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National and Regional Contests
Bosnia Herzegovina Contests
JBMO TST - Bosnia and Herzegovina
2012 Bosnia and Herzegovina Junior BMO TST
2012 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 2012 Problem 4
If
a
a
a
,
b
b
b
and
c
c
c
are sides of triangle which perimeter equals
1
1
1
, prove that:
a
2
+
b
2
+
c
2
+
4
a
b
c
<
1
2
a^2+b^2+c^2+4abc<\frac{1}{2}
a
2
+
b
2
+
c
2
+
4
ab
c
<
2
1
3
1
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Bosnia and Herzegovina JBMO TST 2012 Problem 3
Internal angles of triangle are
(
5
x
+
3
y
)
∘
(5x+3y)^{\circ}
(
5
x
+
3
y
)
∘
,
(
3
x
+
20
)
∘
(3x+20)^{\circ}
(
3
x
+
20
)
∘
and
(
10
y
+
30
)
∘
(10y+30)^{\circ}
(
10
y
+
30
)
∘
where
x
x
x
and
y
y
y
are positive integers. Which values can
x
+
y
x+y
x
+
y
get ?
2
1
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Bosnia and Herzegovina JBMO TST 2012 Problem 2
Let
a
b
c
d
‾
\overline{abcd}
ab
c
d
be
4
4
4
digit number, such that we can do transformations on it. If some two neighboring digits are different than
0
0
0
, then we can decrease both digits by
1
1
1
(we can transform
9870
9870
9870
to
8770
8770
8770
or
9760
9760
9760
). If some two neighboring digits are different than
9
9
9
, then we can increase both digits by
1
1
1
(we can transform
9870
9870
9870
to
9980
9980
9980
or
9881
9881
9881
). Can we transform number
1220
1220
1220
to:
a
)
a)
a
)
2012
2012
2012
b
)
b)
b
)
2021
2021
2021
1
1
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Bosnia and Herzegovina JBMO TST 2012 Problem 1
On circle
k
k
k
there are clockwise points
A
A
A
,
B
B
B
,
C
C
C
,
D
D
D
and
E
E
E
such that
∠
A
B
E
=
∠
B
E
C
=
∠
E
C
D
=
4
5
∘
\angle ABE = \angle BEC = \angle ECD = 45^{\circ}
∠
A
BE
=
∠
BEC
=
∠
EC
D
=
4
5
∘
. Prove that
A
B
2
+
C
E
2
=
B
E
2
+
C
D
2
AB^2 + CE^2 = BE^2 + CD^2
A
B
2
+
C
E
2
=
B
E
2
+
C
D
2