MathDB
Problems
Contests
National and Regional Contests
North Macedonia Contests
JBMO TST - Macedonia
2014 JBMO TST - Macedonia
2014 JBMO TST - Macedonia
Part of
JBMO TST - Macedonia
Subcontests
(5)
5
1
Hide problems
Macedonian JBMO TST 2014, Problem 5
Prove that there exist infinitely many pairwisely disjoint sets
A
(
1
)
,
A
(
2
)
,
.
.
.
,
A
(
2014
)
A(1), A(2),...,A(2014)
A
(
1
)
,
A
(
2
)
,
...
,
A
(
2014
)
which are not empty, whose union is the set of positive integers and which satisfy the following condition: For arbitrary positive integers
a
a
a
and
b
b
b
, at least two of the numbers
a
a
a
,
b
b
b
and
G
C
D
(
a
,
b
)
GCD(a,b)
GC
D
(
a
,
b
)
belong to one of the sets
A
(
1
)
,
A
(
2
)
,
.
.
.
,
A
(
2014
)
A(1), A(2),...,A(2014)
A
(
1
)
,
A
(
2
)
,
...
,
A
(
2014
)
.
3
1
Hide problems
Macedonian JBMO TST 2014, Problem 3
Find all positive integers
n
n
n
which are divisible by 11 and satisfy the following condition: all the numbers which are generated by an arbitrary rearrangement of the digits of
n
n
n
, are also divisible by 11.
1
1
Hide problems
Macedonian JBMO TST 2014, Problem 1
Prove that
1
1
×
2013
+
1
2
×
2012
+
1
3
×
2011
+
.
.
.
+
1
2012
×
2
+
1
2013
×
1
<
1
\frac{1}{1\times2013}+\frac{1}{2\times2012}+\frac{1}{3\times2011}+...+\frac{1}{2012\times2}+\frac{1}{2013\times1}<1
1
×
2013
1
+
2
×
2012
1
+
3
×
2011
1
+
...
+
2012
×
2
1
+
2013
×
1
1
<
1
4
1
Hide problems
Macedonian JBMO TST 2014, Problem 4
In a convex quadrilateral
A
B
C
D
ABCD
A
BC
D
,
E
E
E
is the intersection of
A
B
AB
A
B
and
C
D
CD
C
D
,
F
F
F
is the intersection of
A
D
AD
A
D
and
B
C
BC
BC
and
G
G
G
is the intersection of
A
C
AC
A
C
and
E
F
EF
EF
. Prove that the following two claims are equivalent:
(
i
)
(i)
(
i
)
B
D
BD
B
D
and
E
F
EF
EF
are parallel.
(
i
i
)
(ii)
(
ii
)
G
G
G
is the midpoint of
E
F
EF
EF
.
2
1
Hide problems
Macedonian JBMO TST 2014, Problem 2
Point
M
M
M
is an arbitrary point in the plane and let points
G
G
G
and
H
H
H
be the intersection points of the tangents from point M and the circle
k
k
k
. Let
O
O
O
be the center of the circle
k
k
k
and let
K
K
K
be the orthocenter of the triangle
M
G
H
MGH
MG
H
. Prove that
∠
G
M
H
=
∠
O
G
K
{\angle}GMH={\angle}OGK
∠
GM
H
=
∠
OG
K
.