From Circle to Hyperbola in Taxicab Geometry
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Abstract: The purpose of this article is to provide insight into the shape of circles and related geometric figures in taxicab geometry. 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 Circle 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 line, but a person is confined by the street grid. The geometry obtained from measuring the distance 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 plane, 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 distances. 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 set 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 radius 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 Ball 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 Euclidean geometry: A-Circle. B-Intersection of circles. C-Set of points equidistant to two points. D-Hyperbola. E- Ellipse. 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 perpendicular 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 “tangent” 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 real number 푘 < 퐷, a hyperbola is the set of all points P such that the absolute value of the difference of its distances from the focus 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.