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Mathematically-oriented sculptors have created three dimensional representations of higher dimensional polytopes; for example George Hart. The 5-dimensional polytope $CQHRL_5$ is described in this question and is diagrammed below (where same-labeled vertices are actually the same vertex).

CQHRL(5) diagram

A realization of $CQHRL_5$ has the following vertices and facets, where the facets are given by f ◦ x = 1:

Vertices:

-5: (2, 4, 5, 8, 11)

-4: (2, 4, 6, 8, 11)

-3: (2, 4, 6, 9, 12)

-2: (2, 5, 6, 9, 12)

-1: (3, 6, 8, 12, 16)

0: (1, 2, 3, 4, 5)

1: (1, 3, 4, 6, 8)

2: (2, 3, 4, 6, 8)

3: (2, 4, 5, 7, 10)

4: (2, 4, 5, 8, 10)

Facets:

1: (0, -1, 1, 0, 0)

2: (0, 1, 0, 1, -1)

3: (0, 1, 1, -1, 0)

4: (-1, 0, 0, -1, 1)

5: (-1, 0, -1, 1, 0)

6: (-1, 0, 1, 1, -1)

7: (1, 1, -1, 0, 0)

8: (1, 1, 1, 0, -1)

9: (1, -1, 0, -1, 1)

10: (1, -1, -1, 1, 0)

11: (1, -1, 1, 1, -1)

12: (1, -1, -1, 1, 0)

$CQHRL_5$ has five quadrilateral faces, each of which intersect each of the other four quadrilateral faces at a vertex. Is it possible to construct a three dimensional model of $CQHRL_5$ including only the quadrilateral faces (i.e., preserving the ten vertices of intersection)? Is it possible if some or all of the quadrilaterals are non-convex?

Note: A related question of whether a 3-d representation of $CQHRL_5$ could be constructed from 5 (genus 0) curved surfaces each intersecting each other at one vertex each, is answered in the affirmative. If you have five crosses made of string with appropriately numbered ends, it is a straightforward job to connect the corresponding ends together. So, the question asked above is a geometry question, not a topology question.

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Begin with the tetrahedron with vertices: $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

Now, take the midpoints of all 6 edges, which subtend four central triangles; one on each face of the tetrahedron: $(1/2, 0,0)$, $(0,1/2,0)$, $(0,0,1/2)$, $(1/2,1/2,0)$, $(1/2,0,1/2)$, $(0,1/2,1/2)$. The four central triangles lie in the planes $x = 0$, $y = 0$, $z = 0$, and $x + y + z = 1$.

These midpoints will comprise six of the ten vertices of the 3-d model. We will select four coplanar points, each lying within the interior of one of the three outer triangular regions of each tetrahedron face. We will then take the convex hull of each central triangle and the selected point on that tetrahedron face to get four of the convex quadrilaterals. The fifth quadrilateral will be the convex hull of the selected coplanar points.

We select the points:

$A$:$(5/8, 1/12, 7/24)$, $B$:$(3/5, 0, 1/10)$, $C$:$(0, 3/10, 1/10)$, $D$:$(3/10, 1/10, 0)$;

all lying in the plane $x + 2y – z = ½$. Note that $12/35$ * $A$ + $23/35$ * $D$ = $24/35$ * $B$ + $11/35$ * $C$, so $ABDC$ is a convex quadrilateral. We can now assign vertex labels that are consistent with the $CQHRL_5$ diagram:

Vertex Coordinates

$-5$: $(3/10, 1/10, 0)$

$-4$: $(1/2, 1/2, 0)$

$-3$: $(0, 1/2, 0)$

$-2$: $(0, 0, 1/2)$

$-1$: $(0, 1/2, 1/2)$

$0$: $(5/8, 1/12, 7/24)$

$1$: $(0, 3/10, 1/10)$

$2$: $(3/5, 0, 1/10)$

$3$: $(1/2, 0, 0)$

$4$: $(1/2, 0, 1/2)$

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