SOAL FINAL ARUTMIN MATHEMATIC CHALLENGE (AMC) 2009
UNIVERSITAS LAMBUNG MANGKURAT
1. Jika a2 = 8b + 58 dan b2 = 8a + 58 dengan a dan b bilangan real berbeda, tentukan hasil kali ab.
Jawaban : 6
Solusi :
a2 – b2 = 8(b - a)
(a - b)(a + b) = 8(b - a)
Karena a ≠ b maka a + b = -8
a2 + b2 = 8(a + b) + 116
(a + b)2 - 2ab = 8(a + b) + 116
( -8)2 - 2ab = 8( -8) + 116
2ab = 12
ab = 6
1
1
1
2. Four consecutive integers p, q, r, s with p < q < r < s satisfy 2 𝑝 + 3 𝑞 + 4 𝑟 = 𝑠. What is the value
of s?
Answer : 29
Solution :
Since p, q, r, s are four consecutive integers with p < q < r < s, then r = s − 1, q = s − 2, and
p = s − 3.
Thus,
1
1
1
𝑝+ 𝑞+ 𝑟 =𝑠
2
3
4
1
1
1
(𝑠 − 3) + (𝑠 − 2) + (𝑠 − 1) = 𝑠
2
3
2
6(𝑠 − 3) + 4(𝑠 − 2) + 3(𝑠 − 1) = 12𝑠
(multiplying through by 12)
13𝑠 − 18 − 8 − 3 = 12𝑠
𝑠 = 29
1
1
1
(𝐶ℎ𝑒𝑐𝑘𝑖𝑛𝑔, (26) + (27) + (28) = 13 + 9 + 7 = 29. )
2
3
4
3. Sebuah mobil mempunyai 4 tempat duduk. Berapa banyak cara 3 orang didudukkan jika
diandaikan satu orang harus duduk di kursi supir?
Jawaban : 9 cara
Solusi:
Ada tiga pilihan orang untuk menjadi supir.
Jika satu orang sudah duduk di tempat supir, maka dua orang lagi akan didudukkan pada 3 posisi
tempat duduk yang lain. Ini dilakukan dalam P (3,2) cara.
1|AMC 2009 UNLAM
Jumlah cara mendudukan 3 orang seluruhnya adalah
3 × 𝑃(3,2) = 3 ×
3!
3 × 2!
=3×
= 3 × 3 = 9 𝑐𝑎𝑟𝑎
2!
2!
4. Determine the units digit of the integer equal to 9 + 9 2 + 93 + 94 + 95 + 96 + 97 + 98. (The units digit
of an integer is its rightmost digit. For example, the units digit of the integer 1234 is 4).
Answer : 0
Solusi :
We note that 9 + 92 + 93 + 94 + 95 + 96 + 97 + 98 = 9(1 + 91) + 93(1 + 91) + 95(1 + 91) + 97(1 + 91) = (9 +
93 + 95 + 97)(1 + 9) = 10(9 + 93 + 95 + 97).
Therefore, 9 + 92 + 93 + 94 is an integer that is divisible by 10, so its units digit is 0.
5. Tentukan bilangan kuadrat 4 angka dengan angka pertama sama dengan angka kedua dan angka
ketiga sama dengan angka keempat.
Answer : 7744
Solusi :
Misal bilangan tersebut adalah aabb.
Nilai b yang memenuhi adalah 0, 1, 4, 5, 6, atau 9. Tetapi 11, 55, 99 jika dibagi 4 bersisa 3
sedangkan 66 jika dibagi 4 bersisa 2 yang membuat aabb tidak mungkin merupakan bilangan
kuadrat. Jadi nilai b yang mungkin adalah 0 atau 4.
Jika b = 0 maka aa00 = 100(10a + a) yang berakibat 10a + a harus bilangan kuadrat. Tetapi 11, 22,
33, ..., 99 tidak ada satupun yang merupakan bilangan kuadrat. Sehingga tidak mungkin b = 0.
Jika b = 4 maka aa44 = 11(100a + 4). Karena aa44 bilangan kuadrat maka 100a + 4 habis dibagi 11.
Sesuai dengan sifat bilangan habis dibagi 11 maka a + 4 . 0 habis dibagi 11. Nilai a yang memenuhi
hanya 7.
Jadi bilangan tersebut adalah 7744.
6. If the average of four different positive integers is 8, what is the largest possible valueof any one
of these integers?
Answer : 26
Solution:
Suppose the four positive integers are a, b, c, d.
Since the average of the four positive integers is 8, then
𝑎 + 𝑏 + 𝑐 + 𝑑
= 8
4
𝑜𝑟
𝑎 + 𝑏 + 𝑐 + 𝑑 = 32.
We try to find the maximum possible value of d.
We know that d = 32 − a − b − c = 32 − (a + b + c).
2|AMC 2009 UNLAM
To make d as large as possible, we make a + b + c as small as possible.
Since a, b and c are different positive integers, then the smallest possible value of a + b + c is
1 + 2 + 3 or 6.
Thus, the largest possible value of d is 32 − 6 = 26.
7. Pembilang dan penyebut suatu bilangan memiliki selisih 8. Jika pembilang dan penyebut masingmasing ditambah 11, maka perbandingan pembilang dan penyebut adalah ¾, Berapakah jumlah
pembilang dan penyebut bilangan tersebut?
Jawaban : 34
Solusi:
Pembilang – penyebut = 8
y-x =8
x  11 3

y  11 4
4x + 44 = 3y + 33
4x + 44 = 3(8+x)+33
x = 13, y = 21
x + y = 34
8. If ax3 + bx2 + cx + d = (x2 + x − 2)(x − 4) − (x + 2)(x2 − 5x + 4) for all values of x, what is the value of a
+ b + c + d?
Answer :
Solution :
Solution 1 :
We use the fact that a + b + c + d = a(13) + b(12) + c(1) + d, so a + b + c + d must be equal to
the right side of the given equation with x set equal to 1.
Thus,
a + b + c + d = (12 + 1 − 2)(1 − 4) − (1 + 2)(12 − 5 + 4) = 0(−3) − 3(0) = 0
Solution 2 :
We simplify the right side of the given equation by factoring the two quadratic polynomials:
(x2 + x − 2)(x − 4) − (x + 2)(x2 − 5x + 4) = (x − 1)(x + 2)(x − 4) − (x + 2)(x − 1)(x − 4) = 0
Therefore, ax3 +bx2 +cx+d = 0 for all values of x. (In other words, ax3 +bx2 +cx+d = 0 is
the zero polynomial, so all of its coefficients are equal to 0.)
Therefore, a = b = c = d = 0, so a + b + c + d = 0.
9. Sebuah fungsi f didefinisikan pada bilangan bulat yang memenuhi f(1) + f(2) + ... + f(n) = n2f(n) dan
f(1) = 2009 untuk semua n > 1. Berapakah nilai f(2009) ?
3|AMC 2009 UNLAM
𝟏
Jawaban : 𝟏𝟎𝟎𝟓
Solusi :
f(1) + f(2) + ... + f(n − 1) = (n − 1)2f(n − 1)
f(1) + f(2) + ... + f(n) = n2f(n)
(n − 1)2f(n − 1) + f(n) = n2f(n)
(n − 1)2f(n − 1) = (n2 − 1) f(n)
Karena n ≠ 1 maka :
=
.
.
....
.
=
.
.
.... .
=
f(2009) =
=
10. Augustus, Benedict, Claudio, and Diana have been accused of stealing the golden mean. It is
known that one of these four people must have done it. Augustus says “Benedict did it”. Benedict
says “Diana did it”. Claudio says “I didn’t do it”. Diana says “Benedict is lying when he says I did
it”. If it is known that exactly one of them is lying, which one did it?
Answer: Benedict
Solution:
Since Benedict and Diana contradict each other, one of them must be lying. We know that only
one person is lying, therefore Augustus (and Claudio, but that is not important) must be telling
the truth.
11. Jika a679b adalah bilangan lima angka yang habis dibagi 72, tentukan nilai a dan b!
Jawab: 36792
Solusi:
72 = 9 . 8
Karena 9 dan 8 relatif prima maka a679b harus habis dibagi 8 dan 9
Karena a679b habis dibagi 8 maka 79b habis dibagi 8.
Agar 790 + b habis diagi 8 maka b = 2
Karena a679b habis dibagi 9 maka a + 6 + 7 + 9 + 2 habis dibagi 9.
Nilai a yang memenuhi hanya 3
Jadi, a = 3 dan b = 2. sehingga bilangan terseebut adalah 36792
4|AMC 2009 UNLAM
12. A bag contains a mix of red and blue marbles. If one red marble is removed, then one-seventh of
the remaining marbles are red. If two blue marbles are removed instead of one red, then onefifth of the remaining marbles are red. How many marbles were in the bag originally?
Answer: 22
Solution:
Let r be the initial number of red marbles in the bag, and let n be the total number of marbles the
bag contained initially. Then
r 1 1

n 1 7
and
r
1

n2 5
This will give us a system of equations:
7r - 7 = n - 1
5r = n - 2
or
7r - n = 6
5r - n = -2
Solving this system will give us r = 4 and n = 22.
13. Jika a3- a – 1 = 0 maka a4 + a3 – a2- 2a + 1 = . . .
Jawaban : 2
Solusi:
Diketahui : a3- a – 1 = 0 .....................................................(1)
Kita lihat yang ditanyakan mengandung a4. Agar persamaan (1) mengandung a4, maka haruslah
persamaan tersebut dikali dengan a sehingga didapat:
a4- a2 – a = 0 ....................................................................(2)
agar persamaan (2) mengandung a3 , jumlahkan (2) dan (1)
(a4- a2 – a ) + (a3- a – 1 ) = 0
a4 + a 3 – a2 - 2 a – 1 = 0
a4 + a 3 – a2 - 2 a + 1- 2 = 0
a4 + a 3 – a2 - 2 a +1 = 2
14. An office building has 50 storeys, 25 of which are painted black and the other 25 of which are
painted gold. If the number of gold storeys in the top half of the building is added to the
number of black storeys in the bottom half of the building, the sum is 28. How many gold
storeys are there in the top half of the building?
Answer : 14
Solution :
Let G and g be the number of gold storeys in the top and bottom halfs of the building, and
5|AMC 2009 UNLAM
B and b the number of black storeys in the top and bottom halfs of the building.
Then G + B = 25 and g + b = 25, looking at the top and bottom halfs of the building.
Also, G + g = 25 and B + b = 25, since 25 of the storeys are painted in each colour.
Also, G + b = 28 from the given information, or b = 28 − G.
Since B + b = 25, then B + 28 − G = 25, so B = G − 3.
Since G + B = 25, then G + G − 3 = 25 or 2G = 28 or G = 14.
Thus, there are 14 gold storeys in the top half of the building.
15.
Untuk n anggota bilangan asli.
Jawaban :
2n
n 1
Solusi:
1
1
1
1
1
1
1
1
+
+ 1  2  3 + .... +
=
+
+
+ ....+ 1 n(n  1)
1 2
123
n
3
6
1
1
2
=2(
1
1
1 1



+ 6
12
n
(
n

1
))
2
1 1 1
1
  


= 2 (1
)
.
22
.
33
.
4
n
(
n

1
)
Karena penyebutnya tidak membentuk derat geometri, maka kita upayakan agar saling
menghilangkan
2(
1
1
1
1
1
1 11
1
2n








 ) = 2 (1
)=
1
2
2
3
3
4 n
n

1
n 1
n 1
16. Determine the value of 102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12.
Answer : 55
Solution :
Solution 1
Using differences of squares,
102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12
= (10 − 9)(10 + 9) + (8 − 7)(8 + 7) + (6 − 5)(6 + 5) + (4 − 3)(4 + 3) + (2 − 1)(2 + 1)
= 1(10 + 9) + 1(8 + 7) + 1(6 + 5) + 1(4 + 3) + 1(2 + 1)
= 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= 55
(We can get the answer 55 either by computing the sum directly, or by using the fact that
1
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 2(10)(11) = 55.)
Solution 2
Computing directly,
6|AMC 2009 UNLAM
102 − 92 + 82 − 72 + 62 − 52 + 42 − 32 + 22 − 12
= 100 − 81 + 64 − 49 + 36 − 25 + 16 − 9 + 4 − 1
= 19 + 15 + 11 + 7 + 3 (computing difference of each pair)
= 55
17. Bilangan real 2,525252... adalah bilangan rasional, sehingga dapat ditulis dalam bentuk
m
,
n
dimana m, n bilangan bulat,n ≠ 0. Jika dipilih m dan n yang relatif prima, berapakah m + n?
Jawaban : 349
Solusi:
Misalkan
A
=
2,525252....
100A = 252,525252....
-
- 99A = -250
A=
250
m
=
99
n
Maka, m = 250 dan n = 99
Jadi, m + n = 250 + 99 = 349
18. Rectangle ABCD intersects a circle at points
E, F,G, and H, as shown. If AH = 4, HG = 5
and BE = 3, determine the length of EF!
Answer : 7
Solution :
Let O be the centre of the circle and join O to the midpoints M and N of EF and HG, respectively.
Since O is the centre, then O, M and N are collinear and ON is perpendicular to both EF and HG.
1
5
5
Since N is the midpoint of HG, then HN = 2 HG = 2 , so AN = AH + HN = 4 + 2 =
13
2
.
Since ON is perpendicular to AN and BM, and AN and BM are parallel, then ANMB is a
rectangle.
Therefore, BM = AN =
13
2
.
Thus,
13
7
EF = 2(EM) = 2(BM − BE) = 2( 2 − 3) = 2(2 ) = 7 .
7|AMC 2009 UNLAM
19. Tinjau persamaan yang berbentuk x2 + bx + c = 0. berapa banyakkah persamaan demikian yang
memiliki akar-akar real jika koefisien b dan c hanya boleh dipilih dari himpunan { 1, 2, 3, 4, 5, 6}?
Jawaban : 19 persamaan
Solusi:
Akar real  D  0
b2 – 4ac > 0
b2 – 4c > 0
b2 > 4c
untuk c = 1, maka b = 2,3,4,5,6 ( 5 persamaan)
c = 2, maka b = 3,4,5,6 ( 4 persamaan)
c = 3, maka b = 4,5,6 ( 3 persamaan)
c = 4, maka b = 4,5,6 ( 3 persamaan)
c = 5, maka b = 5,6 ( 2 persamaan)
c = 5, maka b = 5,6 ( 2 persamaan)
c = 6, maka b = 5,6 ( 2 persamaan)
Jadi, banyaknya persamaan yang memiliki akar-akar persamaan real
= (5 + 4 + 3 + 3+ 2 + 2) = 19 persamaan
20. A quarter circle is inscribed in a square. A rectangle with dimensions x and 2x (as shown) is
inscribed in the region of the square outside of the circle, touching the circle, as shown. What is
the radius of the circle if the diagonal of the rectangle equals 15 units?
Answer : 15 5
Solution :
First we will find x using Pythagorean theorem: x2 + 4x2 = 225, or x2 = 45, or x = 3
5 . Then, using Pythagorean theorem again, r 2 = (r – x)2 + (r - 2x)2 = 2r 2 - 6xr +
5x2. Plugging in x = 3 5 , we get 0 = r 2 - 18 5 r + 225. Using quadratic formula,
18
5

1620

900
r


9 5 6 5
2
Obviously the solution r = 3 5 = x does not fit the described situation (the quarter circle is
inscribed inside of the rectangle), so the correct solution must be r = 15 5
8|AMC 2009 UNLAM
CADANGAN
1. Segitiga ABC siku-siku di C dan AC = 15. garis tinggi CH membagi AB dalam segmen AH dan HB
dengan HB = 16. Tentukan luas ∆ ABC?
Jawaban : 150
Solusi:
Misalkan AH = x, BC = y dan CH = t
Perhatikan ∆ ABC:
(x + 16)2= y2 + 152
y2 = (x + 16)2 – 15............... (1)
Perhatikan ∆ AHC
t2 = 152 – x2 ........................ (2)
Perhatikan ∆ BHC
t2 = y2 -162 .......................... (3)
Pers. (2) dan (3)
y2 -162 = 152 – x2 ...............(4)
(x + 16)2 - 152 - 162 = 152 – x2
Subs. (1) ke (4) :
x2 + 32x + 162 – 152 - 162 = 152 – x2
2x2 + 32x – 450 = 0
x2 + 16x -225 = 0
(x + 25) (x – 9) = 0
x1 = - 25 ( tidak memenuhi)
x2 = 9
t2 = 152 – 92 = 225 – 81
Dari (1) didapat
t2 = 144 maka t = 12
Jadi, L∆ ABC
=
1
. (x + 16) t
2
=
1
(9 + 16) . 12 = 150
2
2. Jika a, b dan c adalah akar persamaan kubik x3 + 2x2 – 3x – 5 = 0. Tentukan persamaan kubik yang
ketiga akarnya
1 1
1
, dan
a b
c
Jawaban : 5x3 + 3x2 - 2x – 1 = 0
Solusi :
x3 + 2x2 – 3x – 5 = 0
a + b + c = (-1)1 .
a2
= -2
a3
9|AMC 2009 UNLAM
ab + ac + bc = (-1)2 .
abc = (-1)3 .
a1
= -3
a3
a0
=5
a3

ac

ab
1
11bc
3




abc
abc
5

b

a 2
1 1 1c
 


ab
ac
bc
abc
5
Persamaan kubik yang akar-akarnya
1 1
1
, dan adalah:
a b
c
 1

 1

 1

x

x

x


0






 a

 b

 c

1
1 1
1
1
1
1

 )x x3 – ( a  b  c)x2 + (
+0
ab ac bc abc
x3 – ( 
3
2
1
)x2 + (  )x =0
5
5
5
5x3 + 3x2 - 2x – 1 = 0
3. Andi’s house as far as 1
1
1
km from his office, if the his mean walking is 4 km/hour, how many
2
2
time was needed to coming and go in a week if he working at Monday to Saturday and never
lunch at home in work’s day?
Jawaban: 3 hours
Solution:
1
Time for a walking = 2
1
4
2
1

Sum of time that needed =
1
hour
4
1
x 2 x 6  3 hours
4
4. k adalah bilangan bulat positif yang memenuhi 36 + k, 300 + k, 596 + k adalah kuadrat dari tiga
bilangan yang membentuk barisan aritmatika. Tentukan k !
Jawaban : 925
Solusi :
Misal ketiga barisan aitmatika tersebut adalah a-b, a, a + b.
Kuadratnya adalah (a - b)2, a2 ,(a + b)2 .
a2 + b2 - 2ab = 36 + k, a2 = 300 + k, dan a2 + b2 + 2ab = 596 + k
a2 - (a2 + b2 - 2ab )= 300 + k - (36 + k) = 264
10 | A M C 2 0 0 9 U N L A M
b(2a - b) = 264
......(1)
a2 + b2 + 2ab – a2 = 596 + k - (300 + k)
b (2a + b) = 296
.... (2)
296(2a - b) = 264(2a + b)
592a - 296b = 528a + 264b
64a = 560b
4a = 35b
Dari persamaan (1) didapat
b(4a - 2b) = 528
b=±4
a = ± 35
(a - b)2 = 312 = 36 + k
k = 925
5. Mary and John wrote 100 numbers each. Mary’s sequence starts with 5, 8, 11, 14, … , John’s
sequence starts with 3, 7, 11, 15, … . How many common numbers are there in both sequences?
Answer : 25
Solution :
Mary’s sequence is 5+3k1, 0  k1  99, John’s sequence is 3+4k2, 0  k2  99. We need to find
when 5 +3k1 = 3 + 4k2 for 0  k1, k2  99.
5 +3k1 = 3 + 4k2
2 +3k1 = 4k2
23k1
k2
4
2k1
k1
k2
4
2k1
k2 k1
4
2  k1
must be an integer. Then k1 = - 4t + 2 and k2 = k1 + t = -3t+2. Since 0
4
which means that t =
 k1, k2  99, we have
0  -4t + 2  99
1
2
1
 t   24
4
0  -3t+2  99
2
3
1
 -3t   32
3
There ate 25 integers t that satisfy both of the compound inequalities: t = -24,-23,…,-1, 0.
Therefore there are 25 common numbers in both sequences.
11 | A M C 2 0 0 9 U N L A M
6. Let S1,S2,S3, . . . be a sequence of nested circles such that Si+1 is inside Si and the radius of Si+1 is
half of that of Si . Let r denote the radius of S1. For i = 1, 3, 5, . . . (i.e. for odd i ’s), the region
inside Si and outside Si+1 is shaded. Find the sum of the shaded areas.
4r 2
Answer:
5
Solution:
The circle Sk has radius
r
2 k 1
and its area is
r 2
2 2 k 1
.
Thus
the
shaded
region has area equal to the sum of the infinite series






4
2
2


 l
 l




1

1

1

2
2
r

r

r

1
 k
2
2

k

1

k

1
2
l
l

0
l

0
Clearly, the last infinite sum is a geometric series with first term1 and ratio -1/4; hence the
shaded area converges to
r2
1
4r2

1
5
1
4
7. The square ABCD, with a side a, is inscribed in a circle. Outside of each side of the square, there
are semicircles as shown. What is the area of the shaded region?
Answer: a2
Solution:
Shaded area = A1 – A2 =
 

2
2
2


 2a
1
a
2
a
a


2



a

4


a



a






2
2
2
22





2
2
8. A circular table has exactly 60 chairs around it. There are N people seated around the table. The
next person coming to the table will have to be seated next to an occupied seat. Find the smallest
possible value of N.
Answer : 20
Solution:
For the next person to have to sit next to an occupied seat, there cannot be three consecutive
chairs currently unoccupied (otherwise the next person would simply sit in the middle of the
three empty chairs). Therefore for every three consecutive chairs at least one of them has to be
occupied. Since we are looking for the smallest N, exactly one of the three will have to be
occupied, and each two people will have to have two empty seats between them. Therefore the
number of people sitting at the table is 1= 3 of the number of seats, or 20 people.
12 | A M C 2 0 0 9 U N L A M
9. Sebuah surat selebaran AMC’09 memuat 50 cm persegi bahan cetak. Jalur bebas cetak di atas dan
di bawah selebar 4 cm dan disamping kiri dan kanan selebar 2 cm. Berapakah ukuran surat
selebaran tersebut yang memerlukan kertas seminimum mungkin?
Jawaban : 9cm kali 18cm
Solusi :
A = xy
Kita bermaksud meminimumkan A.
Ukuran bahan cetakan adalah x – 4 dan y – 8 dan luasnya adalah 50 cm2 sehingga:
(x – 4)(y – 8) = 50 dan diperoleh y  50  8
x4
Dengan memasukkan nilai y ke dalam A, maka:
A
50 x
 8x
x4
Nilai x berada pada selang 4 < x < ∞
8 x 2  64 x  72 8( x  1)( x  9) = 0
dA
( x  4)50  50 x


8


dx
( x  4) 2
( x  4) 2
( x  4) 2
Titik kritis : x =9 dan x = -1
x= -1 tidak berada dalam selang (4,∞) maka digunakan x = 9
dA
dA
 0 untuk x dalam (4,9) dan
 0 untuk x dalam (9, )
dx
dx
Maka nilai minimum untuk x adalah 9, sehingga nilai minimum y adalah 18
Jadi, ukuran surat edaran yang akan memakai paling sedikit kertas adalah 9 cm kali 18 cm.
13 | A M C 2 0 0 9 U N L A M

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SECTION B (60 marks)

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