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Contests
National and Regional Contests
Romania Contests
Romania Team Selection Test
2010 Romania Team Selection Test
2010 Romania Team Selection Test
Part of
Romania Team Selection Test
Subcontests
(5)
5
1
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Divisibility by $n!$
Let
a
a
a
and
n
n
n
be two positive integer numbers such that the (positive) prime factors of
a
a
a
be all greater than
n
n
n
. Prove that
n
!
n!
n
!
divides
(
a
−
1
)
(
a
2
−
1
)
⋯
(
a
n
−
1
−
1
)
(a - 1)(a^2 - 1)\cdots (a^{n-1} - 1)
(
a
−
1
)
(
a
2
−
1
)
⋯
(
a
n
−
1
−
1
)
. AMM Magazine
4
3
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Concyclic points
Two circles in the plane,
γ
1
\gamma_1
γ
1
and
γ
2
\gamma_2
γ
2
, meet at points
M
M
M
and
N
N
N
. Let
A
A
A
be a point on
γ
1
\gamma_1
γ
1
, and let
D
D
D
be a point on
γ
2
\gamma_2
γ
2
. The lines
A
M
AM
A
M
and
A
N
AN
A
N
meet again
γ
2
\gamma_2
γ
2
at points
B
B
B
and
C
C
C
, respectively, and the lines
D
M
DM
D
M
and
D
N
DN
D
N
meet again
γ
1
\gamma_1
γ
1
at points
E
E
E
and
F
F
F
, respectively. Assume the order
M
M
M
,
N
N
N
,
F
F
F
,
A
A
A
,
E
E
E
is circular around
γ
1
\gamma_1
γ
1
, and the segments
A
B
AB
A
B
and
D
E
DE
D
E
are congruent. Prove that the points
A
A
A
,
F
F
F
,
C
C
C
and
D
D
D
lie on a circle whose centre does not depend on the position of the points
A
A
A
and
D
D
D
on the respective circles, subject to the assumptions above. ***
Lattice points inside a convex body
Let
n
n
n
be an integer number greater than or equal to
2
2
2
, and let
K
K
K
be a closed convex set of area greater than or equal to
n
n
n
, contained in the open square
(
0
,
n
)
×
(
0
,
n
)
(0, n) \times (0, n)
(
0
,
n
)
×
(
0
,
n
)
. Prove that
K
K
K
contains some point of the integral lattice
Z
×
Z
\mathbb{Z} \times \mathbb{Z}
Z
×
Z
. Marius Cavachi
Additive combinatorics (re Cauchy-Davenport)
Let
X
X
X
and
Y
Y
Y
be two finite subsets of the half-open interval
[
0
,
1
)
[0, 1)
[
0
,
1
)
such that
0
∈
X
∩
Y
0 \in X \cap Y
0
∈
X
∩
Y
and
x
+
y
=
1
x + y = 1
x
+
y
=
1
for no
x
∈
X
x \in X
x
∈
X
and no
y
∈
Y
y \in Y
y
∈
Y
. Prove that the set
{
x
+
y
−
⌊
x
+
y
⌋
:
x
∈
X
and
y
∈
Y
}
\{x + y - \lfloor x + y \rfloor : x \in X \textrm{ and } y \in Y\}
{
x
+
y
−
⌊
x
+
y
⌋
:
x
∈
X
and
y
∈
Y
}
has at least
∣
X
∣
+
∣
Y
∣
−
1
|X| + |Y| - 1
∣
X
∣
+
∣
Y
∣
−
1
elements. ***
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6
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6
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