For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$?

Question

Let $ABCD$ be a regular tetrahedron with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-3\overrightarrow{MO}.$ Prove or disprove that $$NA+NB+NC+ND\geq MA+MB+MC+MD$$

I tried to use CS in the Euclidean space $E_3$, but it does not help, because the minoration is too wide.

Answer 1

This is just the $2$ dimensional variant of the same problem.

Let $ABC$ be a equilateral triangle with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-2\overrightarrow{MO}.$ Prove that: $$NA+NB+NC\geq MA+MB+MC.$$

Let $A',B',C'$ be the midpoints of the segments $BC, CA$ and $AB$ respectively. Notice that the map $$M\mapsto N$$ is actually a homothety with respect to $O$ and factor $-2$.

So this homothety takes $X'\mapsto X$. We thus have $$NX = 2MX'$$ for each $X\in \{A,B,C\}$ and we have to prove $$2(MA'+MB'+MC')\geq MA+MB+MC.$$

Now observe that in a quadrilateral $AB'NC'$, by Ptolemy’s inequality, we have $$AB'\cdot MC'+AC'\cdot MB'\geq B'C'\cdot MA.$$

Notice that $AB'= AC' = B'C'$, so we have $$MC'+MB'\geq MA.$$ And similarly we have $$MC'+MA'\geq MB$$ and $$MA'+MB'\geq MC.$$ Adding all three equations we get what we want to prove.

Answer 2

Considering the distance squared, the problem becomes trivial. I leave this as a curiosity, not as an answer.

Given a regular tetrahedron centered at the origin as

$$ \cal{T}=r\left( \begin{array}{ccc} \frac{2 \sqrt{2}}{3} & 0 & -\frac{1}{3} \\ -\frac{\sqrt{2}}{3} & \sqrt{\frac{2}{3}} & -\frac{1}{3} \\ -\frac{\sqrt{2}}{3} & -\sqrt{\frac{2}{3}} & -\frac{1}{3} \\ 0 & 0 & 1 \\ \end{array} \right) $$

a point $M = (m_x,m_y,m_z)$ and a rotation matrix $R(\alpha,\beta,\gamma)$ we have

$$ \sum_{k=1}^4(M-R\cdot {\cal{T}}(k))^{\dagger}\cdot (M-R\cdot {\cal{T}}(k)) =4||M||^2-2\sum_{k=1}^4M^{\dagger}\cdot R\cdot{\cal{T}}(k)+ \sum_{k=1}^4 ||{\cal{T}}(k)||^2 $$

but for the regular tetrahedron

$$ \sum_{k=1}^4M^{\dagger}\cdot R\cdot{\cal{T}}(k)=0\\ \sum_{k=1}^4 ||{\cal{T}(k)}||^2 = 4r^2 $$

then

$$ \sum_{k=1}^4(M-R\cdot {\cal{T}}(k))^{\dagger}\cdot (M-R\cdot {\cal{T}}(k)) = 4\left(||M||^2+r^2\right) $$

independent of $\{\alpha,\beta,\gamma\}$

NOTE

Considering now the sum of distances as required, we have

$$ ||\vec{MO}(\alpha,\beta,\gamma)||=\sum_{k=1}^4||M-R\cdot {\cal{T}}(k))|| =\sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2} $$

and now the question

Is

$$ \sum_{k=1}^4\sqrt{9||M||^2+6M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}\ge \sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2} $$

for all $\{\alpha, \beta, \gamma\}$ ?

Follows a MATHEMATICA script which minimizes

$$ \Delta(\alpha,\beta,\gamma)=\sum_{k=1}^4\sqrt{9||M||^2+6M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2}-\sum_{k=1}^4\sqrt{||M||^2-2M^{\dagger}\cdot R(\alpha,\beta,\gamma)\cdot{\cal{T}}(k)+ r^2} $$

for values of $\{\alpha,\beta,\gamma\},\ \rho = ||\vec{MO}||/r$ characterizing numerically the assumption

$$NA+NB+NC+ND\geq MA+MB+MC+MD$$

M = rho {1, 0, 0};
R = EulerMatrix[{alpha, beta, gamma}];
tetraR = Table[R.tetra[[k]], {k, 1, 4}];
sd = Sum[Sqrt[(-3 M - tetraR[[k]]).(-3 M - tetraR[[k]])], {k, 1, 4}] - Sum[Sqrt[(M - tetraR[[k]]).(M - tetraR[[k]])], {k, 1, 4}];
list = Table[{rhok, First[NMinimize[(sd /. {rho -> rhok}), {alpha, beta, gamma}]]}, {rhok, 0, 0.5, 0.005}];
ListLinePlot[list, PlotStyle -> {Blue, Thick}]

Follows the plot of $\Delta(\alpha^*,\beta^*,\gamma^*)\times\rho$

Follows the relative positions for given $M$, $N$, and $\{\alpha^*,\beta^*,\gamma^*\}$

Answer 3

In Aqua's proof, we have to show that $MB' + MC' \geq MA$. I want to introduce another way :

In equilateral triangle $\Delta ABC$, we have mid points $A'$ in $[BC]$ and we have $B',\ C'$ similarly.

When $O$ is in $\Delta A'B'C'$, then we have $O'$ s.t. $\Delta O'B'C'$ is congruent to $\Delta OA'C'$.

Hence $$ |A-O' | + |O'-O| = |B'-O| + |O-C'| \geq |A-O| $$

This completes the proof.

Answer 4

Another solution for two-dimensional problem. We will change notations to make them more logical.

Problem. Let $ABC$ be an equailateral triangle with center $O$. Consider two points $P$ and $K$, such that $\overrightarrow{OP}=-2\overrightarrow{OK}$. Prove that $PA+PB+PC\geq KA+KB+KC$.

Solution. Construct an equilateral triangle $PQR$ centered at $O$. Then, $K$ is the midpoint of $QR$ (due to $\overrightarrow{OP}=-2\overrightarrow{OK}$). Denote midpoints of $PQ$ and $RP$ as $M$ and $L$, respectively. Now our construction is "symmetric" (i. e. invariant under rotations by $\frac{2\pi k}{3}$, $k=0,1,2$). Hence, we have the following equalities $$ PB=QC=RA, \\ PC=QA=RB, $$ and $$ KB=LC=MA, \\ KC=LA=MB. $$ Thus, our inequality can be rewritten as $$ PA+QA+RA\geq KA+LA+MA. $$ Finally, note that $K$ is the midpoint of $QR$, so $PA+QA\geq 2KA$. Adding similar inequalities gives the desired inequality.

Further thoughts. I tried to generalize this approach for the original problem but didn't succeed. The problem is in constructing regular tetrahedron $PQRS$. (It seems to me that it's connected to existence of normal subgroup of order 4 in $A_4$)

---

Remark. The solution was taken from the post on AoPS (it was the problem on the China West MO 2014, by the way). There are two solutions one of which is similar to Aqua's one. https://artofproblemsolving.com/community/c6h602780p3578945

Answer 5

HINT:

The function $\phi\colon M\mapsto AM+BM+CM+DM$ is convex ( strictly). The tetrahedron $ABCD$ has a symmetry group $G$ ( $G \simeq S_4$) generated by reflections. For every $g$ in that group we have $\phi(gN) \equiv \phi(N)$. Therefore, if $M$ is in the convex hull of $\{g N\}_{g \in G}$ then $\phi(M) \le \phi(N)$.

Now one checks that $ON \ge 3 OM$ ensures that $M$ is in the convex hull of $\{g N\}_{g \in G}$.

Notes:

1. For this particular function "sum of distances", the constant $3$ can be improved to $\gamma_3\colon =\frac{4 + \sqrt{6}}{5}\simeq 1.289\ldots$. (the point opposite $A$ at distance $R/\gamma_3$ from the center $O$ will have the same sum of distances to the vertices as $A$).

2. The regular tetrahedron can be thought of having vertices $(4,0,0,0)$, $(0,4,0,0)$, $(0,0,4,0)$, and $(0,0,0,4)$, in the hyperplane $\mathcal{H}$ of $\mathbb{R}^4$ given by $x_1 +x_2+x_3+x_4=4$. The center is $(1,1,1,1)$. The groups of symmetries is given by permutations of coordinates. For points $N$, $M$ in $\mathcal{H}$, $M$ lies in the convex hull of permutations of $M$ if and only if $N\succeq M$ in the majorization order. Now, $N$ larger than $M$ in this order means $N$, $M$ have the same average, but $N$ is more spread around. So it should be that if the variance of $N$ is "way larger" than the variance of $M$, then $N \succeq M$. The constant $3= 4-1$ is the optimal one.

3. One can generalize the result following 1. or 2.