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

Question

Tetrahedron $ABCD$ has $AD=BC, AC=BD,$ and $AB=CD$. Find the point $X$ that minimizes the sum of the distances to the four vertices (i.e. it minimizes $f(X) = AX+BX+CX+DX$).

I found the following proof online, but I can't understand how to justify some of its claims:

Claim: $X$ is the centroid, or $X = \frac{A+B+C+D}4$, where $X,A,B,C,D$ are considered as vectors from some arbitrary point $O$.

Proof: Let $M$ and $N$ be the midpoints of $AB$ and $CD$. Observe that $\Delta ABD\cong \Delta BAC$ and $\Delta CDA\cong \Delta DCB$. Thus $MC=MD$ and $NA=NB$.

Now treating all variables as vectors (e.g. $MN$ represents $\vec{MN}$ and $A^2 = A\cdot A$ for a vector $A$), we have $MN = \frac{C+D}2 - \frac{A+B}2$. Also $MC^2 = MD^2 \Rightarrow (C - \frac{A+B}2)^2 = (D-\frac{A+B}2)^2$ and $NA^2 = NB^2\Rightarrow (A - \frac{C+D}2)^2 = (B- \frac{C+D}2)^2$). Thus, it suffices to show that $MN\cdot AB = 0$. We have $C^2 - C\cdot (A+B) + \frac{(A+B)^2}4 = D^2 - D\cdot (A+B) + \frac{(A+B)^2}4\Rightarrow C(C - A-B) = D\cdot (D-A-B)$ and thus $(D-C)\cdot (A+B) = D^2 - C^2\tag{1}$

Similarly, we have $(B-A)\cdot (C+D) = B^2 - A^2$. Thus, $MN\cdot AB = \frac{1}2 ((C+D)\cdot (B-A) - (B^2 - A^2)) = 0$, as required. Similarly, using $(1)$ we get $MN\cdot CD=0$.

But why is it the case that if $X$ is not on the line $MN$, then its orthogonal projection onto line $MN$ results in a smaller value of $f$ (claim (*))?

Assuming claim (*) holds, by similar reasoning, $X$ must lie on the line joining the midpoints of $AC$ and $BD$.

Now let $J$ and $K$ be the midpoints of $AC$ and $BD$ respectively.

Assuming the lines intersect, let $X$ be their intersection point, so that $MX = c_1MN$ for some scalar $c_1$ and $JX = c_2 JK$ for some scalar $c_2$. Then using the definitions of $M,N,J,K$ we have $X - \frac{A+B}2 = c_1(\frac{C+D}2 - \frac{A+B}2)$ and $X - \frac{B+C}2 = c_2(\frac{A+D}2 - \frac{B+C}2)$. Then $X= (1-c_1)\frac{A+B}2 + c_1\frac{C+D}2 = (1-c_2)\frac{B+C}2 + c_2\frac{A+D}2.$

Why is it that since $A,B,C,D$ can have an arbitrary origin (but are fixed as points), we must "equate coefficients"?

Equating coefficients gives $1-c_1 = c_2$ for $A$ and $1-c_1 = 1-c_2$ for $B$, so $c_1 = c_2 = \frac{1}2$, from which we obtain $X = \frac{A+B+C+D}4$, as desired.

For this question, to clarify, it suffices to justify/prove the questions highlighted in gray using the ">" symbol.

Answer 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.