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Problems
Contests
National and Regional Contests
Austria Contests
Austrian MO National Competition
2013 Federal Competition For Advanced Students, Part 1
2013 Federal Competition For Advanced Students, Part 1
Part of
Austrian MO National Competition
Subcontests
(4)
4
1
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Varying circle - constant ratio
Let
A
A
A
,
B
B
B
and
C
C
C
be three points on a line (in this order). For each circle
k
k
k
through the points
B
B
B
and
C
C
C
, let
D
D
D
be one point of intersection of the perpendicular bisector of
B
C
BC
BC
with the circle
k
k
k
. Further, let
E
E
E
be the second point of intersection of the line
A
D
AD
A
D
with
k
k
k
. Show that for each circle
k
k
k
, the ratio of lengths
B
E
‾
:
C
E
‾
\overline{BE}:\overline{CE}
BE
:
CE
is the same.
3
1
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Placing integers in two lines
Arrange the positive integers into two lines as follows: \begin{align*} 1 3 \qquad 6 \qquad\qquad 11 \qquad\qquad\qquad\qquad \ 19\qquad\qquad32\qquad\qquad 53\ldots\\ \mbox{\ \ } 2 4\ \ 5 7\ \ 8\ \ 9\ \ 10 \ 12\ 13\ 14\ 15\ 16\ 17\ 18 \ 20 \mbox{ to } 31 \ 33 \mbox{ to } 52 \ \ldots\end{align*} We start with writing
1
1
1
in the upper line,
2
2
2
in the lower line and
3
3
3
again in the upper line. Afterwards, we alternately write one single integer in the upper line and a block of integers in the lower line. The number of consecutive integers in a block is determined by the first number in the previous block. Let
a
1
a_1
a
1
,
a
2
a_2
a
2
,
a
3
a_3
a
3
,
…
\ldots
…
be the numbers in the upper line. Give an explicit formula for
a
n
a_n
a
n
.
2
1
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Cyclic Equations in x,y,z
Solve the following system of equations in rational numbers:
(
x
2
+
1
)
3
=
y
+
1
,
(
y
2
+
1
)
3
=
z
+
1
,
(
z
2
+
1
)
3
=
x
+
1.
(x^2+1)^3=y+1,\\ (y^2+1)^3=z+1,\\ (z^2+1)^3=x+1.
(
x
2
+
1
)
3
=
y
+
1
,
(
y
2
+
1
)
3
=
z
+
1
,
(
z
2
+
1
)
3
=
x
+
1.
1
1
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Product is not an integer power
Show that if for non-negative integers
m
m
m
,
n
n
n
,
N
N
N
,
k
k
k
the equation
(
n
2
+
1
)
2
k
⋅
(
44
n
3
+
11
n
2
+
10
n
+
2
)
=
N
m
(n^2+1)^{2^k}\cdot(44n^3+11n^2+10n+2)=N^m
(
n
2
+
1
)
2
k
⋅
(
44
n
3
+
11
n
2
+
10
n
+
2
)
=
N
m
holds, then
m
=
1
m = 1
m
=
1
.