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Problems
Contests
National and Regional Contests
Bosnia Herzegovina Contests
Bosnia Herzegovina Team Selection Test
2013 Bosnia Herzegovina Team Selection Test
2013 Bosnia Herzegovina Team Selection Test
Part of
Bosnia Herzegovina Team Selection Test
Subcontests
(6)
5
1
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Minimum of $F_n$
Let
x
1
,
x
2
,
…
,
x
n
x_1,x_2,\ldots,x_n
x
1
,
x
2
,
…
,
x
n
be nonnegative real numbers of sum equal to
1
1
1
. Let
F
n
=
x
1
2
+
x
2
2
+
⋯
+
x
n
2
−
2
(
x
1
x
2
+
x
2
x
3
+
⋯
+
x
n
x
1
)
F_n=x_1^{2}+x_2^{2}+\cdots +x_n^{2}-2(x_1x_2+x_2x_3+\cdots +x_nx_1)
F
n
=
x
1
2
+
x
2
2
+
⋯
+
x
n
2
−
2
(
x
1
x
2
+
x
2
x
3
+
⋯
+
x
n
x
1
)
. Find: a)
min
F
3
\min F_3
min
F
3
; b)
min
F
4
\min F_4
min
F
4
; c)
min
F
5
\min F_5
min
F
5
.
1
1
Hide problems
Right triangle parallel lines
Triangle
A
B
C
ABC
A
BC
is right angled at
C
C
C
. Lines
A
M
AM
A
M
and
B
N
BN
BN
are internal angle bisectors.
A
M
AM
A
M
and
B
N
BN
BN
intersect altitude
C
H
CH
C
H
at points
P
P
P
and
Q
Q
Q
respectively. Prove that the line which passes through the midpoints of segments
Q
N
QN
QN
and
P
M
PM
PM
is parallel to
A
B
AB
A
B
.
2
1
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Sequence with each term perfect square
The sequence
a
n
a_n
a
n
is defined by
a
0
=
a
1
=
1
a_0=a_1=1
a
0
=
a
1
=
1
and
a
n
+
1
=
14
a
n
−
a
n
−
1
−
4
a_{n+1}=14a_n-a_{n-1}-4
a
n
+
1
=
14
a
n
−
a
n
−
1
−
4
,for all positive integers
n
n
n
. Prove that all terms of this sequence are perfect squares.
4
1
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Primes and divisibility
Find all primes
p
,
q
p,q
p
,
q
such that
p
p
p
divides
30
q
−
1
30q-1
30
q
−
1
and
q
q
q
divides
30
p
−
1
30p-1
30
p
−
1
.
6
1
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IO Euler line of triangle PQR
In triangle
A
B
C
ABC
A
BC
,
I
I
I
is the incenter. We have chosen points
P
,
Q
,
R
P,Q,R
P
,
Q
,
R
on segments
I
A
,
I
B
,
I
C
IA,IB,IC
I
A
,
I
B
,
I
C
respectively such that
I
P
⋅
I
A
=
I
Q
⋅
I
B
=
I
R
⋅
I
C
IP\cdot IA=IQ \cdot IB=IR\cdot IC
I
P
⋅
I
A
=
I
Q
⋅
I
B
=
I
R
⋅
I
C
. Prove that the points
I
I
I
and
O
O
O
belong to Euler line of triangle
P
Q
R
PQR
PQR
where
O
O
O
is circumcenter of
A
B
C
ABC
A
BC
.
3
1
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n+1 people knowing each other
Prove that in the set consisting of
(
2
n
n
)
\binom{2n}{n}
(
n
2
n
)
people we can find a group of
n
+
1
n+1
n
+
1
people in which everyone knows everyone or noone knows noone.