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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1982 Poland - Second Round
1982 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
Hide problems
red and green sections by points in space
Given a finite set
B
B
B
of points in space, any two distances between the points of this set are different. Each point of the set
B
B
B
is connected by a line segment to the closest point of the set
B
B
B
. This way we will get a set of sections, one of which (any chosen one) we paint red, all the remaining sections we paint green. Prove that there are two points of the set
B
B
B
that cannot be connected by a line composed of green segments.
5
1
Hide problems
q^{(q+1)^n}+1 is divisible by (q + 1)^{n+1} but not (q + 1)^{n+2}
Let
q
q
q
be an even positive number. Prove that for every natural number
n
n
n
number
q
(
q
+
1
)
n
+
1
q^{(q+1)^n}+1
q
(
q
+
1
)
n
+
1
is divisible by
(
q
+
1
)
n
+
1
(q + 1)^{n+1}
(
q
+
1
)
n
+
1
but not divisible by
(
q
+
1
)
n
+
2
(q + 1)^{n+2}
(
q
+
1
)
n
+
2
.
4
1
Hide problems
there is a sphere passing through all points of set A
Let
A
A
A
be a finite set of points in space having the property that for any of its points
P
,
Q
P, Q
P
,
Q
there is an isometry of space that transforms the set
A
A
A
into the set
A
A
A
and the point
P
P
P
into the point
Q
Q
Q
. Prove that there is a sphere passing through all points of the set
A
A
A
.
3
1
Hide problems
\log_n 2 \cdot \log_n 4 \cdot \log_n 6 \ldots \log_n (2n - 2) \leq 1.
Prove that for every natural number
n
≥
2
n \geq 2
n
≥
2
the inequality holds
log
n
2
⋅
log
n
4
⋅
log
n
6
…
log
n
(
2
n
−
2
)
≤
1.
\log_n 2 \cdot \log_n 4 \cdot \log_n 6 \ldots \log_n (2n - 2) \leq 1.
lo
g
n
2
⋅
lo
g
n
4
⋅
lo
g
n
6
…
lo
g
n
(
2
n
−
2
)
≤
1.
2
1
Hide problems
each point of plane belongs to at most 5 circles.
The plane is covered with circles in such a way that the center of each circle does not belong to any other circle. Prove that each point of the plane belongs to at most five circles.
1
1
Hide problems
x^3 - 3cx^2 - dx + c = 0 has no more than one rational root.
Prove that if
c
,
d
c, d
c
,
d
are integers with
c
≠
d
c \neq d
c
=
d
,
d
>
0
d > 0
d
>
0
then the equation
x
3
−
3
c
x
2
−
d
x
+
c
=
0
x^3 - 3cx^2 - dx + c = 0
x
3
−
3
c
x
2
−
d
x
+
c
=
0
has no more than one rational root.