Buffon Type Problems in Archimedean Tilings II

Transcription

1 pplied Mathematical Sciences, Vol. 1, 16, no. 7, HIKRI Ltd, Buffon Type Problems in rchimedean Tilings II Salvatore Vassallo Università attolica del S. uore, Largo Gemelli 1, I-1 Milan, Italy opyright c 16 Salvatore Vassallo. This article is distributed under the reative ommons ttribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. bstract In this paper we consider the snub square tiling of the plane, 4,, 4 rchimedean tiling and compute the probability that a random circle or a random segment intersects a side of the lattice. Moreover we compute the same probability if also the diameter of the circle or the length of the segment is a random variable. Mathematcs Subject lassification: Primary 6D5; Secondary 5 Keywords: Geometric probability, stochastic geometry, random sets, random convex sets and integral geometry 1 Introduction tiling or tessellation in the plane is a collection of disjoint closed sets the tiles that can intersect only on the boundary, which cover the plane. tiling is said to be polygonal if the tiles are polygon, a polygonal tiling is said to be edgeto-edge if two non disjoint tiles have in common or a vertex or a segment that is an edge for both the polygons. In this case we call any edge of a tile an edge of the tiling. n edge-to-edge tiling is called regular if it is composed of congruent copies of a single regular polygon. n rchimedean tessellation semi-regular or uniform tessellation is an edge-to-edge tessellation of the plane made of more than one type of regular polygon so that the same polygons surround each vertex. There are eight different rchimedean tilings and we can classify them giving the types of polygons as they come together at the vertex [5]. The snub square tiling is a tiling such that three triangles and two squares come together in any vertex in the order triangle, triangle, square, triangle, square so it can be called a, 4,, 4 rchimedean tiling see Figure 1a. Many

2 1 Salvatore Vassallo a Snub square tiling b The fundamental cell T Figure 1: The tiling R authors studied Buffon type problems for different lattices of figures or tilings and different test bodies: See for example [8], [], [], [4], [1].In particular the case of the, 4 rchimedean tiling elongated triangular tiling is studied in [9]. Let E be the Euclidean plane and let R be the snub square tiling of E given in Figure 1a. Let us denote by K a convex body which here means a compact convex set that we shall call test body. We will study this generalized problem of Buffon type: Which is the probability p K,R that the random convex body K, or more precisely, a random congruent copy of K, meets some edge of the tiling R? We denote by T the fundamental cell of R as in Figure 1b in the following we will use the same notation for the vertices of the tiles: The probability p K,R can be calculated as the probability that the body K meets an edge of T supposing that a fixed point of K is in T. We will study Buffon type problems for two special test bodies: circle of diameter D and a line segment of length l. In addition we will study the same problem if the diameter of K itself i.e. the diameter of the circle or the length of the segment is a random variable with first and second moment known. We denote by M the set of all test bodies K whose centroid M is in the interior of T and by N the set of all test bodies K that are completely contained in one of the triangles or in one of the squares that forms T. We also assume that the convex test bodies are uniformly distributed, i.e. that the coordinates of M are a bidimensional random variable with uniform distribution in T, and that the random variable ϕ is uniformly distributed in [, π], M and ϕ stochastically independent. We have 1 µn µm 1 where µ is the Lebesgue measure.

3 Buffon type problems in rchimedean tilings II 11 The test body is a circle Let us suppose that the test body K is a circle of diameter D. Easy geometrical considerations lead us to distinguish between the cases D < a, a D < a and D a. It is obvious that in the third case 1, so we have to study only the first two cases. Proposition.1. The probability that the circle K of diameter D intersects the tiling R is given by D[1a + D] + if D < a a a +4aD D + if a D < a a Proof. We compute the measures µm e µn with the help of the elementary kinematic measure dk = dx dy dφ of E see [6], [7] where x and y are the coordinates of the center of K or the components of a translation, and φ is the angle of rotation. We have µm = π dφ dxdy = π areat = πa x,y T + Let N 1 be the set of circles of diameter D that are contained in the triangle B and N be the set of circles of diameter D that are contained in the square EF. From equation 1 we obtain 1 4µN 1 + µn πa + From Figure a it is easy to see that µn 1 is π times the area of the triangle B F F E E B a Triangle tile b Square tile Figure : The case K = circle with D < a B whose sides are parallel to the sides of the triangle B at distance D/ from them is the center of a disk interior to the triangle B and tangent to the sides B and and so on. Since the side of the triangle is

4 1 Salvatore Vassallo a D we have µn 1 = π 4 a D and in the same way we obtain that µn = π a D. Then in the case D < a we have [ µn = π a D + a D ] and so: D [ 1a + D ] + a Let a D a. If the center of the circle K is in the triangle B, the circle always intersects one of the side of the triangle so that µn 1 =. If the center of the circle is in the square DF the circle does not intersect the side of the square if its center is in the square E F the figure is similar to Figure b with a smaller interior square; since the side of this square is a D we have µn = π a D and so in this case: a + 4aD D + a Let us now consider the problem of the intersection with an edge of R of a random circle K whose diameter is a bounded random variable with upper bound D a Proposition.. The probability that a circle K whose diameter is a random variable, with D a, and known moments E and E, intersects the tiling R is given by 1 E + + a + a E 4 Proof. The upper boundary of assures that K is always small when compared to the elementary tile of R. Let fδ be the density of and pr δ the probability that K intersects a cell of R given the condition = δ. Then the probability pr δ that K intersects at least one of the edges of the tiling can be computed as D pr δfδdδ

5 Buffon type problems in rchimedean tilings II 1 From equation we know that pr δ = δ[1a + δ] + a and so D 1aδ + δ fδdδ = + a 1 D + a δfδdδ + D δ fδdδ = + a 1 a E + + a E The test body is a line segment Let us now consider the case in which K is a line segment of length l. lso in this case easy geometrical considerations give us four cases: l < a, a l < a, a l < a and l a. In the last case the segment always intersects an edge of R, so we have to study the other cases. Proposition.1. The probability that the line segment K of length l intersects the tiling R is given by l[6a 15+π l] π+ a 6al 7a 4l a 15l πl +6a +l arccos a l if l < a + if a l < a πa 4a 8a l a +l + πa +8a arccosa/l + πa if a l < a Proof. We use the same notation as in proof of Proposition.1. i Let us consider the case l < a. First we compute the measure µ N 1 of the set N 1 of all line segments of length l contained in the triangle B. With reference to Figure a, for a fixed angle φ [, π [, we denote by the midpoint of the line segment of length l with one endpoint in and φ = Â B, B the midpoint of the line segment of length l with endpoints on B and B that makes an angle φ with B, and the midpoint of the line segment of length l with endpoints on and B that makes an angle φ with the direction of B. 5

6 14 Salvatore Vassallo α= π 6 arccos a l φ B a l < a, K in the triangle B α B B b a l < a, K in the triangle F E F E F E F E φ c l < a, K in the square arccosa/l d a l < a, K in the square Since area B = 4 µ N 1 = π/ Figure : The case K = line segment [ a l sin π φ] we obtain, by symmetry, area B dφ = π/ = πa 6al π l 1 4 [ a l ] sin π φ dφ In the same way, if ψ [, π [, we obtain for the set N of the line segments contained in the square EF see Figure c area E F = a l sin ψ a l cos ψ, and so, by symmetry, we have µ N = π/ Then µn = + πa al + a l cos ψ a l sin ψ dψ = πa 4al + l 5 + π l and hence if l < a 6 l [ 6a 15 + π l ] π + a 7 ii Let now a l < a. With reference to Figure b it is easy to see that the line segment can be contained in the triangle B only if the angle φ [, π/[ between the line segment and the side B satisfies φ < π 6 arccos a l or π 6 + arccos a l < φ < π see Figure b.

7 Buffon type problems in rchimedean tilings II 15 Hence the measure of the line segments completely contained in the triangle B is, by symmetry, π 6 arccos a [ l µ N 1 = 6 a l ] sin 4 π φ dφ = 9a 4l a 4l + πa π l 6 a + l arccos a l 1 The measure of the line segment completely contained in the square EF is the same as in the case above: µ N = πa 4al + l. Hence if a l < a we have 6al 7a 4l a 15l πl + 6a + l arccos a + πa iii Let now a l < a. With reference to Figure d it is easy to see that if the centroid of the line segment is in the triangle B the line segment always meets one of the side of the triangle and can be contained in the square EF only if the angle φ [, π/[ between the line segment and the side satisfies arccos a l < φ π arccos a l. Therefore the measure of the line segments completely contained in the square EF is given by: µ N = π arccos a l arccosa/l Hence we have if a l < a a l cos φ a l sin φ dφ = = 4a l a l + π a 4a arccosa/l 4a 8a l a + l + πa + 8a arccosa/l + πa l Let now K be a line segment whose length is a bounded random variable with D a. With a proof similar to the proof of Proposition. we obtain easily Proposition.. The probability that a segment K whose length is a random variable, with D a, and known moments E and E, intersects the tiling R is given by π + a E 15 + π π + a E 8

8 16 Salvatore Vassallo References [1] G. aristi and M. I. Stoka, Laplace type problem for a regular lattices with irregular hexagonal cell, Far East J. Math. Sci., 5 11, no. 1, 6. []. Duma and M. I. Stoka, Geometric probabilities for non-regular lattices. I, 5th Italian onference on Integral Geometry, Geometric Probability Theory and onvex Bodies, Italian, Milano, 1995, Rend. irc. Mat. Palermo Suppl., , []. Duma and M. I. Stoka, Problems of geometric probability for nonregular lattices. II, nd International onference in Stochastic Geometry, onvex Bodies and Empirical Measures grigento, 1996, Rend. irc. Mat. Palermo Suppl., , [4]. Duma and M. I. Stoka, Problems of geometric probability for nonregular lattices. III, nd International onference in Stochastic Geometry, onvex Bodies and Empirical Measures grigento, 1996, Rend. irc. Mat. Palermo Suppl., , [5] G.. Grünbaum, B. Shephard, Tilings and Patterns, Freeman, [6] L.. Santaló, Integral Geometry and Geometric Probability, ddison- Wesley, [7] M. I. Stoka, Geometria Integrala, Ed. cademice Rep. Soc. Romania, [8] M. I. Stoka, Probabilités géométriques de type Buffon dans le plan euclidien, tti ccad. Sci. Torino l. Sci. Fis. Mat. Natur., , [9] S. Vassallo, Buffon type problems in archimedean tilings I, Universal Journal of Mathematics and Mathematical Sciences, 4 1, no., Received: February 1, 16; Published: pril 4, 16