MathDB
Problems
Contests
International Contests
European Mathematical Cup
2012 European Mathematical Cup
2012 European Mathematical Cup
Part of
European Mathematical Cup
Subcontests
(4)
1
2
Hide problems
Easy angle equality
Let
A
B
C
ABC
A
BC
be a triangle and
Q
Q
Q
a point on the internal angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
. Circle
ω
1
\omega_1
ω
1
is circumscribed to triangle
B
A
Q
BAQ
B
A
Q
and intersects the segment
A
C
AC
A
C
in point
P
≠
C
P \neq C
P
=
C
. Circle
ω
2
\omega_2
ω
2
is circumscribed to the triangle
C
Q
P
CQP
CQP
. Radius of the cirlce
ω
1
\omega_1
ω
1
is larger than the radius of
ω
2
\omega_2
ω
2
. Circle centered at
Q
Q
Q
with radius
Q
A
QA
Q
A
intersects the circle
ω
1
\omega_1
ω
1
in points
A
A
A
and
A
1
A_1
A
1
. Circle centered at
Q
Q
Q
with radius
Q
C
QC
QC
intersects
ω
1
\omega_1
ω
1
in points
C
1
C_1
C
1
and
C
2
C_2
C
2
. Prove
∠
A
1
B
C
1
=
∠
C
2
P
A
\angle A_1BC_1 = \angle C_2PA
∠
A
1
B
C
1
=
∠
C
2
P
A
.Proposed by Matija Bucić.
Easy diophantine
Find all positive integers
a
a
a
,
b
b
b
,
n
n
n
and prime numbers
p
p
p
that satisfy
a
2013
+
b
2013
=
p
n
.
a^{2013} + b^{2013} = p^n\text{.}
a
2013
+
b
2013
=
p
n
.
Proposed by Matija Bucić.
2
2
Hide problems
GCD's implying GCD's
Let
S
S
S
be the set of positive integers. For any
a
a
a
and
b
b
b
in the set we have
G
C
D
(
a
,
b
)
>
1
GCD(a, b)>1
GC
D
(
a
,
b
)
>
1
. For any
a
a
a
,
b
b
b
and
c
c
c
in the set we have
G
C
D
(
a
,
b
,
c
)
=
1
GCD(a, b, c)=1
GC
D
(
a
,
b
,
c
)
=
1
. Is it possible that
S
S
S
has
2012
2012
2012
elements? Proposed by Ognjen Stipetić.
Yet another cyclic quadrilateral
Let
A
B
C
ABC
A
BC
be an acute triangle with orthocenter
H
H
H
. Segments
A
H
AH
A
H
and
C
H
CH
C
H
intersect segments
B
C
BC
BC
and
A
B
AB
A
B
in points
A
1
A_1
A
1
and
C
1
C_1
C
1
respectively. The segments
B
H
BH
B
H
and
A
1
C
1
A_1C_1
A
1
C
1
meet at point
D
D
D
. Let
P
P
P
be the midpoint of the segment
B
H
BH
B
H
. Let
D
′
D'
D
′
be the reflection of the point
D
D
D
in
A
C
AC
A
C
. Prove that quadrilateral
A
P
C
D
′
APCD'
A
PC
D
′
is cyclic.Proposed by Matko Ljulj.
4
2
Hide problems
Existence of Hamilton Cycle in Directed Graph
Let
k
k
k
be a positive integer. At the European Chess Cup every pair of players played a game in which somebody won (there were no draws). For any
k
k
k
players there was a player against whom they all lost, and the number of players was the least possible for such
k
k
k
. Is it possible that at the Closing Ceremony all the participants were seated at the round table in such a way that every participant was seated next to both a person he won against and a person he lost against. Proposed by Matija Bucić.
Game with n numbers smaller than n-th prime
Olja writes down
n
n
n
positive integers
a
1
,
a
2
,
…
,
a
n
a_1, a_2, \ldots, a_n
a
1
,
a
2
,
…
,
a
n
smaller than
p
n
p_n
p
n
where
p
n
p_n
p
n
denotes the
n
n
n
-th prime number. Oleg can choose two (not necessarily different) numbers
x
x
x
and
y
y
y
and replace one of them with their product
x
y
xy
x
y
. If there are two equal numbers Oleg wins. Can Oleg guarantee a win? Proposed by Matko Ljulj.
3
2
Hide problems
Non-linear System of Equations with 3 Variables
Are there positive real numbers
x
x
x
,
y
y
y
and
z
z
z
such that
x
4
+
y
4
+
z
4
=
13
,
x^4 + y^4 + z^4 = 13\text{,}
x
4
+
y
4
+
z
4
=
13
,
x
3
y
3
z
+
y
3
z
3
x
+
z
3
x
3
y
=
6
3
,
x^3y^3z + y^3z^3x + z^3x^3y = 6\sqrt{3} \text{,}
x
3
y
3
z
+
y
3
z
3
x
+
z
3
x
3
y
=
6
3
,
x
3
y
z
+
y
3
z
x
+
z
3
x
y
=
5
3
?
x^3yz + y^3zx + z^3xy = 5\sqrt{3} \text{?}
x
3
yz
+
y
3
z
x
+
z
3
x
y
=
5
3
?
Proposed by Matko Ljulj.
Inequality in 6 Variables.
Prove that the following inequality holds for all positive real numbers
a
a
a
,
b
b
b
,
c
c
c
,
d
d
d
,
e
e
e
and
f
f
f
a
b
c
a
+
b
+
d
3
+
d
e
f
c
+
e
+
f
3
<
(
a
+
b
+
d
)
(
c
+
e
+
f
)
3
.
\sqrt[3]{\frac{abc}{a+b+d}}+\sqrt[3]{\frac{def}{c+e+f}} < \sqrt[3]{(a+b+d)(c+e+f)} \text{.}
3
a
+
b
+
d
ab
c
+
3
c
+
e
+
f
d
e
f
<
3
(
a
+
b
+
d
)
(
c
+
e
+
f
)
.
Proposed by Dimitar Trenevski.