MathDB
Problems
Contests
National and Regional Contests
Indonesia Contests
Indonesia Regional
2010 Indonesia Regional
2010 Indonesia Regional
Part of
Indonesia Regional
Subcontests
(1)
2
Hide problems
Indonesia Regional MO 2010 Part A 20 problems 90' , answer only
Indonesia Regional also know as provincial level, is a qualifying round for National Math Olympiad Year 2010 [hide=Part A]Part B consists of 5 essay / proof problems, posted [url=https://artofproblemsolving.com/community/c4h2685201p23295229]hereTime: 90 minutes
∙
\bullet
∙
Write only the answers to the questions given.
∙
\bullet
∙
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.
∙
\bullet
∙
Each question is worth 1 (one) point.
∙
∙
\bullet \bullet
∙
∙
to be more exact:
⊳
\rhd
⊳
in years 2002-08 time was 90' for part A and 120' for part B
⊳
\rhd
⊳
since years 2009 time is 210' for part A and B totally
⊳
\rhd
⊳
each problem in part A is 1 point, in part B is 7 points p1. Calculate
∑
j
=
0
n
(
(
n
j
)
(
∑
i
=
0
n
(
j
i
)
8
i
)
)
\sum_{j=0}^{n}\left( {n \choose j}\left( \sum_{i=0}^{n} {j \choose i}8^i\right)\right)
j
=
0
∑
n
(
(
j
n
)
(
i
=
0
∑
n
(
i
j
)
8
i
)
)
p2. In triangle ABC, let
a
a
a
,
b
b
b
, and
c
c
c
be the side lengths of
B
C
BC
BC
,
C
A
CA
C
A
, and
A
B
AB
A
B
, respectively. If
2
a
tan
A
=
b
tan
B
\frac{2a}{\tan A}=\frac{b}{\tan B}
t
a
n
A
2
a
=
t
a
n
B
b
then the value of
sin
2
A
−
sin
2
B
cos
2
A
+
cos
2
B
\frac{\sin^2 A-\sin^2 B}{\cos^2 A+\cos^2 B}
c
o
s
2
A
+
c
o
s
2
B
s
i
n
2
A
−
s
i
n
2
B
is ...p3. Given a polynomial
P
(
x
)
=
x
4
+
a
x
3
+
b
x
2
+
c
x
+
d
P(x) = x^4 + ax^3 + bx^2 + cx + d
P
(
x
)
=
x
4
+
a
x
3
+
b
x
2
+
c
x
+
d
with
a
,
b
,
c
a, b, c
a
,
b
,
c
, and
d
d
d
constants. If
P
(
1
)
=
10
P(1) = 10
P
(
1
)
=
10
,
P
(
2
)
=
20
P(2) = 20
P
(
2
)
=
20
, and
P
(
3
)
=
30
P(3) = 30
P
(
3
)
=
30
, then the value of
P
(
12
)
+
P
(
−
8
)
10
\frac{P(12)+P(-8)}{10}
10
P
(
12
)
+
P
(
−
8
)
is ...p4. Let
S
=
{
1
,
2
,
3
,
4
,
5
}
S = \{1, 2, 3, 4, 5\}
S
=
{
1
,
2
,
3
,
4
,
5
}
. The number of functions
f
:
S
→
S
f : S\to S
f
:
S
→
S
that satisfies f(f(x)) = x for all
x
∈
S
x\in S
x
∈
S
is ..p5. If
a
,
b
a, b
a
,
b
, and
c
c
c
represent the lengths of the sides of a triangle that satisfies
(
a
+
b
+
c
)
(
a
+
b
−
c
)
=
3
a
b
(a + b + c)(a + b-c) = 3ab
(
a
+
b
+
c
)
(
a
+
b
−
c
)
=
3
ab
, then the measure of the angle opposite the side of length
c
c
c
is ...p6. The number of six-digit numberz
a
b
c
d
e
f
‾
\overline{abcdef}
ab
c
d
e
f
with
a
>
b
>
c
≥
d
>
e
>
f
a > b > c\ge d > e > f
a
>
b
>
c
≥
d
>
e
>
f
is ...p7. The prime number
p
p
p
so that
p
2
+
73
p^2 + 73
p
2
+
73
is a perfect cube is ...p8. Given triangle
A
B
C
ABC
A
BC
is right-angled at
C
C
C
,
A
C
=
3
AC = 3
A
C
=
3
, and
B
C
=
4
BC = 4
BC
=
4
. Triangle
A
B
D
ABD
A
B
D
is right-angled at
A
A
A
,
A
D
=
12
AD = 12
A
D
=
12
and points
C
C
C
and
D
D
D
are opposite to side
A
B
AB
A
B
. Parallel line
A
C
AC
A
C
through
D
D
D
cut the extension of
C
B
CB
CB
at
E
E
E
. If
D
E
D
B
=
m
n
\frac{DE}{DB}=\frac{m}{n}
D
B
D
E
=
n
m
where
m
m
m
and
n
n
n
are relatively prime positive integers, then
m
+
n
=
.
.
.
m + n = ...
m
+
n
=
...
p9. On a circle there are
12
12
12
distinct points. By using these
12
12
12
points we will make
6
6
6
non-intersecting chord. There are ... ways to do it.p10. The number of members of the set
S
=
{
g
c
d
(
n
3
+
1
,
n
2
+
3
n
+
9
)
∣
n
∈
Z
}
S = \{gcd (n^3 + 1, n^2 + 3n + 9)|n\in Z\}
S
=
{
g
c
d
(
n
3
+
1
,
n
2
+
3
n
+
9
)
∣
n
∈
Z
}
is ...p11. The quadratic equation
x
2
−
p
x
−
2
p
=
0
x^2-px-2p = 0
x
2
−
p
x
−
2
p
=
0
has two real roots
a
a
a
and
b
b
b
. If
a
3
+
b
3
=
16
a^3 + b^3 = 16
a
3
+
b
3
=
16
, then the sum of satisfying values of
p
p
p
is ...p12. In a plane, there are
n
n
n
points with coordinates a pair of integers. The smallest value of
n
n
n
so that there are two points whose midpoints also have both coordinate integer pairs is ...p13. The natural number
n
n
n
such that the equation
x
[
1
x
]
+
1
x
[
x
]
=
n
n
+
1
x\left[\frac{1}{x}\right]+\frac{1}{x} \left[x\right]=\frac{n}{n+1}
x
[
x
1
]
+
x
1
[
x
]
=
n
+
1
n
has exactly
2010
2010
2010
positive real solution is ⋅⋅⋅⋅⋅⋅Note: For any real number
x
x
x
is defined
[
a
]
[a]
[
a
]
as the largest integer less than or equal to with
x
x
x
. p14. Two circles (not equally large) intersect on the outside. Points
A
A
A
and
A
1
A_1
A
1
are located on the small circle, while
B
B
B
and
B
1
B_1
B
1
are on the large circle. The lines
P
A
B
PAB
P
A
B
and
P
A
1
B
1
PA_1B_1
P
A
1
B
1
, are common tangent lines of the two circles. If
P
A
=
A
B
=
4
PA = AB = 4
P
A
=
A
B
=
4
, then the area of the small circle is ...p15. Twenty -seven students in a class will be made into six discussion groups each consisting of four or five σtudents. The number of ways is ...p16. Someone wrote a chain letter to
6
6
6
people. The recipient of this letter is instructed to sent letters to
6
6
6
other people. All recipients of the letter read the contents of the letter and then some people carry out the orders written in the letter, the rest do not continue the chain letter this. If there are
366
366
366
people who do not continue this chain letter, then the number of people that resides in this chain mail system is ...p17. The sum of the constant terms of
(
x
5
−
2
x
3
)
8
\left(x^5 - \frac{2}{x^3}\right)^8
(
x
5
−
x
3
2
)
8
is ...p18. The number of positive integers
n
<
100
n < 100
n
<
100
, so the equation
3
x
y
−
1
x
+
y
=
n
\frac{3xy-1}{x+y}=n
x
+
y
3
x
y
−
1
=
n
has a solution pair of integers
(
x
,
y
)
(x, y)
(
x
,
y
)
is ...p19. It is known that
x
,
y
x, y
x
,
y
, and
z
z
z
are real numbers that satisfy the system of equations:
x
+
y
+
z
=
1
x
+
1
y
+
1
z
x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
x
+
y
+
z
=
x
1
+
y
1
+
z
1
x
y
z
=
1
xyz = 1
x
yz
=
1
The smallest value of
∣
x
+
y
+
z
∣
|x + y + z|
∣
x
+
y
+
z
∣
is ...p20. Triangle
A
B
C
ABC
A
BC
has side lengths
B
C
=
5
BC = 5
BC
=
5
,
A
C
=
12
AC = 12
A
C
=
12
, and
A
B
=
13
AB = 13
A
B
=
13
. Point
D
D
D
is on
A
B
AB
A
B
and point
E
E
E
on
A
C
AC
A
C
. If
D
E
DE
D
E
divides triangle ABC into two equal parts, then the minimum length of
D
E
DE
D
E
is ...
Indonesia Regional MO 2010 Part B
[url=https://artofproblemsolving.com/community/c6h2371620p19389503]p1. Given triangle
A
B
C
ABC
A
BC
. Suppose
P
P
P
and
P
1
P_1
P
1
are points on
B
C
,
Q
BC, Q
BC
,
Q
lies on
C
A
,
R
CA, R
C
A
,
R
lies on
A
B
AB
A
B
, such that
A
R
R
B
=
B
P
P
C
=
C
Q
Q
A
=
C
P
1
P
1
B
\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP_1}{P_1B}
RB
A
R
=
PC
BP
=
Q
A
CQ
=
P
1
B
C
P
1
Let
G
G
G
be the centroid of triangle
A
B
C
ABC
A
BC
and
K
=
A
P
1
∩
R
Q
K = AP_1 \cap RQ
K
=
A
P
1
∩
RQ
. Prove that points
P
,
G
P,G
P
,
G
, and
K
K
K
are collinear.p2. It is known that
k
k
k
is the largest positive integer, such that we can find the integer
n
n
n
positive prime numbers (not necessarily different)
q
1
,
q
2
,
q
3
,
.
.
.
,
q
k
q_1, q_2, q_3,... , q_k
q
1
,
q
2
,
q
3
,
...
,
q
k
, and different prime numbers
p
1
,
p
2
,
p
3
,
.
.
.
,
p
k
p_1, p_2,p_3, ..., p_k
p
1
,
p
2
,
p
3
,
...
,
p
k
that satisfy
1
p
1
+
1
p
2
+
.
.
.
+
1
p
k
=
7
+
n
q
1
g
2
⋅
⋅
⋅
q
k
2010
\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_k}=\frac{7+nq_1g_2\cdot\cdot\cdot q_k}{2010}
p
1
1
+
p
2
1
+
...
+
p
k
1
=
2010
7
+
n
q
1
g
2
⋅
⋅
⋅
q
k
Determine the number of
n
n
n
that satisfy.p3. Determine the values of
k
k
k
and
d
d
d
so that no pair of real numbers
(
x
,
y
)
(x, y)
(
x
,
y
)
satisfies the system of equations:
x
3
+
y
3
=
2
x^3 + y^3 = 2
x
3
+
y
3
=
2
y
=
k
x
+
d
y = kx + d
y
=
k
x
+
d
p4. It is known that n is a natural multiple of
2010
2010
2010
. Show that the equation
x
+
2
y
+
3
z
=
2
n
x + 2y + 3z = 2n
x
+
2
y
+
3
z
=
2
n
has exactly
1
+
n
2
+
n
2
12
1+ \frac{n}{2}+ \frac{n^2}{12}
1
+
2
n
+
12
n
2
solution of triples
(
x
,
y
,
z
)
(x, y, z)
(
x
,
y
,
z
)
where
x
x
x
,
y
y
y
, and
z
z
z
are not negative integers .p5. Given a chessboard as shown in the picture. Can a horse chess piece depart from one tile passes through every other tile only once and returns to its original place ? Explain your answer! https://cdn.artofproblemsolving.com/attachments/6/3/0bd6b71a0c09cd3aec2f49b31ff5b4d141de03.png Explanation: The horse chess move is
L
L
L
-shaped, that is, from the original box: (a)
2
2
2
squares to the right/left and
1
1
1
box to the front/back; or (b)
2
2
2
squares to the front/back and
1
1
1
box to the right/left.