AREA OF A TRIANGLE IN TERMS OF THE TAXICAB DISTANCE

Münevver Özcan and Rüstem Kaya

Abstract. In this study, we use the taxicab distance function to calculate the area of a triangle, and give the taxicab version of Heron’s Formula.

1. Introduction. The taxicab plane ℝ_{T}^{2} is almost the same as the Euclidean analytical plane ℝ^{2}. The points are the same, the lines are the same, and angles are measured in the same way. However, the distance function is different. Taxicab distance between the points P and Q is the length of a shortest path from P to Q composed of line segments parallel to the coordinate axes. That is, if P = (x_{1}, y_{1}) and Q = (x_{2}, y_{2}) then the taxicab distance from P to Q is d_{T} (P, Q) = |x_{1} − x_{2}| + |y_{1} − y_{2}|.

The taxicab plane geometry is non–Euclidean since it fails to satisfy the side– angle–side axiom but satisfies all the remaining twelve axioms of the Euclidean plane geometry [5]. Since the taxicab plane geometry has a different distance function it seems interesting to study the taxicab analogues of the topics that include the concept of distance in the Euclidean geometry. A few of such topics have been studied by some authors [1, 2, 3, 4, 6, 7, 9, 10]. The group of isometries that preserve taxicab distance is determined in [8].

Here, we study the following problem: How can one compute the area of a triangle in the analytical plane by using the taxicab distance? Clearly, a well– known formula

area of a triangle = (base × height)/2

is not, in general, valid in ℝ_{T}^{2} . It is valid if and only if the base is parallel to any one of the coordinate axes. (In this case, Euclidean and taxicab lengths of the base and height are the same.) The area of a triangle can also be computed using the three sides of the triangle. Let the sides of a triangle have lengths a, b, c. Introduce the semiperimeter p = (a + b + c)/2 and the area A. Then

A^{2} = p(p − a)(p − b)(p − c),

is known as Heron’s Formula. The aim of this work is to give a taxicab version of Heron’s Formula.

2. Extension of Heron’s Formula to the Taxicab Plane. Let the sides of a triangle ABC, in the taxicab plane, have lengths a_{T} = d_{T} (B, C), b_{T} = d_{T} (A, C) and c_{T} = d_{T} (A, B), and denote the taxicab–semiperimeter p_{T} = (a_{T} + b_{T} + c_{T})/2. The following two propositions give the taxicab versions of Heron’s formula in some special cases.

\underline{Proposition 1}. If one side of a triangle ABC, say BC, is parallel to one of the coordinate axes and none of the angles B and C is an obtuse angle, then for the area A of ABC,

A = \frac{1}{2}a_{T} (p_{T} − a_{T}) = \frac{1}{4}a_{T} (b_{T} + c_{T} − a_{T}).

\underline{Proof}. Consider the triangle ABC. Let h_{T} = d_{T} (A, BC) and a′_{T} = d_{T} (B, A′), where A′ denotes the foot of the altitude from A, (Figure 1).

Figure 1. BC is parallel to a coordinate axis. B and C are not obtuse angles.

For the triangles ABA′ and AA′C we get

a′_{T} = c_{T} − h_{T} and b_{T} = h_{T} + (a_{T} − a′_{T}),

respectively. From these equalities;

b_{T} = h_{T} + a_{T} − c_{T} + h_{T} and h_{T} = (b_{T} + c_{T} − a_{T})/2,

and consequently,

A = \frac{1}{2}a_{T} h_{T} = (\frac{a_{T}}{2} )(\frac{−a_{T} + b_{T} + c_{T}}{2} ).

Using the taxicab–semiperimeter p_{T}, in the above statement one obtains,

A = \frac{1}{2}a_{T} (p_{T} − a_{T})

which completes the proof. Unfortunately, in all the remaining cases, one needs a new distance parameter a′_{T} to determine the area of a given triangle in terms of the taxicab lengths of sides.

\underline{Proposition 2}. If one side of a triangle ABC, say BC, is parallel to one of the coordinate axes and one of the angles B and C is not an acute angle, then for the area A of ABC

A = \frac{1}{2}a_{T} (p_{T} − (a_{T} + a′_{T})),

where a′_{T} = d_{T} (A′, B) or a′_{T} = d_{T} (A′, C) depending on whether B or C is not an acute angle, respectively; and A′ denotes the foot point of the altitude from A.

\underline{Proof}. Consider any of the triangles ABC in Figure 2a or Figure 2b and let d_{T} (A, A′) = h_{T}, where the side BC is parallel to one of the coordinate axes.

Figure 2a. BC is parallel to a coordinate axis and B is not an acute angle.

Figure 2b. BC is parallel to a coordinate axis and C is not an acute angle. It is easily seen, from the triangles AA′C and AA′B that

h_{T} = b_{T} − (a_{T} + a′_{T}) and h_{T} = c_{T} − a′_{T}

if B is not an acute angle. Thus, b_{T} = a_{T} + c_{T}. Then for the area A of the triangle ABC,

\begin{aligned}A & = \frac{1}{2}a_{T} h_{T} = \frac{1}{2}a_{T} (b_{T} − (a_{T} + a′_{T})) = \frac{1}{2}a_{T} (\frac{b_{T} + b_{T}}{2} − (a_{T} + a′_{T})) \\ & = \frac{1}{2}a_{T} (\frac{a_{T} + b_{T} + c_{T}}{2} − (a_{T} + a′_{T})). \end{aligned}

Using p_{T} we obtain, A = \frac{1}{2}a_{T} (p_{T} − (a_{T} + a′_{T})). Similarly, in the case where C is not an acute angle then,

h_{T} = c_{T} − (a_{T} + a′_{T}) and h_{T} = b_{T} − a′_{T}

from the triangles ABA′ and ACA′. Thus, c_{T} = b_{T} + a_{T}. Now, it can be easily shown that the above formula is still valid.

\underline{Corollary}. Let the side BC of a triangle ABC in the taxicab plane, be parallel to one of the coordinates axes. Then if the angle B ≥ π/2 or C ≥ π/2 then

b_{T} = a_{T} + c_{T} or c_{T} = a_{T} + b_{T} ,

respectively.

Now, we introduce some new concepts in order to find a general taxicab version of Heron’s Formula for triangles such that none of their sides is parallel to the coordinate axes. Let ABC be any triangle in the taxicab plane. Clearly, there exists a pair of lines passing through every vertex of the triangle, each of which is parallel to a coordinate axis. A line l is called a base line of ABC if and only if 1. l passes through a vertex,

2. l is parallel to a coordinate axis,

3. l intersects the opposite side (as a line segment) to the vertex in Condition 1. Clearly, at least one of the vertices of the triangle always has one or two base lines. Such a vertex of a triangle is called a basic vertex. A base segment is a line segment on a base line, which is bounded by a basic vertex and its opposite side.

The following theorem gives the general taxicab version of Heron’s Formula.

\underline{Theorem 3}. Let α_{T} denote the length of a base segment of a triangle. Then for the area A of a triangle described above,

A = \{ \begin{matrix}α_{T} (p_{T} − (α_{T} + α′_{T}))/2 , & if there exists only one base line \\ & passing through the basic vertex \\ α_{T} (p_{T} − (α_{T} + α′_{T} + α′′_{T}))/2 , & if there exist two base \\ & lines passing through the basic vertex \end{matrix}

where a′_{T} = d_{T} (D, H), α′′_{T} = d_{T} (basic vertex, H′) and D = Intersection point of the base line and the opposite side, H = The point of orthogonal projection of one of the remaining two vertices on the

base line but not on the base segment, H′= The point of orthogonal projection of the third vertex on the same base line

but not on the base segment.

\underline{Proof}. Let ABC be a triangle with the basic vertex C, and l_{C} denote a base line through C. Then D = l_{C} ∩ [AB] where [AB] is the side AB of triangle ABC. Let H be the orthogonal projection of the vertex B to l_{C} but not on [CD]. This choice of B is always possible since A and B have symmetrical roles (see Figures 3 and 4). Now, the triangle ABC is a combination of the triangles ADC and BDC. If c′_{T} = d_{T} (A, D) and c′′_{T} = d_{T} (B, D) then the taxicab–semiperimeters of ADC and BDC are

p′_{T} = (α_{T} + b_{T} + c′_{T})/2 and p′′_{T} = (α_{T} + a_{T} + c′′_{T})/2,

respectively. Consequently, using p_{T} = (a_{T} + b_{T} + c_{T})/2,

p′_{T} + p′′_{T} = p_{T} + α_{T} .

Now two cases are possible.

i. If l_{C} is the only base line through the basic vertex C (Figure 3) then,

Figure 3. A triangle ABC with a base line on the base vertex

the area A_{1} of ADC is given by

A_{1} = α_{T} (p′_{T} − α_{T})/2

by Proposition 1, and the area A_{2} of BDC is given by

A_{2} = α_{T} (p′′_{T} − (α_{T} + α′_{T}))/2

by Proposition 2. Then using the equality p′_{T} + p′′_{T} = p_{T} + α_{T},

A = A_{1} + A_{2} = (α_{T} (p′_{T} − α_{T})/2) + (α_{T} (p′′_{T} − (α_{T} + α′_{T}))/2)
= α_{T} ((p′_{T} + p′′_{T}) − α_{T} − (α_{T} + α′_{T}))/2
= α_{T} (p_{T} − (α_{T} + α′_{T}))/2.

ii. If there exist two base lines through the basic vertex C (Figure 4) then,

Figure 4. A triangle ABC with two base lines on the base vertex

the area A_{1} of ADC is given by

A_{1} = α_{T} (p′_{T} − (α_{T} + α′′_{T}))/2

and the area A_{2} of BDC is given by

A_{2} = α_{T} (p′′_{T} − (α_{T} + α′_{T}))/2

by Proposition 2. Then using the equality p′_{T} + p′′_{T} = p_{T} + α_{T},

A = A_{1} + A_{2} = (α_{T} (p′_{T} − (α_{T} + α′′_{T}))/2) + (α_{T} (p′′_{T} − (α_{T} + α′_{T}))/2)
= α_{T} ((p′_{T} + p′′_{T}) − (α_{T} + α′′_{T}) − (α_{T} + α′_{T}))/2
= α_{T} (p_{T} − (α_{T} + α′_{T} + α′′_{T}))/2.

Notice that here α_{T} + α′_{T} + α′′_{T} = d_{T} (H, H′).

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Münevver Özcan Department of Mathematics University of Osmangazi 26480, Eskişehir Turkey email: mozcan@ogu.edu.tr

Rüstem Kaya Department of Mathematics University of Osmangazi 26480, Eskişehir Turkey email: rkaya@ogu.edu.tr