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National and Regional Contests
Albania Contests
Albania-Balkan MO TST
2014 BMO TST
2014 BMO TST
Part of
Albania-Balkan MO TST
Subcontests
(5)
5
1
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Albanian TST 2014 5th problem
Find all non-negative integers
k
,
n
k,n
k
,
n
which satisfy
2
2
k
+
1
+
9
⋅
2
k
+
5
=
n
2
2^{2k+1} + 9\cdot 2^k+5=n^2
2
2
k
+
1
+
9
⋅
2
k
+
5
=
n
2
.
4
1
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Albanian TST 2014 4th problem
Find all functions
f
:
R
→
R
f:\mathbb{R}\to\mathbb{R}
f
:
R
→
R
such that
f
(
x
)
f
(
y
)
=
f
(
x
+
y
)
+
x
y
f(x)f(y)=f(x+y)+xy
f
(
x
)
f
(
y
)
=
f
(
x
+
y
)
+
x
y
for all
x
,
y
∈
R
x,y\in \mathbb{R}
x
,
y
∈
R
.
3
1
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Albanian TST 2014 3rd problem
From the point
P
P
P
outside a circle
ω
\omega
ω
with center
O
O
O
draw the tangents
P
A
PA
P
A
and
P
B
PB
PB
where
A
A
A
and
B
B
B
belong to
ω
\omega
ω
.In a random point
M
M
M
in the chord
A
B
AB
A
B
we draw the perpendicular to
O
M
OM
OM
, which intersects
P
A
PA
P
A
and
P
B
PB
PB
in
C
C
C
and
D
D
D
. Prove that
M
M
M
is the midpoint
C
D
CD
C
D
.
2
1
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Albanian TST 2014 2nd problem
Solve the following equation in
R
\mathbb{R}
R
:
(
x
−
1
x
)
1
2
+
(
1
−
1
x
)
1
2
=
x
.
\left(x-\frac{1}{x}\right)^\frac{1}{2}+\left(1-\frac{1}{x}\right)^\frac{1}{2}=x.
(
x
−
x
1
)
2
1
+
(
1
−
x
1
)
2
1
=
x
.
1
1
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Albanian TST 2014 1st problem
Prove that for
n
≥
2
n\ge 2
n
≥
2
the following inequality holds:
1
n
+
1
(
1
+
1
3
+
…
+
1
2
n
−
1
)
>
1
n
(
1
2
+
…
+
1
2
n
)
.
\frac{1}{n+1}\left(1+\frac{1}{3}+\ldots +\frac{1}{2n-1}\right) >\frac{1}{n}\left(\frac{1}{2}+\ldots+\frac{1}{2n}\right).
n
+
1
1
(
1
+
3
1
+
…
+
2
n
−
1
1
)
>
n
1
(
2
1
+
…
+
2
n
1
)
.