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Contests
National and Regional Contests
Mathlinks Contests.
MathLinks Contest 6th
1.3
1.3
Part of
MathLinks Contest 6th
Problems
(1)
0613 vector problem 6th edition Round 1 p3
Source:
5/3/2021
Introductory part We call an
n
n
n
-tuple
x
=
(
x
1
,
x
2
,
.
.
.
,
x
n
)
x = (x_1, x_2, ... , x_n)
x
=
(
x
1
,
x
2
,
...
,
x
n
)
, with
x
k
∈
R
x_k \in R
x
k
∈
R
(or respectively with all
x
k
∈
Z
x_k \in Z
x
k
∈
Z
) a real vector (or respectively an integer vector). The set of all real vectors (respectively all integer vectors) is usually denoted by
R
n
R^n
R
n
(respectively
Z
n
Z^n
Z
n
). A vector
x
x
x
is null if and only if
x
k
=
0
x_k = 0
x
k
=
0
, for all
k
∈
{
1
,
2
,
.
.
.
,
n
}
k \in \{1, 2,... , n\}
k
∈
{
1
,
2
,
...
,
n
}
. Also let
U
n
U_n
U
n
be the set of all real vectors
x
=
(
x
1
,
x
2
,
.
.
.
,
x
n
)
x = (x_1, x_2, ... , x_n)
x
=
(
x
1
,
x
2
,
...
,
x
n
)
, such that
x
1
2
+
x
2
2
+
.
.
.
+
x
n
2
=
1
x^2_1 + x^2_2 + ...+ x^2_n = 1
x
1
2
+
x
2
2
+
...
+
x
n
2
=
1
. For two vectors
x
=
(
x
1
,
.
.
.
,
x
n
)
,
y
=
(
y
1
,
.
.
.
,
y
n
)
x = (x_1, ... , x_n), y = (y_1, ..., y_n)
x
=
(
x
1
,
...
,
x
n
)
,
y
=
(
y
1
,
...
,
y
n
)
we define the scalar product as the real number
x
⋅
y
=
x
1
y
1
+
x
2
y
2
+
.
.
.
+
x
n
y
n
x\cdot y = x_1y_1 + x_2y_2 +...+ x_ny_n
x
⋅
y
=
x
1
y
1
+
x
2
y
2
+
...
+
x
n
y
n
. We define the norm of the vector
x
x
x
as
∣
∣
x
∣
∣
=
x
1
2
+
x
2
2
+
.
.
.
+
x
n
2
||x|| =\sqrt{x^2_1 + x^2_2 + ...+ x^2_n}
∣∣
x
∣∣
=
x
1
2
+
x
2
2
+
...
+
x
n
2
The problem Let
A
(
k
,
r
)
=
{
x
∈
U
n
∣
A(k, r) = \{x \in U_n |
A
(
k
,
r
)
=
{
x
∈
U
n
∣
for all
z
∈
Z
n
z \in Z^n
z
∈
Z
n
we have either
∣
x
⋅
z
∣
≥
k
∣
∣
z
∣
∣
r
|x \cdot z| \ge \frac{k}{||z||^r}
∣
x
⋅
z
∣
≥
∣∣
z
∣
∣
r
k
or
z
z
z
is null
}
\}
}
. Prove that if
r
>
n
−
1
r > n - 1
r
>
n
−
1
the we can find a positive number
k
k
k
such that
A
(
k
,
r
)
A(k, r)
A
(
k
,
r
)
is not empty, and if
r
<
n
−
1
r < n - 1
r
<
n
−
1
we cannot find such a positive number
k
k
k
.
vector
algebra
6th edition