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# South-East-North

Consider the Earth as a perfect, rigid, sphere with radius R=6378 km. There exist some points on the surface of the Earth from which we can travel first 6378 km South, then 6378 km East, and after that 6378 km North, and as a result return to the original point of departure. Find such points and paths. Calculate the geographic coordinates of the turning points of your solutions and draw the paths. For the sake of simplicity measure the geographic longitude from 0° to +360° eastward from Greenwich, the geographic latitude from 0° to +90° north of the Equator and from 0° to −90° south of the Equator. Solutions resulting from rotational symmetry should not be considered to be different.

(IOAA 2019)

in this problem we just need to find the latitude because the situation remains same for the small circle of a certain lattitude

@arman_hassan Your approach was quite nice and ok.

In summary, you give a special case and a general case.

As a special case, you said one solution is when A is north pole. Then one can go south along a meridian, then go east along a latitude circle, then go north again along another meridian and end up at north pole again.

As a general case, you said that one possible set of solutions will be the points (A, B and C), If,

1. B and C are the same points and
2. they all lie in the same Meridian.

Because, in that case, One will go south from A to B and then he will go east and end up coming to the same point covering the whole perimeter of that latitude and then go north to the same point A along the same path BA.

But there are some problems with the final answers:

First, when I go east from point B, I am traveling across a latitude line, so it must be parallel to the equator. This means in your picture, BM should be parallel to the equator. So A cannot be the north pole in the picture.

Second, you have already said one potential solution is when A is north pole. Let's see what happens if we follow this general approach for the north pole.

I will travel R distance (which means 57.29 degree angular distance) south from the north pole to point B. So the latitude of B is 90-57.29 = 32.71 degree. Then I start traveling across the latitude circle for distance R.

Radius at a certain latitude $\theta$,

$r=R\cos\theta$

If I end up in the same point B after traveling R distance,

$path BC=2\pi R\cos\theta = R$

$\Rightarrow 2\pi R\cos(32.71^{\circ}) = R$

$\Rightarrow 2\pi \cos(32.71^{\circ}) = 1$

But,

$2\pi \cos(32.71^{\circ}) = 5.28$

Third, qualitatively, try to think if my perimeter of the latitude circle at point B is 5.28R when A is north pole, is it possible to find any point B where I can cover the whole latitude circle by only traveling a distance of R?

actually I didnt mention correctly but I set up the general solution in a way so that north doesnt get include in the solution

i defined  $\theta$ and the general solution a different way

the 1st solution is  latitude =  $\arcsin{\frac{1}{2\pi}} -32.71=-23.55$,

after going downwards along the longitude line by 57.29 degree well reach lattitude -80.84 in which case  $\theta = 9.16$,

now  $2\pi R\sin\theta=2\pi \sin{9.16} *R =R$,

so the solution holds

Irrespective of the problem-solution-- I would like to share with you guys how important and easy looking "critical" this problem is. When I first saw this problem it took me over 30 mins to visualize the situation (our IBM problem presentation was still going on and I was working on a single problem the whole time). Most of my fellow team leaders said this is one of the most remarkable questions this year (IOAA 2019)!
We're talking about the question over dinner, one guy from Czech said he solved this problem in their elementary Olympiad which was very astonishing to me. (Later I realized this is a simple and very famous riddle) - you go one mile south, one mile west, and one mile north, and you end up where you started - where are you?

Btw this problem has infinite solution points. Let's see you guys figure it out!
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Hint:
There are actually four versions of the question: N/W/S, N/E/S, S/W/N, S/E/N. Obviously (or sort of obviously), East or West doesn’t matter (you’re walking around the earth). So there is actually symmetry: the answer is that whatever your starting direction is, on the opposite hemisphere there is one solution and on the same hemisphere there is an infinite.
For the general/trivial solution, you must state the coordinates in TCS otherwise a tiny mark will be penalized.