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( s/ 2) 2 cot 12 =3 s 2 (2 + 3) = 3( a + b 3) 2 (2 + 3)
= 3(2 a 2 +6 b 2 +6 ab ) + 3(4 ab + a 2 +3 b 2 ) 3 .
On the other hand, since T tiles P , the area of P is an integral multiple of 3 / 2,
12
×
which cannot be equal to 3(2 a 2 +6 b 2 +6 ab )+3(4 ab + a 2 +3 b 2 ) 3, a contradiction.
Now, suppose that the ratios of the sides of T are all rationals and T has an
angle equal to ˀ/ 3or2 ˀ/ 3. If ʱ denotes the smallest angle of T , then cos ʱ is
a rational by the law of cosines. Hence ʱ/ˀ is irrational or ʱ = ˀ/ 3 by Lemma
2.1. If ʱ = ˀ/ 3, then T is an equilateral triangle, and no angle ( n
2) ˀ/n ( n
=
6) is an integral multiple of ˀ/ 3. If ʱ<ˀ/ 3, then ʱ/ˀ is irrational, and by
Theorem 2.2 (ii), T must have 2 ˀ/ 3 as its largest angle. In this case, no angle
( n
= 6) can be constructed as a linear combination of the angles of
T with nonnegative integral coecients.
(ii) Let T be a triangle that generates a square, and let ʱ be the smallest
angle of T . By Theorem 2.2 (i), T is a right triangle and tan ʱ is a rational.
Hence ʱ/ˀ is irrational or ʱ = ˀ/ 4 by Lemma 2.1. If T generates a regular
n -gon for n
2) ˀ/n ( n
2) ˀ/n must be represented as a linear
combination of the angles of T with nonnegative integral coecients, which is
possib le only when ʱ = ˀ/ 4and n = 8. So, assume that
=3 , 4, then the angle ( n
( 4 , 4 , 2 ) with sides
1 , 1 , 2 can tile a regular octagon
{
}
, and le t s denote the side-length of the
8
octagon. This s must be written as s = a + b 2 with some nonnegative integers
a, b .Let r be the circum-radiu s of the octagon. By the c os ine law, we have
s 2 =2 r 2
2 ), and hence r 2 = s 2 / (2
2). T h e area of the
2 r 2 cos 4
= r 2 (2
=4 s 2 / (2 2
2) = 2 s 2 / ( 2
1) = 2( a + b 2) 2 / ( 2
octagon is 4 r 2 sin 4
1).
Since a, b are nonnegative integers, this is an irrational number. On the other
hand, since we assumed that the right isosceles triangle tiles the octagon, the
area of the octagon must be an integral multiple of 1 / 2, a rational number, which
is a contradiction.
Example 2.3. e (
{
3
}
,
{
4
}
,
{
6
}
,
{
12
}
)=2 .
This can be seen as follows. Since
{
3
}
,
{
4
}
are independent, the simp li cial element
number is at least 2. Let T 1 be the right triangle w i th sides 1 , 2 , 3, and T 2 be
the right triangle with perpendicular sides 1 , 2
3. Now, T 1 can generate
{
3
}
and
{
6
}
; T 2 can generate
{
12
}
, see Fig. 2 . Since
{
T 1 ,T 2 }
can generate a rectangle
1
×
2,
{
T 1 ,T 2 }
can generate a square. Hence the simplicial element number is at
most 2.
If m, n are suciently large and m
= n , then
{
m
}
,
{
n
}
are independent.
Theorem 2.4. If 420 <m<n , then e (
{
m
}
,
{
n
}
)=2 .
Proof. It is obvious that e (
{
m
}
,
{
n
}
)
2. By [ 8 , Theorem 3.4 (iii)], if k> 420
( ( k 2) ˀ
2 k
, ( k 2) ˀ
2 k
( ( k 2) ˀ
2 k
, 2 k )or
, k , 2 ). From
and T
ₒ{
k
}
, then T is either
this we can deduce that there is no triangle T such that T
ₒ{
m
}
and T
ₒ{
n
}
.
Therefore e (
{
m
}
,
{
n
}
)
2.
Conjecture 2.1. e (
{
m
}
,
{
n
}
)=2for3
m<n and ( m, n )
=(3 , 6).
 
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