It’s been such a long time since my last post. Since then, I’ve graduated from Carleton College, moved to Brooklyn, and starting developing software for a firm over in Manhattan. Now that I’ve found my new groove, and now that I have some down time due to the holidays, I figure it’s time to write a new post!
One awesome perk about my job is that on Fridays, the developers get sent these brain teasers to solve. Today I’d like to share one particularly interesting problem.
The Problem: How many points are there on Earth such that you can walk one mile south, one mile east, and one mile north only to end up at the same starting point? Describe these points. Assume that the Earth is a perfect sphere.
The Answer: There are an infinite number of points on the Earth where one can walk one mile south, one mile east, and one mile north only to end up at the same starting point. We distinguish these points by dividing them up into three classes.
The first class of points: The North Pole is one such point.
The second class of points: Any point one mile away from the South Pole is such a point that will work.
Here, we travel one mile south and reach the South Pole. Moving east from the South Pole requires us to just stay put, since any movement away from the South Pole is technically moving north. Therefore after traveling one mile east, we still remain at the South Pole. Then we traverse one mile north back to our starting point.
The third class of points: This class is a little tricky to nail down. Any point approximately
.8408 miles away, or 1.159 miles away from the South Pole is such a point that will work.
The special part about this class of points is that when we travel east for one mile, we end up at the same point where we started traveling east; that is, the circle we just traversed has the circumference equal to one mile. Let us call this circle the East Circle. So, what points on the globe exist such that walking one mile south from it will land us on a point on the East Circle? In order to answer this question, we need to measure the arc, call it The Arc, that hangs over the East Circle as shown in the following picture.
Since The Arc is symmetrical, I’m going to measure one half of it starting from the center of the East Circle and going to the edge of the East Circle — a distance of 1/(2π) miles in the x direction (since the circumference is 1 mile). Call this half arc A. Then when we want to know The Arc’s entire length, we just double the length of A. Once we have the length of A, we know that a starting point that will work is
1-A miles away from the South Pole, or 1+A miles away.
Notice that A lies on a circle whose positive curve is
f(x) = √ r2 – x2, where r is the radius of the globe. Specifically, if we imagine the East Circle on the x-y-plane of a 3-D plot, centered at the origin, A is the section of f(x) ranging from x = 0 to x = 1/(2π). Note that A is continuous on this interval x = [0, 1/(2π)]. Notice further that the derivative of A is also continuous on this interval. Then,
A = ∫01/(2π) √ 1 + ((-x)2 / (r2 – x2)) dx, where r is the radius of the globe. I will talk about how I got this equation more in my next post. Plugging in the approximate radius of the Earth (3,959 miles) and solving the integral, we get that A = 0.159 miles.
any point 1-A = 1 – 0.159 = 0.8408 miles away from the South Pole, or any point 1+A = 1 + 0.159 = 1.159 miles away from the South Pole will be such a point that works.
Notice that 1/(2π) = 0.15915494309. So, rather than measuring A, we could have just used the radius of the East Circle as an approximation since the radius of the Earth is so large. However, the more general way to think about it is always more fun 🙂
Now, I’ve thought of these three classes. I’m not sure if there are any more but if you think of something I didn’t find, let me know!
EDIT: One case was found to violate the condition of the problem and has since been removed.