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Contests
National and Regional Contests
Germany Contests
Bundeswettbewerb Mathematik
2005 Bundeswettbewerb Mathematik
2005 Bundeswettbewerb Mathematik
Part of
Bundeswettbewerb Mathematik
Subcontests
(4)
3
2
Hide problems
3A + 2B = 180° ==> a^2 + bc = c^2
Let
A
B
C
ABC
A
BC
be a triangle with sides
a
a
a
,
b
b
b
,
c
c
c
and (corresponding) angles
A
A
A
,
B
B
B
,
C
C
C
. Prove that if
3
A
+
2
B
=
18
0
∘
3A + 2B = 180^{\circ}
3
A
+
2
B
=
18
0
∘
, then
a
2
+
b
c
=
c
2
a^2+bc=c^2
a
2
+
b
c
=
c
2
. Additional problem: Prove that the converse also holds, i. e. prove the following: Let
A
B
C
ABC
A
BC
be an arbitrary triangle. Then,
3
A
+
2
B
=
18
0
∘
3A + 2B = 180^{\circ}
3
A
+
2
B
=
18
0
∘
if and only if
a
2
+
b
c
=
c
2
a^2+bc=c^2
a
2
+
b
c
=
c
2
. Similar problem: Let
A
B
C
ABC
A
BC
be an arbitrary triangle. Then,
3
A
+
2
B
=
36
0
∘
3A + 2B = 360^{\circ}
3
A
+
2
B
=
36
0
∘
if and only if
a
2
−
b
c
=
c
2
a^2-bc=c^2
a
2
−
b
c
=
c
2
.
Similar triangles ACD, AEF, AMN
Two circles
k
1
k_1
k
1
and
k
2
k_2
k
2
intersect at two points
A
A
A
and
B
B
B
. Some line through the point
B
B
B
meets the circle
k
1
k_1
k
1
at a point
C
C
C
(apart from
B
B
B
), and the circle
k
2
k_2
k
2
at a point
E
E
E
(apart from
B
B
B
). Another line through the point
B
B
B
meets the circle
k
1
k_1
k
1
at a point
D
D
D
(apart from
B
B
B
), and the circle
k
2
k_2
k
2
at a point
F
F
F
(apart from
B
B
B
). Assume that the point
B
B
B
lies between the points
C
C
C
and
E
E
E
and between the points
D
D
D
and
F
F
F
. Finally, let
M
M
M
and
N
N
N
be the midpoints of the segments
C
E
CE
CE
and
D
F
DF
D
F
. Prove that the triangles
A
C
D
ACD
A
C
D
,
A
E
F
AEF
A
EF
and
A
M
N
AMN
A
MN
are similar to each other.
2
2
Hide problems
3a = x^2 + 2y^2 ==> a = u^2 + 2v^2
Let
a
a
a
be such an integer, that
3
a
3a
3
a
can be written in the form
x
2
+
2
y
2
x^2 + 2y^2
x
2
+
2
y
2
, with integers
x
x
x
and
y
y
y
. Prove that the number
a
a
a
can also be written in this form. Additional problems: a) Find a general (necessary and sufficent) criterion for an integer
n
n
n
to be of that form. b) In how many ways can the integer
n
n
n
be represented in that way?
Finite number of triples with given properties
Let be
x
x
x
a rational number. Prove: There are only finitely many triples
(
a
,
b
,
c
)
(a,b,c)
(
a
,
b
,
c
)
of integers with
a
<
0
a<0
a
<
0
and
b
2
−
4
a
c
=
5
b^2-4ac=5
b
2
−
4
a
c
=
5
such that
a
x
2
+
b
x
+
c
ax^2+bx+c
a
x
2
+
b
x
+
c
is positive.
1
2
Hide problems
Moving a dice across a chessboard (easy)
In the centre of a
2005
×
2005
2005 \times 2005
2005
×
2005
chessboard lies a dice that is to be moved across the board in a sequence of moves. One move consists of the following three steps: - The dice has to be turned with an arbitrary side on top, - then it has to be moved by the shown number of points to the right or left - and finally moved by the concealed number of points upwards or downwards. The attained square is the starting square for the next move. Which squares of the chessboard can be reached in a finite sequence of such moves?
A game on a 100x100 chessboard
Two players
A
A
A
and
B
B
B
have one stone each on a
100
×
100
100 \times 100
100
×
100
chessboard. They move their stones one after the other, and a move means moving one's stone to a neighbouring field (horizontally or vertically, not diagonally). At the beginning of the game, the stone of
A
A
A
lies in the lower left corner, and the one of
B
B
B
in the lower right corner. Player
A
A
A
starts. Prove: Player
A
A
A
is, independently from that what
B
B
B
does, able to reach, after finitely many steps, the field
B
B
B
's stone is lying on at that moment.
4
2
Hide problems
finite set of integers can be arranged without intersection
Prove that each finite set of integers can be arranged without intersection.
Self-intersections of closed broken lines
For any integer
n
≥
3
n\geq 3
n
≥
3
, let
A
(
n
)
A\left(n\right)
A
(
n
)
denote the maximal number of self-intersections a closed broken line
P
1
P
2
.
.
.
P
n
P
1
P_1P_2...P_nP_1
P
1
P
2
...
P
n
P
1
can have; hereby, we assume that no three vertices of the broken line
P
1
P
2
.
.
.
P
n
P
1
P_1P_2...P_nP_1
P
1
P
2
...
P
n
P
1
are collinear. Prove that (a) if n is odd, then
A
(
n
)
=
n
(
n
−
3
)
2
A\left(n\right)=\frac{n\left(n-3\right)}{2}
A
(
n
)
=
2
n
(
n
−
3
)
; (b) if n is even, then
A
(
n
)
=
n
(
n
−
4
)
2
+
1
A\left(n\right)=\frac{n\left(n-4\right)}{2}+1
A
(
n
)
=
2
n
(
n
−
4
)
+
1
. Note. A self-intersection of a broken line is a (non-ordered) pair of two distinct non-adjacent segments of the broken line which have a common point.