Generating Polygons with Triangles - Discrete and Computational Geometry and Graphs - page 116

T 2

√ 3

2

T 1

1

− √ 3 tan 12

= 2

1

Fig. 2. Triangles T 1 and T 2

3 Triangles

Since the minimum angle in a triangle is at most ˀ/ 3, we have the following.

Example 3.1. An equilateral triangle is terminal.
The following result is obtained by Laczkovich [ 7 , Theorem 5.3].
Theorem 3.1 (Laczkovich 1995). Suppose that all angles of T are rational mul-
tiples of ˀ . (i) If the three angles of T are all different, then T is original. (ii)
If T is an isosceles (not an equilateral) triangle with base angle ʸ , then ˃

ₒ T

( ʸ, 2 − ʸ, 2 ) or ˃

implies either ˃ = ∼ T .

As a corollary, we have the following.
Corollary 3.1. Suppose that all angles of T are rational multiples of ˀ and all
angles are different. If T is not a right triangle, then T is terminal.

Proof. If T ˄ , then the angles of ˄ are all rational multiples of ˀ . If all angles

of ˄ are different, then ˄ is original by Theorem 3.1 (i), and hence T

ₒ

˄ . If just
two angles of ˄ are equal, then since T is not a right triangle, we have T

∼

˄
by Theorem 3.1 (ii). This is impossible. Now, suppose that ˄ is an equilateral
triangle. By Theorem 2.2 (ii), the ratios of the sides of T are all rational numbers.
Let ʱ be the minimum angle of T . Since T is not an equilateral triangle, it follows
that ʱ<ˀ/ 3. Since the ratios of side-lengths of T are all rational numbers,
cos ʱ

∼

by the cosine law. This implies either ʱ/ˀ is an irrational or ʱ = ˀ/ 3
by Lemma 2.1, which is a contradiction.

∈ Q

The following was essentially proved in Soifer [ 11 ].
Theorem 3.2. If the three interior angles ʱ, ʲ, ʳ of a triangle T are linearly
independent over Q , then T is original and terminal.

Proof. Suppose that T can be generated by a triangle ˄ with three interior angles
x, y, z . Then, for some nonnegative integers l i ,m i ,n i ,
⊧
⊨ ʱ

= 1 x + m 1 y + n 1 z

ʲ = 2 x + m 2 y + n 2 z

⊩

ʳ = 3 x + m 3 y + n 3 z.