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Contests
International Contests
Middle European Mathematical Olympiad
2014 Middle European Mathematical Olympiad
2014 Middle European Mathematical Olympiad
Part of
Middle European Mathematical Olympiad
Subcontests
(8)
8
1
Hide problems
diophantine 2014
Determine all quadruples
(
x
,
y
,
z
,
t
)
(x,y,z,t)
(
x
,
y
,
z
,
t
)
of positive integers such that
2
0
x
+
1
4
2
y
=
(
x
+
2
y
+
z
)
z
t
.
20^x + 14^{2y} = (x + 2y + z)^{zt}.
2
0
x
+
1
4
2
y
=
(
x
+
2
y
+
z
)
z
t
.
7
1
Hide problems
meanly set arithmetic mean
A finite set of positive integers
A
A
A
is called meanly if for each of its nonempy subsets the arithmetic mean of its elements is also a positive integer. In other words,
A
A
A
is meanly if
1
k
(
a
1
+
⋯
+
a
k
)
\frac{1}{k}(a_1 + \dots + a_k)
k
1
(
a
1
+
⋯
+
a
k
)
is an integer whenever
k
≥
1
k \ge 1
k
≥
1
and
a
1
,
…
,
a
k
∈
A
a_1, \dots, a_k \in A
a
1
,
…
,
a
k
∈
A
are distinct.Given a positive integer
n
n
n
, determine the least possible sum of the elements of a meanly
n
n
n
-element set.
6
1
Hide problems
incircle excenter midpoints
Let the incircle
k
k
k
of the triangle
A
B
C
ABC
A
BC
touch its side
B
C
BC
BC
at
D
D
D
. Let the line
A
D
AD
A
D
intersect
k
k
k
at
L
≠
D
L \neq D
L
=
D
and denote the excentre of
A
B
C
ABC
A
BC
opposite to
A
A
A
by
K
K
K
. Let
M
M
M
and
N
N
N
be the midpoints of
B
C
BC
BC
and
K
M
KM
K
M
respectively.Prove that the points
B
,
C
,
N
,
B, C, N,
B
,
C
,
N
,
and
L
L
L
are concyclic.
5
1
Hide problems
intersections of the angle bisector with intouch triangle
Let
A
B
C
ABC
A
BC
be a triangle with
A
B
<
A
C
AB < AC
A
B
<
A
C
. Its incircle with centre
I
I
I
touches the sides
B
C
,
C
A
,
BC, CA,
BC
,
C
A
,
and
A
B
AB
A
B
in the points
D
,
E
,
D, E,
D
,
E
,
and
F
F
F
respectively. The angle bisector
A
I
AI
A
I
intersects the lines
D
E
DE
D
E
and
D
F
DF
D
F
in the points
X
X
X
and
Y
Y
Y
respectively. Let
Z
Z
Z
be the foot of the altitude through
A
A
A
with respect to
B
C
BC
BC
.Prove that
D
D
D
is the incentre of the triangle
X
Y
Z
XYZ
X
Y
Z
.
4
2
Hide problems
bibinomial coefficient, double factorial
For integers
n
≥
k
≥
0
n \ge k \ge 0
n
≥
k
≥
0
we define the bibinomial coefficient
(
(
n
k
)
)
\left( \binom{n}{k} \right)
(
(
k
n
)
)
by
(
(
n
k
)
)
=
n
!
!
k
!
!
(
n
−
k
)
!
!
.
\left( \binom{n}{k} \right) = \frac{n!!}{k!!(n-k)!!} .
(
(
k
n
)
)
=
k
!!
(
n
−
k
)!!
n
!!
.
Determine all pairs
(
n
,
k
)
(n,k)
(
n
,
k
)
of integers with
n
≥
k
≥
0
n \ge k \ge 0
n
≥
k
≥
0
such that the corresponding bibinomial coefficient is an integer.Remark: The double factorial
n
!
!
n!!
n
!!
is defined to be the product of all even positive integers up to
n
n
n
if
n
n
n
is even and the product of all odd positive integers up to
n
n
n
if
n
n
n
is odd. So e.g.
0
!
!
=
1
0!! = 1
0
!!
=
1
,
4
!
!
=
2
⋅
4
=
8
4!! = 2 \cdot 4 = 8
4
!!
=
2
⋅
4
=
8
, and
7
!
!
=
1
⋅
3
⋅
5
⋅
7
=
105
7!! = 1 \cdot 3 \cdot 5 \cdot 7 = 105
7
!!
=
1
⋅
3
⋅
5
⋅
7
=
105
.
Happy City
In Happy City there are
2014
2014
2014
citizens called
A
1
,
A
2
,
…
,
A
2014
A_1, A_2, \dots , A_{2014}
A
1
,
A
2
,
…
,
A
2014
. Each of them is either happy or unhappy at any moment in time. The mood of any citizen
A
A
A
changes (from being unhappy to being happy or vice versa) if and only if some other happy citizen smiles at
A
A
A
. On Monday morning there were
N
N
N
happy citizens in the city.The following happened on Monday during the day: the citizen
A
1
A_1
A
1
smiled at citizen
A
2
A_2
A
2
, then
A
2
A_2
A
2
smiled at
A
3
A_3
A
3
, etc., and, finally,
A
2013
A_{2013}
A
2013
smiled at
A
2014
A_{2014}
A
2014
. Nobody smiled at anyone else apart from this. Exactly the same repeated on Tuesday, Wednesday and Thursday. There were exactly
2000
2000
2000
happy citizens on Thursday evening.Determine the largest possible value of
N
N
N
.
3
2
Hide problems
concurrent lines incenter
Let
A
B
C
ABC
A
BC
be a triangle with
A
B
<
A
C
AB < AC
A
B
<
A
C
and incentre
I
I
I
. Let
E
E
E
be the point on the side
A
C
AC
A
C
such that
A
E
=
A
B
AE = AB
A
E
=
A
B
. Let
G
G
G
be the point on the line
E
I
EI
E
I
such that
∠
I
B
G
=
∠
C
B
A
\angle IBG = \angle CBA
∠
I
BG
=
∠
CB
A
and such that
E
E
E
and
G
G
G
lie on opposite sides of
I
I
I
.Prove that the line
A
I
AI
A
I
, the line perpendicular to
A
E
AE
A
E
at
E
E
E
, and the bisector of the angle
∠
B
G
I
\angle BGI
∠
BG
I
are concurrent.
MEMOrable rectangles
Let
K
K
K
and
L
L
L
be positive integers. On a board consisting of
2
K
×
2
L
2K \times 2L
2
K
×
2
L
unit squares an ant starts in the lower left corner square and walks to the upper right corner square. In each step it goes horizontally or vertically to a neighbouring square. It never visits a square twice. At the end some squares may remain unvisited. In some cases the collection of all unvisited squares forms a single rectangle. In such cases, we call this rectangle MEMOrable.Determine the number of different MEMOrable rectangles.Remark: Rectangles are different unless they consist of exactly the same squares.
2
2
Hide problems
bicoloured triangulation
We consider dissections of regular
n
n
n
-gons into
n
−
2
n - 2
n
−
2
triangles by
n
−
3
n - 3
n
−
3
diagonals which do not intersect inside the
n
n
n
-gon. A bicoloured triangulation is such a dissection of an
n
n
n
-gon in which each triangle is coloured black or white and any two triangles which share an edge have different colours. We call a positive integer
n
≥
4
n \ge 4
n
≥
4
triangulable if every regular
n
n
n
-gon has a bicoloured triangulation such that for each vertex
A
A
A
of the
n
n
n
-gon the number of black triangles of which
A
A
A
is a vertex is greater than the number of white triangles of which
A
A
A
is a vertex.Find all triangulable numbers.
functional inequality
Determine all functions
f
:
R
→
R
f : \mathbb{R} \to \mathbb{R}
f
:
R
→
R
such that
x
f
(
x
y
)
+
x
y
f
(
x
)
≥
f
(
x
2
)
f
(
y
)
+
x
2
y
xf(xy) + xyf(x) \ge f(x^2)f(y) + x^2y
x
f
(
x
y
)
+
x
y
f
(
x
)
≥
f
(
x
2
)
f
(
y
)
+
x
2
y
holds for all
x
,
y
∈
R
x,y \in \mathbb{R}
x
,
y
∈
R
.
1
2
Hide problems
not so hard functional equation
Determine all functions
f
:
R
→
R
f:\mathbb{R} \to \mathbb{R}
f
:
R
→
R
such that
x
f
(
y
)
+
f
(
x
f
(
y
)
)
−
x
f
(
f
(
y
)
)
−
f
(
x
y
)
=
2
x
+
f
(
y
)
−
f
(
x
+
y
)
xf(y) + f(xf(y)) - xf(f(y)) - f(xy) = 2x + f(y) - f(x+y)
x
f
(
y
)
+
f
(
x
f
(
y
))
−
x
f
(
f
(
y
))
−
f
(
x
y
)
=
2
x
+
f
(
y
)
−
f
(
x
+
y
)
holds for all
x
,
y
∈
R
x,y \in \mathbb{R}
x
,
y
∈
R
.
lowest possible value of the expression
Determine the lowest possible value of the expression
1
a
+
x
+
1
a
+
y
+
1
b
+
x
+
1
b
+
y
\frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y}
a
+
x
1
+
a
+
y
1
+
b
+
x
1
+
b
+
y
1
where
a
,
b
,
x
,
a,b,x,
a
,
b
,
x
,
and
y
y
y
are positive real numbers satisfying the inequalities
1
a
+
x
≥
1
2
\frac{1}{a+x} \ge \frac{1}{2}
a
+
x
1
≥
2
1
1
a
+
y
≥
1
2
\frac{1}{a+y} \ge \frac{1}{2}
a
+
y
1
≥
2
1
1
b
+
x
≥
1
2
\frac{1}{b+x} \ge \frac{1}{2}
b
+
x
1
≥
2
1
1
b
+
y
≥
1.
\frac{1}{b+y} \ge 1.
b
+
y
1
≥
1.