MathDB
Problems
Contests
International Contests
Czech-Polish-Slovak Match
2014 Czech-Polish-Slovak Match
2014 Czech-Polish-Slovak Match
Part of
Czech-Polish-Slovak Match
Subcontests
(6)
6
1
Hide problems
union of mutually disjoint 3-element subsets >= n-5
Let
n
≥
6
n \ge 6
n
≥
6
be an integer and
F
F
F
be the system of the
3
3
3
-element subsets of the set
{
1
,
2
,
.
.
.
,
n
}
\{1, 2,...,n \}
{
1
,
2
,
...
,
n
}
satisfying the following condition: for every
1
≤
i
<
j
≤
n
1 \le i < j \le n
1
≤
i
<
j
≤
n
there is at least
⌊
1
3
n
⌋
−
1
\lfloor \frac{1}{3} n \rfloor -1
⌊
3
1
n
⌋
−
1
subsets
A
∈
F
A\in F
A
∈
F
such that
i
,
j
∈
A
i, j \in A
i
,
j
∈
A
. Prove that for some integer
m
≥
1
m \ge 1
m
≥
1
exist the mutually disjoint subsets
A
1
,
A
2
,
.
.
.
,
A
m
∈
F
A_1, A_2 , ... , A_m \in F
A
1
,
A
2
,
...
,
A
m
∈
F
also, that
∣
A
1
∪
A
2
∪
.
.
.
∪
A
m
∣
≥
n
−
5
|A_1\cup A_2 \cup ... \cup A_m |\ge n-5
∣
A
1
∪
A
2
∪
...
∪
A
m
∣
≥
n
−
5
(Poland)PS. just in case my translation does not make sense, I leave the original in Slovak, in case someone understands something else
5
1
Hide problems
two binomials leave the same remainder when divided by 2
Let all positive integers
n
n
n
satisfy the following condition: for each non-negative integers
k
,
m
k, m
k
,
m
with
k
+
m
≤
n
k + m \le n
k
+
m
≤
n
, the numbers
(
n
−
k
m
)
\binom{n-k}{m}
(
m
n
−
k
)
and
(
n
−
m
k
)
\binom{n-m}{k}
(
k
n
−
m
)
leave the same remainder when divided by
2
2
2
.(Poland) PS. The translation was done using Google translate and in case it is not right, there is the original text in Slovak
2
1
Hide problems
x_n = ax_{n-1} + b, m | n => x_m | x_n
For the positive integers
a
,
b
,
x
1
a, b, x_1
a
,
b
,
x
1
we construct the sequence of numbers
(
x
n
)
n
=
1
∞
(x_n)_{n=1}^{\infty}
(
x
n
)
n
=
1
∞
such that
x
n
=
a
x
n
−
1
+
b
x_n = ax_{n-1} + b
x
n
=
a
x
n
−
1
+
b
for each
n
≥
2
n \ge 2
n
≥
2
. Specify the conditions for the given numbers
a
,
b
a, b
a
,
b
and
x
1
x_1
x
1
which are necessary and sufficient for all indexes
m
,
n
m, n
m
,
n
to apply the implication
m
∣
n
⇒
x
m
∣
x
n
m | n \Rightarrow x_m | x_n
m
∣
n
⇒
x
m
∣
x
n
.(Jaromír Šimša)
4
1
Hide problems
concurrent lines by a triangle ,a circle and 3 products
Let
A
B
C
ABC
A
BC
be a triangle, and let
P
P
P
be the midpoint of
A
C
AC
A
C
. A circle intersects
A
P
,
C
P
,
B
C
,
A
B
AP, CP, BC, AB
A
P
,
CP
,
BC
,
A
B
sequentially at their inner points
K
,
L
,
M
,
N
K, L, M, N
K
,
L
,
M
,
N
. Let
S
S
S
be the midpoint of
K
L
KL
K
L
. Let also
2
⋅
∣
A
N
∣
⋅
∣
A
B
∣
⋅
∣
C
L
∣
=
2
⋅
∣
C
M
∣
⋅
∣
B
C
∣
⋅
∣
A
K
∣
=
∣
A
C
∣
⋅
∣
A
K
∣
⋅
∣
C
L
∣
.
2 \cdot | AN |\cdot |AB |\cdot |CL | = 2 \cdot | CM | \cdot| BC | \cdot| AK| = | AC | \cdot| AK |\cdot |CL |.
2
⋅
∣
A
N
∣
⋅
∣
A
B
∣
⋅
∣
C
L
∣
=
2
⋅
∣
CM
∣
⋅
∣
BC
∣
⋅
∣
A
K
∣
=
∣
A
C
∣
⋅
∣
A
K
∣
⋅
∣
C
L
∣.
Prove that if
P
≠
S
P\ne S
P
=
S
, then the intersection of
K
N
KN
K
N
and
M
L
ML
M
L
lies on the perpendicular bisector of the
P
S
PS
PS
. (Jan Mazák)
3
1
Hide problems
perpendiculars on a convex ABCD with 2 angles 135^o
Given is a convex
A
B
C
D
ABCD
A
BC
D
, which is
∣
∠
A
B
C
∣
=
∣
∠
A
D
C
∣
=
13
5
∘
|\angle ABC| = |\angle ADC|= 135^\circ
∣∠
A
BC
∣
=
∣∠
A
D
C
∣
=
13
5
∘
. On the
A
B
,
A
D
AB, AD
A
B
,
A
D
are also selected points
M
,
N
M, N
M
,
N
such that
∣
∠
M
C
D
∣
=
∣
∠
N
C
B
∣
=
9
0
∘
|\angle MCD| = |\angle NCB| = 90^ \circ
∣∠
MC
D
∣
=
∣∠
NCB
∣
=
9
0
∘
. The circumcircles of the triangles
A
M
N
AMN
A
MN
and
A
B
D
ABD
A
B
D
intersect for the second time at point
K
≠
A
K \ne A
K
=
A
. Prove that lines
A
K
AK
A
K
and
K
C
KC
K
C
are perpendicular.(Irán)
1
1
Hide problems
trigonometric existence of triangle given equation w sides
Prove that if the positive real numbers
a
,
b
,
c
a, b, c
a
,
b
,
c
satisfy the equation
a
4
+
b
4
+
c
4
+
4
a
2
b
2
c
2
=
2
(
a
2
b
2
+
a
2
c
2
+
b
2
c
2
)
,
a^4 + b^4 + c^4 + 4a^2b^2c^2 = 2 (a^2b^2 + a^2c^2 + b^2c^2),
a
4
+
b
4
+
c
4
+
4
a
2
b
2
c
2
=
2
(
a
2
b
2
+
a
2
c
2
+
b
2
c
2
)
,
then there is a triangle
A
B
C
ABC
A
BC
with internal angles
α
,
β
,
γ
\alpha, \beta, \gamma
α
,
β
,
γ
such that
sin
α
=
a
,
sin
β
=
b
,
sin
γ
=
c
.
\sin \alpha = a, \qquad \sin \beta = b, \qquad \sin \gamma= c.
sin
α
=
a
,
sin
β
=
b
,
sin
γ
=
c
.