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National and Regional Contests
Kosovo Contests
Kosovo Team Selection Test
2020 Kosovo Team Selection Test
2020 Kosovo Team Selection Test
Part of
Kosovo Team Selection Test
Subcontests
(4)
4
1
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\pi(m)-\pi(n)<= (m-1)\varphi(n)/n
Prove that for all positive integers
m
m
m
and
n
n
n
the following inequality hold:
π
(
m
)
−
π
(
n
)
≤
(
m
−
1
)
φ
(
n
)
n
\pi(m)-\pi(n)\leq\frac{(m-1)\varphi(n)}{n}
π
(
m
)
−
π
(
n
)
≤
n
(
m
−
1
)
φ
(
n
)
When does equality hold?Proposed by Shend Zhjeqi and Dorlir Ahmeti, Kosovo
3
1
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Angle bisectors on cyclic quadrilateral
Let
A
B
C
D
ABCD
A
BC
D
be a cyclic quadrilateral with center
O
O
O
such that
B
D
BD
B
D
bisects
A
C
.
AC.
A
C
.
Suppose that the angle bisector of
∠
A
B
C
\angle ABC
∠
A
BC
intersects the angle bisector of
∠
A
D
C
\angle ADC
∠
A
D
C
at a single point
X
X
X
different than
B
B
B
and
D
.
D.
D
.
Prove that the line passing through the circumcenters of triangles
X
A
C
XAC
X
A
C
and
X
B
D
XBD
XB
D
bisects the segment
O
X
.
OX.
OX
.
Proposed by Viktor Ahmeti and Leart Ajvazaj, Kosovo
2
1
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Winning if it is divisible by p
Let
p
p
p
be an odd prime number. Ana and Ben are playing a game with alternate moves as follows: in each move, the player which has the turn choose a number, which was not choosen before by any of the player, from the set
{
1
,
2
,
.
.
.
,
2
p
−
3
,
2
p
−
2
}
\{1,2,...,2p-3,2p-2\}
{
1
,
2
,
...
,
2
p
−
3
,
2
p
−
2
}
. This process continues until no number is left. After the end of the process, each player create the number by taking the product of the choosen numbers and then add 1. We say a player wins if the number that did create is divisible by
p
p
p
, while the number that did create the opponent it is not divisible by
p
p
p
, otherwise we say the game end in a draw. Ana start first move. Does it exist a strategy for any of the player to win the game?Proposed by Dorlir Ahmeti, Kosovo
1
1
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f(x+yf(x+y))=y^2+f(x)f(y)
Find all functions
f
:
R
→
R
f:\mathbb{R}\rightarrow\mathbb{R}
f
:
R
→
R
such that, for all real numbers
x
x
x
and
y
y
y
satisfy,
f
(
x
+
y
f
(
x
+
y
)
)
=
y
2
+
f
(
x
)
f
(
y
)
f\left(x+yf(x+y)\right)=y^2+f(x)f(y)
f
(
x
+
y
f
(
x
+
y
)
)
=
y
2
+
f
(
x
)
f
(
y
)
Proposed by Dorlir Ahmeti, Kosovo