# On the Plane Geometry with Generalized Absolute Value Metric

Authors: A. Bayar; S. Ekmekçi; Z. Akça
Journal: Mathematical Problems in Engineering
 (2008)

---

## Abstract

Metric spaces are among the most important widely studied topics
in mathematics. In recent years, Mathematicians began to investigate using other metrics
different from Euclidean metric. These metrics also find their place computer age in addition to their importance in geometry. In this paper, we consider the plane geometry with the generalized absolute value metric and define trigonometric functions and norm and then give a plane tiling  example for engineers underlying Schwarz's inequality in this plane.


---

## Body

## 1. Introduction

In the late 1800's, Minkowski [1] published a whole family of metrics  providing new insight into the
study of plane geometry. In recent years, mathematicians and engineers begin to
investigate non-Euclidean metrics used in real-life applications. For example,  Eisenberg and Khabbaz [2] discussed the
applications of taxicab metric in geometry and network theory. Similarly,
Burman et al. [3] utilized the concept of λ-geometry and discussed its applications in
the problem of global routing of multiterminal nets.Our goal is to further develop the plane geometry with
thegeneralized absolute value metric
d
g suggested by Kaya et al. [4], and to see how it
can lead to future researches in related areas. In ℝ
2,
the d
g-distances between X
=
(
x
1
,
y
1
) and Y
=
(
x
2
,
y
2
) are defined
by(1.1)
d
g
(
X
,
Y
)
=
d
g
(
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
)
=
a
⁢
max
⁡
{
|
x
1
−
x
2
|
,
|
y
1
−
y
2
|
}
+
b
⁢
min
⁡
{
|
x
1
−
x
2
|
,
|
y
1
−
y
2
|
}for all a
,
b
∈
ℝ
,
a
≥
b
≥
0
,
a
⁢
≠
0.
According to the definition of d
g-distances, the shortest way between two
points X and Y is the union of a vertical or a horizontal
line segment and a line segment with the slope ±
2
a
b
/
(
a
2
−
b
2
).We will show that thed
g-distance defined above is a metric. First, it
is positive definite and symmetric, and it can be shown that d
g satisfies the triangle inequality as follows.
For X
=
(
x
1
,
y
1
)
,
Y
=
(
x
2
,
y
2
),
and Z
=
(
x
3
,
y
3
),
since(1.2)
|
x
1
−
x
2
|
=
|
x
1
−
x
3
+
x
3
−
x
2
|
≤
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
,
|
y
1
−
y
2
|
=
|
y
1
−
y
3
+
y
3
−
y
2
|
≤
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
,one obtains(1.3)
d
g
(
X
,
Y
)
=
a
⁢
max
⁡
{
|
x
1
−
x
2
|
,
|
y
1
−
y
2
|
}
+
b
⁢
min
⁡
{
|
x
1
−
x
2
|
,
|
y
1
−
y
2
|
}
=
a
⁢
max
⁡
{
|
x
1
−
x
3
+
x
3
−
x
2
|
,
|
y
1
−
y
3
+
y
3
−
y
2
|
}
+
b
⁢
min
⁡
{
|
x
1
−
x
3
+
x
3
−
x
2
|
,
|
y
1
−
y
3
+
y
3
−
y
2
|
}
≤
a
⁢
max
⁡
{
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
,
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
}
+
b
⁢
min
⁡
{
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
,
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
}
=
k
.Let us examine the following four cases:(i)
|
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≥
|
y
3
−
y
2
|;
(ii)
|
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≤
|
y
3
−
y
2
|;
(iii)
|
x
1
−
x
3
|
≤
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≤
|
y
3
−
y
2
|
;
(iv)
|
x
1
−
x
3
|
≤
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≥
|
y
3
−
y
2
|.We will only prove the first two cases since the
remaining cases can be proved similarly.If|
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≥
|
y
3
−
y
2
|
, then(1.4)
d
g
(
X
,
Y
)
≤
k
=
a
(
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
)
+
b
(
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
)
=
a
|
x
1
−
x
3
|
+
b
|
y
1
−
y
3
|
+
a
|
x
3
−
x
2
|
+
b
|
y
3
−
y
2
|
=
d
g
(
X
,
Z
)
+
d
g
(
Z
,
Y
)
.If |
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≤
|
y
3
−
y
2
|
, then there are two possible situations.(i) Let|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
≥
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
. Then(1.5)
d
g
(
X
,
Y
)
≤
k
=
a
(
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
)
+
b
(
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
)
.Using |
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≤
|
y
3
−
y
2
|
,
(1.6)
d
g
(
X
,
Z
)
=
a
|
x
1
−
x
3
|
+
b
|
y
1
−
y
3
|
,
d
g
(
Z
,
Y
)
=
a
|
y
3
−
y
2
|
+
b
|
x
3
−
x
2
|
.Also, d
g
(
X
,
Y
)
≤
d
g
(
X
,
Z
)
+
d
g
(
Y
,
Z
) if and only if(1.7)
d
g
(
X
,
Y
)
≤
k
=
a
(
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
)
+
b
(
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
)
k
≤
a
(
|
x
1
−
x
3
|
+
|
y
3
−
y
2
|
)
+
b
(
|
y
1
−
y
3
|
+
|
x
3
−
x
2
|
)
⟺
a
(
|
x
3
−
x
2
|
−
|
y
3
−
y
2
|
)
+
b
(
|
y
3
−
y
2
|
−
|
x
3
−
x
2
|
)
≤
0
⟺
(
a
−
b
)
(
|
x
3
−
x
2
|
−
|
y
3
−
y
2
|
)
≤
0
⟺
|
x
3
−
x
2
|
≤
|
y
3
−
y
2
|
.(ii) Let|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
≤
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
. Then(1.8)
d
g
(
X
,
Y
)
≤
k
=
a
(
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
)
+
b
(
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
)
.Using |
x
1
−
x
3
|
≥
|
y
1
−
y
3
| and |
x
3
−
x
2
|
≤
|
y
3
−
y
2
|
,
(1.9)
d
g
(
X
,
Z
)
=
a
|
x
1
−
x
3
|
+
b
|
y
1
−
y
3
|
,
d
g
(
Z
,
Y
)
=
a
|
y
3
−
y
2
|
+
b
|
x
3
−
x
2
|
,
d
g
(
X
,
Y
)
≤
d
g
(
X
,
Z
)
+
d
g
(
Y
,
Z
)
⟺
d
g
(
X
,
Y
)
≤
k
=
a
(
|
y
1
−
y
3
|
+
|
y
3
−
y
2
|
)
+
b
(
|
x
1
−
x
3
|
+
|
x
3
−
x
2
|
)
k
≤
a
(
|
x
1
−
x
3
|
+
|
y
3
−
y
2
|
)
+
b
(
|
y
1
−
y
3
|
+
|
x
3
−
x
2
|
)
⟺
a
(
|
y
1
−
y
3
|
−
|
x
1
−
x
3
|
)
+
b
(
|
x
1
−
x
3
|
−
|
y
1
−
y
3
|
)
≤
0
⟺
(
a
−
b
)
(
|
y
1
−
y
3
|
−
|
x
1
−
x
3
|
)
≤
0
⟺
|
y
1
−
y
3
|
≤
|
x
1
−
x
3
|
.Note that the generalized absolute value metricd
g is a generalization of the taxicab metric d
T,
maximum metric d
max
⁡,
and Chinese checker metric d
C in the following manner: d
g
=
d
T if a
=
b
=
1
;
d
g
=
d
max
⁡ if a
=
1 and b
=
0
;
d
g
=
d
C if a
=
1 and b
=
2
−
1,
respectively. Furthermore, the unit circle in (
ℝ
2
,
d
g
) is the set of points (
x
,
y
) in the plane which satisfies(1.10)
a
⁢
max
⁡
{
|
x
|
,
|
y
|
}
+
b
⁢
min
⁡
{
|
x
|
,
|
y
|
}
=
1
,
a
≥
b
≥
0
,
a
≠
0.It is well known that Schwarz's inequality is a fundamental
inequality of the Euclidean plane, which states that the magnitude of the inner
product of two vectors cannot exceed the product of the lengths of these two
vectors in the plane. Our main goal is to show the validity of Schwarz's
inequality in(
ℝ
2
,
d
g
).
To achieve this goal, our presentation is organized as follows. After the
definitions of trigonometric functions in (
ℝ
2
,
d
g
) are given, we define the norm, prove Schwarz's
inequality, and give an area formula of a triangle in this plane (Section 3).
Finally, we give a plane tiling example underlying Schwarz's inequality in
Section 4.


## 2. The Trigonometric Functions

We know that ifX
=
(
x
,
y
) is a point on the Euclidean unit circle, then x
=
cos
⁡
θ and y
=
sin
⁡
θ,
where θ is the angle with the positive x-axis as the initial side and the radial line
passing through point (
x
,
y
) as the terminal side. Now, we can define the
trigonometric functions cosine and sine of the angle θ in (
ℝ
2
,
d
g
) in the same way as the Euclidean cosine and
sine functions. Let (
x
,
y
) be a point on the unit circle in (
ℝ
2
,
d
g
).
Since the unit circle is an octagon in (
ℝ
2
,
d
g
) (see [4]), trigonometric functions can be
defined according to eight quadrants. If we show the cosine and sine functions
by cos
⁡
g
θ and sin
⁡
g
θ,
they can be given as follows with the method in
[5]:(2.1)
cos
⁡
g
θ
=
{
cos
⁡
θ
a
|
cos
⁡
θ
|
+
b
|
sin
⁡
θ
|
i
f
⁢
|
tan
⁡
θ
|
≤
1
,
cos
⁡
θ
b
|
cos
⁡
θ
|
+
a
|
sin
⁡
θ
|
i
f
⁢
|
tan
⁡
θ
|
≥
1
,
sin
⁡
g
θ
=
{
sin
⁡
θ
a
|
cos
⁡
θ
|
+
b
|
sin
⁡
θ
|
i
f
⁢
|
tan
⁡
θ
|
≤
1
,
sin
⁡
θ
b
|
cos
⁡
θ
|
+
a
|
sin
⁡
θ
|
i
f
⁢
|
tan
⁡
θ
|
≥
1.One can easily see that if θ
' is the related angle in first or second
quadrant of given angle θ in quadrant i
,
(
π
/
4
)
(
i
−
1
)
≤
θ
≤
(
π
/
4
)
i
,
i
=
1
,
…
,
8.
The relationship between cos
⁡
g
θ and cos
⁡
g
θ
' or sin
⁡
g
θ and sin
⁡
g
θ
' is the same as
the Euclidean one.We note that there is the nonuniform increment in arc length when the angle is incremented by a fixed amount whereas on the Euclidean unit circle. So,
we want to
define trigonometric functions any angleθ between any two lines by using reference angle α of θ [6].
Definition 2.1.
Letθ be an angle with the reference angle α which is the angle between θ and the positive direction of the x-axis in d
g-unit circle. Then the cosine and sine
functions of angle θ with the reference
angles α,
g
⁢
cos
⁡
θ,
and g
⁢
sin
⁡
θ are defined by(2.2)
g
cos
θ
:=
cos
⁡
g
(
α
+
θ
)
⋅
cos
⁡
g
α
+
sin
⁡
g
(
α
+
θ
)
⋅
sin
⁡
g
α
,
g
sin
θ
:=
sin
⁡
g
(
α
+
θ
)
⋅
cos
⁡
g
α
−
cos
⁡
g
(
α
+
θ
)
⋅
sin
⁡
g
α
.One can easily see that if α
=
0,
then g
⁢
cos
⁡
θ
=
cos
⁡
g
θ and g
⁢
sin
⁡
θ
=
sin
⁡
g
θ.


### 2.1. Variation ofd
g-Lengths Under Rotations

It is well known that all translations and rotations
of the plane preserve the Euclidean distance. It is obvious thatd
g-distances are invariant under all translations
. Now, we want to investigate the change of d
g-lengths of a line segment after rotations as
in [7].
Theorem 2.2.
Let any two points beA and B in (
ℝ
2
,
d
g
),
and let the line segment A
B be not parallel to the x-axis and the angle α between the line segment A
B and the positive direction of x-axis. If A
'
B
' is the image of A
B under the rotations with the angle θ,
then(2.3)
d
g
(
A
'
,
B
'
)
=
cos
⁡
g
2
α
+
sin
⁡
g
2
α
cos
⁡
g
2
(
α
+
θ
)
+
sin
⁡
g
2
(
α
+
θ
)
⋅
d
g
(
A
,
B
)
.
Proof.
We can translate the line segmentA
B to the line segment O
X such that(2.4)
d
g
(
A
,
B
)
=
d
g
(
O
,
X
)
=
ksince all the translations
preserve the d
g-distance. X
=
(
k
⁢
⋅
cos
⁡
g
α
,
k
⋅
sin
⁡
g
α
) and under the rotation θ, O
X transforms to O
X
', X
'
=
(
k
⋅
cos
⁡
g
(
α
+
θ
)
,
k
⋅
sin
⁡
g
(
α
+
θ
)
).
If we calculate d
g
(
O
,
X
'
)
=
k
' by using d
E
(
O
,
X
)
=
d
E
(
O
,
X
'
),
we obtain(2.5)
k
'
=
k
⋅
cos
⁡
g
2
α
+
sin
⁡
g
2
α
cos
⁡
g
2
(
α
+
θ
)
+
sin
⁡
g
2
(
α
+
θ
)
.
Corollary 2.3.
If the segmentA
B is parallel to the x-axis, then(2.6)
d
g
(
A
'
,
B
'
)
=
d
g
(
A
,
B
)
cos
⁡
g
2
θ
+
sin
⁡
g
2
θ
.
Proof.
In this
situation, sinceα
=
0,
the proof is obvious.


## 2.1. Variation ofd
g-Lengths Under Rotations

It is well known that all translations and rotations
of the plane preserve the Euclidean distance. It is obvious thatd
g-distances are invariant under all translations
. Now, we want to investigate the change of d
g-lengths of a line segment after rotations as
in [7].
Theorem 2.2.
Let any two points beA and B in (
ℝ
2
,
d
g
),
and let the line segment A
B be not parallel to the x-axis and the angle α between the line segment A
B and the positive direction of x-axis. If A
'
B
' is the image of A
B under the rotations with the angle θ,
then(2.3)
d
g
(
A
'
,
B
'
)
=
cos
⁡
g
2
α
+
sin
⁡
g
2
α
cos
⁡
g
2
(
α
+
θ
)
+
sin
⁡
g
2
(
α
+
θ
)
⋅
d
g
(
A
,
B
)
.
Proof.
We can translate the line segmentA
B to the line segment O
X such that(2.4)
d
g
(
A
,
B
)
=
d
g
(
O
,
X
)
=
ksince all the translations
preserve the d
g-distance. X
=
(
k
⁢
⋅
cos
⁡
g
α
,
k
⋅
sin
⁡
g
α
) and under the rotation θ, O
X transforms to O
X
', X
'
=
(
k
⋅
cos
⁡
g
(
α
+
θ
)
,
k
⋅
sin
⁡
g
(
α
+
θ
)
).
If we calculate d
g
(
O
,
X
'
)
=
k
' by using d
E
(
O
,
X
)
=
d
E
(
O
,
X
'
),
we obtain(2.5)
k
'
=
k
⋅
cos
⁡
g
2
α
+
sin
⁡
g
2
α
cos
⁡
g
2
(
α
+
θ
)
+
sin
⁡
g
2
(
α
+
θ
)
.
Corollary 2.3.
If the segmentA
B is parallel to the x-axis, then(2.6)
d
g
(
A
'
,
B
'
)
=
d
g
(
A
,
B
)
cos
⁡
g
2
θ
+
sin
⁡
g
2
θ
.
Proof.
In this
situation, sinceα
=
0,
the proof is obvious.


## 3. Norm in(
ℝ
2
,
d
g
)

In this section, we will define the norm in(
ℝ
2
,
d
g
) by using the standard inner product in ℝ
2 as in [8] .
Definition 3.1.
LetA
=
(
x
,
y
) be any vector in (
ℝ
2
,
d
g
).
Then the norm of A is defined by(3.1)
∥
A
∥
g
=
〈
A
,
A
〉
cos
⁡
g
2
α
+
sin
⁡
g
2
α
,where α is the reference angle of A.
Corollary 3.2.
∥
A
∥
g
=
d
g
(
O
,
A
).
Proof.
IfA
=
(
x
,
y
)
=
(
d
g
(
O
,
A
)
cos
⁡
g
α
,
d
g
(
O
,
A
)
sin
⁡
g
α
) is used,(3.2)
d
E
(
O
,
A
)
=
cos
⁡
g
2
α
+
sin
⁡
g
2
α
⋅
d
g
(
O
,
A
)
.If we use d
E
(
O
,
A
)
=
〈
A
,
A
〉 and Definition 3.1,(3.3)
∥
A
∥
g
=
d
g
(
O
,
A
)is obtained.As in Euclidean geometry, it is easily seen that the
norm function∥
⋅
∥
g satisfies the following properties. So, the
following proposition is given without proof.
Proposition 3.3.
LetA
=
(
x
1
,
y
1
), B
=
(
x
2
,
y
2
),
and C
=
(
x
3
,
y
3
) be any three vectors in (
ℝ
2
,
d
g
) and r
∈
ℝ.
Then
(i)
∥
A
∥
g
≥
0 and (
∥
A
∥
g
=
0
⇔
A
=
0
),
(ii)
∥
r
A
∥
g
=
|
r
|
⋅
∥
A
∥
g,
(iii)
∥
A
+
B
∥
g
≤
∥
A
∥
g
+
∥
B
∥
g,
(iv)
∥
A
−
B
∥
g
≥
∥
A
∥
g
−
∥
B
∥
g,
(v)
∥
A
−
B
∥
g
≤
∥
A
−
C
∥
g
+
∥
C
−
B
∥
g.It is important that Schwarz's inequality is valid for
a metric. For example, matched filters are found in nearly every communication
device, such as cell phones and television broadcasts. The detector utilizes
Cauchy-Schwarz's inequality to compare functions. When the inner product of the
functions is “large,” then there are significant similarities between
the two signals. This detector will help us better select which tile is
appropriate for a small section of the final image. If a quantized block of the
target image is mostly one color on one side but another color on the opposite
side, then the matched filter detector will find a tile with similar
attributes.The following proposition shows that Schwarz's
inequality is valid for some restricted cases ofd
g.
Proposition 3.4 (Schwarz's inequality).
IfA
=
(
x
1
,
y
1
) and B
=
(
x
2
,
y
2
) are any two vectors in (
ℝ
2
,
d
g
) and b
≥
1,
then(3.4)
|
〈
A
,
B
〉
|
≤
∥
A
∥
g
⋅
∥
B
∥
g
.
Proof.
(i) IfA
=
0,
then 〈
A
,
B
〉
=
0 and |
〈
A
,
B
〉
|
=
0.
Furthermore, ∥
A
∥
g
=
0 and ∥
A
∥
g
⋅
∥
B
∥
g
=
0.
(ii) LetA
≠
0,
then we know that(3.5)
|
〈
A
,
B
〉
|
=
|
x
1
x
2
+
y
1
y
2
|
≤
|
x
1
x
2
|
+
|
y
1
y
2
|
,
(3.6)
∥
A
∥
g
⋅
∥
B
∥
g
=
(
a
max
⁡
{
|
x
1
|
,
|
y
1
|
}
+
b
min
⁡
{
|
x
1
|
,
|
y
1
|
}
)
⋅
(
a
max
⁡
{
|
x
2
|
,
|
y
2
|
}
+
b
min
⁡
{
|
x
2
|
,
|
y
2
|
}
)
=
a
2
|
x
1
x
2
|
+
a
b
(
|
x
1
y
2
|
+
|
y
1
x
2
|
)
+
b
2
|
y
1
y
2
|
.
In this case, there are four subcases.(1)
|
x
1
|
≥
|
y
1
| and |
x
2
|
≥
|
y
2
|.
(2)
|
x
1
|
≥
|
y
1
| and |
x
2
|
≤
|
y
2
|.
(3)
|
x
1
|
≤
|
y
1
| and |
x
2
|
≥
|
y
2
|.
(4)
|
x
1
|
≤
|
y
1
| and |
x
2
|
≤
|
y
2
|.
Here, subcase (1) is proved; the others can be similarly
proved.
If the
right-hand sides of (3.4) and (3.5) are
observed,(3.7)
a
b
|
x
1
y
2
|
+
a
b
|
y
1
x
2
|
+
a
2
|
x
1
x
2
|
+
b
2
|
y
1
y
2
|
≥
|
x
1
x
2
+
y
1
y
2
|is valid since b
≥
1
,
|
x
1
|
≥
|
y
1
|,
and |
x
2
|
≥
|
y
2
|.
This completes the proof.Now, we will give the area of the triangleA
B
C
Δ
.
Theorem 3.5.
LetA
B
C
Δ be any triangle in (
ℝ
2
,
d
g
) and let θ be the angle between A
C and B
C.
Then the area  of the triangle A
B
C
Δ  is(3.8)

=
1
2
⋅
∥
A
C
∥
g
⋅
∥
B
C
∥
g
⋅
g
sin
θ
.
Proof.
One can take thatC is the origin since all the translations
preserve d
g-distance. One can easily obtain the following
relation between  g
⁢
sin
⁡
θ and sin
⁡
θ:(3.9)
g
sin
θ
=
∥
O
A
∥
⋅
∥
O
B
∥
∥
O
A
∥
g
⋅
∥
O
B
∥
g
⋅
sin
⁡
θ
.
From (3.9) and 2

=
∥
O
A
∥
⋅
∥
O
B
∥
⋅
sin
⁡
θ for 0
≤
θ
≤
π,(3.10)

=
1
2
⋅
∥
A
C
∥
g
⋅
∥
B
C
∥
g
⋅
g
sin
θ
.


## 4. Application

In Latin, tessella was a small cubical piece of clay,
stone, or glass used to make mosaics as mentioned in [9, 10]. The word
“tessella” means “small square” (from “tessera,”
square, derived from the Greek word for “four”). It also corresponds
to the everyday term “tiling” which refers to the applications of
tessellation, often made of glazed clay.Certain shapes of tiles, most obviously rectangles,
can be replicated to cover a surface with no gaps. These shapes are said to
tessellate (from the Latin tessella, “tile”). Over  the centuries, many ceramic engineers and
artists have employed plane tiling in their work. They used tiles for floors
and walls because they are durable, waterproof, and beautiful. If you are
mathematically inclined, no doubt you see some mathematics in the tiling. There
are some examples where plane tiling on floors, walls, and in paintings
underlies proofs of some well-known (and some not so well-known) theorems.
However, tilings provide “picture proofs” for many theorems. They
also illustrate inequalities such as the arithmetic mean-geometric mean inequality. Squares and
rectangles are some of the most popular tiling used by engineers and artists to
tile the plane. The design of geometric shapes, which fit together to cover a
surface without gaps or overlapping, has a long history. Now, we will give a
plane tiling example satisfying Schwarz's
inequality in(
ℝ
2
,
d
g
) as an application.Suppose that the road is tiled with bricks of two
different sizes as illustrated in Figure1. Let A
=
(
x
1
,
y
1
) and B
=
(
x
2
,
y
2
) be any two vectors in (
ℝ
2
,
d
g
) and the side lengths of two bricks are |
x
1
|
,
|
x
2
| and |
y
1
|
,
|
y
2
|,
respectively; let d
g satisfy the condition b
≥
1 such that C
=
(
x
2
,
y
1
) and D
=
(
x
1
,
y
2
).
Then the following tiling forms part of a proof of some special cases of
Schwarz's inequality in two dimensions.



---

*Source: 673275-2008-08-25.xml*