MathDB
Problems
Contests
National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2005 Moldova Team Selection Test
2005 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(4)
1
2
Hide problems
alpha+beta+gamma=1
In triangle
A
B
C
ABC
A
BC
,
M
∈
(
B
C
)
M\in(BC)
M
∈
(
BC
)
,
B
M
B
C
=
α
\frac{BM}{BC}=\alpha
BC
BM
=
α
,
N
∈
(
C
A
)
N\in(CA)
N
∈
(
C
A
)
,
C
N
C
A
=
β
\frac{CN}{CA}=\beta
C
A
CN
=
β
,
P
∈
(
A
B
)
P\in(AB)
P
∈
(
A
B
)
,
A
P
A
B
=
γ
\frac{AP}{AB}=\gamma
A
B
A
P
=
γ
. Let
A
M
∩
B
N
=
{
D
}
AM\cap BN=\{D\}
A
M
∩
BN
=
{
D
}
,
B
N
∩
C
P
=
{
E
}
BN\cap CP=\{E\}
BN
∩
CP
=
{
E
}
,
C
P
∩
A
M
=
{
F
}
CP\cap AM=\{F\}
CP
∩
A
M
=
{
F
}
. Prove that
S
D
E
F
=
S
B
M
D
+
S
C
N
E
+
S
A
P
F
S_{DEF}=S_{BMD}+S_{CNE}+S_{APF}
S
D
EF
=
S
BM
D
+
S
CNE
+
S
A
PF
iff
α
+
β
+
γ
=
1
\alpha+\beta+\gamma=1
α
+
β
+
γ
=
1
.
2 triangles
Let
A
B
C
ABC
A
BC
and
A
1
B
1
C
1
A_{1}B_{1}C_{1}
A
1
B
1
C
1
be two triangles. Prove that
a
a
1
+
b
b
1
+
c
c
1
≤
3
R
2
r
1
\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\leq\frac{3R}{2r_{1}}
a
1
a
+
b
1
b
+
c
1
c
≤
2
r
1
3
R
, where
a
=
B
C
a = BC
a
=
BC
,
b
=
C
A
b = CA
b
=
C
A
,
c
=
A
B
c = AB
c
=
A
B
are the sidelengths of triangle
A
B
C
ABC
A
BC
, where
a
1
=
B
1
C
1
a_{1}=B_{1}C_{1}
a
1
=
B
1
C
1
,
b
1
=
C
1
A
1
b_{1}=C_{1}A_{1}
b
1
=
C
1
A
1
,
c
1
=
A
1
B
1
c_{1}=A_{1}B_{1}
c
1
=
A
1
B
1
are the sidelengths of triangle
A
1
B
1
C
1
A_{1}B_{1}C_{1}
A
1
B
1
C
1
, where
R
R
R
is the circumradius of triangle
A
B
C
ABC
A
BC
and
r
1
r_{1}
r
1
is the inradius of triangle
A
1
B
1
C
1
A_{1}B_{1}C_{1}
A
1
B
1
C
1
.
4
3
Hide problems
function over set of polynoms
n
n
n
is a positive integer,
K
K
K
the set of polynoms of real variables
x
1
,
x
2
,
.
.
.
,
x
n
+
1
x_1,x_2,...,x_{n+1}
x
1
,
x
2
,
...
,
x
n
+
1
and
y
1
,
y
2
,
.
.
.
,
y
n
+
1
y_1,y_2,...,y_{n+1}
y
1
,
y
2
,
...
,
y
n
+
1
, function
f
:
K
→
K
f:K\rightarrow K
f
:
K
→
K
satisfies f(p+q)=f(p)+f(q), f(pq)=f(p)q+pf(q), (\forall)p,q\in K. If f(x_i)=(n-1)x_i+y_i, f(y_i)=2ny_i for all
i
=
1
,
2
,
.
.
.
,
n
+
1
i=1,2,...,n+1
i
=
1
,
2
,
...
,
n
+
1
and
∏
i
=
1
n
+
1
(
t
x
i
+
y
i
)
=
∑
i
=
0
n
+
1
p
i
t
n
+
1
−
i
\prod_{i=1}^{n+1}(tx_i+y_i)=\sum_{i=0}^{n+1}p_it^{n+1-i}
i
=
1
∏
n
+
1
(
t
x
i
+
y
i
)
=
i
=
0
∑
n
+
1
p
i
t
n
+
1
−
i
for any real
t
t
t
, prove, that for all
k
=
1
,
.
.
.
,
n
+
1
k=1,...,n+1
k
=
1
,
...
,
n
+
1
f
(
p
k
−
1
)
=
k
p
k
+
(
n
+
1
)
(
n
+
k
−
2
)
p
k
−
1
f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1}
f
(
p
k
−
1
)
=
k
p
k
+
(
n
+
1
)
(
n
+
k
−
2
)
p
k
−
1
largest p
Find the largest positive
p
p
p
(
p
>
1
p>1
p
>
1
) such, that
∀
a
,
b
,
c
∈
[
1
p
,
p
]
\forall a,b,c\in[\frac1p,p]
∀
a
,
b
,
c
∈
[
p
1
,
p
]
the following inequality takes place
9
(
a
b
+
b
c
+
c
a
)
(
a
2
+
b
2
+
c
2
)
≥
(
a
+
b
+
c
)
4
9(ab+bc+ca)(a^2+b^2+c^2)\geq(a+b+c)^4
9
(
ab
+
b
c
+
c
a
)
(
a
2
+
b
2
+
c
2
)
≥
(
a
+
b
+
c
)
4
f(n)=n
Given functions
f
,
g
:
N
∗
→
N
∗
f,g:N^*\rightarrow N^*
f
,
g
:
N
∗
→
N
∗
,
g
g
g
is surjective and
2
f
(
n
)
2
=
n
2
+
g
(
n
)
2
2f(n)^2=n^2+g(n)^2
2
f
(
n
)
2
=
n
2
+
g
(
n
)
2
,
∀
n
>
0
\forall n>0
∀
n
>
0
. Prove that if
∣
f
(
n
)
−
n
∣
≤
2005
n
|f(n)-n|\le2005\sqrt n
∣
f
(
n
)
−
n
∣
≤
2005
n
,
∀
n
>
0
\forall n>0
∀
n
>
0
, then
f
(
n
)
=
n
f(n)=n
f
(
n
)
=
n
for infinitely many
n
n
n
.
3
2
Hide problems
circles and points
Does there exist such a configuration of 22 circles and 22 point, that any circle contains at leats 7 points and any point belongs at least to 7 circles?
n^k divides [A]
A
=
3
∑
m
=
1
n
2
(
1
2
−
{
m
}
)
A=3\sum_{m=1}^{n^2}(\frac12-\{\sqrt{m}\})
A
=
3
m
=
1
∑
n
2
(
2
1
−
{
m
})
where
n
n
n
is an positive integer. Find the largest
k
k
k
such that
n
k
n^k
n
k
divides
[
A
]
[A]
[
A
]
.
2
2
Hide problems
\sum1/{4-ab}
Let
a
a
a
,
b
b
b
,
c
c
c
be positive reals such that a^4 \plus{} b^4 \plus{} c^4 \equal{} 3. Prove that \sum\frac1{4 \minus{} ab}\leq1, where the
∑
\sum
∑
sign stands for cyclic summation. Alternative formulation: For any positive reals
a
a
a
,
b
b
b
,
c
c
c
satisfying a^4 \plus{} b^4 \plus{} c^4 \equal{} 3, prove the inequality \frac{1}{4\minus{}bc}\plus{}\frac{1}{4\minus{}ca}\plus{}\frac{1}{4\minus{}ab}\leq 1.
2005 divides smth
Let
m
∈
N
m\in N
m
∈
N
and
E
(
x
,
y
,
m
)
=
(
72
x
)
m
+
(
72
y
)
m
−
x
m
−
y
m
E(x,y,m)=(\frac{72}x)^m+(\frac{72}y)^m-x^m-y^m
E
(
x
,
y
,
m
)
=
(
x
72
)
m
+
(
y
72
)
m
−
x
m
−
y
m
, where
x
x
x
and
y
y
y
are positive divisors of 72. a) Prove that there exist infinitely many natural numbers
m
m
m
so, that 2005 divides
E
(
3
,
12
,
m
)
E(3,12,m)
E
(
3
,
12
,
m
)
and
E
(
9
,
6
,
m
)
E(9,6,m)
E
(
9
,
6
,
m
)
. b) Find the smallest positive integer number
m
0
m_0
m
0
so, that 2005 divides
E
(
3
,
12
,
m
0
)
E(3,12,m_0)
E
(
3
,
12
,
m
0
)
and
E
(
9
,
6
,
m
0
)
E(9,6,m_0)
E
(
9
,
6
,
m
0
)
.