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National and Regional Contests
Bosnia Herzegovina Contests
Bosnia and Herzegovina EGMO Team Selection Test
2018 Bosnia and Herzegovina EGMO TST
2018 Bosnia and Herzegovina EGMO TST
Part of
Bosnia and Herzegovina EGMO Team Selection Test
Subcontests
(4)
4
1
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Bosnia and Herzegovina EGMO TST 2018 Problem 4
It is given positive integer
n
n
n
. Let
a
1
,
a
2
,
.
.
.
,
a
n
a_1, a_2,..., a_n
a
1
,
a
2
,
...
,
a
n
be positive integers with sum
2
S
2S
2
S
,
S
∈
N
S \in \mathbb{N}
S
∈
N
. Positive integer
k
k
k
is called separator if you can pick
k
k
k
different indices
i
1
,
i
2
,
.
.
.
,
i
k
i_1, i_2,...,i_k
i
1
,
i
2
,
...
,
i
k
from set
{
1
,
2
,
.
.
.
,
n
}
\{1,2,...,n\}
{
1
,
2
,
...
,
n
}
such that
a
i
1
+
a
i
2
+
.
.
.
+
a
i
k
=
S
a_{i_1}+a_{i_2}+...+a_{i_k}=S
a
i
1
+
a
i
2
+
...
+
a
i
k
=
S
. Find, in terms of
n
n
n
, maximum number of separators
3
1
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Bosnia and Herzegovina EGMO TST 2018 Problem 3
Let
O
O
O
be a circumcenter of acute triangle
A
B
C
ABC
A
BC
and let
O
1
O_1
O
1
and
O
2
O_2
O
2
be circumcenters of triangles
O
A
B
OAB
O
A
B
and
O
A
C
OAC
O
A
C
, respectively. Circumcircles of triangles
O
A
B
OAB
O
A
B
and
O
A
C
OAC
O
A
C
intersect side
B
C
BC
BC
in points
D
D
D
(
D
≠
B
D \neq B
D
=
B
) and
E
E
E
(
E
≠
C
E \neq C
E
=
C
), respectively. Perpendicular bisector of side
B
C
BC
BC
intersects side
A
C
AC
A
C
in point
F
F
F
(
F
≠
A
F \neq A
F
=
A
). Prove that circumcenter of triangle
A
D
E
ADE
A
D
E
lies on
A
C
AC
A
C
iff
F
F
F
lies on line
O
1
O
2
O_1O_2
O
1
O
2
2
1
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Bosnia and Herzegovina EGMO TST 2018 Problem 2
Prove that for every pair of positive integers
(
m
,
n
)
(m,n)
(
m
,
n
)
, bigger than
2
2
2
, there exists positive integer
k
k
k
and numbers
a
0
,
a
1
,
.
.
.
,
a
k
a_0,a_1,...,a_k
a
0
,
a
1
,
...
,
a
k
, which are bigger than
2
2
2
, such that
a
0
=
m
a_0=m
a
0
=
m
,
a
1
=
n
a_1=n
a
1
=
n
and for all
i
=
0
,
1
,
.
.
.
,
k
−
1
i=0,1,...,k-1
i
=
0
,
1
,
...
,
k
−
1
holds
a
i
+
a
i
+
1
∣
a
i
a
i
+
1
+
1
a_i+a_{i+1} \mid a_ia_{i+1}+1
a
i
+
a
i
+
1
∣
a
i
a
i
+
1
+
1
1
1
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Bosnia and Herzegovina EGMO TST 2018 Problem 1
a
)
a)
a
)
Prove that there exists
5
5
5
nonnegative real numbers with sum equal to
1
1
1
, such that no matter how we arrange them on circle, two neighboring numbers exist with product not less than
1
9
\frac{1}{9}
9
1
a
)
a)
a
)
Prove that for every
5
5
5
nonnegative real numbers with sum equal to
1
1
1
, we can arrange them on circle, such that product of every two neighboring numbers is not greater than
1
9
\frac{1}{9}
9
1