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International Contests
European Mathematical Cup
2020 European Mathematical Cup
2020 European Mathematical Cup
Part of
European Mathematical Cup
Subcontests
(4)
1
2
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Geometry with parallel lines and cyclic quadrilaterals
Let
A
B
C
ABC
A
BC
be an acute-angled triangle. Let
D
D
D
and
E
E
E
be the midpoints of sides
A
B
‾
\overline{AB}
A
B
and
A
C
‾
\overline{AC}
A
C
respectively. Let
F
F
F
be the point such that
D
D
D
is the midpoint of
E
F
‾
\overline{EF}
EF
. Let
Γ
\Gamma
Γ
be the circumcircle of triangle
F
D
B
FDB
F
D
B
. Let
G
G
G
be a point on the segment
C
D
‾
\overline{CD}
C
D
such that the midpoint of
B
G
‾
\overline{BG}
BG
lies on
Γ
\Gamma
Γ
. Let
H
H
H
be the second intersection of
Γ
\Gamma
Γ
and
F
C
FC
FC
. Show that the quadrilateral
B
H
G
C
BHGC
B
H
GC
is cyclic. \\ \\ Proposed by Art Waeterschoot.
concurrency related to parallelogma and a circle
Let
A
B
C
D
ABCD
A
BC
D
be a parallelogram such that
∣
A
B
∣
>
∣
B
C
∣
|AB| > |BC|
∣
A
B
∣
>
∣
BC
∣
. Let
O
O
O
be a point on the line
C
D
CD
C
D
such that
∣
O
B
∣
=
∣
O
D
∣
|OB| = |OD|
∣
OB
∣
=
∣
O
D
∣
. Let
ω
\omega
ω
be a circle with center
O
O
O
and radius
∣
O
C
∣
|OC|
∣
OC
∣
. If
T
T
T
is the second intersection of
ω
\omega
ω
and
C
D
CD
C
D
, prove that
A
T
,
B
O
AT, BO
A
T
,
BO
and
ω
\omega
ω
are concurrent.Proposed by Ivan Novak
4
2
Hide problems
An inequality with square roots and minima
Let
a
,
b
,
c
a,b,c
a
,
b
,
c
be positive real numbers such that
a
b
+
b
c
+
a
c
=
a
+
b
+
c
ab+bc+ac = a+b+c
ab
+
b
c
+
a
c
=
a
+
b
+
c
. Prove the following inequality:
a
+
b
c
+
b
+
c
a
+
c
+
a
b
≤
2
⋅
min
{
a
b
+
b
c
+
c
a
,
b
a
+
c
b
+
a
c
}
.
\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot \min \left\{ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right\}.
a
+
c
b
+
b
+
a
c
+
c
+
b
a
≤
2
⋅
min
{
b
a
+
c
b
+
a
c
,
a
b
+
b
c
+
c
a
}
.
\\ \\ Proposed by Dorlir Ahmeti.
interesting function
Let
R
+
\mathbb{R^+}
R
+
denote the set of all positive real numbers. Find all functions
f
:
R
+
→
R
+
f: \mathbb{R^+}\rightarrow \mathbb{R^+}
f
:
R
+
→
R
+
such that
x
f
(
x
+
y
)
+
f
(
x
f
(
y
)
+
1
)
=
f
(
x
f
(
x
)
)
xf(x + y) + f(xf(y) + 1) = f(xf(x))
x
f
(
x
+
y
)
+
f
(
x
f
(
y
)
+
1
)
=
f
(
x
f
(
x
))
for all
x
,
y
∈
R
+
.
x, y \in\mathbb{R^+}.
x
,
y
∈
R
+
.
Proposed by Amadej Kristjan Kocbek, Jakob Jurij Snoj
3
2
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game on sequence
Let
p
p
p
be a prime number. Troy and Abed are playing a game. Troy writes a positive integer
X
X
X
on the board, and gives a sequence
(
a
n
)
n
∈
N
(a_n)_{n\in\mathbb{N}}
(
a
n
)
n
∈
N
of positive integers to Abed. Abed now makes a sequence of moves. The
n
n
n
-th move is the following:
Replace
Y
currently written on the board with either
Y
+
a
n
or
Y
⋅
a
n
.
\text{ Replace } Y \text{ currently written on the board with either } Y + a_n \text{ or } Y \cdot a_n.
Replace
Y
currently written on the board with either
Y
+
a
n
or
Y
⋅
a
n
.
Abed wins if at some point the number on the board is a multiple of
p
p
p
. Determine whether Abed can win, regardless of Troy’s choices, if
a
)
p
=
1
0
9
+
7
a) p = 10^9 + 7
a
)
p
=
1
0
9
+
7
;
b
)
p
=
1
0
9
+
9
b) p = 10^9 + 9
b
)
p
=
1
0
9
+
9
. Remark: Both
1
0
9
+
7
10^9 + 7
1
0
9
+
7
and
1
0
9
+
9
10^9 + 9
1
0
9
+
9
are prime.Proposed by Ivan Novak
Tiling a board with F tiles and Z tiles
Two types of tiles, depicted on the figure below, are given. https://wiki-images.artofproblemsolving.com//2/23/Izrezak.PNGFind all positive integers
n
n
n
such that an
n
×
n
n\times n
n
×
n
board consisting of
n
2
n^2
n
2
unit squares can be covered without gaps with these two types of tiles (rotations and reflections are allowed) so that no two tiles overlap and no part of any tile covers an area outside the
n
×
n
n\times n
n
×
n
board. \\ Proposed by Art Waeterschoot
2
2
Hide problems
Fibby numbers
A positive integer
k
⩾
3
k\geqslant 3
k
⩾
3
is called fibby if there exists a positive integer
n
n
n
and positive integers
d
1
<
d
2
<
…
<
d
k
d_1 < d_2 < \ldots < d_k
d
1
<
d
2
<
…
<
d
k
with the following properties: \\
∙
\bullet
∙
d
j
+
2
=
d
j
+
1
+
d
j
d_{j+2}=d_{j+1}+d_j
d
j
+
2
=
d
j
+
1
+
d
j
for every
j
j
j
satisfying
1
⩽
j
⩽
k
−
2
1\leqslant j \leqslant k-2
1
⩽
j
⩽
k
−
2
, \\
∙
\bullet
∙
d
1
,
d
2
,
…
,
d
k
d_1, d_2, \ldots, d_k
d
1
,
d
2
,
…
,
d
k
are divisors of
n
n
n
, \\
∙
\bullet
∙
any other divisor of
n
n
n
is either less than
d
1
d_1
d
1
or greater than
d
k
d_k
d
k
. Find all fibby numbers. \\ \\ Proposed by Ivan Novak.
permutations and k-mutations
Let
n
n
n
and
k
k
k
be positive integers. An
n
n
n
-tuple
(
a
1
,
a
2
,
…
,
a
n
)
(a_1, a_2,\ldots , a_n)
(
a
1
,
a
2
,
…
,
a
n
)
is called a permutation if every number from the set
{
1
,
2
,
.
.
.
,
n
}
\{1, 2, . . . , n\}
{
1
,
2
,
...
,
n
}
occurs in it exactly once. For a permutation
(
p
1
,
p
2
,
.
.
.
,
p
n
)
(p_1, p_2, . . . , p_n)
(
p
1
,
p
2
,
...
,
p
n
)
, we define its
k
k
k
-mutation to be the
n
n
n
-tuple
(
p
1
+
p
1
+
k
,
p
2
+
p
2
+
k
,
.
.
.
,
p
n
+
p
n
+
k
)
,
(p_1 + p_{1+k}, p_2 + p_{2+k}, . . . , p_n + p_{n+k}),
(
p
1
+
p
1
+
k
,
p
2
+
p
2
+
k
,
...
,
p
n
+
p
n
+
k
)
,
where indices are taken modulo
n
n
n
. Find all pairs
(
n
,
k
)
(n, k)
(
n
,
k
)
such that every two distinct permutations have distinct
k
k
k
-mutations.Remark: For example, when
(
n
,
k
)
=
(
4
,
2
)
(n, k) = (4, 2)
(
n
,
k
)
=
(
4
,
2
)
, the
2
2
2
-mutation of
(
1
,
2
,
4
,
3
)
(1, 2, 4, 3)
(
1
,
2
,
4
,
3
)
is
(
1
+
4
,
2
+
3
,
4
+
1
,
3
+
2
)
=
(
5
,
5
,
5
,
5
)
(1 + 4, 2 + 3, 4 + 1, 3 + 2) = (5, 5, 5, 5)
(
1
+
4
,
2
+
3
,
4
+
1
,
3
+
2
)
=
(
5
,
5
,
5
,
5
)
.Proposed by Borna Šimić