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2011 Indonesia Regional
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Indonesia Regional MO 2011 Part A
Indonesia Regional also know as provincial level, is a qualifying round for National Math Olympiad Year 2011 [hide=Part A]Part B consists of 5 essay / proof problems, posted [url=https://artofproblemsolving.com/community/c4h2685237p23295278]hereTime: 90 minutes
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Write only the answers to the questions given.
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Some questions can have more than one correct answer. You are asked to provide the most correct or exact answer to a question like this. Scores will only be given to the giver of the most correct or most exact answer.
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Each question is worth 1 (one) point.
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to be more exact:
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in years 2002-08 time was 90' for part A and 120' for part B
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since years 2009 time is 210' for part A and B totally
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each problem in part A is 1 point, in part B is 7 points p1. Given an isosceles triangle
A
B
C
ABC
A
BC
with
A
B
=
A
C
AB = AC
A
B
=
A
C
. Let the bisector of the angle
A
B
C
ABC
A
BC
intersect
A
C
AC
A
C
at point
D
D
D
so
B
C
=
B
D
+
A
D
BC = BD+AD
BC
=
B
D
+
A
D
. The measure of angle
∠
C
A
B
\angle CAB
∠
C
A
B
is ... p2. If
n
n
n
is natural and
1
2
+
1
3
+
1
5
+
1
n
\frac12 + \frac13 + \frac15+ \frac{1}{n}
2
1
+
3
1
+
5
1
+
n
1
is an integer, then the positive divisor of
n
n
n
is as much as ... p3. If
a
≥
b
>
1
a\ge b > 1
a
≥
b
>
1
, then the largest possible value for
a
log
(
a
b
)
+
b
log
(
b
a
)
^a \log_{\left( \frac{a}{b}\right)}+^b \log_{\left( \frac{b}{a}\right)}
a
lo
g
(
b
a
)
+
b
lo
g
(
a
b
)
is ... p4. Given the quadrilateral
A
B
C
D
ABCD
A
BC
D
. All points
A
,
B
,
C
A, B, C
A
,
B
,
C
and
D
D
D
will be numbered
1
,
2
,
3
,
4
,
5
1, 2, 3, 4, 5
1
,
2
,
3
,
4
,
5
or
6
6
6
so that every two points that lie on one side of the number
4
4
4
are different. The number of ways of numbering in this way there are as many as ... p5. Given a function
f
f
f
with
f
(
x
)
=
a
x
2
+
x
f(x) = \sqrt{ax^2 + x}
f
(
x
)
=
a
x
2
+
x
. All possible values of
a
a
a
such that the domain and the range of
f
f
f
are the same, are ... p6. The number of different natural numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
and
d
d
d
that are less than
10
10
10
and satisfies the equation
a
+
b
=
c
+
d
a + b = c + d
a
+
b
=
c
+
d
are as many as ... p7. If the two roots of the equation
x
2
−
2013
x
+
k
=
0
x^2-2013x + k = 0
x
2
−
2013
x
+
k
=
0
are prime numbers, then the possible value of the
k
k
k
is ... p8. If
(
1
−
tan
2
x
2
2011
)
(
1
−
tan
2
x
2
2010
)
.
.
.
(
1
−
tan
2
x
2
)
=
2
2011
3
tan
x
2
2011
\left(1- \tan^2\frac{x}{2^{2011}}\right) \left(1- \tan^2\frac{x}{2^{2010}}\right) ... \left(1- \tan^2\frac{x}{2}\right)=2^{2011}\sqrt3 \tan\frac{x}{2^{2011}}
(
1
−
tan
2
2
2011
x
)
(
1
−
tan
2
2
2010
x
)
...
(
1
−
tan
2
2
x
)
=
2
2011
3
tan
2
2011
x
then
sin
2
x
\sin 2x
sin
2
x
is ... p9. In Cartesian space, we want to move from point
(
2
,
0
,
11
)
(2, 0, 11)
(
2
,
0
,
11
)
to point
(
20
,
1
,
1
)
(20, 1, 1)
(
20
,
1
,
1
)
always at coordinates
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
where at least two of
x
,
y
x, y
x
,
y
and
z
z
z
are integer numbers and the shortest path. How many ways are there to do this? ... p10. Let
x
,
y
x, y
x
,
y
and
z
z
z
be positive real numbers with the property
x
y
z
=
1
xyz = 1
x
yz
=
1
. The smallest value of
(
x
+
2
y
)
(
y
+
2
z
)
(
x
z
+
1
)
(x + 2y)(y + 2z)(xz + 1)
(
x
+
2
y
)
(
y
+
2
z
)
(
x
z
+
1
)
is reached when when
x
+
y
+
z
x + y + z
x
+
y
+
z
is ... p11. In the figure below, the lengths of
A
E
=
x
AE = x
A
E
=
x
,
E
C
=
y
EC = y
EC
=
y
and
D
C
=
2
B
D
DC = 2BD
D
C
=
2
B
D
. The ratio of lengths of
B
F
BF
BF
and
F
E
FE
FE
expressed in terms of
x
x
x
and
y
y
y
is ... https://cdn.artofproblemsolving.com/attachments/f/3/29a3755bcd481159f144c0058c08a0b6c52a11.png p12. How may three-digit numbers are there such that all digits are different and the last digit is the sum of the other two digits ? ... p13. Given a sequence of rational numbers
{
a
k
}
k
∈
N
\{a_k\}_{k\in N}
{
a
k
}
k
∈
N
defined by
a
1
=
2
a_1 = 2
a
1
=
2
and
a
n
+
1
=
a
n
−
1
a
n
+
1
,
n
∈
N
.
a_{n+1} = \frac{a_n-1}{a_n + 1},\,\,\, n \in N.
a
n
+
1
=
a
n
+
1
a
n
−
1
,
n
∈
N
.
The value of
a
2011
a_{2011}
a
2011
is ... p14. Let
Γ
\Gamma
Γ
be the circumcircle of triangle
A
B
C
ABC
A
BC
. The chord
A
D
AD
A
D
is bisects
∠
B
A
C
\angle BAC
∠
B
A
C
and intersects intersects
B
C
BC
BC
at point
L
L
L
. The chord
D
K
DK
DK
is perpendicular to
A
C
AC
A
C
and intersects it at point
M
M
M
. If
B
L
B
C
=
12
\frac{BL}{BC} = 12
BC
B
L
=
12
, then the ratio
A
M
M
C
=
\frac{AM}{MC} =
MC
A
M
=
... p15. Two dice have numbers
1
1
1
through
6
6
6
that can be removed from the dice. The twelfth number is removed from the dice and put into a bag. One number is randomly drawn and placed on one of the two dice. After all the numbers are matched, both dice are thrown simultaneously. The probability of the number seven appearing as the sum of the numbers on the top of the second dice is ... p16. The number of natural numbers
n
n
n
such that each point with the coordinates of natural numbers that lies on the line
x
+
y
=
n
x+y = n
x
+
y
=
n
has a prime number distance from the point center
(
0
,
0
)
(0, 0)
(
0
,
0
)
is ... p17. The natural number
n
n
n
that satisfies
(
−
2004
)
n
−
190
0
n
+
2
5
n
−
12
1
n
(-2004)^n-1900^n + 25^n -121^n
(
−
2004
)
n
−
190
0
n
+
2
5
n
−
12
1
n
and divides
2000
2000
2000
is ... p18. Ten students sit in a row. All students get up and sit again in the row with the rules that each student can sit back on the the same chair or in the seat next to the old one. How many ways are there that all of the students can sit back in the row ? ... p19. The largest natural number
n
≤
123456
n\le 123456
n
≤
123456
so that there is a natural number x with the property the sum of all digits of
x
2
x^2
x
2
equals
n
n
n
is... p20. Suppose
A
B
C
ABC
A
BC
is a triangle and
P
P
P
is a point in the triangle. Let points
D
,
E
,
F
D, E, F
D
,
E
,
F
lie on sides
B
C
BC
BC
,
C
A
CA
C
A
,
A
B
AB
A
B
respectively such that
P
D
PD
P
D
is perpendicular to
B
C
BC
BC
,
P
E
PE
PE
is perpendicular to
C
A
CA
C
A
, and
P
F
PF
PF
is perpendicular to
A
B
AB
A
B
. If the triangle
D
E
F
DEF
D
EF
is equilateral and
∠
A
P
B
=
7
0
o
\angle APB = 70^o
∠
A
PB
=
7
0
o
, then
∠
A
C
B
=
\angle ACB =
∠
A
CB
=
..
Indonesia Regional MO 2011 Part B
p1. Determine all possible values of
k
k
k
so that there are no real number pairs
(
x
,
y
)
(x, y)
(
x
,
y
)
which satisfies the system of equations:
x
2
−
y
2
=
0
x^2-y^2 = 0
x
2
−
y
2
=
0
(
x
−
k
)
2
+
y
2
=
1
(x-k)^2 + y^2 = 1
(
x
−
k
)
2
+
y
2
=
1
p2. A number is said to be beautiful if it satisfies the following two conditions: (a) It is a perfect square, that is, the square of a natural number. (b) If the rightmost digit in the decimal writing is moved its position becomes the leftmost digit, then the number formed is still a perfect square .For example,
441
441
441
is a beautiful number consisting of
3
3
3
digits, because
441
=
2
1
2
441 =21^2
441
=
2
1
2
and
144
=
1
2
2
144 = 12^2
144
=
1
2
2
. While
144
144
144
is not a pretty number because
144
=
1
2
2
144 = 12^2
144
=
1
2
2
but
414
414
414
is not a perfect square. Prove that there are beautiful numbers whose decimal representation consists of exactly
2011
2011
2011
digits..p3. Let
A
A
A
be the set of all positive divisors of
1
0
9
10^9
1
0
9
. If two are chosen, any number
x
x
x
and
y
y
y
in
A
A
A
(may be the same), find the probability that
x
x
x
divides
y
y
y
.[url=https://artofproblemsolving.com/community/c6h2371634p19389707]p4. Given a rectangle
A
B
C
D
ABCD
A
BC
D
with
A
B
=
a
AB = a
A
B
=
a
and
B
C
=
b
BC = b
BC
=
b
. Point
O
O
O
is the intersection of the two diagonals. Extend the side of the
B
A
BA
B
A
so that
A
E
=
A
O
AE = AO
A
E
=
A
O
, also extend the diagonal of
B
D
BD
B
D
so that
B
Z
=
B
O
.
BZ = BO.
BZ
=
BO
.
Assume that triangle
E
Z
C
EZC
EZC
is equilateral. Prove that (i)
b
=
a
3
b = a\sqrt3
b
=
a
3
(ii)
E
O
EO
EO
is perpendicular to
Z
D
ZD
Z
D
p5. Let
M
M
M
be a subset of
{
1
,
2
,
3
,
.
.
.
,
12
,
13
}
\{1, 2, 3, ..., 12, 13\}
{
1
,
2
,
3
,
...
,
12
,
13
}
and there are no three member of
M
M
M
whose product is a perfect square. Specify the maximum number of possible
M
M
M
members.