This is the 100th Mathematical Lens column! Ron Lancaster and Brigitte Bentele wish to thank the Mathematics Teacher editorial staff and the many readers who have submitted material that has been published in this department over the years. To celebrate this milestone, Lancaster and Bentele decided to walk 100 blocks (5 miles) through Manhattan. They started at the corner of East 1st Street and 1st Avenue; seven hours later, they finished at the corner of East 101st Street and 1st Avenue. All along the way they found mathematics. They documented their findings by taking hundreds of photographs, and they discussed mathematical questions related to what they saw during this journey along 1st Avenue. 1. (a) What was Lancaster and Bentele’s average walking speed? How long did it take them to walk one block? (b) Photographs 1–10 show ten streets that they encountered during their walk along 1st Avenue. Express the street number s that appears in the ith photograph as a function of i. If Lancaster and Bentele had continued to walk beyond 101st Street, what other street signs would they have encountered that are part of this pattern? (c) Calculate the distance that Lancaster and Bentele walked between the streets in the photographs. How are these distances related to the graph of y = x2? 2. (a) The ellipse design with dog silhouettes in the window of Reme’s Oggi Pets (see photograph 11) is comparable to the ellipse design shown in figure 1. Assume that the ellipses in figure 1 are the same size. (A careful look at the window shows that the ellipses in the top row are larger than those in the bottom row.) Two ellipses are shown in figure 2. The equation of ellipse 1 (red) is x2/16 + y2/25 = 1. Let the equation of ellipse 2 (blue) be (x – 4)2/16 + (y – t)2/25 = 1. Find the value of t for which the two ellipses intersect at exactly one point. (b) Use the value of t from part (a) to determine the equations of two other complete ellipses shown in figure 1. 3. The design of the Daily Bagel logo (see photograph 12) is shown on a set of coordinate axes in figure 3. This presentation could be useful if the designer is asked to submit technical specifications for the design. Let P1P3P5P7 be the points (7, 0), (0, 7), (–7, 0), and (0, –7), respectively. Let P1P2P3P4P5P6P7P8 be a regular octagon with center O. (a) Determine the coordinates of P2 and P10. (b) Describe how the coordinates of P2 and P10 can be used to determine the coordinates of P6 and P9. (c) Describe how the equation of the line that passes through the points O and P10 could be used to find the equation of the line that passes through the points O and P9. 4. (a) Looking at the Second Avenue Deli clock (see photograph 13), determine the next time that the hour and minute hands will be in the same position. (b) Study the Hebrew symbols on the clock for the numbers 1, 2, 10, 11, and 12. How are the symbols for 11 and 12 related to the symbols for 1 and 2? What would you expect the symbols for 13, 14, 15, 16, 17, 18, and 19 to be in Hebrew? 5. (a) Calculate the surface area and volume for each of the four boxes on display in the window (see photograph 14). Record your answers in table 1. (b) Make a scatter plot of the cost versus the surface area. Find a mathematical model for these data. (c) Use your model from part (b) to suggest a price for a box whose dimensions are 18 in. × 18 in. × 28 in. When a company such as Big John’s determines the price for a box, do you think that the price is based on the surface area or some other quantity? Why? solutions 1. (a) Lancaster and Bentele’s average speed was 5/7 ≈ 0.714 miles per hour. Equivalently, they walked 7 • 60/100 = 4.2 minutes per block. Teachers may want students to find typical walking speeds and explain any discrepancy. (b) s = i2. According to this formula, if Lancaster and Bentele had continued to walk along 1st Avenue, they would have expected to encounter East 121st Street, East 144th Street, East 169th Street, and so on. As it turns out, these streets do not intersect 1st Avenue, so they would not have encountered any more street signs with a square number. (c) There are 20 blocks in a mile (5280 feet), so the distance between two successive streets is 5280/20 = 264 feet. Table 2 shows the distance between pairs of streets with a square number. The numbers 3, 5, 7, . . . form a linear sequence and are the first differences of a quadratic function y = x2 (see table 3). The distances between streets form a linear sequence, but the total distance traveled is a cumulative sum represented by a quadratic function y = 264x2, in which x is the street number. 2. (a) Symmetry suggests that the point of intersection would be the midpoint of the segment between the centers of the two ellipses—namely, (2, t/2). Because this point would necessarily be contained in the first ellipse, we can solve for t: Choosing t so that the blue ellipse will be in the first quadrant, we find that The following method for determining the value of t makes use of calculus. At the point of intersection P (see fig. 4), the ellipses share a common tangent, meaning that their slopes are the same. The slope at any point on the red ellipse is found using implicit differentiation: Similarly, the slope at any point on the blue ellipse can be determined: At a point of tangency, these two slopes are equal: Substituting into the equation of the first ellipse, we have the following: Substituting into the equation of the second ellipse gives the following: Comparing the two expressions equal to 16, we see that x2 = (x – 4)2, so x = 2. This value of x substituted into the equation x2/16 + y2/25 = 1 yields y = (again, selecting a point in the first quadrant). Therefore, y = xt/4 becomes = 2t/4, so (b) Ellipse 1 can be translated by 8 units to the right to obtain an adjacent ellipse in the bottom row of figure 1, as (x – 8)2/16 + y2/25 = 1. Similarly, ellipse 2 can be translated by 8 units to the right to obtain (x –12)2/16 + = 1. 3. (a) In the isosceles right triangle of figure 5a, we have a2 + a2 = 72; therefore, a = . Thus, point P2 is (, ). Placing point Q at the foot of the perpendicular from P10 to segment OP1 in figure 5b, we find that m∠P10OQ = 67.5°and OQ = 3.5 (because OP1 = 7). Therefore, b = 3.5/tan(67.5°) ≈ 8.45. Thus, point P10 is approximately located at (3.5, 8.45). (b) The points P6 and P2 are symmetric about the origin. So if P2 is the point (u, v), then P6 is the point (–u, –v). The points P10 and P9 are symmetric about the line y = x. So if P10 is the point (u, v), then P9 is the point (v, u). (c) Because of the symmetry about the line y = x, if the slope of is m, then the slope of is (1/m). Thus, the equation of is y = (1/m)x. 4. (a) The hour and minute hands coincide 11 times in a 12-hour period. The hands are on top of each other at noon. The next time they coincide will be 1/11th of 12 hours later. This is 12/11 hours after noon, or 5 5/11 minutes after 1 p.m. The next time the hands coincide will be at 10 10/11 minutes after 2 p.m. This is a popular puzzle, and a variety of solutions can be found online (see such sites as http://puzzles.nigelcoldwell.co.uk/thirtyfive.htm and http://astarmathsandphysics.com/gcse-maths-notes/689-when-hour-and-minute-hands-coincide.html). (b) Note that the number 11 is formed by combining the symbols for 1 and 10, with the symbol for 10 on the right. Similarly, the number 12 is formed by combining the symbols for 2 and 10. The symbols for 13, 14, 15, … 19 are formed in the same manner. 5. (a) See table 4. (b) The window, scatter plot, and quadratic regression obtained using a TI-84 graphing calculator are given in figure 6. (c) The surface area of a box whose dimensions are 18 in. × 18 in. × 28 in. is 2664 in.2. If we use the quadratic regression equation from part (b), the price of this box would be $7.09 (see fig. 6d). Extrapolation is not appropriate using this model. It seems reasonable that the company would base its prices on surface area, but perhaps box volume, cardboard thickness, or other quantities may be important.