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Contests
National and Regional Contests
Moldova Contests
Moldova Team Selection Test
2018 Moldova Team Selection Test
2018 Moldova Team Selection Test
Part of
Moldova Team Selection Test
Subcontests
(10)
6
1
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folklore
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be positive real numbers such that
a
+
b
+
c
=
3
a+b+c=3
a
+
b
+
c
=
3
. Show that
a
1
+
b
2
+
b
1
+
c
2
+
c
1
+
a
2
≥
3
2
.
\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}.
1
+
b
2
a
+
1
+
c
2
b
+
1
+
a
2
c
≥
2
3
.
11
1
Hide problems
Geometry problem
Let
Ω
\Omega
Ω
be the circumcincle of the quadrilateral
A
B
C
D
ABCD
A
BC
D
, and
E
E
E
the intersection point of the diagonals
A
C
AC
A
C
and
B
D
BD
B
D
. A line passing through
E
E
E
intersects
A
B
AB
A
B
and
B
C
BC
BC
in points
P
P
P
and
Q
Q
Q
. A circle ,that is passing through point
D
D
D
, is tangent to the line
P
Q
PQ
PQ
in point
E
E
E
and intersects
Ω
\Omega
Ω
in point
R
R
R
, different from
D
D
D
. Prove that the points
B
,
P
,
Q
,
B,P,Q,
B
,
P
,
Q
,
and
R
R
R
are concyclic .
10
1
Hide problems
Intersting inequality
The positive real numbers
a
,
b
,
c
,
d
a,b, c,d
a
,
b
,
c
,
d
satisfy the equality
1
a
+
1
+
1
b
+
1
+
1
c
+
1
+
1
d
+
1
=
3
\frac {1}{a+1} + \frac {1}{b+1} + \frac {1}{c+1} + \frac{ 1}{d+1} = 3
a
+
1
1
+
b
+
1
1
+
c
+
1
1
+
d
+
1
1
=
3
. Prove the inequality
a
b
c
3
+
b
c
d
3
+
c
d
a
3
+
d
a
b
3
≤
4
3
\sqrt [3]{abc} + \sqrt [3]{bcd} + \sqrt [3]{cda} + \sqrt [3]{dab} \le \frac {4}{3}
3
ab
c
+
3
b
c
d
+
3
c
d
a
+
3
d
ab
≤
3
4
.
9
1
Hide problems
Number Theory problem
The positive integers
a
a
a
and
b
b
b
satisfy the sistem
{
a
10
+
b
10
=
a
a
11
+
b
11
=
b
\begin {cases} a_{10} +b_{10} = a \\a_{11}+b_{11 }=b \end {cases}
{
a
10
+
b
10
=
a
a
11
+
b
11
=
b
where
a
1
<
a
2
<
…
a_1 <a_2 <\dots
a
1
<
a
2
<
…
and
b
1
<
b
2
<
…
b_1 <b_2 <\dots
b
1
<
b
2
<
…
are the positive divisors of
a
a
a
and
b
b
b
. Find
a
a
a
and
b
b
b
.
8
1
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Combinatorics-Number Theory problem
Let the set
A
=
A=
A
=
{
1
,
2
,
3
,
…
,
48
n
+
24
1,2,3, \dots ,48n+24
1
,
2
,
3
,
…
,
48
n
+
24
} , where
n
∈
N
∗
n \in \mathbb {N^*}
n
∈
N
∗
. Prove that there exist a subset
B
B
B
of
A
A
A
with
24
n
+
12
24n+12
24
n
+
12
elements with the property : the sum of the squares of the elements of the set
B
B
B
is equal to the sum of the squares of the elements of the set
A
A
A
\
B
B
B
.
7
1
Hide problems
Cute geometry
Let the triangle
A
B
C
ABC
A
BC
with
m
(
∠
A
B
C
)
=
6
0
∘
m (\angle ABC)=60^{\circ}
m
(
∠
A
BC
)
=
6
0
∘
and
m
(
∠
B
A
C
)
=
4
0
∘
m (\angle BAC)=40^{\circ}
m
(
∠
B
A
C
)
=
4
0
∘
. Points
D
D
D
and
E
E
E
are on the sides
(
A
B
)
(AB)
(
A
B
)
and
(
A
C
)
(AC)
(
A
C
)
such that
m
(
∠
D
C
B
)
=
7
0
∘
m (\angle DCB )=70^{\circ}
m
(
∠
D
CB
)
=
7
0
∘
and
m
(
∠
E
B
C
)
=
4
0
∘
m (\angle EBC)=40^{\circ}
m
(
∠
EBC
)
=
4
0
∘
.
B
E
BE
BE
and
C
D
CD
C
D
intersect in
F
F
F
. Prove that
B
C
BC
BC
and
A
F
AF
A
F
are perpendicular.
5
1
Hide problems
Polynomial problem
Let
n
,
∈
N
∗
,
n
≥
3
n, \in \mathbb {N^*} , n\ge 3
n
,
∈
N
∗
,
n
≥
3
a) Prove that the polynomial
f
(
x
)
=
X
2
n
−
1
−
1
X
−
1
−
X
n
f (x)=\frac {X^{2^n-1}-1}{X-1}-X^n
f
(
x
)
=
X
−
1
X
2
n
−
1
−
1
−
X
n
has a divisor of form
X
p
+
1
X^p +1
X
p
+
1
where
p
∈
N
∗
p\in\mathbb {N^*}
p
∈
N
∗
b) Show that for
n
=
7
n=7
n
=
7
the polynomial
f
(
X
)
f (X)
f
(
X
)
has three divisors with integer coefficients .
4
1
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Combinatorial problem
A pupil is writing on a board positive integers
x
0
,
x
1
,
x
2
,
x
3
.
.
.
x_0,x_1,x_2,x_3...
x
0
,
x
1
,
x
2
,
x
3
...
after the following algorithm which implies arithmetic progression
3
,
5
,
7
,
9...
3,5,7,9...
3
,
5
,
7
,
9...
.Each term of rank
k
≥
2
k\ge2
k
≥
2
is a difference between the product of the last number on the board and the term of arithmetic progression of rank
k
k
k
and the last but one term on the bord with the sum of the terms of the arithemtic progression with ranks less than
k
k
k
.If
x
0
=
0
x_0=0
x
0
=
0
and
x
1
=
1
x_1=1
x
1
=
1
find
x
n
x_n
x
n
according to n.
1
1
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interesting problem
Let
x
,
y
,
z
∈
Q
x,y,z \in\mathbb{Q}
x
,
y
,
z
∈
Q
,such that
(
x
+
y
+
z
)
3
=
9
(
x
2
y
+
y
2
z
+
z
2
x
)
.
(x+y+z)^3=9(x^2y+y^2z+z^2x).
(
x
+
y
+
z
)
3
=
9
(
x
2
y
+
y
2
z
+
z
2
x
)
.
Prove that
x
=
y
=
z
x=y=z
x
=
y
=
z
2
1
Hide problems
Interesting problem
The sequence
(
a
n
)
n
∈
N
\left(a_{n}\right)_{n\in\mathbb{N}}
(
a
n
)
n
∈
N
is defined recursively as
a
0
=
a
1
=
1
a_{0}=a_{1}=1
a
0
=
a
1
=
1
,
a
n
+
2
=
5
a
n
+
1
−
a
n
−
1
a_{n+2}=5a_{n+1}-a_{n}-1
a
n
+
2
=
5
a
n
+
1
−
a
n
−
1
,
∀
n
∈
N
\forall n\in\mathbb{N}
∀
n
∈
N
Prove that
a
n
∣
a
n
+
1
2
+
a
n
+
1
+
1
a_{n}\mid a_{n+1}^{2}+a_{n+1}+1
a
n
∣
a
n
+
1
2
+
a
n
+
1
+
1
for any
n
∈
N
n\in\mathbb{N}
n
∈
N