# How to find the smallest distance from the south pole where this is travelling is possible?

You travel on Earth's surface south $n$ miles, then east $n$ miles, then north $n$ miles and find yourself back where you started, without visiting any point more than twice. What is the closest you could have been to the south pole when you started? Assume Earth is a sphere with radius $R>n$.

My initial thought for this question was that you could actually be on the south pole and make $n = 0$, but this seems like a silly answer. At the moment I am thinking that it is some infinitely small number like $0.000000....1$ because, in my mind, you could travel any value of $n$ miles around the south pole when you are this close to it and it wouldn't matter how far you have travelled south or north, because there are no restrictions to this.

I can't seem to get this into algebra or an actual answer though.

• If you're closer to the south pole than $n$ miles, you'll have to define what you mean by "go south $n$ miles". Nov 18, 2016 at 23:47

You can't travel $n$ miles south unless you start at least $n$ miles north of the south pole.
So what's the closest we can start to the south pole? The real restrictions are not visiting any point more than twice, and ending up back where you started. So we want to find a line of latitude such that the circumference is $n$, then start $n$ miles north of it.
Circumference of $n$ implies radius $n/2\pi$. So the angle from the vertical at the centre of the Earth is $\arcsin(n/(2\pi R))$. So we should start $R \arcsin(n/(2\pi R)) + n$ miles north of the south pole.
• I've never used the $arcsin$ function, but I understand that it is basically the inverse of $sin$ provided $|sin(x)| < 1$. How does $arcsin$ make sense in the situation? Nov 20, 2016 at 14:22
• That's right, it's the inverse of $sin$. We can form a right triangle with the centre of the Earth as one point, a point on the latitude with a radius of $n$ as another, and the point on the line between the poles which is as far south as the latitude as another. The sin = opposite / hypotenuse rule tells us that the sin of the angle at the centre of the Earth is $n/2\pi$ divided by $R$. Nov 21, 2016 at 22:58
The answer is $n+R\arcsin(\frac{n}{2\pi R})\approx n+\frac{n}{2\pi}$ assuming that $n\ll R$, where $R$ is the radius of the earth