A Note on the Proportional Partitioning of Line Segments, Triangles and Tetrahedra

A Note on the Proportional Partitioning

of

Line Segments, Triangles and Tetrahedra

by

Frank J. Attanucci

Scottsdale Community College

Department of Mathematics

[email protected]

ABSTRACT: In the first part of this paper I solve the following

problem: Where can one place a point G inside or on a triangle so

that line segments from G to each of the vertices divide the

triangle into three sub triangles whose areas A1, A2 and A3,

1

respectively, satisfy the proportion: where the

are non-negative constants with positive sum? I then state

and prove an analogous result for tetrahedra. Surprised by the

structural similarity of the results (and preserving my order of

discovery) I “go back” and prove an analogous theorem for line

segments to see if the same pattern appears. I finish with a

theorem concerning the centroid of n! points. Along the way,

everything is made more intriguing by allowing the to be

parameterized functions.

It is generally known that, if line segments are drawn from the

centroid of any triangle to each of its three vertices, then the

three sub triangles so formed all have equal area. This fact

suggested to my mind the following problem:

Where can one place a point G inside or on a triangle so that

line segments from G to each of the vertices divide the

triangle into sub triangles whose areas A1, A2 and A3,

respectively, satisfy the proportion: where

the are non-negative constants with positive sum?

After identifying the notational conventions that will be used in

this paper, I formally state (and prove) a theorem which provides

the answer to this problem (Theorem 1). I then state and prove

an analogous result (Theorem 2) for tetrahedra, answering the

question:

Where can one place a point G inside or on a tetrahedron so

that line segments from G to each of the vertices divide the

2

tetrahedron into four subtetrahedra whose volumes V1, V2, V3

and V4, respectively, satisfy the proportion:

where the are non-negative

constants with positive sum?

Amazed by the structural similarities of my findings in Theorems

1 and 2, I go back to verify that the same pattern holds for line

segments (Theorem 0). Extrapolating from these results, I state

and prove a final theorem (Theorem 3) dealing with the centroid

of n! points.

A Note on Notation

For points P1, P2, P3 and G, I will use boldface to denote vectors:

and where O is the origin. Also,

I will denote the cross product by: the dot product by:

and the scalar triple product by: . Finally, I

will, without comment, make use of properties of these products

(which can be found in any book on vector calculus) as the need

arises. My decision to take a vector approach was motivated by a

desire to obtain results that would be (Cartesian) coordinate

free. It is highly unlikely that the connections that exist

between my results—those I have seen (and any I have yet to see)—

could have revealed themselves had I decided otherwise.

Partitioning a triangle into three sub triangles

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Theorem 1: Let be any three non-collinear points (the

vertices of the triangle ) and let and

Then, for any given nonnegative constants ,

having positive sum one can find a point G inside or on the

triangle such that the areas A1, A2 and A3 of the sub

triangles and , respectively, satisfy the

proportion: Moreover, the number of such

points G for which this is so is given by the number of distinct

permutations of the elements in Finally, the centroid

of these 6 (or 3 or 1) points is the same as the centroid of the

original triangle.

Proof: The figure below will be of help.

P3

A2 c

A1

G

G* b

P1 P2

A3

G

4

O

To simplify the calculations that lie ahead I first translate the

above triangle so that one of its vertices, P1, lies at the

origin. Doing this leads to the triangle :

Q3

A2 c

A1

G*

G* b

O Q2

A3

where and

Let G* = xQ2 + yQ3 be a point inside or on the triangle .

Then ; moreover, from the figure above we see that

b = Q2 – G* and c = Q3 – G*.

Let A1, A2 and A3 be the areas of the indicated sub triangles.

Then, using the cross product, we have A1 = where

5

A1

i.e., A1 ,

(1)

A2

6

(2)

and

A3

i.e., A3 .

(3)

7

Therefore, from (1), (2) and (3), if we want ,

we must solve the linear system:

. (4)

The unique solution to (4) is:

Therefore, we have

(5)

However, since G* = G – P1, Q2 = P2 – P1 and Q3 = P3 – P1,

equation (5) becomes

G

or, finally,

8

G

(6)

By considering all distinguishable permutations of the elements

of we see that the proportions in: is

just one of the (up to) six possible:

(7)

each leading to a different location for G. In fact, the number

N of locations of G is given by the following cases:

Case I: are different. In this case, we have

different locations for G; from (6) and (7) they are

given (in 2 groups of three) by:

9

G1

G2

G3

and (8)

G4

G5

G6

Case II: Exactly two of are equal . In

this case, we have different locations for G; from (6)

and (7) they are given by:

10

(9)

and

Case III: In this case, we have only possible

location for G, namely:

Moreover, in Case I, where the six Gi’s are given in (8), we see

that their centroid H is given by

H

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i.e., H , which coincides with the

centroid of the triangle.

Since the same is also true in Cases II and III, this completes

the proof of Theorem 1.

The following figures (showing all six ways of partitioning a

triangle ) illustrate the result.

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Before leaving Theorem 1 there are three other facts that I wish

to call attention to:

In Case I: Consider the two triangles:

From (8) they are “concentroidal” with the triangle

. If are their respective areas, then it can be

shown that ; in fact, if A is the area of ,

then

(10)

In Case II : Consider the one triangle:

From (9), it too is concentroidal with . And, if

is its area, then

(11)

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NOTE: To obtain (10) and (11), I made use of:

Finally, we note that if we take to be non-negative

parameterized functions that do

not simultaneously vanish on their domain, then the trajectories

of the 6 (or 3 or 1) points can create fascinating patterns—a

family of curves whose union has H as a centroid (and a point of

symmetry if is equilateral)! In the three examples

below on the interval

Remark: for all in the above example; in the next

two this is not the case.

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Partitioning a tetrahedron into four subtetrahedra

15

Theorem 2: Let be any four non-coplanar points (the

vertices of the tetrahedron ) and let

and Then, for any given

nonnegative constants and having positive sum one can

find a point G inside or on the tetrahedron such that

the volumes V1, V2, V3 and V4 of the subtetrahedra

respectively, satisfy the

proportion: Moreover, the number of such

points G for which this is so is given by the number of distinct

permutations of the elements in Finally, the

centroid of these 24 (or 12 or 6 or 4 or 1) points is the same as

the centroid of the original tetrahedron.

Proof: The figure below will be of help.

P4

d

P3 c

G

G* b

G

P1 P2

16

O

Again, to simplify the vector algebra calculations which lie

ahead I first translate the above tetrahedron so that one of its

vertices, P1, lies at the origin. Doing this leads to the

tetrahedron :

Q4

d V1

Q3 c

V2 G*

G* b V3

O Q2

V4

where and

Let G* = xQ2 + yQ3 + zQ4 be a point inside or on

the tetrahedron . Then ; moreover,

from the figure we see that

b = Q2 – G*, c = Q3 – G* and d = Q4 – G*.

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Let V1, V2, V3 and V4 be the volumes of the indicated

subtetrahedra. Then, where from the scalar triple

product,

V1*

18

i.e., V1*

(12)

In entirely similar ways (and after much vector algebra), one can

show that

V2*

(13)

19

V3*

(14)

and

V4*

(15)

NOTE: If V is the volume of the tetrahedron T(OQ2Q3Q4) [or

], then using the triple scalar product

From (12), (13), (14) and (15), in order to have

, we must solve the linear system:

20

(16)

The unique solution to (16) is:

Therefore, substituting into our formula for G*, namely, G* = xQ2

+ yQ3 + zQ4, we have

(17)

However, since G* = G – P1, Q2 = P2 – P1, Q3 = P3 – P1 and Q4 =

P4 – P1, equation (17) becomes

G

21

or, finally,

G

(18)

Again, if we consider all distinguishable permutations of the

elements of we see that the proportions in:

is just one of the (up to) twenty-four

possible:

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each leading to a different location for G. In fact, the number

N of locations of G is given by the following five cases:

Case I: are different. In this case, we have

different locations for G. In light of the claim

made regarding their centroid (see also Theorem 3 below)

it will be helpful to list them all(!) From (18) and (19) they

are (in groups of 4) given by:

23

G1

G2

G3

G4

(20)

G5

G6

G7

G8

G9

G10

24

G11

G12 ;

G13

G14

G15

(20)

G16

G17

G18

G19

G20

and

25

G21

G22

(20)

G23

G24

Case II: Of , exactly two are the same (and different

from the other two). In this case, we have different

locations for G. If we assume , then from (18) and (19)

they are (in groups of 4) given by:

G1

G2

(21)

G3

G4

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G5

G6

G7

G8 ;

and (21)

G9 ;

G10

G11

G12

Case III: Of , exactly two are the same (and so are

the other two). In this case, we have different

locations for G. If we assume that , then from

(18) and (19) they are given by:27

G1

G2

G3

(22)

G4

G5

G6

Case IV: Of , exactly three are the same, the fourth

is different. In this case, there are possible

locations for G. If we assume they are given by

G1

G2

(23)

28

G3

G4 .

Case V: Finally, we have In this case, we have only

possible location for G:

G1

which coincides with the centroid of the original tetrahedron

In Case I, where the twenty-four Gi’s are given in (20), their

centroid H is given by

H

i.e., H ,

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which coincides with the centroid of the original tetrahedron

T(P1P2P3P4) . Since the same is also true in Cases II, III, IV

and V, this completes the proof of Theorem 2.

The following figure (showing all 24 locations of the Gi and one

partition) illustrates the result.

Before leaving Theorem 2, I call attention to one amazing fact:

When in Case I I listed the 24 locations of G, I did so in six

groups, with four points per group [see (20) above].

The centroid of each of the six groups of points:30

{G1,G2,G3,G4}, {G5,G6,G7,G8}, {G9,G10,G11,G12},

{G13,G14,G15,G16}, {G17,G18,G19,G20}, and {G21,G22,G23,G24}

is the same as the centroid of the original tetrahedron

.

Furthermore, when the four points in a group are noncoplanar,

they form the vertices of a tetrahedron; when this holds for

all six groups we have the six tetrahedra:

and ,

where all six tetrahedra have the same centroid as the

original tetrahedron. The following figures illustrate the

result:

NOTE: For two of the six tetrahedra are

degenerate (with V = 0).

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Taking one step back…leading to n steps forward?

Having proven the main results of the paper, the “structural

similarities” between the two formulas:

G

(6)

—from which the six possible locations of G can be identified

(Theorem 1) and

G

(18)

—from which the twenty-four possible locations of G can be

identified (Theorem 2)—can certainly catch one’s attention. It

was enough for me to wonder if…

Theorem 0: Let be any two non-coinciding points (the

endpoints of the line segment ) and let

Then, for any given nonnegative constants having

positive sum, one can find a point G inside or on the line

segment such that the lengths L1 and L2 of the sub segments

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, respectively, satisfy the proportion:

Moreover, the number of such points G for which this is so is

given by the number of distinct permutations of the elements in

Finally, the centroid of these 2 (or 1) points is the

same as the centroid of the original line segment.

Proof: The figure below will be of help.

P1 G* G b P2

L2 L1

G

O

As in the proofs of Theorems 1 and 2, I first translate the above

line segment so that one of its vertices, P1, lies at the origin.

Doing this leads to the line segment :

O G* G* b Q2

33

L2 L1

where and

Let G* = xQ2 be a point inside or on the line segment . Then

; moreover,

b = Q2 – G*

= Q2 – xQ2

b = (1 – x)Q2.

Let L1 and L2 be the lengths of the indicated sub segments. Then

L1 and L2 .

(24)

Therefore, from (24), if we want , we must solve the

linear equation

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which has the unique solution: Therefore, we have

G*

(25)

However, since G* = G – P1, equation (25) becomes

G

or, finally,

G

(26)

By considering all distinguishable permutations of the elements

of we see that the proportions in: is just one

of the (up to) two possible:

35

(27)

each leading to a different location for G. In fact, the number

N of locations of G is given by the following two cases:

Case I: are different. In this case, we have

different locations for G; from (21) and (22) they are:

G1

and (28)

G2

Case II: In this case, we have only location for

G; from (21) and (22) it is:

G1

Moreover, in Case I, where the Gi’s are given in (23), we see

that their centroid H is given by

H

36

,

which coincides with the centroid (= midpoint) of the line

segment: . Since the same was also true in Case II, this

completes the proof of Theorem 0.

Because of their geometric significance (the proportional

partitioning of line segments, triangles, and tetrahedra), the

connection between our Theorems 0, 1 and 2 is, of course, much

more than the structural similarity of the formulas:

G

G

and G

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The question is: How much more? This is undoubtedly an instance

where I am hamstrung by my

limited knowledge of geometry and ignorance of projective

geometry and…, and….

On the other hand, if G. H. Hardy was correct in describing

mathematics—or, at least, modern mathematics [see M. Fried

(2004)]—as “the science of patterns,” then, inspired by the above

pattern, I state and prove a final theorem (whose possible

geometric implications are to me not yet fully known).

Theorem 3: Let P1, P2, …, Pn be n distinct points whose centroid is

G. For i = 1, 2,…, n, let

be constants whose sum is Let , and let S

be the set of all permutations (not necessarily distinct) of the

elements of A. If , then for i = 1, 2, …, n, let denote

its element. Finally, for k = 1, 2, 3, …, n!, define the

points by

(29)38

where Then if H is the centroid of the (not

necessarily distinct) points , then H coincides with the

centroid G of the original points: P1, P2, …, Pn.

Proof: By definition,

i.e., .

(30)

In (30), the “inner sum”: , since there are

(n - 1)! permutations (not necessarily distinct) in S whose

term is . Substituting this into (30), gives:

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= G

,

and, hence, H coincides with G. This completes the proof of

Theorem 3.

[I regret that when , I cannot specify in advance how many of

the n! points , as defined in (29), will be distinct.]

The following two figures illustrate Theorem 3 when P1, P2, P3 and

P4 are taken to be the vertices of a square (whose centroid is at

its center) and the non-negative constants.

40

NOTE: In the figure to the above right the four chosen were

relatively prime.

The following figures show what can happen when one allows the

to be functions that do not simultaneously vanish on their

common domain—the become trajectories:

41

(In the following note the successive changes in the )

42

In the last two examples P1, P2, P3 and P4 are taken to be the

vertices of a tetrahedron.

Work cited:

Fried, Michael N. (2004). “Mathematics as the Science of Patterns.”

Convergence (Internet Journal of the Mathematics Association of America (MAA)),

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http://mathdl.maa.org/mathDL/46/?

pa=content&sa=viewDocument&nodeId=2167. (Last accessed October 11,

2012.)

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