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Contests
National and Regional Contests
Turkey Contests
Turkey MO (2nd round)
2001 Turkey MO (2nd round)
2001 Turkey MO (2nd round)
Part of
Turkey MO (2nd round)
Subcontests
(3)
3
2
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Turkey NMO 2001 Problem 3
One wants to distribute
n
n
n
same sized cakes between
k
k
k
people equally by cutting every cake at most once. If the number of positive divisors of
n
n
n
is denoted as
d
(
n
)
d(n)
d
(
n
)
, show that the number of different values of
k
k
k
which makes such distribution possible is
n
+
d
(
n
)
n+d(n)
n
+
d
(
n
)
Turkey NMO 2001 Problem 6, Coloring Problem
We wish to color the cells of a
n
×
n
n \times n
n
×
n
chessboard with
k
k
k
different colors such that for every
i
∈
{
1
,
2
,
.
.
.
,
n
}
i\in \{1,2,...,n\}
i
∈
{
1
,
2
,
...
,
n
}
, the
2
n
−
1
2n-1
2
n
−
1
cells on
i
i
i
. row and
i
i
i
. column have all different colors.a) Prove that for
n
=
2001
n=2001
n
=
2001
and
k
=
4001
k=4001
k
=
4001
, such coloring is not possible.b) Show that for
n
=
2
m
−
1
n=2^{m}-1
n
=
2
m
−
1
and
k
=
2
m
+
1
−
1
k=2^{m+1}-1
k
=
2
m
+
1
−
1
, such coloring is possible.
2
2
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Turkey NMO 2001 Problem 2
(
x
n
)
−
∞
<
n
<
∞
(x_{n})_{-\infty<n<\infty}
(
x
n
)
−
∞
<
n
<
∞
is a sequence of real numbers which satisfies
x
n
+
1
=
x
n
2
+
10
7
x_{n+1}=\frac{x_{n}^2+10}{7}
x
n
+
1
=
7
x
n
2
+
10
for every
n
∈
Z
n \in \mathbb{Z}
n
∈
Z
. If there exist a real upperbound for this sequence, find all the values
x
0
x_{0}
x
0
can take.
Turkey NMO 2001 Problem 5, locus of point P_G
Two nonperpendicular lines throught the point
A
A
A
and a point
F
F
F
on one of these lines different from
A
A
A
are given. Let
P
G
P_{G}
P
G
be the intersection point of tangent lines at
G
G
G
and
F
F
F
to the circle through the point
A
A
A
,
F
F
F
and
G
G
G
where
G
G
G
is a point on the given line different from the line
F
A
FA
F
A
. What is the locus of
P
G
P_{G}
P
G
as
G
G
G
varies.
1
2
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Turkey NMO 2001 Problem 1
Let
A
B
C
D
ABCD
A
BC
D
be a convex quadrilateral. The perpendicular bisectors of the sides
[
A
D
]
[AD]
[
A
D
]
and
[
B
C
]
[BC]
[
BC
]
intersect at a point
P
P
P
inside the quadrilateral and the perpendicular bisectors of the sides
[
A
B
]
[AB]
[
A
B
]
and
[
C
D
]
[CD]
[
C
D
]
also intersect at a point
Q
Q
Q
inside the quadrilateral. Show that, if
∠
A
P
D
=
∠
B
P
C
\angle APD = \angle BPC
∠
A
P
D
=
∠
BPC
then
∠
A
Q
B
=
∠
C
Q
D
\angle AQB = \angle CQD
∠
A
QB
=
∠
CQ
D
Turkey NMO 2001 Problem 4, Diophant Equation
Find all ordered triples of positive integers
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
such that
3
x
+
1
1
y
=
z
2
3^{x}+11^{y}=z^{2}
3
x
+
1
1
y
=
z
2