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It is easy to see that the placement of (i) forms a regular
p
-gon; the difference
of the angles is all
4
mˀ
n
=
2
p
.Now
k
≥
p
is assumed, the regular
p
-gon is always
completed.
One can see that in (ii), every pair of angles taken from the head forms a
diameter. Since
k
and
p
are odd, it does not occur that at the end a diameter is
left uncompleted.
Besides, it is seen that in (ii), each iteration with respect to
j
consists of
m
1 distinct diameters. What remains is to claim that any angle in (ii) does
not coincide with the angles in (i). Note that
2((
j
−
1)
m
+
l
)
ˀ
n
−
+
ˀ
=
2((
j
−
1)
m
+
l
+
pm
)
ˀ
n
.
For
l
=1
,
2
,
1)
m
+
l
+
pm
are indivisible
by
m
, which implies that none of the angles in (ii) has appeared in (i).
···
,m
−
1, both (
j
−
1)
m
+
l
and (
j
−
See Fig.
4
for the behavior of the strategy for
n
= 40,
p
=5,and
k
= 15. Together
with Lemma
4
, we have a corollary.
Corollary 1.
For
n
divisible by six,
C
opt
(
k
)=0
holds if and only if
2
≤
k
≤
n
−
2
or
k
=
n
.
For the case that
n
is a prime number, we show that
C
opt
(
k
) cannot be zero
unless
k
=
n
through algebraic arguments. Let
B
:=
cos
2
ˀ
,
cos
4
ˀ
,...,
cos
2(
n
,
n
,
sin
2
ˀ
n
,
sin
4
ˀ
−
1)
ˀ
,
sin
2(
n
−
1)
ˀ
n
n
n
n
2
as the
which is the set of the destinations other than (1
,
0). We here regard
R
complex plane. Then, the
n
1 positions in
B
stand for the roots of the equation
z
n
= 1 except
z
= 1. Letting
ʶ
be cos
2
n
+
i
sin
2
n
, these positions are expressed
as
ʶ,ʶ
2
,...,ʶ
n−
1
. The following lemma is a special case of Theorem 12.13 in [
8
],
where
−
Q
[
X
] denotes the set of polynomials of
X
with the coecients being
rational.
Lemma 5.
([
8
]) For
n
prime, every element in
A
:=
{
f
(
ʶ
)
|
f
∈
Q
[
X
]
}
can be
uniquely expressed in a linear combination of
ʶ,ʶ
2
,...,ʶ
n−
1
with
a
j
∈
Q
,
a
1
ʶ
+
a
2
ʶ
2
+
+
a
n−
1
ʶ
n−
1
.
···
Theorem 5.
For
n
prime,
C
opt
(
k
)=0
holds if and only if
k
=
n
.
Proof.
Apply Lemma
5
with the element to be expressed being 0. Then, the
lemma states that we have to set
a
1
,a
2
,...,a
n−
1
to all zero. Back in the context
of
R
2
, this fact implies that wherever we place
k
points with 1
≤
k
≤
n
−
1
at some positions in
B
, the center of mass does not come to the origin. This
is based on the observation that the linear combination for
a
j
0
,
1
∈{
k
}
with
n−
1
j
=1
a
j
= 1 represents the center of mass of the
k
points placed at
k
distinct
positions in
B
. By rotational symmetry, even if we use the position (1
,
0), we
know that any placement of
n
−
1 or fewer points cannot let the center come to
the origin.
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