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Czech-Polish-Slovak Match
2018 Czech-Polish-Slovak Match
2018 Czech-Polish-Slovak Match
Part of
Czech-Polish-Slovak Match
Subcontests
(7)
Source
1
Hide problems
Czech-Polish-Slovak Match 2018
[url=https://artofproblemsolving.com/community/c678145]Czech-Polish-Slovak Match 2018 Austria, 24 - 27 June 2018[url=http://artofproblemsolving.com/community/c6h1667029p10595005]Problem 1. Determine all functions
f
:
R
→
R
f : \mathbb R \to \mathbb R
f
:
R
→
R
such that for all real numbers
x
x
x
and
y
y
y
,
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
Proposed by Walther Janous, Austria[url=http://artofproblemsolving.com/community/c6h1667030p10595011]Problem 2. Let
A
B
C
ABC
A
BC
be an acute scalene triangle. Let
D
D
D
and
E
E
E
be points on the sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively, such that
B
D
=
C
E
BD=CE
B
D
=
CE
. Denote by
O
1
O_1
O
1
and
O
2
O_2
O
2
the circumcentres of the triangles
A
B
E
ABE
A
BE
and
A
C
D
ACD
A
C
D
, respectively. Prove that the circumcircles of the triangles
A
B
C
,
A
D
E
ABC, ADE
A
BC
,
A
D
E
, and
A
O
1
O
2
AO_1O_2
A
O
1
O
2
have a common point different from
A
A
A
.Proposed by Patrik Bak, Slovakia[url=http://artofproblemsolving.com/community/c6h1667031p10595016]Problem 3. There are
2018
2018
2018
players sitting around a round table. At the beginning of the game we arbitrarily deal all the cards from a deck of
K
K
K
cards to the players (some players may receive no cards). In each turn we choose a player who draws one card from each of the two neighbors. It is only allowed to choose a player whose each neighbor holds a nonzero number of cards. The game terminates when there is no such player. Determine the largest possible value of
K
K
K
such that, no matter how we deal the cards and how we choose the players, the game always terminates after a finite number of turns.Proposed by Peter Novotný, Slovakia[url=http://artofproblemsolving.com/community/c6h1667033p10595021]Problem 4. Let
A
B
C
ABC
A
BC
be an acute triangle with the perimeter of
2
s
2s
2
s
. We are given three pairwise disjoint circles with pairwise disjoint interiors with the centers
A
,
B
A, B
A
,
B
, and
C
C
C
, respectively. Prove that there exists a circle with the radius of
s
s
s
which contains all the three circles. Proposed by Josef Tkadlec, Czechia[url=http://artofproblemsolving.com/community/c6h1667034p10595023]Problem 5. In a
2
×
3
2 \times 3
2
×
3
rectangle there is a polyline of length
36
36
36
, which can have self-intersections. Show that there exists a line parallel to two sides of the rectangle, which intersects the other two sides in their interior points and intersects the polyline in fewer than
10
10
10
points.Proposed by Josef Tkadlec, Czechia and Vojtech Bálint, Slovakia[url=http://artofproblemsolving.com/community/c6h1667036p10595032]Problem 6. We say that a positive integer
n
n
n
is fantastic if there exist positive rational numbers
a
a
a
and
b
b
b
such that
n
=
a
+
1
a
+
b
+
1
b
.
n = a + \frac 1a + b + \frac 1b.
n
=
a
+
a
1
+
b
+
b
1
.
(a) Prove that there exist infinitely many prime numbers
p
p
p
such that no multiple of
p
p
p
is fantastic. (b) Prove that there exist infinitely many prime numbers
p
p
p
such that some multiple of
p
p
p
is fantastic.Proposed by Walther Janous, Austria
6
1
Hide problems
Fantastic integers n = a+1/a+b+1/b [Czech-Polish-Slovak Match 2018]
We say that a positive integer
n
n
n
is fantastic if there exist positive rational numbers
a
a
a
and
b
b
b
such that
n
=
a
+
1
a
+
b
+
1
b
.
n = a + \frac 1a + b + \frac 1b.
n
=
a
+
a
1
+
b
+
b
1
.
(a) Prove that there exist infinitely many prime numbers
p
p
p
such that no multiple of
p
p
p
is fantastic. (b) Prove that there exist infinitely many prime numbers
p
p
p
such that some multiple of
p
p
p
is fantastic.Proposed by Walther Janous, Austria
5
1
Hide problems
Polyline and intersections [Czech-Polish-Slovak Match 2018]
In a
2
×
3
2 \times 3
2
×
3
rectangle there is a polyline of length
36
36
36
, which can have self-intersections. Show that there exists a line parallel to two sides of the rectangle, which intersects the other two sides in their interior points and intersects the polyline in fewer than
10
10
10
points.Proposed by Josef Tkadlec, Czechia and Vojtech Bálint, Slovakia
4
1
Hide problems
A circle of radius s containing all three points [Czech-Polish-Slovak 2018]
Let
A
B
C
ABC
A
BC
be an acute triangle with the perimeter of
2
s
2s
2
s
. We are given three pairwise disjoint circles with pairwise disjoint interiors with the centers
A
,
B
A, B
A
,
B
, and
C
C
C
, respectively. Prove that there exists a circle with the radius of
s
s
s
which contains all the three circles. Proposed by Josef Tkadlec, Czechia
3
1
Hide problems
2018 players in a game [Czech-Polish-Slovak Match 2018]
There are
2018
2018
2018
players sitting around a round table. At the beginning of the game we arbitrarily deal all the cards from a deck of
K
K
K
cards to the players (some players may receive no cards). In each turn we choose a player who draws one card from each of the two neighbors. It is only allowed to choose a player whose each neighbor holds a nonzero number of cards. The game terminates when there is no such player. Determine the largest possible value of
K
K
K
such that, no matter how we deal the cards and how we choose the players, the game always terminates after a finite number of turns.Proposed by Peter Novotný, Slovakia
2
1
Hide problems
Circumcircles of ABC, ADE, AO_1O_2 have a common point
Let
A
B
C
ABC
A
BC
be an acute scalene triangle. Let
D
D
D
and
E
E
E
be points on the sides
A
B
AB
A
B
and
A
C
AC
A
C
, respectively, such that
B
D
=
C
E
BD=CE
B
D
=
CE
. Denote by
O
1
O_1
O
1
and
O
2
O_2
O
2
the circumcentres of the triangles
A
B
E
ABE
A
BE
and
A
C
D
ACD
A
C
D
, respectively. Prove that the circumcircles of the triangles
A
B
C
,
A
D
E
ABC, ADE
A
BC
,
A
D
E
, and
A
O
1
O
2
AO_1O_2
A
O
1
O
2
have a common point different from
A
A
A
.Proposed by Patrik Bak, Slovakia
1
1
Hide problems
f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y) [Czech-Polish-Slovak Match 2018]
Determine all functions
f
:
R
→
R
f : \mathbb R \to \mathbb R
f
:
R
→
R
such that for all real numbers
x
x
x
and
y
y
y
,
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
f(x^2 + xy) = f(x)f(y) + yf(x) + xf(x+y).
f
(
x
2
+
x
y
)
=
f
(
x
)
f
(
y
)
+
y
f
(
x
)
+
x
f
(
x
+
y
)
.
Proposed by Walther Janous, Austria