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Proof.
Let
a
1
,a
2
,...,a
n
be the side-lengths of
P
that are linearly independent
over
. Suppose that a set of
k
triangles tiles
P
, and let
b
1
,b
2
,...,b
3
k
be the
side-lengths of the
k
triangles. Each
a
i
can be represented as a linear combination
of
b
1
,...,b
3
k
with integral coecients. Since
a
1
,...,a
n
are linearly independent
over
Q
Q
,wemusthave3
k
≥
n
, and hence
k
≥
n/
3.
Corollary 2.1.
For any integer
n>
0
, there is a polygon
P
with
e
(
P
)
≥
n
.
Exa
m
p
le
2
.1
.
The simplicial element number of a quadrilateral with sides
1
,
√
2
,
√
3
,
√
5
is
2
.
The following theorem is proved by Laczkovich [
8
, Theorems 3.3, 3.6].
Theorem 2.2
(Laczkovich 2012).
b
if and only if
T
is a right triangle
in which the ratio of perpendicular sides is a rational multiple of
a/b
or
b/a
.
(ii) A triangle
T
can generate an equilateral triangle if and only if one of the
following holds.
-
T
is
(i) A triangle
T
can generate a rectangle
a
×
(
6
,
3
,
2
)
.
(
6
,
6
,
2
3
)
.
- The ratios of sides of
T
are all rationals, and one angle of
T
is
ˀ/
3
or
2
ˀ/
3
.
-
T
is
Example 2.2.
If the side-lengths of
T
are 7
,
8
,
13 (in this case, the largest angle
of
T
becomes 2
ˀ/
3), then the equilateral triangle of side 11760 can be dissected
into 2469600 triangles each congruent to
T
, see Laczkovich [
7
] p.86.
Let us state a well-known fact (see Appendix D of Niven [
10
]) as a lemma for
later use.
Lemma 2.1.
Let
ʱ
be an acute angle of a triangle. If
cos
ʱ
∈
Q
, then either
ʱ/ˀ
is irrational or
ʱ
=
ˀ/
3
.If
tan
ʱ
∈
Q
, then either
ʱ/ˀ
is irrational or
ʱ
=
ˀ/
4
.
From Theorem 2.2, we have the following.
Theorem 2.3.
(i)
For any
n>
3
,n
=6
, an equilateral triangle and a regular
n
-
gon are independent.
(ii)
A square and any other regular polygon are independent.
Proof.
(i) First, note that the interior angle of a regular
n
-gon is (
n
2)
ˀ/n
.
Let
T
be a triangle that generates an equilateral triangle. We use Theorem 2.2
(ii). If
T
is
−
(
6
,
6
,
2
3
)or
(
6
,
3
,
2
), then, except the cases
n
=4
,
6
,
12, no
angle (
n
2)
ˀ/n
can be constructed as a linear combination of the angles of
T
with nonnegative integral coecients. By Theorem 2.2 (i),
T
cannot generate
{
−
(
6
,
3
,
2
) generates
(
6
,
6
,
2
3
), we
4
}
. Suppose that
T
generates
{
12
}
. Since
may suppose that
T
is a triangle with sides 1
,
√
3
,
2 (which is a
(
6
,
3
,
2
)) and
T
tiles a re
g
ular 12-gon
P
. The side-length
s
of
P
can be then represented as
s
=
a
+
b
√
3 (where
a, b
are nonnegative integers), and the area of
P
is
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