Let $A,B,C$ be three points on the surface of a unit sphere with centre $O$, such that they are not all three on the same maximal circle. There is a plane $P$ that contains all three points and on it they define a triangle. Let $X$ be a point (to be determined) on the surface of the sphere and let $XA$ be the angle between the line segments $OA$ and $OX$ and similarly for $XB$ and $XC$.

Q1: Find the point $X$ on the surface of the sphere such that the sum of the three angles $XA+XB+XC$ is minimal.

Q2: Is the problem of Q1 equivalent to finding a point on the plane $P$ that minimises the sum of the (Euclidean) distances from the vertices to that point? (Eg if that point on the plane is found, call it $Y$, then one can extend the line segment from $O$ to $Y$ until the surface of the sphere is intersected)

If one can show that Q2 is equivalent to Q1 then the point of minimal sum can be determined by first finding the Fermat point on the plane.

Thanks in advance!

  • $\begingroup$ You are right of course; I meant maximal circle as you suggested. I will edit the question to clarify that. Thanks! $\endgroup$
    – AG1123
    Aug 5, 2016 at 14:25

1 Answer 1


Since the angle $\widehat{AOX}$ is proportional to the length of the shortest path (geodetic) between $A$ and $X$ on the sphere, the point $X$ is the spherical analogue of Fermat's point. The euclidean construction of the Fermat point just relies on the properties of geodetics and their interactions with isometries, so, in order to find $X$, one is tempted to follow these steps:

  1. Let $B'$ be the image of $C$ under a counter-clockwise rotation around $A$ with amplitude $60^\circ$;
  2. Let $A'$ be the image of $B$ under a counter-clockwise rotation around $C$ with amplitude $60^\circ$;
  3. $X$ is ($\color{red}{\large ?}$) given by the intersection between the $BB'$ geodetic and the $AA'$ geodetic.

However, this approach does not really work, because if $X_A$ is the image of $X$ under a counter-clockwise rotation around $A$ with amplitude $60^\circ$, while the geodetic distance between $A$ and $X$ is the same as the geodetic distance between $A$ and $X_A$, the geodetic distance between $X$ and $X_A$ may be different: in spherical geometry an isosceles triangle with a vertex angle of $60^\circ$ is not always a equilateral triangle, and that is a key point of the Fermat-Torricelli construction.

I do not even expect some straightforward work-arounds, since

  1. Due to different Gaussian curvatures, there is no isometric embedding in the plane of a spherical triangle;
  2. The locus of points $P$ on a sphere such that the sum between the geodetic distances $PA,PB$ is constant is a quite complicated object.

At the very least, we have a negative answer to Q2: our problem is not completely equivalent to its euclidean analogue. Anyway, it was recently solved through Lagrange's multipliers: for sufficiently small spherical triangles $ABC$, the spherical Fermat point $X$ is the point such that the angles $\widehat{AXB},\widehat{BXC},\widehat{CXA}$, intended as angles between geodetics, all equal $120^\circ$: that property is shared also by the euclidean Fermat point.

  • $\begingroup$ Thank you for the very informative answer. I haven't carefully read the paper you mentioned yet but could you please explain a bit more 1 and 2 from the second part of your answer? $\endgroup$
    – AG1123
    Aug 6, 2016 at 11:01

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