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National and Regional Contests
Iran Contests
Iran MO (2nd Round)
2008 Iran MO (2nd Round)
2008 Iran MO (2nd Round)
Part of
Iran MO (2nd Round)
Subcontests
(3)
3
2
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Iran NMO 2008 (Second Round) - Problem6
In triangle
A
B
C
ABC
A
BC
,
H
H
H
is the foot of perpendicular from
A
A
A
to
B
C
BC
BC
.
O
O
O
is the circumcenter of
Δ
A
B
C
\Delta ABC
Δ
A
BC
.
T
,
T
′
T,T'
T
,
T
′
are the feet of perpendiculars from
H
H
H
to
A
B
,
A
C
AB,AC
A
B
,
A
C
, respectively. We know that
A
C
=
2
O
T
AC=2OT
A
C
=
2
OT
. Prove that
A
B
=
2
O
T
′
AB=2OT'
A
B
=
2
O
T
′
.
Iran NMO 2008 (Second Round) - Problem3
Let
a
,
b
,
c
,
a,b,c,
a
,
b
,
c
,
and
d
d
d
be real numbers such that at least one of
c
c
c
and
d
d
d
is non-zero. Let
f
:
R
→
R
f:\mathbb{R}\to\mathbb{R}
f
:
R
→
R
be a function defined as
f
(
x
)
=
a
x
+
b
c
x
+
d
f(x)=\frac{ax+b}{cx+d}
f
(
x
)
=
c
x
+
d
a
x
+
b
. Suppose that for all
x
∈
R
x\in\mathbb{R}
x
∈
R
, we have
f
(
x
)
≠
x
f(x) \neq x
f
(
x
)
=
x
. Prove that if there exists some real number
a
a
a
for which
f
1387
(
a
)
=
a
f^{1387}(a)=a
f
1387
(
a
)
=
a
, then for all
x
x
x
in the domain of
f
1387
f^{1387}
f
1387
, we have
f
1387
(
x
)
=
x
f^{1387}(x)=x
f
1387
(
x
)
=
x
. Notice that in this problem,
f
1387
(
x
)
=
f
(
f
(
⋯
(
f
(
x
)
)
)
⋯
)
⏟
1387 times
.
f^{1387}(x)=\underbrace{f(f(\cdots(f(x)))\cdots)}_{\text{1387 times}}.
f
1387
(
x
)
=
1387 times
f
(
f
(
⋯
(
f
(
x
)))
⋯
)
.
Hint. Prove that for every function
g
(
x
)
=
s
x
+
t
u
x
+
v
g(x)=\frac{sx+t}{ux+v}
g
(
x
)
=
ux
+
v
s
x
+
t
, if the equation
g
(
x
)
=
x
g(x)=x
g
(
x
)
=
x
has more than
2
2
2
roots, then
g
(
x
)
=
x
g(x)=x
g
(
x
)
=
x
for all
x
∈
R
−
{
−
v
u
}
x\in\mathbb{R}-\left\{\frac{-v}{u}\right\}
x
∈
R
−
{
u
−
v
}
.
2
2
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Phone Numbers - Iran NMO 2008 - Problem5
We want to choose telephone numbers for a city. The numbers have
10
10
10
digits and
0
0
0
isn’t used in the numbers. Our aim is: We don’t choose some numbers such that every
2
2
2
telephone numbers are different in more than one digit OR every
2
2
2
telephone numbers are different in a digit which is more than
1
1
1
. What is the maximum number of telephone numbers which can be chosen? In how many ways, can we choose the numbers in this maximum situation?
Iran NMO 2008 (Second Round) - Problem2
Let
I
a
I_a
I
a
be the
A
A
A
-excenter of
Δ
A
B
C
\Delta ABC
Δ
A
BC
and the
A
A
A
-excircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
be tangent to the lines
A
B
,
A
C
AB,AC
A
B
,
A
C
at
B
′
,
C
′
B',C'
B
′
,
C
′
, respectively.
I
a
B
,
I
a
C
I_aB,I_aC
I
a
B
,
I
a
C
meet
B
′
C
′
B'C'
B
′
C
′
at
P
,
Q
P,Q
P
,
Q
, respectively.
M
M
M
is the meet point of
B
Q
,
C
P
BQ,CP
BQ
,
CP
. Prove that the length of the perpendicular from
M
M
M
to
B
C
BC
BC
is equal to
r
r
r
where
r
r
r
is the radius of incircle of
Δ
A
B
C
\Delta ABC
Δ
A
BC
.
1
2
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Diagonals of a n-gon - Iran NMO 2008 - Problem1
In how many ways, can we draw
n
−
3
n-3
n
−
3
diagonals of a
n
n
n
-gon with equal sides and equal angles such that:
i
)
i)
i
)
none of them intersect each other in the polygonal.
i
i
)
ii)
ii
)
each of the produced triangles has at least one common side with the polygonal.
Prove that a=1 if 4(a^n+1) is a cube for all n (Iran 2008)
N
\mathbb{N}
N
is the set of positive integers and
a
∈
N
a\in\mathbb{N}
a
∈
N
. We know that for every
n
∈
N
n\in\mathbb{N}
n
∈
N
,
4
(
a
n
+
1
)
4(a^n+1)
4
(
a
n
+
1
)
is a perfect cube. Prove that
a
=
1
a=1
a
=
1
.