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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
2010 Poland - Second Round
2010 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(3)
3
2
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Polish 2nd stage 2010, 3rd problem (consecutive numbers)
Positive integer numbers
k
k
k
and
n
n
n
satisfy the inequality
k
>
n
!
k > n!
k
>
n
!
. Prove that there exist pairwisely different prime numbers
p
1
,
p
2
,
…
,
p
n
p_1, p_2, \ldots, p_n
p
1
,
p
2
,
…
,
p
n
which are divisors of the numbers
k
+
1
,
k
+
2
,
…
,
k
+
n
k+1, k+2, \ldots, k+n
k
+
1
,
k
+
2
,
…
,
k
+
n
respectively (i.e.
p
i
∣
k
+
i
p_i|k+i
p
i
∣
k
+
i
).
Polish 2nd stage 2010, 6th problem (arithmetic means)
The
n
n
n
-element set of real numbers is given, where
n
≥
6
n \geq 6
n
≥
6
. Prove that there exist at least
n
−
1
n-1
n
−
1
two-element subsets of this set, in which the arithmetic mean of elements is not less than the arithmetic mean of elements in the whole set.
2
2
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Polish 2nd stage 2010, 2nd problem (tetrahedron)
The orthogonal projections of the vertices
A
,
B
,
C
A, B, C
A
,
B
,
C
of the tetrahedron
A
B
C
D
ABCD
A
BC
D
on the opposite faces are denoted by
A
′
,
B
′
,
C
′
A', B', C'
A
′
,
B
′
,
C
′
respectively. Suppose that point
A
′
A'
A
′
is the circumcenter of the triangle
B
C
D
BCD
BC
D
, point
B
′
B'
B
′
is the incenter of the triangle
A
C
D
ACD
A
C
D
and
C
′
C'
C
′
is the centroid of the triangle
A
B
D
ABD
A
B
D
. Prove that tetrahedron
A
B
C
D
ABCD
A
BC
D
is regular.
Polish 2nd stage 2010, 5th problem (functional equation)
Find all monotonic functions
f
:
R
→
R
f: \mathbb{R} \rightarrow \mathbb{R}
f
:
R
→
R
satisfying
f
(
f
(
x
)
−
y
)
+
f
(
x
+
y
)
=
0
,
f(f(x) - y) + f(x+y) = 0,
f
(
f
(
x
)
−
y
)
+
f
(
x
+
y
)
=
0
,
for every real
x
,
y
x, y
x
,
y
. (Note that monotonic means that function is not increasing or not decreasing)
1
2
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Polish 2nd stage 2010, 1st problem (system of equations)
Solve in the real numbers
x
,
y
,
z
x, y, z
x
,
y
,
z
a system of the equations:
{
x
2
−
(
y
+
z
+
y
z
)
x
+
(
y
+
z
)
y
z
=
0
y
2
−
(
z
+
x
+
z
x
)
y
+
(
z
+
x
)
z
x
=
0
z
2
−
(
x
+
y
+
x
y
)
z
+
(
x
+
y
)
x
y
=
0.
\begin{cases} x^2 - (y+z+yz)x + (y+z)yz = 0 \\ y^2 - (z + x + zx)y + (z+x)zx = 0 \\ z^2 - (x+y+xy)z + (x+y)xy = 0. \\ \end{cases}
⎩
⎨
⎧
x
2
−
(
y
+
z
+
yz
)
x
+
(
y
+
z
)
yz
=
0
y
2
−
(
z
+
x
+
z
x
)
y
+
(
z
+
x
)
z
x
=
0
z
2
−
(
x
+
y
+
x
y
)
z
+
(
x
+
y
)
x
y
=
0.
Polish 2nd stage 2010, 4th problem (pentagon with one angle)
In the convex pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
all interior angles have the same measure. Prove that the perpendicular bisector of segment
E
A
EA
E
A
, the perpendicular bisector of segment
B
C
BC
BC
and the angle bisector of
∠
C
D
E
\angle CDE
∠
C
D
E
intersect in one point.