MathDB
Problems
Contests
National and Regional Contests
Greece Contests
Greece National Olympiad
2005 Greece National Olympiad
2005 Greece National Olympiad
Part of
Greece National Olympiad
Subcontests
(4)
2
1
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Sequence
The sequence
(
a
n
)
(a_n)
(
a
n
)
is defined by
a
1
=
1
a_1=1
a
1
=
1
and
a
n
=
a
n
−
1
+
1
n
3
a_n=a_{n-1}+\frac{1}{n^3}
a
n
=
a
n
−
1
+
n
3
1
for
n
>
1.
n>1.
n
>
1.
(a) Prove that
a
n
<
5
4
a_n<\frac{5}{4}
a
n
<
4
5
for all
n
.
n.
n
.
(b) Given
ϵ
>
0
\epsilon>0
ϵ
>
0
, find the smallest natural number
n
0
n_0
n
0
such that
∣
a
n
+
1
−
a
n
∣
<
ϵ
{\mid a_{n+1}-a_n}\mid<\epsilon
∣
a
n
+
1
−
a
n
∣<
ϵ
for all
n
>
n
0
.
n>n_0.
n
>
n
0
.
4
1
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Fixed point
Let
O
X
1
,
O
X
2
OX_1 , OX_2
O
X
1
,
O
X
2
be rays in the interior of a convex angle
X
O
Y
XOY
XO
Y
such that
∠
X
O
X
1
=
∠
Y
O
Y
1
<
1
3
∠
X
O
Y
\angle XOX_1=\angle YOY_1< \frac{1}{3}\angle XOY
∠
XO
X
1
=
∠
Y
O
Y
1
<
3
1
∠
XO
Y
. Points
K
K
K
on
O
X
1
OX_1
O
X
1
and
L
L
L
on
O
Y
1
OY_1
O
Y
1
are fixed so that
O
K
=
O
L
OK=OL
O
K
=
O
L
, and points
A
A
A
,
B
B
B
are vary on rays
(
O
X
,
(
O
Y
(OX , (OY
(
OX
,
(
O
Y
respectively such that the area of the pentagon
O
A
K
L
B
OAKLB
O
A
K
L
B
remains constant. Prove that the circumcircle of the triangle
O
A
B
OAB
O
A
B
passes from a fixed point, other than
O
O
O
.
1
1
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Easy polynomial
Find the polynomial
P
(
x
)
P(x)
P
(
x
)
with real coefficients such that
P
(
2
)
=
12
P(2)=12
P
(
2
)
=
12
and
P
(
x
2
)
=
x
2
(
x
2
+
1
)
P
(
x
)
P(x^2)=x^2(x^2+1)P(x)
P
(
x
2
)
=
x
2
(
x
2
+
1
)
P
(
x
)
for each
x
∈
R
x\in\mathbb{R}
x
∈
R
.
3
1
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Equation
We know that
k
k
k
is a positive integer and the equation x^3+y^3-2y(x^2-xy+y^2)=k^2(x-y) (1) has one solution
(
x
0
,
y
0
)
(x_0,y_0)
(
x
0
,
y
0
)
with
x
0
,
y
0
∈
Z
−
{
0
}
x_0,y_0\in \mathbb{Z}-\{0\}
x
0
,
y
0
∈
Z
−
{
0
}
and
x
0
≠
y
0
x_0\neq y_0
x
0
=
y
0
. Prove that i) the equation (1) has a finite number of solutions
(
x
,
y
)
(x,y)
(
x
,
y
)
with
x
,
y
∈
Z
x,y\in \mathbb{Z}
x
,
y
∈
Z
and
x
≠
y
x\neq y
x
=
y
; ii) it is possible to find
11
11
11
addition different solutions
(
X
,
Y
)
(X,Y)
(
X
,
Y
)
of the equation (1) with
X
,
Y
∈
Z
−
{
0
}
X,Y\in \mathbb{Z}-\{0\}
X
,
Y
∈
Z
−
{
0
}
and
X
≠
Y
X\neq Y
X
=
Y
where
X
,
Y
X,Y
X
,
Y
are functions of
x
0
,
y
0
x_0,y_0
x
0
,
y
0
.