--

3. Using the method of difference, we obtain an= 3 - � for n � 3. Given that n 2009 , w have an > 2 + 2010 e 2 20091 3- > 2 + 32010 -:;;, 2010' yielding n > 4020. Hence the least value of m is 4021. for

n

�

14. Answer:

14

We have

8 + 8 sin 10° - 8 sin 10° -.8 sin 50° 9 + 8 sin 10° - 8 (2 sin 30° cos 20°) 9 + 8 sin 10° - 8(1 - 2 sin2 10° )

9 - 8 sin 50°

=

.·. v9 - 8 sin 506 Thus,

a = 1, b=

4 and

c=

16 sin2 106 + 8 sin 10° (1 + 4 sin 10° )2 1 + 4 sin 10°.

10, and hence

b+c

.

54

a

-

=

14.

+1

15. Answer: 3 It can be easily seen that the given equation has exactly one real r oot a, since (1) all polynomial equations of degree 5 has at least one r eal r oot, and (2) the function f(x) = x5- x3 + x- 2 is strictly increasing since f' (x) = 5x4- 3x2 + 1 > 0 for all r eal values of x. 1 It can also be checked that f(�) <0 and f(2) > 0, so that 2
a6 <4 since

a6 <4

-¢::::::} -¢::::::} -¢::::::} -¢::::::}

In addition, we claim that a6

a6

:::=::

:::=::

3

a4- a2 + 2a<4 a5 - a3 + 2a2 <4a 2a2 - 5a + 2 <0 1 2
3 since -¢::::::} -¢::::::} -¢::::::}

a4- a2 + 2a :::=:: 3 a5 - a3 + 2a2 - 3a 2a2 - 4a + 2 :::=:: 0,

:::=:: 0

the last inequality is always tr ue. Hence 3 :::; a6 <4, thereby showing that L a6J

=

3.

16. Answer: 8030

Any odd number greater than 1 can be written as 2k+1, where 2k+1 = (k+1)2-k2 . Hence all odd integers greater than 1 are not "cute" integers. Also, since 4m = (m+1)2-(m-1?, so that all integers of the for m 4m, where m > 1, ar e not "cute". We claim that all integer s of the form 4m + 2 ar e "cute". Suppose 4m + 2 ( for m:::=:: 1) is not "cute", then

for some integers positive integers x and y. However, x + y and x - y have the same parity, so that x - y and x + y must all be even since 4m + 2 is even. Hence 4m + 2 is divisible by 4, which is absurd. Thus, 4m + 2 is "cute". The first few "cute" integers are 1, 2, 4, 6, 10 For n > 3, the nth "cute" integer is 4n- 10. Thus, the 2010th "cute" integer is 4(2010)- 10 = 8030. ·

·

· .

17. Answer: 841

We have f(1) = 3, f(2) = 1, f(3) = 7, f( 4 )

= 36

and

f(x) = (x2 - x- 1)g(x) + (x- 1),

where g(x) is a polynomial in x of degree 3. Hence g(1) g(4) = 3. Thus

f(x)

Thus,

=

f(-1)

( X2 _ X _

[

-29,

so

-3, g(2)

=

0, g(3)

=

1, and

1). (-3) . ( x- 2)(x- 3)(x- 4) + (1). (x- 1)(x- 2)(x- 4)

]

(-1)(-2)(-3) (x- 1)(x- 2)(x- 3) · . + (3). . + (X _ 1) (3)(2)(1)

=

=

that its square is 841.

55

(2)(1)(-1)

•

18. Answer: 18

Solving for b, we get b = 10��t0�00. Since a and b are positive, we must have a > 100. Let a = 100 + m, where m is a positive integer. Thus b = lOO(lOO m)+lOO = 100 + 10!00. Therefore m must be a factor of 10100 = 101 x 2 2 x 5 2• Conversely, each factor r of 10�00) of the equation 100(a + m determines a unique solution (a, b) = (100 + r, 100 + b) = ab -100. There are 18 = (1 + 1)(2 + 1)(2 + 1) factors of 10100. Consequently there are 18 solutions of the given equation. In fact, these 18 solutions can be found to be (a, b) = (101, 10200), (102, 5150), (104, 2625), (105, 2120), (110, 1110), (120, 605), (125, 504), (150, 302), (200, 201), (201, 200), . (302, 150), (504, 125), (605, 120), (1110, 110), (2120, 105), (2625, 104), (5150, 102), (10200, 101).

�

19. Answer: 17 If both a, b are odd prime numbers, then p is even and p that p is prime. Thus a =2 or b=2.

�

4, contradicting the condition

Assume that a =2. Then b i= 2; otherwise, p =8 which is not prime. Thus b is an odd prime number. Let b

We shall show that k=1.

=

2k + 1, where k is an integer greater than 1. Thus

(

Suppose that k � 2. If k = 1 mod 3), then b > 3 and b = 2k + 1

=

(

O mod 3),

contradicting the condition that b is prime. If k ¢. 1(mo d 3), then

a contradiction too. Thus

k =1

and

b= 3.

Therefore

20. Answer: 10045 The number of grid points (x, y) inside the rectangle with four corner vertices (0, 0), (11, 0), (0, 2010) and (11, 2010) is (11 -1)

X

(2010 -1) =20090.

There are no grid points on the diagonal between (0, 0) and (11, 2010) excluding these two points, since for any point (x, y) on this segment between (0, 0) and (11, 2010), we have 2010x

y=--·-uBut for an integer x with 1 ::; x ::; 10,

0l�x is not an integer.

2

Hence the number of grid points (x, y) inside the triangle formed by corner points (0, 0), (0, 2010) and (11, 2010) is (11 -1)

X

(2010 -1)/2 56

=

20090/2 =10045.

(11, 2010),

For any grid point (x, y) inside the tr iangle formed by corner points we have

1

1

::::::

y ::::::

2009, 1

:S: X <

11y 2010'

(0, 0), (0, 2010)

and

Thus, for any fixed positive integer y, there ar e the number of grid points satisfying the is condition that :S: x <

.,i-�{0

11 l2 o :o J ' 1�{0 1 (0,2009. 0), (0, 2010) (11, 2010) 11 11 2 11 3 11 4 2009 . 11 l2010J l 2010 J l 2010 J l 2010 J l 2010 J 10045. 21. 2007 3. 3 3. 3. 3, 0. 3 3. as 2 is not an integer when :S: y :S: the tr iangle for med by cor ner points X

+

Thus the number of grid points (x, y) inside and is

X

+

X

+

+. . . +

X

Therefore the answer is

Answer:

Consider n distinct numbers where n 2: of fr iendly pairs is always n- for n 2: also n-

We shall show by induction that the number Hence the minimal number of fr iendly pairs is

If n = then there are no fr iendly pairs as each number is adjacent to the other two. Thus the number of fr iendly pairs is Assume that the number of fr iendly pairs for any arr angement of n distinct numbers on the cir cle is n- for some n 2: Consider n+ distinct number s on the cir cle. Letm be the largest number. Now consider the n numbers on cir cle after deletingm. The two numbers adjacent tom which originally for m a fr iendly pair do not for m a fr iendly pair anymore. Any other fr iendly pair r emains a fr iendly pair when m is deleted. On the other hand, any fr iendly pair aft er m is deleted was originally a fr iendly pair. By induction hypothesis, there are n- fr iendly pairs after m is deleted. Therefore, there are (n+ fr iendly pairs originally.

22.

Answer:

3

1) - 3

12345

1

We are given the equation f(x2+yf(z))

0 (1), 0 (1),

=

xf(x) +zf(y).

(1) 0. (2)

Substituting x = y = into we get zf(O) = f(O) for all real number z. Hence f(O) = Substituting y = into we get Similarly, substituting x = Substituting y =

1 (3) into

0 (1) in

xf(x) we get f(yf(z))

(3)

zf(1)

(4)

we get f(f(z))

for all r eal z. Now, using

zf(y).

(2) (4), and

=

we obtain

57

(5)

Substituting y =z =x into (3) also yields f(xf(x)) = xf(x).

(6)

Comparing (5) and (6), it follows that x2f(1) = xf(x), so that if x is non-zero, then f(x) =ex for some constant c. Since f(O) = 0, we also have f(x) =ex for all real values of x. Substituting this into (1), we have that c(x2 + cyz) = cx2 + cyz. This implies that c2 = c, forcing c = 0 or c = 1. Since f is non-zero, we must have c = 1, so that f(x) = x for all real values of x. Hence f(12345) =12345. 23. Answer: 86422 The number of ways of choosing r objects from n different types of objects of which + repetition is allowed is (n � 1 . In particular, if we write r-digit numbers using n digits allowing for repetitions with the condition that the digits appear in a non-increasing + order, there are (n �- 1 ways of doing so. Grouping the given numbers into different categories, we have the following tabulation. In order to track the enumeration of these elements, the cumulative sum is also computed:

)

)

+ n r (n �

Numbers Beginning with

Digits used other than the fixed part

1

1,0

2

5

2

2,1,0

3

5

3

3,2,1,0

4

5

4

4,3,2,1,0

5

5

5

5,4,3,2,1,0

6

5

6

6,5,4,3,2,1,0

7

5

7

7,6,5,4,3,2,1,0

8

5

From 800000 to

5,4,3,2,1,0

6

5

3,2,1,0

4

4

855555 From 860000 to 863333

-

1

)

�� =6 � �) =21 )

l!

0

I

5

27 83

=126

209

=252

11 =462 5 1 =792 o

6

=56

: �) 1 ;) c) (:) l

Cumulative

= 252

=35

461 923 1715 1967

2002

The next 6-digit numbers are: 864000,864100,864110,864111,864200,864210,864211,864220. Hence, the 2010th 6-digit number is 864220. Therefore, x =864220 so that l J =86422.

�

24. Answer: 309 LetS be the set of permutations of the six integers from 1 to 6. Then lSI=6! =720. Define P( i) to be the subset of S such that the digit i + 1 follows immediately i, i =1,2,3,4,5. 58

Then =

� IP(i)l (�)5! _I; 1P(i1)nP(i2)1 2

21 <22

(� )4!

=

. �. 1P(i1) nP(i2) nP(i3)1 (� )3! . _I; 1P(i1)nP(i2)nP(i3)nP(i4)1 (!) 2! . IP(1)nP(2)nP(3)nP(4)nP(5)1 (�) 1!.

21 <22<23

=

=

21 <22<23<24

=

By the Principle of Inclusion and Exclusion, the r equired number is

25. Answer: 2011 Note that A =

(COo10) - (2��0))2+ ((2010)- (2010))2+ ((2010) - C010)) 2 2 0 1 1 +...+ (G��) G��) ) 2 � f ((2010k ) _ (��))2 . k=O =

_

Obser ve that

is the coefficient of

x2010 in the expansion of the following expression:

We also have

2011 ((2010) - ( 2010 )) k 2011 (2010) k - 2011 ( 2010 ) k L L k L k k 1 k 1 k=O k=O k=O (x+1)2010 _ x(x+1)2010 (1 _ x)(x+1)2010 2010 ((2010)- ( 2010 )) k 2010 (2010)xk-! 2009 (2010 )xk+1 + L1 k k 1 x k=O L k k 1 L1 + k=k=(x+1)2010 _-(1 x+1)2010 (1 _ 1/x)(x+1)2010. 59 X =

=

and

X

=

=

=

X

X

=

X

The coefficient of x2010 in the expansion of the following expression: (1 - x)(x + 1)2010 (1 - 1/x)(x + 1)2010 = (2 - 1/x - x)(x + 1)4020 is 2 Hence

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4020 2010

- 4020 2011

A=

- 4020 2009 4020 2010

Consider the inequality:

sA 2:

=

_

2

4020 2009 ·

4020 2010 '

s(1 - 2010/2011) s

2:

Hence the answer is 2011.

60

4020 - 4020 2 2010 2009 .

2011.

2:

1,


2010

(Open Section, Round 2) Saturday, 26 June 2010

0900-1330


1.

2. 3.

4.


No

10

mark.

calculators are allowed.

1.

Let CD be a chord of a circle r1 and AB a diameter of r1 perpendicular to CD at N with AN> NB. A circle r2 centred at C with radius CN intersects r1 at points P and Q. The line PQ intersects CD at M and AC at K; and the extension of NK meets r2 at L. Prove that PQ is perpendicular to AL.

2.

Let an, bn, n = 1, 2, . . . be two sequences of integers defined by a1 for n;::: 1,

an+1 bn+1

=

=

=

1, b1

=

0

and

7 an + 12bn + 6 4an + 7bn + 3.

Prove that a� is the difference of two consecutive cubes. 3.

4.

5.

Suppose that a1, ..., a15 are prime numbers forming an arithmetic progression with common differenced> 0. If a1> 15, prove thatd> 30, 000. Let n be a positive integer. Find the smallest positive integer k with the property that for any colouring of the squares of a 2n x k chessboard with n colours, there are 2 columns and 2 rows such that the 4 squares in their intersections have the same colour. Let p be a prime number and let x, y, z be positive integers so that 0 < x < y < z < p. Suppose that x3, y3 and z3 have the same remainder when divided by p, 2 show that x + y2 + z2 is divisible by x + y + z.

61

Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Open Section, Round

2

solutions)

Extend DC meeting r2 at H. Let the radius of r2 be r. Note that DN = NC = CH =r. Since M is of equal power with respect to r1 and r2. Thus MN MH = MC ·MD. That is MN(MC + r) =MC(MN + r) giving MN=MG. Thus M is the midpoint of NC. 1.

·

A

As K lies on the radical axis of r1 and r2, the points C, N, A, L are concyclic. Thus LALC=LANG=goo so that AL is tangent to r2. It.follows that AC is perpendicular to NL at K, and hence MN=MC=MK. Now let PQ intersect AL at T. We have LTAK = LKNM = LNKM = LLKT and similarly LTLK = LAKT. Consequently, 2LKTL = 2(LTAK + LAKT) LTAK + LAKT + LLKT + LTLK= 180°, which means LKTL=goo. First we shall prove that a� is the difference of two consecutive cubes. To do so, we shall prove by induction that a� = (bn + 1)3- b�. When n = 1, this is true. Suppose for n 2: 1, this is true. We have 2.

(bn+l + 1)3- b�+l =3b;+l + 3bn+l + 1 =3(4an + 7bn + 3)2 + 3(4an + 7bn + 3) + 1 =48a; + 147b; + 168anbn + 84an + 147bn + 37 =(7an + 12bn + 6)2 + (3b; + 3bn + 1)- a; a;+1 + (bn + 1)3- b�- a�=a;+l =

where the last equality is by induction hypothesis.

62

Lemma: Suppose p is prime and a1, ... , ap are primes forming an A. P. with common difference d. If a1 > p, we claim that p I d.

3.

Proof Since p is prime and every

ai is a prime > p, p does not divide ai for any

the pigeonhole principle, there exist 1 � i < j � p so that ai aj aj - ai = (j -i)d, and p does not divide j - i. So p must divide d.

( mod p) .

i. By Now

Apply the Lemma to the sequences at, ... , ak fork= 2,3, 5, 7,11 and 13. Then all such k's are factors of d. So d > 2 3 5 7 11 13 > 30, 000. ·

4.

·

·

·

·

The answer is 2n2 -n + 1.

Consider an n-colouring of the 2n x k chessboard. A vertical-pair is a pair of squares in the same column that are coloured the same. In every column there are at least n vertical-pairs. Let P be the total number of vertical-pairs and Pi be the number of vertical-pairs with colour i. Then P = P1+ + Pn � nk. Thus there is colour i with Pi � k. There are = 2n2 -n pairs of rows. Thus if k � 2n2 -n + 1, there is a pair of rows that contains two vertical-pairs with colour i.

e;)

·

·

·

Next for k = 2n2 - n, exhibit an n-colouring where no such sets of 4 squares exists. Note that it suffices to find such ann-colouring for the 2n x (2n-1) board. We can then rotate the colours to obtain n of these boards which can then be put together to obtain the requiring n-colouring of the 2n x (2n2 -n) board. For each i = 1,2,...,2n- 1, let Ai = {(i,2n-1+i),(i+1,2n-2+i),...,(n-1+i,n+i)}, where 2n+k, k > 0, is taken to bek. Note that the pairs in each Ai give a partition of {1,2,...,2n}. Moreover, each pair of elements appears in exactly one Ai. Now colour the squares of column i using n colours so that the two squares in each pair of Ai receive the same colour and the colours the 2n pairs are mutually distinct. This gives ann-colouring of the 2n x 2n -1 board with the required property and we are done.

5.

It is given that

p I x3-y3= (x -y)(x2 + xy + y2).

Moreover,-p < x-y < Oand sop f x-y. Thusp I x2+xy+y2. Similarly,p I y2+yz+z2 and p I x2 + xz + z2. Hence

p I x2 + xy -yz -z2 = (x -z)(x + y + z). Since p f x -z,p I x + y + z. Note that 0 < x + y + z < 3p. So x + y + z = p or 2p. We will show that p I x2 + y2 + z2. Assuming this for the moment, the proof is complete if x+y+z = p. If x+y+z = 2p, then x+y+z is even and hence x2+y2+z2 is even. So both 2 and p divide x2 + y2 + z2. Moreover, p > 2 and so 2 and p are relatively prime. Thus 2p divides x2 + y2 + z2 and the proof is also complete in this case.

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Singapore Mathematical Olympiad (2005-2013)

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