ALPHA–DISTANCE – A GENERALIZATION OF CHINESE CHECKER DISTANCE AND TAXICAB DISTANCE

Songlin Tian

The Euclidean distance measures the shortest distance between two points. The taxicab distance [2] measures the distance between two points when only moves along axis–directions are permitted. If, in addition, the diagonal moves are also permitted, the distance between A and B is given by the Chinese checker distance [1]. Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be two points in ℝ^{2}. Denote

Δ_{AB} = max{|x_{2} − x_{1}|, |y_{2} − y_{1}|} and δ_{AB} = min{|x_{2} − x_{1}|, |y_{2} − y_{1}|}.

The Euclidean distance between A and B is

d(A, B) = \sqrt{Δ_{AB}^{2} + δ_{AB}^{2}} .

The taxicab distance between A and B is

d_{T} (A, B) = Δ_{AB} + δ_{AB}.

The Chinese checker distance between A and B is

d_{C}(A, B) = Δ_{AB} + (\sqrt{2} − 1)δ_{AB}.

In this paper we introduce a family of distances which include both Chinese checker distance and taxicab distance as special cases. For each α ∈ [0, π/4], we define the α–distance between A and B by

d_{α}(A, B) = Δ_{AB} + (sec α − tan α)δ_{AB}.

Then, d_{0}(A, B) = d_{T} (A, B) and d_{π/4}(A, B) = d_{C}(A, B). Observe that if δ_{AB} > 0, then

d_{C}(A, B) < d_{α}(A, B) < d_{T} (A, B) for all α ∈ (0, π/4).

When δ_{AB} = 0, A and B lie on a horizontal or vertical line, and it follows that d_{C}(A, B) = d_{α}(A, B) = d_{T} (A, B) = d(A, B) for all α ∈ [0, π/4].

Clearly, d_{α}(A, B) = 0 if and only if A = B, and d_{α}(A, B) = d_{α}(B, A) for all A, B ∈ ℝ^{2}. The main result of this paper will be that

d_{α}(A, B) ≤ d_{α}(A, C) + d_{α}(C, B),

for all A, B, C ∈ ℝ^{2}, and α ∈ [0, π/4].

For two points A(x_{1}, y_{1}), B(x_{2}, y_{2}) ∈ ℝ^{2}, we denote by R_{AB} the rectangular region bounded by the lines x = x_{1}, y = y_{1}, x = x_{2}, and y = y_{2}. The following propositions follow directly from the definition of the α–distance.

\underline{Proposition 1}. Let A, B, C, and D be four points in ℝ^{2}. If R_{AB} and R_{CD} are congruent, then

d_{α}(A, B) = d_{α}(C, D) for all α ∈ [0, π/4].

As a consequence of Proposition 1, the α–distance between two points is an invariant under translations and reflections about the x–axis, the y–axis, or both.

\underline{Proposition 2}. Let A and B be two points in ℝ^{2}. Then

d_{α}(A, B) ≥ d_{α}(C, D) for all C, D ∈ R_{AB} and α ∈ [0, π/4].

The following lemma is useful in establishing the main result.

\underline{Lemma 3}. If α ∈ [0, π/4], then

Δ_{AB} + (sec α − tan α)δ_{AB} ≥ δ_{AB} + (sec α − tan α)Δ_{AB}

for all A, B ∈ ℝ^{2}.

\underline{Proof}. If Δ_{AB} = δ_{AB}, the result holds. Assume that Δ_{AB} > δ_{AB}. Since α ∈ [0, π/4], it follows that cos α > 0 and sin α+cos α ≥ 1. Therefore, tan α+1 ≥ sec α, i.e., 1 ≥ sec α − tan α. Multiplying both sides by Δ_{AB} − δ_{AB}, we have

Δ_{AB} − δ_{AB} ≥ (Δ_{AB} − δ_{AB})(sec α − tan α), i.e.,
Δ_{AB} + (sec α − tan α)δ_{AB} ≥ δ_{AB} + (sec α − tan α)Δ_{AB}.

\underline{Theorem 4}. Let A and B be two points in ℝ^{2} and α ∈ [0, π/4]. Then,
d_{α}(A, B) ≤ d_{α}(A, C) + d_{α}(C, B),

for all C ∈ ℝ^{2}.

\underline{Proof}. It is obvious that the result holds when δ_{AB} = 0. Assume δ_{AB} > 0. By Proposition 1, without loss of generality, assume that A lies at the origin, and B(x_{2}, y_{2}), with x_{2} > y_{2} > 0, lies in the first quadrant. Identify points D(x_{2} −y_{2}, 0), E(x_{2} − y_{2} tan α, 0), and F (x_{2}, 0). Let R_{1} = ΔABD, R_{2} = ΔDBE, R_{3} = ΔEBF, and

R_{4} = {(x, y) | x ≥ 0 and y < 0, or x > x_{2} and y ≥ 0}

(see Figure 1).

Figure 1.

By Proposition 1, it suffices to prove the result for C ∈ R_{1} ∪ R_{2} ∪ R_{3} ∪ R_{4}.

\underline{Case 1}. Assume that C(x, y) ∈ R_{1}.
In this case, it follows that x ≥ y and x_{2} − x ≥ y_{2} − y. Thus,

d_{α}(A, C) = x + (sec α − tan α)y and d_{α}(C, B) = x_{2} − x + (sec α − tan α)(y_{2} − y).

Therefore,

d_{α}(A, C) + d_{α}(C, B) = x_{2} + (sec α − tan α)y_{2} = d_{α}(A, B).

\underline{Case 2}. Assume C(x, y) ∈ R_{2}.

Draw a line segment parallel to the x–axis through C to intersect line segment \overline{BE} at G. Draw a line segment parallel to \overline{BE} through C to intersect the x–axis at point H (see Figure 2).

Since the quadrilateral CGEH is a parallelogram, it follows that |\overline{CG}| = |\overline{HE}| and |\overline{HC}| = |\overline{EG}|. Note that d_{α}(A, C) = |\overline{AH}| + |\overline{HC}| and

d_{α}(A, B) = |\overline{AE}| + |\overline{EB}|

= |\overline{AH}| + |\overline{HE}| + |\overline{EG}| + |\overline{GB}| = |\overline{AH}| + |\overline{HC}| + |\overline{CG}| + |\overline{GB}| = d_{α}(A, C) + |\overline{CG}| + |\overline{GB}|.

Figure 2.

By Lemma 3,

|\overline{CG}| + |\overline{GB}| ≤ d_{α}(C, B).

Therefore,

d_{α}(A, B) ≤ d_{α}(A, C) + d_{α}(C, B).

\underline{Case 3}. Assume that C(x, y) ∈ R_{3}.

Let G and H be the points obtained as in Case 2. As in Case 2, it follows that |\overline{GC}| = |\overline{EH}| and |\overline{HC}| = |\overline{EG}|. Observe that,

d_{α}(A, B) = |\overline{AE}| + |\overline{EB}|
= |\overline{AE}| + |\overline{EG}| + |\overline{GB}|,
d_{α}(A, C) = |\overline{AH}| + |\overline{HC}|
= |\overline{AE}| + |\overline{EH}| + |\overline{HC}|
= |\overline{AE}| + |\overline{EG}| + |\overline{GC}|,

and

|\overline{CB}| ≤ d_{α}(C, B).

Therefore,

d_{α}(A, B) = |\overline{AE}| + |\overline{EG}| + |\overline{GB}|
≤ |\overline{AE}| + |\overline{EG}| + |\overline{GC}| + |\overline{CB}|
= d_{α}(A, C) + d_{α}(C, B).

\underline{Case 4}. Assume that C(x, y) ∈ R_{4}.

Let h = min{max{0, y}, y_{2}} and k = min{x, x_{2}}. Denote by D the point (h, k). Then D lies on the line segment \overline{AF} or \overline{BF} (see Figure 1). By Proposition 2,

d_{α}(A, D) ≤ d_{α}(A, C) and d_{α}(D, B) ≤ d_{α}(C, B).

Based on the results from Case 2 and Case 3,

d_{α}(A, B) ≤ d_{α}(A, D) + d_{α}(D, B) ≤ d_{α}(A, C) + d_{α}(C, B).

We know that geometric figures such as lines, circles, parabolas, ellipses, and hyperbolas can be defined based on a distance. It is a good undergraduate/graduate research topic to characterize these figures under the α–distance.

\underline{References}

1. G. Chen, Lines and Circles in Taxicab Geometry, Master’s Thesis, Central Missouri State University, Warrensburg, MO, 1992.

2. E. Krause, Taxicab Geometry, Addison–Wesley Publishing Company, Menlo Park, CA, 1975.

Songlin Tian Department of Mathematics and Computer Science Central Missouri State University Warrensburg, MO 64093 email: slt8884@cmsu2.cmsu.edu