Taxicab Geometry: Geometry without SAS
Congruence
September 4, 2011
Taxicab Geometry: Geometry without SAS Congruence
Definitions
Points: Points are defined as in the case of 2-dimensional
geometry, P(x, y ), where x, y are real.
Lines and Planes: The lines and planes will be defined exactly as
in coordinate geometry, a plane is represented by the linear
equation ax + by = c, for a, b, c not all zero.
Distance Formula: Let P and Q be the two points (x1 , y1 ) and
(x2 , y2 ) respectively.
p The distance formula under the Euclidean
Metric is, PQ = (x1 − x2 )2 + (y1 − y2 )2
but under the new metric distance is defined in the following way:
PQ ∗ = |x1 − x2 | + |y1 − y2 |
When restricted to the plane, this metric is commonly known as
the Taxicab (or Manhattan) Metric.
Taxicab Geometry: Geometry without SAS Congruence
Properties
1: Definition of betweenness will be the same as that for Euclidean:
If AB ∗ + BC ∗ = AC ∗ , then A ∗ B ∗ C ∗ .
2: Adopt the angle measure for this geometry, then all the axioms
on angle measure (Protractor Postulate) hold.
Taxicab Geometry: Geometry without SAS Congruence
Example
Consider the grids in a city. Now look at the route between 2
points A, B on the grid.
F
B
E
D
A
C
From the picture, AB ∗ = AC + CB. Now, AD ∗ = AD, and,
DB ∗ = DE + EF + FB, we have AB ∗ = AD ∗ + DB ∗ .
Taxicab Geometry: Geometry without SAS Congruence
Another Example
The red, blue and yellow lines represent distance under the taxicab
metric while, the green line represents distance under Euclidean
√
metric. The Euclidean distance shown is given by 6 × 2 ≈ 8.48
whereas the distance under the taxicab metric (or length of the
red, blue or yellow lines) is 12.
Taxicab Geometry: Geometry without SAS Congruence
Lines under the Taxicab Metric
Under the Euclidean Metric the equation of a line is y = mx + b,
let us call this line l. Let P(x1 , y1 ), Q(x2 , y2 ) be two points on l.
Therefore,
PQ ∗ = |x1 − x2 | + |y1 − y2 | = |x1 − x2 | + |(mx1 + b) − (mx2 + b)|
⇒ |x1 − x2 | + |m||x1 − x2 | = (1 + |m|)|x1 − x2 |, so
PQ ∗ = k|x1 − x2 | where k = 1 + |m|, (a constant).
To prove the Ruler Postulate from here, proceed in the following
way:
Consider the coordinate system for the Taxicab metric where
P(x, y ) ↔ kx, for a point P on the line l. Let a = kx1 and b = kx2
be the coordinates of two points A, B on l. So, (from the above
computation of PQ ∗ ), AB ∗ = k|x1 − x2 | = |kx1 − kx2 | = |a − b|,
so the Ruler Postulate is valid under the Taxicab Metric too.
Taxicab Geometry: Geometry without SAS Congruence
Results
Theorem 1: On any given line l, the betweenness under the
Euclidean and the Taxicab Metric coincide.
Proof: Let l : y = mx + b, let P,p
Q be two points (x1 , y1 ),
(x1 − x2 )2 + (y1 − y2 )2 =
(x
,
y
),
(respectively)
on
l.
d
=
p2 2
(1 + m2 )(x1 − x2 )2 = k 0 |x1 − x2 |, where k 0 = 1 + m2 . So,
PQ = k 0 |x1 − x2 |. Now from above we have PQ ∗ = k|x1 − x2 |. So
even though k and k 0 are different real numbers, the points have
the same order on line l.
Note: This shows that the geometric betweenness for the
Euclidean and the Taxicab Metrics are the same but, metric
betweenness is entirely different. For example, we could have
AB ∗ = AD ∗ + DB ∗ under the Taxicab metric while,
AB < AD + DB, under the Euclidean metric.
Taxicab Geometry: Geometry without SAS Congruence
Results
Theorem 2: In the Taxicab Geometry, the SAS hypothesis does
not imply that the triangles are congruent.
Taxicab Geometry: Geometry without SAS Congruence
Example
Find the lengths of the sides of right triangles 4ABC and 4XYZ
using Taxicab Metric. Do these triangles satisfy SAS hypothesis?
(Are AB ∼
= XY , ∠B ∼
= ∠Y , BC ∼
= YZ true?) Are they congruent?
y
Y(8,8)
A
X(5,5)
6
B
6
C
Z(5,11)
x
Taxicab Geometry: Geometry without SAS Congruence
Examples
The analogue of the Euclidean Parabola y = 14 x 2 in Taxicab
geometry, having ‘focus’ F (0, 1) and ‘directrix’ y = −1, is
illustrated in Figure below. Verify this by finding conditions for the
conditions for the coordinates x and y of a variable point P(x, y )
on this parabola using the defining property PF ∗ = PQ ∗ and the
distance formula.
y
F(0,1)
P(x,y)
x
y=−1
Q(x,−1)
PF ∗ = |x − 0| + |y − 1|, PQ ∗ = |y + 1|. If |x| + |y − 1| = |y + 1|,
then we have 3 cases,
Taxicab Geometry: Geometry without SAS Congruence
Examples
y < −1, then y + 1 < 0 and y − 1 < 0 and so the above
equation becomes,
|x| − (y − 1) = −(y + 1) ⇒ |x| = −2......this is a contradiction.
−1 ≤ y < 1, then y + 1 ≥ 0 and y − 1 < 0, the equation
becomes,
|x| − y + 1 = y + 1 ⇒ |x| = 2y ⇒ y ≥ 0,
Hence P lies on either of the 2 lines y = ± 12 x, 0 ≤ y < 1.
y ≥ 1, then y − 1 ≥ 0 and y + 1 > 0, in this case the
equation becomes,
|x| + y − 1 = y + 1 ⇒ |x| = 2
So, P lies on either of the 2 lines x = ±2, for y ≥ 1.
Taxicab Geometry: Geometry without SAS Congruence