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Contests
National and Regional Contests
Vietnam Contests
Vietnam National Olympiad
2010 Vietnam National Olympiad
2010 Vietnam National Olympiad
Part of
Vietnam National Olympiad
Subcontests
(5)
4
1
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VMO 2010-Pro 4:Equation has at least n root
Prove that for each positive integer n,the equation
x
2
+
15
y
2
=
4
n
x^{2}+15y^{2}=4^{n}
x
2
+
15
y
2
=
4
n
has at least
n
n
n
integer solution
(
x
,
y
)
(x,y)
(
x
,
y
)
3
1
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VMO 2010-Circle geometry
In plane,let a circle
(
O
)
(O)
(
O
)
and two fixed points
B
,
C
B,C
B
,
C
lies in
(
O
)
(O)
(
O
)
such that
B
C
BC
BC
not is the diameter.Consider a point
A
A
A
varies in
(
O
)
(O)
(
O
)
such that
A
≠
B
,
C
A\neq B,C
A
=
B
,
C
and
A
B
≠
A
C
AB\neq AC
A
B
=
A
C
.Call
D
D
D
and
E
E
E
respective is intersect of
B
C
BC
BC
and internal and external bisector of
B
A
C
^
\widehat{BAC}
B
A
C
,
I
I
I
is midpoint of
D
E
DE
D
E
.The line that pass through orthocenter of
△
A
B
C
\triangle ABC
△
A
BC
and perpendicular with
A
I
AI
A
I
intersects
A
D
,
A
E
AD,AE
A
D
,
A
E
respective at
M
,
N
M,N
M
,
N
.1/Prove that
M
N
MN
MN
pass through a fixed point2/Determint the place of
A
A
A
such that
S
A
M
N
S_{AMN}
S
A
MN
has maxium value
5
1
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VMO 2010-Number of coloring a 3.3 square
Let a positive integer
n
n
n
.Consider square table
3
∗
3
3*3
3
∗
3
.One use
n
n
n
colors to color all cell of table such thateach cell is colored by exactly one color.Two colored table is same if we can receive them from other by a rotation through center of
3
∗
3
3*3
3
∗
3
tableHow many way to color this square table satifies above conditions.
2
1
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VMO 2010-Prove sequence is decreasing
Let
{
a
n
}
\{a_{n}\}
{
a
n
}
be a sequence which satisfy
a
1
=
5
a_{1}=5
a
1
=
5
and
a
n
=
a
n
−
1
n
−
1
+
2
n
−
1
+
2.
3
n
−
1
n
∀
n
≥
2
a_{n=}\sqrt[n]{a_{n-1}^{n-1}+2^{n-1}+2.3^{n-1}} \qquad \forall n\geq2
a
n
=
n
a
n
−
1
n
−
1
+
2
n
−
1
+
2.
3
n
−
1
∀
n
≥
2
(a) Find the general fomular for
a
n
a_{n}
a
n
(b) Prove that
{
a
n
}
\{a_{n}\}
{
a
n
}
is decreasing sequences
1
1
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VMO 2010-System equation
Solve the system equations
{
x
4
−
y
4
=
240
x
3
−
2
y
3
=
3
(
x
2
−
4
y
2
)
−
4
(
x
−
8
y
)
\left\{\begin{array}{cc}x^{4}-y^{4}=240\\x^{3}-2y^{3}=3(x^{2}-4y^{2})-4(x-8y)\end{array}\right.
{
x
4
−
y
4
=
240
x
3
−
2
y
3
=
3
(
x
2
−
4
y
2
)
−
4
(
x
−
8
y
)