MathDB
Problems
Contests
National and Regional Contests
Poland Contests
Poland - Second Round
2013 Poland - Second Round
2013 Poland - Second Round
Part of
Poland - Second Round
Subcontests
(6)
3
1
Hide problems
Arrange square with polyomino
We have tiles (which are build from squares of side length 1) of following shapes: [asy] unitsize(0.5 cm); draw((1,0)--(2,0)); draw((1,1)--(2,1)); draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((0,1)--(1,1)); draw((0,2)--(1,2)); draw((0,1)--(0,2)); draw((1,1)--(1,2)); draw((0, 0)--(1, 0)); draw((0, 0)--(0, 1));draw((5,0)--(6,0)); draw((5,1)--(6,1)); draw((5,0)--(5,1)); draw((6,0)--(6,1)); draw((4,1)--(5,1)); draw((5,2)--(6,2)); draw((5,1)--(5,2)); draw((6,1)--(6,2)); draw((4, 0)--(5, 0)); draw((4, 0)--(4, 1)); draw((6,2)--(7,2)); draw((7,1)--(7,2)); draw((6,1)--(7,1));draw((11,0)--(12,0)); draw((11,1)--(12,1)); draw((11,0)--(11,1)); draw((12,0)--(12,1)); draw((10,1)--(11,1)); draw((10,2)--(11,2)); draw((10,1)--(10,2)); draw((11,1)--(11,2)); draw((10, 0)--(11, 0)); draw((10, 0)--(10, 1)); draw((9, 2)--(9, 1)); draw((9,1)--(10, 1)); draw((9,2)--(10,2)); [/asy] For each odd integer
n
≥
7
n \ge 7
n
≥
7
, determine minimal number of these tiles needed to arrange square with side of length
n
n
n
. (Attention: Tiles can be rotated, but they can't overlap.)
2
1
Hide problems
Two circles
Circles
o
1
o_1
o
1
and
o
2
o_2
o
2
with centers in
O
1
O_1
O
1
and
O
2
O_2
O
2
, respectively, intersect in two different points
A
A
A
and
B
B
B
, wherein angle
O
1
A
O
2
O_1AO_2
O
1
A
O
2
is obtuse. Line
O
1
B
O_1B
O
1
B
intersects circle
o
2
o_2
o
2
in point
C
≠
B
C \neq B
C
=
B
. Line
O
2
B
O_2B
O
2
B
intersects circle
o
1
o_1
o
1
in point
D
≠
B
D \neq B
D
=
B
. Show that point
B
B
B
is incenter of triangle
A
C
D
ACD
A
C
D
.
1
1
Hide problems
Divisibility of product, trinomial
Let
b
b
b
,
c
c
c
be integers and
f
(
x
)
=
x
2
+
b
x
+
c
f(x) = x^2 + bx + c
f
(
x
)
=
x
2
+
b
x
+
c
be a trinomial. Prove, that if for integers
k
1
k_1
k
1
,
k
2
k_2
k
2
and
k
3
k_3
k
3
values of
f
(
k
1
)
f(k_1)
f
(
k
1
)
,
f
(
k
2
)
f(k_2)
f
(
k
2
)
and
f
(
k
3
)
f(k_3)
f
(
k
3
)
are divisible by integer
n
≠
0
n \neq 0
n
=
0
, then product
(
k
1
−
k
2
)
(
k
2
−
k
3
)
(
k
3
−
k
1
)
(k_1 - k_2)(k_2 - k_3)(k_3 - k_1)
(
k
1
−
k
2
)
(
k
2
−
k
3
)
(
k
3
−
k
1
)
is divisible by
n
n
n
too.
6
1
Hide problems
Similar tetrahedrons
Decide, whether exist tetrahedrons
T
T
T
,
T
′
T'
T
′
with walls
S
1
S_1
S
1
,
S
2
S_2
S
2
,
S
3
S_3
S
3
,
S
4
S_4
S
4
and
S
1
′
S_1'
S
1
′
,
S
2
′
S_2'
S
2
′
,
S
3
′
S_3'
S
3
′
,
S
4
′
S_4'
S
4
′
, respectively, such that for
i
=
1
,
2
,
3
,
4
i = 1, 2, 3, 4
i
=
1
,
2
,
3
,
4
triangle
S
i
S_i
S
i
is similar to triangle
S
i
′
S_i'
S
i
′
, but despite this, tetrahedron
T
T
T
is not similar to tetrahedron
T
′
T'
T
′
.
5
1
Hide problems
Polynomial of integer coefficients
Let
W
(
x
)
W(x)
W
(
x
)
be a polynomial of integer coefficients such that for any pair of different rational number
r
1
r_1
r
1
,
r
2
r_2
r
2
dependence
W
(
r
1
)
≠
W
(
r
2
)
W(r_1) \neq W(r_2)
W
(
r
1
)
=
W
(
r
2
)
is true. Decide, whether the assuptions imply that for any pair of different real numbers
t
1
t_1
t
1
,
t
2
t_2
t
2
dependence
W
(
t
1
)
≠
W
(
t
2
)
W(t_1) \neq W(t_2)
W
(
t
1
)
=
W
(
t
2
)
is true.
4
1
Hide problems
Equation in real numbers
Solve equation
(
x
4
+
3
y
2
)
∣
x
+
2
∣
+
∣
y
∣
=
4
∣
x
y
2
∣
(x^4 + 3y^2)\sqrt{|x + 2| + |y|}=4|xy^2|
(
x
4
+
3
y
2
)
∣
x
+
2∣
+
∣
y
∣
=
4∣
x
y
2
∣
in real numbers
x
x
x
,
y
y
y
.