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National and Regional Contests
Turkey Contests
Turkey MO (2nd round)
2011 Turkey MO (2nd round)
2011 Turkey MO (2nd round)
Part of
Turkey MO (2nd round)
Subcontests
(6)
2
1
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Turkey NMO 2011 Problem 2
Let
A
B
C
ABC
A
BC
be a triangle
D
∈
[
B
C
]
D\in[BC]
D
∈
[
BC
]
(different than
A
A
A
and
B
B
B
).
E
E
E
is the midpoint of
[
C
D
]
[CD]
[
C
D
]
.
F
∈
[
A
C
]
F\in[AC]
F
∈
[
A
C
]
such that
F
E
C
^
=
90
\widehat{FEC}=90
FEC
=
90
and
∣
A
F
∣
.
∣
B
C
∣
=
∣
A
C
∣
.
∣
E
C
∣
.
|AF|.|BC|=|AC|.|EC|.
∣
A
F
∣.∣
BC
∣
=
∣
A
C
∣.∣
EC
∣.
Circumcircle of
A
D
C
ADC
A
D
C
intersect
[
A
B
]
[AB]
[
A
B
]
at
G
G
G
different than
A
A
A
.Prove that tangent to circumcircle of
A
G
F
AGF
A
GF
at
F
F
F
is touch circumcircle of
B
G
E
BGE
BGE
too.
4
1
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Turkey NMO 2011 Problem 4
a
1
=
5
a_{1}=5
a
1
=
5
and
a
n
+
1
=
a
n
3
−
2
a
n
2
+
2
a_{n+1}=a_{n}^{3}-2a_{n}^{2}+2
a
n
+
1
=
a
n
3
−
2
a
n
2
+
2
for all
n
≥
1
n\geq1
n
≥
1
.
p
p
p
is a prime such that
p
=
3
(
m
o
d
4
)
p=3(mod 4)
p
=
3
(
m
o
d
4
)
and
p
∣
a
2011
+
1
p|a_{2011}+1
p
∣
a
2011
+
1
. Show that
p
=
3
p=3
p
=
3
.
6
1
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Turkey NMO 2011 q6
Let
A
A
A
and
B
B
B
two countries which inlude exactly
2011
2011
2011
cities.There is exactly one flight from a city of
A
A
A
to a city of
B
B
B
and there is no domestic flights (flights are bi-directional).For every city
X
X
X
(doesn't matter from
A
A
A
or from
B
B
B
), there exist at most
19
19
19
different airline such that airline have a flight from
X
X
X
to the another city.For an integer
k
k
k
, (it doesn't matter how flights arranged ) we can say that there exists at least
k
k
k
cities such that it is possible to trip from one of these
k
k
k
cities to another with same airline.So find the maximum value of
k
k
k
.
5
1
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Turkey NMO 2011 q5
Let
M
M
M
and
N
N
N
be two regular polygonic area.Define
K
(
M
,
N
)
K(M,N)
K
(
M
,
N
)
as the midpoints of segments
[
A
B
]
[AB]
[
A
B
]
such that
A
A
A
belong to
M
M
M
and
B
B
B
belong to
N
N
N
. Find all situations of
M
M
M
and
N
N
N
such that
K
(
M
,
N
)
K(M,N)
K
(
M
,
N
)
is a regualr polygonic area too.
1
1
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Turkey NMO 2011 Problem1
n
≥
2
n\geq2
n
≥
2
and
E
=
{
1
,
2
,
.
.
.
,
n
}
.
A
1
,
A
2
,
.
.
.
,
A
k
E=\left \{ 1,2,...,n \right \}. A_1,A_2,...,A_k
E
=
{
1
,
2
,
...
,
n
}
.
A
1
,
A
2
,
...
,
A
k
are subsets of
E
E
E
, such that for all
1
≤
i
<
j
≤
k
1\leq{i}<{j}\leq{k}
1
≤
i
<
j
≤
k
Exactly one of
A
i
∩
A
j
,
A
i
′
∩
A
j
,
A
i
∩
A
j
′
,
A
i
′
∩
A
j
′
A_i\cap{A_j},A_i'\cap{A_j},A_i\cap{A_j'},A_i'\cap{A_j'}
A
i
∩
A
j
,
A
i
′
∩
A
j
,
A
i
∩
A
j
′
,
A
i
′
∩
A
j
′
is empty set. What is the maximum possible
k
k
k
?
3
1
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Hard inequality Turkey 2011
x
,
y
,
z
x,y,z
x
,
y
,
z
positive real numbers such that
x
y
z
=
1
xyz=1
x
yz
=
1
Prove that:
1
x
+
y
20
+
z
11
+
1
y
+
z
20
+
x
11
+
1
z
+
x
20
+
y
11
≤
1
\frac{1}{x+y^{20}+z^{11}}+\frac{1}{y+z^{20}+x^{11}}+\frac{1}{z+x^{20}+y^{11}}\leq1
x
+
y
20
+
z
11
1
+
y
+
z
20
+
x
11
1
+
z
+
x
20
+
y
11
1
≤
1