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Contests
National and Regional Contests
Turkey Contests
Turkey MO (2nd round)
1998 Turkey MO (2nd round)
1998 Turkey MO (2nd round)
Part of
Turkey MO (2nd round)
Subcontests
(3)
3
2
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Colored Points on a Circle
The points of a circle are colored by three colors. Prove that there exist infinitely many isosceles triangles inscribed in the circle whose vertices are of the same color.
Colored Points on a nxn Chessboard, finding a limit
Some of the vertices of unit squares of an
n
×
n
n\times n
n
×
n
chessboard are colored so that any
k
×
k
k\times k
k
×
k
(
1
≤
k
≤
n
1\le k\le n
1
≤
k
≤
n
) square consisting of these unit squares has a colored point on at least one of its sides. Let
l
(
n
)
l(n)
l
(
n
)
denote the minimum number of colored points required to satisfy this condition. Prove that \underset{n\to \infty }{\mathop \lim }\,\frac{l(n)}{{{n}^{2}}}=\frac{2}{7}.
2
2
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An Easy Inequality on three variables
If
0
≤
a
≤
b
≤
c
0\le a\le b\le c
0
≤
a
≤
b
≤
c
real numbers, prove that
(
a
+
3
b
)
(
b
+
4
c
)
(
c
+
2
a
)
≥
60
a
b
c
(a+3b)(b+4c)(c+2a)\ge 60abc
(
a
+
3
b
)
(
b
+
4
c
)
(
c
+
2
a
)
≥
60
ab
c
.
Geometric locus problem about a constant sum
Variable points
M
M
M
and
N
N
N
are considered on the arms
[
O
X
\left[ OX \right.
[
OX
and
[
O
Y
\left[ OY \right.
[
O
Y
, respectively, of an angle
X
O
Y
XOY
XO
Y
so that
∣
O
M
∣
+
∣
O
N
∣
\left| OM \right|+\left| ON \right|
∣
OM
∣
+
∣
ON
∣
is constant. Determine the locus of the midpoint of
[
M
N
]
\left[ MN \right]
[
MN
]
.
1
2
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Turkish NMO 1998, 1. Problem, points in a isosceles triangle
Let
D
D
D
be the point on the base
B
C
BC
BC
of an isosceles
△
A
B
C
\vartriangle ABC
△
A
BC
triangle such that
∣
B
D
∣
∣
D
C
∣
=
2
\frac{\left| BD \right|}{\left| DC \right|}=\text{ }2
∣
D
C
∣
∣
B
D
∣
=
2
, and let
P
P
P
be the point on the segment
[
A
D
]
\left[ AD \right]
[
A
D
]
such that
∠
B
A
C
=
∠
B
P
D
\angle BAC=\angle BPD
∠
B
A
C
=
∠
BP
D
. Prove that
∠
D
P
C
=
1
2
∠
B
A
C
\angle DPC=\frac{1}{2}\angle BAC
∠
D
PC
=
2
1
∠
B
A
C
.
solving equation x^3+3367=2^n
Find all positive integers
x
x
x
and
n
n
n
such that
x
3
+
3367
=
2
n
{{x}^{3}}+3367={{2}^{n}}
x
3
+
3367
=
2
n
.