nemo@rochester.UUCP (Wolfe) (10/29/85)
> > The sequel: (1) From how many points on Earth (assuming it's spherical) > > can you ... walk 1 mile south, 1 > > mile west, 1 mile north, and be back where you started? > > > > (2) Describe all of them. > > > > Judith Abrahms > > {ucbvax,ihnp4}!dual!proper!judith > > How about a more difficult sequel like the following: > > Where on the earth can one walk 1 mile south, 1 mile west, 1 mile > north, AND 1 mile east, and end up at the starting point? > > If you think you have a solution, there should be more... It would be nice > if some mathematically inclined readers could contribute exact and complete > solutions (to both sequels). Sequel : Aside from the north pole the concentric circles at distance 1+x north of the south pole work, where the latitude at distance x from the south pole has circumference 1/n miles for positive integers n. To get the exact distances for x, let R be the radius of the earth, r the radius of the disk perpendicular to the earth's axis at arc distance x from the south pole, and let A be the angle from the center of the earth to the latitude in question. Then we want 2*pi*r = 1/n miles, so r = 1/(2*pi*n). The information we get from the angle A is sine A = r/R, so A = arcsine (r/R). Finally, x = A * R (where A is in radians) x = R * arcsine(1/(2*pi*R*n). Since A is a very small angle, we can make the approximation that x ~= r, which gives us x ~= 1/(2*pi*n). The points we want are at distance 1+x from the south pole, for all positive integers n = 1, 2, .... Sequel's Sequel : First, the points along the latitude 1/2 mile north of the equator all work. The interesting points are near the poles. Let's look at the points near the north pole. Here we want the points to be at a distance x from the north pole such that we after we have walked the first three legs of our trip, we are at a latitude arc distance of m from our starting point, and walking one mile will take us around the circumference of the latitude line n times, plus the extra distance of m. Pull out your globes again, and let the starting point be on the latitude of distance x from the north pole, with angle A1 from the center of the earth as before, and (horizontal slice of the globe) radius r1. Let r2 be the radius of the latitude at distance x+1 from the pole (the latitude we use for the second leg of our trip), and A2 be its angle. So x = R * A1, and x+1 = R * A2 = R * A1 + 1, so A2 = A1 + 1/R. We can express r2 as a (grody) function of r1, r2 = R * sine A2 = R * sine (A1 + 1/R) = R * sine (1/R + arcsine(r1/R)) By similarity, we have the distance m from the end of the third leg to the starting point m/r1 = 1/r2, so m = r1/r2 = r1/[R * sine (1/R + arcsine(r1/R))] = f(r1,R). Our constraint on the radius r1 is that (*) 1 = 2n*pi*r1 + m, so that we can walk m miles back to the starting point and then n complete times around the world at that latitude. Given R, you can find the zeros of 2n*pi*r1 + r1/[R * sine (1/R + arcsine(r1/R))] if you really have to. But barf, who wants to do that! Since R >> x+1, we can approximate r2 ~= r1 + 1, so m ~= r1/(r1 + 1). Plugged into the constraint (*) on r1, this yields the quadratic 1 = 2n*pi*r1 + r1/(r1 + 1), 2n*pi*r1^2 + 2n*pi*r1 - 1 = 0. Solving for r1 we obtain r1 = (1/4n*pi)(-2n*pi + sqrt(4n^2*pi^2 + 8n*pi)) = (1/2) (sqrt((n*pi + 2)/(n*pi)) - 1) = F(n). So the latitudes we want to start at are x ~= r1 ~= F(n) for n = 1,2,... from the north pole. There are ones near the south pole, as well. They are all 1 + F(n) miles from the south pole. In this situation, you walk south, then east n times around the world, plus another m miles, then north, then west to your starting point. Ne (it's good for you) mo -- Internet: nemo@rochester.arpa UUCP: {decvax, allegra, seismo, cmcl2}!rochester!nemo Phone: [USA] (716) 275-5766 school 232-4690 home USMail: 104 Tremont Circle; Rochester, NY 14608 School: Department of Computer Science; University of Rochester; Rochester, NY 14627