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Oct 10, 2013 (3 years and 9 months ago)

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Spatial Analogues of Ceva’s Theorem and its Applications



Nadav Goldberg

12020 Montrose Village Terrace

Rockville, MD 20852 USA

nadav2@hotmail.com



Abstract

Some interesting and useful theorems of planar geometry have interesting analogues
in three
-
dimen
sional geometry as well. Within the framework of this project, such an
analogue for Ceva’s theorem was found. It was then used to solve problems in three
-
dimensional space that correspond to planar corollaries of the original Ceva’s theorem.


Introducti
on


Many important planar theorems have solid geometric analogues. For example, the
famous Pythagorean theorem has many analogues in solid geometry. Other lesser
-
known but
important theorems also have analogues. These analogies between theorems on the p
lane and
theorems in space are not only elegant, but are many times quite useful. If a theorem on the
plane is useful in problem solving, then perhaps the spatial analogue may be used to formulate
and solve similar problems in space.




A concrete example
, which will be used later, can be provided. There is a theorem
that an angle bisector in a triangle divides the opposite side into lengths proportional to the
adjacent sides. A proof of the stereometric analogue is presented here.


Theorem 1 (Dihedral
angle bisector theorem)

Let
ABCD
be a tetrahedron (see figure 1). Point
X
is on

and plane
AXD
is the dihedral angle
bisector of dihedral angle
. Then

.

The proof is not long. The rati
o

is
equal to

because both tetrahedrons have a
common height from
A

to (
CBD
). The ratio of areas can also be simplified further to

because both triangles have a common height from
D

t
o
. Now, the original ratio of
volumes can be rewritten in another way. The height from
X
to (
ABD
) is equal to the height
from
X

to (
ACD
). This is a property of the dihedral angle bisector. This means that in the
ratio of volumes
, the height cancels out from numerator and denominator, leaving the ratio of
areas of triangle
ABD
to triangle
ACD
. To recapitulate,


D

C

B

A

X

Figure 1

. Q.E.D.


This is an elegant example of an extension of a planar theorem to space. Whereas on
the plane the angle bisector divides the opposite side into lengths proportional to the lengths
of the adjacent sides, in space, the dihedral angle bisector divides the opposite edge into
lengths proportional to the areas of the adjacent faces. It is a ve
ry direct analogy.



An aim of this project was to extend two other useful planar theorems to space: the
theorem of Menelaus and Ceva’s theorem. Before learning about these two closely connected
statements, one should know a bit about their discoverers.


Giovanni Ceva


Giovanni Ceva was born in 1647 and learned at a Jesuit college in his hometown of
Milan. He proceeded to study at the University of Pisa and began teaching there until his
appointment as professor of mathematics at the University of Mantua
in 1686 for the
remainder of his life. Ceva supported the local rulers until the area was taken over by Austria,
which he then quickly moved to support. Ceva’s great discoveries were mostly in the field of
geometry. In his 1678 work,
De lineis rectis
, h
e first published an important geometric
theorem overlooked by the Greeks that is now named “Ceva’s theorem.” He also
rediscovered and published the theorem of Menelaus. In his other works, he applied
geometry to mechanics and hydraulics and even anticip
ated calculus to some extent. His
1711 work,
De Re Nummeraria
, was one of the first works in mathematical economics
*
.


Theorem 2 (
Ceva’s theorem)

Let
ABC

be a triangle (see figure 2) with p
oint
X

on
,
Y

on
,
and
Z

on
.
,

,

and

are concurrent

if and only if


.


For proofs of this well
-
known statement, the reader is referred to [1] or [4].
Ceva’s

theorem provides an easy and useful tool for proving concurrency of segments, such as in the
following famous corollary, which will be later used.


Theorem 3

(Gergonne point)

Let a circle be inscribed in triangle. The segments drawn from each vertex of t
he
triangle to the point of tangency on the opposite side are concurrent.



This theorem may be easily proven with the aid of Ceva’s theorem. For specifics, see
[4].





*

The biographical information presented here is from [6].

C

A

X

Y

Z

B

Figure 2

Menelaus of Alexandria


The mathematician Menelaus was born around 70 AD in Alexandria.
He lived in
Rome for a time, as evidenced by a description by Plutarch of a conversation he had with the
philosopher Lucius in Rome. Both Pappus and Proclus mention Menelaus and there is a
reference to his astronomical observations in the records of Ptole
my. Tenth century Arab
records indicate that he wrote a great number of books. Unfortunately, only one survives and
it has been heavily edited by Arabic authors. His work,
Sphaerica,
deals with spherical
triangles and their application to astronomy. It

was the first detailed exposition of spherical
geometry. In this work, he proved a spherical version of the planar theorem of Menelaus.
Interestingly, the planar version was well known already
*
.


Theorem 4 (Theorem of Menelaus)

Let
ABC
be a triangle (s
ee figure 3) with
point
X

on
,
Y

on
,
and

Z
on
.
X
,
Y
, and
Z

are collinear if and only if


.




The exact proof may be found in [1] or [3]. Notice that both the
theorems of Ceva and
Menelaus have the same equation. Furthermore, they speak of similar but opposte concepts.
Ceva’s theorem relates to the concurrentness of segments in a triangle, and Menelaus’s
theorem relates to collinearity of points on a triangle.

This idea of similarity between the two
theorems is called “duality.” In fact, the two theorems may even be shown to be equivalent:
one may use one to prove the other [4].


Theorem of Menelaus in Space


First attempts to formulate an analogue of this th
eorem by examining the ratios of
areas when a plane passes through a tetrahedron failed. While doing background research on
solid geometry, however, an interesting three
-
dimensional extension of the theorem of
Menelaus was found in [2].


Theorem 5

Let
AB
CD
be a space quadrilateral (see figure 4).
Point
X

lies on
,
Y

lies on
,
Z
is on
, and

W
is
on
.
Points
X
,
Y
,
Z
, and
W
are coplanar if and only
if

.



This theorem is a direct analogue of the planar
original. Sides of a triangle become edges of a space
quadrilateral and a line intersecting the triangle becomes
a plane intersecting the space quadrilateral. The equation is still nearly iden
tical.
This
extension, like the original, uses ratios of distances. If one wishes to make the extension more
“three
-
dimensional,” then the ratios of distances can be replaced with ratios of areas by



*

This information was taken from [5].

A

X

B

Y

C

D

Z

W

Figure 3

Figure 4

A

X

Y

Z

B

multiplying the numerator and denominator of each ratio
by one
-
half a height common to the
two bases.


Results

Just as Ceva’s theorem on the plane has a duality with the theorem of Menelaus, it was
hoped that there would be a similar duality in the spatial analogues. This was the one of the
guiding principles
in the development of a three
-
dimensional Ceva’s theorem. Such a duality
was found. Instead of a plane cutting through the edges of a quadrilateral at four points, four
planes are constructed from each edge to to the opposite one. In both cases, the equ
ation is
the same and quite similar to the planar version.



Theorem 6 (Ceva’s theorem in
space)

Let
ABCD
be a space
quadrilateral (see figure 5).
Point
X

lies on
,
Y

lies on
,
Z
is on
, and

W
is on
.
Four planes
AZB
,
BWC
,
CXD
, and
DYA
intersect at exactly one point
if and only if


(1)




In other words, four planes drawn from the edges of a quadrilateral to a point on the
opposite edge int
ersect at exactly one point if and only if equation (1) is true.



Let us introduce this theorem in symbolic form.


Given
:



Space quadrilateral
ABCD



,
,
,



(
AZB
)



(
BWC
)



(
CXD
)


(
DYA
)

=

P



,



(
AZB
) ∩

(
AYD
)

=
, (
CXD
)


(
DYA
)

=


Prove:



A

X

B

Y

C

D

Z

W

P



A

B

C

D

X

Y

Z

W

T

P

Figure 6

Figure 5

Construct auxiliary segment

(see figure 6
)
. Next, construct the plane that
contains both

and
. This plane can be constructed because

and

are
coplanar, since they intersect at
P
. The plane will inters
ect

at a point
T
, so it may be
referred to as “(
ATC
).”

and
will pass through
and
, respectively, because of
the definition of (
ATC
).



App
lying Ceva’s theorem to

and
, we get the following two equations:


(2a.)



(2b.)




Multiply the two equations and the result is equation (1). Q.E.D.


Now the conv
erse must be proven.


Given
:



Space quadrilateral
ABCD



,
,
,






,



(
AZB
)


(
AYD
)

=
, (
CXD
)


(
DYA
)

=


Prove:

(
AZB
)



(
BWC
)



(
CXD
)


(
DYA
)

=

P



Construct a segment from
A
to a point
T
on auxiliary segment

such that

passes through
. Ceva’s theorem then states that equation (2a) is true. Multiply equation
(2a) with equation (1), which is given, and the result is equation (2b). Using Ceva’s theorem
once more, the result is that
passes throu
gh
. This result leads to the conclusion that
both
and

are on the plane (
ATC
), because the endpoints of each segment are on the
plane. Furthermore,

and

intersect at a point
P

because
C

lies in the half
-
plane
opposite from
if
is the boundary. However, because
and

were defined as the
intersection li
nes of (
ADY
) and (
AZB
), (
BWC
) and (
DCX
), respectively, then in fact all four
planes intersect at point
P.
Q.E.D


Corollary of Ceva’s Theorem in Space


Theorem 7

On each face of a tetrahedron, let there be three concurrent Cevians and let every
Cevian on a

face intersect a Cevian from another face on the common edge. Then the
segments connecting each vertex of the tetrahedron with the intersection point of the
Cevians on the opposite face are concurrent.



This corollary is similar to the stereometric anal
ogy of Ceva’s theorem that was
proven above. Using the theorem, it is easy to prove this statement if one realizes that Cevians
of adjacent faces meeting on the common edge are basically intersections of planes with the
faces of the tetrahedron. Equation
(1) is correct in this situation by the same logic by which it
was derived. Therefore, according to theorem 6, (
AZB
)



(
BWC
)



(
CXD
)


(
DYA
)

=

P
. It
must be shown that the intersection of all these planes and (
DUB
)


(
ATC
)

is still
P
(see
figure 7). Geo
metrically, this means that it must be proven that
, the intersection of
(
DUB
) and (
ATC
), passes through
P
.


The stereometric extension of the
theorem of Menelaus applied to
quadrilateral
ABCT

as it is cut by plane
XDY

states that


.


Divide this equation by the equation from
Ceva’s theorem for
,
, and the result proves
the concurrency of
,
, and


at
P
. This is equivalent to saying that all six
planes intersect at
P
. Therefore, the lines of intersection of the planes intersect at
P
.
Segments with vertices of the tetrahedron and points of intersection of the Cevians as
endpoints are parts of lin
es of intersection of the planes, meaning that the segments also
intersect at
P.

Q.E.D.


Applications of Ceva’s Theorem in Space and its Corollary


There are important consequences of theorems two and three. Just as Ceva’s theorem
on the plane is used to

prove the concurrency of several important segments in the triangle,
Ceva’s theorem in space and its corollary can be used to prove similar things in the
tetrahedron.


Theorem 8

On each face of a tetrahedron, let the three medians be drawn. Then the segme
nts
connecting each vertex of the tetrahedron with the intersection point of the medians on
the opposite face are concurrent.



All the conditions of theorem 7 are easily met. The medians on each face are
concurrent and the medians from adjacent faces cle
arly intercept the common edge at the
same point. Therefore, theorem 8 is proven by theorem 7. It is interesting to add that this
formulation is equivalent to saying that the planes drawn from each edge of the tetrahedron
and bisecting the opposite edge
intersect at one point.


Theorem 9

Let a sphere intersect a tetrahedron in such a way so that it is tangent to every edge. On
every face, draw Cevians connecting to the points of tangency with the sphere on the
opposite edge. Connect the points of inters
ection of the Cevians on each face with the
vertex opposite the face. These new segments will intersect at one point.



It is important first of all to notice that this theorem is a three
-
dimensional analogue of
the Gergonne point theorem. Since the inte
rsection of the sphere with each face is a circle
tangent to the edges, the segments connecting the vertex of the face with the point of tangency
on the opposite side of the face are concurrent. This meets the first condition of theorem 7.


A

B

C

D

X

Y

Z

W

T

P

U

Figure 7

Secondly, beca
use the sphere can only be tangent to each edge at one point, Cevians to the
same edge of the tetrahedron will intersect at the same point on the edge. Therefore, both
conditions for theorem 7 are met and so the lines are concurrent.


Theorem 10

The dihed
ral angle bisectors of a tetrahedron meet at one point, which is the center of
the sphere inscribed in the tetrahedron.



By the stereometric angle bisector theorem, the following equations are true (figure 7
may be used):

,

,

,

and
.


When these equations are multiplied, the result is equation (1). Hence theorem 6 states that
the four planes
AZB
,
BWC
,
CXD
, and
DYA
intersect in one point. It is now necessary to sho
w
that planes
DUB
and
ATC

intersect at that point as well. To do that, the same reasoning as in
theorem 7 may be used. However, its assumption that the Cevians of
are concurrent
must first be proven:



Ceva’s

theorem then states that these segments are concurrent and thus the logic in theorem 7
shows that all six dihedral angle bisectors intersect at one point. Furthermore, because the
dihedral angle bisectors are the loci of all points equidistant from two f
aces of a tetrahedron,
the point of intersection of the bisectors is the point equidistant from all four faces. This
means that a sphere centered at that point may be inscribed in the tetrahedron.


Theorem 11

The altitudes of a tetrahedron are concurrent
if and only if the tetrahedron is
orthocentric
*
.



Before proceeding to prove this theorem, a lemma is necessary.


Lemma

If a tetrahedron is orthocentric then

I.

A segment is an altitude of the
tetrahedron when the segment is
drawn from a vertex perpendicul
ar
to the height on the opposite face.

II.

The heights of two faces to the same
base are concurrent.

III.

The heights of a face intersect at the
point where the altitude from the
opposite vertex intercepts the face.

A.

Segment

is perpendicular

to

by construction (See figure 8) and

is
perpendicular to it by the definition of an orthocentric tetrahedron. Because

is
perpendicular to two segments on (
ADY
), it is perpendicular
to the whole plane. Since



*

In an orthocentric tetrahedron, opposite edges are perpendicular.

A

B

Y

C


D

W

Figure 8


is on this plane,

is also perpendicular to
. By construction,

is also
perpendicular to
. Both of these segment
s are on (
ABC
), so

is perpendicular to
all of (
ABC
), meaning it is the altitude.


B.

The three perpendiculars theorem directly proves this based on the result of part A and
the definition of
.


C.

The three perpendi
culars theorem directly proves this based on the result of part A and
the fact that

is perpendicular to
. Q.E.D.



Parts B and C of the lemma in combination with theorem 7 now easily prove the
theorem. Simpl
y construct heights on every face of the tetrahedron and construct the altitude
from the intersection of the heights to the opposite vertex. Theorem 7 states, then, that the
altitudes intersect. Q.E.D.


Discussion


As can be seen, Ceva’s theorem in space

has a great number of applications similar to
those of Ceva’s theorem on the plane. On the plane, Ceva’s theorem is used to prove that
certain interesting lines in a triangle intersect at a special point. Similarly, in a tetrahedron in
space, certain in
teresting planes

or segments from vertices to the opposite plane, depending
on one’s frame of reference

intersect at a special point. Hence Ceva’s theorem is useful in
exploring interesting points in the tetrahedron like the center or Gergonne point. The
re are
many other fascinating applications that are extensions of the ways Ceva’s theorem is used on
the plane. With more time, it would have been possible to apply it to these other interesting
three
-
dimensional extensions.


In our examination of the lit
erature, we found less applications of the theorem of
Menelaus than of the Ceva’s theorem, though it does have a few interesting ones.
Unfortunately, there was not sufficient time to apply the three
-
dimensional extension of the
theorem of Menelaus found d
uring research to generalized spatial problems.


Conclusions


The purpose of this project was to generalize two closely related, interesting theorems,
Ceva’s theorem and Menelaus’s theorem, to three
-
dimensions and then to use them in
problem solving. An e
xtension of the theorem of Menelaus was found during examination of
the literature. The main result of the project was discovering an extension of Ceva’s theorem.
This endeavor was aided by the duality between the two theorems. The last part of the proj
ect
was to apply the newfound Ceva’s theorem. The power of this spatial extension was shown
by its simple solutions to these somewhat difficult problems. Interesting properties of the
tetrahedron were easily proven with the help of the spatial Ceva’s theo
rem and its corollary.
These properties included the fact that planes from each edge of the tetrahedron that bisect the
opposite edge intersect at one point, and that planes bisecting each dihedral angle intersect at
one point. Also, if a sphere is tange
nt to each edge of the tetrahedron, then all the planes from
each edge to the point of tangency on the opposite edge intersect at one point. Lastly, in an
orthocentric tetrahedron, the altitudes meet at one point. All of these applications are
interestin
g stereometric analogues in themselves of planar properties of triangles.




Acknowledgements


I would like to greatly thank my mentor for this project, Mr. Boris Koichu, for his
guidance and assistance throughout the project. His ideas were always useful
. I would also
like to extend my gratitude to Ralph Tandetsky for his constant encouragement, and to the
organizers of SciTech 2001 for making this project possible.


References

1. Coxeter, H.S.M and Greitzer, S.L.
Geometry Revisited
, Mathematical Associ
ation of
America: 1967.


2. Lines, L.
Solid Geometry, with chapters on space
-
lattices, sphere
-
packs and crystals.

Dover Publications, New York: 1965.


3. Bogomolny, Alexander.
Cut the Knot!
Website. <http://www.cut
-
the
-
knot.com/Generalization/Menelau
s.html>.


4.
----

<http://www.cut
-
the
-
knot.com/Generalization/ceva.html>.


5. O'Connor, J.J. and Robertson E.F.
Menelaus of Alexandria
.
The MacTutor History of
Mathematics Archive.

University of St. Andrews, Scotland: 2000. Website.
<http://www
-
history
.mcs.st
-
and.ac.uk/~history/Mathematicians/Menelaus.html>.


6.
----

Giovanni Ceva. The MacTutor History of Mathematics Archive.

University of St.
Andrews, Scotland: 2000. Website. <http://www
-
history.mcs.st
-
and.ac.uk/~history/Mathematicians/Giovanni
-
Ceva.h
tml>.