Applied Mathematical Sciences, Vol. 11, 2017, no. 22, 1049 - 1056
HIKARI Ltd, www.m-hikari.com
https://doi.org/10.12988/ams.2017.73113
A Buffon-Laplace Type Problem for a Lattice
with Cell Composed by Three Triangles and
One Octagon
Alfio Puglisi
Department of Economics, University of Messina
Via dei Verdi, 75
98122 Messina, Italy
Marius Stoka
Sciences Academy of Turin
Via Maria Vittoria, 3
10123 Torino, Italy
c 2017 Alfio Puglisi and Marius Stoka. This article is distributed under the
Copyright 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 consider a lattice with the cell represented in fig. 1
and we compute the probability that a segment of random position and
of constant length intersects a side of the lattice.
Mathematics Subject Classification: 60D05, 52A22
Keywords: Geometric Probability, stochastic geometry, random sets, random convex sets and integral geometry
1
Introduction
The geometric probability problems have an important role in the field of mathematics and its applications. Starting from the results obtained by Poincaré
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Alfio Puglisi and Marius Stoka
[8] and Stoka [9] in recent years many authors have considered several BuffonLapalce type problems for particular fundamental cells: [1], [2], [3], [4], [5], [6]
and [7]. Considering these results, in this paper we study as fundamental cell
a lattice composed by three triangles and an octagon and the Buffon-Laplace
type problem was solved, considering the geometric point view also. We computed the probability that a random segment of constant length intersects the
fundamental cell represented in fig. 1.
2
Main Result
Let < (a, b; α), where a < b and α < π/4, be the lattice with fundamental cell
C0 represented in fig. 1
b−c
A
H
c
D
α
c
C 01
G
a−c
C 03
E
a−c
C 02
c
B
α
c
C
b−c
F
fig.1
From this figure we obtain the following
c=
a − btgα
,
1 − tgα
\ = π − α,
AEH
2
|EH| = |F G| =
a
c< ;
2
\ = π − α;
BEH
b−a
;
cos α − sin α
areaC0 = ab.
(1)
(2)
(3)
(4)
We want to compute the probability that the segment s with random position and of constant length l < a2 intersects a side of lattice <, i.e. the
probability Pint that a segment s intersects a side of the fundamental cell C0 .
1051
Buffon-Laplace type problem
The position of the segment s is determinated by its centre and by the
angle ϕ formed with the line BC (o AD).
To compute the probability Pint we consider the limiting positions of segment s, for a specified value of ϕ, of segment s .
Thus we obtain the fig. 2
H2
A2
A
a5
a1
H
a4
D1
H4
Cˆ01 (ϕ )
A1
D
c7
H1
c6
H3
a2
G4
c5
G2
G
c8
G3
a3
G1
Cˆ03 (ϕ )
E1
E c
1
E3
b3
E2
E4
c2
b2
c4
Cˆ02 (ϕ )
F3
F1
B1
c3
B
C3
F4
b5
b4
ϕ
F
C1
b1
C2
F2
C
fig.2
and
b01 (ϕ) = areaC
b02 (ϕ) = areaC01 −
areaC
5
X
areaai (ϕ) ,
(5)
i=1
b03 (ϕ) = areaC03 −
areaC
8
X
areaci (ϕ) .
i=1
By fig. 2 we have:
areaa1 (ϕ) =
areaa2 (ϕ) = (a − c)
areaa4 (ϕ) =
areaa5 (ϕ) =
l2
sin 2ϕ,
4
l
l2
cos ϕ − sin 2ϕ,
2
4
l2 sin ϕ sin (ϕ − α)
,
2 sin α
b−a l
l2
l2 sin ϕ sin (ϕ − α)
sin ϕ − sin 2ϕ −
,
1 − tgα 2
4
2 sin α
(6)
1052
Alfio Puglisi and Marius Stoka
areaa3 (ϕ) =
l2 sin ϕ sin (ϕ − α)
(b − a) l
sin (ϕ − α) −
.
2 (cos α − sin α)
2 sin α
We obtain:
A1 (ϕ) =
5
X
areaai (ϕ) =
i=1
b−a
1
l sin ϕ − ctgα (1 − cos 2ϕ) l2 ,
1 − tgα
4
(7)
then
b01 (ϕ) = areaC01 − A1 (ϕ) .
areaC
(8)
In the same way we have:
areac1 (ϕ) = −
l2 cos ϕ cos (ϕ + α)
,
2 sin α
cl
l2 cos ϕ cos (ϕ + α)
areac2 (ϕ) = cos ϕ +
,
2
2 sin α
areac3 (ϕ) =
areac5 (ϕ) =
areac4 (ϕ) =
a − btgα
l sin ϕ,
2 (1 − tgα)
l2 cos ϕ cos (ϕ − α)
,
2 sin α
l2 cos ϕ sin (ϕ − α)
b−a
l sin (ϕ − α) −
,
2 (cos α − sin α)
2 cos α
a − btgα
l2
l cos ϕ − [sin 2ϕ − tgα (1 + cos 2ϕ)] ,
areac6 (ϕ) =
2 (1 − tgα)
4
areac7 (ϕ) =
areac8 (ϕ) =
a − btgα
l sin ϕ,
2 (1 − tgα)
(b − a) l
l2
(cos α sin ϕ − sin α cos ϕ)− (ctgα sin 2ϕ − 1 − cos 2ϕ) .
2 (cos α − sin α)
4
We obtain:
1053
Buffon-Laplace type problem
8
X
A3 (ϕ) =
areaci (ϕ) =
i=1
l [a (cos α + sin α) − 2b sin α]
(sin ϕ + cos ϕ) −
cos α − sin α
l2
[(1 + ctgα) sin 2ϕ − (1 + tgα) (1 + cos 2ϕ)] .
4
With this expression formula (6) becomes
(9)
b03 (ϕ) = areaC03 − A3 (ϕ) .
areaC
(10)
Denoting by Mi (i = 1, 2, 3) the set of the segments s which have their
centre in C0i and by Ni the set of segments s all contained in the cell C0i we
have (cf. [9]):
P3
µ (Ni )
,
(11)
Pint = 1 − P3i=1
i=1 µ (Mi )
where µ is the Lebesgue measure in the Euclidean plane.
To compute the above measure µ (Mi ) and µ (Ni ) we use the Poincaré
kinematic measure (cf. [8]):
dk = dx ∧ dy ∧ dϕ,
where x, y are the coordinates of the centre of s and ϕ the fixed angle.
Considering formulas (14) and figure 2, it follows
h πi
.
ϕ ∈ α,
2
We have
π
2
Z
µ (M1 ) = µ (M2 ) =
Z Z
dϕ
dxdy =
{(x,y)C01 }
α
Z
µ (M3 ) =
π
2
Z Z
π
2
(areaC01 ) dϕ =
α
π
2
Z
dϕ
α
Z
dxdy =
(areaC03 ) dϕ =
{(x,y)C03 }
α
Then
3
X
i=1
i.e.
µ (Mi ) =
π
2
− α areaC0 ,
π
2
π
2
− α areaC01 ,
− α areaC03 .
1054
Alfio Puglisi and Marius Stoka
3
X
µ (Mi ) =
π
− α ab.
2
i=1
(12)
In the same way we can write:
π
2
Z
Z Z
µ (N1 ) = µ (N2 ) =
dϕ
dxdy =
{(x,y)Cb01 (ϕ)}
α
π
2
Z
α
Z
µ (N3 ) =
π
2
π
2
Z
− α areaC01 −
2
[A1 (ϕ)] dϕ,
α
Z Z
Z
dϕ
dxdy =
{(x,y)Cb03 }
α
π
2
[areaC01 − A1 (ϕ)] dϕ =
α
π
Z
π
2
Z
b
areaC0i dϕ =
[areaC03 − A3 (ϕ)] dϕ =
π
2
α
π
2
h
i
b03 (ϕ) dϕ =
areaC
α
Z
− α areaC03 −
π
2
[A3 (ϕ)] dϕ.
α
Therefore,
3
X
µ (Ni ) =
π
2
i=1
Z
− α ab −
π
2
[2A1 (ϕ) + A3 (ϕ)] dϕ.
(13)
α
Relations (11), (12), and (13) give
Pint =
π
2
1
− α ab
Z
π
2
[2A1 (ϕ) + A3 (ϕ)] dϕ.
(14)
α
Considering (7) and (9), we obtain
2A1 (ϕ) + A3 (ϕ) = (2b − a) l sin ϕ +
a (cos α + sin α) − 2b sin α
l cos ϕ−
cos α − sin α
l2
[(1 + ctgα) sin 2ϕ − (1 + tgα + 2ctgα) cos 2ϕ + 1 − tgα + 2ctgα] .
4
Then
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Buffon-Laplace type problem
Z
π
2
[2A1 (ϕ) + A3 (ϕ)] dϕ =
α
b (2 cos 2α − sin 2α) + a (sin 2α − cos 2α)
l−
cos α − sin α
π
i
l2 h
(1 + ctgα) sin 2α − (1 + tgα + 2ctgα) cos 2α +
− α (2ctgα − tgα + 1) .
4
2
With this value, (14) become
Pint =
h
π
2
1
− α ab
b (2 cos 2α − sin 2α) + a (sin 2α − cos 2α)
l−
cos α − sin α
(1 + ctgα) sin 2α − (1 + tgα + 2ctgα) cos 2α +
π
2
− α (2ctgα − tgα + 1)
i l2 4
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Alfio Puglisi and Marius Stoka
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Received: June 15, 2016; Published: April 12, 2017