Multipe Choice Questions 1 alculate the following sum: 1
2
3
10
4
- + + - + + + 2 2 4 8 16 ()
50 3 256'
505 , 256
B
507 256 ;
D 250569 ,
E ) None of the above
2 It is kown that the roots of the equation 44118x 121323 x + 3x 4 044118x 12132362 62x x 121323 12132363 63xx 20 1 1= 0 are all integers How many distinc roots does the equatio have? ()
B 2; 3; D 4 ;
1;
(E ) 5
3 fair dice is thrown three times The results of of the rst, second and third throw are are recorded recorded as
x,
y
x,
y
and
()
and
1 12 ;
z
z,
respectively Suppose x + y=
z.
What is the probability that at least one of
is ?
B 8� ,
8 15 ;
D 3;1
! . 13
(E )
4 Let X=
000 000 1 000 000 50 1 times times ·
·
·
·
·
Which of the following is a perfect square? ()
x 7 5;
B x 25; C x ; D x + 25;
(E ) x + 7 5
5. Su ppose N 1, N 2, , N 2o11 are positive itegers. Let
X=
(N + N +
Y=
(N + N +
·
·
·
·
+ N ) (N + N + + N ) (N + N +
·
·
·
+ N ), + )
Which one of the following relatioships always holds? () X=Y X=Y
B X>Y ( ) X
(E ) None of the above
6 onside onsiderr the the folo folowi wing ng egg egg shaped curve A BC D is a circle of of radius arc A is centred at C , CF is centred at A and EF is centred c entred at D D 2
1 centred at 0 he
(A) 1680;
B
712;
(C) 3696; D 3760;
(E ) None of the above.
0. In the set {, 6, 7, 9}, which of the numbes apear as the last digit of fo innitely many positive integers ? (A) 1, 6, 7 only ;
B , 6, 9 ony ; (C) 1, 7, 9 ony ; D 6, 7, 9 only ;
(E ) , 6, 7, 9.
Short Questions x y z 11. Suppose-++= a b c
2. Suppose x=
13 +
a b c . . and++= 0 Fd x y
sv
z
. Find the exact value of x - 6x - 2x +18x + 23 x - 8x + 15
13. Let a= 3, and dene al=
a J a +
3
fo all ositive integes
14. Let a, b, c be positive eal numbers such that a + ab + b = 25, b + be+ c = 49, c +ca + a= 64
4
n.
Find a·
dstct ways of usg two tles of sze 1
1, two tles of sze 1
2 ad oe tle of sze 1
x
4.
It s ot ecessary to use all the hree kds of les. )
24. A 4 2
4 Sudoku grd s lled wt dgts so that each coum, eac row ad eac of te four
2 sub-grids tat composes the grd contas all of te dgts from 1 to 4. For example,
Fd te total umber of possble 4
4
3
1
2
2
1
3
4
1
2
4
3
3
4
2
1
4 Sudoku grds.
25. If the 13th of ay partcular moth fals o a day, we call t Friday the 13th. It s kow tat Frday e 13th occurs at east oce every caledar year. If the ogest terval betwee two cosecuve occurreces of day the 13th s
moths, d
.
26. How may ways are there to put 7 detca apples to 4 detca packages so tat each package has at leas oe apple?
27 . At a fu far, coupos ca be used to purchase food. Eac coupo s worth $5 $8 or $12. For example, for a $15 purchase you ca use three coupos of $5 or use oe coupo of $5 ad oe coupo of $8 ad pay $2 by cash. Suppose the prices the fu far are all whole dollars. What s the largest amout that you caot purchase usg oly coupos?
28. Fd the legth of the spragle the followg dagram, were the gap betwee adjacet parallel lies s 1 ut.
6
E
D
34
c
onsider an equilateral triangle ABC, where AB= AB= BC = CA = 2011 Let P be b e a oint inside inside A BC . raw line segments passing through P such that DE I BC, G CA and 7 : 1 H II AB Suppose DE : G : H= 8 : 7 : 1 ind ind D +G + H A
c
3 5
n the following diagram, AB
j
BC . D and E are points points on segments segments A B and BC respecrespec-
tive tively, ly, such that that BA + AE = D + DC . t is is known known that A D = 2, BE = Find Find BA + AE .
8
3
and EC EC =
4.
Note that = ( 10 +1)
10 4 +50= 1046 + 104 +50 = (103 ) +2 103 X 5 +50= ( 103 +5) +25 X
So x- 25 is a perfect square.
5 nswer: ( B ). Let K= N + · · +N Then X= (N +K ) (K +N on) andY = (N +K +N o)K
6 nswer: ( ) .
,
The area enclosed by A D , DE and AE is
The area of the wedge E DF So the area of the egg is:
7 (2 )
-
(- ) (�- 1) (�- )
IS (2- ) = 3
� +1 +2
4
x
1
1= 2- 1
+
J
1 .
J = (3- J) - 1
7 Answer: ( B ). The le shows that 3 colours are not enou gh. The right is a painting using 4 colours.
8
.
Answer: () Since 5 I (24- 1), 7 (36 1), 1 I 5- 1), 13 I (7 - 1),
n
is divisible by 5, 7, 11 and 13.
9 Answer: 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.
et (x)=
3x J X+ 3 v x -1
Then f (J (x))= _J
= J (J (J (x)))= x+v -1 + 1 vx x+v
J
v-1 _1
3x--x- 3x -1 +J x- 1 + V x+ 3 x +v
x- x
3 _ - 1 = . So J (J (J (J (J (J (x))))))= x. X 3x + x+
Since 2010= 6 x 335, = (J (J··· (J (3)) ·· ))= 3. ' copies ·
14. nswer: 12. Consider the following picture, where LAOB= LBO= LOA= 120°, OA= , OB= b and O= A
B
c
Then B = 5, I A = 7 and A B = 8. The area of the triangle A B is 5) (10- 7 ) (10- 8)= 10 .
1 2 (b+b )= 10. So b+b+ = 40. 2( + b+ )= ( + b b ) (b +b+ ) ( + + ) 3 (b+b+ )= 258.
Then
Thus, (+ b+ )= 129. 15. nswer: 0. Dene Q (x)= (1 + x)P(x) x. Then Q (x) is a polynomial of degree 2011. Since Q() = Q(1)= Q(2= ···= Q(2010= 0, e 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, (2012) 2012 and P(2012= = o. 2013 16. nswer: 924 12
Suppose x
x
=
M
=
=
=
···
=
X
=
<
2 � Xl � · · · � X · Let M
=
x · · · Xl · Then
XX . . . XQlOX ll (Xl- )x · · · X X +X . . . X X
2
(Xl- ) 2 +X · · · X
>
. . . . . . . . . . . . . . . . . . . . .
2
(x l- ) 2 + · · · + (x - 1)2 + XX
2
(x l- ) 2 + · · · + (x - )2 + (x o- ) 2 + (x o- ) 2
=
2 (x + · · · +x o- (20- k))
=
2 (M- 20) .
Therefore, M � 4 022. On the other hand, ( , , . . . , , 2, 20) is a soution to the equaion. So the maximum value is 4 022.
2. Answer: 0. If n 2 02, then M (n)
M(9)
=
M(M(02))
For each k
=
=
=
n- 0 2 92.
M(92)
=
M(M(03))
, . . . , 0, M(80 + k)
=
=
···
=
M( 0 )
M(9)
=
9, and thus
M (93)
M(M(9 + k))
=
=
M (7 0 + k)
=
M(M(8 + k))
=
M(9)
=
9,
M(k)
=
M(M( + k))
=
M(91)
=
9.
Hence, all integrs from to 0 are olutions to M(n)
=
=
9.
9.
22. Answer: 9.
A nl
1
A
n +
Then A l
<
2onl +11nl 20 +
A if n
Note tha t +
>
+
0 +
)
20 +11 . (M) n ; and A l + ) w
0 +9
If 0 � n � 8, then n � 0 +8
n
�
18 implies A
A l ·
=
9. So
0 +
2
A if n
>
<
9 im plies A
9 . ; 1f n + ( )
0 + >
+
) w
.
A l ·
0, then n
0 +
9 Hence, + ( ) .
23. Ans wer: 69. ys to pave a block o 1 Let a b e the nu ber of wa
4
X
n. Then a
=
a +a +a w ith
Case 2
The rst year is a leap year. Jan 0 2
Case 3:
Feb 3 5
Mar 4 5
Mar 2 3
Apr 0 1
Jun 5 6
Jul
Aug 3 4
Sep Oct 6 1 2
Nov 4 5
Dec 6 0
Jun 4
Jul Aug [ 2 4 1
Sep Oct 5 @ @ 2
Nov 3 5
Dec 5 0
1
The second year is a leap year. an 0 1
Feb Mar 3 3 4 5
Mar 1 3
Apr 6 1
]
om these tables we see that the answer is 14 The longest time period occurs when the iday of 13th falls in July of the rst 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: ) (4, , , ) 2) (3, 2, 1, 1 ) . 3) (2, 2, 2, l).
7 6 5 4 (47) 432 7 6 5 37 (4 2) 32
=
=
=
5 3 ) ( 7 3! 2 2 2
35
=
�
6
=
X
72
4 3
X
2 6
1
X
=
35
5
X
2
X
4
1
X
X
6 = 210. 3
X
2
2
X
1
=
l05.
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 n 25, write n = 5 + m where 20 m 24 and is a positive integer. So any amount 25 can be paid exactly using coupons. However, 1 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 The last L " is 100 + 101 = 201 Then the total length is 2(1 + 2 +3 + + 100) + 201 = 10301. ·
29 Answer: 15. 16
·
·
So c= 1 or c=-0 (ignored ) . Substitute y=
-
+ 1 into the parabola: 2=
+ 1
-
=
= 3,-4 So A is (3, 9) and B is (4, 16) Then
A B 2= (3 (4)) 2+ (9 16) 2= 98. 33 nswer: 30 J raw BF . CE , where F is on CE . If A B = 1, then BF = 2 and BE = J. Thus LE= 30°
E
D
34 nswer: 40 Set D P= GP= , P= FP= , E P= HP= c Then DE +FG+H= (+ c)+ (+ )+ (+ c)= (+ + c)= 35 nswer: 10 y given, B D + +AE= B D +D C So +AE= D C Note that A B 2+ BE 2= AE 2 and BD 2+ BC 2 (+ D ) 2+ 32= AE 2, 4(AE + B D )= 3 Then A + BA
32 =
=
DC 2. Then
B D 2+ 7 2= (AE + ) 2
4 + = 10
18
x
011= 40
Singapore Mathematical Society Od (O)
211
(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 x2y2 (a2 + b2)(c2 + d2) (ac)2 + (bd)2 + a2d2 + b2c2 (ac)2 + (bd)2 + (ac)(bd) AM-GM =
(ac + bd)2.
Let the tangent at P meet the tangent at R at the point Let 0 be the centre of r1 Then OR is a square. Hence LKPR L R P 45° Als o LNPS LNKP = LPMS LMPK. Thus LMPR LRPN. 2.
K
3 Let ai max,bi min and s uppos e that t1 0, then a1 b j Therefore a1 E j
min{} For each j if 81 j
Note: Problem 4 in the Senior Section is the general vers ion 4. eplace 0 by any pos itive odd integer
rs t s how by induction that am 3mn-m for m 0, , ..., n- . This i certainly true for m 0. uppose it's true for s ome m < n . Then 3am + 3m +ln-m . Since n- m > , the odd part is 3m +ln-m-l which is am +l· Now an-1 3n-11 . Thus
0
n. We
Singapore Mathemaical Society Singapore Mathematical Olympiad (SMO) 0 (Senior Section)
Tesda, May 0
090-00 hrs
Instructions to contestants 1. Answer ALL 35 questions. 2 Enter your answers on the answer shee 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 appropriat bubble below your answer. 5 No steps are needed to justify your answers. 6 Each question carries 1 mark. 7 o calculators are allowed.
PLAS DO NOT TUR OVER UNTIL YOU AR TOLD TO DO SO 22
6. Determine the value of
2
1 1 + 0+ -V+ A 4 - E 4 -2
B 2 -2
(C) 4 +
(D) 2 +
1 7. Let x hich of the following stateents 1+ 1+ 1. W lo � 2 log 4 log 8 s true? =
15 < < 2 (D) 3 << 35
A
B < x < 5 E 35 <<4
8. Determine the last two digits of 756• A B 7 9
2.5 < < 3
() 43
E 49
9 It is given that x and y are two real nuers such that y > 1. Fin the smallest possile value of
x >
nd
logx21 +loy2 ly 11 A
B 6
4
8
E 1
()
1. It s gvethat , ban are three eal nuers such that +b c 1 an b 2- 7 +14. Find te lagest possile value of 2 + b2. =
=
A 5
B 6
8
() 9
E 1
Short Questions 11. Find the value of 211 +211-2 X 211 X 211 25 24
sies of the triange are 18 In the iagram below, the lengts of thea2three 2 a m, b m an m. t is givethat +2 b = 011 Fin the value
cot C of cot A +cotE
a
19 Su p pose there are a total of 01 partici pants in a mathematics competitio, an at least 1000 of the are female . Moreover, given ay 011 partici pats, at least 11 of the are male. How many male partici pants are there in his co petition?
0 Let : Q \ {0, 1}
-
Q be a function such that
2xx+ (X -)= x2 x for all rational nubers x # 0, 1. Here Q enotes the set of ratioal �
numbers. Fin the value of( ).
1 In the iagram below, ABCD is a convex quadrilateral such that AC
intersects BD at the mipoint of BD. The point is the foot of the per penicular from A oto D, an lie in the interior of the segmet D. Su p pose BCA = C = m, = m, A= m an CD= m. Fin the value of
40
90°,
x
1
A
c
26
x
15
28 It is given that
b, nd d are four positive prie nuers such that the product of these four prie nuers is e qual to the u of 55 consecutive positive inteer Find the sallest possile value of a+ b+ c +d. ( Reark: The four nuers a, b, d are not necessarily distinct )
29 n the iara elow, ABC is a triange with AB= 39 , B 45 and CA= 42 The tangents at A and B t the circucirce of DABC eet at the point P. The point lies on BC such that P is parallel to AC. It is given that the area of DAB is x c 2 Find the value of x
30 t is given that and b are positive integers such that has exactly 9 poitive divisors and b has exactly 10 positive divisor If the east comon ulti ple ( LM ) of an b is 4400, nd the value of a + b. 31 In te diagra elow, ABC is a quadrilteral such that ABC= 35° and BC= 120° Moreover, it is given that AB = 2 J , = 4 - 2 - , C= 4 - m an A= x Find the value of x- 4x c
28
Singapore Mathemaical Socey Singapore Mathematical Oympiad (SO)
2011
(Senior Section Soutions) 1. Ans wer: (B) < Sice , and are nonzero real numbers and b 2+ 2 2 + 2 that + + a 2 +b 2
<
, w e s e e 2 +b 2
>
>
Adding 1 throughout, we obtain
T h u s
1 1 2>
b
>
1 , w hi ch im p lie s th at <
b < . So we ha ve I I < b
2. A n s w e r: (A ) N o te t ha t (s in+co s ) = sin +co s + 2 s in co s = 1+ s in 2= 1+ . S in c e 0 � ' w e ha ve s in+ co s > 0 S o s in c os = v . 3 A n s w er : (D) W e ha v e � [( a- b) 2 +(b- e ) 2 +(c- a ) 2 ] 2 [(-1) +(-1) +2 ] = 3 ·
�.
4 A ns w er : C 1 1 1 X 2y 2x+y
2x+y :
-
:
:
N ow w e h av e
3 0
X
-
2x+y 2y
=1
y X 1 2+- -= 1 X y 2 1 y X 2 X y
< l e i ·
oreove 2x
SO X >
(log 3+log 5 og 7 ) log 2 log 9 5)+log(49 ) log(45 27 ) og 2 og( 45 3) log(128) = 7' > �g2 �2
3 5
8. ns wr: B Note tha 74-1 = 2400 s o that 74n-1 is divis ible by 100 for any 7 56 77 56-1 +1 77 561+7
n
E
z + . Now
77 4n1+7 , where
5-1 4 Since 74n-1 is divis ible by 100 its las t two digits are 00. It follows that the ast two digits of 7 56 are 07 =
n
9 ns wer: () log 2011+log 2011 log 2011
>
2011 log 2011 log x + ) ( ) log x log log 2011 1 1 ( + ) (log x +log) log x og ogx og 2+ log og x 4 (us ing A M 2 G M),
and the equality is attained when og x = og or equivaenty x= y.
10. ns wer (C) The roots of th equation x 2 c 1x+c 2- 7c+ 14 = 0 are and which are rea. Thus the dis criminant of the equaton is nonnegaive In other words
c-12 c2- 7c+14)= -32+ 26- 55= -3c+11c- 5) 2 0. 11 < < c - 5. Together wih the equaities So we have 3 (+?- 2 c 2c2- 7c+ 14) -c2+1c- 27= 9 - c 2, 32
15. ns wer: 8001 Note that J-
4 < 0. 01 if and only if 4 > 400. y+ 4= J 5 4 < 400. If = 8000, then J+ - 4= 40000+ If = 8001, then J+ 4= 40005+ 40001 > 400. So the answer is 8001 -
16 ns wer: 1006 The series can be paired as 1 1 1 1 1 1 ) + )+ + ( ( + )+( + 1+11 1+11 1+11 9 1+11 9 · · · 1+11 1+ 11 ach pair of terms is of the form 1 1 1 1+ There are 1006 pairs of such terms , and thus the sum of the s eries is 1006. 1 7 . ns wer: 54 X
.
Si
+cos
7 7 +cos 8 +sin (8)
sin (i) +cos (i + sin (i) +cos (i . . + COS ) - 4Si 2 (Si 4 = 32 · 8 COS 8 = 2- Si Thus 36x = 54. 18. ns wer: 1005 y the laws of sine and cosine, we have s in A s inE sinG and Then 1 cosC cot cotA+cot B sinC smAsmB s in A s in B cos C sin(A+B) s inC .
.
Si
cos C
( ab jc 2 ) s in 2 C a 2 +b - 2 ) )( ( 2 ab s in C a +b - c 2c
011 1 = 1005. 2 34
22 Ans wer: 3 Note that 1 1 + X
1 211
=
= -
x- 211x- 211 0 = (x- 211)(- 211) 211 =
=
Since 211 is a prime number, the factors of 211 are 1, 211, 211 , -1, -211, 211 Thus the pairs of inteers (x, ) s atis fying the las t equation are iven by: (x- 211, - 211) (1, 211 ), (21, 211), (211 , 1), (-1, -211 ), (-211, -211), (-211 , -1) =
Euivalently, (x, ) are iven by (212, 211+211 ), (422, 422), (21+ 211 , 212), (210, 211- 211 ), (0, 0), (211 211 , 21) Note that (0, 0) does not s atis the rs t equation. Amon the remainin 5 pairs which satis the rs t equation, three of tem satis the inequality x � , and they are iven by (x, ) (422, 422), (21+211 , 212), (210, 211- 211 ) =
23 Ans wer: 93 By lon division, we ave x + x +x+
=
( x + 3x- 1) (x - 3x+ (+10))+ (-3-33 )x+ (++ 10)
Let , be the two roots of the equation x + 3x- 1 0 Note that , s ince the dis criminant of the above quadratic equation is 3 -4 · 1 (-1) 13 0 Since , als o s atis fy the equation x + x +x+ 0, it follows that and als o s atis the equation =
=
=
(- 3 33)x+( ++ 10) 0 =
Thus we have (- 3- 33) + ( ++ 10) 0, =
and (- 3- 33) + ( ++ 10) 0 =
Sinc , it follows that -3-33 0 and ++O 0 Hence we ave 3+33 and --10 Thus ++4+100 +(3+33)+4(--10)+100 93 =
=
=
=
=
=
24 Ans wer: 1120 Let and be positive integers s atis in te iven equation. Then 3( -) 2011 Since 2011 is a prime, 3 divides By lettin 3k, we have (3k ) + 2011k This implies that k divides Let k Then 9k k+2011k s o that 9k +2011 The s mallest pos itive inteer s uch that +2011 is divis ible by 9 is r 5 Thus k (5 +2011)/ 224 The correspondin values of and are 1120 and 672. =
=
=
=
=
=
=
=
=
36
=
=
Since GN is parallel to AD and G is the centroid of the triangle A, we �= hae MD/MN= 3. It follows that += Thus , AB - 15-6= 9. 28. ns wer 28 The sum of 55 positive consecutive integers is at least (55 x 56)/2 1540. et te middle number of these consecutive pos itive integers be x. Then the product = 55x 5 11 x. So we have 55x 2 1540 and thus x 2 28. The least value of + + + is attained when x= 5(7). Tus the ans wer is 5+11+5+ 7 28. ·
·
29. ns wer: 168 irst LD P= LCA= LAP so that P, , D, A are concyclic. Thus LACD= L PA= L P DA= LDAC s o that DA= DC
y cosine rule, cos C = 3/5. Thus C = � ACjcos C 21 x 5/3 35. Hence D = 10 and BC = 10 + 35= 45. Thus area(A) � area(AC). y Heron's formula, area(ABC)= 756. Tus area(AD)= x 756= 168. 30. ns wer: 276 Since te number of positive divisors of is odd, a must be a perfect s quare s is a divis or of 4400 2 x 5 x 11 and a has eactly 9 positive divis ors , we see that = 2 x 5 . ow the least common multiple of and is 4400 implies tat mus t have 2 x 11 as a divis or. Since 2 x 11 has eactly 10 positive divis ors , we deduce that 2 x 11 176. Hence + 276. 31. ns wer: 2 0 irst we let / be the line which etends C in both directions. et E be the point on / s uch tat AE is perpendicular o f Similarly, we let F be the point on / s uch tat DF is perpendicular to f Then, it is easy to see that E= AE = v, CF = 2J and DF = 2. Thus EF + 4 2J +2J = 4 +. ow we lt G be the point on DF s uch that AG is parallel to £. Then AG F 4 + and
38
which is a contradiction to the given fact that each element of S is les s than or equal to 2011. 1 is in A. We may let a16 = 1. Then , , s in Case 1, we have Case 2.
·
·
·
are comps ite numbers .
,
which is a contradiction. Thus we have shown that every 16-element s ubs et A of S s uc that all elements in A are pairwise relatively prime mus t contain a prime number. Hence the s malls k is 16. 35. ns wer: 12 Let the extens ions of A Q and BP meet at the point R. Then P R Q= P AB= QPR s o that QP = QR. ince QA = QP, the point Q is the midpoint of A R. s A R is parallel to P, the triangles A RB and B are s imilar s o that M is the midpoint of P. Therefore, N is the centroid of the triangle P B, and 3 N= BM.
Let ABP= . Thus tan= A R/AB= 2A QjAB= 5/6. Then B= P cos= AB cos . ls o BM/B = B QBA s o that 3 MN = EM = = 12. = cos (Q + 3 MN). olving 'for MN, we have MN=
40
Singapore Mathematcal Socety Od (O)
2011
(Senior Section, Round 2 solutions)
1 There is an error in this proble. The triangle is not necess arily equilatera. In fact we s hall prove that the altitude at A , the bis ector of B and the edian at C meet at a coon point if and only if cos B= where BC = a, CA = an A B= 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 B and CF is the edian at C. Suppos e that AD, BE, CF meet at a comon point. The point of concurrence of AD, BE and C F us t le ins ide the triange A BC. ince F i s the idpoint of A B, by eva's theore CE : EA = CD : DB. Us ing the angle bis ector theore, CE : EA =a : c Thus CD= a2/(ac) and D B= acj(ac). Thus cos B= = A
B
onvers ey, if coB= then B is acute and BD= cos B= acj(ac)
2. Yes , in fact, for any k E N, there is a s et 8k having k elements s atis fying the given condition. or k = , let 81 be any s ingleton s et. or k = , let 8 2 = { ,3} Suppose that 8k = {a1, , ak} s atis es the given conditions . Let
b1 = a1a2 ak b = b1a -1, j k . ·
·
Let 8k + 1 = {b1, 2, , bk +1} Then the fact that 8k +1 s atis es the required property can be veried by obs erving that lm- nl = (m, n) if and only if (m- n) divides m. 42
Note that a1 a2 ··· an
=
1 It suces to show that
sice it is equvalet to 1 1 +a1
1 1 �n- 1. +···+ 1+a2 1+an
We sha show that ( * ) s true for n 2. Th case n 2 s ovious We wl ow prove it y iductio Suppose ( * ) holds for n k. Now assume a1 ··· ak+l 1, ai > 0 for all i. To prove the iductive step, it suces to show that
=
which c be veried directly Note: This is a extesio of the proem :
44
=
Throughout ths paper, let xJ denote the greatest integer less than or equal to x Fo eample, 1J = , 39j = 3 ( This notaton s used n Quetons 7 , 9, 19 and 0)
1 A crcular con A s rolled, wthout slng, along the crcumference of another stationary crcular con B wth radus twce the radius of con A et x be the number of· degrees that the con A makes around ts cntre untl t rst returns to ts ntal positon. Find the value of x Three towns X,Y and Z le on a plane with coordnates (0, 0), (00, 0) and (0, 300) respec tvely. There are 100, 00 and 300 stuents n towns X,Y and Z respectvely. A school s to be bult on a grd pont (x, ) , where and are both ntegers, such that the overall dstance travelled by all he students is inmzed. Fnd the value of x+. 3 Fnd the last non-zero dgt n 30! ( For eample, 5! = 10; the last non-zero dgt s 2.)
4 . The dagram below shows �A BC , whch s soceles wth A B = AC and LA ° The pont D les on AC such that A D = BC The segment E D s constructed as shown. etermne LABD n degrees. A =
cos a sin a cos sn . = 1, evaluate 5 Gven that �( + + cos
s
cos a
s a
6. T e nu mb er 2 s e pressed a s th e su m of postive in teg ers XI, x 2, X, w he re k : 5 W a t is th e m a im um v al u e of t e produ ct of XI, x 2, X3, a nd X ? ·
·
46
·
·
,
·
·
,
18
A collection of 2011 circles divide he 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 he last four digits of N . 19 If a positive integer N can be epressed as x + 2x + l3x for some real numbers x, then we say that N is visible"; otherwise, we say that N is invisible" For eample, 8 is visible since 8= 15 + 2 (15 )J + 3 ( 15 )J , whereas 10 is invisible If we arrange all the invisible positive integers in increasing order, nd the 2011 invisible" integer 20 et A be the sum of all non-negative integers n satisfying
Determine A . 21
A triangle whose angles are A,
B, C satises the following conditions sin A+ sinE+ sin cos A+ cos B + cosC
12 7'
and
12 25 Given that sin C takes on three possible values , and , nd the value of 100 sin A sin B sin C =
22 et x > 1,
y
> 1 and z > 1 be positive integers for which the following equation
1! + 2! + 3! + + x!= yz is satised Find the largest possible value of x + y+ z. 23 et A BC be a nonisosceles acute-angled triangle with circumcentre , orthocentre H and C = 41o. Suppose the bisector of LA passes through the midpoint M of OH . Find LHA i degrees 24 Te circle ' centred at intersects te circle ' centred at at two points P and Q The tangent to ' at P intersects ' at the point A and the tangent to ' at P intersects ' at the point B where A and B are distinct from P Suppose PQ · = P P and LAP B is acute Determine the size of LAP B in degrees ·
25 Determine
n 1 lim . n-oL n i=O �
( ote:
ere
(�) enotes 1 (n
" ' for= 0, 1, 2, 3, · · · , n ) )
8
Thus 30! 30! 10
2 . 3 4 . 5 . 7 4 . 112 . 132 . 17 . 19 . 3 . 9 . 3 4 . 7 4 . 112 . 132 . 17 . 19 . 3 . 9 6 4 . . 7 4 . 112 . 132 . 17 . 19 . 3 . 9 6 () (1) (1)(9) (7)(9) (3)(9)(mod 10) (-1) (-3) (-1)(3)(-1) (mod 10) 8(mod 10)
showing hat he las nonzero digi is 8. D
10
4 Answer.
et E be he poin inside fABC such tha fE BC is equilaeral. onnect A and D o E respecively . Solution.
It is clear tha fAE B and fAEC are congruent since AE = AE , AB = AC and BE= CE implies hat L BAE = LCAE= O . Since A D = BC = BE , LE BA = L D A B = 0 and A B = BA, we have A BE and fBAD are congruen implying ha LAB �= L BAE = O .
D
5.
Answer.
1
Solutn
cos4 a sin 4 a cos2 a sin 2 a . = 1 se cos= Sce �( + and sin= ( . Then ( cos cos s c
cos (- a )= cos cos a + sin sin a= cos2 a + sin2 a= 1. 50
Now let x +x+ b= (x-x)(x x), where x x Then the set of integer solutions of x+ x + b< 0 is {k k is an integer x< k< x} the given condition {k k is an integer x< k< x}= {k k is an integer -11< k< 6} = { 10,9, ·,5} Thus -11 x < -10 and 5 < x 6 It implies that -6 < x + x < -4 and -66 XX< 50 om x+x+= (x-x)(x-x), we have = -(x+x) and b= x x Thus 4< < 6 and -66 b<50 It follows that 5< b< 72 Thus mLa- b 71 It remains to show that it is possible that - bJ = 71 for some a and b. Let = 5 and b= -66 Then x++= (x+11) (x-6) and the inequalit x++< 0 has solutions {x -11< x< 6} So the set of intger solutions of x + ab + b< 0 is eall the set of integers in A n B D Hence mia- biJ = 71 10
8
Answer. Solution.
We consider the polnomial P()= + + +
Sppose the root of the equation P() = 0 are x, y and z Then b= + y+ Z=
14,
and x +y + z +3= (x+y+ z) (x +y+ z xy xz yz) Solving for b, c ad , we get b= 14, = 30 and =6 Finall since 14+30D 6= 0 implies = 2 or = or 8, we conclude tht ma{ a, , '}= 8 =
11
Answer.
38
Let be an even positive integer Then each of the following epresses as the sm of two odd integers: = ( 15) + 15, - 2) + 25 or - 35) +35 Note that at least one of - 15, - 25, - 35 is divisibl b 3, hence can be epressed as the sum of two composite odd numbers if > 38 Indeed it can be veried that 38 canot be epressed as the sum of two composite odd positive integers. Solution.
12
Answer.
1936 52
We shall show that A ijection from A to B:
=
B
=
= 3420 y showing tha the maping¢ elow is a
First, since { , , } E A, we have +3 � and +4 � , and so - 5, implying that {, -, -5} E B . t is clear that¢ is injective. It is also surjective, as for any {b, b , b } E B with b b < b , we have { b, b +2, b + 5} E A and Hence is a ijection and A = B = 3420. 16.
Answer.
3
It is clear that 8(co 40°) -6 cos 40°+ 1 = 0, since cos 3A = 4 cos A-3 cos A . Oserve that 3 1 + 64 sin 20° sin 20° cos 2° 6 + 3(1-cos 40°) 1- cos 40° 1 + cos 40° 8 cos 40°+ 4 + 32 3 cos 40 1- (cos40°) 8 COS 40° + 4-3 OS 40°+ 32 ( COS 40°) + 3 1- (cos 40°) 1-6cos40°+8(cos40°) 4 +32 1-(cos40°) 3, 0 where the last step ollos from 8(cos 40°) - 6 cs 40°+ 1 = 0. Solution
_
=
17 .
Answer Solution.
609 Given the original equation
f(x+ x) + 2fx-3x + ) = 9x-15x, we replace x y 1-x and otain f(x-3x+ )+ f (x+x) = 9(1-x)-15(1-x) = 9x-3x-6. Elimiating f(x-3x + 2) from the two eqations, we otain 3f(x+ x) = + 9x-1, therey
f(x+ x = 3x + 3x-4 = 3(x+ x)-4, hece f(011) = 3(011)- = 609 .
54
20. Answer. 95004
a, if f(a) denotes the sum of all
Solution. We shall prove that for any positive integer then nonnegative integer solutions to LJ=
f(a)= 16a(a-)(a+ 2). Thus
!(27)= 95004.
Let n be a solution to lso n= for each = 0, 1, . . .
=
aq + r, r
rite n= where 0 Since = we have 0 that is, 1, 1, ca be ayone of the values
(a+ )q+r-q. q , a- r
q,
a-1a1
A=
LL(qa+r) q=O r=q a-1
=
aa1
L(a-q)qa+LLr
= =
q=O r=q a-1 r
q=O a
a-1
q=O a1
q=O a-1
r=O q=O a-1
q=O a-1
q=O a
a1
a-1
q=O
q=O
r=O
=O
aq-aLq+LLr aq-aLq+ r(r+1) r=O
aLq-aL q+Lr+Lr = (a+ 1 ) · �a(a-1 + (1- a) a(2a-)(a-1 2 6 1 = 6a(-)(a + 2). =
D
21. Answer. 48 y using factor formulae and double angle formulae: sin A+ sinE+ siG 4cos cos f cos Q cosA + cos B + cosC 1+4sin sin sin �
4
and
12 7'
B C . . . . . B . C sA s B sC= 8 s s 2 s 2 cos cos 2 cos 2=
Solvig these equations, we obtain . A . B . C sss 0 1
2
2
2
B A C cos cos cos 2
Frthermore, sin
C
2
= cos
(
A
.
0. 6
+ B ) = cos A cos B -sin A sin B 2 2 2 2 2' 56
12 . 25
p
et L be the midpoint of BC . It is known fct tht AH = 2£. To see this extend CO meeting the circumcircle of the triangle ABC the point . Then A H i a parllelogrm . Thus AH = B = 2£. Therefore in the rightngled tringle OLC, OC = OA = AH = 2£. This implies OCL = 30°. Since the tringe ABC is acute the circumcntre lies inside the tring. In fct LA = 60° nd LB = 79° Then OAC = LOCA = 41°-30° = 11°. onsequently HAO = 2LOAM = 2 X (30°-11°) = 38°. D 24 . Answer. 30 et P = nd P = . First note tht intersecs PQ at the midpoint H (not shown in the gure) of PQ perpendiculrly . Next observe tht LAPQ = PQ = LP, nd LBPQ = LPAQ = LP. Threfore LAPB = APQ + LBPQ = LP + LP.
' 1 ' 2 H et LP = nd P = . Then sin = PrQ, cos = 02 r2 and sin = 2 + , cos = Thus sin AP B = sin( + /) = sin cos + cos a sin = f H H . Q E 1 + 02 ) = P·01 02 = 1. Sin ce LA P B is a cu t it is eq ua l to 30°. P (· 0 02 D r2 r1 r1r2 r1r2 ·
=
'
25. Answer. 2 Solution. et
ssume tht
n
� 3. It is clear tht n-1 a n=
2+
()
� ; �=
58
-1 > 2.
·
Singapore Mathematical Society Od (O) 2011 (Open Section, Round 2) Saturday, 2 July 2011
0900-330
1. I the acuteagled oisosceles triagle ABC, 0 is its circucetre, Hs its orthocetre ad AB > AC. Let Q be poit o AC such that the exteso of HQ eets the extesio of BC at the poit Suppose BD D, where Dis te foot of the perpedicular fro A oto BC Prove that ODQ 90° =
=
2. If 46 squares are coored red i a 9 x 9 board, show that there is a 2 x 2 block o the board i which at least 3 of the squares are coored red.
�
3. Let , y, z > 0 such that + + < 2 + +
Show that 2 y
+ y
+
2z < 3 + z
4 Fid al polois () with real coeciets such tat
() E Z iplies that a E Z
5 Fid all pairs of positive itegers (m, n) such tat
m+n-
3mn m+n
60
2011 3
which leads to a contradiction. On the other hand, s uppos e that r1, r3, r5, r7, r9 6. Then the s um of any cons ecutive 's is : 9 gain we get a contradiction as
(r1+r2)+
·
·
·
+(r7+rs)+rg :
4
x
9+9=45
Let r =/x, = /y, t= /z. There exists < s uch that r + + t =2rt or (r + + t) = 3t. Let a=r, b=, c=t. Write a= tan A, b tan, c = tan C, then A+ B C 1 It is clear that 3.
=
=
-
+ + +r 2 + 2 + < + + +a2 b2 + = cos A+ cos B+ cos B <- 3 cos A+B + C =�= 3
LHS=
(
2nd soln:
)
x
R HS
Note that
+ x y z Hence
x +
<
xyz
:
xy+yz+xz
x +xy+xz+ yz
y MGM we have
x
<
< .
x +y)(x z)
X
X
+ -. x+y x+z
-
Similarly, 2y
+ z)y+x)
<-
2z
yz
y x'
Z
+
+
z z <- . Z
+X
Z
y
The des ired inequality then follows by addng p the three inequalities
4.
Let P(x)=anxn + +a1x+a0. Dene Q(x)=P(x + ) - Px). Then Q(x) is of degree n . We'll prove by contradiction that IQ(x)l : 3 for all x. This will imply that n . ssume that IQ(a)i > 3 for s ome a E R Then P(a ) - P(a)l > 3 Thus there ae 3 integers between P(a) and P(a ). Hence there exis ts three values ·
·
6
