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Contests
National and Regional Contests
Bosnia Herzegovina Contests
Bosnia Herzegovina Team Selection Test
2014 Bosnia Herzegovina Team Selection Test
2014 Bosnia Herzegovina Team Selection Test
Part of
Bosnia Herzegovina Team Selection Test
Subcontests
(3)
1
2
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Easy circle and tangents
Let
k
k
k
be the circle and
A
A
A
and
B
B
B
points on circle which are not diametrically opposite. On minor arc
A
B
AB
A
B
lies point arbitrary point
C
C
C
. Let
D
D
D
,
E
E
E
and
F
F
F
be foots of perpendiculars from
C
C
C
on chord
A
B
AB
A
B
and tangents of circle
k
k
k
in points
A
A
A
and
B
B
B
. Prove that
C
D
=
C
E
⋅
C
F
CD= \sqrt {CE \cdot CF}
C
D
=
CE
⋅
CF
Easy sequence
Sequence
a
n
a_n
a
n
is defined by
a
1
=
1
2
a_1=\frac{1}{2}
a
1
=
2
1
,
a
m
=
a
m
−
1
2
m
⋅
a
m
−
1
+
1
a_m=\frac{a_{m-1}}{2m \cdot a_{m-1} + 1}
a
m
=
2
m
⋅
a
m
−
1
+
1
a
m
−
1
for
m
>
1
m>1
m
>
1
. Determine value of
a
1
+
a
2
+
.
.
.
+
a
k
a_1+a_2+...+a_k
a
1
+
a
2
+
...
+
a
k
in terms of
k
k
k
, where
k
k
k
is positive integer.
3
2
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Nice exponential
Find all nonnegative integer numbers such that
7
x
−
2
⋅
5
y
=
−
1
7^x- 2 \cdot 5^y = -1
7
x
−
2
⋅
5
y
=
−
1
Circumcenter and orhocenter
Let
D
D
D
and
E
E
E
be foots of altitudes from
A
A
A
and
B
B
B
of triangle
A
B
C
ABC
A
BC
,
F
F
F
be intersection point of angle bisector from
C
C
C
with side
A
B
AB
A
B
, and
O
O
O
,
I
I
I
and
H
H
H
be circumcenter, center of inscribed circle and orthocenter of triangle
A
B
C
ABC
A
BC
, respectively. If
C
F
A
D
+
C
F
B
E
=
2
\frac{CF}{AD}+ \frac{CF}{BE}=2
A
D
CF
+
BE
CF
=
2
, prove that
O
I
=
I
H
OI = IH
O
I
=
I
H
.
2
2
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Regular polygon
It is given regular
n
n
n
-sided polygon,
n
≥
6
n \geq 6
n
≥
6
. How many triangles they are inside the polygon such that all of their sides are formed by diagonals of polygon and their vertices are vertices of polygon?
Difficult inequality
Let
a
a
a
,
b
b
b
and
c
c
c
be distinct real numbers.
a
)
a)
a
)
Determine value of
1
+
a
b
a
−
b
⋅
1
+
b
c
b
−
c
+
1
+
b
c
b
−
c
⋅
1
+
c
a
c
−
a
+
1
+
c
a
c
−
a
⋅
1
+
a
b
a
−
b
\frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b}
a
−
b
1
+
ab
⋅
b
−
c
1
+
b
c
+
b
−
c
1
+
b
c
⋅
c
−
a
1
+
c
a
+
c
−
a
1
+
c
a
⋅
a
−
b
1
+
ab
b
)
b)
b
)
Determine value of
1
−
a
b
a
−
b
⋅
1
−
b
c
b
−
c
+
1
−
b
c
b
−
c
⋅
1
−
c
a
c
−
a
+
1
−
c
a
c
−
a
⋅
1
−
a
b
a
−
b
\frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b}
a
−
b
1
−
ab
⋅
b
−
c
1
−
b
c
+
b
−
c
1
−
b
c
⋅
c
−
a
1
−
c
a
+
c
−
a
1
−
c
a
⋅
a
−
b
1
−
ab
c
)
c)
c
)
Prove the following ineqaulity
1
+
a
2
b
2
(
a
−
b
)
2
+
1
+
b
2
c
2
(
b
−
c
)
2
+
1
+
c
2
a
2
(
c
−
a
)
2
≥
3
2
\frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2}
(
a
−
b
)
2
1
+
a
2
b
2
+
(
b
−
c
)
2
1
+
b
2
c
2
+
(
c
−
a
)
2
1
+
c
2
a
2
≥
2
3
When does eqaulity holds?