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Problems
Contests
National and Regional Contests
Germany Contests
German National Olympiad
2009 German National Olympiad
2009 German National Olympiad
Part of
German National Olympiad
Subcontests
(6)
4
1
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1 /sqrt{x}} 1/ \sqrt{a+b-x} < 1/\sqrt{a} + 1 / \sqrt{b}
Let
a
a
a
and
b
b
b
be two fixed positive real numbers. Find all real numbers
x
x
x
, such that inequality holds
1
x
+
1
a
+
b
−
x
<
1
a
+
1
b
\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{a+b-x}} < \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}}
x
1
+
a
+
b
−
x
1
<
a
1
+
b
1
1
1
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Cube roots in an equation
Find all non-negative real numbers
a
a
a
such that the equation
1
+
x
3
+
1
−
x
3
=
a
\sqrt[3]{1+x}+\sqrt[3]{1-x}=a
3
1
+
x
+
3
1
−
x
=
a
has at least one real solution
x
x
x
with
0
≤
x
≤
1
0 \leq x \leq 1
0
≤
x
≤
1
. For all such
a
a
a
, what is
x
x
x
?
6
1
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Find the "best" |a;b|
Let a sequences: x_0\in [0;1],x_{n\plus{}1}\equal{}\frac56\minus{}\frac43 \Big|x_n\minus{}\frac12\Big|. Find the "best"
∣
a
;
b
∣
|a;b|
∣
a
;
b
∣
so that for all
x
0
x_0
x
0
we have
x
2009
∈
[
a
;
b
]
x_{2009}\in [a;b]
x
2009
∈
[
a
;
b
]
5
1
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Half of circle with diameter EF is tangent with BC,CA
Let a triangle
A
B
C
ABC
A
BC
.
E
,
F
E,F
E
,
F
in segment
A
B
AB
A
B
so that
E
E
E
lie between
A
F
AF
A
F
and half of circle with diameter
E
F
EF
EF
is tangent with
B
C
,
C
A
BC,CA
BC
,
C
A
at
G
,
H
G,H
G
,
H
.
H
F
HF
H
F
cut
G
E
GE
GE
at
S
S
S
,
H
E
HE
H
E
cut
F
G
FG
FG
at
T
T
T
. Prove that
C
C
C
is midpoint of
S
T
ST
ST
.
2
1
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Perfect square number
Find all positive interger
n
n
n
so that n^3\minus{}5n^2\plus{}9n\minus{}6 is perfect square number.
3
1
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A nice property of circumscribed quadrangles
Let
A
B
C
D
ABCD
A
BC
D
be a (non-degenerate) quadrangle and
N
N
N
the intersection of
A
C
AC
A
C
and
B
D
BD
B
D
. Denote by
a
,
b
,
c
,
d
a,b,c,d
a
,
b
,
c
,
d
the length of the altitudes from
N
N
N
to
A
B
,
B
C
,
C
D
,
D
A
AB,BC,CD,DA
A
B
,
BC
,
C
D
,
D
A
, respectively. Prove that \frac{1}{a}\plus{}\frac{1}{c} \equal{} \frac{1}{b}\plus{}\frac{1}{d} if
A
B
C
D
ABCD
A
BC
D
has an incircle. Extension: Prove that the converse is true, too. [If this has already been posted, I humbly apologize. A quick search turned up nothing.]