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You can not start 1 km away from the South Pole, if you did, you would not be able to do the second leg of the journey.
Short answer: These points are at a distance from the South Pole roughly equal to (1 + 1/2(pi)n) km , for some positive integer n.
Long answer: On a sphere of radius R, the distance d measured on the surface is along an arc of a great circle (that's indeed the shortest possible distance). If the "radius" d of a circle is measured in that way, its circumference is only 2pi[R sin(d/R)] (since the bracketed quantity is the actual radius of that circle in 3D space). For small values of x=d/R, we have sin(x) » x (1-x ^2/6), and 1/sin(x) » 1/x (1+x ^2/6).
Before we apply this to the above situation, we may stop to consider what is theoretically the best value to use for R around the South Pole (this may seem like a ludicrous concern --and it is-- but we are already into ludicrous precision at this point, so we may as well learn something from this): First, consider the Reference Ellipsoid (the defined regular shape with respect to which professionals are charting the irregularities of the Earth's so-called "sea-level", which is an equipotential surface averaged over time at each point of the Globe). Its "equatorial radius" is defined as precisely equal to a=6378137 m, whereas the "polar radius" (i.e. half the distance between the North Pole and the South Pole) is about 6356752.3141404 m. This makes the radius of curvature of the Meridian at the South Pole equal to a 2/b or 6399593.6258639 m, for this surface of reference. Now, the South Pole is not at sea level, but on an elevated plateau at about 2835 m of altitude (with an unbelievably thick layer of about 8800' feet of ice, which accounts for almost 95% of that altitude). This altitude is to be essentially added to the radius of curvature of the "sea-level" reference meridian. Therefore, if the surface at the South Pole was perfectly level and smooth, it would be extremely close to the surface of a sphere with a radius R of about 6402428 m. This would make the tiny quantity 1/(6 R ^2 ) (used in the "final" result below) equal to about 4.066 10^-9
All told, we may now state with more (ludicrous) precision that "home" is at a distance from the South Pole very close to:
[ 1 + 1/2(pi)n (1 + (4.066 10^-9)/n ^2 ) ] km ,
for some positive integer n.
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