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Problems
Contests
National and Regional Contests
North Macedonia Contests
JBMO TST - Macedonia
2021 Junior Macedonian Mathematical Olympiad
2021 Junior Macedonian Mathematical Olympiad
Part of
JBMO TST - Macedonia
Subcontests
(5)
Problem 1
1
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No matter what you do, there will always be friends in different rooms
At this year's Olympiad, some of the students are friends (friendship is symmetric), however there are also students which are not friends. No matter how the students are partitioned in two contest halls, there are always two friends in different halls. Let
A
A
A
be a fixed student. Show that there exist students
B
B
B
and
C
C
C
such that there are exactly two friendships in the group
{
A
,
B
,
C
}
\{ A,B,C \}
{
A
,
B
,
C
}
.Proposed by Mirko Petrushevski
Problem 2
1
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Intersection of lines in tangential quadrilateral lies on diameter
Let
A
B
C
D
ABCD
A
BC
D
be a tangential quadrilateral with inscribed circle
k
(
O
,
r
)
k(O,r)
k
(
O
,
r
)
which is tangent to the sides
B
C
BC
BC
and
A
D
AD
A
D
at
K
K
K
and
L
L
L
, respectively. Show that the circle with diameter
O
C
OC
OC
passes through the intersection point of
K
L
KL
K
L
and
O
D
OD
O
D
.Proposed by Ilija Jovchevski
Problem 5
1
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Ratio of distances from incenters
Let
A
B
C
ABC
A
BC
be an acute triangle and let
X
X
X
and
Y
Y
Y
be points on the segments
A
B
AB
A
B
and
A
C
AC
A
C
such that
B
X
=
C
Y
BX = CY
BX
=
C
Y
. If
I
B
I_{B}
I
B
and
I
C
I_{C}
I
C
are centers of inscribed circles in triangles
A
B
Y
ABY
A
B
Y
and
A
C
X
ACX
A
CX
, and
T
T
T
is the second intersection point of the circumcircles of
A
B
Y
ABY
A
B
Y
and
A
C
X
ACX
A
CX
, show that:
T
I
B
T
I
C
=
B
Y
C
X
.
\frac{TI_{B}}{TI_{C}} = \frac{BY}{CX}.
T
I
C
T
I
B
=
CX
B
Y
.
Proposed by Nikola Velov
Problem 3
1
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Diophantine equation with squares and powers of 17
Find all positive integers
n
n
n
and prime numbers
p
p
p
such that
1
7
n
⋅
2
n
2
−
p
=
(
2
n
2
+
3
+
2
n
2
−
1
)
⋅
n
2
.
17^n \cdot 2^{n^2} - p =(2^{n^2+3}+2^{n^2}-1) \cdot n^2.
1
7
n
⋅
2
n
2
−
p
=
(
2
n
2
+
3
+
2
n
2
−
1
)
⋅
n
2
.
Proposed by Nikola Velov
Problem 4
1
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Cyclic inequality with fractions and tricky condition
Let
a
a
a
,
b
b
b
,
c
c
c
be positive real numbers such that
1
a
2
+
1
b
2
+
1
c
2
=
27
4
.
\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} = \frac{27}{4}.
a
2
1
+
b
2
1
+
c
2
1
=
4
27
.
Show that:
a
3
+
b
2
a
2
+
b
2
+
b
3
+
c
2
b
2
+
c
2
+
c
3
+
a
2
c
2
+
a
2
≥
5
2
.
\frac{a^3+b^2}{a^2+b^2} + \frac{b^3+c^2}{b^2+c^2} + \frac{c^3+a^2}{c^2+a^2} \geq \frac{5}{2}.
a
2
+
b
2
a
3
+
b
2
+
b
2
+
c
2
b
3
+
c
2
+
c
2
+
a
2
c
3
+
a
2
≥
2
5
.
Proposed by Nikola Velov