MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
2011 Poland - Second Round
2011 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(3)
3
2
Hide problems
Perfect square
Prove that
∀
x
1
,
x
2
,
…
,
x
2011
,
y
1
,
y
2
,
…
,
y
2011
∈
Z
+
\forall x_{1},x_{2},\ldots,x_{2011},y_{1},y_{2},\ldots,y_{2011}\in\mathbb{Z_{+}}
∀
x
1
,
x
2
,
…
,
x
2011
,
y
1
,
y
2
,
…
,
y
2011
∈
Z
+
product:
(
2
x
1
2
+
3
y
1
2
)
(
2
x
2
2
+
3
y
2
2
)
…
(
2
x
2011
2
+
3
y
2011
2
)
(2x_{1}^{2}+3y_{1}^{2})(2x_{2}^{2}+3y_{2}^{2})\ldots(2x_{2011}^{2}+3y_{2011}^{2})
(
2
x
1
2
+
3
y
1
2
)
(
2
x
2
2
+
3
y
2
2
)
…
(
2
x
2011
2
+
3
y
2011
2
)
is not a perfect square.
Divisibility of polynomials
There are two given different polynomials
P
(
x
)
,
Q
(
x
)
P(x),Q(x)
P
(
x
)
,
Q
(
x
)
with real coefficients such that
P
(
Q
(
x
)
)
=
Q
(
P
(
x
)
)
P(Q(x))=Q(P(x))
P
(
Q
(
x
))
=
Q
(
P
(
x
))
. Prove that
∀
n
∈
Z
+
\forall n\in \mathbb{Z_{+}}
∀
n
∈
Z
+
polynomial:
P
(
P
(
…
P
(
P
⏟
n
(
x
)
)
…
)
)
−
Q
(
Q
(
…
Q
(
Q
⏟
n
(
x
)
)
…
)
)
\underbrace{P(P(\ldots P(P}_{n}(x))\ldots))- \underbrace{Q(Q(\ldots Q(Q}_{n}(x))\ldots))
n
P
(
P
(
…
P
(
P
(
x
))
…
))
−
n
Q
(
Q
(
…
Q
(
Q
(
x
))
…
))
is divisible by
P
(
x
)
−
Q
(
x
)
P(x)-Q(x)
P
(
x
)
−
Q
(
x
)
.
2
2
Hide problems
Cyclic quadrilateral
The convex quadrilateral
A
B
C
D
ABCD
A
BC
D
is given,
A
B
<
B
C
AB<BC
A
B
<
BC
and
A
D
<
C
D
AD<CD
A
D
<
C
D
.
P
,
Q
P,Q
P
,
Q
are points on
B
C
BC
BC
and
C
D
CD
C
D
respectively such that
P
B
=
A
B
PB=AB
PB
=
A
B
and
Q
D
=
A
D
QD=AD
Q
D
=
A
D
.
M
M
M
is midpoint of
P
Q
PQ
PQ
. We assume that
∠
B
M
D
=
9
0
∘
\angle BMD=90^{\circ}
∠
BM
D
=
9
0
∘
, prove that
A
B
C
D
ABCD
A
BC
D
is cyclic.
Maximal length of a sequence
∀
n
∈
Z
+
−
{
1
,
2
}
\forall n\in \mathbb{Z_{+}}-\{1,2\}
∀
n
∈
Z
+
−
{
1
,
2
}
find the maximal length of a sequence with elements from a set
{
1
,
2
,
…
,
n
}
\{1,2,\ldots,n\}
{
1
,
2
,
…
,
n
}
, such that any two consecutive elements of this sequence are different and after removing all elements except for the four we do not receive a sequence in form
x
,
y
,
x
,
y
x,y,x,y
x
,
y
,
x
,
y
(
x
≠
y
x\neq y
x
=
y
).
1
2
Hide problems
System of equations
For
x
,
y
∈
R
x,y\in\mathbb{R}
x
,
y
∈
R
, solve the system of equations
{
(
x
−
y
)
(
x
3
+
y
3
)
=
7
(
x
+
y
)
(
x
3
−
y
3
)
=
3
\begin{cases} (x-y)(x^3+y^3)=7 \\ (x+y)(x^3-y^3)=3 \end{cases}
{
(
x
−
y
)
(
x
3
+
y
3
)
=
7
(
x
+
y
)
(
x
3
−
y
3
)
=
3
Points on semicircle
Points
A
,
B
,
C
,
D
,
E
,
F
A,B,C,D,E,F
A
,
B
,
C
,
D
,
E
,
F
lie in that order on semicircle centered at
O
O
O
, we assume that
A
D
=
B
E
=
C
F
AD=BE=CF
A
D
=
BE
=
CF
.
G
G
G
is a common point of
B
E
BE
BE
and
A
D
AD
A
D
,
H
H
H
is a common point of
B
E
BE
BE
and
C
D
CD
C
D
. Prove that:
∠
A
O
C
=
2
∠
G
O
H
.
\angle AOC=2\angle GOH.
∠
A
OC
=
2∠
GO
H
.