sign in sign up
<<
Home , Hyperbola, Taxicab geometry

Abstract: The purpose of this article is to provide insight into the shape of and related geometric figures in taxicab . This will enable teachers to guide student explorations by designing interesting worksheets, leading knowledgeable class room discussions, and be aware of the different results that students can obtain if they decide to delve deeper into this fascinating subject.

From to Hyperbola in Taxicab Geometry

Ruth I. Berger, Luther College

How far is the shortest path from Grand Central Station to the Empire State building? A pigeon could fly there in a straight , but a person is confined by the street grid. The geometry obtained from measuring the between points by the actual distance traveled on a square grid is known as taxicab geometry. This topic can engage students at all levels, from plotting points and observing surprising shapes, to examining the underlying reasons for why these figures take on this appearance. Having to work with a new distance measurement takes everyone out of their comfort zone of routine memorization and makes them think, even about definitions and facts that seemed obvious before. It can also generate lively group discussions. There are many good resources on taxicab geometry. The ideas from Krause’s classic book [Krause 1986] have been picked up in recent NCTM publications [Dreiling 2012] and [Smith 2013], the latter includes an extensive pedagogy discussion. [House 2005] provides nice worksheets.

Definition: Let A and B be points with coordinates (푥1, 푦1) and (푥2, 푦2), respectively. The taxicab distance from A to B is defined as 푑푖푠푡(퐴, 퐵) = |푥1 − 푥2| + |푦1 − 푦2|.

This applies to all points in the , unlike on a city grid where at least one coordinate needs to be an integer in order to be located on a street. To compute the distance between two points in taxicab geometry we need to add their horizontal and vertical . The distance between A(-2,1) and B(3,5) is: |−2 − 3| + |1 − 5| = 5 + 4 = 9. Rather than formally computing this, just note that to get from A to B we move a total of 5 right and 4 up. Interestingly, the shortest path is not unique, just alternate the up and right segments, see figure 1.

This change in how distance is measured has huge implications for the shapes of well known geometric figures. Working in taxicab geometry requires careful attention to definitions. Nothing can be assumed or intuitively derived. What does a circle look like in this geometry? The of all points of distance 5 from the origin is shown in figure 2. By definition, this circle consists of all points (x,y) with |x|+|y|=5. From this we derive the four straight line segments that make up the circle: 푦 = ±푥 ± 5. Using a different center point or still results in this diamond shape.

Fact 1: In Taxicab geometry a circle consists of four congruent segments of slope ±1.

Check your student’s understanding: Hold a pen of length 5 inches vertically, so it extends from (0,0) to (0,5). Now tilt it so the tip is at (3,4). The taxicab distance from base to tip is 3+4=7, the pen became longer! This can be confirmed by holding it over the circle of radius 5 and noting that the tilted pen sticks out beyond the circle. Keep rotating, it will be longest at 45°, then shrink back to length 5 in the horizontal position. This activity shows that an object’s length changes when it is tilted, so be careful to never use a ruler to measure distances in taxicab geometry. Always count the number of horizontal and vertical moves to get from A to B.

***********************************

Grid City Worksheet Adam, Brenna, Carl, Dana, and Erik live in Grid City where each city block is exactly 300 feet wide. They were in Middle school last year, but now they attend Grid City High school. The Middle school (MS) is located 6 blocks east and 4 blocks north of the High school (HS). Using the information below, plot all possible locations for everyone’s homes on separate sheets. A. Adam lives 5 blocks (1500ft) from HS.

B. Last year Brenna walked 7 blocks (2100ft) to MS, this year she only has to walk 3 blocks (900ft) to HS.

C. Carl uses his skateboard to get to school. He says that it is exactly the same distance from his apartment to HS as it was from his apartment to MS last year.

D. Dana rides her bike to school. She complains that this year she has to ride 4 blocks (1200ft) further every morning than when she biked to MS last year. The exact opposite is true for her best friend Denise, her bike ride to school is now 1200ft shorter.

E. Last year Erik’s best friend Earl moved to the suburbs, but finished the school year in the city. After school he walked from MS to Erik’s house, and later he walked to HS to catch a ride home with his sister after her Basket practice. Earl’s total walking distance was 14 blocks (4200ft).

Repeat the above if MS is located 5 blocks east and 5 blocks north of HS.

Note to teachers: You might want to first consider an easier case where MS and HS are 8 blocks apart on the same street.

**************************************

The questions on the Grid City worksheet lead to well known figures in : A-Circle. B-Intersection of circles. C-Set of points equidistant to two points. D-Hyperbola. E- .

To facilitate investigations we use the following notation:

 Two points on the same horizontal or vertical line will be called h/v-aligned.  The rectangle with diagonal vertex points A and B will be referred to as AB-box. If A and B are h/v-aligned, the AB-box collapses to segment AB.

We already observed that in taxicab geometry there can be many paths of minimal length between points A and B. Fact 2: For any point Q in or on an AB-box dist(A,Q)+dist(Q,B)= dist(A,B). Proof: There exists a shortest path from A to B that passes through Q, see fig.1.

Fact 3: a) Let P be at distance r from any point Q on an AB-box, as shown in figure 3. Then dist(P,A)=r+dist(Q,A) and dist(P,B)=r+dist(Q,B). b) The same holds true for points P on a quarter-circle segment of radius r centered at vertex Q. Proof: There exists a shortest path from P to A, or B, that passes through Q.

Let’s examine the Grid City problems.

A. Adam lives on a circle of radius 5 around HS, see figure 2.

B. Brenna lives at the intersection of circles of radius 3 and 7 around HS and MS, respectively. Two circles can be “” in a segment of slope 푚 = ±1, instead of one point, see figure 4. Let 퐷 = 푑푖푠푡(퐶1, 퐶2) be the distance of the centers of circles of radius 푟1 and 푟2, respectively.

 If 푟1 + 푟2 < 퐷: no intersection.  If 푟1 + 푟2 = 퐷: they intersect in a segment inside the 퐶1퐶2-box, or a point if the centers are h/v- aligned.  If 푟1 + 푟2 > 퐷 the circles intersect in two points except for the following cases: o If |푟1 − 푟2| = 퐷 the intersection is one segment, or two adjacent segments if the centers are h/v-aligned. o If |푟1 − 푟2| = 푡 − 푠, where 0 < 푠 < 푡 are the sides of the 퐶1퐶2-box, they intersect in a segment and a point. o If the 퐶1퐶2-box is a square (s=t) and 푟1 = 푟2 they intersect in two segments. On the worksheet MS and HS are located 푟1 + 푟2 = 퐷 blocks apart on different streets.

C. For Carl’s apartment we need to find the set of all points P that are equidistant from two points A and B: 푑푖푠푡(푃, 퐴) = 푑푖푠푡(푃, 퐵). Draw circles of varying radii around A and B and find their intersection. If A and B are h/v- aligned this leads to a familiar figure: the points on the perpendicular bisector of segment AB. For other arrangements of A and B this set of points has surprising shapes, see figure 5. The smallest such 푑푖푠푡(퐴,퐵) circles have radius 푟 = , they intersect in a segment of slope ±1 in the interior of the AB-box. 2 Once points Q on the longer sides of the box are reached, the set of equidistant points P continue on in rays perpendicular to the sides. Proof: Since Q is equidistant from A and B, so is P by fact 3a. If the AB-box is square the interior segments end at vertex points Q, so by fact 3b the set of points P equidistant to A and B contains not only vertical and horizontal components, but also all points in between!

Definition: Given two points 퐹1, 퐹2 (foci) of distance D, and a positive 푘 < 퐷, a hyperbola is the set of all points P such that the absolute value of the difference of its distances from the points is k. |푑푖푠푡(푃, 퐹1) − 푑푖푠푡(푃, 퐹2)| = 푘

D. Dana and Denise live on different branches of a hyperbola with foci HS & MS and constant difference k=4. Dana’s house is located on the branch closer to focus point MS. To construct the hyperbola, look for intersections of circles of varying radii r and r+k respectively around the foci. For k=0 we obtain the set of points equidistant to the foci, so the hyperbola has similar features, see figure 6. If the foci are h/v-aligned the hyperbola consists of two vertical or horizontal lines a distance of k apart. For other arrangements, consider the 퐹1퐹2-box. On the inside there are segments of slope ±1 퐷−푘 퐷+푘 from intersections of circles of radius and around the foci. Once the sides of the box are 2 2 reached at Q, the hyperbola continues with points P on rays perpendicular to the side, because by fact 3a: 푑푖푠푡(푃, 퐹1) − 푑푖푠푡(푃, 퐹2) = [푟 + 푑푖푠푡(푄, 퐹1)] − [푟 + 푑푖푠푡(푄, 퐹2)] = 푑푖푠푡(푄, 퐹1) − 푑푖푠푡(푄, 퐹2) = 퐷+푘 퐷−푘 − = 푘. 2 2 Each pair of points Q between 퐹1and 퐹2 is a distance of k apart. Their location depends on the 퐹1퐹2- 퐷−푘 box. Let s be the length of the smaller side of the box. If 푠 < all Q will be on the two longer 2 퐷−푘 sides, else there will be one on each side. If 푠 = two Q’s are vertices of the box, from there the 2 hyperbola contains not only vertical and horizontal rays, but also all points in between! For street-grid problems the hyperbola should contain many grid points. Choosing D and k both even 퐷±푘 or both odd makes an integer. 2

Definition: Given two points 퐹1, 퐹2 (foci) of distance D, and a real number 푘 > 퐷, an ellipse is the set of all points P such that the sum of its distances from the focus points is k. 푑푖푠푡(푃, 퐹1) + 푑푖푠푡(푃, 퐹2) = 푘

E. Erik’s house is located on an ellipse with foci HS and MS. Construct the ellipse around the 퐹1퐹2-box: Consider any point Q on the box, and let P be a point of 푘−퐷 perpendicular distance 푟 = from Q, as shown in figure 7. 2 Since 푑푖푠푡(푃, 퐹푖) = 푟 + 푑푖푠푡(푄, 퐹푖) and 푑푖푠푡(푄, 퐹1) + 푑푖푠푡(푄, 퐹2) = 퐷, we have 푑푖푠푡(푃, 퐹1) + 푘−퐷 푑푖푠푡(푃, 퐹 ) = 퐷 + 2 ∗ = 푘. 2 2 푘−퐷 If Q is a vertex of the 퐹 퐹 -box, all points P on a quarter-circle of radius 푟 = around Q also satisfy 1 2 2 this distance requirement, by fact 3b.

An ellipse consists of 8 segments (6 if the foci are h/v-aligned). The four slanted segments have slope ±1. The others are horizontal and vertical segments of the same length as the sides of the 퐹1퐹2-box. To 푘−퐷 ensure that the ellipse contains grid points, choose D and k both even or both odd, so is an integer. 2

Further exploration: By introducing a subway line which does not follow the street grid, the work sheet could be extended to include . This will first require a discussion of the definition of the distance of a point to a line!

When looking at the definition of taxicab distance measure without the motivation of a city grid, students might get the impression that mathematicians can make up arbitrary methods for measuring distance in the Cartesian plane. There are in fact many different distance functions, also called metrics, but they are all subject to a few basic requirements. For all points A, B and C:  푑푖푠푡(퐴, 퐵) ≥ 0 , with 푑푖푠푡(퐴, 퐵) = 0 iff 퐴 = 퐵.  : 푑푖푠푡(퐴, 퐵) = 푑푖푠푡(퐵, 퐴)  : 푑푖푠푡(퐴, 퐶) + 푑푖푠푡(퐶, 퐵) ≥ 푑푖푠푡(퐴, 퐵)

Note that in the usual Euclidean the triangle inequality is an equality iff point C is on segment AB. For the taxicab metric we saw in fact 2 that equality holds iff C is in or on the AB-box. The above properties required for a distance are minimal. One could define 푑푖푠푡(퐴, 퐴) = 0 and 푑푖푠푡(퐴, 퐵) = 2 for all 퐵 ≠ 퐴, but then we would not have any meaningful circles or other geometric figures. A more interesting metric on the Cartesian plane is given by: 1 푝 푝 ⁄푝 푑푖푠푡((푥1, 푦1), (푥2, 푦2)) ≔ (|푥1 − 푥2| + |푦1 − 푦2| ) This is a metric for any positive real number p. For p=1 and p=2 we obtain the taxicab and functions, respectively. In each of these metrics we can solve |푥|푝 + |푦|푝 = 푟푝 for y and use a graphing calculator to visualize the graph of a “circle” of radius r centered at the origin. Since these metrics are invariant under , this circle is representative of all other circles. The pen-tilting activity illustrates that the taxicab distance is not invariant under . None of the above metrics are, except for the Euclidean metric (p=2). Working with values other than 푝 = 1 requires calculations, and geometric figures can’t be obtained by simple geometric observation. So while the distance function used in taxicab geometry is just one of many possible metrics, it is the one best suited to visual explorations.

References

Berger, Ruth. Parabolas in Taxicab Geometry. Preprint available online from author web site.

Dreiling, Keith M. 2012. Triangle Construction in Taxicab Geometry. Mathematics Teacher (Delving Deeper) 105 (6): 474-478.

House, Peggy. 2005. Hey, Taxi. NCTM Student Math Notes (Student Explorations in Mathematics). http://www.nctm.org/publications/article.aspx?id=22334

Krause, Eugene F. 1986. Taxicab Geometry: an Adventure in Non-Euclidean Geometry. New York: Dover Publications.

Smith, Christopher E. 2013. Is That Square Really a Circle? Mathematics Teacher 106 (8): 615-619.