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National and Regional Contests
Serbia Contests
Serbia National Math Olympiad
2019 Serbia National Math Olympiad
2019 Serbia National Math Olympiad
Part of
Serbia National Math Olympiad
Subcontests
(6)
5
1
Hide problems
2019 Serbia MO Day 2 P5
In the spherical shaped planet
X
X
X
there are
2
n
2n
2
n
gas stations. Every station is paired with one other station , and every two paired stations are diametrically opposite points on the planet. Each station has a given amount of gas. It is known that : if a car with empty (large enough) tank starting from any station it is always to reach the paired station with the initial station (it can get extra gas during the journey). Find all naturals
n
n
n
such that for any placement of
2
n
2n
2
n
stations for wich holds the above condotions, holds: there always a gas station wich the car can start with empty tank and go to all other stations on the planet.(Consider that the car consumes a constant amount of gas per unit length.)
2
1
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2019 Serbia MO Day 1 P2
For the sequence of real numbers
a
1
,
a
2
,
…
,
a
k
a_1,a_2,\dots ,a_k
a
1
,
a
2
,
…
,
a
k
we say it is invested on the interval
[
b
,
c
]
[b,c]
[
b
,
c
]
if there exists numbers
x
0
,
x
1
,
…
,
x
k
x_0,x_1,\dots ,x_k
x
0
,
x
1
,
…
,
x
k
in the interval
[
b
,
c
]
[b,c]
[
b
,
c
]
such that
∣
x
i
−
x
i
−
1
∣
=
a
i
|x_i-x_{i-1}|=a_i
∣
x
i
−
x
i
−
1
∣
=
a
i
for
i
=
1
,
2
,
3
,
…
k
i=1,2,3,\dots k
i
=
1
,
2
,
3
,
…
k
. A sequence is normed if all its members are not greater than
1
1
1
. For a given natural
n
n
n
, prove :a)Every normed sequence of length
2
n
+
1
2n+1
2
n
+
1
is invested in the interval
[
0
,
2
−
1
2
n
]
\left[ 0, 2-\frac{1}{2^n} \right ]
[
0
,
2
−
2
n
1
]
. b) there exists normed sequence of length
4
n
+
3
4n+3
4
n
+
3
wich is not invested on
[
0
,
2
−
1
2
n
]
\left[ 0, 2-\frac{1}{2^n} \right ]
[
0
,
2
−
2
n
1
]
.
6
1
Hide problems
2019 Serbia MO Day 2 P6
Sequences
(
a
n
)
n
=
0
∞
(a_n)_{n=0}^{\infty}
(
a
n
)
n
=
0
∞
and
(
b
n
)
n
=
0
∞
(b_n)_{n=0}^{\infty}
(
b
n
)
n
=
0
∞
are defined with recurrent relations :
a
0
=
0
,
a
1
=
1
,
a
n
+
1
=
2018
n
a
n
+
a
n
−
1
for
n
≥
1
a_0=0 , \;\;\; a_1=1, \;\;\;\; a_{n+1}=\frac{2018}{n} a_n+ a_{n-1}\;\;\; \text {for }\;\;\; n\geq 1
a
0
=
0
,
a
1
=
1
,
a
n
+
1
=
n
2018
a
n
+
a
n
−
1
for
n
≥
1
and
b
0
=
0
,
b
1
=
1
,
b
n
+
1
=
2020
n
b
n
+
b
n
−
1
for
n
≥
1
b_0=0 , \;\;\; b_1=1, \;\;\;\; b_{n+1}=\frac{2020}{n} b_n+ b_{n-1}\;\;\; \text {for }\;\;\; n\geq 1
b
0
=
0
,
b
1
=
1
,
b
n
+
1
=
n
2020
b
n
+
b
n
−
1
for
n
≥
1
Prove that :
a
1010
1010
=
b
1009
1009
\frac{a_{1010}}{1010}=\frac{b_{1009}}{1009}
1010
a
1010
=
1009
b
1009
4
1
Hide problems
2019 Serbia MO Day 2 P4
For a
△
A
B
C
\triangle ABC
△
A
BC
, let
A
1
A_1
A
1
be the symmetric point of the intersection of angle bisector of
∠
B
A
C
\angle BAC
∠
B
A
C
and
B
C
BC
BC
, where center of the symmetry is the midpoint of side
B
C
BC
BC
, In the same way we define
B
1
B_1
B
1
( on
A
C
AC
A
C
) and
C
1
C_1
C
1
(on
A
B
AB
A
B
). Intersection of circumcircle of
△
A
1
B
1
C
1
\triangle A_1B_1C_1
△
A
1
B
1
C
1
and line
A
B
AB
A
B
is the set
{
Z
,
C
1
}
\{Z,C_1 \}
{
Z
,
C
1
}
, with
B
C
BC
BC
is the set
{
X
,
A
1
}
\{X,A_1\}
{
X
,
A
1
}
and with
C
A
CA
C
A
is the set
{
Y
,
B
1
}
\{Y,B_1\}
{
Y
,
B
1
}
. If the perpendicular lines from
X
,
Y
,
Z
X,Y,Z
X
,
Y
,
Z
on
B
C
,
C
A
BC,CA
BC
,
C
A
and
A
B
AB
A
B
, respectively are concurrent , prove that
△
A
B
C
\triangle ABC
△
A
BC
is isosceles.
3
1
Hide problems
2019 Serbia MO Day 1 P3
Let
k
k
k
be the circle inscribed in convex quadrilateral
A
B
C
D
ABCD
A
BC
D
. Lines
A
D
AD
A
D
and
B
C
BC
BC
meet at
P
P
P
,and circumcircles of
△
P
A
B
\triangle PAB
△
P
A
B
and
△
P
C
D
\triangle PCD
△
PC
D
meet in
X
X
X
. Prove that tangents from
X
X
X
to
k
k
k
form equal angles with lines
A
X
AX
A
X
and
C
X
CX
CX
.
1
1
Hide problems
2019 Serbia MO Day 1 P1
Find all positive integers
n
,
n
>
1
n, n>1
n
,
n
>
1
for wich holds : If
a
1
,
a
2
,
…
,
a
k
a_1, a_2 ,\dots ,a_k
a
1
,
a
2
,
…
,
a
k
are all numbers less than
n
n
n
and relatively prime to
n
n
n
, and holds
a
1
<
a
2
<
⋯
<
a
k
a_1<a_2<\dots <a_k
a
1
<
a
2
<
⋯
<
a
k
, then none of sums
a
i
+
a
i
+
1
a_i+a_{i+1}
a
i
+
a
i
+
1
for
i
=
1
,
2
,
3
,
…
k
−
1
i=1,2,3,\dots k-1
i
=
1
,
2
,
3
,
…
k
−
1
are divisible by
3
3
3
.