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Problems
Contests
National and Regional Contests
Austria Contests
Austrian MO National Competition
2018 Federal Competition For Advanced Students, P1
2018 Federal Competition For Advanced Students, P1
Part of
Austrian MO National Competition
Subcontests
(4)
4
1
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subset of pos. integers with 3 properties , prove that it is the set {1,2,3,...}
Let
M
M
M
be a set containing positive integers with the following three properties: (1)
2018
∈
M
2018 \in M
2018
∈
M
. (2) If
m
∈
M
m \in M
m
∈
M
, then all positive divisors of m are also elements of
M
M
M
. (3) For all elements
k
,
m
∈
M
k, m \in M
k
,
m
∈
M
with
1
<
k
<
m
1 < k < m
1
<
k
<
m
, the number
k
m
+
1
km + 1
km
+
1
is also an element of
M
M
M
. Prove that
M
=
Z
≥
1
M = Z_{\ge 1}
M
=
Z
≥
1
.(Proposed by Walther Janous)
2
1
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austrian collinear, incircle and midpoints related
Let
A
B
C
ABC
A
BC
be a triangle with incenter
I
I
I
. The incircle of the triangle is tangent to the sides
B
C
BC
BC
and
A
C
AC
A
C
in points
D
D
D
and
E
E
E
, respectively. Let
P
P
P
denote the common point of lines
A
I
AI
A
I
and
D
E
DE
D
E
, and let
M
M
M
and
N
N
N
denote the midpoints of sides
B
C
BC
BC
and
A
B
AB
A
B
, respectively. Prove that points
M
,
N
M, N
M
,
N
and
P
P
P
are collinear.(Proposed by Karl Czakler)
3
1
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Alice and Bob choose digits from a number with 2018 digits, different mod3
Alice and Bob determine a number with
2018
2018
2018
digits in the decimal system by choosing digits from left to right. Alice starts and then they each choose a digit in turn. They have to observe the rule that each digit must differ from the previously chosen digit modulo
3
3
3
. Since Bob will make the last move, he bets that he can make sure that the final number is divisible by
3
3
3
. Can Alice avoid that?(Proposed by Richard Henner)
1
1
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Determine the greatest real number
Let
α
\alpha
α
be an arbitrary positive real number. Determine for this number
α
\alpha
α
the greatest real number
C
C
C
such that the inequality
(
1
+
α
x
2
)
(
1
+
α
y
2
)
(
1
+
α
z
2
)
≥
C
(
x
z
+
z
x
+
2
)
\left(1+\frac{\alpha}{x^2}\right)\left(1+\frac{\alpha}{y^2}\right)\left(1+\frac{\alpha}{z^2}\right)\geq C\left(\frac{x}{z}+\frac{z}{x}+2\right)
(
1
+
x
2
α
)
(
1
+
y
2
α
)
(
1
+
z
2
α
)
≥
C
(
z
x
+
x
z
+
2
)
is valid for all positive real numbers
x
,
y
x, y
x
,
y
and
z
z
z
satisfying
x
y
+
y
z
+
z
x
=
α
.
xy + yz + zx =\alpha.
x
y
+
yz
+
z
x
=
α
.
When does equality occur? (Proposed by Walther Janous)