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Problems
Contests
National and Regional Contests
Vietnam Contests
Vietnam National Olympiad
2016 Vietnam National Olympiad
2016 Vietnam National Olympiad
Part of
Vietnam National Olympiad
Subcontests
(4)
4
1
Hide problems
Plant tree on tabular grid
Let
m
m
m
and
n
n
n
be positive integers. A people planted two kind of different trees on a plot tabular grid size
m
×
n
m \times n
m
×
n
(each square plant one tree.) A plant called inpressive if two conditions following conditions are met simultaneously: i) The number of trees in each of kind is equal; ii) In each row the number of tree of each kind is diffrenent not less than a half of number of cells on that row and In each colum the number of tree of each kind is diffrenent not less than a half of number of cells on that colum.a) Find an inpressive plant when
m
=
n
=
2016
m=n=2016
m
=
n
=
2016
; b) Prove that if there at least a inpressive plant then
4
∣
m
4|m
4∣
m
and
4
∣
n
4|n
4∣
n
.
2
2
Hide problems
two sequences
a) Let
(
a
n
)
(a_n)
(
a
n
)
be the sequence defined by
a
n
=
ln
(
2
n
2
+
1
)
−
ln
(
n
2
+
n
+
1
)
∀
n
≥
1.
a_n=\ln (2n^2+1)-\ln (n^2+n+1)\,\,\forall n\geq 1.
a
n
=
ln
(
2
n
2
+
1
)
−
ln
(
n
2
+
n
+
1
)
∀
n
≥
1.
Prove that the set
{
n
∈
N
∣
{
a
n
}
<
1
2
}
\{n\in\mathbb{N}|\,\{a_n\}<\dfrac{1}{2}\}
{
n
∈
N
∣
{
a
n
}
<
2
1
}
is a finite set; b) Let
(
b
n
)
(b_n)
(
b
n
)
be the sequence defined by
a
n
=
ln
(
2
n
2
+
1
)
+
ln
(
n
2
+
n
+
1
)
∀
n
≥
1
a_n=\ln (2n^2+1)+\ln (n^2+n+1)\,\,\forall n\geq 1
a
n
=
ln
(
2
n
2
+
1
)
+
ln
(
n
2
+
n
+
1
)
∀
n
≥
1
. Prove that the set
{
n
∈
N
∣
{
b
n
}
<
1
2016
}
\{n\in\mathbb{N}|\,\{b_n\}<\dfrac{1}{2016}\}
{
n
∈
N
∣
{
b
n
}
<
2016
1
}
is an infinite set.
Vietnam MO 2016 P6
Given a triangle
A
B
C
ABC
A
BC
inscribed by circumcircle
(
O
)
(O)
(
O
)
. The angles at
B
,
C
B,C
B
,
C
are acute angle. Let
M
M
M
on the arc
B
C
BC
BC
that doesn't contain
A
A
A
such that
A
M
AM
A
M
is not perpendicular to
B
C
BC
BC
.
A
M
AM
A
M
meets the perpendicular bisector of
B
C
BC
BC
at
T
T
T
. The circumcircle
(
A
O
T
)
(AOT)
(
A
OT
)
meets
(
O
)
(O)
(
O
)
at
N
N
N
(
N
≠
A
N\ne A
N
=
A
).a) Prove that
∠
B
A
M
=
∠
C
A
N
\angle{BAM}=\angle{CAN}
∠
B
A
M
=
∠
C
A
N
.b) Let
I
I
I
be the incenter and
G
G
G
be the foor of the angle bisector of
∠
B
A
C
\angle{BAC}
∠
B
A
C
.
A
I
,
M
I
,
N
I
AI,MI,NI
A
I
,
M
I
,
N
I
intersect
(
O
)
(O)
(
O
)
at
D
,
E
,
F
D,E,F
D
,
E
,
F
respectively. Let
P
=
D
F
∩
A
M
,
Q
=
D
E
∩
A
N
{P}=DF\cap AM, {Q}=DE\cap AN
P
=
D
F
∩
A
M
,
Q
=
D
E
∩
A
N
. The circle passes through
P
P
P
and touches
A
D
AD
A
D
at
I
I
I
meets
D
F
DF
D
F
at
H
H
H
(
H
≠
D
H\ne D
H
=
D
).The circle passes through
Q
Q
Q
and touches
A
D
AD
A
D
at
I
I
I
meets
D
E
DE
D
E
at
K
K
K
(
K
≠
D
K\ne D
K
=
D
). Prove that the circumcircle
(
G
H
K
)
(GHK)
(
G
HK
)
touches
B
C
BC
BC
.
1
2
Hide problems
A system of equations
Solve the system of equations
{
6
x
−
y
+
z
2
=
3
x
2
−
y
2
−
2
z
=
−
1
e
m
s
p
;
e
m
s
p
;
(
x
,
y
,
z
∈
R
.
)
6
x
2
−
3
y
2
−
y
−
2
z
2
=
0
\begin{cases}6x-y+z^2=3\\ x^2-y^2-2z=-1   (x,y,z\in\mathbb{R}.)\\ 6x^2-3y^2-y-2z^2=0\end{cases}
⎩
⎨
⎧
6
x
−
y
+
z
2
=
3
x
2
−
y
2
−
2
z
=
−
1
6
x
2
−
3
y
2
−
y
−
2
z
2
=
0
e
m
s
p
;
e
m
s
p
;
(
x
,
y
,
z
∈
R
.
)
.
VMO 2016 Problem 5
Find all
a
∈
R
a\in\mathbb{R}
a
∈
R
such that there is function
f
:
R
→
R
f:\mathbb{R}\to\mathbb{R}
f
:
R
→
R
i)
f
(
1
)
=
2016
f(1)=2016
f
(
1
)
=
2016
ii) f(x+y+f(y))=f(x)+ay \forall x,y\in\mathbb{R}
3
2
Hide problems
A point is on a fixed line
Let
A
B
C
ABC
A
BC
be an acute triange with
B
,
C
B,C
B
,
C
fixed. Let
D
D
D
be the midpoint of
B
C
BC
BC
and
E
,
F
E,F
E
,
F
be the foot of the perpendiculars from
D
D
D
to
A
B
,
A
C
AB,AC
A
B
,
A
C
, respectively. a) Let
O
O
O
be the circumcenter of triangle
A
B
C
ABC
A
BC
and
M
=
E
F
∩
A
O
,
N
=
E
F
∩
B
C
M=EF\cap AO, N=EF\cap BC
M
=
EF
∩
A
O
,
N
=
EF
∩
BC
. Prove that the circumcircle of triangle
A
M
N
AMN
A
MN
passes through a fixed point; b) Assume that tangents of the circumcircle of triangle
A
E
F
AEF
A
EF
at
E
,
F
E,F
E
,
F
are intersecting at
T
T
T
. Prove that
T
T
T
is on a fixed line.
VMO 2016 Problem 7
a) Prove that if
n
n
n
is an odd perfect number then
n
n
n
has the following form
n
=
p
s
m
2
n=p^sm^2
n
=
p
s
m
2
where
p
p
p
is prime has form
4
k
+
1
4k+1
4
k
+
1
,
s
s
s
is positive integers has form
4
h
+
1
4h+1
4
h
+
1
, and
m
∈
Z
+
m\in\mathbb{Z}^+
m
∈
Z
+
,
m
m
m
is not divisible by
p
p
p
. b) Find all
n
∈
Z
+
n\in\mathbb{Z}^+
n
∈
Z
+
,
n
>
1
n>1
n
>
1
such that
n
−
1
n-1
n
−
1
and
n
(
n
+
1
)
2
\frac{n(n+1)}{2}
2
n
(
n
+
1
)
is perfect number