Taxicab Geometry; Dispatching the Closest CabDate: 09/06/98 at 22:47:41 From: Melanie T. Nguyen Subject: Taxicab Geometry I have two problems in Taxicab Geometry. -- Jane and Dick are looking for an apartment to share. They have agreed to locate one that is equidistant from each of their job sites and such that the total of the distances that they commute does not exceed ten miles. If Dick's job site is six miles east and two miles north from Jane's, draw a map showing where they could look for an apartment. Explain. I worked out the problem, but I wasn't sure whether I did it correctly or not. I set up a graph where Jane's job is at point (0,0) and Dick's job at point (6,2) (since Dick's job is six miles east and two miles north from Jane's). I then found points on the graph where the distance from Jane's job and Dick's job was equal to 10 miles. For example, the points I found on the graph are (0,3), (1,3), (2,3), (3,3), (4,3), (5,3), (6,3), (7,0), (7,1), (7,2), (0,-1), (1,-1), (2,-1), (3,-1), (4,-1), (5,-1), (6,-1), (-1,0), (-1,2), (-1,3). I then connect all of the points and it becomes an eight-sided figure. I shaded inside the connected dots, because it represents "...such that the total of the distances that they commute does not exceed ten miles." And so did I do it correctly? -- Greta dispatches cabs for Bayou Taxi Co. which has cabs at three garages. Garage B is six blocks east and six blocks north of garage A, and garage C is sixteen blocks east of garage A. When possible Greta dispatches a cab from the garage nearest to the caller. Draw a map for her showing the regions closest to each garage. Explain. I set up the problem by plotting the points on the graph where garage A is at point (0,0), garage B at point (6,6), and garage C at point (16,0). And that is far as I can go. I really don't know what to do next. Please help me on these two problems. It would be really appreciated. Thank you. Date: 09/08/98 at 16:09:03 From: Doctor Rob Subject: Re: Taxicab Geometry Hi Greta, You did the first problem perfectly. In the second problem, proceed in pairs. Let a be the distance to point A, b the distance to B, and c the distance to C. First consider just the points A(0,0) and C(16,0). There are two regions. The points with x < 8 have a < c, and the points with x > 8 have a > c. Now consider just the points A(0,0) and B(6,6). There are four regions. In one of them, a < b. It is determined by the three inequalities x < 6, y < 6, and x + y < 6. In a second one, b < a. It is determined by the three inequalities x > 0, y > 0, and x + y > 6. In the other two, a = b. They are determined by x < 0 and y > 6, and by x > 6 and y < 0. Now consider just the points B(6,6) and C(16,0). There are two regions. In the one in which b < c, either x < 8, or both x < 14 and y > x - 8. In the other, b > c, and either x > 14, or both x > 8 and y < x - 8. The boundaries described cut the plane into seven regions: 1) x < 0 and y > 6: a = b < c. Greta should send either A or B. 2) x < 6, y < 6, and x + y < 6: a < b and a < c. Greta should send A. 3) 6 < x < 8 and y < 0: a = b < c. Greta should send either A or B. 4) x > 8 and y < 0: c < a = b. Greta should send C. 5) y > 0 and either x > 14 or x > y + 8: c < b < a. Greta should send C. 6) 6 < x < 14 and y > x - 8: b < c < a. Greta should send B. 7) 0 < x < 8, y > 0, and x + y > 6: b < a and b < c. Greta should send B. For points on the boundaries, pick the option of any of the regions meeting there. You supply the explanations. This was a very interesting problem. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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