MathDB
Problems
Contests
National and Regional Contests
Vietnam Contests
Vietnam Team Selection Test
2003 Vietnam Team Selection Test
2003 Vietnam Team Selection Test
Part of
Vietnam Team Selection Test
Subcontests
(3)
3
2
Hide problems
Challenging functional equation
Let
f
(
0
,
0
)
=
5
2003
,
f
(
0
,
n
)
=
0
f(0, 0) = 5^{2003}, f(0, n) = 0
f
(
0
,
0
)
=
5
2003
,
f
(
0
,
n
)
=
0
for every integer
n
≠
0
n \neq 0
n
=
0
and
f
(
m
,
n
)
=
f
(
m
−
1
,
n
)
−
2
⋅
⌊
f
(
m
−
1
,
n
)
2
⌋
+
⌊
f
(
m
−
1
,
n
−
1
)
2
⌋
+
⌊
f
(
m
−
1
,
n
+
1
)
2
⌋
\begin{array}{c}\ f(m, n) = f(m-1, n) - 2 \cdot \Bigg\lfloor \frac{f(m-1, n)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n-1)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n+1)}{2}\Bigg\rfloor \end{array}
f
(
m
,
n
)
=
f
(
m
−
1
,
n
)
−
2
⋅
⌊
2
f
(
m
−
1
,
n
)
⌋
+
⌊
2
f
(
m
−
1
,
n
−
1
)
⌋
+
⌊
2
f
(
m
−
1
,
n
+
1
)
⌋
for every natural number
m
>
0
m > 0
m
>
0
and for every integer
n
n
n
.Prove that there exists a positive integer
M
M
M
such that
f
(
M
,
n
)
=
1
f(M, n) = 1
f
(
M
,
n
)
=
1
for all integers
n
n
n
such that
∣
n
∣
≤
(
5
2003
−
1
)
2
|n| \leq \frac{(5^{2003}-1)}{2}
∣
n
∣
≤
2
(
5
2003
−
1
)
and
f
(
M
,
n
)
=
0
f(M, n) = 0
f
(
M
,
n
)
=
0
for all integers n such that
∣
n
∣
>
5
2003
−
1
2
.
|n| > \frac{5^{2003}-1}{2}.
∣
n
∣
>
2
5
2003
−
1
.
2n + 1 has no prime divisor of the form 8k - 1
Let
n
n
n
be a positive integer. Prove that the number
2
n
+
1
2^n + 1
2
n
+
1
has no prime divisor of the form
8
⋅
k
−
1
8 \cdot k - 1
8
⋅
k
−
1
, where
k
k
k
is a positive integer.
2
2
Hide problems
incircle with center I of triangle ABC touches the side BC
Given a triangle
A
B
C
ABC
A
BC
. Let
O
O
O
be the circumcenter of this triangle
A
B
C
ABC
A
BC
. Let
H
H
H
,
K
K
K
,
L
L
L
be the feet of the altitudes of triangle
A
B
C
ABC
A
BC
from the vertices
A
A
A
,
B
B
B
,
C
C
C
, respectively. Denote by
A
0
A_{0}
A
0
,
B
0
B_{0}
B
0
,
C
0
C_{0}
C
0
the midpoints of these altitudes
A
H
AH
A
H
,
B
K
BK
B
K
,
C
L
CL
C
L
, respectively. The incircle of triangle
A
B
C
ABC
A
BC
has center
I
I
I
and touches the sides
B
C
BC
BC
,
C
A
CA
C
A
,
A
B
AB
A
B
at the points
D
D
D
,
E
E
E
,
F
F
F
, respectively. Prove that the four lines
A
0
D
A_{0}D
A
0
D
,
B
0
E
B_{0}E
B
0
E
,
C
0
F
C_{0}F
C
0
F
and
O
I
OI
O
I
are concurrent. (When the point
O
O
O
concides with
I
I
I
, we consider the line
O
I
OI
O
I
as an arbitrary line passing through
O
O
O
.)
Vietnam TST 2003 set of all permutations
Let
A
A
A
be the set of all permutations
a
=
(
a
1
,
a
2
,
…
,
a
2003
)
a = (a_1, a_2, \ldots, a_{2003})
a
=
(
a
1
,
a
2
,
…
,
a
2003
)
of the 2003 first positive integers such that each permutation satisfies the condition: there is no proper subset
S
S
S
of the set
{
1
,
2
,
…
,
2003
}
\{1, 2, \ldots, 2003\}
{
1
,
2
,
…
,
2003
}
such that
{
a
k
∣
k
∈
S
}
=
S
.
\{a_k | k \in S\} = S.
{
a
k
∣
k
∈
S
}
=
S
.
For each
a
=
(
a
1
,
a
2
,
…
,
a
2003
)
∈
A
a = (a_1, a_2, \ldots, a_{2003}) \in A
a
=
(
a
1
,
a
2
,
…
,
a
2003
)
∈
A
, let
d
(
a
)
=
∑
k
=
1
2003
(
a
k
−
k
)
2
.
d(a) = \sum^{2003}_{k=1} \left(a_k - k \right)^2.
d
(
a
)
=
∑
k
=
1
2003
(
a
k
−
k
)
2
.
I. Find the least value of
d
(
a
)
d(a)
d
(
a
)
. Denote this least value by
d
0
d_0
d
0
. II. Find all permutations
a
∈
A
a \in A
a
∈
A
such that
d
(
a
)
=
d
0
d(a) = d_0
d
(
a
)
=
d
0
.
1
2
Hide problems
Path with binomials in the coordinate plane
Let be four positive integers
m
,
n
,
p
,
q
m, n, p, q
m
,
n
,
p
,
q
, with
p
<
m
p < m
p
<
m
given and
q
<
n
q < n
q
<
n
. Take four points
A
(
0
;
0
)
,
B
(
p
;
0
)
,
C
(
m
;
q
)
A(0; 0), B(p; 0), C (m; q)
A
(
0
;
0
)
,
B
(
p
;
0
)
,
C
(
m
;
q
)
and
D
(
m
;
n
)
D(m; n)
D
(
m
;
n
)
in the coordinate plane. Consider the paths
f
f
f
from
A
A
A
to
D
D
D
and the paths
g
g
g
from
B
B
B
to
C
C
C
such that when going along
f
f
f
or
g
g
g
, one goes only in the positive directions of coordinates and one can only change directions (from the positive direction of one axe coordinate into the the positive direction of the other axe coordinate) at the points with integral coordinates. Let
S
S
S
be the number of couples
(
f
,
g
)
(f, g)
(
f
,
g
)
such that
f
f
f
and
g
g
g
have no common points. Prove that
S
=
(
n
m
+
n
)
⋅
(
q
m
+
q
−
p
)
−
(
q
m
+
q
)
⋅
(
n
m
+
n
−
p
)
.
S = \binom{n}{m+n} \cdot \binom{q}{m+q-p} - \binom{q}{m+q} \cdot \binom{n}{m+n-p}.
S
=
(
m
+
n
n
)
⋅
(
m
+
q
−
p
q
)
−
(
m
+
q
q
)
⋅
(
m
+
n
−
p
n
)
.
ratio which is not less than 1 / sqrt{3}
On the sides of triangle
A
B
C
ABC
A
BC
take the points
M
1
,
N
1
,
P
1
M_1, N_1, P_1
M
1
,
N
1
,
P
1
such that each line
M
M
1
,
N
N
1
,
P
P
1
MM_1, NN_1, PP_1
M
M
1
,
N
N
1
,
P
P
1
divides the perimeter of
A
B
C
ABC
A
BC
in two equal parts (
M
,
N
,
P
M, N, P
M
,
N
,
P
are respectively the midpoints of the sides
B
C
,
C
A
,
A
B
BC, CA, AB
BC
,
C
A
,
A
B
). I. Prove that the lines
M
M
1
,
N
N
1
,
P
P
1
MM_1, NN_1, PP_1
M
M
1
,
N
N
1
,
P
P
1
are concurrent at a point
K
K
K
. II. Prove that among the ratios
K
A
B
C
,
K
B
C
A
,
K
C
A
B
\frac{KA}{BC}, \frac{KB}{CA}, \frac{KC}{AB}
BC
K
A
,
C
A
K
B
,
A
B
K
C
there exist at least a ratio which is not less than
1
3
\frac{1}{\sqrt{3}}
3
1
.