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National and Regional Contests
Bosnia Herzegovina Contests
JBMO TST - Bosnia and Herzegovina
2019 Bosnia and Herzegovina Junior BMO TST
2019 Bosnia and Herzegovina Junior BMO TST
Part of
JBMO TST - Bosnia and Herzegovina
Subcontests
(4)
3
1
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JBMO TST Bosnia and Herzegovina
3.
3.
3.
Let
S
S
S
be the set of all positive integers from
1
1
1
to
100
100
100
included. Two players play a game. The first player removes any
k
k
k
numbers he wants, from
S
S
S
. The second player's goal is to pick
k
k
k
different numbers, such that their sum is
100
100
100
. Which player has the winning strategy if :
a
)
a)
a
)
k
=
9
k=9
k
=
9
?
b
)
b)
b
)
k
=
8
k=8
k
=
8
?
4
1
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JBMO TST Bosnia and Herzegovina P4
4.
4.
4.
Let there be a variable positive integer whose last two digits are
3
′
s
3's
3
′
s
. Prove that this number is divisible by a prime greater than
7
7
7
.
2
1
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JBMO TST Bosnia and Herzegovina Problem 2
2.
2.
2.
Let
A
B
C
ABC
A
BC
be a triangle and
A
D
AD
A
D
the angle bisector (
D
∈
B
C
D\in BC
D
∈
BC
). The perpendicular from
B
B
B
to
A
D
AD
A
D
cuts the circumcircle of triangle
A
B
D
ABD
A
B
D
at
E
E
E
. If
O
O
O
is the center of the circle around
A
B
C
ABC
A
BC
, prove
A
,
O
,
E
A,O,E
A
,
O
,
E
are collinear.https://artofproblemsolving.com/community/c6h605458p3596629 https://artofproblemsolving.com/community/c6h1294020p6857833
1
1
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Bosnia and Herzegovina JBMO TST 2019 Q1
Let
x
,
y
,
z
x,y,z
x
,
y
,
z
be real numbers (
x
≠
y
x \ne y
x
=
y
,
y
≠
z
y\ne z
y
=
z
,
x
≠
z
x\ne z
x
=
z
) different from
0
0
0
. If
x
2
−
y
z
x
(
1
−
y
z
)
=
y
2
−
x
z
y
(
1
−
x
z
)
\frac{x^2-yz}{x(1-yz)}=\frac{y^2-xz}{y(1-xz)}
x
(
1
−
yz
)
x
2
−
yz
=
y
(
1
−
x
z
)
y
2
−
x
z
, prove that the following relation holds:
x
+
y
+
z
=
1
x
+
1
y
+
1
z
.
x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.
x
+
y
+
z
=
x
1
+
y
1
+
z
1
.