/^creative ^commor
ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 14 (2018) 25-38
https://doi.org/10.26493/1855-3974.1022.a6a
(Also available at http://amc-journal.eu)
Right quadruple convexity*
Dandan Li, Liping Yuan t
College of Mathematics and Information Science, Hebei Normal University, 050024 Shijiazhuang, People's Republic of China
Tudor Zamfirescu
Fachbereich Mathematik, TU Dortmund, 44221 Dortmund, Germany Institute of Mathematics "Simion Stoilow", Roumanian Academy, Bucharest, Roumania
Received 31 January 2016, accepted 16 November 2016, published online 5 April 2017 Abstract
A set of four points w, x,y,z G IRd (always d > 2) form a rectangular quadruple if their convex hull is a non-degenerate rectangle. The set M is called rq-convex if for every pair of its points we can find another pair in M, such that the four points form a rectangular quadruple. In this paper we start the investigation of rq-convexity in Euclidean spaces.
Keywords: rq-convex sets, parallelotopes, finite sets, Platonic solids. Math. Subj. Class.: 53C45, 53C22
1 Introduction
Let F be a family of sets in IRd. A set M c IRd is called F-convex if for any pair of distinct points x, y G M there is a set F G F such that x, y G F and F c M.
The third author proposed at the 1974 meeting on Convexity in Oberwolfach the investigation of this very general kind of convexity. Usual convexity, affine linearity, arc-wise connectedness, polygonal connectedness, are just some examples of F-convexity (for suitably chosen families F).
Blind, Valette and the third author [1], and also Boroczky Jr [2], investigated rectangular convexity. Magazanik and Perles dealt with staircase connectedness [5]. The third author
*The work is supported by NNSF of China (11471095), NSF of Hebei Province (A2013205189), Program for Excellent Talents in University, Hebei Province (GCC2014043) and the High-end Foreign Experts Recruitment Program of People's Republic of China. t Corresponding author.
E-mail address: ddli68@163.com (Dandan Li), lpyuan@hebtu.edu.cn (Liping Yuan), tuzamfirescu@gmail.com (Tudor Zamfirescu)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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ArsMath. Contemp. 14 (2018) 25-38
studied right convexity [8]; then the second and the third author generalized the latter type of convexity and investigated the right triple convexity (see [7] and [6]). All these concepts are particular cases of F-convexity. The rectangular convexity is obtained if F is the family of all non-degenerate rectangles in IRd.
In this paper we present a discretization of rectangular convexity, the right quadruple convexity, which constitutes a generalization of rectangular convexity. As usual, for M c IRd, bdM denotes its boundary, intM its interior, diamM = supx£M \\x — y\\ its diameter, and convM its convex hull. A set of four points w, x, y, z G IRd (always d > 2) form a rectangular quadruple if conv{w, x, y, z} is a non-degenerate rectangle. Let R be the family of all rectangular quadruples. Here, we shall choose F to be this family R.
Let M c IRd. A pair of points x, y G M is said to enjoy the rq-property in M if there exists another pair of points z, w G M, such that {w, x, y, z} is a rectangular quadruple. The set M is called rq-convex, if every pair of its points enjoys the rq-property in M. This property is the right quadruple convexity.
Let A c IRd. We call A* an rq-convex completion of A, if A* is rq-convex, A* D A and card(A*\A) is minimal (but possibly infinite). Let y(A) = card(A*\A), which is called the rq-convex completion number of A, in case A is finite. For finite n, let y (n) = sup{y(A) : cardA = n}.
For distinct x, y G IRd, let xy be the line through x, y, xy the line-segment from x to y, Hxy the hyperplane through x orthogonal to xy, and Cxy the hypersphere of diameter xy. For Si, S2 C IRd, let d(Si, S2) = inf{d(x, y) | x G Si, y G S2} denote the distance between S1 and S2. The d-dimensional unit ball (centred at 0) is denoted by Bd (d > 2). Let us remark that every open set in IRd is rq-convex.
2 Not simply connected rq-convex sets
In IR2, all compact rectangularly convex sets are conjectured to be extremely circular and symmetric. A planar convex set is extremely circular if its set of extreme points lies on a circle. Analogously, it is reasonable to conjecture that all compact rq-convex sets have an extremely circular and symmetric convex hull. Consequently, when investigating a compact connected rq-convexset M, we may reasonably start by assuming that convM is extremely circular and symmetric. We shall now take bdconvM to be a circle. If M is simply connected we get the disc. So, assume (convM) \ M = 0.
Theorem 2.1. If convM is a disc and (convM) \ M lies in a circular disc of radius r at distance at least (%/3 — 1)r from bdconvM, then M is rq-convex.
This theorem gives a useful sufficient condition for the rq-convexity of a set M which is not simply connected, regardless the shape of (convM) \ M. Notice that it allows both M and its complement to have arbitrarily many components.
Proof. Let Q be a square circumscribed to the disc D of radius r including (convM) \ M. We may suppose that the origin 0 is the centre of D, so D = rB2.
We have Q c convM. Indeed, Q is obviously included in the disc concentric with D of radius %/2r, which in turn must be included in convM, since the distance from D to bdconvM is at least (V3 — 1)r > (V2 — 1)r.
Let x, y G M. We verify the rq-convexity of M at these two points.
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Figure 1: x G D, y G D.
Case 1: x,y G D. Choose Q to have a side s parallel to xy. Then x,y and their orthogonal projections on s are vertices of a rectangle. Moreover, these latter vertices lie
in (convM) \ D.
Case 2: x G D, y G D. Let r2, r3 be the two circles concentric with bdD, of radii ri/2, rV3. Let a be the point of xy n r2 such that x G ay. In case y G ax, consider the rectangle xyyx, such that xy separates 0 from x, y (if 0 G xy) and xy is tangent to bdD. See Figure 1. Let o be the orthogonal projection of 0 onto yy. We have
llyll2 = yy - 5||2 + l|5||2 = r2 + ||o||2 < r2 + ||y||2 < r2 + ||a||2 = 3r2.
Hence, y, and of course x too, lie in M, and the rq-property is satisfied in x, y.
We reconsider the case y = a. Choose b,cG bdD such that ab0c be a square and b, 0 be not separated by xy. Let jx', y'} = Cax n bdD, where x' is closer than y' from b. Let x" be the point of Cax diametrally opposite to x'. For any position of x,
Z0ax' < Z0ab.
Hence, Z0ax'' > Z0ac, whence x'' G D. (This confirms the rq-property in x, a.) It follows that Zx'oy' < n, where o is the centre of Cax and the angle is measured towards 0.
In case a G xy, consider the circle Cxy with centre o', which cuts bdD in x*, y* (the former being closer than the latter from b). We have
Zxo'x* < Zxo'x' < Zxox'
and
Zxo'y* < Zxo'y' < Zxoy',
whence
Zx o y < Zx oy < n,
where both angles are taken towards 0. Consequently, the points x+, y+ G Cxy diametrally opposite to x*, y* (respectively) lie outside D. They also lie in different half-circles
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Ars Math. Contemp. 14 (2018) 97-116
determined by x, y on Cxy. Of these two half-circles, at least one is contained in the disc
convM.
So, either {x,y, x*,x+} c M or {x,y, y*,y+} c M, and the rq-property is again satisfied at x, y.
Case 3: x, y G D. Besides the trivial cases x, y G intM and x, y G bdconvM, we only have the simple situation x G intM, y G bdconvM. In that situation, the circle Cxy has necessarily two opposite arcs in M starting at x, respectively y. This proves the rq-property at x, y.	□
Conjecture 2.2. Each simply connected rq-convex set in IR2 is convex. 3 Unbounded rq-convex sets
An infinite family K of closed convex sets is said to be uniformly bounded below if, for some A > 0, each of the sets contains a translate of the disc AB2.
Theorem 3.1. Let K be a family ofpairwise disjoint closed convex sets in IRd. If K is finite or uniformly bounded below, then the closure of the complement of[j K is rq-convex.
Proof. We may assume that all sets in K possess interior points, because the case of empty interior is irrelevant. Let M be the closure of IRd \ |J K, and choose x, y G M. Clearly, the only interesting case is when x, y G bdM.
The condition of uniform boundedness below for infinite K guarantees that x G bdM only if x is a boundary point of some member of K.
Let M' be the intersection of M with an arbitrary 2-dimensional plane n 3 x, y. For some Kx, Ky G K, x G bdKx, y G bdKy. Consider the supporting hyperplane Hx of Kx at x, the line H'x = Hx n n, and analogously Hy and Hy. If H'x, Hy are not orthogonal to xy, there are six different situations in the neighbourhood of x and y, depicted in Figure 2 (subfigures (a)-(f)). (In the figure only the generic case is depicted, when Kx n n and Ky n n are not degenerate; but the proof works in all cases.)
In the situations of Figure 2 (subfigures (a),(c) and (e)), the circle Cxy has two opposite arcs inside M, so M has the rq-property at x, y. In the cases of Figure 2 (subfigures (b),(d) and (f)), a thin rectangle with xy as a side has its short sides in M, so the rq-property is again verified.
The only remaining case is that of at least one of the lines Hx, H'y, say the first, being orthogonal to xy. In this case, there is a short line-segment x(x + v) c M in any direction v orthogonal to xy. Now, if y + v G M, we found the right quadruple {x, y, y + v, x + v}.
If y + v G M, i.e. y + v G intKy, then y - v G Ky. Thus, {x, y, y - v, x - v} is a suitable rectangular quadruple.	□
We can drop the convexity condition if the considered sets are bounded.
Theorem 3.2. The complement of any bounded set in IRd is rq-convex. The easy proof is left to the reader.
A plane tiling T is a countable family {Ti,T2,...} of closed sets with non-empty interiors, which cover the plane without gaps or overlaps. Every closed set T G T is called a tile of T. We consider the special case in which each tile is a polygon. If the corners and sides of a polygon coincide with the vertices and edges of the tiling, we call the tiling
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b	c
g
Figure 2: Illustration for the proof of Theorem 3.1.
edge-to-edge. A so-called type describes the neighbourhood of any vertex of the tiling. If, for example, in some cyclic order around a vertex there are a triangle, then another triangle, then a square, next a third triangle, and last another square, then its type is (32.4.3.4). We consider plane edge-to-edge tilings in which all tiles are regular polygons, and all vertices are of the same type. Thus, the vertex-type defines our tiling up to similarity.
There exist precisely eleven such tilings [3]. These are (36), (34.6), (33.42), (32.4.3.4), (3.4.6.4), (3.6.3.6), (3.122), (44), (4.6.12), (4.82),and (63). They are called Archimedean tilings.
Theorem 3.3. The Archimedean tilings (44), (36), (63), (3.6.3.6), (34.6), (3.3.4.3.4), (4.8.8) have rq-convex vertex sets.
Theorem 3.4. The vertex sets of the Archimedean tilings (3.3.3.4.4), (3.4.6.4), (4.6.12), (3.12.12) are not rq-convex.
The proofs of Theorems 3.3 and 3.4 are also left to the reader. 4 rq-convex skeleta of parallelotopes
As already remarked in [1], for d > 3, there is not even any conjectured characterization of rectangularly convex sets in IRd. Among the sets mentioned in [1] as rectangularly convex we find the cylinder K x [0,1] with a (d - 1)-dimensional compact convex set K as basis. In particular, any right parallelotope, i.e. the cartesian product of d pairwise orthogonal line-segments, is rectangularly convex and, a fortiori, rq-convex.
Theorem 4.1. The 1-skeleton of any right parallelotope is rq-convex.
Proof. Let P = Ii x I2 x ... x Id be our parallelotope, where I = [0, a] (i = 1,..., d). We verify the rq-property at the points x, y belonging to the 1-skeleton of P.
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Ars Math. Contemp. 14 (2018) 97-116
Case 1: x, y belong to parallel edges of P. We have without loss of generality
x = (xi, 0,..., 0),
y = (yi,ai,..., a®, 0,..., 0).
Then we choose as third and fourth point
u = (yi, 0,..., 0), v = (xi, ai, .. ., a®, 0,.. ., 0).
Indeed, xuvy is a rectangle.
Case 2: x, y belong to two edges of P having a common endpoint. Say without loss of generality that
x = ( xi , 0, . . . , 0) , y = (0, y2, 0,..., 0).
Then take
u = (xi, 0, a3, 0, .. ., 0), v = (0, y2,a3, 0,. .., 0),
completing the vertex set of a rectangle xuvy.
Case 3: x, y belong to two non-parallel disjoint edges of P. If
x = (xi, 0,..., 0), y = (0, .. ., 0, y®, a®+i, .. ., ad),
then we choose
u = (xi, 0, ..., 0, a®+i, .. ., ad), v = (0,..., 0,y®, 0,..., 0),
and again we get the vertices of a rectangle.
Case 4: x, y belong to the same edge of P. This is immediate.	□
Contrary to the case of an arbitrary cylinder, the following is true.
Theorem 4.2. The boundary of any right parallelotope is rq-convex.
Proof. Take x, y on the boundary of the parallelotope P. We show that they have the rq-property.
If x, y belong to the same facet F, choose their orthogonal projections onto the opposite facet F'; the four points are vertices of a rectangle.
If x € F, y € F', choose the projection x' of x onto F' and the projection y' of y onto F; we get the rectangle xx'yy'.
If x, y belong to two adjacent facets F, F*, respectively, take the orthogonal projections x* and y* of x and y (respectively) onto F n F*. We complete the rectangles xx*y*y and yy*x*x. Then, clearly, {x, x, y, y} C bdP is a rectangular quadruple.	□
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Theorem 4.3. Not every convex cylinder has an rq-convex boundary.
Proof. Take a cylinder Z = E x [0,1] C IR3, where E c IR2 is convex and bdE is a long ellipse. Choose x on the long axis of bdE x {1}, close to one of its endpoints {e} = {ep} x {1}, and let {e'} = {e^}x{1} be the other endpoint. Let {y£} = {e^}x{e}, where e > 0. See Figure 3.
Figure 3: A convex cylinder without rq-convex boundary.
The plane Hxye cuts (bdE) x [0,1] along an arc a£ of an ellipse. Let f (e) = d({x}, a£). The function f is increasing, and f (0) > 0.
The plane ByeX cuts bdZ along a closed curve (reduced to a single point if e = 0), of diameter g(e). This function g is also increasing, and g(0) = 0. Therefore, for e > 0 small enough,
g(e) < f (0) < f(e).
Choose y = y£. The above inequalities show that there is no rectangle xyy'x' with x', y' G bdZ.
Consider now the sphere Cxy. The set Cxy n Z has four components: a component Zi containing x, another one Z2 containing y, a third Z3 containing the point e* diametrically opposite to e' in Cxy, and a fourth, {e'}. It is easily seen that the only pairs of diametrically opposite points in Z1 U Z2 U Z3 U {e'} are (x, y) and (e', e*). But e* G intZ, so bdZ is not rq-convex.	□
5 rq-convexity of finite sets
In these last two sections, we shall use the following notation. For x, y G IRd, we set ^^xy Hxy U Hyx U Cxy. Let A be the family of all finite point sets in IR .
Theorem 5.1. For any set A G A with cardA = n > 3, we have y(A) < n2 — 2n.
Proof. If A is included in a line L, consider a line L' parallel to L and the orthogonal projection A' of A onto L'. Then obviously AuA' is rq-convex and cardA' = n < n2 —2n, since n > 3.
If A is not included in any line, let A = {a1, a2,..., a„}, and assume that a1a2 is a side of the polygon convA. Obviously there are at most n — 2 lines L1,..., Ln-2 passing through the remaining points of A and parallel to a1a2. Also, there are at most n lines L1,..., Ln passing through the points of A orthogonally onto L0 = ala^.
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Ars Math. Contemp. 14 (2018) 97-116
The set
A' = U (Li n Lj)
0<i<n-2;1<j<n
is obviously rq-convex and has at most n(n - 1) points, whence y(A) < n2 - 2n. □
Thus, for any n > 3, Y(n) < n2 — 2n. In particular, 7(3) = 3.
Theorem 5.2. There are precisely two kinds of 6-point rq-convex sets in A, shown in Figure 4.
Figure 4: 6-point rq-convex sets.
Proof. Let F = {a, b, c, d, e, f} be a 6-point rq-convex set. We assume without loss of generality that {a, b, c, d} G R, where Zabc = | .By the definition of rq-convexity, e, f must meet one of the following seven conditions.
Ci.	{{e, f, a, b}, {e,f,c,d}} CR;
C2.	{{e, f, a, c}, {e,f,b,d}} CR;
C3.	{{e,f,a,d}, {e,f,b,c}} CR;
C4.	{{e, f, a, b}, {e, f, a, c}, {e, f, a, d}} C R;
C5.	{{e, f, b, a}, {e, f, b, c}, {e, f, b, d}} C R;
C6. {{e f, c, a}, {e, fc, b}, {e, f, c d}} C R;
C7. {{e, f, d, a}, {e, f, d, b}, {e, f, d, c}} C R.
Clearly, C1 and C3 generate the same kind of set F, and so do C4, C5, C6 and C7.
Case 1: e, f satisfy C1. By the definition of rq-convexity, e, f G Wab n Wcd, and so e, f G (Hab U Hba) \ {a, b, c, d} and ef, ab are parallel. Without loss of generality, we may suppose e G Hab, f G Hba, which leads to the three solutions depicted in Figure 5, all of them providing a 6-point set of the first type.
ex a	e2 d e3
fx b	f c f$
Figure 5: e, f satisfy C1.
Case 2: e, f satisfy C2. By the definition of rq-convexity, we have e, f G Wac n Wbd, so e, f are antipodal points of Cac; see Figure 6.
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Figure 6: e, f satisfy C2.
Case 3: e, f satisfy C4. By the definition of rq-convexity, (e, f) G Wab n Wac n Wad.
But Wab n Wac n Wad = {a, b, d}, so we obtain no solution in this case. See Figure 7.
□
Figure 7: e, f satisfy C3. Theorem 5.3. There are precisely three kinds of 8-point rq-convex sets, shown in Figure 8.
Figure 8: 8-point rq-convex sets.
The proof is ten pages long, so we decided not to include it into the paper. It is a case-by-case examination, treating separately those sets which contain a 6-point rq-convex subset and those which do not. It can be read in [4].
Theorem 5.4. The smallest odd cardinality of an rq-convex set in IR2 is 9.
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Ars Math. Contemp. 14 (2018) 97-116
Proof. It is obvious that every rq-convex set contains a rectangular quadruple and quickly seen that no fifth point can be added to a rectangular quadruple keeping rq-convexity. Similarly, knowing what a 6-point rq-convex set looks like (Theorem 5.2), it is an easy exercise to establish that there is no 7-point rq-convex set containing a 6-point rq-convex set.
Next, we will consider the case that a 7-point rq-convex set does not contain any 6-point rq-convex subset. Let F = {a, b, c, d, e, f, g} be such a set. Suppose a, c realise the diameter of F. Since F is rq-convex, there is another pair of antipodal points of Cac in F, say {b, d}. Hence {a, b, c, d} G R, and put conv{a, b, c, d} = R.
For the set of points {x, y} C {e, f, g}, if there exist z, w G {a, b, c, d} such that {w, x, y, z} G R, then we say that {x, y} is rq-good. Next we will prove that for any two points x, y G {e, f, g}, {x, y} is not rq-good. Suppose {e, f} is rq-good. Then there exist z, w G {a, b, c, d}, such that {e, f, z, w} G R.
Case 1: zw is a diagonal of R. Without loss of generality, we assume zw = ac, so e, f G Wac. Since {a, c} realise the diameter of F, e, f are antipodal on Cac. But so we obtain a 6-point rq-convex subset of F, which contradicts our assumption about F.
Case 2: zw is an edge of R. We assume without loss of generality that zw = ab. Then e, f G Wab. Clearly, e, f must be antipodal points of Cab. Take a diameter a0b0 of Cab orthogonal to ab, such that ab separates a0 from cd. We may suppose that e belongs to the (smaller) arc of Cab from a to a0; see Figure 9. If ||a-b|| < ||b-c||,then ||c-e|| > ||c-a||,
Figure 9: zw is an edge of R.
which is impossible. So, ||a - b|| > ||b - c||.
As F is rq-convex, {e, d} enjoys the rq-property and there exist p, q G {a, b, c, f, g} such thatp, q G Wed. Clearly, a, b, c G Hde. Since d G Cab, we have f G Hde. Further, we easily verify that a, b, c, f G Hed U Hde. It follows that p, q are two antipodal points of Ced.
Since Zdae > f, Zdce < f, we get a, c G Ced. Hence, p = b G Ced or p = f G Ced, and q = g.
(i)	p = b G Ced. We only can choose g such that b, g are antipodal points of Ced. As Zebf = n, b, f, d are collinear. But then we get a 6-point rq-convex set {e, b, a, f, g, d} C F, contradicting our choice of F; see Figure 10.
(ii)	p = f G Ced. Now, f, g are antipodal points of Ced. Hence, ef dg is a rectangle. The points g and b are separated by ad, or a G dg if e = a0. Also,
||d - g|| = ||e - f | = ||a - b|| > |b - c| = ||a - d|.
It follows that ||c - g| > ||c - a|, and a contradiction is obtained. See Figure 11. Hence,
L. Yuan et al.: Right quadruple convexity	35
Figure 10: p = b e Ced.
Figure 11: p = f e Ced.
{e, f} is not rq-good.
Since F is rq-convex, we must have {e, f, g, a}, {e, f, g, b}, {e, f, g, c}, {e, f, g, d} e R, which is true only for a = b = c = d, which is impossible. Thus, there is no 7-point rq-convex set.
On the other hand, a 9-point rq-convex set is easily produced: just take the intersection L n L', where L is the union of three horizontal lines and L' the union of three vertical lines.	□
Consider now the square lattice ZZ2, and the usual norm
ll(x,y)||m = max{|x|, |y|}, defining in ZZ2 the discs of radius n e ZZ
Q(n) = {(x,y): ||(x,y)|m < n},
centred at the origin 0. The subset Q(n) \ Q(n - 1) will be called boundary of Q(n), and Q(n - 1) its interior. Obviously, Q(n) is rq-convex, for any n > 1, and so is its boundary too. Does Q(n) remain rq-convex if one deletes parts of its interior (but not all of it)?
Theorem 5.5. The set Q(n) \ {0} is rq-convex.
Proof. For any pair of points x = (xi,x2), y = (yi,y2) in Q(n) \ {0}, consider the points (x1, y2), (y1, x2) e ZZ2. If none of them is 0, the rq-propertyis verified at (x, y), as
(xi,y2), (yi,x2) e Q(n).
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ArsMath. Contemp. 14 (2018) 25-38
Otherwise, assume without loss of generality that x1 = y2 =0. We can also assume that both x2, y1 are positive, the other cases being symmetrical. Consider the points x' =
(—x2, x2 - y1) and y' = (y1 - x2, -y1). Then x', y' G Q(n) \ {0} and x, y, y', x' are the vertices of a square.	□
Perhaps removing several layers of the boundary, thereby giving a set Q(n) \ Q(m) for m < n — 1, will provide an rq-convex set?
Theorem 5.6. The set Q(n) \ Q(n — 2) is not rq-convex, for any n > 3.
Proof. Assume first that n > 3. Consider the points x = (n, 2 — n) and y = (n — 3, n — 1). The point (n — 3,2 — n) does not belong to Q(n) \ Q(n — 2). The line Hyx meets Q(n) again at y' = (—n, n — 4). But the fourth vertex x' = (3 — n, —n — 1) of the square xyy'x' lies outside Q(n), whence Q(n) \ Q(n — 2) is not rq-convex.
Now, consider the set Q(3) \ Q(1). In this case take the points x = (3, —1), y = ( —1, 2). As ( —1, —1) G Q(3) \ Q(1) and Hyx n Q(3) \ Q(1) = 0, the result is proven. □
Figure 12: rq-convex proper subsets of Q(n).
It seems that Q(n) \ {0} is the only rq-convex proper subset of Q(n) containing the boundary of Q(n) but different from it. However, this is not proven. Other proper subsets of Q(n) which are rq-convex abound. For some examples, see Figure 12, where the solid black dots form rq-convex proper subsets of Q(n).
6 rq-convexity of the vertex sets of Platonic solids
Due to their symmetry, the vertex sets of the cube, regular octahedron, regular dodecahedron, and regular icosahedron are all rq-convex. Among the Platonic solids, only the
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regular tetrahedron lacks this property. But what is the rq-convex completion number of the vertex set of the regular tetrahedron?
Theorem 6.1. The rq-convex completion number of the vertex set of the regular tetrahedron is 3.
Proof. Let T = {a, b, c, d} denote the vertex set of a regular tetrahedron in IR3. Obviously, for any x, y G T, we have T n Wxy = {x, y}. Also, it is easily seen that there is no 5-point rq-convex set containing T. Suppose there is a 6-point rq-convex set {a, b, c, d, x, y}. The only suitable pair of points x, y G Wab n Wcd is obtained when {x, y} = (Hab n Ccd) U (Hba n Ccd). But then b, c do not enjoy the rq-property in {a, b, c, d, x, y}. Hence Y(T) > 3.
Next, we only need to prove y(T) < 3. Let ai, bi, ci denote the midpoints of ad, bd, cd, respectively. See Figure 13. The line La 3 a1 parallel to bc and the anal-
d
ogous lines Lb and Lc are coplanar. Put
{a'} = Lb n Lc, {b'} = Lc n La, {c'} = La n Lb.
Obviously, ab'dc', bc'da', ca'db' are squares, while a'b'ab, b'c'bc, c'a'ca are rectangles. Thus, {a, b, c, d, a', b', c'} is a 7-point rq-convex set, and y(T) = 3.	□
Theorem 6.1 reveals the existence of 7-point rq-convex sets in IR3, in contrast with the inexistence of such sets in IR2. What happens in higher dimensions?
References
[1]	R. Blind, G. Valette and T. Zamfirescu, Rectangular convexity, Geom. Dedicata 9 (1980), 317— 327, doi:10.1007/bf00181177.
[2]	K. Boroczky, Jr., Rectangular convexity of convex domains of constant width, Geom. Dedicata 34 (1990), 13-18, doi:10.1007/bf00150684.
[3]	B. Grunbaum and G. C. Shephard, Tilings and Patterns, Dover Publications, New York, 2nd edition, 2016.
[4]	D. Li, L. Yuan and T. Zamfirescu, 8-point rq-convex sets, 2016, http://sxxy.hebtu. edu.cn/upload/fckeditor/2015-10/30112750.pdf.
[5]	E. Magazanik and M. A. Perles, Staircase connected sets, Discrete Comput. Geom. 37 (2007), 587-599, doi:10.1007/s00454-007-1308-9.
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Ars Math. Contemp. 14 (2018) 97-116
[6]	L. Yuan and T. Zamfirescu, Right triple convex completion, J. Convex Anal. 22 (2015), 291-301, http://www.heldermann.de/JCA/JCA2 2/JCA221/jca22014.htm.
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