MathDB
Problems
Contests
International Contests
European Mathematical Cup
2023 European Mathematical Cup
2023 European Mathematical Cup
Part of
European Mathematical Cup
Subcontests
(4)
2
2
Hide problems
Game with convex hull
Let
n
>
5
n>5
n
>
5
be an integer. There are
n
n
n
points in the plane, no three of them collinear. Each day, Tom erases one of the points, until there are three points left. On the
i
i
i
-th day, for
1
<
i
<
n
−
3
1<i<n-3
1
<
i
<
n
−
3
, before erasing that day's point, Tom writes down the positive integer
v
(
i
)
v(i)
v
(
i
)
such that the convex hull of the points at that moment has
v
(
i
)
v(i)
v
(
i
)
vertices. Finally, he writes down
v
(
n
−
2
)
=
3
v(n-2) = 3
v
(
n
−
2
)
=
3
. Find the greatest possible value that the expression
∣
v
(
1
)
−
v
(
2
)
∣
+
∣
v
(
2
)
−
v
(
3
)
∣
+
…
+
∣
v
(
n
−
3
)
−
v
(
n
−
2
)
∣
|v(1)-v(2)|+ |v(2)-v(3)| + \ldots + |v(n-3)-v(n-2)|
∣
v
(
1
)
−
v
(
2
)
∣
+
∣
v
(
2
)
−
v
(
3
)
∣
+
…
+
∣
v
(
n
−
3
)
−
v
(
n
−
2
)
∣
can obtain among all possible initial configurations of
n
n
n
points and all possible Tom's moves.Remark. A convex hull of a finite set of points in the plane is the smallest convex polygon containing all the points of the set (inside it or on the boundary).Ivan Novak, Namik Agić
Interesting Right triangle geometry
Let
A
B
C
ABC
A
BC
be a triangle such that
∠
B
A
C
=
9
0
∘
\angle BAC = 90^{\circ}
∠
B
A
C
=
9
0
∘
. The incircle of triangle
A
B
C
ABC
A
BC
is tangent to the sides
B
C
‾
\overline{BC}
BC
,
C
A
‾
\overline{CA}
C
A
,
A
B
‾
\overline{AB}
A
B
at
D
,
E
,
F
D,E,F
D
,
E
,
F
respectively. Let
M
M
M
be the midpoint of
E
F
‾
\overline{EF}
EF
. Let
P
P
P
be the projection of
A
A
A
onto
B
C
BC
BC
and let
K
K
K
be the intersection of
M
P
MP
MP
and
A
D
AD
A
D
. Prove that the circumcircles of triangles
A
F
E
AFE
A
FE
and
P
D
K
PDK
P
DK
have equal radius. Kyprianos-Iason Prodromidis
3
2
Hide problems
Configuration with median and right angles
Consider an acute-angled triangle
A
B
C
ABC
A
BC
with
A
B
<
A
C
AB < AC
A
B
<
A
C
. Let
M
M
M
and
N
N
N
be the midpoints of segments
B
C
BC
BC
and
A
B
AB
A
B
, respectively. The circle with diameter
A
B
AB
A
B
intersects the lines
B
C
,
A
M
BC, AM
BC
,
A
M
and
A
C
AC
A
C
at
D
,
E
D, E
D
,
E
, and
F
F
F
, respectively. Let
G
G
G
be the midpoint of
F
C
FC
FC
. Prove that the lines
N
F
,
D
E
NF, DE
NF
,
D
E
and
G
M
GM
GM
are concurrent.Michal Pecho
You cannot have common descendants!
Let
n
n
n
be a positive integer. Let
B
n
B_n
B
n
be the set of all binary strings of length
n
n
n
. For a binary string s_1\hdots s_n, we define it's twist in the following way. First, we count how many blocks of consecutive digits it has. Denote this number by
b
b
b
. Then, we replace
s
b
s_b
s
b
with
1
−
s
b
1-s_b
1
−
s
b
. A string
a
a
a
is said to be a descendant of
b
b
b
if
a
a
a
can be obtained from
b
b
b
through a finite number of twists. A subset of
B
n
B_n
B
n
is called divided if no two of its members have a common descendant. Find the largest possible cardinality of a divided subset of
B
n
B_n
B
n
.Remark. Here is an example of a twist:
101100
→
101000
101100 \rightarrow 101000
101100
→
101000
because
1
∣
0
∣
11
∣
00
1\mid 0\mid 11\mid 00
1
∣
0
∣
11
∣
00
has
4
4
4
blocks of consecutive digits. Viktor Simjanoski
1
2
Hide problems
Cute NT at EMC
Suppose
a
,
b
,
c
a,b,c
a
,
b
,
c
are positive integers such that
gcd
(
a
,
b
)
+
gcd
(
a
,
c
)
+
gcd
(
b
,
c
)
=
b
+
c
+
2023
\gcd(a,b)+\gcd(a,c)+\gcd(b,c)=b+c+2023
g
cd
(
a
,
b
)
+
g
cd
(
a
,
c
)
+
g
cd
(
b
,
c
)
=
b
+
c
+
2023
Prove that
gcd
(
b
,
c
)
=
2023
\gcd(b,c)=2023
g
cd
(
b
,
c
)
=
2023
.Remark. For positive integers
x
x
x
and
y
y
y
,
gcd
(
x
,
y
)
\gcd(x,y)
g
cd
(
x
,
y
)
denotes their greatest common divisor.Ivan Novak
Easy Algebra at EMC
Determine all sets of real numbers
S
S
S
such that: [*]
1
1
1
is the smallest element of
S
S
S
, [*] for all
x
,
y
∈
S
x,y\in S
x
,
y
∈
S
such that
x
>
y
x>y
x
>
y
,
x
2
−
y
2
∈
S
\sqrt{x^2-y^2}\in S
x
2
−
y
2
∈
S
Adian Anibal Santos Sepcic
4
2
Hide problems
Sweet and Smooth!
We say that a
2023
2023
2023
-tuple of nonnegative integers (a_1,\hdots,a_{2023}) is sweet if the following conditions hold: [*] a_1+\hdots+a_{2023}=2023 [*] \frac{a_1}{2}+\frac{a_2}{2^2}+\hdots+\frac{a_{2023}}{2^{2023}}\le 1 Determine the greatest positive integer
L
L
L
so that a_1+2a_2+\hdots+2023a_{2023}\ge L holds for every sweet
2023
2023
2023
-tuple (a_1,\hdots,a_{2023})Ivan Novak
Conditional squares imply injectivity
Let
f
:
N
→
N
f\colon\mathbb{N}\rightarrow\mathbb{N}
f
:
N
→
N
be a function such that for all positive integers
x
x
x
and
y
y
y
, the number
f
(
x
)
+
y
f(x)+y
f
(
x
)
+
y
is a perfect square if and only if
x
+
f
(
y
)
x+f(y)
x
+
f
(
y
)
is a perfect square. Prove that
f
f
f
is injective.Remark. A function
f
:
N
→
N
f\colon\mathbb{N}\rightarrow\mathbb{N}
f
:
N
→
N
is injective if for all pairs
(
x
,
y
)
(x,y)
(
x
,
y
)
of distinct positive integers,
f
(
x
)
≠
f
(
y
)
f(x)\neq f(y)
f
(
x
)
=
f
(
y
)
holds.Ivan Novak