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Contests
National and Regional Contests
Brazil Contests
Brazil Team Selection Test
2010 Brazil Team Selection Test
2010 Brazil Team Selection Test
Part of
Brazil Team Selection Test
Subcontests
(3)
4
1
Hide problems
exist k + 1 people where any two of them have not played each other
6
k
+
2
6k+2
6
k
+
2
people play in odd or even championship. In each odd or even match they participate exactly two people. Six rounds have been arranged so that in each round there are
3
k
+
1
3k + 1
3
k
+
1
simultaneous matches, and no player participates in two games of the same round. It is known that two people do not play with each other more than one turn. Prove that there are
k
+
1
k + 1
k
+
1
people where any two of them have not played each other.[hide=original wording] 6k+2 pessoas jogam em campeonato de par ou impar. Em cada partida de par ou impar participam exatamente duas pessoas. Seis rodadas foram organizadas, de modo que, em cada rodada, ha 3k + 1 partidas simultaneas, e nenhum jogador participa de dois jogos da mesma rodada. Sabe-se que duas pessoas nao jogam entre si mais de uma vez. Prove que existem k + 1 pessoas em que quaisquer duas delas nao jogaram entre si.
2
1
Hide problems
lcm (n, n + 1, n + 2, ... , n + k) > lcm (n + 1, n + 2, n + 3,... , n + k + 1)
Let
k
>
1
k > 1
k
>
1
be a fixed integer. Prove that there are infinite positive integers
n
n
n
such that
l
c
m
(
n
,
n
+
1
,
n
+
2
,
.
.
.
,
n
+
k
)
>
l
c
m
(
n
+
1
,
n
+
2
,
n
+
3
,
.
.
.
,
n
+
k
+
1
)
.
lcm \, (n, n + 1, n + 2, ... , n + k) > lcm \, (n + 1, n + 2, n + 3,... , n + k + 1).
l
c
m
(
n
,
n
+
1
,
n
+
2
,
...
,
n
+
k
)
>
l
c
m
(
n
+
1
,
n
+
2
,
n
+
3
,
...
,
n
+
k
+
1
)
.
1
1
Hide problems
OD _ |_ AB wanted, 2 circumrcircles
Let
A
B
C
ABC
A
BC
be an acute triangle and
D
D
D
a point on the side
A
B
AB
A
B
. The circumcircle of triangle
B
C
D
BCD
BC
D
cuts the side
A
C
AC
A
C
again at
E
E
E
.The circumcircle of triangle
A
C
D
ACD
A
C
D
cuts the side
B
C
BC
BC
again at
F
F
F
. If
O
O
O
is the circumcenter of the triangle
C
E
F
CEF
CEF
. Prove that
O
D
OD
O
D
is perpendicular to
A
B
AB
A
B
.