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Problems
Contests
National and Regional Contests
Brazil Contests
Brazil National Olympiad
1991 Brazil National Olympiad
1991 Brazil National Olympiad
Part of
Brazil National Olympiad
Subcontests
(5)
3
1
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This sequence is a polynomial? a rational function?
Given
k
>
0
k > 0
k
>
0
, the sequence
a
n
a_n
a
n
is defined by its first two members and
a
n
+
2
=
a
n
+
1
+
k
n
a
n
a_{n+2} = a_{n+1} + \frac{k}{n}a_n
a
n
+
2
=
a
n
+
1
+
n
k
a
n
a)For which
k
k
k
can we write
a
n
a_n
a
n
as a polynomial in
n
n
n
? b) For which
k
k
k
can we write
a
n
+
1
a
n
=
p
(
n
)
q
(
n
)
\frac{a_{n+1}}{a_n} = \frac{p(n)}{q(n)}
a
n
a
n
+
1
=
q
(
n
)
p
(
n
)
? (
p
,
q
p,q
p
,
q
are polynomials in
R
[
X
]
\mathbb R[X]
R
[
X
]
).
4
1
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Very nice and very famous too! 1991|1999999....91
Show that there exists
n
>
2
n>2
n
>
2
such that
1991
∣
1999
…
91
1991 | 1999 \ldots 91
1991∣1999
…
91
(with
n
n
n
9's).
5
1
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Find the common point!
P
0
=
(
1
,
0
)
,
P
1
=
(
1
,
1
)
,
P
2
=
(
0
,
1
)
,
P
3
=
(
0
,
0
)
P_0 = (1,0), P_1 = (1,1), P_2 = (0,1), P_3 = (0,0)
P
0
=
(
1
,
0
)
,
P
1
=
(
1
,
1
)
,
P
2
=
(
0
,
1
)
,
P
3
=
(
0
,
0
)
.
P
n
+
4
P_{n+4}
P
n
+
4
is the midpoint of
P
n
P
n
+
1
P_nP_{n+1}
P
n
P
n
+
1
.
Q
n
Q_n
Q
n
is the quadrilateral
P
n
P
n
+
1
P
n
+
2
P
n
+
3
P_{n}P_{n+1}P_{n+2}P_{n+3}
P
n
P
n
+
1
P
n
+
2
P
n
+
3
.
A
n
A_n
A
n
is the interior of
Q
n
Q_n
Q
n
. Find
∩
n
≥
0
A
n
\cap_{n \geq 0}A_n
∩
n
≥
0
A
n
.
2
1
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Yatpp(yet another triangle-point problem)- parallels by p
P
P
P
is a point inside the triangle
A
B
C
ABC
A
BC
. The line through
P
P
P
parallel to
A
B
AB
A
B
meets
A
C
AC
A
C
A
0
A_0
A
0
and
B
C
BC
BC
at
B
0
B_0
B
0
. Similarly, the line through
P
P
P
parallel to
C
A
CA
C
A
meets
A
B
AB
A
B
at
A
1
A_1
A
1
and
B
C
BC
BC
at
C
1
C_1
C
1
, and the line through P parallel to BC meets
A
B
AB
A
B
at
B
2
B_2
B
2
and
A
C
AC
A
C
at
C
2
C_2
C
2
. Find the point
P
P
P
such that
A
0
B
0
=
A
1
B
1
=
A
2
C
2
A_0B_0 = A_1B_1 = A_2C_2
A
0
B
0
=
A
1
B
1
=
A
2
C
2
.
1
1
Hide problems
Boys and girls in a party
At a party every woman dances with at least one man, and no man dances with every woman. Show that there are men M and M' and women W and W' such that M dances with W, M' dances with W', but M does not dance with W', and M' does not dance with W.