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Problems
Contests
National and Regional Contests
Poland Contests
Poland - First Round
1992 Poland - First Round
1992 Poland - First Round
Part of
Poland - First Round
Subcontests
(12)
12
1
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Factoring x^n+4
Prove that the polynomial
x
n
+
4
x^n+4
x
n
+
4
can be expressed as a product of two polynomials (each with degree less than
n
n
n
) with integer coefficients, if and only if
n
n
n
is divisible by
4
4
4
.
11
1
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Six pawns on a chessboard
Given is a
n
×
n
n \times n
n
×
n
chessboard. With the same probability, we put six pawns on its six cells. Let
p
n
p_n
p
n
denotes the probability that there exists a row or a column containing at least two pawns. Find
lim
n
→
∞
n
p
n
\lim_{n \to \infty} np_n
lim
n
→
∞
n
p
n
.
10
1
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Cubes, surjections, and isometries
Let
C
C
C
be a cube and let
f
:
C
⟶
C
f: C \longrightarrow C
f
:
C
⟶
C
be a surjection with
∣
P
Q
∣
≥
∣
f
(
P
)
f
(
Q
)
∣
|PQ| \geq |f(P)f(Q)|
∣
PQ
∣
≥
∣
f
(
P
)
f
(
Q
)
∣
for all
P
,
Q
∈
C
P,Q \in C
P
,
Q
∈
C
. Prove that
f
f
f
is an isometry.
9
1
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Sum of squares vs the sum squared
Prove that for all real numbers
a
,
b
,
c
a,b,c
a
,
b
,
c
the inequality
(
a
2
+
b
2
−
c
2
)
(
b
2
+
c
2
−
a
2
)
(
c
2
+
a
2
−
b
2
)
≤
(
a
+
b
−
c
)
2
(
b
+
c
−
a
)
2
(
c
+
a
−
b
)
2
(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2) \leq (a+b-c)^2(b+c-a)^2(c+a-b)^2
(
a
2
+
b
2
−
c
2
)
(
b
2
+
c
2
−
a
2
)
(
c
2
+
a
2
−
b
2
)
≤
(
a
+
b
−
c
)
2
(
b
+
c
−
a
)
2
(
c
+
a
−
b
)
2
holds.
8
1
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Maximizing sum of cubes
Given is a positive integer
n
≥
2
n \geq 2
n
≥
2
. Determine the maximum value of the sum of natural numbers
k
1
,
k
2
,
.
.
.
,
k
n
k_1,k_2,...,k_n
k
1
,
k
2
,
...
,
k
n
satisfying the condition:
k
1
3
+
k
2
3
+
⋯
+
k
n
3
≤
7
n
k_1^3+k_2^3+ \dots +k_n^3 \leq 7n
k
1
3
+
k
2
3
+
⋯
+
k
n
3
≤
7
n
.
7
1
Hide problems
Polyhedra and an overrightarrow equation.
Given are the points
A
0
=
(
0
,
0
,
0
)
,
A
1
=
(
1
,
0
,
0
)
,
A
2
=
(
0
,
1
,
0
)
,
A
3
=
(
0
,
0
,
1
)
A_0 = (0,0,0), A_1 = (1,0,0), A_2 = (0,1,0), A_3 = (0,0,1)
A
0
=
(
0
,
0
,
0
)
,
A
1
=
(
1
,
0
,
0
)
,
A
2
=
(
0
,
1
,
0
)
,
A
3
=
(
0
,
0
,
1
)
in the space. Let
P
i
j
(
i
,
j
∈
0
,
1
,
2
,
3
)
P_{ij} (i,j \in 0,1,2,3)
P
ij
(
i
,
j
∈
0
,
1
,
2
,
3
)
be the point determined by the equality:
A
0
P
i
j
→
=
A
i
A
j
→
\overrightarrow{A_0P_{ij}} = \overrightarrow{A_iA_j}
A
0
P
ij
=
A
i
A
j
. Find the volume of the smallest convex polyhedron which contains all the points
P
i
j
P_{ij}
P
ij
.
6
1
Hide problems
Sum in a recursive series in a sum
The sequence
(
x
n
)
(x_n)
(
x
n
)
is determined by the conditions:
x
0
=
1992
,
x
n
=
−
1992
n
⋅
∑
k
=
0
n
−
1
x
k
x_0=1992,x_n=-\frac{1992}{n} \cdot \sum_{k=0}^{n-1} x_k
x
0
=
1992
,
x
n
=
−
n
1992
⋅
∑
k
=
0
n
−
1
x
k
for
n
≥
1
n \geq 1
n
≥
1
. Find
∑
n
=
0
1992
2
n
x
n
\sum_{n=0}^{1992} 2^nx_n
∑
n
=
0
1992
2
n
x
n
.
5
1
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Halfplane
Given is a halfplane with points
A
A
A
and
C
C
C
on its edge. For every point
B
B
B
on this halfplane consider the squares
A
B
K
L
ABKL
A
B
K
L
and
B
C
M
N
BCMN
BCMN
lying outside the triangle
A
B
C
ABC
A
BC
. Prove that all the lines
L
M
LM
L
M
(as the point
B
B
B
varies) have a common point.
4
1
Hide problems
Functional equation
Determine all functions
f
:
R
⟶
R
f: R \longrightarrow R
f
:
R
⟶
R
such that
f
(
x
+
y
)
−
f
(
x
−
y
)
=
f
(
x
)
∗
f
(
y
)
f(x+y)-f(x-y)=f(x)*f(y)
f
(
x
+
y
)
−
f
(
x
−
y
)
=
f
(
x
)
∗
f
(
y
)
for
x
,
y
∈
R
x,y \in R
x
,
y
∈
R
3
1
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Hexagon with a center of symmetry
Given is a hexagon
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
with a center of symmetry. The lines
A
B
AB
A
B
and
E
F
EF
EF
meet at the point
A
′
A'
A
′
, the lines
B
C
BC
BC
and
A
F
AF
A
F
meet at the point
B
′
B'
B
′
, and the lines
A
B
AB
A
B
and
C
D
CD
C
D
meet at the point
C
′
C'
C
′
. Prove that
A
B
⋅
B
C
⋅
C
D
=
A
A
′
⋅
B
B
′
⋅
C
C
′
AB \cdot BC \cdot CD = AA' \cdot BB' \cdot CC'
A
B
⋅
BC
⋅
C
D
=
A
A
′
⋅
B
B
′
⋅
C
C
′
.
2
1
Hide problems
Trigonometric system of equations
Given is a natural number
n
≥
3
n \geq 3
n
≥
3
. Solve the system of equations: $ \begin{cases} \tan (x_1) + 3 \cot (x_1) &= 2 \tan (x_2) \\ \tan (x_2) + 3 \cot (x_2) &= 2 \tan (x_3) \\ & \dots \\ \tan (x_n) + 3 \cot (x_n) &= 2 \tan (x_1) \\ \end{cases} $
1
1
Hide problems
Diophantine equation with sgn
Solve the following equation in real numbers:
(
x
2
−
1
)
(
∣
x
∣
+
1
)
x
+
s
g
n
x
=
[
x
+
1
]
.
\frac{(x^2-1)(|x|+1)}{x+sgnx}=[x+1].
x
+
s
g
n
x
(
x
2
−
1
)
(
∣
x
∣
+
1
)
=
[
x
+
1
]
.