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Problems
Contests
National and Regional Contests
Germany Contests
German National Olympiad
2002 German National Olympiad
2002 German National Olympiad
Part of
German National Olympiad
Subcontests
(6)
6
1
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A combinatorics problem where the currency is important
Theo Travel, who has
5
5
5
children, has already visited
8
8
8
countries of the eurozone. From every country, he brought
5
5
5
not necessarily distinct coins home. Moreover, among these
40
40
40
coins there are exactly
5
5
5
of every value (
1
,
2
,
5
,
10
,
20
,
1,2,5,10,20,
1
,
2
,
5
,
10
,
20
,
and
50
50
50
ct,
1
1
1
and
2
2
2
euro). He wants to give each child
8
8
8
coins such that they are from different countries and that each child gets the same amount of money. Is this always possible?
4
1
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Inequality with sequences
Given a positive real number
a
1
a_1
a
1
, we recursively define
a
n
+
1
=
1
+
a
1
a
2
⋯
⋅
a
n
.
a_{n+1} = 1+a_1 a_2 \cdots \cdot a_n.
a
n
+
1
=
1
+
a
1
a
2
⋯
⋅
a
n
.
Furthermore, let
b
n
=
1
a
1
+
1
a
2
+
⋯
+
1
a
n
.
b_n = \frac{1}{a_1 } + \frac{1}{a_2 } +\cdots + \frac{1}{a_n }.
b
n
=
a
1
1
+
a
2
1
+
⋯
+
a
n
1
.
Prove that
b
n
<
2
a
1
b_n < \frac{2}{a_1}
b
n
<
a
1
2
for all positive integers
n
n
n
and that this is the smallest possible bound.
1
1
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Typical German system of equations
Find all real numbers
a
,
b
a,b
a
,
b
satisfying the following system of equations \begin{align*} 2a^2 -2ab+b^2 &=a\\ 4a^2 -5ab +2b^2 & =b. \end{align*}
2
1
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2001 GER Minimal distance
Minimal distance of a finite set of different points in space is length of the shortest segment, whose both ends belong to this set and segment has length greater than
0
0
0
. a) Prove there exist set of
8
8
8
points on sphere with radius
R
R
R
, whose minimal distance is greater than
1
,
15
R
1,15R
1
,
15
R
. b) Does there exist set of
8
8
8
points on sphere with radius
R
R
R
, whose minimal distance is greater than
1
,
2
R
1,2R
1
,
2
R
?
3
1
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Fractions, prime, floor, sum
Prove that for all primes
p
p
p
true is equality
∑
k
=
1
p
−
1
⌊
k
3
p
⌋
=
(
p
−
2
)
(
p
−
1
)
(
p
+
1
)
4
\sum_{k=1}^{p-1}\left\lfloor\frac{k^3}{p}\right\rfloor=\frac{(p-2)(p-1)(p+1)}{4}
k
=
1
∑
p
−
1
⌊
p
k
3
⌋
=
4
(
p
−
2
)
(
p
−
1
)
(
p
+
1
)
5
1
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Try out this geometry.
Show that the triangle whose angles satisfy the equality
sin
2
A
+
sin
2
B
+
sin
2
C
cos
2
A
+
cos
2
B
+
cos
2
C
=
2
\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2
cos
2
A
+
cos
2
B
+
cos
2
C
sin
2
A
+
sin
2
B
+
sin
2
C
=
2
is right angled