Find the point in a tetrahedron that minimizes the sum of distances between vertices of a tetrahedron

Solution 1:

Hint: the desired point $X$ is required to have the following property - from the point $X$ each of the faces $ABC, ABD, ACD, BCD$ of the tetrahedron should be viewed by the same solid angle (which is equal to $\pi$). This is equivalent to the following property of the required $X$: if one draws unit vectors from $X$ toward each of the vertices $A,B,C,D$, then the ends of those unit vectors form a regular tetrahedron.

The proof is analogues to the construction of Fermat-Toriccelli point for triangles.

Solid angle:

When there are 3 points $A,B,C$ on a plane and a point $X$ putside that plane, a solid angle is formed between $X$ and the other points. If we draw a sphere with $X$ as its center, and each of the lines $XA,XB,XC$ intersect that sphere at points $A',B',C'$ respectively, the solid angle is defined to be the fraction of the area of the spherical triangle $A'B'C'$ relative to the area of the whole sphere, multiplied by $4\pi$.

Fermat point of a triangle A famous result of Fermat states that for any triangle $ABC$, the point for which $XA+XB+XC$ is minimal is the point from which one views each the sides with equal angle 120 degrees. There is a nice physical construction that proves it: suppose we hang three weights of equal mass by wires, and those three wires connect in a point inside the triangle. The system stabilizes itself in a state of minimal total potential energy of the weights; and this is also the state where the sum of total wire lengths from $X$ to $A,B,C$ is minimal. Since the tension force is the same at each wire, the only possible configuration for this to occur is when all the view angles are 120.