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Since
ʱ, ʲ, ʳ
are linearly independent over
,soare
x, y, z
.From
x
+
y
+
z
=
ʱ
+
ʲ
+
ʳ
=(
i
)
x
+(
m
i
)
y
+(
n
i
)
z
it follows that 1 =
i
=
m
i
=
n
i
. Changing the suxes suitably, we may
assume
1
=1
,m
2
=1
,n
3
= 1. Then,
ʱ
=
x, ʲ
=
y, ʳ
=
z
,and
˄
is similar to
T
.
The latter part follows similarly.
Q
Theorem 3.3.
(i)
All right triangles are original, and
(ii)
isosceles triangles
other than
(
6
,
6
,
2
3
)
are terminal.
Proof.
(i) Let
T
be a right triangle and suppose
˄
T
. Since
T
can generate a
rectangle,
˄
can generate a rectangle, and hence
˄
is a right triangle by Theorem
2.2. Let
ʱ, ʲ
(
ʱ
ₒ
ʲ
) be two acute angles of
T
,and
ʱ
,ʲ
(
ʱ
≤
ʲ
) be two acute
≤
angles of
˄
.Wehave
ʱ
≤
ʱ
(and
ʲ
≥
ʲ
), for otherwise,
˄
cannot generate
T
.
If
ʱ
=
ʲ
(i.e.
ʱ
=
ʲ
=
ˀ/
4), then
T
can generate a square, and hence
˄
can
generate a square. In this case, tan
ʱ
is a rational by Theorem 2.2 (i), and since
ˀ/
4 is an integral multiple of
ʱ
,
ʱ
is also a rational multiple of
ˀ
. This implies
ʱ
=
ˀ/
4 by Lemma 2.1, and hence
˄
T
. So, we may suppose
ʱ<ʲ
.If
ʱ
is
a rational multiple of
ˀ
, then so is
ʲ
,and
T
is original by Theorem 3.1 (i), and
˄
∼
∼
T
. So, we suppose
ʱ/ˀ
is not a rational. In this case,
ʱ
and
ʲ
(=
ˀ/
2
−
ʱ
)
.If
ʱ
<ʱ
, then
ʲ
>ʲ
, and hence both
ʱ, ʲ
must be multiples of
ʱ
, but this is impossible since
ʱ, ʲ
are linearly independent
over
are linearly independent over
Q
. Therefore,
ʱ
=
ʱ
and
˄
T
.
(ii) Since an equilateral triangle is terminal, we consider the case that
T
is
an isosceles triangle that is neither equilateral nor
Q
∼
(
6
,
6
,
2
3
). Let
ʱ, ʱ, ʲ
be
the three angles of
T
.If
ʱ
or
ʲ
is a rational multiple of
ˀ
, then both
ʱ, ʲ
are
rational multiples of
ˀ
,and
T
is terminal by Theorem 3.1 (ii). So, suppose that
one of
ʱ, ʲ
are irrational multiples of
ˀ
. Then, since 2
ʱ
+
ʲ
=
ˀ
, it follows that
ʱ, ʲ
are both irrational multiple of
ˀ
,and
ʱ, ʲ
are linearly independent over
Q
.
Suppose that
T
ₒ
˄
, and let
x, y, z
be the angles of
˄
.Wehave
⊧
⊨
x
=
m
1
ʱ
+
n
1
ʲ
y
=
m
2
ʱ
+
n
2
ʲ
z
=
m
3
ʱ
+
n
3
ʲ.
⊩
where
m
i
,n
i
are all nonnegative integers. Hence
2
ʱ
+
ʲ
=
ˀ
=
x
+
y
+
z
=
m
i
ʱ
+
n
i
ʲ,
and
m
i
=2
,
n
i
= 1. Hence, after suitable change of notations, we can
deduce that
x
=
ʱ, y
=
ʱ, z
=
ʲ
. Therefore
T
∼
˄
,and
T
is terminal.
Angles of a triangle are
commensurable
if they are all rational multiples of
ˀ
.
Corollary 3.2.
(i)
Among the triangles with commensurable angles, the triangle
(
6
,
6
,
2
3
)
is a unique intermediate triangle.
(ii)
The generating chain (
∗
)isa
(
6
,
6
,
2
3
)
.
≥
unique chain of length
3
that contains
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