Similar Reflection Tetrahedron

Question

With a few intersection points of heptagon diagonals, a heptagon with double the area can be constructed. The reflection triangle of the heptagonal triangle (in green) is another heptagonal triangle (in cyan) with twice the area.

There is one other triangle with similarity to the reflected triangle, with area ratio 2/φ.

pp = {{0, 0}, {1, 0}, {Root[-1 + 52 # - 36 #^2 - 288 #^3 + 528 #^4 - 320 #^5 + 64 #\
^6& , 1, 0], Root[-31 + 744 #^2 + 496 #^4 - 17664 #^6 - 24832 #^8 + 10240 #\
^10 + 4096 #^12& , 5, 0]}};

Is there a tetrahedron where the reflected tetrahedron is similar? The following comes close, but not quite there.

tet = {{0,0,0},{1,0,0},{0.8029,2.8552,-2.6304},{2.2197,1.0925,1.0949}};

The reflection tetrahedron here isn't quite similar, but close. Does a perfect solution exist?

Answer 1

$\newcommand{\ssin}{\overline}$ $\newcommand{\ccos}{\ddot}$

I'll use labeling conventions from an answer of mine to one of OP's previous questions.

Let $OABC$ be a tetrahedron with edge-lengths $a$, $b$, $c$, $d$, $e$, $f$ and face-areas $W$, $X$, $Y$, $Z$ as indicated below. Without too much danger of confusion, I give the dihedral angles the upper-case letter of the corresponding edge. The volume is $V$.

Also, to compactify equations, I'll use the trigonometric shorthand $\ssin{\theta}:=\sin\theta$ and $\ccos{\theta}:=\cos\theta$, as well as $$\sigma_{PQR}:=\ssin{P}^2+\ssin{Q}^2+\ssin{R}^2 \qquad \tau_{PQR} := 1+\ccos{P}\ccos{Q}\ccos{R}$$ which allows me to state a couple of known results thusly: $$\begin{align} W &= X\ccos{D}+Y\ccos{E}+Z\ccos{F} \\ X &= W\ccos{D}+Y\ccos{C}+Z\ccos{B} \\ Y &= X\ccos{C}+W\ccos{E}+Z\ccos{A} \\ Z &= X\ccos{B}+Y\ccos{A}+W\ccos{F} \end{align} \tag{1}$$ $$ \frac{W^2}{\sigma_{ABC}-2\tau_{ABC}} = \frac{X^2}{\sigma_{AEF}-2\tau_{AEF}} = \frac{Y^2}{\sigma_{DBF}-2\tau_{DBF}} = \frac{Z^2}{\sigma_{DEC}-2\tau_{DEC}} \tag{2}$$

---

To the problem at hand ...

Let $O'A'B'C'$ be the reflection tetrahedron of $OABC$, with $O'$ the reflection of $O$ in the plane of $ABC$, etc. Barycentric coordinates of the new vertices are strangely reminiscent of $(1)$: $$\begin{align} O' &=\; -W\phantom{\ccos{T}} : 2X\ccos{D} : 2Y\ccos{E} : 2Z\ccos{F} \qquad \text{sum}=W\\ A' &=\; 2W\ccos{D} : -X\phantom{\ccos{T}} : 2Y\ccos{C} : 2Z\ccos{B} \qquad \text{sum}=X \\ B' &=\; 2W\ccos{E} : 2X\ccos{C} : -Y\phantom{\ccos{T}} : 2Z\ccos{A} \qquad \text{sum}=Y \\ C' &=\; 2W\ccos{F} : 2X\ccos{B} : 2Y\ccos{A} : - Z\phantom{\ccos{T}} \qquad \text{sum}=Z \end{align}$$ and, for $a':=|O'A'|$, etc, we can calculate $$(a')^2 - a^2 = 72\,\frac{V^2 \ccos{D}}{WX} \qquad \text{etc} \tag3$$

Suppose the $OABC$ and $O'A'B'C'$ are similar, with linear scale factor $\kappa = a'/a=b'/b=\cdots$. If $\kappa=1$, then $(3)$ implies either $V=0$ or $A=B=C=D=E=F=90^\circ$. The latter is impossible; the former "works" if $OABC$ is a degenerate "flat" tetrahedron with four coplanar vertices (but with no three collinear), since then $O=O'$, $A=A'$, $B=B'$, $C=C'$. Going forward, we'll assume $OABC$ is non-degenerate, so that $\kappa\neq1$.

So, for $OABC$ to have a similar reflection tetrahedron, the right-hand sides in $(3)$ must be proportional to respective values $a^2$, $b^2$, $c^2$, $\ldots$. Since we "know", for instance, $a^2 = 2 Y Z\ssin{A}/(3V)$, we can write

$$ \frac{\ssin{A}^2}{WX\ccos{D}} = \frac{\ssin{B}^2}{WY\ccos{E}} = \frac{\ssin{C}^2}{WZ\ccos{F}} = \frac{\ssin{D}^2}{YZ\ccos{A}} = \frac{\ssin{E}^2}{ZX\ccos{B}} = \frac{\ssin{F}^2}{XY\ccos{C}} = \lambda \tag4 $$ where $\lambda := 32V^4/((\kappa^2-1)W^2X^2Y^2Z^2)$. Note the immediate consequence that all dihedral cosines must match in sign; since no tetrahedron has six obtuse dihedral angles, we conclude that they must all be (strictly) acute.

Note also that the first three denominators sum to $WX\ccos{D}+WY\ccos{E}+WZ\ccos{F}=W^2$ (by $(1)$) which, considering the numerators, is equal to $\sigma_{ABC}/\lambda$; likewise with other sums, and we find ourselves with a proportion that we can re-write as $$\frac{\sigma_{ABC}}{\tau_{ABC}}=\frac{\sigma_{AEF}}{\tau_{AEF}}=\frac{\sigma_{DBF}}{\tau_{DBF}}=\frac{\sigma_{DEC}}{\tau_{DEC}} \tag5$$ Ths gives us a system in $\ccos{D}$, $\ccos{E}$, $\ccos{F}$ that we can solve to get $$\begin{align} (\ccos{D},\ccos{E},\ccos{F}) &= \,\left(\phantom{-}\ccos{A},\phantom{-}\ccos{B},\phantom{-}\ccos{C}\right) \\ \text{or} &\;\quad \left(-\ccos{A},-\ccos{B},-\ccos{C}\right) \\ \text{or} &\;\quad \left(\phantom{-}\ccos{A},\phantom{-}\ccos{B},-\ccos{C}-\ccos{A}\ccos{B}\,\frac{\sigma_{ABC}}{\tau_{ABC}}\right) \\ \text{or} &\;\quad \left(-\ccos{A},-\ccos{B},\phantom{-}\ccos{C}+\ccos{A}\ccos{B}\,\frac{\sigma_{ABC}}{\tau_{ABC}}\right) \\ \text{or} &\;\quad \left(\text{cyclic variants of the previous two}\right) \end{align} \tag6$$ Since all cosines must have the same (positive) sign, and since $\sigma_{ABC}/\tau_{ABC}$ is strictly positive for non-degenerate tetrahedra, we can dismiss all options but the first.

Thus, all three opposing pairs of dihedral angles are congruent. One can show that this implies that the opposing edges are then also congruent, making all four faces congruent. From, say, the first equality in $(3)$, we deduce that $A=B$; repeating the argument shows that all dihedral angles are equal. Consequently,

A non-degenerate tetrahedron has a similar reflection tetrahedron iff it is regular.

$\square$