INTERNATIONAL JOURNAL OF Vol. 7 (2018), No. 2, 12 - 24

SOME DISTANCE PROPERTIES OF TWO SPACES INDUCED BY DUAL CONVEX POLYHEDRA ZEYNEP ÇOLAK and ZEYNEP CAN

Abstract.Dual convex spaces have an important place in mathematics. In geometry, some polyhedra are associated into pairs called duals, where the vertices of one correspond to the faces of the other. Arranging regular polyhedra into dual pairs needs some simple operations but it is not that easy to do this for semi-regular polyhedra. In further studies it has seen that there are some relations between polyhedra and metrics. For example, it has been shown that , and deltoidal are maximum, taxicab and Chinese Checker’sunit , respectively. In this paper some distance formulae of two spaces which are induced by two dual convex polyhedra has been researched by a new method di¤erent from [1] and [8], and by this method distance formulae in Minkowski can be generalized. This polyhedra are which is an and rhombic a which is the dual of cuboc- tahedron.

1. Introduction

Molecules, galaxies, sculpture, viruses, crystals, architecture and more, polyhedra are always in nature and art. It is at the same time hands- on, mind-turned-on introduction to one of the oldest and most fascinating branches of mathematics. Although it is so ancient it is not that easy to describe what a is. To describe a polyhedron the terms faces, edges and vertices would be used. Polygonal parts of the polyhedron are called faces, line segments that along faces come together are called edges and points where several edges and faces come together are called vertex. So, simply, a polyhedron can be de…ned as a closed, three-dimensional …gure which faces are polygons. For example, any closed box is a polyhedron, for its faces are rectangles. ————————————– Keywords and phrases: Polyhedra, Dual polyhedra, Convex polyhe- dra, Metric geometry, , Cuboctahedron, Distance Formulae. (2010)Mathematics Subject Classi…cation: 51P99, 60A99 Received: 27.07.2018. In revised form: 30.09.2018. Accepted: 20.08.2018. Some distance properties of two spaces induced by dual convex polyhedra 13

A polyhedron is called convex if a joining any two points in the interior of the …gure lies completely within the …gure. There are some classi…cations of convex polyhedra. The regular polyhedra (for each of these polyhedra, every face, every vertex and every edge is like every other) are also known as the "Platonic Solids" because the Greek philoso- pher Plato (427-347 B.C.E.) immortalized them in his dialogue Timaeus. Platonic solids are only …ve. An other class of polyhedra is Archimedean solids because they were …rst discovered by Archimedes (287-212 B.C.E.). Vertices of the Archimedean polyhedra are all alike, but their faces, which are regular polygons, are of two or more di¤erent kinds. For this reason they are often called semiregular [13]. And irregular polyhedra are de…ned by polygons that are composed of elements that are not all equal and they are called Catalan solids or Archimedean duals. There are thirteen Cata- lan solids just like Archimedean solids. They are named for the Belgian mathematician, Eugéne Catalan, who …rst described them in 1865. Two called dual if either of a pair of polyhedra in which the faces of one are equivalent to the vertices of the other. Starting with any regular polyhedra, the dual can be constructed in the following manner: (1) Place a point in the center of each face of the original polyhedron; (2) Connect each new point with the new points of its neighboring faces; (3) Erase the original polyhedron. This is an operation "of order 2" meaning that taking the dual of the dual of x gives back the original x. For example, take the dual of the octahedron and see that it is a cube and take the dual of the cube and see that is an octahedron. The relative sizes of the two dual polyhedra can be adjusted as shown here, so that their edges are the same distance from their common center, and so cross through each other. In geometry, polyhedra are associated into pairs called duals, where the vertices of one correspond to the faces of the other. The dual of the dual is the original polyhedron. The dual of a polyhedron with equivalent vertices is one with equivalent faces, and of one with equivalent edges is another with equivalent edges. So the regular polyhedra — the Platonic solids and Kepler-Poinsot polyhedra, — are arranged into dual pairs. But the same simple operations are not valid for semi-regular polyhedrons. Because of the centers of surfaces around a corner is not on the same . Thereby we have to put the semi-regular polyhedra into and make a plane tangent to each hill. The dual polyhedra of a or Archimedean solid can be also drawn by constructing polyhedra edges tangent to the which are perpendicular to the original polyhedra edges. 14 ZeynepÇolakandZeynepCan

Figure 1 is lightly rotated, showing the edges of rhombic dodecahedron (yellow), octahedron (green) and cube (blue). To measure the distance between two points in a space commonly Euclid- ean distance is used. Despite of it is so popular and ancient, sometimes it is not practical in the real world. For example taxicab distance is useful instead of Euclidean distance in an urban life where streets are planned par- allel or perpendicular to each other, as it is in Manhattan. Especially the maximum distance is a very useful model in the real world and its applica- tions are so much. For example, urban planning at the macro dimensions can be used for nearly calculations. Minkowski geometry is a non-Euclidean geometry in a …nite number of dimentions. Here the linear structure is same as the Euclidean one but distance is not uniform in all directions. That is, the points, lines and planes are the same, and the angles are measured in the same way, but the distance function is di¤erent. Instead of the usual sphere in Euclidean space, the unit ball is a general symmetric [15]. In studies about metric space geometry it had seen that unit spheres of these geometries are associated with convex polyhedra. When a metric is given it is easy to …nd its unit sphere in the related space geometry. In [9 and 4] authors found cuboctahedron and rhombic dodecahedron metrics by a contrary question as if sphere is known how the metric would be found and by these metrics new geometries are induced. 3 For two points P1 = (x1; y1; z1) and P2 = (x2; y2; z2) in R , Cuboctahe- dron and Rhombic Dodecahedron metrics are de…ned as 1 dCO(P ;P ) = max x1 x2 ; y1 y2 ; z1 z2 ; ( x1 x2 + y1 y2 + z1 z2 ) 1 2 j j j j j j 2 j j j j j j   and dRD(P1;P2) = max x1 x2 + y1 y2 ; x1 x2 + z1 z2 ; y1 y2 + z1 z2 fj j j j j j j j j j j jg 3 3 3 respectively. RCO , RRD and RE are denote analytic 3-spaces which are furnished by Cuboctahedron metric, Rhombic Dodecahedron metric and 3 3 Euclidean metric, respectively. Linear structure of the RCO , RRD are al- 3 most the same of RE, there is only one di¤erence, this di¤erence is that its distance function. It is so important to work on notions concerned to Some distance properties of two spaces induced by dual convex polyhedra 15 distance in geometric studies, because change of metric can bring out in- teresting results. In this study formulae of distance of a point to a plane, distance of a point to line and distance between two lines in two spaces which are induced by two dual convex polyhedra has been researched by a new method which is di¤erent from further calculations done in [1] and [8], and which enables generalizing distance formulae for Minkowski geometries. This convex polyhedra are cuboctahedron which is an Archimedean solid and rhombic dodecahedron a Catalan solid which is the dual of cuboctahe- dron.

2. DISTANCE FORMULAE OF CUBOCTAHEDRON AND RHOMBIC· DODECAHEDRON Cuboctahedron The cuboctahedron is a uniform polyhedron bounded by 6 squares and 8 . A cuboctahedron has 14 faces, 12 identical vertices, with two triangles and two squares meeting at each, and 24 identical edges, each sep- arating a from a square. As such it is a quasiregular polyhedron, i.e. an Archimedean solid, being vertex-transitive and edge-transitive. Its is the rhombic dodecahedron. It is edge-uniform, and its two kinds of faces alternate around each vertex, so it is also a quasi-. It is, also called the heptaparallelohedron or dymaxion. The distance between the center of a cuboctahedron and each vertex is equal to the length of its edges. The regular hexagon and the cuboctahedron are the only uniform of their respective dimensions with this property. The cuboctahedron has the Oh octahedral group of . According to Heron, Archimedes ascribed the cuboctahedron to Plato (Heath 1981; Coxeter 1973, p. 30). The Cartesian coordinates of vertex of the cuboctahe- dron, centered on the origin and having edge length 2, are all permutations of coordinates and changes of sign of (0; 1; 1):

Figure 2 Cuboctahedron

De…nition 1. Let P1 = (x1; y1; z1) and P2 = (x2; y2; z2) be two points in 3 R . The cuboctahedron distance between P1 and P2 is 1 dCO(P1;P2) = max x1 x2 ; y1 y2 ; z1 z2 ; ( x1 x2 + y1 y2 + z1 z2 ) j j j j j j 2 j j j j j j   In accordance with dCO-metric, the shortest way between points P1 and 1 P2 is a line segment that is parallel to a coordinate axis or 2 times of sum of three line segments each of them is parallel to a coordinate axis. 16 ZeynepÇolakandZeynepCan

Figure 3 the shortest way between points P1and P2 3 3 Lemma 2.1. dCO : R xR [0; ) de…ned by ! 1 1 dCO(P1;P2) = max x1 x2 ; y1 y2 ; z1 z2 ; ( x1 x2 + y1 y2 + z1 z2 ) j j j j j j 2 j j j j j j   where P1 = (x1; y1; z1) and P2 = (x2; y2; z2) forms a metric space, that is 3 3 RCO := (R ; dCO)

Lemma 2.2. Let l be a line through the points P1 and P2. If l has direction vector (p; q; r) then d (P ;P ) d (P ;P ) E 1 2 = CO 1 2 p2 + q2 + r2 max p ; q ; r ; 1 ( p + q + r ) j j j j j j 2 j j j j j j The followingp theorem is given from without proof [12] Theorem 2.1. The Cuboctahedron distance is invariant under all transla- tions in analytic 3 space. That is T : R3 R3 de…ned by T (x; y; z) = ! (a + x; y + b; z + c) for every a; b; c R does not change the distance between 3 2 any two points in RCO:

Proposition 2.2. dCO distance of a point P = (xo; yo; zo) to a plane : Ax + By + Cz + D = 0 is P

Axo + Byo + Czo + D dCO(P; ) = j j P max A + B ; A + C ; B + C ; A B ; A C ; B C fj j j j j j j j j j j jg where it is apparently each of the mentions in the denominators is not zero. Proof. Set of points at equal distance from a point P is a sphere. This sphere is Cuboctahedron in our space. If we enlarge this P -centered sphere, then the …rst intersection point of the sphere and the plane is the closest point of plane to the point P . P P 3 In the CO-space RCO, the distance from a point P to a plane would be de…ned as P dCO(P; ) = min dCO(P;X) X P f j 2 Pg To …nd this distance the Cuboctahedron which center is P would be con- sidered. The sphere consists of equidistant points from the …xed point. So dCO(P; ) = dCO(P;Q) where Q is the intersection point of the plane and the cuboctahedronP while enlarging the cuboctahedron, and Q must be a cor- ner of the cuboctahedron. Thus if we consider lines li, i = 1; 2; 3; 4; 5; 6 pass- ing through P and a corner of the cuboctahedron, each of li has a direction vector which is an element of  = (1; 1; 0) ; (1; 0; 1) ; (0; 1; 1) ; (1; 1; 0) ; f (1; 0; 1) ; (0; 1; 1) . Let Pi = li ; i = 1; 2; 3; 4; 5; 6. So, g \P Some distance properties of two spaces induced by dual convex polyhedra 17 dCO(P; ) = min dCO(P;Pi): i = 1; 2; 3; 4; 5; 6 . For example let P f g (p; q; r) = (1; 1; 0) be the direction vector l1. Thus, any point on the line l1 which is passing through P is

x xo y yo = = t1; z zo = 0 1 1

So coordinates of the any point P1 = (x; y; z) on the line l1 is x = x0 + t1; y = yo + t1; z = z0. Also this point provides plane equation, since it is on the plane. Thus,

A(x0 + t1) + B(y0 + t1) + C(z0) + D = 0

Ax0+By0+Cz0+D t1 = ( A B) ) Similarly, If (p; q; r) = (1; 0; 1) ; (0; 1; 1) ; (1; 1; 0) ; (1; 0; 1) or (0; 1; 1) then

Ax0 + By0 + Cz0 + D dCO(P;Pi) = i where 2 = A + C, 3 = B + C, 4 = A B, 5 = A C, 6 = B C. Therefore the required distance from the point to the plane is obtained.

Proposition 2.3. CO-distance of a point P = (xo; yo; zo) to a line l given by x a y b z c = = p q r is

max 1; 2; 3; 4; 5; 6 d (P; l) = f g CO  where 1 = r (x0 a) r (y0 b) (p q)(z0 c) , 2 = q (x0 a) (p r)(y0 b) q (z0 c) , j j j j 3 = (q r)(x0 a) p (y0 b) + p (z0 c) , 4 = r (x0 a) + r (y0 b) (p + q)(z0 c) , j j j j 5 = q (x0 a) (p + r)(y0 b) + q (z0 c) , 6 = (q + r)(x0 a) p (y0 b) p (z0 c) and j j j j

max p q + r ; p q r if 1 i , i = 2; 3; 4; 5; 6 fj j j jg  max p + q r ; p q r if 2 i , i = 1; 3; 4; 5; 6 8 fj j j jg  max p + q r ; p q + r if 3 i , i = 1; 2; 4; 5; 6  = > fj j j jg  > max p + q + r ; p + q r if 4 i , i = 1; 2; 3; 5; 6 > fj j j jg  < max p + q + r ; p q + r if 5 i , i = 1; 2; 3; 4; 6 fj j j jg  > max p + q + r ; p q r if 6 i , i = 1; 2; 3; 4; 5 > fj j j jg  > Proof. We:> would consider a cuboctahedron which center is a point P to x a y b z c …nd distance of a point P = (xo; yo; zo) to a line l given by p = q = r : If we enlarge this P -centered sphere, then the …rst intersection point of the line l and the cuboctahedron is the closest point of the line l to the point P . If we enlarge the radius of the cuboctahedron, then the cuboctahedron distance between P and l is same with the cuboctahedron distance between 18 ZeynepÇolakandZeynepCan

P and the intersection point of the cuboctahedron and the line l.

Figure 4 Intersection of the P-centered cuboctahedron and the line l

For example let’stake the cuboctahedron which center is point P = (xo; yo; zo) and the length of the radius of this cuboctahedron be k when the cubocta- hedron and the line l intersects while enlarging the radius of the cubocta- hedron. So eight of the cuboctahedron’s vertices are P1 = (xo + k; yo; zo + k);P2 = (xo k; yo; zo + k);P3 = (xo + k; yo k; zo);P4 = (xo k; yo k; zo);P5 = (xo k; yo; zo k);P6 = (xo k; yo +k; zo);P7 = (xo +k; yo +k; zo) and P8 = (xo +k; yo; zo k). Line l and the cuboctahedron would intersect at an edge that is on a line which direction vector is (1; 1; 0) ; (1; 0; 1) ; (0; 1; 1) ; (1; 1; 0) ; (1; 0; 1)or (0; 1; 1), and passing through P1;P2;P3;P4;P5;P6;P7 or P8. For example let’s consider the edge on the line that direction vec- tor is (1; 1; 0) and passing through P2. Hence this edge is on the line x = x0 k+; y = y0 +; z = z0 +k and line l is x = p+a; y = q+b; z = r+c. Now we can …nd the intersection point of line l and the cuboctahedron So r(x0 a) r(y0 b) (p q)(z0 c) we obtain k = p q+r as the cuboctahedron distance be- tween P and l. Similarly the other cases can be easily found. Proposition 2.4. CO-distance between any two lines given by

x a y b z c l::: = = =  p q r x a y b z c l::: = = =  p q r 3 in RCO can be expressed as follows: If l is parallel to l, then dCO(l; l) = dCO(A; l);where A = (a; b; c): If l is not parallel to lthen 3 in RCO can be expressed as follows: If l is parallel to l, then dCO(l; l) = dCO(A; l);where A = (a; b; c): If l is not parallel to lthen (rq rq)(a a)+ (pr rp)(b b)+ (pq pq)(c c) d (l; l)= j j CO max + ; ; + ; ; + ; fj j j j j j j j j j j jg where = pq pq, = pr pr and = rq rq. Proof. dCO(l; l) = min dCO(X;X) where X l, X l:If l is parallel to l then it can be taken as (p; q; r) = (p; q; r) and2P = (a2+ p; b+ q; c+ r) Some distance properties of two spaces induced by dual convex polyhedra 19 for any point P on l, without loss of generality. Then it can be eas- ily computed using the formula given by proposition 2 that dCO(l; P ) = dCO(A; l) where A = (a; b; c):If l is not parallel to l; then at least one of pq qp; qr rq; rp pris not zero. Otherwise these lines would be parallel to each other. Let’sconsider the points P = (p + a; q + b; r + c) and P= (p + a; q + b; r + c) which are on l and lrespectively. Thus if dCO(P;P) is minimum then direction vector of the line lwhich passes through P and P is an element of (1; 1; 0) ; (1; 0; 1) ; (0; 1; 1) ; (1; 1; 0) ; (1; 0; 1) ; (0; 1; 1) . For example let directonf vector of lbe (p; q; r)= (1; 1; 0). Since g

p + a p a q + b q b r + c r c = = 1 1 0 then

r (a a) r (b b) (p q)(c c)  = r (p q) r(p q) and

r(a a)+ r(b b) (p q)(c c)  = r (p q) r(p q) So

p + a p a ; q + b q b ; r + c r c ; dCO(P;P) = max 1 j j j j j j 2 ( p + a p a + q + b q b + r + c r c) (rq rq)(a ja)+(pr rp)(b b)+( pjq pqj)(c c) j j j  = (pq pq)+(rq rq)

Other cases can be given by the similiar way. And as a subcase if (p; q; r) is an element of (1; 0; 0) ; (0; 1; 0) ; (0; 0; 1) then f g

(rq rq)(a a)+ (pr rp)(b b)+ (pq pq)(c c) d (l; l)= j j CO max pq pq ; pr rp ; rq rq fj j j j j jg

Rhombic Dodecahedron The rhombic dodecahedron is a very interesting polyhedron. It has 12 faces, 14 vertices, 24 sides or edges. Its sometimes also called the rhomboidal dodecahedron. The rhombic dodecahedron can be built up by a placing six on the faces of a seventh.The rhombic dodecahedron is a and a space-…lling polyhedron (Steinhaus 1999, p. 185). The vertices are given by (+/-1, +/-1, +/-1), (+/-2, 0, 0), (0, +/-2, 0), (0, 0, +/-2). 20 ZeynepÇolakandZeynepCan

Figure 5 Rhombic Dodecahedron

De…nition 2. Let P1 = (x1; y1; z1) and P2 = (x2; y2; z2) be two points in 3 R . The rhombic dodecahedron distance between two points P1 and P2 is dRD(P1;P2) = max x1 x2 + y1 y2 ; x1 x2 + z1 z2 ; y1 y2 + z1 z2 fj j j j j j j j j j j jg In accordance with dRD-metric, the shortest way between points P1 and P2 is union of two line segments that each one is parallel to a coordinate axis

Figure 6 the shortest way between points P1and P2

Lemma 2.3. Let l be a line through the points P1 and P2. If l has direction vector (p; q; r) then it is easy to see that

d (P ;P ) d (P ;P ) E 1 2 = RD 1 2 p2 + q2 + r2 max p + q ; p + r ; q + r fj j j j j j j j j j j jg The following theorem is given from without proof [14]: p Lemma 2.4. Let l be a line through the points P1 and P2. If l has direction vector (p; q; r) then it is easy to see that

d (P ;P ) d (P ;P ) E 1 2 = RD 1 2 p2 + q2 + r2 max p + q ; p + r ; q + r fj j j j j j j j j j j jg The following theorem is given from without proof [14]: p Theorem 2.5. The Rhombic dodecahedron distance is invariant under all translations in analytic 3 space. That is 3 3 T : R R • T (x; y; z) = (a + x; y + b; z + c); a; b; c R ! 2 3 does not change the distance between any two points in RRD: Some distance properties of two spaces induced by dual convex polyhedra 21

Proposition 2.6. RD distance of a point P = (xo; yo; zo) to a plane : Ax + By + Cz + D = 0 is P Axo + Byo + Czo + D dRD(P; ) = j j P max A ; B ; C ; A + B + C ; A + B C ; A B + C ; A B C fj j j j j j j j j j j j j jg where it is apparently each of the mentions in the denominators is not zero. Proof. In our space sphere is Rhombic Dodecahedron. The similar way to Cuboctahedron; if we enlarge this P -centered sphere, then the …rst intersec- tion point of the plane is the closest point of plane to the point P . In P P the RD-space R3 , the distance from a point P to a plane is de…ned as RD P dRD(P; ) = min dRD(P;X) X P f j 2 Pg The rhombic dodecahedron is the sphere of the rhombic dodecahedron space geometry and the sphere consists of equidistant points from the …xed point. So dRD(P; ) = dRD(P;Q) where Q is the intersection point of the plane and the rhombicP dodecahedron while enlarging the rhombic dodecahedron and Q must be a corner of the rhombic dodecahedron. Thus if we consider lines li, i = 1; 2; 3; 4; 5; 6; 7 passing through P and a corner of the rhom- bic dodecahedron, each of li has a direction vector which is an element of  = (1; 0; 0) ; (0; 1; 0) ; (0; 0; 1) ; (1; 1; 1) ; (1; 1; 1) ; (1; 1; 1) ; (1; 1; 1) . f g Let Pi = li ; i = 1; 2; 3; 4; 5; 6; 7 So, dRD(P; ) = min dRD(P;Pi): i = 1; 2; 3; 4; 5; 6; 7 . \P P f g For example let (p; q; r) = (1; 1; 1) be the direction vector of l1. Thus, any point on the line l1 which is passing through P is x xo y yo z zo = = = t 1 1 1 1 So coordinates of the any point P1 = (x; y; z) on the line l1 is x = x0 +t1; y = yo + t1; z = z0 + t1. Also this point provides plane equation, since it is on the plane. Thus,

A(x0 + t1) + B(y0 + t1) + C(z0 + t1) + D = 0

(Ax0+By0+Cz0+D) t1 = ) A+B+C Similarly, If (p; q; r) = (1; 0; 0) ; (0; 1; 0) ; (0; 0; 1) ; (1; 1; 1) ; (1; 1; 1) or (1; 1; 1) then Ax0 + By0 + Cz0 + D dRD(P;Pi) = i where 2 = A, 3 = B, 4 = C, 5 = A + B C, 6 = A B + C, 7 = A B C. Therefore the required formula for distance from the point to the plane is obtained.RD distance of a point P = (xo; yo; zo) to a plane : Ax + By + Cz + D = 0 is P Axo + Byo + Czo + D dRD(P; ) = j j P max A ; B ; C ; A + B + C ; A + B C ; A B + C ; A B C fj j j j j j j j j j j j j jg where it is apparently each of the mentions in the denominators is not zero.RD distance of a point P = (xo; yo; zo) to a plane : Ax + By + Cz + D = 0 is P Axo + Byo + Czo + D dRD(P; ) = j j P max A ; B ; C ; A + B + C ; A + B C ; A B + C ; A B C fj j j j j j j j j j j j j jg where it is apparently each of the mentions in the denominators is not zero. 22 ZeynepÇolakandZeynepCan

Proposition 2.7. RD-distance of a point P = (xo; yo; zo) to a line l given by

x a y b z c = = p q r is

max 1; 2; 3; 4; 5; 6 d (P; l) = f g CO  where 1 = (q + r)(x0 a) (p + r)(y0 b) (p q)(z0 c) , j j 2 = (q r)(x0 a) (p + r)(y0 b) + (p + q)(z0 c) , j j 3 = (q + r)(x0 a) (p r)(y0 b) (p + q)(z0 c) , j j 4 = (q r)(x0 a) (p r)(y0 b) + (p q)(z0 c) and j j max p q ; p + r ; q + r if 1 i , i = 2; 3; 4 fj j j j j jg  max p + q ; p + r ; q r if 2 i , i = 1; 3; 4;  = 8 fj j j j j jg  max p + q ; p r ; q + r if 3 i , i = 1; 2; 4 > fj j j j j jg  < max p q ; p r ; q r if 4 i , i = 1; 2; 3 fj j j j j jg  Proof. If we> enlarge this P -centered sphere, then the …rst intersection :> point of the line l and the sphere is the closest point of line l to the point P . We would consider a rhombic dodecahedron which center is point a x a P to …nd distance of a point P = (xo; yo; zo) to a line l given by p = y b z c q = r : If we enlarge the radius of the rhombic dodecahedron, then the rhombic dodecahedron distance between P and l is same with the rhombic dodecahedron distance between P and the intersection point of the rhombic dodecahedron and the line l.

Figure 7 Intersection of the P centered rhombic dodecahedron and the line l For example let’stake the rhombic dodecahedron which center is the point P = (xo; yo; zo) and the length of the radius of this rhombic dodecahedron be k when the rhombic dodecahedron and the line l intersects while enlarging the radius of the rhombic dodecahedron. So six of the rhombic dodecahe- dron’svertices are P1 = (xo; yo; zo + k);P2 = (xo; yo + k; zo); P3 = (xo; yo k; zo);P4 = (xo + k; yo; zo);P5 = (xo k; yo; zo) and P6 = (xo; yo; zo k). Line l and the rhombic dodecahedron would intersect at an edge that is on a line which direction vector is (1; 1; 1) ; (1; 1; 1) ; (1; 1; 1) or ( 1; 1; 1), and passing through P1;P2;P3;P4;P5 or P6. For example let’s consider the edge on the line that direction vector is (1; 1; 1) and passing through P1. Hence this edge is on the line x = x0+; y = y0+; z = z0+k  and line l is x = p + a; y = q + b; z = r + c. Now we can …nd the Some distance properties of two spaces induced by dual convex polyhedra 23 intersection point of line l and the rhombic dodecahedron So we obtain (q+r)(x0 a) (p+r)(y0 b) (p q)(z0 c) k = p q as the rhombic dodecahedron distance between P and l. Similarly the other cases can be easily found. Proposition 2.8. RD-distance between any two lines given by

x a y b z c l::: = = =  p q r x a y b z c l::: = = =  p q r 3 in RRD can be expressed as follows: If l is parallel to l, then dRD(l; l) = dRD(A; l);where A = (a; b; c): If l is not parallel to lthen (rq rq)(a a)+ (pr rp)(b b)+ (pq pq)(c c) dRD(l; l)= j j + + + + + + max ; ; ; j j ; j j ; j j ; j j j j j j j j 2 2 2 2 where = pq pq, n= pr pr and = rq rq. o Proof. dRD(l; l) = min dRD(X;X) where X l, X l:If l is parallel to l then it can be taken as (p; q; r) = (p; q; r) and2P = (a2+ p; b+ q; c+ r) for any point P on l, without loss of generality. Then it can be eas- ily computed using the formula given by proposition 2 that dRD(l; P ) = dRD(A; l) where A = (a; b; c):If l is not parallel to l; then at least one of pq qp; qr rq; rp pris not zero. Otherwise these lines would be parallel to each other. Let’sconsider the points P = (p + a; q + b; r + c) and P= (p + a; q + b; r + c) which are on l and lrespectively. Thus if dRD(P;P) is minimum then direction vector of the line lwhich passes through P and P is an element of (1; 0; 0) ; (0; 1; 0) ; (0; 0; 1) ; (1; 1; 1) ; ( 1; 1; 1) ; (1; 1; 1) ; (1; 1; 1) . For example letf directon vector of lbe (p; q; r)= (1;1; 1). Since g p + a p a q + b q b r + c r c = = 1 1 1 then (q r)(a a)+ (r p)(b b)+ (p q)(c c)  = (pq pq)+ (pr pr) + (rq qr) and (r+ q)(a a)+ (r p)(b b)+ (p q)(c c)  = (pq pq)+ (pr pr) + (rq qr) So

p + a p a + q + b q b ; j j j j dRD(P;P) = max p + a p a + r + c r c ; 8 j j j j 9 < q + b q b + r + c r c = (rq rqj)(a a)+(pr rp)(b jb)+(jpq pq)(cc) j = 2 (pq pq)+(pr pr)+( rq rq) : ;

Other cases can be given by the similiar way.

24 ZeynepÇolakandZeynepCan

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DEPARTMENT OF ECONOMETRY, CANAKKALE UNIVERSITY, 17270, CANAKKALE, TURKEY. E-mail address: [email protected]

DEPARTMENT OF MATHEMATICS FACULTY OF SCIENCE· AND LETTERS, AKSARAY UNIVERSITY, 68000, AKSARAY, TURKEY. E-mail address: [email protected]