PROCEEDINGS OF THE YEREVAN STATE UNIVERSITY
Physical and Mathematical Sciences
2012, № 3, p. 3–8
Mathematics
MEAN DISTANCE BETWEEN TWO POINTS IN A DOMAIN
N. G. AHARONYAN∗
Chair of Probability Theory and Mathematical Statistics YSU, Armenia
Let D be a bounded convex domain in the Euclidean plane and we choose
uniformly and independently two points in D. How large is the mean distance
m(D) between these two points? Up to now, there were known explicit
expressions for m(D) only in three cases, when D is a disc, an equilateral triangle
and a rectangle. In the present paper a formula for calculation of mean distance
m(D) by means of the chord length density function of D is obtained. This
formula allows to find m(D) for those domains D, for which the chord length
distribution is known. In particular, using this formula, we derive explicit forms of
m(D) for a disc, a regular triangle, a rectangle, a regular hexagon and a rhombus.
Keywords: chord length distribution function, mean distance, convex domain
geometry.
Introduction. Let D be a bounded convex domain in the Euclidean plane
\ (with area || D || and perimeter | D | ), and we choose uniformly and independently
two points P1 and P2 from D . How large is the mean distance m(D) between these
two points? In other words, we want to calculate the following quantity [1]:
m(D) = 1/ || D ||2 ∫ ∫ ρ ( P1 , P2 )dPdP
(1.1)
1
2,
2
DD
where ρ ( P1 , P2 ) is the Euclidean distance between P1 and P2 , dPi (i = 1, 2) is the
two dimensional Lebesgue measure. Up to now, there were known explicit
expressions for m(D) only in three cases when D is the disc K r of a radius r [2]:
m( K r ) = 128r 45π
(1.2)
for an equilateral triangle Δa of side a:
m(Δ a ) = a(1/ 5 + 3ln/20),
(1.3)
and for a rectangle Ra, b with sides a ≤ b [3]:
m( Ra, b ) =
1 ⎡ a 3 b3
a 2 b2 ⎞
2
2 ⎛
+
+
+
−
− ⎟+
a
b
3
⎢
⎜
15 ⎣⎢ b 2 a 2
b2 a2 ⎠
⎝
5 ⎛ b2 a + a 2 + b2 a2 b + a 2 + b2
+ ⎜ ln
+ ln
2⎜ a
b
b
a
⎝
∗
E-mail: [email protected]
⎞⎤
⎟⎥ .
⎟⎥
⎠⎦
(1.4)
Proc. of the Yerevan State Univ. Phys. and Mathem. Sci., 2012, № 3, p. 3–8.
4
In the present paper a formula for calculation of mean distance m(D) by
means of the chord length density function of D is obtained, which allows to find
an explicit form of mean distance m(D) for those domains D, for which the chord
length distribution is known. In particular, using this formula, we derive explicit forms
of m(D) for a disc, a regular triangle, a rectangle, a regular hexagon and a rhombus.
2. The Main Formula. In this section we obtain a formula for calculation
m(D) in terms of chord length density function for domain D .
Let χ ( g ) = g ∩ D, g ∈[D], is a chord in D , where [D] = {g ∈ G : g ∩ D ≠ ∅}
is the set of lines intersecting domain D ( G is the space of all undirected lines in the
plane \ 2 ). It is well known the following formula between m(D) and the integral
I4 =
4
∫ χ ( g ) dg :
[D]
I 4 = 6 || D ||2 m(D),
(2.1)
where dg is the locally finite measure in the space G , which is invariant with respect
to the group of all Euclidean motions in the plane [1, 4].
By definition, FD ( y ) = 1/ | ∂D | ∫ dg is the chord length distribution
χ ( g )≤ y
function for D [5, 6].
To transform the formula for m(D), we write
I 4 =| ∂D |
+∞
∫
x 4 f D ( x)dx,
(2.2)
0
where f D ( x) is the chord length density function of D (that is f D ( x) = FD′ ( x) is the
first derivative of the distribution function). Finally, from (2.2) and (2.1), we get
m(D) =| ∂D | / 6 || D ||2
+∞
∫
x 4 f D ( x)dx,
(2.3)
0
Therefore, if we know the explicit form of f D ( x) for domain D , using (2.3)
we can calculate m(D) . Further we demonstrate the efficiency of formula (2.3).
3. The Case of a Disc. Consider a disc of radius r, D = K r . In this case the
chord length density function has the following form [5]:
if x ∉ (0, 2r ),
⎧⎪0,
f Kr ( x) = ⎨
2
2
2
⎪⎩ x /(4r 1 − x / 4r ), if x ∈ (0, 2r ).
2r
Therefore, we get m( K r ) = 1/ 6π r 4 ∫ x 5 / 4r 2 − x 2 dx = 128r / 45π , i.e. (1.2) [1, 5, 7].
0
4. The Case of a Regular Triangle. The chord length density function for a
regular triangle Δa with side a has the following form [5, 8, 9]:
⎧
⎛ a 3⎤
1
π
, if x ∈ ⎜⎜ 0,
+
⎪0, if x ∉ [0, a ];
⎥;
2a 3a 3
2 ⎦
⎪
⎝
(4.1)
f Δa ( x) = ⎨
⎛a 3 ⎤
4 x 2 − 3a 2
2
a 3
2π
⎪1
+
−
arcsin
,
if x ∈ ⎜⎜
, a⎥.
⎪ 2a −
2 x 3a 3
2 x2
a 3
⎝ 2
⎦
⎩
Aharonyan N. G. Mean Distance Between Two Points in a Domain.
5
Substituting the values of f Δ a ( x) from (4.1) into (2.3), we obtain
m( Δ a ) =
8 ⎡1 a 4
π
x dx +
∫
3 ⎢
2
a
3a ⎣ 0
3a 3
3a / 2
∫
x 4 dx −
0
2π
3a 3
a
∫
x 4 dx −
3a / 2
(4.2)
a 3 ⎤
∫ x 4 x − 3a dx + a 3 ∫ x arcsin 2 x dx ⎥ .
⎥⎦
3a / 2
3a / 2
Using some standard formulas we obtain m(Δ a ) = a (1/ 5 + 3ln 3/ 20) (1.3) [1–3, 5]).
5. The Case of a Rectangle. It is well known that the chord length density
function for a rectangle Ra, b with sides a ≤ b has the following form [5–7, 9, 10]:
1
−
2
a
2
2
2
2
a
4
⎧
⎪
2
2
x / (a + b), if x ∈ (0, a];
⎪0, if x ∉[0, a + b );
⎪ 2
f Ra, b ( x) = ⎨a b /(a + b) x 2 x2 − a 2 ,
if x ∈ (a, b];
(5.1)
⎪
⎛
⎞
ab
a
b
1
⎪
⎜
⎟−
+
, if x ∈ (b, a 2 + b2 ].
⎪ ( a + b) x 2 ⎜ 2
2
2
2 ⎟ a+b
x −b ⎠
⎝ x −a
⎩
Substituting the values of f Ra, b ( x) from (5.1) into (2.3), we obtain
1 ⎡ a5
m( Ra ,b ) = 2 2 ⎢ + a 2b
3a b ⎣ 5
a 2 + b2
∫
a
x2
x2 − a2
dx +
(5.2)
b5 ( a 2 + b 2 ) 2 a 2 + b 2 ⎤
⎥.
+ ab ∫
−
dx +
5
5
b
x2 − b2
⎦⎥
After some calculations we obtain (1.4) [1, 2]. In the case a = b we get the mean
distance for a square with side a
m(,a ) = a /15(2 + 2 + 5ln(1 + 2)).
(5.3)
6. The Case of a Regular Hexagon. For a regular hexagon Ha with side a
we have: | H a |= 6a, || H a ||= 3 3a 2 / 2. The density function has the following form
(see [6, 8–10]):
⎧
1⎛1
π ⎞
−
, if y ∈ (0, a];
⎪0, if y ∉ (0, 2a ];
⎜
a ⎝ 2 6 3 ⎟⎠
⎪
⎪
4 y 2 − 3a 2
a 3
2
⎪ π
−
+
f Ha ( y) = ⎨
arcsin
,
if y ∈ (a, a 3]; (6.1)
2y
2 y2
⎪ 2a 3 a 3
⎪ π
a 3
1
1
6a 2 − y 2
⎪
−
−
+
arccos
, if y ∈ (a 3, 2a ].
2
2
2
y
⎪ 6a 3 2a a 3
−
y
y
a
3
⎩
4 ∞ 4
Therefore, m( H a ) =
∫ x f H a ( x)dx . Substituting (6.1) form, we get
27a 3 0
2
m( H a ) =
a2 +b2
x2
4
1 a⎛ 1
π ⎞ 4
π ⎞
⎛1
I + I 2 + I 3 ] , where I1 = ∫ ⎜ −
x dx = a 4 ⎜ −
⎟
⎟;
3[ 1
a 0⎝ 2 6 3 ⎠
27 a
⎝ 10 30 3 ⎠
Proc. of the Yerevan State Univ. Phys. and Mathem. Sci., 2012, № 3, p. 3–8.
6
I2 =
π
a 3
2a 3
a
∫
x 4 dx −
2
a 3
a 3
a
∫
x 4 arcsin
a 3 2
a 3
x 4 x 2 − 3a 2
dx + ∫
dx;
2x
2
a
2a
1 ⎞
1 2a 4
a 3
6a 2 − x 2 2
⎛ π
− ⎟x 4 dx −
x arccos
dx + ∫
x dx .
∫
⎜
2a ⎠
x
a 3a 3
x 2 − 3a 2
a 3 ⎝ 6a 3
a 3
Calculating I2 and I3, we obtain finally:
⎛ 4π 227
⎞
478π
113 3
3
2 + 3 14
−
+
+
−
+ ln 3 ⎟⎟ .
m( H a ) = a ⎜⎜
ln
240
160
15
3
⎝ 15 2160 3645 3
⎠
7. The Case of Rhombus. In terms of elementary functions the chord length
distribution function for rhombus R(a,γ) for π / 3 < γ ≤ π / 2 has the form [10]:
I3 =
2a
∫
(1 + (π / 2 − γ )cot γ )/ 2a, if x ∈[0, a sin γ );
⎧0, if x ∉[0, 2a cosγ / 2];
⎪
2
2
2
⎪ 1 ⎛1 − ⎛ π + γ − 2arcsin a sin γ ⎞ cot γ ⎞ + a sin γ + x cosγ , if x ∈[a sin γ , a);
⎟
⎟
⎪ 2a ⎜⎝ ⎜⎝ 2
x ⎠
⎠ x2 x2 − a2 sin2 γ
⎪
⎪⎪ 1 ⎛ ⎛ π
⎞
a2 sin2 γ
γ⎞
⎡
⎞
fR( a,γ ) ( x) = ⎨− ⎜1 + ⎜ − γ ⎟ cot γ ⎟ +
, if x ∈ ⎢a, 2a sin ⎟;
2
2
2
2
2⎠
⎣
⎠ x x − a sin γ
⎪ 2a ⎝ ⎝ 2 ⎠
⎪
⎪ −1 + ⎛ γ − 2arcsin a sin γ ⎞ cot γ
⎜
⎟
⎪
a2 sin2 γ + x2 cosγ
γ
γ⎞
x ⎠
⎡
⎝
+
, if x ∈⎢2a sin , 2a cos ⎟.
⎪
4a
2
2⎠
⎣
2x2 x2 − a2 sin2 γ
⎪⎩
Since |R(a,γ)| = 4a, ||R(a,γ)|| = a2sinγ, using (2.3) we obtain
∞
2
m(R(a, γ )) = 3 2 ∫ x4 f R(a,γ ) ( x)dx.
3a sin γ 0
m 2 ( R ( a , γ )) =
m3 ( R ( a , γ )) =
and
⎤
⎛π
⎞
⎢1 + ⎜ 2 − γ ⎟ cot γ ⎥dx ,
⎝
⎠
⎦
a sin γ
x4 ⎡
0
2a ⎣
We denote by m1 ( R ( a , γ )) = ∫
a
∫
a sin γ
x 4 ⎡⎛ ⎛ π
a sin γ
⎢ ⎜ 1 − ⎜ + γ − 2 arcsin
2 a ⎣⎝ ⎝ 2
x
a
a 2 x 2 sin 2 γ + x 4 cos γ
a sin γ
x 2 x 2 − a 2 sin 2 γ
∫
⎞⎤
⎞
⎟ cot γ ⎟ ⎥ dx
⎠
⎠⎦
dx .
We have m2 (R(a, γ )) =
a4 a4 sin5 γ a4 (π + 2γ )cot γ a4 (π + 2γ )cosγ sin4 γ cot γ
−
−
+
+
G1,
10
10
20
20
a
where G1 = a5 sin 5 γ
∫
1
(1/ x 6 ) arcsin xdx . Calculating G1 and m3(R(a,γ)), we get
sin γ
⎡π
γ
1 ⎡ cos γ
3cos γ 3 1 + cos γ
− ⎢
+
+ ln
G1 = − a 5 sin 5 γ ⎢ −
5
4
2
sin γ
⎣10 5sin γ 5 ⎣ 4sin γ 8sin γ 8
⎡ sin 2 γ cos γ
m3 ( R ( a , γ )) = a 4 ⎢
⎣
Further, we have
2
+
⎤⎤
⎥ ⎥ and
⎦⎦
⎤
sin 4 γ
3
1 + cos γ cos 2 γ 3 2
(1 + cos γ ) ln
+
+ sin γ cos 2 γ ⎥ .
2
4
sin γ
4
8
⎦
Aharonyan N. G. Mean Distance Between Two Points in a Domain.
⎡ a4
m4 ( R ( a, γ )) = ⎢
−
7
16a 4 sin 5 (γ / 2) ⎤ ⎡
⎤
π
− γ ) cot γ ⎥ + M1a 2 sin 2 γ
⎥ ⎢1 + (
2
⎦⎣
⎦
5
⎣ 10
8
m5 ( R(a, γ )) = ⎡⎣ a 4 sin 5 (γ / 2) − a 4 cos5 (γ / 2) + γ a 4 cot γ cos5 (γ / 2) − γ a 4 cot γ sin 5 (γ / 2) ⎤⎦ −
5
− K1 cot γ / 2a,
where M 1 =
2 a sin(γ / 2)
∫
a
x2
x − a sin γ
2
and at least m6 ( R ( a, γ )) =
2
2
dx
and
K1 =
2 a cos(γ / 2)
∫
x 4 arcsin
2 a sin(γ / 2)
a sin γ
dx ,
x
1 2 a cos(γ / 2) a 2 x 2 sin 2 γ + x 4 cos γ
dx .
∫
2 2 a sin(γ / 2)
x 2 − a 2 sin 2 γ
Using a standard integral, we obtain
γ cos γ
⎛
M 1 = a 2 ⎜ 2sin 3 −
2
2
⎝
2
⎞ a
2
⎟ + sin γ ln
⎠ 2
γ
γ
2sin (1 + sin )
2
2 , besides we get
1 + cos γ
⎡
γ
π −γ
cos(γ / 2)
sin(γ / 2)
K1 = − a5 sin 5 γ ⎢ −
+
−
+
−
5
5
4
4
⎣ 10sin (γ / 2) 10cos (γ / 2) 20sin (γ / 2) 20cos (γ / 2)
−
3cos(γ / 2)
3sin(γ / 2)
3 cos(γ / 2) + cos 2 (γ / 2) ⎤
+
−
ln
⎥.
40sin 2 (γ / 2) 40cos 2 (γ / 2) 40 sin(γ / 2) + sin 2 (γ / 2) ⎦
Since m( R ( a, γ )) =
m(R(a, γ )) =
2
3a sin 2 γ
3
6
∑ mi ( R (a, γ )),
finally we have
i =1
2a ⎡1 sinγ cosγ 3 3
1 4
1+ cosγ cos2 γ
+
+
sin
γ
cos
γ
+
sin
γ
(10
+
9cos
γ
)ln
+
+
20
40
20
sinγ
4
3sin2 γ ⎢⎣5
3
8
γ 8π
γ
γ sin4 γ
+ sin2 γ cos2 γ − sin5 + sin5 cot γ + sin2 γ sin3 +
ln
8
5
2 5
2
2
2
γ⎛
γ⎞
2sin ⎜1+ sin ⎟
2⎝
2⎠
−
1+ cosγ
8
γ 8
γ
γ ⎞ sin4 γ cosγ ⎡
γ
π −γ
⎛
− cos5 + γ cot γ ⎜ cos5 − sin5 ⎟ +
+
−
⎢−
5
5
5
2 5
2
2⎠
2
⎝
⎣ 10sin (γ /2) 10cos (γ /2)
γ
γ
γ ⎤
sin
3cos
3sin ⎥
2
2
2
2 + sin2 γ cos3 γ + 2cosγ ⎛ cos5 γ − sin5 γ ⎞ +
−
+
−
+
⎜
⎟
γ
γ
γ
γ⎥
2
2
2⎠
⎝
20sin4
20cos4
40sin2
40cos2 ⎥
2
2
2
2⎦
cos
γ
γ
γ⎤
cos + cos2 ⎥
3 2
γ
3
γ
1
2 .
+ sin γ cos3 cosγ − sin2 γ sin3 cosγ + sin4 γ (20 + 3cosγ )ln 2
⎥
γ
4
2
4
2
80
2γ
sin + sin ⎥
2
2⎦
Therefore, m( R (a, γ )) = C (γ )a, were C (γ ) is a constant depending only on
the angel γ ∈ [π / 3, π / 2].
8
Proc. of the Yerevan State Univ. Phys. and Mathem. Sci., 2012, № 3, p. 3–8.
In the case γ = π / 2, C (π / 2) coincides with the rectangle (5.3)
C (π / 2) = 1/15 ⎡⎣ 2 + 2 + 5ln(1 + 2) ⎤⎦ .
Similarly we can consider the case γ ∈ [0, π / 3).
8. Conclusion. We have calculated explicit values for m(D) for some
particular cases, for which there exist explicit form of the chord length density
function. We would like to stress that in [8] there exist an explicit form for the
chord length distribution function for any regular polygon. In particular, for a
regular pentagon the explicit form for chord length distribution you can find in [5].
Therefore, using the result of [8], we can calculate m(D) for any regular polygon.
Received 18.06.2012
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Exp. Techn. Phys., 1988, v. 36, p. 197–208.
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Contemporary Mathem. Analysis (Armenian Academy of Sciences), 2005, v. 40, № 4, p. 43–56.
6. Harutyunyan H.S. Chord Length Distribution Functions for Regular Hexagon. // Uchenie
Zapiski EGU, 2007, № 1, p. 17–24 (in Russian).
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International Journal of Mathematical Science Education, 2009, v. 2, № 1, p. 1–12.
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Polygons. // Sutra: International Journal of Mathem. Science Education, 2011, v. 4, № 2, p. 1–15.
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Aharonyan N. G. Mean Distance Between Two Points in a Domain.
9
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Среднее расстояние между двумя точками области
Независимо и с равномерным распределением выбраны две точки из
ограниченной выпуклой области (D) в евклидовой плоскости. В настоящее
время существуют явные выражения для среднего расстояния m(D) между
этими точками для круга, равностороннего треугольника и прямоугольника. В
статье приведена формула для вычисления m(D) с помощью плотности длины
хорды области D. Она позволяет находить m(D) для тех D, для которых
распределение длины хорды известно. В частности с использованием этой
формулы выведены явные выражения m(D) для круга, правильного треугольника, прямоугольника, правильного шестиугольника и ромба.