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Problems
Contests
International Contests
Middle European Mathematical Olympiad
2019 Middle European Mathematical Olympiad
2019 Middle European Mathematical Olympiad
Part of
Middle European Mathematical Olympiad
Subcontests
(8)
6
1
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right triangle and a midpoint
Let
A
B
C
ABC
A
BC
be a right-angled triangle with the right angle at
B
B
B
and circumcircle
c
c
c
. Denote by
D
D
D
the midpoint of the shorter arc
A
B
AB
A
B
of
c
c
c
. Let
P
P
P
be the point on the side
A
B
AB
A
B
such that
C
P
=
C
D
CP=CD
CP
=
C
D
and let
X
X
X
and
Y
Y
Y
be two distinct points on
c
c
c
satisfying
A
X
=
A
Y
=
P
D
AX=AY=PD
A
X
=
A
Y
=
P
D
. Prove that
X
,
Y
X, Y
X
,
Y
and
P
P
P
are collinear.Proposed by Dominik Burek, Poland
5
1
Hide problems
perpendiculars and midpoints
Let
A
B
C
ABC
A
BC
be an acute-angled triangle such that
A
B
<
A
C
AB<AC
A
B
<
A
C
. Let
D
D
D
be the point of intersection of the perpendicular bisector of the side
B
C
BC
BC
with the side
A
C
AC
A
C
. Let
P
P
P
be a point on the shorter arc
A
C
AC
A
C
of the circumcircle of the triangle
A
B
C
ABC
A
BC
such that
D
P
∥
B
C
DP \parallel BC
D
P
∥
BC
. Finally, let
M
M
M
be the midpoint of the side
A
B
AB
A
B
. Prove that
∠
A
P
D
=
∠
M
P
B
\angle APD=\angle MPB
∠
A
P
D
=
∠
MPB
.Proposed by Dominik Burek, Poland
8
1
Hide problems
existential NT
Let
N
N
N
be a positive integer such that the sum of the squares of all positive divisors of
N
N
N
is equal to the product
N
(
N
+
3
)
N(N+3)
N
(
N
+
3
)
. Prove that there exist two indices
i
i
i
and
j
j
j
such that
N
=
F
i
F
j
N=F_iF_j
N
=
F
i
F
j
where
(
F
i
)
n
=
1
∞
(F_i)_{n=1}^{\infty}
(
F
i
)
n
=
1
∞
is the Fibonacci sequence defined as
F
1
=
F
2
=
1
F_1=F_2=1
F
1
=
F
2
=
1
and
F
n
=
F
n
−
1
+
F
n
−
2
F_n=F_{n-1}+F_{n-2}
F
n
=
F
n
−
1
+
F
n
−
2
for
n
≥
3
n\geq 3
n
≥
3
.Proposed by Alain Rossier, Switzerland
7
1
Hide problems
Divisibility inequality
Let
a
,
b
a,b
a
,
b
and
c
c
c
be positive integers satisfying
a
<
b
<
c
<
a
+
b
a<b<c<a+b
a
<
b
<
c
<
a
+
b
. Prove that
c
(
a
−
1
)
+
b
c(a-1)+b
c
(
a
−
1
)
+
b
does not divide
c
(
b
−
1
)
+
a
c(b-1)+a
c
(
b
−
1
)
+
a
.Proposed by Dominik Burek, Poland
2
2
Hide problems
Polynomial inequality
Let
α
\alpha
α
be a real number. Determine all polynomials
P
P
P
with real coefficients such that
P
(
2
x
+
α
)
≤
(
x
20
+
x
19
)
P
(
x
)
P(2x+\alpha)\leq (x^{20}+x^{19})P(x)
P
(
2
x
+
α
)
≤
(
x
20
+
x
19
)
P
(
x
)
holds for all real numbers
x
x
x
.Proposed by Walther Janous, Austria
Bohemian vertices of a convex polygon
Let
n
≥
3
n\geq 3
n
≥
3
be an integer. We say that a vertex
A
i
(
1
≤
i
≤
n
)
A_i (1\leq i\leq n)
A
i
(
1
≤
i
≤
n
)
of a convex polygon
A
1
A
2
…
A
n
A_1A_2 \dots A_n
A
1
A
2
…
A
n
is Bohemian if its reflection with respect to the midpoint of
A
i
−
1
A
i
+
1
A_{i-1}A_{i+1}
A
i
−
1
A
i
+
1
(with
A
0
=
A
n
A_0=A_n
A
0
=
A
n
and
A
1
=
A
n
+
1
A_1=A_{n+1}
A
1
=
A
n
+
1
) lies inside or on the boundary of the polygon
A
1
A
2
…
A
n
A_1A_2\dots A_n
A
1
A
2
…
A
n
. Determine the smallest possible number of Bohemian vertices a convex
n
n
n
-gon can have (depending on
n
n
n
).Proposed by Dominik Burek, Poland
3
2
Hide problems
4 concyclic points
Let
A
B
C
ABC
A
BC
be an acute-angled triangle with
A
C
>
B
C
AC>BC
A
C
>
BC
and circumcircle
ω
\omega
ω
. Suppose that
P
P
P
is a point on
ω
\omega
ω
such that
A
P
=
A
C
AP=AC
A
P
=
A
C
and that
P
P
P
is an interior point on the shorter arc
B
C
BC
BC
of
ω
\omega
ω
. Let
Q
Q
Q
be the intersection point of the lines
A
P
AP
A
P
and
B
C
BC
BC
. Furthermore, suppose that
R
R
R
is a point on
ω
\omega
ω
such that
Q
A
=
Q
R
QA=QR
Q
A
=
QR
and
R
R
R
is an interior point of the shorter arc
A
C
AC
A
C
of
ω
\omega
ω
. Finally, let
S
S
S
be the point of intersection of the line
B
C
BC
BC
with the perpendicular bisector of the side
A
B
AB
A
B
. Prove that the points
P
,
Q
,
R
P, Q, R
P
,
Q
,
R
and
S
S
S
are concyclic.Proposed by Patrik Bak, Slovakia
n boys and n girls
There are
n
n
n
boys and
n
n
n
girls in a school class, where
n
n
n
is a positive integer. The heights of all the children in this class are distinct. Every girl determines the number of boys that are taller than her, subtracts the number of girls that are taller than her, and writes the result on a piece of paper. Every boy determines the number of girls that are shorter than him, subtracts the number of boys that are shorter than him, and writes the result on a piece of paper. Prove that the numbers written down by the girls are the same as the numbers written down by the boys (up to a permutation).Proposed by Stephan Wagner, Austria
4
2
Hide problems
smallest no. of consecutive numbers
Determine the smallest positive integer
n
n
n
for which the following statement holds true: From any
n
n
n
consecutive integers one can select a non-empty set of consecutive integers such that their sum is divisible by
2019
2019
2019
.Proposed by Kartal Nagy, Hungary
2019 no's as an arithmetic expression
Prove that every integer from
1
1
1
to
2019
2019
2019
can be represented as an arithmetic expression consisting of up to
17
17
17
symbols
2
2
2
and an arbitrary number of additions, subtractions, multiplications, divisions and brackets. The
2
2
2
's may not be used for any other operation, for example, to form multidigit numbers (such as
222
222
222
) or powers (such as
2
2
2^2
2
2
).Valid examples:
(
(
2
×
2
+
2
)
×
2
−
2
2
)
×
2
=
22
,
(
2
×
2
×
2
−
2
)
×
(
2
×
2
+
2
+
2
+
2
2
)
=
42
\left((2\times 2+2)\times 2-\frac{2}{2}\right)\times 2=22 \;\;, \;\; (2\times2\times 2-2)\times \left(2\times 2 +\frac{2+2+2}{2}\right)=42
(
(
2
×
2
+
2
)
×
2
−
2
2
)
×
2
=
22
,
(
2
×
2
×
2
−
2
)
×
(
2
×
2
+
2
2
+
2
+
2
)
=
42
Proposed by Stephan Wagner, Austria
1
2
Hide problems
a FE on MEMO
Find all functions
f
:
R
→
R
f:\mathbb{R} \to \mathbb{R}
f
:
R
→
R
such that for any two real numbers
x
,
y
x,y
x
,
y
holds
f
(
x
f
(
y
)
+
2
y
)
=
f
(
x
y
)
+
x
f
(
y
)
+
f
(
f
(
y
)
)
.
f(xf(y)+2y)=f(xy)+xf(y)+f(f(y)).
f
(
x
f
(
y
)
+
2
y
)
=
f
(
x
y
)
+
x
f
(
y
)
+
f
(
f
(
y
))
.
Proposed by Patrik Bak, Slovakia
Minimum and maximum
Determine the smallest and the greatest possible values of the expression
(
1
a
2
+
1
+
1
b
2
+
1
+
1
c
2
+
1
)
(
a
2
a
2
+
1
+
b
2
b
2
+
1
+
c
2
c
2
+
1
)
\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)
(
a
2
+
1
1
+
b
2
+
1
1
+
c
2
+
1
1
)
(
a
2
+
1
a
2
+
b
2
+
1
b
2
+
c
2
+
1
c
2
)
provided
a
,
b
a,b
a
,
b
and
c
c
c
are non-negative real numbers satisfying
a
b
+
b
c
+
c
a
=
1
ab+bc+ca=1
ab
+
b
c
+
c
a
=
1
.Proposed by Walther Janous, Austria