8
$\begingroup$

The largest (by edge length) regular simplex inscribed in a unit cube is well known in $\mathbb{R}^2$ and $\mathbb{R}^3$:


 
  Image sources: left: NMSU, right: Mathworld.
A recent Amer Math Monthly problem posed by Ionascu & Strong and solved by Yuri Ionin (Problem 11693, 122:2, 178-181, 2015), showed that the largest equilateral triangle inscribed in the unit $d$-cube has edge length of $\sqrt{\frac{2}{3}d}$ for $d \equiv 0 \pmod 3$, and slightly more complex expressions for mod $1$ and $2$ (but still the same $\frac{2}{3}$ dominating fraction). Ionin proved that a maximal equilateral triangle has all three corners on edges of the cube, and exactly one triangle corner at a vertex of the cube.

The proof is not straightforward and suggests that, e.g., finding the largest regular $3$-dimensional tetrahedron inscribed in a $4$-dimensional unit cube might not be simple. Although, that the simple $\frac{2}{3}$ fraction emerges through all the proof complexities gives hope that there might be an analogously simple characterization in answer to this question:

Q. What is the largest $k$ simplex inscribed in a unit $d$-cube, for $k < d$? Is there a characterization of how the tetrahedron corners sit with respect to the vertices/edges/facets of the $d$-cube? Is there an analogous $d \bmod (k+1)$ splitting of the results? Are there known results for some $(k,d)$ pairs?


Incidentally, the "slightly more complex expression" for $d \equiv 2 \pmod 3$ is $\sqrt{\frac{2}{3}d + \frac{20}{3} - 4 \sqrt{3}}$, which for $d=2$, evaluates to $\sqrt{6}-\sqrt{2} = \sec(15^\circ)$, in accord with the figure above.

$\endgroup$
7
$\begingroup$

Allow me look at one aspect, or special case, of your question, namely "finding the largest regular 3-dimensional tetrahedron inscribed in a d-dimensional unit cube".

I can find the following largest regular $3$-simplex inscribed in a 4-cube. Essentially I used the same methods as described in the paper Computing Maximal Copies of Polyhedra Contained in a Polyhedron, Experimental Mathematics Vol. 24 (2015), Issue 1, pp.98-105 or in this MathOverflow answer: On maximal regular polyhedra inscribed in a regular polyhedron. First I solve a quadratically-constrained non-linear program, then I use Newton's method to find high precision solutions, convert them to algebraic solutions using integer relations, and finally check the solutions using calculations in extensions of $\mathbb{Q}$.

Since the tetrahedron is 3-dimensional, we can visualize the configuration by taking the intersection of the affine span of the tetrahedron with the cube. This gives combinatorially a prism over a hexagon and looks like this:

3-simplex in 4-cube

Click here for an animation

If the $4$-cube is given as $[0, 1]^4$, then the coordinates of the tetrahedron are $$ (a, 0, 0, 1) \\ (b, 1, 0, 0) \\ (0, c, 1, 0) \\ (1, d, 1, e) \\ $$ where $a,b,c,d$ and $e$ are some algebraic numbers of degree 8. in fact, we have $$\begin{align} a &= \text{ zero near }0.2417346828\text{ of }\\ &128x^8 - 128x^7 + 464x^6 - 296x^5 + 959x^4 - 1568x^3 + 958x^2 - 248x + 23\\ b &= \text{ zero near }0.5021530192\text{ of }\\ &128x^8 - 384x^7 + 208x^6 + 328x^5 - 361x^4 - 16x^3 + 170x^2 - 56x - 1\\ c &= \text{ zero near }0.09686099894\text{ of }\\ &128x^8 - 768x^7 + 2224x^6 - 3848x^5 + 4119x^4 - 2452x^3 + 542x^2 + 60x - 9\\ d &= \text{ zero near }0.6856472601\text{ of }\\ &512x^8 - 3072x^7 + 9408x^6 - 17632x^5 + 19388x^4 - 10480x^3 + 500x^2 + 2000x - 625\\ e &= \text{ zero near }0.8492045976\text{ of }\\ &8x^8 - 16x^7 + 16x^6 - 16x^5 + 6x^4 - 4x^3 + 6x^2 - 1\\ \end{align}$$ The edge length of this regular tetrahedron is the square root of the number $$\begin{align} s &= \text{ zero near }2.067817710\text{ of} \\ &64x^8 - 704x^7 + 2944x^6 - 5720x^5 + 4929x^4 - 1456x^3 + 608x^2 - 768x + 256\\ \end{align}.$$

From the case $(3,4)$, we can observe the following: In an optimal solution, there might not be a vertex of the simplex coincident with a vertex of the cube (as one might expect from examining the cases $(2,2)$ and $(3,3)$) For the $(3,4)$ configuration, three vertices of the tetrahedron lie on edges of the cube and one on a 2-face. As you guessed the solution is at least somewhat "not simple": All coordinates of the optimal solution lie in the number field with defining polynomial $y^8 - 2y^7 + y^6 + 2y^5 - 10y^4 + 6y^3 + 9y^2 - 6y + 1$, which is not Galois.

Curiously, the algebraic numbers satisfy the relation $a+c+e = b+d$.


Edit: Next let's look at the case 3-simplex in 7-cube: Here the situation is similar: If the $7$-cube is given as $[0, 1]^7$, then the coordinates are $$ (0, 0, 0, a, 1, 1, 1) \\ (b, 0, 0, 1, 0, 0, 0) \\ (0, 1, 1, 0, c, 0, 0) \\ (1, 1, 1, d, e, 1, 1) \\ $$ where $a,b,c,d$ and $e$ are some algebraic numbers of degree 8. in fact, we have $$\begin{align} a &= \text{ zero near }0.2386303477\text{ of }\\ &x^8 - 12x^7 + 34x^6 - 36x^5 + 57x^4 - 32x^3 - 8x^2 - 64x + 16\\ b &= \text{ zero near }0.7308933837\text{ of }\\ &x^8 + 4x^7 - 42x^6 + 52x^5 - 103x^4 + 384x^3 - 424x^2 + 576x - 320\\ c &= \text{ zero near }0.7613696523\text{ of }\\ &x^8 + 4x^7 - 22x^6 + 28x^5 + 37x^4 - 152x^3 + 164x^2 - 44\\ d &= \text{ zero near }0.8560025785\text{ of }\\ &4x^8 - 20x^6 + 32x^5 - 87x^4 + 16x^3 + 124x^2 - 96x + 20\\ e &= \text{ zero near }0.1439974215\text{ of }\\ &4x^8 - 32x^7 + 92x^6 - 136x^5 + 53x^4 + 188x^3 - 218x^2 + 76x - 7\\ \end{align}$$ The edge length of this regular tetrahedron is the square root of the number $$\begin{align} s &= \text{ zero near }4.113888886\text{ of }\\ &x^8 - 184x^7 + 6328x^6 - 79264x^5 + 443152x^4 - 1091328x^3 + 683520x^2 + 634880x + 1183744. \end{align}$$ Curiously, the algebraic numbers satisfy the relation $a+c =d+e$


For all other cases I tried, the following patters seems to emerge: if $s_{(3,d)}$ is the maximal side length of the regular 3-simplex inside the $d$-cube, then we have $$s^2_{(3, d)}\geq\begin{cases}\frac{2}{3}d &\text{ if }d=0 \pmod{3}\\ \frac{2}{3}d - \frac{13}{24} = \frac{2}{3}(d-1)+\frac{1}{8}&\text{ if }d=1 \pmod{3} \text{ and }d\geq 10\\ \frac{2}{3} d- \frac{5}{6} = \frac{2}{3}(d-2)+\frac{1}{2}&\text{ if }d=2 \pmod{3} \end{cases}$$

  • For $d=3n$, coordinates are given by ($n\geq 1$) $$ n\cdot(0,0,0)\\ n\cdot(0,1,1)\\ n\cdot (1,0,1)\\ n\cdot (1,1,0),$$ where '$n\cdot$' signifies $n$-fold concatenation.
  • For $d=3n+1$, coordinates are given by ($n\geq 3$) $$ ( 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , \tfrac{1}{4} ) + (n-3)\cdot(0,0,0)\\ ( 0 , 1 , \tfrac{3}{4} , 0 , 1 , 0 , 1 , 1 , 1 , 1) + (n-3)\cdot(0,1,1)\\ ( 1 , \tfrac{1}{4}, 0, 1 , 1 , 1 , 1 , 1 , 0 , 0) +(n-3)\cdot(1,0,1)\\ ( 1 , 1 , 1 , 1 , 0 , 1 , 0 , 0 , \tfrac{3}{4} , 1 ,)+(n-3)\cdot(1,1,0),$$ where '$+$' signifies concatenation.
  • For $d=3n+2$, coordinates are given by ($n\geq 1$) $$ (0,0,\tfrac{1}{2}, 1, 1) + (n-1)\cdot(0,0,0)\\ (0,1,1,0,\tfrac{1}{2})+(n-1)\cdot(0,1,1)\\ (1,\tfrac{1}{2}, 1, 1, 0)+ (n-1)\cdot (1,0,1)\\ (\tfrac{1}{2},0,0,0,0) +(n-1)\cdot (1,1,0).$$

I conjecture that these values together with the values above for $s_{(3,4)}$ and $s_{(3,7)}$, are indeed the optimal values. It looks like there is again a $d\pmod{3}$ splitting of the result, and not a splitting $d\pmod{4}$, which is what you expected to see. Also the factor $\frac{3}{2}$ stays prominent.

Perhaps this would make a good new question for the Amer Math Monthly, after the case $k=2$ was solved.

$\endgroup$
  • $\begingroup$ Beautiful! And nice animation. Can you add something about how you found this solution? $\endgroup$ – Joseph O'Rourke Aug 9 '17 at 12:22
  • $\begingroup$ @JosephO'Rourke : Added some more info. some more cases are also probably doable this way. $\endgroup$ – Moritz Firsching Aug 9 '17 at 21:13
  • $\begingroup$ Those are compelling conjectures! $\endgroup$ – Joseph O'Rourke Aug 25 '17 at 11:44
5
$\begingroup$

There are a lot of results on this (and related questions in the paper by Hudelson, Klee, and Larman (1996). They are concerned primarily with the largest simplex in the cube (without assumption of regularity), but there are a number of results in the paper on when such a simplex is, indeed, regular. The paper also shows many connections to Hadamard matrices.

$\endgroup$
  • 1
    $\begingroup$ Full reference: Hudelson, Matthew, Victor Klee, and David Larman. "Largest $j$-simplices in $d$-cubes: some relatives of the Hadamard maximum determinant problem." Linear algebra and its applications. 241 (1996): 519-598. $\endgroup$ – Joseph O'Rourke Apr 22 '15 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.