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National and Regional Contests
Vietnam Contests
Vietnam Team Selection Test
1994 Vietnam Team Selection Test
1994 Vietnam Team Selection Test
Part of
Vietnam Team Selection Test
Subcontests
(3)
3
2
Hide problems
given a polynomial of degree 4
Let
P
(
x
)
P(x)
P
(
x
)
be given a polynomial of degree 4, having 4 positive roots. Prove that the equation
(
1
−
4
⋅
x
)
⋅
P
(
x
)
x
2
+
(
x
2
+
4
⋅
x
−
1
)
⋅
P
′
(
x
)
x
2
−
P
′
′
(
x
)
=
0
(1-4 \cdot x) \cdot \frac{P(x)}{x^2} + (x^2 + 4 \cdot x - 1) \cdot \frac{P'(x)}{x^2} - P''(x) = 0
(
1
−
4
⋅
x
)
⋅
x
2
P
(
x
)
+
(
x
2
+
4
⋅
x
−
1
)
⋅
x
2
P
′
(
x
)
−
P
′′
(
x
)
=
0
has also 4 positive roots.
sum over long fraction
Calculate
T
=
∑
1
n
1
!
⋅
n
2
!
⋅
⋯
n
1994
!
⋅
(
n
2
+
2
⋅
n
3
+
3
⋅
n
4
+
…
+
1993
⋅
n
1994
)
!
T = \sum \frac{1}{n_1! \cdot n_2! \cdot \cdots n_{1994}! \cdot (n_2 + 2 \cdot n_3 + 3 \cdot n_4 + \ldots + 1993 \cdot n_{1994})!}
T
=
∑
n
1
!
⋅
n
2
!
⋅
⋯
n
1994
!
⋅
(
n
2
+
2
⋅
n
3
+
3
⋅
n
4
+
…
+
1993
⋅
n
1994
)!
1
where the sum is taken over all 1994-tuples of the numbers
n
1
,
n
2
,
…
,
n
1994
∈
N
∪
{
0
}
n_1, n_2, \ldots, n_{1994} \in \mathbb{N} \cup \{0\}
n
1
,
n
2
,
…
,
n
1994
∈
N
∪
{
0
}
satisfying
n
1
+
2
⋅
n
2
+
3
⋅
n
3
+
…
+
1994
⋅
n
1994
=
1994.
n_1 + 2 \cdot n_2 + 3 \cdot n_3 + \ldots + 1994 \cdot n_{1994} = 1994.
n
1
+
2
⋅
n
2
+
3
⋅
n
3
+
…
+
1994
⋅
n
1994
=
1994.
2
2
Hide problems
x^2 + y^2 + z^2 + t^2 - N * x * y * z * t - N = 0
Consider the equation
x
2
+
y
2
+
z
2
+
t
2
−
N
⋅
x
⋅
y
⋅
z
⋅
t
−
N
=
0
x^2 + y^2 + z^2 + t^2 - N \cdot x \cdot y \cdot z \cdot t - N = 0
x
2
+
y
2
+
z
2
+
t
2
−
N
⋅
x
⋅
y
⋅
z
⋅
t
−
N
=
0
where
N
N
N
is a given positive integer. a) Prove that for an infinite number of values of
N
N
N
, this equation has positive integral solutions (each such solution consists of four positive integers
x
,
y
,
z
,
t
x, y, z, t
x
,
y
,
z
,
t
), b) Let
N
=
4
⋅
k
⋅
(
8
⋅
m
+
7
)
N = 4 \cdot k \cdot (8 \cdot m + 7)
N
=
4
⋅
k
⋅
(
8
⋅
m
+
7
)
where
k
,
m
k,m
k
,
m
are no-negative integers. Prove that the considered equation has no positive integral solutions.
Vietnam TST 1994 functional equation
Determine all functions
f
:
R
↦
R
f: \mathbb{R} \mapsto \mathbb{R}
f
:
R
↦
R
satisfying
f
(
2
⋅
x
)
+
f
(
4
+
3
⋅
2
⋅
x
)
=
2
⋅
f
(
(
2
+
2
)
⋅
x
)
f\left(\sqrt{2} \cdot x\right) + f\left(4 + 3 \cdot \sqrt{2} \cdot x \right) = 2 \cdot f\left(\left(2 + \sqrt{2}\right) \cdot x\right)
f
(
2
⋅
x
)
+
f
(
4
+
3
⋅
2
⋅
x
)
=
2
⋅
f
(
(
2
+
2
)
⋅
x
)
for all
x
x
x
.
1
1
Hide problems
three sides of which are equal to BM, MN, ND
Given a parallelogram
A
B
C
D
ABCD
A
BC
D
. Let
E
E
E
be a point on the side
B
C
BC
BC
and
F
F
F
be a point on the side
C
D
CD
C
D
such that the triangles
A
B
E
ABE
A
BE
and
B
C
F
BCF
BCF
have the same area. The diaogonal
B
D
BD
B
D
intersects
A
E
AE
A
E
at
M
M
M
and intersects
A
F
AF
A
F
at
N
N
N
. Prove that: I. There exists a triangle, three sides of which are equal to
B
M
,
M
N
,
N
D
BM, MN, ND
BM
,
MN
,
N
D
. II. When
E
,
F
E, F
E
,
F
vary such that the length of
M
N
MN
MN
decreases, the radius of the circumcircle of the above mentioned triangle also decreases.