A Laplace Type Problem for Lattice with Cell Composed by Four Isoscele Triangles and the Test Body Rectangle

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1 Applied Mathematical Sciences Vol no HIKARI Ltd A Laplace Type Problem for Lattice with Cell Composed by Four Isoscele Triangles and the Test Body Rectangle Ersilia Saitta Department of Economics University of Messina Via dei Verdi Messina Italy Marius Stoka Sciences Academy of Turin Via Maria Vittoria Torino Italy Copyright c 2016 Ersilia Saitta 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 study a Laplace type problem for a lattice composed by four isoscele triangles considering a rectangle as test body. We determine the probability that a random rectangle intersects a side of the fundamental cell represented in fig. 1. Mathematics Subject Classification: 60D0 2A22 Keywords: Geometric Probability stochastic geometry random sets random convex sets and integral geometry 1 Introduction Considering the several results obtained in recent years by different papers 1] 2] 3] and 7] the authors studied some Buffon-Laplace type problems for particular lattices 4] ] and 6]. In particular Caristi and Stoka 6] considered a

2 362 Ersilia Saitta and Marius Stoka Laplace type problem for three different lattices with non-convex fundamental cells. Starting from these results in this paper we consider as the fundamental cell a lattice composed by four isoscele triangles introducing as test body a rectangle and the Laplace type problems were solved. In other words we computed the probability that a random rectangle intersects the fundamental cell represented in fig Main Results Let R a be the lattice with the fundamental cell C 0 represented in fig.1 A B E C 01 C 03 C 02 C 04 C D F fig.1 With the notations of this figure we have: and C 0 = C 01 C 02 C 03 C 04 AC = AD 2a cos π DF = a 2 cos π 1 areac 0 = a2 2 sin π 2 cos π We want to compute the probability that a rectangle r with a random position and with sides of constant length l m with 0 m < l < α 4 cos π intersects a side of the lattice i.e. the probability P int that r intersects a side of

3 Isoscele triangles test body rectangle 363 the the fundamental cell C 01. The position of the rectangle r is determinated by centre and by the angle ϕ definite in fig. 2 In order to compute the probability P int we consider the limiting positions of rectangle r for a fixed value of ϕ in the cell C 0i i = So we have fig. 2 A a B 1 2 A' b 1 A 1 E 1 c 1 B a 2 B' E '' c E Ĉ 01 ϕ b 2 Ĉ 03 ϕ a 3 C 2 C ' a A 2 A'' Ĉ 02 ϕ E E 2 3 E ' d 1 E '' b c 2 c 4 D 4 D 2 E 4 d 2 Ĉ 04 ϕ d F 2 C 1 a 4 C '' b 3 b 4 D' D'' D 3 c 3 F ' d 3 d 4 C D F D 1 fig.2 F 1 and the following: areaĉ01 ϕ = areac 01 areaaj ϕ 3 By fig.1 and fig. 2 we have: j=1 areaĉ02 ϕ = areac 02 areac j ϕ 4 j=1 areac 03 ϕ = areac 03 areab j ϕ j=1 areac 04 = areac 04 aread j ϕ. 6 j=1 areab ϕ = l2 2 tgϕ

4 364 Ersilia Saitta and Marius Stoka areab 4 ϕ = m2 cos ϕ sin π ϕ 2 sin 2π areaa 8 ϕ = l2 sin 2π ϕ sin 2π ϕ 2 sin π areaa 7 ϕ = m2 π 2 tg ϕ areaa ϕ = m2 sin π ϕ cos π ϕ 2 sin 2π areaa 6 ϕ = α 2 π ϕ m sin π ϕ ] 2 m2 π π π ] 2 tg ϕ ϕ m sin ϕ areaa 1 ϕ = areaa 3 ϕ = a 2 l sin 2π ϕ sin π m cos π ϕ ] sin 2π areaa 4 ϕ = l2 2 tg π ϕ areaa 2 ϕ = m2 2 tg π ϕ π ϕ m sin π ϕ] 2 sin π ϕ 2 sin π cos π ϕ l sin π ϕ m cos 1 π π ] l sin 2 ϕ m cos ϕ π ϕ m sin π ϕ sin π l 2 π 2 tg ϕ 2 π ϕ ] m sin π ϕ ] sin 2π areaa 9 ϕ = areaa 2 ϕ = m2 2 tg π ϕ

5 Isoscele triangles test body rectangle 36 We obtain: areaa ϕ = a cos π π π ] ϕ m sin ϕ l sin 2π ϕ π ϕ m sin π ϕ] 2 sin π π ϕ m sin π ϕ] 2 sin π ϕ 2 sin π cos π ϕ π m 2 tg ϕ. areaĉ01 ϕ = areac 01 { a cos π π π ] ϕ m sin ϕ π π ] a l sin ϕ m cos ϕ 1 4 sin π sin π cos 2ϕ cos π sin 2ϕ cos π l 2 2 cos π sin 2ϕ sin π cos 2ϕ cos π ] m 2 4 sin 2π sin π 6 cos π cos 2ϕ 6 sin π sin 2ϕ 2 cos π cos π 6 sin π ]}. 7 To compute areaĉ03 ϕ we have: areab 8 = m2 2 tg π ϕ areab 7 ϕ = π ϕ m sin π ϕ] 2 sin ϕ 2 sin 2π cos π ϕ areab 6 ϕ = a l sin ϕ m cos ϕ 2 l sin ϕ m cos ϕ π ϕ m sin π ϕ m sin π ϕ] 2 sin 2π

6 366 Ersilia Saitta and Marius Stoka We obtain: 2 l2 2 tgϕ areab 1 ϕ = areaa 8 ϕ = l2 sin 2π ϕ sin 2π ϕ 2 sin π areab 2 ϕ = areaa 8 ϕ = m2 π 2 tg ϕ areab 3 ϕ = a cos π π ϕ m sin π ϕ ] π ϕ π ϕ m sin π ϕ] 2 sin π m cos ϕ π ϕ m sin π ϕ] 2 sin 2π 2 m2 π 2 tg ϕ areab ϕ = areab 8 ϕ = m2 2 tg π ϕ areab 9 ϕ = a cos π π ϕ m sin π ϕ ] π ϕ π ϕ m sin π ϕ] 2 sin π π ϕ m sin π ϕ] 2 sin ϕ 2 sin 2π cos π ϕ π m 2 tg ϕ. areaĉ03 ϕ = areac 03 a 3l2 l sin ϕ m cos ϕ 2 4 sin π l 2 4 sin 2π a cos π cos π ϕ m sin ϕ cos 2ϕ cos π cos π sin 2ϕ sin π cos 2ϕ sin π

7 Isoscele triangles test body rectangle 367 m 2 2 sin 2π sin π 1 cos 2ϕ 2 sin 2π cos π cos 2ϕ sin π ] sin 2ϕ. 8 To compute areaĉ02 ϕ we have: areac ϕ = l2 4 sin π cos 2ϕ cos π areac 6 ϕ = areaa 2 ϕ = m2 2 tg π ϕ areac 2 ϕ = areac 4 ϕ = areaa 7 = m2 2 tg π ϕ areac 1 ϕ = π ϕ m sin π ϕ] 2 sin π ϕ 2 sin π cos π ϕ areac ϕ = m2 2 ctg π ϕ areac 8 ϕ = m2 sin π ϕ cos π ϕ 2 sin 2π areac 3 ϕ = a cos π π ϕ m sin π ϕ ] l sin 2π ϕ π ϕ m sin π ϕ] 2 sin π π ϕ m sin π ϕ] 2 sin π ϕ 2 sin π cos π ϕ π m 2 tg ϕ areac 7 ϕ = a 2 π π ] ϕ m sin ϕ m cos π ϕ π ϕ m sin π ϕ] 2 sin 2π l sin 2π ϕ π ϕ m sin π ϕ] 2 sin π

8 368 Ersilia Saitta and Marius Stoka 2 m2 π 2 tg ϕ areac 9 ϕ = a π π ] l sin 2 ϕ m cos ϕ m sin π ϕ l sin π ϕ m cos π ϕ] 2 sin 2π l sin π ϕ m cos π ϕ] π ϕ m sin π ϕ] 2 sin π We obtain: 2 m2 2 ctg π ϕ. areaĉ02 ϕ = areac 02 = cos π cos ϕ sin π sin ϕ a cos π a π m sin π l sin π m cos π sin ϕ π m sin π ] cos ϕ l 2 4 sin π 1 cos 2π cos 2ϕ sin 2π sin 2ϕ 2 cos π ] cos π m 2 2 sin 2π cos π 1 sin π cos 2ϕ sin π cos π cos π sin 2ϕ cos π π ] cos2 4 sin 2π sin π 3 sin π 2 sin 2π cos π cos π 4 cos 2π To compute areaĉ04 ϕ we have: { sin π cos π sin π sin 2ϕ cos π cos π 2 cos 2π sin π 4 cos π ] cos 2ϕ }. 9

9 Isoscele triangles test body rectangle 369 aread ϕ = areab ϕ = l2 2 tgϕ aread 7 ϕ = areab 7 ϕ = π ϕ m sin π ϕ] 2 sin ϕ 2 sin 2π cos π ϕ aread 6 ϕ = a l sin ϕ m cos ϕ 4 cos π m sin π ϕ l sin ϕ m cos ϕ 2 sin 2π l sin ϕ m cos ϕ π ϕ m sin π ϕ] 2 sin 2π 2 l2 2 tgϕ aread 2 ϕ = m2 2 tg π ϕ aread 1 ϕ = l2 sin 2π ϕ sin 2π ϕ 2 sin π aread 3 ϕ = a 2 π ϕ m sin π ϕ ] l sin 2π ϕ π ϕ m sin π ϕ] 2 sin π m cos ϕ π ϕ m sin π ϕ] 2 sin 2π 2 m2 π 2 tg ϕ aread 9 ϕ = a π π ] 2 ϕ m sin ϕ We obtain: l sin 2π ϕ π ϕ m sin π ϕ] 2 sin π π ϕ m sin π ϕ] 2 sin ϕ 2 sin 2π cos π ϕ π m 2 tg ϕ.

10 370 Ersilia Saitta and Marius Stoka areaĉ04 ϕ = areac 04 cos ϕ m 4 cos π { sin ϕ a l 4 cos π sin π ] sin ϕ l2 4 sin 2π cos π cos ϕ 2 cos π cos π cos 2ϕ sin π sin 2ϕ 2 π cos2 cos π ] sin π 2 sin 2π m 2 1 sin 2ϕ 2 sin 2π cos π sin π cos 2ϕ 3 cos π ] sin π }. Denoting with M i i = the set of all rectangles r that have their centre in cell C 01 and with N i the set of all rectangles r entirely contained in C 0i we have cf. 9] P int = 1 4 i=1 µ N i 4 i=1 µ M i 11 where µ is the Lebesgue measure in the Euclidean plane. To compute the measure µ M i and µ N i we use the kinematic measure of Poincarè cf. 8]: dk = dx dy dϕ where x y are the coordinates of the the centre of rectangle r and ϕ the fixed angle. We can write: µ M i = π/ 0 dϕ dxdy = {xyɛc 01 } then π/ 0 areac 0i dϕ = π areac 0i i = In view of we can write 4 µ M i = π areac i=1 A i ϕ = areac 0i areaĉ0i ϕ

11 Isoscele triangles test body rectangle 371 then µ N i = π/ 0 dϕ {xyɛĉ01ϕ} dxdy = π areac 0i π/ 4 µ N i = π areac 0 i=1 0 π/ 0 A i ϕ] dϕ π/ The formulas and give us i=1 0 ] areaĉ01 ϕ dϕ = 4 ] A i ϕ dϕ. 13 i=1 4 { A i ϕ = a l 4 cos π cos π cos π sin π m cos π cos π 2 cos π sin π 1 ] cos ϕ 2 π sin π cos π cos π 2m cos π cos π ] } sin ϕ l 2 4 sin 2π 2 cos 2π cos π 2 cos π sin π cos π cos π sin π cos 2ϕ 2 cos π sin 2π 2 cos π cos π cos π sin π sin 2ϕ 12 cos 2 π 2 π cos2 cos π sin π ] m2 2 sin 2π {sin π cos 2ϕ sin 2ϕ 2 2 cos π 1 sin π cos 2ϕ cos π cos π sin π sin 2ϕ cos 2 π cos π ] 2 cos π cos π sin 2ϕ sin π cos 2ϕ cos π } 4 sin 2π { cos π 3 sin π cos π sin π 3 sin π 2 sin 2π

12 372 Ersilia Saitta and Marius Stoka cos π cos π cos π 2 cos 2π ] cos 2ϕ sin π cos π sin π 8 sin 2ϕ 2 sin π 2 cos π sin π cos π Consequently π/ 0 i=1 2 cos π 12 sin π cos π 4 cos 2π }. 4 ] { A i ϕ dϕ = a l 4 cos π cos π cos π sin π m cos π cos π 2 cos π sin π 1 ] sin π 2 π sin π cos π cos π 2m cos π cos π ] 2 cos 2π cos π 2 sin π cos π 1 cos π } l2 8 sin 2π cos π cos π sin π sin π 1 cos π 1 2 cos π 2 sin 2π cos π sin π π 12 cos 2 π 2 cos 2π cos π sin π ] m 2 2 sin 2π cos π 1 2 sin π sin π cos π cos π 3 2 sin π 1 cos π cos 2 π cos π sin π ] π 8 sin 2π { cos π 3 sin π cos π sin π sin π 3 4 cos π

13 Isoscele triangles test body rectangle 373 cos π sin π cos π 2 cos 2π ] sin π cos π cos π cos π sin π 8 1 cos π π 2 cos π 2 sin π sin π 2 cos π 12 sin π cos π 4 cos 2π ]}. 14 By formulas and 1 we find probability P int wanted and in particular for m = 0 we have the probability calculated in paper 2]. References 1] U. Baesel A. Duma A Laplace type problem for a lattice of rectangles with triangular obstacles Applied Mathematical Sciences no ] D. Barilla M. Bisaia G. Caristi and A. Puglisi On Laplace type problems I Journal of Pure and Applied Mathematics: Advances and Applications no ] D. Barilla M. Bisaia G. Caristi and A. Puglisi On Laplace type problems II Far East Journal of Mathematical Sciences no ] D. Barilla G. Caristi M. Stoka A Laplace Type Problem for an Irregular Lattice with Cell Composed by Two Isoscele Triangles and an Isoscele Trapezium International Journal of Mathematical Analysis no ] D. Barilla G. Caristi E. Saitta M. Stoka A Laplace type problem for lattice with cell composed by two quadrilaterals and one triangle Applied Mathematical Sciences no ] G. Caristi M. Pettineo M. Stoka A Laplace type problem for three lattices with non-convex cell J. Nonlinear Sci. Appl ] G. Caristi and M. Stoka Some extension of the Laplace problem Rend Cric. Mat. di Palermo

14 374 Ersilia Saitta and Marius Stoka 8] H. Poincaré Calcul des Probabilités ed. 2 Gauthier Villars Paris ] M. Stoka Probabilités géométriques de type Buffon dans le plan euclidien Atti Acc. Sci. Torino Received: July ; Published: January