A Laplace Type Problem for a Lattice of Rectangles with Triangular Obstacles

In this paper, we calculate the probability that a small rectangle hits a lattice of rectangles with triangular obstacles. Mathematics Subject Classification: 60D05, 52A22


Introduction
A rectangle D with side lengths L and l, L ≥ l, is randomly thrown onto a plane lattice R a, b, m of rectangles with side lengths 4a and b, and 2a and b, respectively (see Fig. 1).In every corner of the rectangles there is an obstacle that is an isosceles right triangle with legs of length m/2 and hypotenuse of length m √ 2 with m < min(2a, b).The fundamental cell F of R a, b, m is shaded in Fig. 1.Using the Cartesian coordinate system in Fig. 2, we have The area of F is equal to 3ab.After attaching a fixed reference point C to D, we define the random throw of D onto R a, b, m as follows: The coordinates x and respectively; the angle φ between the x-axis and the longer sides of D is a random variable uniformly distributed in [0, 2π].All three random variables are stochastically independent.We only consider events with C ∈ F.
In the following, we assume D to be small with respect to R a, b, m ; this means We denote by d = √ L 2 + l 2 the length of the diagonals of D, and by α the angle between a diagonal and the longer sides of D, hence sin α = l/d and cos α = L/d.Known results for lattices with obstacles are for instance to be found in [1], [2], [3] and [4].
2 Hitting probability Theorem 2.1.If inequality (1) holds, then the probability p that D hits R a, b, m is given by Proof.As reference point C of D we choose its barycentre.Due to the symmetry of F with respect to the y-axis, it is sufficient to consider φ only in the interval [0, π/2] in order to get all positions of D exactly once.Let F(D, φ) be the set of all points (x, y) ∈ F where the rectangle D with coordinates (x, y) and angle φ does not intersect the lattice R a, b, m .Let M be the set of all rectangles D with barycentre in F, and N ⊂ M the set of all rectangles D with barycentre in F, not intersecting R a, b, m .Then, the probability p is given by Stoka's formula where µ is the Lebesgue measure in the Euclidean plane.We compute µ(M) and µ(N ) by means of the elementary kinematic density dx ∧ dy ∧ dφ = dK for the group of motions in the plane (see [5, pp. 126-128], [6, p. 85]), and get Therefore, for the calculation of p it is sufficient to know the continuous function [0, π/2] → R, φ → area F(D, φ), and to integrate it from 0 to π/2 with respect to φ.The set F(D, φ) consists of two hexagons.In order to calculate the dimensions of these hexagons, we have to distinguish the cases φ ∈ [0, π/4) and φ ∈ [π/4, π/2], see Figures 2 and 3.
In the case φ ∈ [0, π/4), F(D, φ) consists of Therefore, we have area F(D, φ) hence Hence area F(D, φ) As plausibility check, from (4) and ( 6) one finds that area F(D, • ) is continuous in φ = π/4 with area F(D, π/4) Remark 2. From ( 2) and (8) it follows that in the set of all rectangles with constant perimeter u = 2(l + L), the maximum of π/2 0 area F(D, φ) dφ and hence the minimum of p is obtained only for squares with side length u/4.

Fig. 1 :
Fig. 1: Lattice R a, b, m with its fundamental cell F