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Let $A_1, A_2, A_3, A'_1, A'_2, A'_3$ be the vertices of a (not-necessarily convex) octahedron; here $X'$ is the vertex not on an edge with $X$. Suppose that the four non-adjacent triangular faces $A_1 A_2 A_3$, $A_1 A'_2 A'_3$,$A'_1 A_2 A'_3$,$A'_1 A'_2 A_3$ are equilateral with sides of lengths $x_0, x_1, x_2, x_3$.

I am interested in knowing the conditions on $x_0,x_1,x_2,x_3$ that must be satisfied for such a configuration to exist; and how unique the configuration is. This is a preliminary step to understanding this Question on Octahedra and the answer by @Blue.

Clearly it is necessary that $x_i>0$ for each $i$; and by using the triangle inequality on the other four faces, that $x_i+x_j>x_k$ for every 3-set $\{i,j,k\}\subset \{0,1,2,3\}$.

Question 1: are these conditions sufficient for existence? I think so, but have only a hand-waving argument and would like something better.

It seems to me that if such a configuration exists for some $x_0,x_1,x_2,x_3$ then in general if we fix $A_2, A_3, A'_1, A'_2, A'_3$ there are two points $A_1$ which satisfy the right conditions, giving a convex octahedron and a non-convex one. Again I can only wave my hands, and my intuition may be wrong. So,

Question 2: what can be proved about uniqueness?

Lastly, to get some feeling for which $(x_0,x_1, x_2,x_3)$ yield an octahedron, I would like to understand their geometry.

Question 3: Describe geometrically the (subset of) the solid bounded by

\begin{align*} x_0+x_1+x_2+x_3&=1;&\\ x_i&=0, &i=0,\dots,3; \\ x_i+x_j&=x_k &\textrm{ for every 3-set } \{i,j,k\}\subset \{0,1,2,3\} \end{align*}

which yield solutions.

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2 Answers 2

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This is not an answer, but just a conjecture about Question 1.

Consider your octahedron "unfolded" on a plane, and in particular its equilateral face $ABC$ of side $x_0$ and its three neighbouring (not equilateral) faces $ABE$, $BCD$, $ACF$ (see diagrams below). We can imagine to "fold" these neighbouring faces (all on the same side of the equilateral face) until $DE=DB$, $DF=CF$ and $FE=AE$. We have then three equations for three unknown parameters (the positions of points $D$, $E$, $F$ along the three circles where they are constrained to lie).

But if you try that with GeoGebra, it turns out that the construction succeeds only if, in the unfolded planar graph, $DE\ge DB$, $DF\ge CF$ and $FE\ge AE$. Thus the first diagram shows a constructible octahedron, while the second one is not constructible, because $DE<DB$. I haven't a proof for that (yet) and would like to know other opinions on this conjecture.

enter image description hereenter image description here

A short comment on Question 3: I think the first condition can always be met by a suitable rescaling, the second one gives a single point and the third one (as written) implies the second one. But maybe I just don't understand what you are asking.

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  • $\begingroup$ Yes, I think I was mentally folding something that wouldn't join up, I must look again. As to Qn 3: sure, the interesting set is really projective, and eqn 1 is just a way of getting into $\mathbb{R}^3$: then we have $4+12$ bounding plane "faces". Plus whatever other restrictions there are. $\endgroup$ Commented Jan 2, 2019 at 7:53
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Without loss of generality, let me assume the lengths are scaled so the shortest is 1. I'll label the four lengths $(1, a, b, c)$ withere $1 \leq a \leq b \leq c$. Then the condition $c < a + 1$ is sufficient to ensure all the triangle inequalities are met.

Approaching it a different way from Intelligenti pauca, I came to a similar obstruction for $(1, 1, 1.8, 1.8)$. If you consider the wedge formed by two equilateral triangles of side length 1.8 meeting at a point, joined by two isosceles triangles with side lengths 1.8, 1.8, and 1, the base is a 1 by 1.8 rectangle; the two equilateral triangles of side length 1 cannot reach across it (since the height of the triangle is $\sqrt{3}/2 \approx 0.87$).

This also fails Intelligenti pauca's condition, which we can rephrase by saying that when we unfold the three non-equilateral triangles adjacent to the equilateral triangle of side length 1, the angles in the gaps must be greater than $60^\circ$. Thus the sums of the angles in the two non-equilateral triangles at a vertex must be less than $240^\circ$.

Let me crudely edit Intelligenti pauca's picture:

enter image description here

The marked angle must be greater than $60^\circ$, so $\beta+ \gamma$ must be less than $240^\circ$. If $a = 1$ and $b = c = 1.8$ as in my example, then the angles $\beta$ and $\gamma$ are about $128.3^\circ$, so we see the criterion fails.

To apply this more generally, the law of cosines tells us the angle $\beta$ is $\cos^{-1}(\frac{1+a^2-b^2}{2a})$ and $\gamma$ is $\cos^{-1}(\frac{1+a^2-c^2}{2a})$. Since $\gamma \geq \beta$, if $\beta$ is more than $120^\circ$ we already know the situation is impossible. Hence $$ \frac{1+a^2-b^2}{2a} > \frac{-1}{2}, \text{ or } \ a^2 + a - b^2 > -1. $$ This fails for $1^2 + 1 - 1.8^2 = -1.24$ in my example, but not for $1.1^2 + 1.1 - 1.6^2 = -0.25$ in Intelligenti pauca's example.

If $\gamma$ is more than $120^\circ$, we can't automatically say if the total angle is too big without knowing $\beta$. Using the idenity $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}(AB - \sqrt{(1 - A^2)(1-B^2)})$, we get the condition $AB - \sqrt{(1 - A^2)(1-B^2)} < \frac{-1}{2}$ where $A = \frac{1+a^2-b^2}{2a}$ and $B = \frac{1+a^2-c^2}{2a}$.

We can rearrange that condition to $$ A^2 + AB + B^2 < \frac{3}{4}. $$ In terms of the triangle side lengths $a$, $b$, and $c$, that expands to the truly horrendous expression $$ 3 a^4 + 3a^2 + b^4 + b^2c^2 +c^4 + 3 < 3a^2b^2 + 3a^2c^2 + 3b^4 + 3c^2. $$ When combined with $a + 1 > c \geq b \geq a \geq 1$, this is a necessary and sufficient condition for the four lengths to appear as sides of equilateral triangles on an octahedron.

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