MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
2003 Poland - Second Round
2003 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
6
1
Hide problems
Integer function, number of solutions of equation
Each pair
(
x
,
y
)
(x, y)
(
x
,
y
)
of nonnegative integers is assigned number
f
(
x
,
y
)
f(x, y)
f
(
x
,
y
)
according the conditions:
f
(
0
,
0
)
=
0
f(0, 0) = 0
f
(
0
,
0
)
=
0
;
f
(
2
x
,
2
y
)
=
f
(
2
x
+
1
,
2
y
+
1
)
=
f
(
x
,
y
)
f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y)
f
(
2
x
,
2
y
)
=
f
(
2
x
+
1
,
2
y
+
1
)
=
f
(
x
,
y
)
,
f
(
2
x
+
1
,
2
y
)
=
f
(
2
x
,
2
y
+
1
)
=
f
(
x
,
y
)
+
1
f(2x + 1, 2y) = f(2x, 2y + 1) = f(x ,y) + 1
f
(
2
x
+
1
,
2
y
)
=
f
(
2
x
,
2
y
+
1
)
=
f
(
x
,
y
)
+
1
for
x
,
y
≥
0
x, y \ge 0
x
,
y
≥
0
. Let
n
n
n
be a fixed nonnegative integer and let
a
a
a
,
b
b
b
be nonnegative integers such that
f
(
a
,
b
)
=
n
f(a, b) = n
f
(
a
,
b
)
=
n
. Decide how many numbers satisfy the equation
f
(
a
,
x
)
+
f
(
b
,
x
)
=
n
f(a, x) + f(b, x) = n
f
(
a
,
x
)
+
f
(
b
,
x
)
=
n
.
1
1
Hide problems
Sequence of binomial coefficients
Prove that exists integer
n
>
2003
n > 2003
n
>
2003
that in sequence
(
n
0
)
\binom{n}{0}
(
0
n
)
,
(
n
1
)
\binom{n}{1}
(
1
n
)
,
(
n
2
)
\binom{n}{2}
(
2
n
)
, ...,
(
n
2003
)
\binom{n}{2003}
(
2003
n
)
each element is a divisor of all elements which are after him.
3
1
Hide problems
Polynomial equation
Let
W
(
x
)
=
x
4
−
3
x
3
+
5
x
2
−
9
x
W(x) = x^4 - 3x^3 + 5x^2 - 9x
W
(
x
)
=
x
4
−
3
x
3
+
5
x
2
−
9
x
be a polynomial. Determine all pairs of different integers
a
a
a
,
b
b
b
satisfying the equation
W
(
a
)
=
W
(
b
)
W(a) = W(b)
W
(
a
)
=
W
(
b
)
.
4
1
Hide problems
Sum of two squares, equation
Prove that for any prime number
p
>
3
p > 3
p
>
3
exist integers
x
,
y
,
k
x, y, k
x
,
y
,
k
that meet conditions:
0
<
2
k
<
p
0 < 2k < p
0
<
2
k
<
p
and
k
p
+
3
=
x
2
+
y
2
kp + 3 = x^2 + y^2
k
p
+
3
=
x
2
+
y
2
.
5
1
Hide problems
Construction of similar triangle
Point
A
A
A
lies outside circle
o
o
o
of center
O
O
O
. From point
A
A
A
draw two lines tangent to a circle
o
o
o
in points
B
B
B
and
C
C
C
. A tangent to a circle
o
o
o
cuts segments
A
B
AB
A
B
and
A
C
AC
A
C
in points
E
E
E
and
F
F
F
, respectively. Lines
O
E
OE
OE
and
O
F
OF
OF
cut segment
B
C
BC
BC
in points
P
P
P
and
Q
Q
Q
, respectively. Prove that from line segments
B
P
BP
BP
,
P
Q
PQ
PQ
,
Q
C
QC
QC
can construct triangle similar to triangle
A
E
F
AEF
A
EF
.
2
1
Hide problems
Cyclic Quadrilateral, bisectors, points lie on perpendicular line
The quadrilateral
A
B
C
D
ABCD
A
BC
D
is inscribed in the circle
o
o
o
. Bisectors of angles
D
A
B
DAB
D
A
B
and
A
B
C
ABC
A
BC
intersect at point
P
P
P
, and bisectors of angles
B
C
D
BCD
BC
D
and
C
D
A
CDA
C
D
A
intersect in point
Q
Q
Q
. Point
M
M
M
is the center of this arc
B
C
BC
BC
of the circle
o
o
o
which does not contain points
D
D
D
and
A
A
A
. Point
N
N
N
is the center of the arc
D
A
DA
D
A
of the circle
o
o
o
, which does not contain points
B
B
B
and
C
C
C
. Prove that the points
P
P
P
and
Q
Q
Q
lie on the line perpendicular to
M
N
MN
MN
.