Semi Regular Lattice Polyhedra

The aim of this paper is to explain the probability of the existence of five regular and thirteen semi regular polyhedra; and we indicate that among these thirteen geometrical figures there are only three lattice polyhedra. Also in this work we present a proof of the existence of three regular lattice polygons. Introduction In 1985 two mathematicians, Peter Hilton and Jean Pederson have done a research on the folding regular star polygons. They have taken benefit from a practical method which is, folding paper, for proving the number of existence of regular polygons (Hilton & Pederson, 1985), and the way they have used is similar to the way which we used. In 1988, Bokowski and Wills described some fundamental ideas in the study of regular maps and their polyhedral realizations in the Euclidean 3-space (Bokowski & Wills, 1988). In collecting information, we faced some problems and the greatest problem seems to be the shortage of sources. Lattice's subject includes two aspects, geometric and algebraic. In our district we can find those sources which have been dealt with algebraic aspect. The alert reader will have surmised by now that we are prepared to go along way with this topic. But we restrain ourselves! We will focus our attention, and organize this article, to bring out the special features of this subject. Definition [1]. (Diested,2005). A polygon is a simple closed curve which comes in to existence by union of some intersecting straight parts in which each two parts cut each other, and it is regular if it is both equilateral and equiangular. The polygon is said to be convex if we take any gon, the polygon be in one side of it. Definition [2]. (Fraleigh, 2003 ; Scott, 1987) A point ) , , ( z y x in coordinates 3space 3 R is called a lattice point if Journal of Kirkuk University – Scientific Studies , vol.6, No.1,2011 301 y x, and z are all integers, and so a polygon in 3 R is called lattice polygon if all its vertices are lattice point. Definition [3]. (Diested,2005; Scott, 1987) A polyhedra is a solid which comes to existence by union of some intersecting planes in which any two planes cut each other, and it is regular if its faces are congruent regular polygons; and each vertex has the same number of faces surrounding it. A semiregular polyhedra has regular polygons as faces and all vertices congruent, but they admit a variety of such polygons in one solid. A polyhedra in 3 R is called lattice if all its vertices are lattice points. Some information about the main subject Before talking about our main subjects it is necessary to have some information about: 1) Those regular polygons which can be drawn and those which can not be drawn. 2) The angle between two gones of a regular polygon. The regular polygons which the number of their gons indicated bellow are those which can be drawn : n 2 2× , n 2 3× and n 2 5× , 2 , 1 , 0 n = . For instance 3, 4, 5, 6, .... Also Gauss proved that if a prime number 1 2 P n 2 + = then P gons can be drawn. The prime numbers which are in this form are 3, 5, 17, 257, 65537, .... And the regular 7, 9, 11, 13, 14, 19, 23,... gons can not be drawn. To find the angle btween two gons of regular polygon, we divide regular ngons in to traingles by drawing diagonals from one of the vertex and the polygon will divide into ( n-2 ) traingles. Hence the mesure of each angle btween two gons is n ) 2 n ( 180 α × = for example n= 5 so o 108 5 ) 2 5 ( 180 α = × = . Theorem [4]. (Scott, 1987) A covex lattice n-gon is equiangular if and only if n = 4 or 8. Theorem [5]. (Scott, 1987) A regular n-gon can be embeded is the three dimentional integral lattice if and only if n = 3, 4 or 6. Journal of Kirkuk University – Scientific Studies , vol.6, No.1,2011 301 Regular and Semi Regular Lattice Polyhedra In coordinate 3– space R 3 there are five regular and thirteen semi regular polyhedra. To prove the existence of five regular polyhedra let β be the sum of all angles around a vertex of a regular polyhedra and we shall use n to indicate the type of the faces meeting at a vertex and m is the number of all faces around a vertex . ( it is clear that m ) 3 If n=3 then o 60 α = ( is the angle btween any two gons of n-gon ) and if m=3 then o o 0 360 180 60 3 β < = × = is acceptable case and other cases when 0 360 β < are also acceptable ( which are m= 4, 5 ) and do not acceptable when 0 360 β ≥ which are the cases 5 m > . If n= 4 then 0 90 α = the only acceptable case is m=3 because o 0 360 270 β < = . If n=5 then 0 108 α = here also the only acceptable case is m=3 because o 0 360 324 β < = . For other choice of n 6 the cases are not acceptable and hence there are only five polyhedra [figure 1]. Figure 1 ( The five regular polyhedra) Now we want to prove that there are thirteen semiregular polyhedra First we take the all possible arrangements(orders) of 1 n -gon, 2 n -gon, ..., r n -gon around a vertex. We note that the only regular polygons which are candidates to participate in our proof are 3-gon, 4-gon, 5-gon, 6-gon, 8gon, and 10-gon. Then A the number of all arrangments is given by: For 3 m = we have 50 ( 6 ( A 5 1 6 3 = × + = For 4 m = we have 195 ( ( 6 ( 6 2 ( A 4 1 5 1 5 1 6 4 = × × + × × + = For 5 m = we have 4 1 5 1 5 1 6 5 ( ( 6 2 ( 6 2 ( A × × + × × × + = 666 ( ( ( 6 3 1 4 1 5 1 = × × × + If m=3 we shall use ) n , n , n ( 3 2 1 to indicate the faces meeting at a vertex where = 3 2 1 n , n , n 3,4,5,6,8,10 then the all accept possible arrangments which o 360 β < are: Journal of Kirkuk University – Scientific Studies , vol.6, No.1,2011 301 ) n , n , n ( 3 2 1 β ) n , n , n ( 3 2 1 (3,3,4) o 210 (3,8,8) o 330 (3,3,5) o 228 (3,8,10) o 339 (3,3,6) o 240 (3,10, 10) o 348 (3,3,8) o 255 (4,4,5) o 288 (3,3,10) o 264 (4,4,6) o 300 (3,4,4) o 240 (4,4,8) o 315 (3,4,5) o 258 (4,4,10) o 324 (3,4,6) o 270 (4,5,5) o 306 (3,4,8) o 285 (4,5,6) o 318 (3,4,10) o 294 (4,5,8) o 333 (3,5,5) o 276 (4,5,10) o 342 (3,5,6) o 288 (4,6,6) o 330 (3,5,8) o 303 (4,6,8) o 345 (3,5,10) o 312 (4,6,10) o 354 (3,6,6) o 300 (5,5,6) o 336 (3,6,8) o 315 (5,5,8) o 351 (3,6,10) o 324 (5,6,6) o 348 "Collection 1" By a similar method we shall use ) n , n , n , n ( 4 3 2 1 to indicate the faces meeting at a vertex for m=4, where = 4 3 2 1 n , n , n , n 3,4,5,6,8,10, then the accept arrangements for which o 360 β < are: ) n , n , n , n ( 4 3 2 1 β ) n , n , n , n ( 4 3 2 1 β (3,3,3,4) o 270 (3,4,3,6) o 330 (3,3,3,5) o 285 (3,4,3,8) o 345 (3,3,3,6) o 300 (3,4,3, 10) o 354 (3,3,3,8) o 315 (3,4,4,4) o 330 (3,3,3,10) o 324 (3,5,3,5) o 336 (3,4,3,4) o 300 (3,5,3,6) o 348 (3,4,3,5) o 315 (4,3,4,5) o 348 Journal of Kirkuk University – Scientific Studies , vol.6, No.1,2011