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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
2005 Poland - Second Round
2005 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(3)
1
2
Hide problems
Integer m exists such that W(m)=W(m+1)=0
The polynomial
W
(
x
)
=
x
2
+
a
x
+
b
W(x)=x^2+ax+b
W
(
x
)
=
x
2
+
a
x
+
b
with integer coefficients has the following property: for every prime number
p
p
p
there is an integer
k
k
k
such that both
W
(
k
)
W(k)
W
(
k
)
and
W
(
k
+
1
)
W(k+1)
W
(
k
+
1
)
are divisible by
p
p
p
. Show that there is an integer
m
m
m
such that
W
(
m
)
=
W
(
m
+
1
)
=
0
W(m)=W(m+1)=0
W
(
m
)
=
W
(
m
+
1
)
=
0
.
Both expressions are primes
Find all positive integers
n
n
n
for which
n
n
+
1
n^n+1
n
n
+
1
and
(
2
n
)
2
n
+
1
(2n)^{2n}+1
(
2
n
)
2
n
+
1
are prime numbers.
3
2
Hide problems
Number of segments and triangles inequality
In space are given
n
≥
2
n\ge 2
n
≥
2
points, no four of which are coplanar. Some of these points are connected by segments. Let
K
K
K
be the number of segments
(
K
>
1
)
(K>1)
(
K
>
1
)
and
T
T
T
be the number of formed triangles. Prove that
9
T
2
<
2
K
3
9T^2<2K^3
9
T
2
<
2
K
3
.
Poland 2005
Prove that if the real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
lie in the interval
[
0
,
1
]
[0,1]
[
0
,
1
]
, then
a
b
c
+
1
+
b
a
c
+
1
+
c
a
b
+
1
≤
2
\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le 2
b
c
+
1
a
+
a
c
+
1
b
+
ab
+
1
c
≤
2
2
2
Hide problems
Poland second round 2005
In a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
, point
M
M
M
is the midpoint of the diagonal
A
C
AC
A
C
. Prove that if
∠
B
A
D
=
∠
B
M
C
=
∠
C
M
D
\angle BAD=\angle BMC=\angle CMD
∠
B
A
D
=
∠
BMC
=
∠
CM
D
, then a circle can be inscribed in quadrilateral
A
B
C
D
ABCD
A
BC
D
.
In the rhombus ABCD
A rhombus
A
B
C
D
ABCD
A
BC
D
with
∠
B
A
D
=
6
0
∘
\angle BAD=60^{\circ}
∠
B
A
D
=
6
0
∘
is given. Points
E
E
E
on side
A
B
AB
A
B
and
F
F
F
on side
A
D
AD
A
D
are such that
∠
E
C
F
=
∠
A
B
D
\angle ECF=\angle ABD
∠
ECF
=
∠
A
B
D
. Lines
C
E
CE
CE
and
C
F
CF
CF
respectively meet line
B
D
BD
B
D
at
P
P
P
and
Q
Q
Q
. Prove that
P
Q
E
F
=
A
B
B
D
\frac{PQ}{EF}=\frac{AB}{BD}
EF
PQ
=
B
D
A
B
.