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Problems
Contests
National and Regional Contests
Iran Contests
Iran MO (2nd Round)
2009 Iran MO (2nd Round)
2009 Iran MO (2nd Round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
2
2
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Neighbors in the garden - Iran NMO 2009 - Problem 2
In some of the
1
×
1
1\times1
1
×
1
squares of a square garden
50
×
50
50\times50
50
×
50
we've grown apple, pomegranate and peach trees (At most one tree in each square). We call a
1
×
1
1\times1
1
×
1
square a room and call two rooms neighbor if they have one common side. We know that a pomegranate tree has at least one apple neighbor room and a peach tree has at least one apple neighbor room and one pomegranate neighbor room. We also know that an empty room (a room in which there’s no trees) has at least one apple neighbor room and one pomegranate neighbor room and one peach neighbor room. Prove that the number of empty rooms is not greater than
1000.
1000.
1000.
Prove the inequality - Iran NMO 2009 - Problem 5
Let
a
1
<
a
2
<
⋯
<
a
n
a_1<a_2<\cdots<a_n
a
1
<
a
2
<
⋯
<
a
n
be positive integers such that for every distinct
1
≤
i
,
j
≤
n
1\leq{i,j}\leq{n}
1
≤
i
,
j
≤
n
we have
a
j
−
a
i
a_j-a_i
a
j
−
a
i
divides
a
i
a_i
a
i
. Prove that
i
a
j
≤
j
a
i
for
1
≤
i
<
j
≤
n
ia_j\leq{ja_i} \qquad \text{ for } 1\leq{i}<j\leq{n}
i
a
j
≤
j
a
i
for
1
≤
i
<
j
≤
n
3
2
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Geometric Inequality on Angles - Iran NMO 2009 - Problem 3
Let
A
B
C
ABC
A
BC
be a triangle and the point
D
D
D
is on the segment
B
C
BC
BC
such that
A
D
AD
A
D
is the interior bisector of
∠
A
\angle A
∠
A
. We stretch
A
D
AD
A
D
such that it meets the circumcircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
at
M
M
M
. We draw a line from
D
D
D
such that it meets the lines
M
B
,
M
C
MB,MC
MB
,
MC
at
P
,
Q
P,Q
P
,
Q
, respectively (
M
M
M
is not between
B
,
P
B,P
B
,
P
and also is not between
C
,
Q
C,Q
C
,
Q
). Prove that
∠
P
A
Q
≥
∠
B
A
C
\angle PAQ\geq\angle BAC
∠
P
A
Q
≥
∠
B
A
C
.
Cards - Iran NMO 2009 - Problem 6
11
11
11
people are sitting around a circle table, orderly (means that the distance between two adjacent persons is equal to others) and
11
11
11
cards with numbers
1
1
1
to
11
11
11
are given to them. Some may have no card and some may have more than
1
1
1
card. In each round, one [and only one] can give one of his cards with number
i
i
i
to his adjacent person if after and before the round, the locations of the cards with numbers
i
−
1
,
i
,
i
+
1
i-1,i,i+1
i
−
1
,
i
,
i
+
1
don’t make an acute-angled triangle. (Card with number
0
0
0
means the card with number
11
11
11
and card with number
12
12
12
means the card with number
1
1
1
!) Suppose that the cards are given to the persons regularly clockwise. (Mean that the number of the cards in the clockwise direction is increasing.) Prove that the cards can’t be gathered at one person.
1
2
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Polynomials - Iran NMO 2009 - Problem 1
Let
p
(
x
)
p(x)
p
(
x
)
be a quadratic polynomial for which :
∣
p
(
x
)
∣
≤
1
∀
x
∈
{
−
1
,
0
,
1
}
|p(x)| \leq 1 \qquad \forall x \in \{-1,0,1\}
∣
p
(
x
)
∣
≤
1
∀
x
∈
{
−
1
,
0
,
1
}
Prove that:
∣
p
(
x
)
∣
≤
5
4
∀
x
∈
[
−
1
,
1
]
\ |p(x)|\leq\frac{5}{4} \qquad \forall x \in [-1,1]
∣
p
(
x
)
∣
≤
4
5
∀
x
∈
[
−
1
,
1
]
Soldiers - Iran NMO 2009 - Problem 4
We have a
(
n
+
2
)
×
n
(n+2)\times n
(
n
+
2
)
×
n
rectangle and we’ve divided it into
n
(
n
+
2
)
1
×
1
n(n+2) \ \ 1\times1
n
(
n
+
2
)
1
×
1
squares.
n
(
n
+
2
)
n(n+2)
n
(
n
+
2
)
soldiers are standing on the intersection points (
n
+
2
n+2
n
+
2
rows and
n
n
n
columns). The commander shouts and each soldier stands on its own location or gaits one step to north, west, east or south so that he stands on an adjacent intersection point. After the shout, we see that the soldiers are standing on the intersection points of a
n
×
(
n
+
2
)
n\times(n+2)
n
×
(
n
+
2
)
rectangle (
n
n
n
rows and
n
+
2
n+2
n
+
2
columns) such that the first and last row are deleted and 2 columns are added to the right and left (To the left
1
1
1
and
1
1
1
to the right). Prove that
n
n
n
is even.