projective reduction - a new technique for calculating expected

Projective Reduction - a new technique for

calculating expected values in the field of

geometrical probability

Bertrand F. Evertz

January 22, 2012

Abstract

Projective Reduction, a new technique for calculating expected val-

ues and distribution functions in the field of geometric probability, is

introduced and applied to several fundamental problems. The so far un-

known values of the mean distance of the points and the corresponding

distribution function are calculated for the arbitrary triangle, the regular

pentagon, the regular hexagon and the regular tetrahedron. Additional

results and possible applications are presented.

email-address of the author: [email protected]

Link zur deutschen Originalarbeit

1

mailto:[email protected]

http://evertz-web.de/math/p_de

Bertrand Evertz Projective Reduction

Contents

1 Dedication 3

2 Introduction 4

3 The Mean Distance of the Points in an Arbitrary Triangle 6

3.1 The Principle of Projective Reduction . . . . . . . . . . . . . . . 6

3.2 First Step of Reduction . . . . . . . . . . . . . . . . . . . . . . . 8

3.3 Further Steps of Reduction . . . . . . . . . . . . . . . . . . . . . 11

3.4 Calculating the Expected Value . . . . . . . . . . . . . . . . . . . 13

4 The Mean Distance of the Points of a Convex Polygon 17

4.1 The Regular Pentagon . . . . . . . . . . . . . . . . . . . . . . . . 17

4.2 The Regular Hexagon . . . . . . . . . . . . . . . . . . . . . . . . 23

4.3 Irregular Convex Polygons . . . . . . . . . . . . . . . . . . . . . . 27

5 The Mean Distance of the Points in Non-Convex Areas 30

5.1 The Concave Quadrilateral . . . . . . . . . . . . . . . . . . . . . 30

5.2 Separated Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.3 A Star as a Complex Example . . . . . . . . . . . . . . . . . . . 35

6 Determining the Distribution Functions 39

6.1 The Distribution Function in Two-dimensional Figures . . . . . . 40

6.2 Calculating the Distribution Functions . . . . . . . . . . . . . . . 41

6.3 The Distribution Function within an Arbitrary Triangle . . . . . 52

7 The Mean Distance of Points in the Regular Tetrahedron 56

7.1 The Expected Value . . . . . . . . . . . . . . . . . . . . . . . . . 56

7.2 The Distribution Function . . . . . . . . . . . . . . . . . . . . . . 59

7.3 Expected Value for the Hyper-Tetrahedron . . . . . . . . . . . . 61

8 Afterword 63

9 Literature 64

2


1 Dedication

It is unusual for a physician, even more so for a psychiatrist, to write a longermathematical article. Doing this nonetheless is due to the great joy of scientificand philosophical thinking, instilled in me by my father, the mathematician andcomputer pioneer W. P. Evertz, who encouraged and supported me all my life.This work is therefore dedicated to him.

3


2 Introduction

”Mathematics, rightly viewed, possesses not only truth, but supreme beauty . . . ”– Bertrand Russel

For more than 250 years many mathematicians have been fascinated by thecalculation of geometrical probabilities because of the connection of geometry,stochastics and analysis and even more so because of tangible, even seeminglysimple problem constellations, which however often require complex problemsolving approaches.

Engaging with Sylvester’s 4-point-problem led to significant theoretical progressincluding Crofton’s formula, which made it possible (among other things) tocalculate the mean area of a randomly selected triangle within a regular polygonor the mean volume of a randomly selected n-dimensional hyper-tetrahedronwithin an n-dimensional hyper-ball.

It is therefore surprising that other issues that seem to be less demanding,like calculating the expected value and the distribution of the distance of tworandomly selected points, remained unsolved in many 2-dimensional figures,especially in the most elemental geometrical figure, the arbitrary triangle. Itshows that even in this simple case the usual analytical procedure leads to an4-dimensional integral (considering both coordinates of both points), which canonly be solved with a feasible amount of effort when the boundary conditionsare more or less symmetrical. With the help of computer algebra systems theexpected value of the distance of two arbitrary points could be computed forat least some specific triangles. However, the analytical challenge is increasessignificantly in non-symmetrical scenarios, so no general solution has be foundso far.

The technique of Projective Reduction bypasses this problem with its basicapproach of reducing the degrees of freedom of the randomly selected pointsand thereby simplifying the resulting integrals. This shows some similaritiesto Crofton’s formula, which fixates one of the points onto the boundary of theexamined figure. In contrast however, the technique of Projective Reduction canbe applied repeatedly, thus fixating two or more points onto the boundaries. Inchapter 3 this unfamiliar approach is first illustrated using specific partial stepsand then explained generally.

The high degree of accordance with the approach of Johan Philip is noteworthy;his concept of using “size” and “shape” of the figures formed by the randompoints creates distinctions similar to the “factor of dilation” and “expected valueof the projection” used by Projective Reduction and made it possible to solvesubstantial problems such as finding the expected volume of an arbitrary tetra-hedron within a cube. Projective Reduction in my humble opinion goes beyondPhilip’s approach in some aspects: First, the figure resulting from the randomlyselected points can have a different dimension than the given figure they areselected from, which makes it possible to examine different problems. Second,Projective Reduction generally demonstrates the independence of the expectedvalue of the projection and the dilation factor while Philip demonstrated the in-dependence of “size” and “shape” for each examined problem separately. Third,

4


breaking down the total projection into single, separate steps permits the intro-duction of precise case distinctions, which make it possible to examine concavepolygons, for example. And finally, since the expected values can be determinedwithout the much more demanding calculation of the corresponding distributionfunctions, the Projective Reduction gives you additional control with regard tothe correctness of its results.

5


3 The Mean Distance of the Points in an Arbi-

trary Triangle

In this paper a special notation is being used to describe the examined expectedvalues: For example, E(Q1Q2 |Q1 ∈ M1 ∧ Q2 ∈ M2) represents the expectedvalue of the length of the line segment Q1Q2, assuming that Q1 and Q2 areuniformly distributed in M1 respectively M2. M1 and M2 are always convexsets of points (or can be subdivided into a limited number of convex sets) likeline segments or convex polygons. In the case of M1 being a single point P , thisnotation condenses to E(PQ |Q ∈ M2).

3.1 The Principle of Projective Reduction

A B

C

b

b

Q=⇒

A = Z B

C

Br

Cr

b

b

Q

Q′

⇓ ∗ 23

A B

C

b

b

Q′

Figure 1: Triangle Point Picking

Let us first consider the simplified case of the mean distance of a point Q,selected arbitrarily within the triangle △ABC, to the vertex A, with Q beinguniformly distributed within the area of △ABC. This is usually referred to asTriangle Point Picking (compare figure 1).

The corresponding expected value for the length of the line segment AQ isdenoted as

E1 = E(AQ |Q ∈ △ABC)

6


If you now map the triangle △ABC onto △ABrCr, using a dilation withZ = A being the center of dilation and r < 1 being the dilation factor,the probability that the point Q lies within △ABrCr is given by F (r) =A(△ABrCr) : A(△ABC) = r2, due to the uniform distribution of Q within△ABC. The density function for the event of Q lying on BrCr must thereforebe f(r) = F ′(r) = 2r .

Given that Q is lying on BrCr, the position of Q is uniformly distributed withinBrCr (a fact which does not apply to figures with curved boundaries). The uni-form distribution is evident when you consider a dilation with two differentfactors of dilation r and (r + h). The probability for Q lying within the emerg-ing trapezoid BrCrC(r+h)B(r+h) is uniformly distributed within the trapezoid.

With h → 0 and accordingly BrCrC(r+h)B(r+h) −→ BrCr, the given uniformdistribution within the trapezoid is being transferred to the uniform distributionof Q within BrCr for Q ∈ BrCr.

The new expected value E1r, which depends on r, is now denoted as

E1r = E(AQ |Q ∈ BrCr)

According to the rules for calculating expected values this leads to:

E1 =

1∫

0

f(r)E1r dr =

1∫

0

f(r)E(AQ |Q ∈ BrCr) dr

The term f(r)E(AQ |Q ∈ BrCr) can be further transformed based on followingconsiderations:

If we project △ABrCr back onto △ABC using A as the dilation center and 1r

as the dilation factor, the image Q′ of Q must be lying on BC and is uniformlydistributed within this line segment. We receive AQ′ = 1

rAQ, independent of

the position Q within BrCr. Therefore

E1r = E(AQ |Q ∈ BrCr)

= r E(AQ′ |Q′ ∈ BC)

= r E(AQ |Q ∈ BC)

Since the term E(AQ |Q ∈ BC) in independent of r, it shows that:

E1 =

1∫

0

f(r)E(AQ |Q ∈ BrCr) dr

=

1∫

0

f(r)rE(AQ |Q ∈ BC) dr

=

1∫

0

f(r)r dr E(AQ |Q ∈ BC)

7


=

1∫

0

2r2 dr E(AQ |Q ∈ BC)

=2

3E(AQ |Q ∈ BC)

Therefore we arrive at the following:

Theorem 1 For any triangle △ABC applies:

E(AQ |Q ∈ △ABC) =2

3E(AQ |Q ∈ BC)

(Triangle Point Picking)

By projecting the arbitrary point within the triangle onto one side we hencesucceeded in reducing a two-dimensional expected value to a one-dimensionalexpected value.

3.2 First Step of Reduction

With the assistance of a case differentiation the same approach can be used forcalculating the distance of two arbitrary points within a triangle (see figure 2).

The expected value is now denoted as:

E3 = E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ △ABC)

We once more want to project one of the two random points onto one side ofthe triangle, while preventing the second from being projected outside of thetriangle.

If we continuously contract the triangle △ABC by using a dilation with thecenter being Z = A and the dilation factor r getting continuously smaller, thenthe line segment BrCr being the image of BC will meet Q1 for some r = r1 andmeet Q2 for some other value r = r2.

Depending on the value of r the probability for Q1 to lie within △ABrCr isF1(r) = rd1 , with d1 being the dimension of the convex set of points thatcontains Q1. Accordingly, the density function describing the event of Q1 ∈BrCr is f1(r) = F1

′(r) = d1 rd1−1. Similarly we have F2(r) = rd2 and f2(r) =

F2′(r) = d2 r

d2−1.

The probability that both points Q1 and Q2 lie within △ABrCr is then givenby

F (r) = F1(r)F2(r)

The density function f(r) which represents the case that the “outer” point liesexactly on BrCr results in

f(r) = F ′(r) = f1(r)F2(r) + f2(r)F1(r)

8


1. case: P (r1 ≥ r2) =d1

d1+d2= 1

2 2. case: P (r1 < r2) =d2

d1+d2= 1

2

A B

C

bb

Q1

Q2

Cr1

Br1

⇓

A = Z B

C

bb b

b Q′1

Q′2

Q1

Q2

∗ 45⇓

A B

C

bb Q1

Q2

A B

C

b

b

Q1

Q2

Cr2

Br2

⇓

A = Z B

C

b

b

b

b

Q′1

Q′2

Q1

Q2

⇓ ∗ 45

A B

C

b

b

Q1

Q2

Figure 2: First step of reduction

9


with f1(r)F2(r) representing the partial density of Q1 being exactly on BrCr

while Q2 lies within △ABC. The associated expected value of the distanceQ1Q2 isE(Q1Q2 |Q1 ∈ BrCr∧Q2 ∈ △ABrCr). The same applies to f2(r)F1(r).

The quotient of the two cases is constant for all values of r and is given by

f1(r)F2(r)

f2(r)F1(r)=

d1 rd1+d2−1

d2 rd1+d2−1=

d1d2

This leads to

f1(r)F2(r) =d1

d1 + d2f(r) and f2(r)F1(r) =

d2d1 + d2

f(r)

So we have shown, that the two cases illustrated in figure 2 are stochasticallyindependent of r and have a common basic density function f(r). Based on thisfact we can now rearrange the expected value E3 as follows:

E3 =

1∫

0

f1(r)F2(r)E(Q1Q2 |Q1 ∈ BrCr ∧Q2 ∈ △ABrCr)

+f2(r)F1(r)E(Q1Q2 |Q1 ∈ △ABrCr ∧Q2 ∈ BrCr) dr

=

1∫

0

d1d1 + d2

f(r)E(Q1Q2 |Q1 ∈ BrCr ∧Q2 ∈ △ABrCr) dr

+

1∫

0

d2d1 + d2

f(r)E(Q1Q2 |Q1 ∈ △ABrCr ∧Q2 ∈ BrCr) dr

The term E(Q1Q2 |Q1 ∈ BrCr ∧Q2 ∈ △ABrCr) can be transformed by apply-ing the following consideration:When you project △ABrCr back onto △ABC using a dilation by 1

r with the

dilation center A, you have Q′1 (being the image of Q1) lying on BC with

uniform distribution. Q′2 (being the image of Q2) is lying within the triangle

△ABC with uniform distribution as well, since Q2 has been uniformly dis-tributed within △ABrCr. Furthermore we have Q′

1Q′2 = 1

rQ1Q2, independentof the given position of Q1 and Q2. Therefore we get

E(Q1Q2 |Q1 ∈ BrCr ∧Q2 ∈ △ABrCr)

= rE(Q′1Q

′2 |Q′

1 ∈ BC ∧Q′2 ∈ △ABC)

= rE(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC)

and similarly

E(Q1Q2 |Q1 ∈ △ABrCr ∧Q2 ∈ BrCr)

= rE(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ BC)

10


Thus we have

E3 =

1∫

0

d1d1 + d2

f(r) r E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC) dr

+

1∫

0

d1d1 + d2

f(r) r E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ BC) dr

=

1∫

0

d1d1 + d2

(d1 + d2)rd1+d2 dr E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC)

+

1∫

0

d2d1 + d2

(d1 + d2)rd1+d2 dr E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ BC)

=d1

d1 + d2 + 1E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC)

+d2

d1 + d2 + 1E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ BC) (1)

Inserting the concrete values d1 = d2 = 2 results in:

E3 = E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ △ABC)

=2

5E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC)

+2

5E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ BC)

=4

5E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC) (2)

Thus we have reduced the original 4-dimensional expected value to a 3-dimensionalone.

3.3 Further Steps of Reduction

To shorten the presentation of the following steps we will generalize the deriva-tion above.

We have two (not necessarily identical) convex sets of points called M1 (ofdimension d1) and M2 (of dimension d2), in which Q1 respectively Q2 are ran-domly selected (with uniform distribution). Now we select a common pointZ ∈ M1 ∩ M2 as the center of a dilation with the dilation factor r < 1constituting the images M1r and M2r, depending on r. By selecting Z wealso determine the marginal sets ρM1 and ρM2, which are defined as follows:ρM1 := limr→1(M \M1r) and ρM2 := limr→1(M \M2r). Furthermore we defineρM1r respectively ρM2r as the images of the sets ρM1 and ρM2 being dilatedby r with the dilation center Z (constituting BrCr in the example above).

As we saw in the previous chapter, the only requirement for transforming theexpected value in the way shown above is Q1 being uniformly distributed in

11


ρM1r whenever Q1 ∈ ρM1r (respectively Q2 being uniformly distributed inρM2r whenever Q2 ∈ ρM2r). This is always true if ρM1 and ρM2 are convexsets of points themselves as is explained in the following:

Given the construction of ρM1, this set is a part of the boundary of M1 hav-ing the dimension d1 − 1. Furthermore ρM1 can not be curved into the d1-dimensional set M1 since this would contradict ρM1 being convex.

Therefore ρM1 is a perfectly plane surface of dimension d1 − 1. It follows thatfor each r < 1, ρM1r must be a perfectly plane surface as well, contained inM1. If we let r become smaller in equal steps, we subdivide M1 into equallythick layers, with homogeneous density for the position of Q1. If we then let thesteps become infinitely small we transfer the homogeneous density in regard ofthe volume of the layers into the homogeneous density of the area of the planesurface ρM1r as we wanted to show. The same applies for the distribution ofQ2 within ρM2r.

Therefore the equation (1) is applicable whenever ρM1 and ρM2 are convex setsof points, so that the following formula can be applied for simple line segments,triangles, tetrahedrons and so on. Thus we have:

Theorem 2 Let M1 and M2 be convex sets of points of dimension d1 respec-tively d2. If a point Z ∈ M1 ∩ M2 can be found as a center of dilation by r(r < 1) mapping M1 and M2 onto M1r respectively M2r in a way that the re-sulting sets of points ρM1 := limr→1(M1 \M1r) and ρM2 := limr→1(M2 \M2r)are convex sets of points as well, the following applies:

E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M2)

=d1

d1 + d2 + 1E(Q1Q2 |Q1 ∈ ρM1 ∧Q2 ∈ M2)

+d2

d1 + d2 + 1E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ ρM2)

(Basic Theorem of Projective Reduction)

The first weighting factor d1

d1+d2+1 is composed of the probability d1

d1+d2for the

appearance of the first projection possibility and the actual reduction factord1+d2

d1+d2+1 . The same applies for the second weighting factor.

Theoren 2 is now being applied to the intermediate result given in equation (2).Since Q1 ∈ BC = M1 with d1 = 1 and Q2 ∈ △ABC = M2 with d2 = 2 we canselect Z = B as the center of dilation with the resulting projective sets beingρM1 = C and ρM2 = AC.

Hence we get:

E3 =4

5E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ △ABC)

=4

5(1

4E(Q1Q2 |Q1 = C ∧Q2 ∈ △ABC)

+2

4E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AC))

=1

5E(CQ |Q ∈ △ABC)

12


+2

5E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AC)

The first term is the triangle point picking discussed above and leads, accordingto theorem 1, to:

1

5E(CQ |Q ∈ △ABC) =

1

5

2

3E(CQ |Q ∈ AB)

To be able to apply Theorem 2 for the second term with Q1 ∈ BC = M1

(d1 = 1) and Q2 ∈ AC = M2 (d2 = 1) we have to select Z = C as the onlypossible center of dilation with ρM1 = B and ρM2 = A. This results in

2

5E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AC)

=2

5

1

3E(Q1Q2 |Q1 = B ∧Q2 ∈ AC)

+2

5

1

3E(Q1Q2 |Q1 ∈ BC ∧Q2 = A)

=2

5

1

3E(BQ |Q ∈ AC) +

2

5

1

3E(AQ |Q ∈ BC)

So finally we have

E3 =2

15

(

E(AQ |Q ∈ BC) + E(BQ |Q ∈ AC) + E(CQ |Q ∈ AB))

(3)

The mean distance of two points in an arbitrary triangle is therefore 215 of the

sum of the mean distances of each vertex to the points of the side facing thatvertex. The summarizing schematic illustration of this derivation is shown infigure 4.

3.4 Calculating the Expected Value

The remaining three expected values, representing one-dimensional integrals,can now easily be calculated: Following the labels in figure 3 (with p beingnegative if B ∈ FC) we get:

E2 = E(AQ |Q ∈ BC)

=

a−p∫

−p

1

a

√

c2 − p2 + x2 dx

=1

2a

(

cp+ (a− p)√

a2 + c2 − 2ap

+(c2 − p2)(ln(a− p+√

a2 + c2 − 2ap)− ln(c− p)))

From c2 − p2 = b2 − (a− p)2 follows p = a2−b2+c2

2a . Inserting this into the termabove we finally, after some transformation, arrive at:

13


AB = c

AC = b

BC = a

BF = p

A B

C

Fb

Q

Figure 3: Sketch for calculating E(AQ |Q ∈ BC)

Theorem 3 In a triangle with the side lengths a, b and c, the mean distanceE2 of a randomly selected point on side a to the vertex facing this side is givenby:

E2(a, b, c) =b+ c

4+

(b+ c)(b − c)2

4a2

+(a+ b+ c)(−a+ b+ c)(a− b+ c)(a+ b− c) ln a+b+c

−a+b+c

8a3

(Point To Line Picking)

Applying this to equation (3) finally leads to:

Theorem 4 The mean distance E3 of two randomly selected points within atriangle with the side lengths a, b and c is given by:

E3(a, b, c) =a+ b+ c

15

+(b+ c)(b − c)2

30a2+

(a+ c)(a− c)2

30b2+

(a+ b)(a− b)2

30c2

+1

60(a+ b+ c)(−a+ b+ c)(a− b+ c)(a+ b− c)

(

ln a+b+c−a+b+c

a3+

ln a+b+ca−b+c

b3+

ln a+b+ca+b−c

c3

)

(Triangle Line Picking)

14


Examples

Inserting values for the two known special cases, the equilateral triangle and theright-angled isosceles triangle, we receive the correct expected values:

E3(1, 1, 1) =1

20(4 + 3 ln 3) ≈ 0.364791843 . . .

E3(1, 1,√2) =

1

30(2 + 4

√2 + (4 +

√2) ln(1 +

√2)) ≈ 0.414293302 . . .

However, the mean distance in arbitrary triangles can be calculated. For ex-ample, inserting the side lengths for the smallest Pythagorean Triangle resultsin:

E3(3, 4, 5) =20460 + 9728 ln2 + 5103 ln3

22500≈ 1.458184635 . . .

15


b

b

45

b

b

14

24

b

bb

b

23

13

13

b

b

b

b

b

b

b : Q1 or Q2 : center of dilation Z

Figure 4: The steps of reduction in a Triangle

16


4 The Mean Distance of the Points of a Convex

Polygon

Now we want to apply the results found above to convex polygons consistingof more than three sides. Since the mean distance of two points in a squareis already known, we will go on to higher polygons. Again we will use linesegments and triangles as the basic sets the reduction process will be appliedto, but we will also have to weight the single triangles the polygons are composedof according to their area, since this represents the probability with which Q1

respectively Q2 lie within any of those (sub-)triangles.

4.1 The Regular Pentagon

A B SF

M

D

CE

b

3π

10

π

10

b

4π

10

Figure 5: The regular pentagon

We start with the distance between two points selected randomly within theregular pentagon with a side length of 1 (see figure 5). For this purpose we needthe probability p, with which a randomly selected point will lies within △ABD.p results from the ratio of the corresponding areas:

p =A(△ABD)

5 ∗A(△ABM)=

DF

5FM=

tan 4 π10

5 tan3π10

=

√

1

5

We also need the length of the diagonal:

d = AD = BS =

√5 + 1

2

Now we repeatedly apply the basic theorem of Projective Reduction to theexpected value we are looking for by dividing the pentagon into 3 trianglesby using the two diagonals intersecting at the selected center of dilation Z

17


b

b

45

b

b

b

b

241−p2

b

b

24p

︸︷︷︸

p+15

b

b

241−p2

b

b

14

b

b

13

b

b

13

b

b

231−p2

b

b

231−p2

b

b

23p

︸︷︷︸

4(1−p)15

︸︷︷︸

2 p15

b : Q1 or Q2 : center of dilation Z

Figure 6: The steps of reduction within the regular pentagon

18


(see figure 6). How much each triangle contributes to the expected value isdetermined by the ratio of the area of each triangle to the area of the wholepentagon. Equivalent expected values are summarized intermediately:

E5 = E(Q1Q2 |Q1 ∈ ABCDE ∧Q2 ∈ ABCDE)

(Z=E)=

2

5

(1− p

2E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ ABCDE)

+pE(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ ABCDE)

+1− p

2E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ ABCDE)

)

+2

5

(1− p

2E(Q1Q2 |Q1 ∈ ABCDE) ∧Q2 ∈ AB)

+pE(Q1Q2 |Q1 ∈ ABCDE) ∧Q2 ∈ BC)

+1− p

2E(Q1Q2 Q1 ∈ ABCDE) ∧Q2 ∈ CD)

)

=4

5E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ ABCDE)

(Z=A)=

4

5

1

4E(Q1Q2 |Q1 = B ∧Q2 ∈ ABCDE)

+4

5

2

4

(1− p

2E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ BC)

+pE(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

+1− p

2E(Q1Q2|Q1 ∈ AB ∧Q2 ∈ DE)

)

=1

5E(BQ |Q ∈ ABCDE)

+1− p

5E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ BC)

+1 + p

5E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

(Z=B)=

1

5

2

3

(1− p

2E(BQ |Q ∈ CD) + pE(BQ |Q ∈ DE)

+1− p

2E(BQ |Q ∈ AE)

)

+1− p

5

1

3

(

E(AQ |Q ∈ BC) + E(CQ |Q ∈ AB))

+1 + p

5E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

So we get:

E5 = E(Q1Q2 |Q1 ∈ ABCDE ∧Q2 ∈ ABCDE)

=4

15(1− p)E(AQ |Q ∈ BC) +

2

15pE(AQ |Q ∈ CD)

+1

5(p+ 1)E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD) (4)

The first two expected values can be calculated according to theorem 3. How-ever, this theorem cannot be applied to the last term noted above since no

19


common center of dilation exists. To solve this problem we use the elementarycalculus of probability:

Let’s consider the triangle△ASD in figure 7 in which Q1 and Q2 were randomlyselected within the line segments AS respectively DS. The probability of one ofthe points lying within a specific part of the line segment is given by the ration

of this part to the total length of the line segment. Let p1 = BSAS

and p2 = CSDS

.

A B SQ1

Q2

D

C

Figure 7: The mean distance of points of line segments not being parallel

Hence we arrive at:

E(Q1Q2 |Q1 ∈ AS ∧Q2 ∈ DS)

= (1− p1)(1 − p2)E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

+ p1(1− p2)E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ CD)

+ p2(1− p1)E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CS)

+ p1p2E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ CS)

Furthermore we have

(1− p2)E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ CD)

= E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ DS)

− p2E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ CS)

and

(1− p1)E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CS)

= E(Q1Q2 |Q1 ∈ AS ∧Q2 ∈ CS)

− p1E(Q1Q2 |Q1 ∈ BS ∧Q2 ∈ CS)

20


Since eight of the terms annul each other, we finally get to

E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=1

3(1− p1)E(AQ |Q ∈ DC) +

1

3(1− p2)E(DQ |Q ∈ AB)

− p13(1− p1)

E(BQ |Q ∈ CD)− p23(1− p2)

E(CQ |Q ∈ AB)

By reinserting the original ratio of the given line segments for p1 and p2 we getthe following theorem:

Theorem 5 If you randomly select two points on the line segments AB respec-tively CD not being parallel with S = AB ∩ CD, B ∈ AS and C ∈ DS, thenthe expected value of the distance of the two points is given by

E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=AS

3ABE(AQ |Q ∈ DC) +

DS

3CDE(DQ |Q ∈ AB)

− BS

3ABE(BQ |Q ∈ CD)− CS

3CDE(CQ |Q ∈ AB)

(line to line picking 1)

Now we insert the length of the line segments of the pentagon and get:

E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=d+ 1

3E(AQ |Q ∈ DC) +

d+ 1

3E(DQ |Q ∈ AB)

− d

3E(BQ |Q ∈ CD)− d

3E(CQ |Q ∈ AB)

Inserting this into equation (4) we get:

E5 =4

15(1− p)E(AQ |Q ∈ BC) +

2

15pE(AQ |Q ∈ CD)

+1

15(p+ 1)(d+ 1)E(AQ |Q ∈ DC)

+1

15(p+ 1)(d+ 1)E(DQ |Q ∈ AB)

− 1

15(p+ 1)dE(BQ |Q ∈ CD)− 1

15(p+ 1)dE(CQ |Q ∈ AB)

Adding identical expected values leads to:

E5 =2

15(2(1− p)− d(p+ 1))E(AQ |Q ∈ BC)

+2

15(p+ (p+ 1)(d+ 1))E(AQ |Q ∈ CD)

22


By finally reinserting p =√

15 and d =

√5+12 and using theorem 3 we get:

E5 =2

15(√5 + 2)E(AQ |Q ∈ CD)− 2

15(√5− 1)E(AQ |Q ∈ BC)

=2

15(√5 + 2)E2(1,

√5 + 1

2,

√5 + 1

2)− 2

15(√5− 1)E2(1,

√5 + 1

2, 1)

=1

120

((18

√5 + 40) ln(

√5 + 2)− (11

√5 + 20) ln(5) + 4

√5 + 28

)

≈ 0.675160011193203 . . .

Normalizing the regular pentagon to the area of 1 results in:

E5N ≈ 0.51473325292936 . . .

4.2 The Regular Hexagon

A B S

F

D

C

E

Figure 8: The regular hexagon

Considering the fact that △ABC amounts to 16 and △ACD to 1

3 of the area ofthe hexagon in figure 8 we come to following transformation:

E6 = E(Q1Q2 |Q1 ∈ ABCDEF ∧Q2 ∈ ABCDEF )

(Z=F)=

2

5

(1

6E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ ABCDEF )

+1

3E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ ABCDEF )

+1

3E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ ABCDEF )

+1

6E(Q1Q2 |Q1 ∈ DE ∧Q2 ∈ ABCDEF )

)

+2

5

(1

6E(Q1Q2 |Q1 ∈ ABCDEF ) ∧Q2 ∈ AB)

23


=16

45E(Q1Q2 |Q1 = A ∧Q2 ∈ CD)

− 2

45E(Q1Q2 |Q1 = A ∧Q2 ∈ BC)

+2

15E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ DE)

(Theorem3)=

16

45E2(1, 2,

√3)− 2

45E2(1,

√3, 1)

+2

15E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ DE)

For determining E(Q1Q2 |Q1 ∈ AB ∧ Q2 ∈ DE) Theorem 5 is not applicablesince AB||DE; due to this circumstance the corresponding integral can howeverbe easily calculated. Using the notation of the general case shown in figure 9with AB||DE we get:

A B

D C

E Q1

Q2

b

AB = a

CD = c

DE = h

AE = e

Figure 9: The mean distance of points of parallel line segments

E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=

∫ c

0

∫ a−e

−e

√

(x− y)2 + h2

acdx dy

=

∫ c

0

1

2ac

(

h2 arsinhy + e

h− h2 arsinh

y + e − a

h

+ (y + e)

√

h2 + (y + e)2

− (y + e− a)

√

h2 + (y + e− a)2

)

dy

=1

6 a c

(

3h2(

(c+ e)arsinhc+ e

h− e arsinh

e

h

− (c+ e− a)arsinhc+ e − a

h+ (e− a)arsinh

e − a

h

)

− (3h2 − (h2 + (c+ e)2))√

h2 + (c+ e)2

25


+ (3h2 − (h2 + (c+ e− a)2))√

h2 + (c+ e− a)2

+ (3h2 − (h2 + e2))√

h2 + e2

− (3h2 − (h2 + (e − a)2))√

h2 + (e− a)2)

Inserting the corresponding sides and diagonals of the trapezoid ABCD resultsin

Theorem 6 If you randomly select two points on the line segments AB respec-tively CD with AB||CD, then the expected value of the distance of the two pointsis given by

E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=h2

2ABCD

(√

AC2 − h2 arcsinh

√

AC2 − h2

h−AC

+

√

BD2 − h2 arcsinh

√

BD2 − h2

h−BD

−√

AD2 − h2 arcsinh

√

AD2 − h2

h+AD

−√

BC2 − h2 arcsinh

√

BC2 − h2

h+BC

)

+AC

3+BD

3 −AD3 −BC

3

6ABCD

with h being the distance between AB and CD.

(line-to-line picking 2)

Inserting the length of the given line segments in equation (5) results in:

E6 =16

45(3 ln (3)

4+ 1)− 2

45

6 ln(√

3 + 2)− 3 ln (3) + 12

√3− 4

16

+2

15(3

2ln 3 + 2

√3− 10

3)

=−6 ln

(√3 + 2

)+ 171 ln (3) + 84

√3− 28

360≈ 0.8262589494902 . . .

Normalizing the regular hexagon to an area of 1, we now have:

E6N ≈ 0.5126137288 . . .

The following list of the mean distances of two points within regular polygonsnormalized to an area of 1 indicates a fairly quick convergence when the numberof vertices increase and it shows alignment with results found earlier:

26


Number of vertices k Value of EkN

3 0.5543637207 . . .4 0.5214054331 . . .5 0.5147332529 . . .6 0.5126137288 . . .

∞ 0.5108255918 . . .

4.3 Irregular Convex Polygons

Using the presented theorems you can calculate the mean distance of two pointsin any convex polygon since the two randomly selected points can always be pro-jected with determinated probability onto either two different sides or onto onevertex and one side of the polygon, thus making theorems 3, 5 or 6 applicable.However, the effort required for the calculations will increase roughly propor-tional to the square of the number of vertexes, in contrast to the roughly linearprogression in the case of regular polygons due to their high degree of symmetry.

As a simple example we will calculate the mean distance of two points within aconvex quadrilateral containing no parallel sides (see figure 10).

A B

D

C

Figure 10: The convex quadrilateral

For this we need the quotients of the given areas called p1 = A(△ABC)A(�ABCD) und

p2 = A(△ABD)A(�ABCD) .

Applying theorem 2 repeatedly we get (the selected center of dilation beingindicated each time):

E(Q1Q2 |Q1 ∈ �ABCD ∧Q2 ∈ �ABCD)

(Z=A)=

4

5p1E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ �ABCD)

+4

5(1− p1)E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ �ABCD)

27


(Z=C)=

1

4

4

5p1E(BQ |Q ∈ �ABCD)

+2

4

4

5p21E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AB)

+2

4

4

5p1(1− p1)E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AD)

+1

4

4

5(1− p1)E(DQ |Q ∈ �ABCD)

+2

4

4

5(1− p1)

2E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ AD)

+2

4

4

5p1(1− p1)E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ AB)

Theorem5=

1

5p1

( 2

3p2E(BQ |Q ∈ AD) +

2

3(1 − p2)E(BQ |Q ∈ CD)

)

+2

5p21

( 1

3E(AQ |Q ∈ BC) +

1

3E(CQ |Q ∈ AB)

)

+2

5p1 (1− p1)

( 1

3

p1p1 − (1 − p2)

E(AQ |Q ∈ BC)

+1

3

p2p2 − (1− p1)

E(BQ |Q ∈ AD)

+1

3

1− p1(1− p1)− p2

E(CQ |Q ∈ AD)

+1

3

1− p2(1− p2)− p1

E(DQ |Q ∈ BC))

+1

5(1− p1)

( 2

3p2E(DQ |Q ∈ AB)

+2

3(1− p2)E(DQ |Q ∈ AD)

)

+2

5(1− p1)

2( 1

3E(AQ |Q ∈ CD) +

1

3E(CQ |Q ∈ BC)

)

+2

5p1 (1− p1)

( 1

3

1− p1(1− p1)− (1− p2)

E(AQ |Q ∈ CD)

+1

3

1− p2(1− p2)− (1− p1)

E(BQ |Q ∈ CD)

+1

3

p1p1 − p2

E(CQ |Q ∈ AB)

+1

3

p2p2 − p1

E(DQ |Q ∈ AB))

Thus we have

Theorem 7 The mean distance of two points in a convex quadrilateral �ABCD,with no sides being parallel, is given by:


=2

15

(p21p2

p1 + p2 − 1E(AQ |Q ∈ BC)

+(1− p1)

2p2(1− p1) + p2 − 1

E(AQ |Q ∈ CD)

28


+p22p1

p2 + p1 − 1E(BQ |Q ∈ AD)

+(1− p2)

2p1(1− p2)+p1 − 1

E(BQ |Q ∈ CD)

+p21(1− p2)

p1 + (1− p2)− 1E(CQ |Q ∈ AB)

+(1− p1)

2(1− p2)

(1− p1) + (1− p2)− 1E(CQ |Q ∈ AD)

+p22(1− p1)

p2 + (1− p1)− 1E(DQ |Q ∈ AB)

+(1− p2)

2(1− p1)

(1− p2) + (1− p1)− 1E(DQ |Q ∈ BC)

)

with p1 = A(△ABC)A(�ABCD) and p2 = A(△ABD)

A(�ABCD) .

(convex quadrilateral line picking)

29


5 TheMean Distance of the Points in Non-Convex

Areas

In line with the traditional approach we have so far considered only convexpolygons. However, the technique of Projective Reduction makes it possible tocalculate the mean distance of two points for some specific concave polygonsand even some unconnected areas. We will start with a simple figure and thenmove on to more complex applications.

5.1 The Concave Quadrilateral

A BF

D

CE

Figure 11: The concave quadrilateral

Consider a concav quadrilateral �ABCD, with C lying within △ABD. Byextending the sides BC and CD to their opposite sides we get the intersectionsE and F (see figure 11). Furthermore we need the quotients of the given areas

p1 = A(△ABC)A(�ABCD) and p2 = A(△ABD)

A(�ABCD) .

By selecting the center of dilation appropriately we can start determining theexpected value of the line segment Q1Q2 the same way as in the case of theconvex quadrilateral:


(Z=A)=

4

5p1E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ �ABCD)

+4

5(1 − p1)E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ �ABCD)

(Z=C)=

1

4

4

5p1E(BQ |Q ∈ �ABCD) (5)

30


+2

4

4

5p21E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AB)

+2

4

4

5p1(1− p1)E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AD) (6)

+1

4

4

5(1− p1)E(DQ |Q ∈ �ABCD)

+2

4

4

5(1− p1)

2E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ AD)

+2

4

4

5p1(1− p1)E(Q1Q2 |Q1 ∈ CD ∧Q2 ∈ AB)

The term is formally identical with the corresponding term in the case of theconvex quadrilateral. However, since the relative position of the line segmentsconstituting the expected values in line 5 and line 6 do not allow the use ofTheorem 5 we have to examine them separately.

The expected value in line (5) is found applying the following reasoning:

E(BQ |Q ∈ △ABD)

=1

p2E(BQ |Q ∈ �ABCD) + (1− 1

p2)E(BQ |Q ∈ △BCD)

Thus

E(BQ |Q ∈ �ABCD)

=p22

3E(BQ |Q ∈ AD) + (1− p2)

2

3E(BQ |Q ∈ CD)

The result is identical with the convex case.

The expected value in line (6) can be transformed into:

E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AD)

=AE

ADE(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AE)

+DE

ADE(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ DE)

(Theorem5)=

AE

AD

( AE

3AEE(AQ |Q ∈ BC)

+BE

3BCE(BQ |Q ∈ AE)

− CE

3BCE(CQ |Q ∈ AE)

)

+DE

AD

( DE

3DEE(DQ |Q ∈ BC)

+BE

3BCE(BQ |Q ∈ DE)

− CE

3BCE(CQ |Q∈DE)

)

=AE

3ADE(AQ |Q ∈ BC)

31


+DE

3ADE(DQ |Q ∈ BC)

+BE

3BCE(BQ |Q ∈ AD)

− CE

3BCE(CQ |Q ∈ AD)

By replacing the quotients of the line segments with the corresponding quotientsof the areas we arrive at:

E(Q1Q2 |Q1 ∈ BC ∧Q2 ∈ AD)

=1

3

p1p1 + p2 − 1

E(AQ |Q ∈ BC)

+1

3

p2p1 + p2 − 1

E(BQ |Q ∈ AD)

+1

3

1− p11− p1 − p2

E(CQ |Q ∈ AD)

+1

3

1− p21− p1 − p2

E(DQ |Q ∈ BC)

Again we get exactly the same formal term as in the case of the convex quadrilat-eral. The remaining expected values are all calculated according to the presentedtheorems and are therefore no different from the convex case either. Thus wehave:

Theorem 8 The mean distance of two points in a concave quadrilateral �ABCDwith C ∈ △ABD is given by:


=2

15

(p21p2

p1 + p2 − 1E(AQ |Q ∈ BC)

+(1− p1)

2p2(1− p1) + p2 − 1

E(AQ |Q ∈ CD)

+p22p1

p2 + p1 − 1E(BQ |Q ∈ AD)

+(1− p2)

2p1(1− p2)+p1 − 1

E(BQ |Q ∈ CD)

+p21(1− p2)

p1 + (1− p2)− 1E(CQ |Q ∈ AB)

+(1− p1)

2(1− p2)

(1− p1) + (1− p2)− 1E(CQ |Q ∈ AD)

+p22(1− p1)

p2 + (1− p1)− 1E(DQ |Q ∈ AB)

+(1− p2)

2(1− p1)

(1− p2) + (1− p1)− 1E(DQ |Q ∈ BC)

)

with p1 = A(△ABC)A(�ABCD) and p2 = A(△ABD)

A(�ABCD) .

(concave quadrilateral line picking)

32


5.2 Separated Areas

Consider in the plane a set of points M with the area A, consisting of threedisjunct sets of points M1, M2 and M3 with the areas A1, A2 and A3 (see figure12).

M1

M2

M3

Figure 12: Disjunct Areas Picking

If you select two points at random within M the expected value of the distanceof the two points corresponds to:

E(Q1Q2 |Q1 ∈ M ∧Q2 ∈ M)

=A2

1

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M1)

+A2

2

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

+A2

3

A2E(Q1Q2 |Q1 ∈ M3 ∧Q2 ∈ M3)

+ 2A1 A2

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M2)

+ 2A2 A3

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M3)

+ 2A1 A3

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M3)

33


= −A22

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

+(A2

1

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M1)

+ 2A1 A2

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M2)

+A2

2

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

)

+(A2

2

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

+ 2A2 A3

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M3)

+A2

3

A2E(Q1Q2 |Q1 ∈ M3 ∧Q2 ∈ M3)

)

+ 2A1 A3

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M3)

= −A22

A2E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

+(A1 +A2)

2

A2E(Q1Q2 |Q1 ∈ M1 ∪M2 ∧Q2 ∈ M1 ∪M2)

+(A2 +A3)

2

A2E(Q1Q2 |Q1 ∈ M2 ∪M3 ∧Q2 ∈ M2 ∪M3)

+ 2A1 A3

A2E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M3)

Therefore following theorem results:

Theorem 9 For the set of points M with the area A, consisting of three disjunctsets of points M1, M2 and M3 with the areas A1, A2 and A3, the mean distanceof a point selected randomly in M1 to a point selected randomly in M3 is givenby:

E(Q1Q2 |Q1 ∈ M1 ∧Q2 ∈ M3)

=A2

2A1 A3E(Q1Q2 |Q1 ∈ M ∧Q2 ∈ M)

+A2

2

2A1A3E(Q1Q2 |Q1 ∈ M2 ∧Q2 ∈ M2)

− (A1 +A2)2

2A1 A3E(Q1Q2 |Q1 ∈ M1 ∪M2 ∧Q2 ∈ M1 ∪M2)

− (A2 +A3)2

2A1 A3E(Q1Q2 |Q1 ∈ M2 ∪M3 ∧Q2 ∈ M2 ∪M3)

(disjunct areas picking)

If the terms on the right side of the equation can be calculated then the expectedvalue for the disjunct areas M1 and M3 can be calculated too. This is alwaystrue when M is an convex polygon and M1 and M3 are being cut of from M

34


using two straight lines. If M2 = ∅ the formula above can be simplified byA2 = 0.

5.3 A Star as a Complex Example

A B

D

CE

F

G

I

H

J

Figure 13: The pentagram

The theorems defined so far make it possible to calculate the mean distance oftwo points for a variety of shapes. As an example you see in figure (13) thepentagram that results by extending the sides of the regular pentagon with aside length of AB = 1.

The quotient of the area of the pentagon︷︸︸︷

ABCDE to one of the outer spikes△AGB is given by

A(︷︸︸︷

ABCDE)

A(△AGB)=

√5

By summarizing identical expected values on the run the expected value of the

35


distance of two points selected within the pentagram can be broken down into:

E(Q1Q2 |Q1 ∈︷︸︸︷

AGBHCIDJEF ∧Q2 ∈︷︸︸︷

AGBHCIDJEF )

=

√5√5

(5 +√5)2

E(Q1Q2 |Q1 ∈︷︸︸︷

ABCDE ∧Q2 ∈︷︸︸︷

ABCDE)

+ 25√5

(5 +√5)2

E(Q1Q2 |Q1 ∈︷︸︸︷

ABCDE ∧Q2 ∈ △AGB)

+5

(5 +√5)2

E(Q1Q2 |Q1 ∈ △AGB ∧Q2 ∈ △AGB)

+ 25

(5 +√5)2

E(Q1Q2 |Q1 ∈ △AGB ∧Q2 ∈ △BHC)

+ 25

(5 +√5)2

E(Q1Q2 |Q1 ∈ △AGB ∧Q2 ∈ △CID)

(Th.9)=

3−√5

8E(Q1Q2 |Q1 ∈

︷︸︸︷


ABCDE)

+3√5− 5

4

(

1 +√5)2

2√5

E(Q1Q2 |Q1 ∈ �CDEG ∧Q2 ∈ �CDEG)

− 1

2√5E(Q1Q2 |Q1 ∈ △AGB ∧Q2 ∈ △AGB)

−√5

2E(Q1Q2 |Q1 ∈

︷︸︸︷


ABCDE))

+3−

√5

8E(Q1Q2 |Q1 ∈ △AGB ∧Q2 ∈ △AGB)

+3−

√5

4

(

(3 +√5)2

2E(Q1Q2 |Q1 ∈ �BHJG ∧Q2 ∈ �BHJG)

+(1 +

√5)2

2E(Q1Q2 |Q1 ∈ �ABCJ ∧Q2 ∈ �ABCJ)

− (2 +√5)2

2E(Q1Q2 |Q1 ∈ △CJG ∧Q2 ∈ △CJG)

− (2 +√5)2

2E(Q1Q2 |Q1 ∈ △AHJ ∧Q2 ∈ △AHJ)

)

+3−

√5

4

((2 +√5)2

2E(Q1Q2 |Q1 ∈ △EGI ∧Q2 ∈ △EGI)

+(√5)2

2E(Q1Q2 |Q1 ∈

︷︸︸︷


ABCDE)

− (1 +√5)2

2E(Q1Q2 |Q1 ∈ �CDEG ∧Q2 ∈ �CDEG)

− (1 +√5)2

2E(Q1Q2 |Q1 ∈ �ABIE ∧Q2 ∈ �ABIE)

)

36


=3−

√5

8E(Q1Q2 |Q1 ∈

︷︸︸︷


ABCDE)

− 7 + 3√5

2E(Q1Q2 |Q1 ∈ △AHJ ∧Q2 ∈ △AHJ)

3 +√5

2E(Q1Q2 |Q1 ∈ �BHJG ∧Q2 ∈ �BHJG)

By inserting the known term for the regular pentagon and applying theorem4 for the triangle △AHJ and theorem 8 for the quadrilateral �BHJG using

p1 = 12 and p2 = A(△GHJ)

A(△BGH) =√5+34 we receive

E(Q1Q2 |Q1 ∈︷︸︸︷


AGBHCIDJEF )

=3−

√5

8

( 2

15(√5 + 2)E(AQ |Q ∈ CD)

− 2

15(√5− 1)E(AQ |Q ∈ BC)

)

− 7 + 3√5

8

( 2

15E(AQ |Q ∈ HJ) + 2

2

15E(HQ |Q ∈ AJ)

)

3 +√5

2

(

22

15

1 +√5

8E(JQ |Q ∈ BG)

+ 22

15

2 +√5

4E(GQ |Q ∈ HJ)

− 22

15

√5− 2

4E(GQ |Q ∈ BH)

+ 22

15

3−√5

8E(BQ |Q ∈ GJ)

)

=1

60

(

(1 +√5)E(AQ |Q ∈ CD)− (4

√5− 8)E(AQ |Q ∈ BC)

− (7 + 3√5)E(AQ |Q ∈ HJ)− (14 + 6

√5)E(HQ |Q ∈ AJ)

+ (8 + 4√5)E(JQ |Q ∈ BG) + (22 + 10

√5)E(GQ |Q ∈ HJ)

− (2√5− 2)E(GQ |Q ∈ BH) + 4E(BQ |Q ∈ GJ)

)

=1

60

(

(1 +√5)E2(1,

√5 + 1

2,

√5 + 1

2)

− (4√5− 8)E2(1,

√5 + 1

2, 1)

− (7 + 3√5)E2(

√5 + 2,

√5 + 3

2,

√5 + 3

2)

− (14 + 6√5)E2(

√5 + 3

2,√5 + 2,

√5 + 3

2)

+ (8 + 4√5)E2(

√5 + 1

2,√5 + 2,

√

5 + 2√5)

+ (22 + 10√5)E2(

√5 + 2,

√5 + 2,

√5 + 3

2)

− (2√5− 2)E2(

√

5 + 2√5,√5 + 2,

√5 + 1

2)

37


+ 4E2(√5 + 2,

√

5 + 2√5,

√5 + 1

2))

Inserting the explicit formula for the function E2 according to theorem 3, wearrive at:

E(Q1Q2 |Q1 ∈︷︸︸︷


AGBHCIDJEF )

=1

240

(

(29√5 + 65) ln

((√5 + 3)

√

2√5 + 5− 2

√5− 5

)

+ (3√5− 5) ln

((√5 + 1)

√

2√5 + 5 + 2

√5 + 5

)

− 20 (√5 + 2) ln(5)

− 10 (√5 + 1)

√

2√5 + 5 + 66

√5 + 170

)

≈1.38569603515211 . . .

The plausibility of the result can be checked by conducting a Monte CarloSimulation. A run of 108 arbitrarily selected line segments presented a meanvalue of 1.38570264 . . . , which was - considering the deviance of just ≈ 4.8 ·10−4% - lying within the expected range.

38


6 Determining the Distribution Functions

Applying the technique of Projective Reduction for calculating the expectedvalue of the distance of two points randomly selected within an given figure canbe used to determine the corresponding distribution function as well.

Let us reconsider in figure 2 the first intermediate step of projecting Q1Q2 ontoQ′

1Q′2, using the same notation for the involved sets of points as in chapter 3.

As we showd there, two cases are distinguished with every step od projection,dependent on the dimensions d1 and d2 with the probability p1 = d1

d1+d2respec-

tively p2 = d2

d1+d2, both independent of the dilation factor 1

r . Accordingly the

distribution functions of the distance t = Q′1Q

′2 of the two projected points in the

corresponding projective figures FP1(t) = P (Q′1Q

′2 < t |Q′

1 ∈ ρM1 ∧ Q′2 ∈ M2)

respectively FP2(t) = P (Q′1Q

′2 < t |Q′

1 ∈ M1 ∧Q′2 ∈ ρM2) are independent of r.

Therefore the distribution function of the combined projections

FP (t) = p1 FP1(t) + p2 FP2(t)

must also be independent of r.

Projecting Q′1Q

′2 back on to Q1Q2 is equivalent to a dilation with the dila-

tion factor r. Therefore we can formulate the distribution function of originaldistance s = Q1Q2 depending on a specific dilation factor r = r0 (that is deter-mined by the specific position of Q1 and Q2) as follows:

F(r0)(s) = P (Q1Q2 < s |Q1 ∈ M1 ∧Q2 ∈ M2 ∧ r = r0)

=p1 P (Q1Q2 < s |Q1 ∈ ρM1r ∧Q2 ∈ M2r ∧ r = r0)

+ p2 P (Q1Q2 < s |Q1 ∈ M1r ∧Q2 ∈ ρM2r ∧ r = r0)

=p1 P (Q′1Q

′2 <

s

r0|Q′

1 ∈ ρM1 ∧Q′2 ∈ M2)

+ p2 P (Q′1Q

′2 <

s

r0|Q′

1 ∈ M1 ∧Q′2 ∈ ρM2)

=FP (s

r0)

Thus the original distribution function of s results in

F (s) =

∫ ∞

0

fr(r)F(r)(s) dr =

∫ ∞

0

fr(r)FP (s

r) dr

=p1

∫ ∞

0

fr(r)FP1(s

r) dr + p2

∫ ∞

0

fr(r)FP2(s

r) dr (7)

with fr(r) = (d1 + d2) rd1+d2−1 for r ≤ 1 respectively fr(r) = 0 for r > 1.

Therefore we can derive the distribution function of the distance of the twopoints directly from the distribution function of the distance of the correspond-ing projected points resulting from each step of Projective Reduction and thedensity function of the dilation factor r.

Hence we can determine the original density function stepwise in reverse order,starting with the last projection with the least degrees of freedom. Howeverfairly complex integrals can appear intermediately, so in most cases it is more

39


advisable to compute the original distribution function directly in one singlestep, again starting from the last stage of projection. This is readily possibleas the different steps of Projection constitute, from a geometrical point of view,nothing more than several dilations following each other, so they can be com-bined into one single dilation. Since the distribution of the projected points isindependent of the dilation factor in each step, the same applies for the totaldilation factor.

Furthermore, since not only the basic theorem of Projective Reduction, but allother theorems worked out so far ensure not only the independence of each caseof projection but also the uniform distribution of the projected points withinthe new sets of points, equation (7) can be applied generally. The number ofprojective steps leading to the different projections may differ (like in the caseof the regular hexagon with two one- and one two-dimensional projections), sothe different density functions in respect to the dilation factor r may have to betaken into account. Thus we have

Theorem 10 If the technique of Projective Reduction being applied to a specificgeometrical figure, within which two points have been selected arbitrarily, leadsto n different partial projections with the respective probability of pi, then thedistribution function for the distance s of the two original points is given by

F (s) =

n∑

i=1

pi

∫ ∞

0

fri(ri)FPi(s

ri) dri

with FPi being the respective (partial) distribution function of the distance ofthe projected points in the respective partial projection and fri being the densityfunction of the respective dilation factor ri, which is only dependent on therespective degree of dimensional reduction.

6.1 The Distribution Function in Two-dimensional Fig-

ures

As an example for the application of the last theorem we want to look at theProjective Reduction in the case of the arbitrary triangle, consisting of threestep of projection. The total dilation factor r (that finally compensates the en-largement caused by the individual steps of projection) is given by r = r1 r2 r3,with r1, r2 and r3 being the respective dilation factors of the consecutive projec-tions. The probability of the single cases of projection can be extracted directlyfrom equation (3). Since the correction factor amounts to d1+d2

d1+d2+1 for each step,

we have a total correction factor of 453423 = 2

5 . Reducing the total projection fac-tor of 2

15 by this factor results in the remaining probability of 13 for each case.

While it is trivial in this case, the same procedure applies for more complexfigures as well.

Now we have to determine fr(r) from the distribution functions for the inter-

40


mediate reduction factors. From the given dimensions it follows that:

Fr1(r1) =r41 ; fr1(r1) = 4 r31 ;

Fr2(r2) =r32 ; fr2(r2) = 3 r22 ;

Fr3(r3) =r23 ;

Hence

Fr(r) =P (r1 r2 r3 < r)

=P (r1 < r) + P (r1 > r)P (r1r2 < r | r1 > r)

+ P (r1 > r)P (r1r2 > r | r1 > r)P (r1 r2 r3 < r | r1 > r ∧ r1r2 > r)

=Fr1(r) +

∫ 1

r1=r

fr1(r1)Fr2(r

r1)dr1

+

∫ 1

r1=r

fr1(r1)

∫ 1

r2=r

r1

fr2(r2)Fr3(r

r1r2) dr2 dr1

=r4 +

∫ 1

r1=r

4r13(

r

r1)3dr1

+

∫ 1

r1=r

4r13

∫ 1

r2=r

r1

3r22(

r

r1r2)2 dr2 dr1

=3 r4 − 8 r3 + 6 r2

Thus we have

fr(r) =F ′r(r) =

{

12 r (1 − r)2 for 0 ≤ r ≤ 1

0 for r ≥ 1

With these results we can determine the distribution function for the distanceof two points within a triangle △ABC from:

F (s) =p1

∫ ∞

0

fr(r)FP1(s

r) dr

+ p2

∫ ∞

0

fr(r)FP2(s

r) dr

+ p3

∫ ∞

0

fr(r)FP3(s

r) dr

=1

3

∫ 1

0

12r(1− r)2 P (AQ ≤ s

r|Q ∈ BC) dr

+1

3

∫ 1

0

12r(1− r)2 P (BQ ≤ s

r|Q ∈ AC)dr

+1

3

∫ 1

0

12r(1− r)2 P (CQ ≤ s

r|Q ∈ AB)dr

6.2 Calculating the Distribution Functions

For the concrete calculation of the distribution functions we first of all need theanalytical term of for FP1(s) = P (AQ ≤ t |Q ∈ BC). If we take the case of an

41


arbitrary triangle △ABC with side a, b and c, we can assume b ≤ c without lossof gernerality, since b and c could simply be renamed. Let us further assumefor now ∡ACB < π

2 (see figure 14). For the reason of clarity in later donecalculations we will use G(s) and GP (t) to denote the distribution function forthe original points respectively for the projected points.

B C

A

ht

a

bc

bQb

Figure 14: Distribution function in the triangle, first case

Due to the uniform distribution of Q on the side a the probability P (AQ < t) isrepresented by the coverage of the line segment a by the disk with center A andthe radius t. Using h to denote the height of the triangle, we arrive at followingterm:

GP (s) =

0 for t < h1a2

√t2 − h2 for h ≤ t < b

1a (√t2 − h2 +

√b2 − h2) for b ≤ t ≤ c

1 for c < t

Respectively we have

GP (s

r) =

0 for sh < r

1a2√( sr )

2 − h2 for sb < r ≤ s

h1a (√( sr )

2 − h2 +√b2 − h2) for s

c < r ≤ sb

1 for r < sc

42


Hence we obtain the distribution function G(s) for the distance of the originalpoints:

G(s) =

∫ ∞

0

fr(r)GP (s

r) dr

=

∫ s

c

0

fr(r) dr

+

∫ s

b

s

c

fr(r)1

a(

√

(s

r)2 − h2 +

√

b2 − h2) dr

+

∫ s

h

s

b

fr(r)1

a2

√

(s

r)2 − h2 dr

For s < h we have

G1(s) =

∫ ∞

0

fr(r)GP (s

r) dr

=

∫ s

c

0

fr(r) dr

+

∫ s

b

s

c

fr(r)1

a(

√

(s

r)2 − h2 +

√

b2 − h2) dr

+

∫ s

h

s

b

fr(r)1

a2

√

(s

r)2 − h2 dr

=

∫ s

c

0

12 r (1 − r)2 dr

+

∫ s

b

s

c

12 r (1− r)21

a(

√

(s

r)2 − h2 +

√

b2 − h2) dr

+

∫ s

h

s

b

12 r (1 − r)21

a2

√

(s

r)2 − h2 dr

=

∫ s

c

0

12 r (1 − r)2 dr

+

∫ s

b

s

c

12 (1− r)21

a(√

s2 − h2r2 + r√

b2 − h2) dr

+

∫ s

h

s

b

12 (1− r)21

a2√

s2 − h2r2 dr

For h ≤ s < b we have

G2(s) =

∫ s

c

0

12 r (1 − r)2 dr

+

∫ s

b

s

c

12 (1− r)21

a(√

s2 − h2r2 + r√

b2 − h2) dr

+

∫ 1

s

b

12 (1− r)21

a2√

s2 − h2r2 dr

43


And for b ≤ s < c we have

G3(s) =

∫ s

c

0

12 r (1 − r)2 dr

+

∫ 1

s

c

12 (1− r)21

a(√

s2 − h2r2 + r√

b2 − h2) dr

Moreover∫

1

a12 (1− r)2

√

s2 − h2 r2 dr

=arcsin

(h rs

) (3 s4 + 12 h2 s2

)

2 a h3

+

√s2 − h2 r2

((16− 3 r) s2 + 6 h2 r3 − 16 h2 r2 + 12 h2 r

)

2 a h2

as can be confirmed by differentiation. The remaining integrals can easily becalculated, so by taking into account that

√b2 − h2 = a−

√c2 − h2) we arrive

at the following:

Theorem 11 If two points which have been randomly selected within a two-dimensional figure are being projected in the process of Projective Reductiononto the vertex A respectively the opposite side a of a triangle △ABC with theheight h orthogonal to a, b ≤ c and ∡ABC < π

2 , then the (partial) distributionfunction G(s) for the distance s of the original points for this particular partialprojection is given by:

G(s) =G1(a, b, c, h, s)

=3(π − arcsin

(hc

)− arcsin

(hb

))s2(s2 + 4 h2

)

2 a h3

+

√c2 − h2

((3 b2 − 3 c2

)s4 +

(16 b c2 − 16 b2 c

)s3)

2 a b2 c2 h2

+3 s4 − 16 b s3

2 b2 h2

for s ≤ h

G(s) =G2(a, b, c, h, s)

=3(2 arcsin

(hs

)− arcsin

(hc

)− arcsin

(hb

))s2(s2 + 4 h2

)

2 a h3

+

√s2 − h2

(13 s2 + 2 h2

)

a h2

+

√c2 − h2

((3 b2 − 3 c2

)s4 +

(16 b c2 − 16 b2 c

)s3)

2 a b2 c2 h2

+3 s4 − 16 b s3

2 b2 h2

for h < s ≤ b

44


G(s) =G3(a, b, c, h, s)

=3(arcsin

(hs

)− arcsin

(hc

))s2(s2 + 4 h2

)

2 a h3

+

√s2 − h2

(13 s2 + 2 h2

)

2 a h2

+

√c2 − h2

(3 s4 − 16 c s3 − 2 c2 h2

)

2 a c2 h2

+ 1

for b < s ≤ c

and

G(s) =G4(a, b, c, h, s) = 1

for s > c

(partial distribution: point to line picking 1)

For the case of π2 ≤ ∡ACB (see figure 15) we denote the resulting distribution

functions as H(s) and HP (t).

B C

A

ht

a

bc

bQb

Figure 15: Distribution function in the triangle, 2st case

This leads to following function:

HP (t) =

0 for t < b1a (√t2 − h2 −

√b2 − h2) for b ≤ t ≤ c

1 for c < t

45


Hence we have

HP (s

r) =

0 for sb < r

1a (√( sr )

2 − h2 −√b2 − h2) for s

c < r ≤ sb

1 for r < sc

Therefore the distribution function H(s) for the distance S of the original pointsis given by:

H(s) =

∫ ∞

0

fr(r)HP (s

r) dr

=

∫ s

c

0

fr(r) dr

+

∫ s

b

s

c

fr(r)1

a(

√

(s

r)2 − h2 −

√

b2 − h2) dr

For s < b we have

H1(s) =

∫ s

c

0

12 r (1− r)2 dr

+

∫ s

b

s

c

12 (1− r)21

a(√

s2 − h2r2 − r√

b2 − h2) dr

And for b ≤ s ≤ c

H2(s) =

∫ s

c

0

12 r (1− r)2 dr

+

∫ 1

s

c

12 (1− r)21

a(√

s2 − h2r2 − r√

b2 − h2) dr

Therfore it follow:

Theorem 12 If two points which have been randomly selected within a two-dimensional figure are being projected in the process of Projective Reduction ontothe vertex A respectively onto the opposite side a of a triangle △ABC with theheight h orthogonal to a, b ≤ c and π

2 ≤ ∡ABC, then the (partial) distributionfunction H(s) for the distance s of the original points for this particular partialprojection is given by:

H(s) =H1(a, b, c, h, s)

=3(arcsin

(hb

)− arcsin

(hc

))s2(s2 + 4 h2

)

2 a h3

+

√c2 − h2

((3 b2 − 3 c2

)s4 +

(16 b c2 − 16 b2 c

)s3)

2 a b2 c2 h2

+3 s4 − 16 b s3

2 b2 h2

for s ≤ b

46


H(s) =H2(a, b, c, h, s)

=3(arcsin

(hs

)− arcsin

(hc

))s2(s2 + 4 h2

)

2 a h3

+

√s2 − h2

(13 s2 + 2 h2

)

2 h2

+

√c2 − h2

(3 s4 − 16 c s3 − 2 c2 h2

)

2 a c2 h2

+ 1

for b < s ≤ c

and

H(s) =H3(a, b, c, h, s) = 1

for s > c

(partial distribution: point to line picking 2)

For the sake of completeness we also calculate furthermore the partial distribu-tion function I(s) for the projection of the two original points onto two parallelline segments. In order to minimize case distinctions we will assume the sim-plified case that the two line segments are the two opposite sides of length a ofa rectangle with the other side length being h; this is sufficient to calculate thedistribution function within the regular hexagon, as shown below.

Since the projection onto two parallel line segments requires only two projectivesteps we arrive at a different density function for the total reduction factor r.

In this case we have:

Fr1(r1) =r41 ; fr1(r1) = 4 r31 ;

Fr2(r2) =r32 ; fr2(r2) = 3 r22 ;

Fr(r) =P (r1 r2 < r)

=P (r1 < r) + P (r1 > r)P (r1r2 < r | r1 > r)

=Fr1(r) +

∫ 1

r1=r

fr1(r1)Fr2(r

r1)dr1

=r4 +

∫ 1

r1=r

4r13(

r

r1)3dr1

=4 r3 − 3 r4

Thus we get:

fr(r) =F ′r(s) =

{

12 r2 (1− r) for 0 ≤ r ≤ 1

0 for r > 1

The calculation of the distribution function I(s) of the distance of the originalpoints Q1 and Q2 is illustrated in figure 16.

47


B

CD

A

Q1

F

x

t

a

h

b Q2b

b

Figure 16: Distribution function in the case of parallel line segments

Since Q2 is uniformly distributed within AB, P (Q1Q2 ≤ t) is represented bythe coverage of the line segment a = AB by a disk with center Q1 and the radiust. The (random) position of Q1 is represented by the variable x = DQ1. Thuswe get for h ≤ t ≤

√a2 + h2 :

IP (t) = 2

(∫ √

t2−h2

0

1

a

x

adx+

∫ a

√t2−h2

1

a

√t2 − h2

adx

)

=2 a

√t2 − h2 − t2 + h2

a2

Respectively

IP (t) =

0 for t < h2 a

√t2−h2−t2+h2

a2 for h ≤ t ≤√a2 + h2

1 for√a2 + h2 < t

Therefore we obtain

IP (s

r) =

0 for sh < r

2 a√

( s

r)2−h2−( s

r)2+h2

a2 for s√a2+h2

≤ r ≤ sh

1 for r < s√a2+h2

48


Hence the distribution function I(s) for the distance s of the original points isgiven by:

I(s) =

∫ ∞

0

fr(r)IP (s

r) dr

=

∫ s√a2+h2

0

fr(r) dr

+

∫ s

h

s√a2+h2

fr(r)2 a√

( sr )2 − h2 − ( sr )

2 + h2

a2dr

Thus we have for s ≤ h:

I1(s) =

∫ s√a2+h2

0

12 r2 (1 − r) dr

+

∫ s

h

s√a2+h2

12 (1− r)2 a r

√s2 − h2r2 − s2 + h2r2

a2dr

and for h < s ≤√a2 + h2

I2(s) =

∫ s√a2+h2

0

12 r2 (1 − r) dr

+

∫ 1

s√a2+h2

12 (1− r)2 a r

√s2 − h2r2 − s2 + h2r2

a2dr

From this follows

Theorem 13 If two points which have been selected randomly within a two-dimensional figure are being projected in the process of Projective Reductiononto two opposite sides with length a of a rectangle with the remaining sidelength being h, then the (partial) distribution function I(s) for the distance s ofthe original points for this particular partial projection is given by:

I(s) =I1(a, h, s)

=3 arcsin ( h√

h2+a2)− 3 π

2

a h3s4 +

8√h2 + a2 − 8 h

a2 h2s3

for s ≤ h

I(s) =I2(a, h, s)

=3 a arcsin

(h√

a2+h2

)

− 3 a arcsin(hs

)− 3 h

a2 h3s4

+8√h2 + a2 s3 +

√s2 − h2

(2 a h2 − 5 a s2

)− 6 h2 s2 + h4

a2 h2

for h < s ≤√

a2 + h2

and

I(s) =I3(a, h, s) = 1

for s >√

a2 + h2

(partial distribution: parallel lines picking)

49


Examples

The regular pentagon

For the regular pentagon we have found following term for the expected valueof the mean distance of two points:

E5 =2

15(√5 + 2)E2(1,

√5 + 1

2,

√5 + 1

2)

− 2

15(√5− 1)E2(1, 1,

√5 + 1

2)

By taking intoaccount the total reduction factor of 25 we come to following

expression for the distribution function F (s) (after calculating the respectiveheights of the triangles used in the projections) :

F (s) =1

3(√5 + 2)G(1,

√5 + 1

2,

√5 + 1

2,

√

5 + 2√5

2, s)

− 1

3(√5− 1)H(1, 1,

√5 + 1

2,

√

10 + 2√5

4, s)

So we get for 0 < s ≤ 1

F1(s) =1

3(√5 + 2)G1(1,

√5 + 1

2,

√5 + 1

2,

√

5 + 2√5

2, s)

− 1

3(√5− 1)H1(1, 1,

√5 + 1

2,

√

10 + 2√5

4, s)

=s4

25

(√

5−√5 (2

√10− 6

√2)π + 4

√

5− 2√5 π − 20

√5 + 50

)

− s3

15

(

160− 64√5)

+s2

25

(√

5−√5 (2

√10− 10

√2)π +

√

5− 2√5 (8

√5 + 20)π

)

For 1 < s ≤√

5+2√5

2 we receive

F2(s) =1

3(√5 + 2)G1(1,

√5 + 1

2,

√5 + 1

2,

√

5 + 2√5

2, s)

− 1

3(√5− 1)H2(1, 1,

√5 + 1

2,

√

10 + 2√5

4, s)

=1

60

√2

√

−5−√5 + 8s2

(

5− 5√5− 26

(

−5 + 3√5)

s2)

− 1

6

(

−3 +√5)

+4

25s2π

(√

5(

25− 2√5)

+

√

5(

5− 2√5)

s2

)

− 4

5s2

(

2

√

5(

5− 2√5)

+

√

50− 22√5s2

)

arcsin

√

10 + 2√5

4s

50


And for

√5+2

√5

2 < s ≤√5+12 we have

F3(s) =1

3(√5 + 2)G2(1,

√5 + 1

2,

√5 + 1

2,

√

5 + 2√5

2, s)

− 1

3(√5− 1)H2(1, 1,

√5 + 1

2,

√

10 + 2√5

4, s)

=− 1

6

(

−3 +√5)

+1

15

√

−5− 2√5 + 4s2

(

5(

2 +√5)

+ 26√5s2)

− 1

60

√2

√

−5−√5 + 8s2

(

5(

−1 +√5)

+ 26(

−5 + 3√5)

s2)

− 2

25πs2

(

6

√

5(

5− 2√5)

− 2

√

5(

5 + 2√5)

+

(

3

√

50− 22√5− 2

√

5− 2√5

)

s2

)

− 8

5s2

(√

5(

5 + 2√5)

+

√

5− 2√5s2

)

arcsin

√

−5− 2√5 + 4s2

2s

+4

5s2

(

2

√

5(

5− 2√5)

+

√

50− 22√5s2

)

arcsin

√

−10− 2√5 + 16s2

4s

If you determine the densitiy f(s) = F ′(s), you can at least numerically calculate

the expected value of s being∫√

5+1

0s f(s) ds, that corresponds exactly with

the value determined in chapter 3. All of the following concrete distributionfunctions have been verified in this manner using MAXIMA or Mathematica.

The regular hexagon

In case of the regular hexagon with side length 1 we have found the followingexpression for the expected value:

E6 = 1645E2(1,

√3, 2)− 2

45E2(1, 1,√3) + 2

15E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ DE)

By taking into account the total reduction factor of 25 for the first two partial

projections and 35 for the third one (projecting the two original points onto

to opposite sides of the hexagon) we get following term for the correspondingdistribution F (s):

F (s) =8

9H(1,

√3, 2,

√3, s)− 1

9H(1, 1,

√3,

√3

2, s) +

2

9I(1,

√3, s)

So, for 0 < s ≤ 1 we have

F1(s) =8

9H1(1,

√3, 2,

√3, s)− 1

9H1(1, 1,

√3,

√3

2, s) +

2

9I1(1,

√3, s)

=

(9−

√3π)s4 − 48 s3 + 18

√3 π s2

81

51


For 1 < s ≤√3 we have

F2(s) =8

9H1(1,

√3, 2,

√3, s)− 1

9H2(1, 1,

√3,

√3

2, s) +

2

9I1(1,

√3, s)

=(√3π − 4

√3 arcsin

√3

2 s ) s4 + (10

√3 π − 12

√3 arcsin

√3

2 s ) s2

27

−√4 s2 − 3

(26 s2 + 3

)

54+

1

18

and for√3 < s ≤ 2 we receive

F3(s) =8

9H2(1,

√3, 2,

√3, s)− 1

9H3(1, 1,

√3,

√3

2, s) +

2

9I2(1,

√3, s)

=

(

6√3 arcsin

√3s − 2

√3π − 9

)

s4

81

+

(

144√3 arcsin

√3s − 48

√3π − 108

)

s2

81

+

√s2 − 3

(126 s2 + 108

)+ 45

81

6.3 The Distribution Function within an Arbitrary Trian-

gle

With the help of the results above the distribution function for the distancewithin an arbitrary triangle can easily be calculated. Let us assume withoutloss of generality that a ≤ b ≤ c. Let A be the area of the given triangle.

The following is true for the heights of the triangle : hc = 2Ac ≤ hb = 2A

b ≤ha = 2A

a . Since no more than one height may be longer than one of the sidesof the triangle there are only two possibilities:

Either we have hc ≤ hb ≤ ha ≤ a ≤ b ≤ c or hc ≤ hb ≤ a < ha ≤ b ≤ c, whathas to be taking into account further down.

We will denote the total distribution function of the distance of the points withinthe triangle by D(s). As shown above is it composed of the the three distributionfunctions resulting from the partial projections with one image point beingprojected onto one vertex and the other image point onto its opposite side ofthe triangle. To determine the concrete function, following case distinction isneeded:

Case A of the obtuse-angled triangle

Since hc ≤ a ≤ b < c within the obtuse-angled triangle, the distribution functionDA(s) is composed as follows:

52


DA1(s) =1

3H1(a, b, c,

2A

a, s) +

1

3H1(b, a, c,

2A

b, s) +

1

3G1(c, a, b,

2A

c, s)

for 0 ≤ s < hc =2Ac

DA2(s) =1

3H1(a, b, c,

2A

a, s) +

1

3H1(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for hc =2Ac ≤ s < a

DA3(s) =1

3H1(a, b, c,

2A

a, s) +

1

3H2(b, a, c,

2A

b, s) +

1

3G3(c, a, b,

2A

c, s)

for a ≤ s < b and

DA4(s) =1

3H2(a, b, c,

2A

a, s) +

1

3H2(b, a, c,

2A

b, s) +

1

3

for b ≤ s ≤ c.

Case B of the acute-angled with ha ≤ a

The distribution function DB(s) is given by

DB1(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G1(b, a, c,

2A

b, s) +

1

3G1(c, a, b,

2A

c, s)


DB2(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G1(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for 2Ac ≤ s < hb =

2Ab

DB3(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G2(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for 2Ab ≤ s < ha = 2A

a

DB4(s) =1

3G2(a, b, c,

2A

a, s) +

1

3G2(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for 2Aa ≤ s < a

DB5(s) =1

3G2(a, b, c,

2A

a, s) +

1

3G3(b, a, c,

2A

b, s) +

1

3G3(c, a, b,

2A

c, s)

for a ≤ s < b and

DB6(s) =1

3G3(a, b, c,

2A

a, s) +

1

3G3(b, a, c,

2A

b, s) +

1

3

for b ≤ s ≤ c.

53


Case C of the acute-angled with ha > a

The distribution function DC(s) is given by:

DC1(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G1(b, a, c,

2A

b, s) +

1

3G1(c, a, b,

2A

c, s)


DC2(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G1(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for 2Ac ≤ s < hb =

2Ab

DC3(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G2(b, a, c,

2A

b, s) +

1

3G2(c, a, b,

2A

c, s)

for 2Ab ≤ s < a

DC4(s) =1

3G1(a, b, c,

2A

a, s) +

1

3G3(b, a, c,

2A

b, s) +

1

3G3(c, a, b,

2A

c, s)

for a ≤ s < ha = 2Aa

DC5(s) =1

3G2(a, b, c,

2A

a, s) +

1

3G3(b, a, c,

2A

b, s) +

1

3G3(c, a, b,

2A

c, s)

for 2Aa ≤ s < b and

DC6(s) =1

3G3(a, b, c,

2A

a, s) +

1

3G3(b, a, c,

2A

b, s) +

1

3

for b ≤ s ≤ c.

Examples

The distribution function for the distance s of two points chosen randomlywithin a right-angled isosceles triangle with the catheti length being 1 is givenby

F1(s) =DA1(s) = 21

3H1(1, 1,

√2, 1, s) +

1

3G1(

√2, 1, 1,

1

2

√2, s)

=(9 π + 18) s4 +

(−32

√2− 64

)s3 + 24 π s2

12

for 0 ≤ s < hc =√

12

F2(s) =DA2(s) = 21

3H2(1, 1,

√2, 1, s) +

1

3G2(

√2, 1, 1,

1

2

√2, s)

=

(

24 arcsin 1√2 s

− 3 π + 18)

s4 + 48 s2 arcsin 1√2 s

12

+

√2 s2 − 1

(13 s2 + 1

)−(8√2 + 16

)s3

3

54


for√

12 ≤ s < 1 = a = b and

F3(s) =DA4 = 21

3H2(1, 1,

√2, 1, s) +

1

3

=

(12 arcsin 1

s − 3 π + 6) (

s4 + 4 s2)

12

+−8

√2 s3 +

√s2 − 1

(13 s2 + 2

)− 6 s2 + 1

3

for 1 ≤ s ≤√2 = c.

The equilateral triangle with the side length being one is given by

F1(s) =DB1(s) =

(4 π + 6

√3)s4 − 32

√3 s3 + 12 π s2

3√3

for 0 ≤ s <√32 = hc = hb = ha and

F2(s) =DB4(s) =

(

24√3 arcsin

√3

2 s − 8√3π + 18

) (s4 + 3 s2

)

9

+−32 s3 +

√4 s2 − 3

(26 s2 + 3

)− 18 s2

3

for√32 ≤ s ≤ 1 = a = b = c.

55


7 The Mean Distance of Points in the Regular

Tetrahedron

7.1 The Expected Value

Theorem 2 can be applied without any constraint to the figure of the regulartetrahedron ABCD with side length 1. Using the indicated centers of projectionwe get:

E(Q1Q2 |Q1 ∈ ABCD ∧Q2 ∈ ABCD)

(Z=D)=

3

7E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ ABCD)

+3

7E(Q1Q2 |Q1 ∈ ABCD ∧Q2 ∈ △ABC)

=6

7E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ ABCD)

(Z=C)=

6

7

2

6E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ ABCD)

+6

7

3

6E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ △ABD)

=2

7E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ ABCD)

+3

7E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ △ABD)

(Z=B)=

2

7

1

5E(AQ |Q ∈ ABCD)

+2

7

3

5E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ △ACD)

+3

7

2

5E(Q1Q2 |Q1 ∈ AC ∧Q2 ∈ ABD)

+3

7

2

5E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ AD)

=2

35E(AQ |Q ∈ ABCD)

+18

35E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ △ACD)

(Z=A)=

2

35

3

4E(AQ |Q ∈ △BCD)

+18

35

1

4E(BQ |Q ∈ △ACD)

+18

35

2

4E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

=6

35E(DQ |Q ∈ △ABC)

+9

35E(Q1Q2 |Q1 ∈ AB ∧Q2 ∈ CD)

Thus we have reduced an expected value containing 6 degrees of freedom intothe sum of two expected values each containing merely 2 degrees of freedom.

56


The first expected value can be easily calculated in two ways (see figure 17):

s =

√

DF2+ x2 + y2

B

C

D

A

M1

Fx y

s

Q

Figure 17: The expected value in the first partial projection

So we could simply integrate over the x- and y-coordinates, yielding

Ea =1

A(△AM1F )

∫√

3

6

y=0

∫ √3 y

x=0

√

DF2+ x2 + y2 dx dy

For later computations however it is advisable to integrate over the length ofthe line segment d = FQ. From the uniform distribution of Q in △ABC arisesfollowing distribution function for the size of d, with

Fd(d) =

4√3πs2 for 0 ≤ d ≤

√112√

12d2−1+4d2(π−3arctan√12d2−1)√

3for

√112 < d ≤

√13

1 for√

13 < d

being the quotient of the area of the triangle △ABC that is covered by the

circle with center F and the radius d. Substituting d by s =√

d2 + 23 results in

57


following distribution function for s = DQ:

TPa(s) =

0 for s ≤√

23

4√3π(s2 − 2

3

)for

√23 < s ≤

√34

3√12s2−9+(12s2−8)(π−3arctan

√12s2−9)

3√3

for√

34 < s ≤ 1

1 for 1 < s

and its corresponding density function

tPa(s) =

0 for s ≤√

23

8πs√3

for√

23 < s ≤

√34

8s(π−3arctan√12s2−9)√

3for

√34 < s ≤ 1

0 for 1 < s

Thus the expected value of the first projection is given by:

Ea =

1∫

0

s tPa(s) ds

=1

108

(

36− 64√2 π + 192

√2 arctan(

√2) + 75 ln(3)

)

The second expected value is obtained by simple integration over x and y asindicated in figure 18:

Eb =4

∫ 12

y=0

∫ 12

x=0

√

M1M22+ x2 + y2 dx dy

=4

∫ 12

y=0

∫ 12

x=0

√

1

2+ x2 + y2 dx dy

=

∫ 12

y=0

(4 y2 + 2

)arsinh

(

1√4 y2+2

)

+√

4 y2 + 3

2dy

=

[

− 1

3

√2 arctan

√2 y

√

4 y2 + 3+

2 y3 + 3 y

3arsinh

√2√

2 y2 + 1

4 y2 + 2

+7

12arsinh

(2 y√3

)

+1

3y√

4 y2 + 3

] 12

y=0

=−24

√2 arctan

(√2)+ 21 ln (3) + 6

√2 π + 12

36

58


s =

√

M1M22+ x2 + y2

B

C

D

A

M1

M2

s

x

y

Q1

Q2

Figure 18: The expected value in the second partial projection

Combining both partial results leads to:

E(Q1Q2 |Q1 ∈ ABCD ∧Q2 ∈ ABCD)

=6

35Ea +

9

35Eb

=168

√2 arctan

(√2)+ 339 ln (3)− 74

√2π + 180

1260≈0.3576411722304 . . .

7.2 The Distribution Function

The calculation of the distribution function T (s) for the distance of the pointsis done according to Theorem 10 taking into account the probability of 2

5 and35 for the two partial projections, resulting in

T (s) =2

5

∫ 1

0

fr(r)TPa(s/r)dr +3

5

∫ 1

0

fr(r)TPb(s/r)dr (8)

with fr(r) being the density function for the total reduction factor r and TPa

and TPb being the distribution functions for the partial projections.The distribution function TPa(s) has been deduced above. TPb(s) arises readily

from s =√

x2 + y2 + 12 with x, y being uniformly distributed in [ 0; 12 ] (see

59


figure 18):

TPb(s) =

0 for s ≤√

12

π(s2 − 12 ) for

√12 < s ≤

√34√

4s2 − 3 +(2s2 − 1

)

(arccsc

√4s2 − 2− arctan

√4s2 − 3

)for

√34 < s ≤ 1

1 for 1 < s

Finally we have the density function fr(r)=60 r2(1− r)3 for the total reductionfactor r that is derived in the same way as it was shown for the 2-dimensionalfigures.

Inserting the above into equation (8), we arriva at the distribution function forthe distance of the points within a regular tetrahedron being

T1(s) =− 1

5s3(

π(

−280√2 + 90

√3s− 288

√2s2 + 25

√3s3)

+ 6(

40√2π + 5s3 + 24s2

(

−2 + 2√2π −

√2arctan

√2)))

for 0 ≤ s <√

12

T2(s) =π

(

− 3

10+ 9s2 + 32

√2s3 − 18

(

−3 +√3)

s4 +144

√2s5

5+(

12− 5√3)

s6

)

− 6

5s3(

40√2π + 5s3 + 24s2

(

−2 + 2√2π −

√2arctan

√2))

for√

12 ≤ s <

√23

T3(s) =1

90π(

−27− 32√3 + 90

(

9 + 8√3)

s2 + 1620(

3 +√3)

s4 + 90(

12 +√3)

s6)

− 6

5s3(

40√2π + 5s3 + 24s2

(

−2 + 2√2π −

√2 arctan

√2))

for√

23 ≤ s <

√34 and

T4(s) =6

5(48− 5s)s5 +

1

5

√

−3 + 4s2(5− 258s4

)

+1

90π(

− 27− 32√3 + 90

(

9 + 8√3)

s2 − 4320√2s3

+ 1620(

3 +√3)

s4 − 5184√2s5 + 90

(

12 +√3)

s6)

+144

5

√2s5 arctan

√2 + 6

√3s4(18 + 5s2

)arctan

√

−1 +4s2

3

+96

5

√2s3

(5 + 6s2

)arctan

√

− 32 + 2s2

s

60


− 6

5

(−1 + 30s2 + 180s4 + 40s6

)arctan

√

−3 + 4s2

+48

5

√2s3

(10 + 9s2

)arctan

√−6 + 8s2

s

− 2√3

15

(−8 + 180s2 + 810s4 + 135s6

)arctan

√

−9 + 12s2

for√

34 ≤ s ≤ 1.

7.3 Expected Value for the Hyper-Tetrahedron

In the case of the 4-dimensional hyper-tetrahedron, the mean distance of itspoints could only be calculated numerically since the resulting integrals appar-ently can not be expressed in a closed form.

The regular hyper-tetrahedron can be constucted using a the regular tetrahedronABCD in 3-dimensional space and an additional point E with a 4-dimensionalcomponent allowing E to have the distance of 1 from all other points. Us-

ing the coordinates A(0|0|0|0), B(1|0|0|0), C(12 |√

34 |0|0), D(12 |

√112 |√

23 |0) and

E(12 |√

112 |√

124 |√

58 ) these requirements are fulfilled. Further points needed

for the calculation below are the foot-point F (12 |√

112 |0|0) of the height of the

tetrahedron ABCD and its mass center M(12 |√

112 |√

124 |0). Notably we have

FM =√

124 and EM =

√58 .

After a total of 5 sequential steps of reduction using each of the vertexes ascenter of reduction Theorem 2 leads to

E(Q1Q2 |Q1 ∈ ABCDE ∧Q2 ∈ ABCDE)

=4

63E(EQ |Q ∈ ABCD)

+24

63E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ DE)

To calculate the expected value E(EQ |Q ∈ ABCD) we imagine Q′ to be

the projection of Q seen from the mass center M of the tetrahedron ABCDonto one of its bounding triangles. The reduction factor is characterized by thedistribution function Fr(r) = r3, which delivers fr(r) = 3r2. Since

EQ =

√

MQ2+ EM

2

=

√

(rMQ′)2 + 5/8

=

√

r2FQ′2 + r2FM2+ 5/8

=

√

r2FQ′2 +r2

24+ 5/8 ,

61


using d = FQ′, characterized by Fd(d) calculated in 7.1, leads to followingdouble integral:

E(EQ |Q ∈ ABCD)

=

∫√

1/3

0

∫ 1

0

3r2fd(d)

√

r2d2 +r2

24+ 5/8 dr dd

=

∫√

1/3

0

fd(d)1

32 (1 + 24d2)3/2

(4√

6 + 9d2√

1 + 24d2(17 + 48d2

)

− 225√6 arcsinh

√

1

15+

8d2

5

)dd

≈ 0.83607672222174 . . .

with

fd(d) = F’d(d) =

8√3πs for 0 ≤ d ≤

√112

8d(π−3arctan√12d2−1)√

3for

√112 < d ≤

√13

To calculate the second expected value E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ DE) wewill use the fact that DQ1 = EQ1 for every Q1 ∈ △ABC. Therefor we cancalculate the expected value of Q1Q2 for a fixed value of t = DQ1 by usingTheorem 3:

E2(1, t, t) =t

2+

1

8(−1 + 2t)(1 + 2t) ln

1 + 2t

−1 + 2t

Since t has the density function tPa(t) calculated in 7.1, it follows that

E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ DE)

=

∫ 1

√2/3

tPa (t)

(t

2+

1

8(−1 + 2t)(1 + 2t) ln

1 + 2t

−1 + 2t

)

dt

≈ 0.76110934819675 . . .

Thus we receive numerically

E(Q1Q2 |Q1 ∈ ABCDE ∧Q2 ∈ ABCDE)

=4

63E(EQ |Q ∈ ABCD)

+24

63E(Q1Q2 |Q1 ∈ △ABC ∧Q2 ∈ DE)

≈ 0.343030654692208 . . .

62


8 Afterword

Because I attempted to illustrate the steps of the Projective Reduction usingeidetic examples the stringency of my presentation may have suffered in somepoints. On the other hand I hope to have conveyed a sense of new possibilitiesfor approaching difficult problems in the field of geometrical probability, helpingthe reader to pursue his own calculations.

With increasing dimensions of the examined problems the resulting integralsbecome very demanding and could partly only be solved with the help of com-puter algebra systems (like MAXIMA and Mathematica), which however oftenhad to be guided ”manually” in order to produce useful results. It is comfort-ing to know that Mr. Johan Philip described similar difficulties calculating hisresults, indicating that this is due mostly to the complexity of the tasks ratherthan to my flawed computer knowledge.

The principle of Projective Reduction can be further expanded, for instanceby examining figures with curved boundaries, mostly leading to a non-uniformdistribution of the projected points that has to be taken into account. Anothervariation would be to increase the number of randomly chosen points, leadingto different problems as well. Both extentions exceed the given scope of thispaper and remain reserved for further articles.

63


9 Literature

Since this work is no synopsis and the fundamental idea of Projective Reductionwas developed on my part without any prior knowledge of earlier articles, noextensive list of literature is given.

However two authors gained importance to me as I was examining possibleapplications, so I want to mention them here.

First of all this is Eric W. Weisstein, founder and main author of the mathe-matical online-encyclopaedia ’MathWorld’, who made is possible for me - likemany others - to quickly gain an overview of the current state of research in myfield and who furthermore inspired me by doing some own work on the problemof triangle line picking so I finally completed this monograph and translated itinto English. Here is the corresponding source:

Weisstein, Eric W. ”Triangle Line Picking.”From MathWorld–A Wolfram Web Resource.http://mathworld.wolfram.com/TriangleLinePicking.html

Furthermore it is Johan Philip who spurred my ambition of calculating evencomplex distribution functions down to the details, since his own research Ifound by browsing MathWorld is very masterful and inspiring, especially inhigher dimensions.

Here is a link to an overview of his articles:

http://www.math.kth.se/~johanph/

64

http://mathworld.wolfram.com/about/author.html

http://mathworld.wolfram.com/TriangleLinePicking.html

http://www.math.kth.se/~johanph/

projective reduction - a new technique for calculating expected

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