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IMO Longlists
1992 IMO Longlists
49
4k+2 real numbers
4k+2 real numbers
Source:
September 2, 2010
algebra
recurrence relation
Inequality
Sequence
IMO Shortlist
IMO Longlist
Problem Statement
Given real numbers
x
i
(
i
=
1
,
2
,
⋯
,
4
k
+
2
)
x_i \ (i = 1, 2, \cdots, 4k + 2)
x
i
(
i
=
1
,
2
,
⋯
,
4
k
+
2
)
such that
∑
i
=
1
4
k
+
2
(
−
1
)
i
+
1
x
i
x
i
+
1
=
4
m
(
x
1
=
x
4
k
+
3
)
\sum_{i=1}^{4k +2} (-1)^{i+1} x_ix_{i+1} = 4m \qquad ( \ x_1=x_{4k+3} \ )
i
=
1
∑
4
k
+
2
(
−
1
)
i
+
1
x
i
x
i
+
1
=
4
m
(
x
1
=
x
4
k
+
3
)
prove that it is possible to choose numbers
x
k
1
,
⋯
,
x
k
6
x_{k_{1}}, \cdots, x_{k_{6}}
x
k
1
,
⋯
,
x
k
6
such that
∑
i
=
1
6
(
−
1
)
i
k
i
k
i
+
1
>
m
(
x
k
1
=
x
k
7
)
\sum_{i=1}^{6} (-1)^{i} k_i k_{i+1} > m \qquad ( \ x_{k_{1}} = x_{k_{7}} \ )
i
=
1
∑
6
(
−
1
)
i
k
i
k
i
+
1
>
m
(
x
k
1
=
x
k
7
)
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