What is the maximum number of distinct locations you can select on the surface of the Earth, such that the distance between every pair of locations is the same? Assume that the Earth is a perfect sphere and the distance between two locations is measured along the surface of the Earth.

Bonus question: Can you find actual locations of significance (such as cities) that have this property?

  • $\begingroup$ Surface of the Earth. Updated the last statement to reflect this better. $\endgroup$ – Dmitry Kamenetsky Jan 22 at 1:41

The answer is

4 - arranged as a tetrahedron


In order to select any more than 4, we would have to satisfy 5. But this is impossible, as it would have to begin as the tetrahedral arrangement with an extra point, but there is no point on the surface of a sphere which satisfies this.
A tetrahedron is the only arrangement of four points in 3D space for which the vertices are equidistant (kind of the definition of a tetrahedron). And for any tetrahedron, there exists one sphere on which all vertices lie.
This can also be shown by adding points incrementally.
2 points is trivial, and will always be equidistant, so any two points on the globe will satisfy.
3 points must be arranged as an equilateral triangle to preserve the equidistance. For (almost) any two points, there are only 2 options for the third.
adding the 4th point will only work on certain sizes of triangle, so that it lies on the surface of the globe.

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    $\begingroup$ It would seem the part about the Earth's surface is superfluous to the question. Good answer though. $\endgroup$ – Earlien Jan 22 at 1:58
  • $\begingroup$ Can you provide more information? What are the edge lengths relative to the circumference? Perhaps a diagram. $\endgroup$ – Dmitry Kamenetsky Jan 22 at 2:45
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    $\begingroup$ See also: puzzling.stackexchange.com/a/58763 $\endgroup$ – Rubio Jan 22 at 6:59
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    $\begingroup$ I don't disagree with the result, but the answer seems to be using some arguments that don't really translate from 3d geometry to 2d geometry on curved surfaces. For example, if you choose Toronto and Perth as the first two points, there cannot be a third point at all, because their distance is larger than one third of Earth's circumference. (The same applies to 25% of all possible pairs of points.) Somewhat similarly, the "equilateral triangle" argument works, but only because we are measuring a sphere, which is highly symmetrical; other surface shapes would give a very different result. $\endgroup$ – Bass Jan 22 at 7:38
  • $\begingroup$ @Bass The first part is a bit handwavy, but can be made strict (for a spherical earth) by showing that equidistant points on a sphere are also equidistant in 3D, which is kind of obvious. I think the only bit that is wrong is "(almost) any two points" in the second part. The "almost" is a bit of an understatement, but it does not matter as we are not interested in pairs of points that cannot have at least one other equidistant point. $\endgroup$ – Jaap Scherphuis Jan 22 at 8:48

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