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Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
1999 Poland - Second Round
1999 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
Hide problems
a_1+2^ka_2+3^ka_3+...+n^ka_n is divisible by k!
Suppose that
a
1
,
a
2
,
.
.
.
,
a
n
a_1,a_2,...,a_n
a
1
,
a
2
,
...
,
a
n
are integers such that
a
1
+
2
i
a
2
+
3
i
a
3
+
.
.
.
+
n
i
a
n
=
0
a_1 +2^ia_2 +3^ia_3 +...+n^ia_n = 0
a
1
+
2
i
a
2
+
3
i
a
3
+
...
+
n
i
a
n
=
0
for
i
=
1
,
2
,
.
.
.
,
k
−
1
i = 1,2,...,k -1
i
=
1
,
2
,
...
,
k
−
1
, where
k
≥
2
k \ge 2
k
≥
2
is a given integer. Prove that
a
1
+
2
k
a
2
+
3
k
a
3
+
.
.
.
+
n
k
a
n
a_1+2^ka_2+3^ka_3+...+n^ka_n
a
1
+
2
k
a
2
+
3
k
a
3
+
...
+
n
k
a
n
is divisible by
k
!
k!
k
!
.
5
1
Hide problems
S = {1,2,3,4,5}, no of f : f : S \to S such that f ^{50}(x)= x
Let
S
=
{
1
,
2
,
3
,
4
,
5
}
S = \{1,2,3,4,5\}
S
=
{
1
,
2
,
3
,
4
,
5
}
. Find the number of functions
f
:
S
→
S
f : S \to S
f
:
S
→
S
such that
f
50
(
x
)
=
x
f ^{50}(x)= x
f
50
(
x
)
=
x
for all
x
∈
S
x \in S
x
∈
S
.
4
1
Hide problems
pair of equal angles given, right angle wanted, circumcenter related
Let
P
P
P
be a point inside a triangle
A
B
C
ABC
A
BC
such that
∠
P
A
B
=
∠
P
C
A
\angle PAB = \angle PCA
∠
P
A
B
=
∠
PC
A
and
∠
P
A
C
=
∠
P
B
A
\angle PAC = \angle PBA
∠
P
A
C
=
∠
PB
A
. If
O
≠
P
O \ne P
O
=
P
is the circumcenter of
△
A
B
C
\triangle ABC
△
A
BC
, prove that
∠
A
P
O
\angle APO
∠
A
PO
is right.
3
1
Hide problems
ratio of triangle area indepedent of choices of 2 points, ratios in cyclic
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral and let
E
E
E
and
F
F
F
be the points on the sides
A
B
AB
A
B
and
C
D
CD
C
D
respectively such that
A
E
:
E
B
=
C
F
:
F
D
AE : EB = CF : FD
A
E
:
EB
=
CF
:
F
D
. Point
P
P
P
on the segment EF satsfies
E
P
:
P
F
=
A
B
:
C
D
EP : PF = AB : CD
EP
:
PF
=
A
B
:
C
D
. Prove that the ratio of the areas of
△
A
P
D
\vartriangle APD
△
A
P
D
and
△
B
P
C
\vartriangle BPC
△
BPC
does not depend on the choice of
E
E
E
and
F
F
F
.
2
1
Hide problems
a cube of edge 2^n is divided into 2^{3n} unit cubes , remove unit cube
A cube of edge
2
2
2
with one of the corner unit cubes removed is called a piece. Prove that if a cube
T
T
T
of edge
2
n
2^n
2
n
is divided into
2
3
n
2^{3n}
2
3
n
unit cubes and one of the unit cubes is removed, then the rest can be cut into pieces.
1
1
Hide problems
f(1/n) = (−1)^n, there are no increasing functions such f = g - h
Let
f
:
(
0
,
1
)
→
R
f : (0,1) \to R
f
:
(
0
,
1
)
→
R
be a function such that
f
(
1
/
n
)
=
(
−
1
)
n
f(1/n) = (-1)^n
f
(
1/
n
)
=
(
−
1
)
n
for all n ∈ N. Prove that there are no increasing functions
g
,
h
:
(
0
,
1
)
→
R
g,h : (0,1) \to R
g
,
h
:
(
0
,
1
)
→
R
such that
f
=
g
−
h
f = g - h
f
=
g
−
h
.