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National and Regional Contests
Bosnia Herzegovina Contests
Bosnia Herzegovina Team Selection Test
2000 Bosnia and Herzegovina Team Selection Test
2000 Bosnia and Herzegovina Team Selection Test
Part of
Bosnia Herzegovina Team Selection Test
Subcontests
(6)
6
1
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Bosnia and Herzegovina TST 2000 Day 2 Problem 3
It is given triangle
A
B
C
ABC
A
BC
such that
∠
A
B
C
=
3
∠
C
A
B
\angle ABC = 3 \angle CAB
∠
A
BC
=
3∠
C
A
B
. On side
A
C
AC
A
C
there are two points
M
M
M
and
N
N
N
in order
A
−
N
−
M
−
C
A - N - M - C
A
−
N
−
M
−
C
and
∠
C
B
M
=
∠
M
B
N
=
∠
N
B
A
\angle CBM = \angle MBN = \angle NBA
∠
CBM
=
∠
MBN
=
∠
NB
A
. Let
L
L
L
be an arbitrary point on side
B
N
BN
BN
and
K
K
K
point on
B
M
BM
BM
such that
L
K
∣
∣
A
C
LK \mid \mid AC
L
K
∣∣
A
C
. Prove that lines
A
L
AL
A
L
,
N
K
NK
N
K
and
B
C
BC
BC
are concurrent
5
1
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Bosnia and Herzegovina TST 2000 Day 2 Problem 2
Let
T
m
T_m
T
m
be a number of non-congruent triangles which perimeter is
m
m
m
and all its sides are positive integers. Prove that:
a
)
a)
a
)
T
1999
>
T
2000
T_{1999} > T_{2000}
T
1999
>
T
2000
b
)
b)
b
)
T
4
n
+
1
=
T
4
n
−
2
+
n
T_{4n+1}=T_{4n-2}+n
T
4
n
+
1
=
T
4
n
−
2
+
n
,
(
n
∈
N
)
(n \in \mathbb{N})
(
n
∈
N
)
4
1
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Bosnia and Herzegovina TST 2000 Day 2 Problem 1
Prove that for all positive real
a
a
a
,
b
b
b
and
c
c
c
holds:
b
c
a
2
+
2
b
c
+
a
c
b
2
+
2
a
c
+
a
b
c
2
+
2
a
b
≤
1
≤
a
2
a
2
+
2
b
c
+
b
2
b
2
+
2
a
c
+
c
2
c
2
+
2
a
b
\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab} \leq 1 \leq \frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}
a
2
+
2
b
c
b
c
+
b
2
+
2
a
c
a
c
+
c
2
+
2
ab
ab
≤
1
≤
a
2
+
2
b
c
a
2
+
b
2
+
2
a
c
b
2
+
c
2
+
2
ab
c
2
3
1
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Bosnia and Herzegovina TST 2000 Day 1 Problem 3
We call Pythagorean triple a triple
(
x
,
y
,
z
)
(x,y,z)
(
x
,
y
,
z
)
of positive integers such that
x
<
y
<
z
x<y<z
x
<
y
<
z
and
x
2
+
y
2
=
z
2
x^2+y^2=z^2
x
2
+
y
2
=
z
2
. Prove that for all
n
∈
N
n \in \mathbb{N}
n
∈
N
number
2
n
+
1
2^{n+1}
2
n
+
1
is in exactly
n
n
n
Pythagorean triples
2
1
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Bosnia and Herzegovina TST 2000 Day 1 Problem 2
Let
S
S
S
be a point inside triangle
A
B
C
ABC
A
BC
and let lines
A
S
AS
A
S
,
B
S
BS
BS
and
C
S
CS
CS
intersect sides
B
C
BC
BC
,
C
A
CA
C
A
and
A
B
AB
A
B
in points
X
X
X
,
Y
Y
Y
and
Z
Z
Z
, respectively. Prove that
B
X
⋅
C
X
A
X
2
+
C
Y
⋅
A
Y
B
Y
2
+
A
Z
⋅
B
Z
C
Z
2
=
R
r
−
1
\frac{BX\cdot CX}{AX^2}+\frac{CY\cdot AY}{BY^2}+\frac{AZ\cdot BZ}{CZ^2}=\frac{R}{r}-1
A
X
2
BX
⋅
CX
+
B
Y
2
C
Y
⋅
A
Y
+
C
Z
2
A
Z
⋅
BZ
=
r
R
−
1
iff
S
S
S
is incenter of
A
B
C
ABC
A
BC
1
1
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Bosnia and Herzegovina TST 2000 Day 1 Problem 1
Find real roots
x
1
x_1
x
1
,
x
2
x_2
x
2
of equation
x
5
−
55
x
+
21
=
0
x^5-55x+21=0
x
5
−
55
x
+
21
=
0
, if we know
x
1
⋅
x
2
=
1
x_1\cdot x_2=1
x
1
⋅
x
2
=
1