A Laplace Type Problem for a Lattice with Cell Composed by Two Triangles and a Trapezium and Obstacles

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1 Applied Mathematical Sciences, Vol. 11, 17, no. 9, 5-3 HIKARI Ltd, A Laplace Type Problem for a Lattice with Cell Composed by Two Triangles and a Trapezium and Obstacles Alfio Puglisi Department of Economics, University of Messina Via dei Verdi, Messina, Italy Marius Stoka Sciences Academy of Turin Via Maria Vittoria, Torino, Italy Copyright c 16 Alfio Puglisi and Marius Stoka. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper we studied a Laplace type problem for a particular lattice composed by two triangles and a trapezium with obstacles. In other words, we considered a lattice with the fundamental cell represented in fig. 1 and we computed the probability that a segment of a random position and constant length intersects a side of the lattice. Mathematics Subject Classification: 6D5, 5A Keywords: Geometric Probability, stochastic geometry, random sets, random convex sets and integral geometry 1 Introduction Considering several results obtained in recent years on Buffon-Laplace problems for multiplicity of lattices [1], [], [3], [], [5] and [6], we study as the

2 6 Alfio Puglisi and Marius Stoka fundamental cell a lattice composed by two triangles and a trapezium introducing the obstacles. Starting from the fundamental results obtained by Poincaré [], Stoka [9], Caristi and Stoka [7], we wanted to consider another particular lattice composed by two triangles and an isoscele trapezium since we aim to study the geometric point of view as well. In other words, we obtained a particular lattice see fig. 1) and we solved the Laplace type problem computing the probability that a random segment of constant length intersects the fundamental cell. Main Results Let R a, b, ; m) be the lattice with fundamental cell C represented in figure 1, where is an angle with ɛ], Arctg b a ]. A A b m D D A 1 A 3 D 3 D 1 C 3 a m a m C 1 C B 1 E F C 1 B B E E E 1 F 1 F F C C fig.1 From this figure we have AE = DF = a cos m, BE = CF = atg m, EF = b atg m; 1) areaaa 1 A 3 = areadd 1 D 3 = m sin, areaaa A 3 = areadd D 3 = areaee E = areaf F F = areaee 1 E = areaf F 1 E = m cos,

3 Two triangles and trapezium 7 areabb 1 B = areacc 1 C = m ; ) areac 1 = areac = a m tg sin + cos + 1), areac 3 = a b atg) m cos, areac = ab m sin + 3 cos + 1). 3) Our aim is to compute the probability that a segment s with random position and of constant length l < min atg, b atg m) intersects a side of the lattice R, i.e., the probability P int that the segment s intersects a side of the fundamental cell C. The position of the segment s is determinated by the centre and by the angle ϕ that s formed with the line BC o AD). To compute the probability P int we consider the limiting positions of segment s, for a specified value of ϕ, in the cells C i, i = 1,, 3). Thus we have fig. A A A D D c 1 c A 1 A 3 A 5 D c 7 5 D D 3 D 1 a 1 b 1 A 6 A A 7 c ˆ C 3 ϕ) b a c 6 ˆ C 1 ϕ) a 6 b 6 ˆ C ϕ) B E 6 B 1 B E F 6F7 B a 3 E E 7 F 3 5 c 3 5 b 5 ϕ a a 5 c ϕ B E 3 E E E 1 F 1 F F c 5 F b C C 3 C b 3 C 1 C fig. and the following areaĉ1 ϕ) = areac 1 areaĉ ϕ) = areac 6 areaa i ϕ), ) 6 areab i ϕ), 5)

4 Alfio Puglisi and Marius Stoka areaĉ3 ϕ) = areac 3 areac i ϕ). 6) By fig. we have: areaa 3 ϕ) = lm sin ϕ + cos ϕ), areaa 1 ϕ) = l cos ϕ cos ϕ ) sin areaa ϕ) = a m m sin, ) l cos ϕ l cos ϕ cos ϕ ), sin areaa 5 ϕ) = l sin ϕ sin π ϕ + ) sin π + ), areaa ϕ) = atg m) l sin ϕ l sin ϕ sin π ϕ + ) sin π + ), areaa 6 ϕ) = We obtain: a cos m ) l cos ϕ ) l cos ϕ cos ϕ ). sin A 1 ϕ) = 6 areaa i ϕ) = l m + m sin atg) sin ϕ+ l a m cos ) cos ϕ l m sin ϕ + ctg cos ϕ + ctg) sin. 7) In the same way we have: areab 3 ϕ) = l sin ϕ m, ) areab 1 ϕ) = l m ctg cos ϕ sin ϕ + ctg) sin, areab ϕ) = atg 3m ) l sin ϕ l sin ϕ,

5 Two triangles and trapezium 9 areab ϕ) = a m ) l cos ϕ + l sin ϕ ctg cos ϕ ctg), areab 5 ϕ) = lm cos [ π cos ϕ ctg + ) ] sin ϕ, areab 6 ϕ) = l [a m ) m ) ] cos cos ϕ + sin atg sin ϕ + l sin ϕ ctg cos ϕ ctg). We obtain: A ϕ) = 6 areab i ϕ) = a m) l cos ϕ lm sin ϕ [ π 3 sin + cos ctg + )] l ctg cos ϕ sin ϕ + ctg) In the end we have: m 1 + sin ). 9) areac 1 ϕ) = l sin ϕ cos ϕ ) cos areac ϕ) = al m cos, ml cos ϕaltg sin ϕ cos ϕ ), areac ϕ) = b m ) l sin ϕ l sin ϕ cos ϕ ), cos lm cos areac 3 ϕ) = sin π ) sin ϕ + π ), areac 5 ϕ) = l sin ϕ sin π + ϕ) sin π + ), areac ϕ) = b atg m) l sin ϕ l sin ϕ sin π + ϕ) sin π + ),

6 3 Alfio Puglisi and Marius Stoka We obtain: areac 6 ϕ) = areac 7 ϕ) = a ) l cos m cos ϕ + ), ml cos sin π + ) sin ϕ + π + ). A 3 ϕ) = areac i ϕ) = al cos ϕ + b atg) l sin ϕ lm cos cos ϕ 5lm sin ϕ m cos. 1) Denoting by M i i = 1,, 3) the set of all segments s which have their centre in C i. We denote likewise by N i the set of all segments s completely contained in C i, we have [9]: P int = 1 3 µ N i) 3 µ M i), 11) where µ denotes the Lebesgue measure in the Euclidean plane. To compute the above measure µ M i ) and µ N i ) we use the Poincaré kinematic measure []: dk = dx dy dϕ, where x, y are the coordinates of the centre of s and ϕ the fixed angle. To compute the measures µ M i ) and µ N i ) we must determine the range of variation of ϕ. By figure we have that in the cell C 1 there is ϕɛ [ ], π, in the cell C we have ϕɛ [, π ] and in the cell C 3 we have ϕɛ [, π + ] and, for [, π ] [, π ] we have in the cell C the following: ϕɛ [, π ]. [, π + ], Then µ M i ) = dϕ dxdy = {x,y)ɛc i }

7 Two triangles and trapezium 31 then areac i ) dϕ = π ) areac i, i = 1,, 3), At the same we have 3 µ M i ) = π ) ab. 1) µ N i ) = dϕ dxdy = {x,y)ɛc i } ) areaĉi dϕ, i = 1,, 3), By considering the formulas ), 5), 6) and 7), 9), 1) we have with this expression, the 13) becomes Consequently, µ N i ) = 3 µ N i ) = areaĉi ϕ) = areac i A i ϕ) π ) areac i [A i ϕ)] dϕ. π ) ab From relations 7), 9) and 1) we have 13) [ 3 ] A i ϕ) dϕ. 1) 3 A i ϕ) = 3al cos ϕ+bl sin ϕ lm 13 sin ϕ + cos ϕ + 3 cos cos ϕ sin sin ϕ) then l ctg 1 + cos ϕ) m 1 + cos + sin ), [ 3 ] A i ϕ) dϕ = 3al cos + bl 1 sin ) lm 13 + cos 1 sin + sin ) l ctg π + sin ) m 16 π ) 1 + cos + sin ). 15)

8 3 Alfio Puglisi and Marius Stoka The formulas 11), 1), 1) and 15) give us l References P int = lm 1 π ) [3al cos + bl 1 sin ) ab 13 + cos 1 sin + sin ) ctg π + sin ) m 16 ] π ) 1 + cos + sin ). [1] U. Baesel, A. Duma, A Laplace type problem for a lattice of rectangles with triangular obstacles, Applied Mathematical Sciences, 1), no. 166, [] G. Caristi, E. L. Sorte, M. Stoka, Laplace problems for regular lattices with three different types of obstacles, Applied Mathematical Sciences, 5 11), no. 56, [3] G. Caristi, A. Puglisi, E. Saitta, A Laplace type for regular lattices with convex-concave cell and obstacles rhombus, Applied Mathematical Sciences, 7 13), no., [] G. Caristi, M. Stoka, Laplace type problem for a regular lattice of Dirichlet- Voronoi with different obstacles, Applied Mathematical Sciences, 5 11), no. 3, [5] G. Caristi and M. Stoka, A Laplace type problem for lattice with the axial symmetry and different types obstacles II), Far East Journal of Mathematical Sciences, 6 1), no., [6] G. Caristi and M. Stoka, A Laplace type problem for lattice with axial symmettric and different obstacles type I), Far East Journal of Mathematical Sciences, 5 11), no. 1, [7] G. Caristi and M. Stoka, Some extension of the Laplace problem, Rend. Cric. Mat di Palermo, 6 11), [] H. Poincarè, Calcul des Probabilitès, nd ed., Gauthier-Villard, Paris, 191. [9] M. Stoka, Probabilités géométriques de type Buffon dans le plan euclidien, Atti Accd. Sci. Torino, T ), Received: July 15, 16; Published: February 15, 17