MathDB
Problems
Contests
National and Regional Contests
Germany Contests
Bundeswettbewerb Mathematik
1989 Bundeswettbewerb Mathematik
1989 Bundeswettbewerb Mathematik
Part of
Bundeswettbewerb Mathematik
Subcontests
(4)
3
2
Hide problems
Bundeswettbewerb Mathematik 1989 Problem 1.3
A convex polygon is divided into finitely many quadrilaterals. Prove that at least one of these quadrilaterals must also be convex.
Bundeswettbewerb Mathematik 1989 Problem 2.3
Over each side of a cyclic quadrilateral erect a rectangle whose height is equal to the length of the opposite side. Prove that the centers of these rectangles form another rectangle.
2
2
Hide problems
Bundeswettbewerb Mathematik 1989 Problem 1.2
A trapezoid has area
2
m
2
2\, m^2
2
m
2
and the sum of its diagonals is
4
m
4\,m
4
m
. Determine the height of this trapezoid.
Bundeswettbewerb Mathematik 1989 Problem 2.2
Find all pairs
(
a
,
b
)
(a,b)
(
a
,
b
)
of real numbers such that
∣
1
−
x
2
−
a
x
−
b
∣
≤
2
−
1
2
|\sqrt{1-x^2 }-ax-b| \leq \frac{\sqrt{2} -1}{2}
∣
1
−
x
2
−
a
x
−
b
∣
≤
2
2
−
1
holds for all
x
∈
[
0
,
1
]
x\in [0,1]
x
∈
[
0
,
1
]
.
1
2
Hide problems
Bundeswettbewerb Mathematik 1989 Problem 1.1
For a given positive integer
n
n
n
, let
f
(
x
)
=
x
n
f(x) =x^{n}
f
(
x
)
=
x
n
. Is it possible for the decimal number
0.
f
(
1
)
f
(
2
)
f
(
3
)
…
0.f(1)f(2)f(3)\ldots
0.
f
(
1
)
f
(
2
)
f
(
3
)
…
to be rational? (Example: for
n
=
2
n=2
n
=
2
, we are considering
0.1491625
…
0.1491625\ldots
0.1491625
…
)
Bundeswettbewerb Mathematik 1989 Problem 2.1
Determine the polynomial
f
(
x
)
=
x
k
+
a
k
−
1
x
k
−
1
+
⋯
+
a
1
x
+
a
0
f(x) = x^k + a_{k-1} x^{k-1}+\cdots +a_1 x +a_0
f
(
x
)
=
x
k
+
a
k
−
1
x
k
−
1
+
⋯
+
a
1
x
+
a
0
of smallest degree such that
a
i
∈
{
−
1
,
0
,
1
}
a_i \in \{-1,0,1\}
a
i
∈
{
−
1
,
0
,
1
}
for
0
≤
i
≤
k
−
1
0\leq i \leq k-1
0
≤
i
≤
k
−
1
and
f
(
n
)
f(n)
f
(
n
)
is divisible by
30
30
30
for all positive integers
n
n
n
.
4
2
Hide problems
Bundeswettbewerb Mathematik 1989 Problem 1.4
Let
n
n
n
be an odd positive integer. Show that the equation
4
n
=
1
x
+
1
y
\frac{4}{n} =\frac{1}{x} + \frac{1}{y}
n
4
=
x
1
+
y
1
has a solution in the positive integers if and only if
n
n
n
has a divisor of the form
4
k
+
3
4k+3
4
k
+
3
.
Every integer on a circle divides its neighbor sum
Positive integers
x
1
,
x
2
,
…
,
x
n
x_1, x_2, \dots, x_n
x
1
,
x
2
,
…
,
x
n
(
n
≥
4
n \ge 4
n
≥
4
) are arranged in a circle such that each
x
i
x_i
x
i
divides the sum of the neighbors; that is
x
i
−
1
+
x
i
+
1
x
i
=
k
i
\frac{x_{i-1}+x_{i+1}}{x_i} = k_i
x
i
x
i
−
1
+
x
i
+
1
=
k
i
is an integer for each
i
i
i
, where
x
0
=
x
n
x_0 = x_n
x
0
=
x
n
,
x
n
+
1
=
x
1
x_{n+1} = x_1
x
n
+
1
=
x
1
. Prove that
2
n
≤
k
1
+
k
2
+
⋯
+
k
n
<
3
n
.
2n \le k_1 + k_2 + \dots + k_n < 3n.
2
n
≤
k
1
+
k
2
+
⋯
+
k
n
<
3
n
.