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National and Regional Contests
Serbia Contests
Serbia JBMO TST
2017 Serbia JBMO TST
2017 Serbia JBMO TST
Part of
Serbia JBMO TST
Subcontests
(4)
4
1
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A=\{7,12\}\cup\{8m+3\mid m\in\mathbb{N}\cup\{16m+4\mid m\in\mathbb{N}\}
Positive integer
q
q
q
is the
k
k{}
k
-successor of positive integer
n
n{}
n
if there exists a positive integer
p
p{}
p
such that
n
+
p
2
=
q
2
n+p^2=q^2
n
+
p
2
=
q
2
. Let
A
A{}
A
be the set of all positive integers
n
n{}
n
that have at least a
k
k{}
k
-successor, but every
k
k{}
k
-successor does not have
k
k{}
k
-successors of its own. Prove that
A
=
{
7
,
12
}
∪
{
8
m
+
3
∣
m
∈
N
}
∪
{
16
m
+
4
∣
m
∈
N
}
.
A=\{7,12\}\cup\{8m+3\mid m\in\mathbb{N}\}\cup\{16m+4\mid m\in\mathbb{N}\}.
A
=
{
7
,
12
}
∪
{
8
m
+
3
∣
m
∈
N
}
∪
{
16
m
+
4
∣
m
∈
N
}
.
1
1
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Combinatorics with grid
15 of the cells of a chessboard 8x8 are chosen. We draw the segments which unite the centers of every two of the chosen squares. Prove that among these segments there are four segments which have the same length.
3
1
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Geometry problem
Let ABC be a triangle with angle ACB=60. Let AA' and BB' be altitudes and let T be centroid of the triangle ABC. If A'T and B'T intersect triangle's circumcircle in points M and N respectively prove that MN=AB.
2
1
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Inequality
Let
x
,
y
,
z
x,y,z
x
,
y
,
z
be positive real numbers.Prove that
(
x
y
2
+
y
z
2
+
z
x
2
)
(
x
2
y
+
y
2
z
+
z
2
x
)
(
x
y
+
y
z
+
z
x
)
≥
3
(
x
+
y
+
z
)
2
(
x
y
z
)
2
.
(xy^2+yz^2+zx^2)(x^2y+y^2z+z^2x)(xy+yz+zx)\geq 3(x+y+z)^2(xyz)^2.
(
x
y
2
+
y
z
2
+
z
x
2
)
(
x
2
y
+
y
2
z
+
z
2
x
)
(
x
y
+
yz
+
z
x
)
≥
3
(
x
+
y
+
z
)
2
(
x
yz
)
2
.