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Problems
Contests
National and Regional Contests
Bosnia Herzegovina Contests
JBMO TST - Bosnia and Herzegovina
2014 Bosnia and Herzegovina Junior BMO TST
2014 Bosnia and Herzegovina Junior BMO TST
Part of
JBMO TST - Bosnia and Herzegovina
Subcontests
(4)
4
1
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Bosnia and Herzegovina JBMO TST 2014 Problem 4
It is given
5
5
5
numbers
1
1
1
,
3
3
3
,
5
5
5
,
7
7
7
,
9
9
9
. We get the new
5
5
5
numbers such that we take arbitrary
4
4
4
numbers(out of current
5
5
5
numbers)
a
a
a
,
b
b
b
,
c
c
c
and
d
d
d
and replace them with
a
+
b
+
c
−
d
2
\frac{a+b+c-d}{2}
2
a
+
b
+
c
−
d
,
a
+
b
−
c
+
d
2
\frac{a+b-c+d}{2}
2
a
+
b
−
c
+
d
,
a
−
b
+
c
+
d
2
\frac{a-b+c+d}{2}
2
a
−
b
+
c
+
d
and
−
a
+
b
+
c
+
d
2
\frac{-a+b+c+d}{2}
2
−
a
+
b
+
c
+
d
. Can we, with repeated iterations, get numbers:
a
)
a)
a
)
0
0
0
,
2
2
2
,
4
4
4
,
6
6
6
and
8
8
8
b
)
b)
b
)
3
3
3
,
4
4
4
,
5
5
5
,
6
6
6
and
7
7
7
3
1
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Bosnia and Herzegovina JBMO TST 2014 Problem 3
Let
a
a
a
,
b
b
b
and
c
c
c
be positive real numbers such that
a
+
b
+
c
=
1
a+b+c=1
a
+
b
+
c
=
1
. Prove the inequality:
1
(
a
+
2
b
)
(
b
+
2
a
)
+
1
(
b
+
2
c
)
(
c
+
2
b
)
+
1
(
c
+
2
a
)
(
a
+
2
c
)
≥
3
\frac{1}{\sqrt{(a+2b)(b+2a)}}+\frac{1}{\sqrt{(b+2c)(c+2b)}}+\frac{1}{\sqrt{(c+2a)(a+2c)}} \geq 3
(
a
+
2
b
)
(
b
+
2
a
)
1
+
(
b
+
2
c
)
(
c
+
2
b
)
1
+
(
c
+
2
a
)
(
a
+
2
c
)
1
≥
3
2
1
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Bosnia and Herzegovina JBMO TST 2014 Problem 2
In triangle
A
B
C
ABC
A
BC
, on line
C
A
CA
C
A
it is given point
D
D
D
such that
C
D
=
3
⋅
C
A
CD = 3 \cdot CA
C
D
=
3
⋅
C
A
(point
A
A
A
is between points
C
C
C
and
D
D
D
), and on line
B
C
BC
BC
it is given point
E
E
E
(
E
≠
B
E \neq B
E
=
B
) such that
C
E
=
B
C
CE=BC
CE
=
BC
. If
B
D
=
A
E
BD=AE
B
D
=
A
E
, prove that
∠
B
A
C
=
9
0
∘
\angle BAC= 90^{\circ}
∠
B
A
C
=
9
0
∘
1
1
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Bosnia and Herzegovina JBMO TST 2014 Problem 1
Let
x
x
x
,
y
y
y
and
z
z
z
be nonnegative integers. Find all numbers in form
13
x
y
45
z
‾
\overline{13xy45z}
13
x
y
45
z
divisible with
792
792
792
, where
x
x
x
,
y
y
y
and
z
z
z
are digits.