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Contests
National and Regional Contests
Poland Contests
Polish MO Finals
2000 Polish MO Finals
2000 Polish MO Finals
Part of
Polish MO Finals
Subcontests
(3)
1
2
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Find number of solutions in non-negative reals
Find number of solutions in non-negative reals to the following equations: \begin{eqnarray*}x_1 + x_n ^2 = 4x_n \\ x_2 + x_1 ^2 = 4x_1 \\ ... \\ x_n + x_{n-1}^2 = 4x_{n-1} \end{eqnarray*}
Pyramid and inequality
P
A
1
A
2
.
.
.
A
n
PA_1A_2...A_n
P
A
1
A
2
...
A
n
is a pyramid. The base
A
1
A
2
.
.
.
A
n
A_1A_2...A_n
A
1
A
2
...
A
n
is a regular n-gon. The apex
P
P
P
is placed so that the lines
P
A
i
PA_i
P
A
i
all make an angle
6
0
⋅
60^{\cdot}
6
0
⋅
with the plane of the base. For which
n
n
n
is it possible to find
B
i
B_i
B
i
on
P
A
i
PA_i
P
A
i
for
i
=
2
,
3
,
.
.
.
,
n
i = 2, 3, ... , n
i
=
2
,
3
,
...
,
n
such that
A
1
B
2
+
B
2
B
3
+
B
3
B
4
+
.
.
.
+
B
n
−
1
B
n
+
B
n
A
1
<
2
A
1
P
A_1B_2 + B_2B_3 + B_3B_4 + ... + B_{n-1}B_n + B_nA_1 < 2A_1P
A
1
B
2
+
B
2
B
3
+
B
3
B
4
+
...
+
B
n
−
1
B
n
+
B
n
A
1
<
2
A
1
P
?
3
2
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polynomial of odd degree
Show that the only polynomial of odd degree satisfying
p
(
x
2
−
1
)
=
p
(
x
)
2
−
1
p(x^2-1) = p(x)^2 - 1
p
(
x
2
−
1
)
=
p
(
x
)
2
−
1
for all
x
x
x
is
p
(
x
)
=
x
p(x) = x
p
(
x
)
=
x
bounded sequence
The sequence
p
1
,
p
2
,
p
3
,
.
.
.
p_1, p_2, p_3, ...
p
1
,
p
2
,
p
3
,
...
is defined as follows.
p
1
p_1
p
1
and
p
2
p_2
p
2
are primes.
p
n
p_n
p
n
is the greatest prime divisor of
p
n
−
1
+
p
n
−
2
+
2000
p_{n-1} + p_{n-2} + 2000
p
n
−
1
+
p
n
−
2
+
2000
. Show that the sequence is bounded.
2
2
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Chessboard
In the unit squre For the given natural number
n
≥
2
n \geq 2
n
≥
2
find the smallest number
k
k
k
that from each set of
k
k
k
unit squares of the
n
n
n
x
n
n
n
chessboard one can achoose a subset such that the number of the unit squares contained in this subset an lying in a row or column of the chessboard is even
Midpoint of base in isosceles triangle [<APM + <BPC = 180°]
Let a triangle
A
B
C
ABC
A
BC
satisfy
A
C
=
B
C
AC = BC
A
C
=
BC
; in other words, let
A
B
C
ABC
A
BC
be an isosceles triangle with base
A
B
AB
A
B
. Let
P
P
P
be a point inside the triangle
A
B
C
ABC
A
BC
such that
∠
P
A
B
=
∠
P
B
C
\angle PAB = \angle PBC
∠
P
A
B
=
∠
PBC
. Denote by
M
M
M
the midpoint of the segment
A
B
AB
A
B
. Show that
∠
A
P
M
+
∠
B
P
C
=
18
0
∘
\angle APM + \angle BPC = 180^{\circ}
∠
A
PM
+
∠
BPC
=
18
0
∘
.