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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1994 Poland - Second Round
1994 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
2
1
Hide problems
\sum_{i=1}^n ai=\Pi_{i=1}^n a_i, 0<a_i \le b_i=>\sum_{i=1}^n bi \le\Pi_{i=1}^n
Let
a
1
,
.
.
.
,
a
n
a_1,...,a_n
a
1
,
...
,
a
n
be positive real numbers such that
∑
i
=
1
n
a
i
=
∏
i
=
1
n
a
i
\sum_{i=1}^n a_i =\prod_{i=1}^n a_i
∑
i
=
1
n
a
i
=
∏
i
=
1
n
a
i
, and let
b
1
,
.
.
.
,
b
n
b_1,...,b_n
b
1
,
...
,
b
n
be positive real numbers such that
a
i
≤
b
i
a_i \le b_i
a
i
≤
b
i
for all
i
i
i
. Prove that
∑
i
=
1
n
b
i
≤
∏
i
=
1
n
b
i
\sum_{i=1}^n b_i \le\prod_{i=1}^n b_i
∑
i
=
1
n
b
i
≤
∏
i
=
1
n
b
i
6
1
Hide problems
p | n^2 -n+3 iff p | m^2 -m+25
Let
p
p
p
be a prime number. Prove that there exists
n
∈
Z
n \in Z
n
∈
Z
such that
p
∣
n
2
−
n
+
3
p | n^2 -n+3
p
∣
n
2
−
n
+
3
if and only if there exists
m
∈
Z
m \in Z
m
∈
Z
such that
p
∣
m
2
−
m
+
25
p | m^2 -m+25
p
∣
m
2
−
m
+
25
.
5
1
Hide problems
right angle wanted, incircle and angle bisector related
The incircle
ω
\omega
ω
of a triangle
A
B
C
ABC
A
BC
is tangent to the sides
A
B
AB
A
B
and
B
C
BC
BC
at
P
P
P
and
Q
Q
Q
respectively. The angle bisector at
A
A
A
meets
P
Q
PQ
PQ
at point
S
S
S
. Prove
∠
A
S
C
=
9
0
o
\angle ASC = 90^o
∠
A
SC
=
9
0
o
.
4
1
Hide problems
sum of products of 4 numbers at vertices of a cube, with 1 or -1 assigned
Each vertex of a cube is assigned
1
1
1
or
−
1
-1
−
1
. Each face is assigned the product of the four numbers at its vertices. Determine all possible values that can be obtained as the sum of all the
14
14
14
assigned numbers.
3
1
Hide problems
plane passing center of cube intersects the cube in cyclic hexagon, regular ?
A plane passing through the center of a cube intersects the cube in a cyclic hexagon. Show that this hexagon is regular.
1
1
Hide problems
5-degree polynomial such that (x-1)^3| P(x)+1 and (x+1)^3| P(x)-1
Find all real polynomials
P
(
x
)
P(x)
P
(
x
)
of degree
5
5
5
such that
(
x
−
1
)
3
∣
P
(
x
)
+
1
(x-1)^3| P(x)+1
(
x
−
1
)
3
∣
P
(
x
)
+
1
and
(
x
+
1
)
3
∣
P
(
x
)
−
1
(x+1)^3| P(x)-1
(
x
+
1
)
3
∣
P
(
x
)
−
1
.