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National and Regional Contests
Germany Contests
Bundeswettbewerb Mathematik
2013 Bundeswettbewerb Mathematik
2013 Bundeswettbewerb Mathematik
Part of
Bundeswettbewerb Mathematik
Subcontests
(4)
4
2
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Winning strategy: Prevent B from getting a square
Two players
A
A
A
and
B
B
B
play the following game taking alternate moves. In each move, a player writes one digit on the blackboard. Each new digit is written either to the right or left of the sequence of digits already written on the blackboard. Suppose that
A
A
A
begins the game and initially the blackboard was empty.
B
B
B
wins the game if ,after some move of
B
B
B
, the sequence of digits written in the blackboard represents a perfect square. Prove that
A
A
A
can prevent
B
B
B
from winning.
BWM 2013 P8: Combinatorial identity
Consider the Pascal's triangle in the figure where the binomial coefficients are arranged in the usual manner. Select any binomial coefficient from anywhere except the right edge of the triangle and labet it
C
C
C
. To the right of
C
C
C
, in the horizontal line, there are
t
t
t
numbers, we denote them as
a
1
,
a
2
,
⋯
,
a
t
a_1,a_2,\cdots,a_t
a
1
,
a
2
,
⋯
,
a
t
, where
a
t
=
1
a_t = 1
a
t
=
1
is the last number of the series. Consider the line parallel to the left edge of the triangle containing
C
C
C
, there will only be
t
t
t
numbers diagonally above
C
C
C
in that line. We successively name them as
b
1
,
b
2
,
⋯
,
b
t
b_1,b_2,\cdots,b_t
b
1
,
b
2
,
⋯
,
b
t
, where
b
t
=
1
b_t = 1
b
t
=
1
. Show that
b
t
a
1
−
b
t
−
1
a
2
+
b
t
−
2
a
3
−
⋯
+
(
−
1
)
t
−
1
b
1
a
t
=
1
b_ta_1-b_{t-1}a_2+b_{t-2}a_3-\cdots+(-1)^{t-1}b_1a_t = 1
b
t
a
1
−
b
t
−
1
a
2
+
b
t
−
2
a
3
−
⋯
+
(
−
1
)
t
−
1
b
1
a
t
=
1
. For example, Suppose you choose
(
4
1
)
=
4
\binom41 = 4
(
1
4
)
=
4
(see figure), then
t
=
3
t = 3
t
=
3
,
a
1
=
6
,
a
2
=
4
,
a
3
=
1
a_1 = 6, a_2 = 4, a_3 = 1
a
1
=
6
,
a
2
=
4
,
a
3
=
1
and
b
1
=
3
,
b
2
=
2
,
b
3
=
1
b_1 = 3, b_2 = 2, b_3 = 1
b
1
=
3
,
b
2
=
2
,
b
3
=
1
.
1
1
1
b
3
1
2
b
2
1
1
3
b
1
3
1
1
4
6
a
1
4
a
2
1
a
3
…
…
…
…
…
…
\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & \underset{b_3}{1} & & & & \\ & & & 1 & & \underset{b_2}{2} & & 1 & & & \\ & & 1 & & \underset{b_1}{3} & & 3 & & 1 & & \\ & 1 & & \boxed{4} & & \underset{a_1}{6} & & \underset{a_2}{4} & & \underset{a_3}{1} & \\ \ldots & & \ldots & & \ldots & & \ldots & & \ldots & & \ldots \\ \end{array}
…
1
1
…
1
4
1
b
1
3
…
1
b
2
2
a
1
6
b
3
1
3
…
1
a
2
4
1
…
a
3
1
…
3
2
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BWM 2013 P3: P is a point in ABCD. Find angle PBA
In the interior of the square
A
B
C
D
ABCD
A
BC
D
, the point
P
P
P
lies in such a way that
∠
D
C
P
=
∠
C
A
P
=
2
5
∘
\angle DCP = \angle CAP=25^{\circ}
∠
D
CP
=
∠
C
A
P
=
2
5
∘
. Find all possible values of
∠
P
B
A
\angle PBA
∠
PB
A
.
BWM 2013 P7: Hexagon whose vertices lie on a circle
Let
A
B
C
D
E
F
ABCDEF
A
BC
D
EF
be a convex hexagon whose vertices lie on a circle. Suppose that
A
B
⋅
C
D
⋅
E
F
=
B
C
⋅
D
E
⋅
F
A
AB\cdot CD\cdot EF = BC\cdot DE\cdot FA
A
B
⋅
C
D
⋅
EF
=
BC
⋅
D
E
⋅
F
A
. Show that the diagonals
A
D
,
B
E
AD, BE
A
D
,
BE
and
C
F
CF
CF
are concurrent.
2
2
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BWM'13 P2: Partition the triangle into 5 isosceles triangles
Is it possible to partition a triangle, with line segments, into exactly five isosceles triangles? All the triangles in concern are assumed to be nondegenerated triangles.
BWM 2013 P6: Change the parallelogram to cube
A parallelogram of paper with sides
25
25
25
and
10
10
10
is given. The distance between the longer sides is
6
6
6
. The paper should be cut into exactly two parts in such a way that one can stick both the pieces together and fold it in a suitable manner to form a cube of suitable edge length without any further cuts and overlaps. Show that it is really possible and describe such a fragmentation.
1
2
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BWM 2013 P1: Basic combinatorics
Is it possible to partition the set
S
=
{
1
,
2
,
…
,
21
}
S=\{1,2,\ldots,21\}
S
=
{
1
,
2
,
…
,
21
}
into subsets that in each of these subsets the largest number is equal to the sum of the other numbers?
BWM 2013 P5: mn|(m^2+n^2+m)
Suppose
m
m
m
and
n
n
n
are positive integers such that
m
2
+
n
2
+
m
m^2+n^2+m
m
2
+
n
2
+
m
is divisible by
m
n
mn
mn
. Prove that
m
m
m
is a square number.