> jlid[ 0N]bjbj jjRl",,,,".555~5+++$ٟ >i+G+++>a5~5Taaa+$5p~5a+aaݘ6$~5.`"~,^Ɲ0ڝyya""Spatial Analogues of Cevas 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 threedimensional geometry as well. Within the framework of this project, such an analogue for Cevas theorem was found. It was then used to solve problems in threedimensional space that correspond to planar corollaries of the original Cevas theorem.
Introduction
Many important planar theorems have solid geometric analogues. For example, the famous Pythagorean theorem has many analogues in solid geometry. Other lesserknown but important theorems also have analogues. These analogies between theorems on the plane 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 EMBED Equation.3 and plane AXD is the dihedral angle bisector of dihedral angle EMBED Equation.3 . Then
EMBED Equation.3 .
The proof is not long. The ratio EMBED Equation.3 is equal to EMBED Equation.3 because both tetrahedrons have a common height from A to (CBD). The ratio of areas can also be simplified further to EMBED Equation.3 because both triangles have a common height from D to EMBED Equation.3 . 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,
EMBED Equation.3 . 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 very direct analogy.
An aim of this project was to extend two other useful planar theorems to space: the theorem of Menelaus and Cevas 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. Cevas great discoveries were mostly in the field of geometry. In his 1678 work, De lineis rectis, he first published an important geometric theorem overlooked by the Greeks that is now named Cevas theorem. He also rediscovered and published the theorem of Menelaus. In his other works, he applied geometry to mechanics and hydraulics and even anticipated calculus to some extent. His 1711 work, De Re Nummeraria, was one of the first works in mathematical economics*.
Theorem 2 (Cevas theorem)
Let ABC be a triangle (see figure 2) with point X on EMBED Equation.3 , Y on EMBED Equation.3 , and Z on EMBED Equation.3 . EMBED Equation.3 , EMBED Equation.3 , and EMBED Equation.3 are concurrent if and only if
EMBED Equation.3 .
For proofs of this wellknown statement, the reader is referred to [1] or [4]. Cevas 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 the triangle to the point of tangency on the opposite side are concurrent.
This theorem may be easily proven with the aid of Cevas theorem. For specifics, see [4].
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 Ptolemy. 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 (see figure 3) with point X on EMBED Equation.3 , Y on EMBED Equation.3 , and Z on EMBED Equation.3 . X, Y, and Z are collinear if and only if
EMBED Equation.3 .
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. Cevas theorem relates to the concurrentness of segments in a triangle, and Menelauss 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 theorem by examining the ratios of areas when a plane passes through a tetrahedron failed. While doing background research on solid geometry, however, an interesting threedimensional extension of the theorem of Menelaus was found in [2].
Theorem 5
Let ABCD be a space quadrilateral (see figure 4). Point X lies on EMBED Equation.3 , Y lies on EMBED Equation.3 , Z is on EMBED Equation.3 , and W is on EMBED Equation.3 . Points X, Y, Z, and W are coplanar if and only if
EMBED Equation.3 .
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 identical. This extension, like the original, uses ratios of distances. If one wishes to make the extension more threedimensional, then the ratios of distances can be replaced with ratios of areas by multiplying the numerator and denominator of each ratio by onehalf a height common to the two bases.
Results
Just as Cevas 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 threedimensional Cevas 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 equation is the same and quite similar to the planar version.
Theorem 6 (Cevas theorem in space)
Let ABCD be a space quadrilateral (see figure 5). Point X lies on EMBED Equation.3 , Y lies on EMBED Equation.3 , Z is on EMBED Equation.3 , and W is on EMBED Equation.3 . Four planes AZB, BWC, CXD, and DYA intersect at exactly one point if and only if
(1) EMBED Equation.3
In other words, four planes drawn from the edges of a quadrilateral to a point on the opposite edge intersect at exactly one point if and only if equation (1) is true.
Let us introduce this theorem in symbolic form.
Given:
Space quadrilateral ABCD
EMBED Equation.3 , EMBED Equation.3 , EMBED Equation.3 , EMBED Equation.3
(AZB) )" (BWC) )" (CXD) )" (DYA) = P
EMBED Equation.3 , EMBED Equation.3
(AZB) )" (AYD) = EMBED Equation.3 , (CXD) )" (DYA) = EMBED Equation.3
Prove: EMBED Equation.3
Construct auxiliary segment EMBED Equation.3 (see figure 6). Next, construct the plane that contains both EMBED Equation.3 and EMBED Equation.3 . This plane can be constructed because EMBED Equation.3 and EMBED Equation.3 are coplanar, since they intersect at P. The plane will intersect EMBED Equation.3 at a point T, so it may be referred to as (ATC). EMBED Equation.3 and EMBED Equation.3 will pass through EMBED Equation.3 and EMBED Equation.3 , respectively, because of the definition of (ATC).
Applying Cevas theorem to EMBED Equation.3 and EMBED Equation.3 , we get the following two equations:
(2a.) EMBED Equation.3 (2b.) EMBED Equation.3
Multiply the two equations and the result is equation (1). Q.E.D.
Now the converse must be proven.
Given:
Space quadrilateral ABCD
EMBED Equation.3 , EMBED Equation.3 , EMBED Equation.3 , EMBED Equation.3
EMBED Equation.3 , EMBED Equation.3
(AZB) )" (AYD) = EMBED Equation.3 , (CXD) )" (DYA) = EMBED Equation.3
Prove: (AZB) )" (BWC) )" (CXD) )" (DYA) = P
Construct a segment from A to a point T on auxiliary segment EMBED Equation.3 such that EMBED Equation.3 passes through EMBED Equation.3 . Cevas theorem then states that equation (2a) is true. Multiply equation (2a) with equation (1), which is given, and the result is equation (2b). Using Cevas theorem once more, the result is that EMBED Equation.3 passes through EMBED Equation.3 . This result leads to the conclusion that both EMBED Equation.3 and EMBED Equation.3 are on the plane (ATC), because the endpoints of each segment are on the plane. Furthermore, EMBED Equation.3 and EMBED Equation.3 intersect at a point P because C lies in the halfplane opposite from EMBED Equation.3 if EMBED Equation.3 is the boundary. However, because EMBED Equation.3 and EMBED Equation.3 were defined as the intersection lines of (ADY) and (AZB), (BWC) and (DCX), respectively, then in fact all four planes intersect at point P. Q.E.D
Corollary of Cevas 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 analogy of Cevas 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). Geometrically, this means that it must be proven that EMBED Equation.3 , 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
EMBED Equation.3 .
Divide this equation by the equation from Cevas theorem for EMBED Equation.3 , EMBED Equation.3 , and the result proves the concurrency of EMBED Equation.3 , EMBED Equation.3 , and EMBED Equation.3 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 lines of intersection of the planes, meaning that the segments also intersect at P. Q.E.D.
Applications of Cevas Theorem in Space and its Corollary
There are important consequences of theorems two and three. Just as Cevas theorem on the plane is used to prove the concurrency of several important segments in the triangle, Cevas 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 segments 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 clearly 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 intersection 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 threedimensional analogue of the Gergonne point theorem. Since the intersection 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. Secondly, because 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 dihedral 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):
EMBED Equation.3 , EMBED Equation.3 , EMBED Equation.3 , and EMBED Equation.3 .
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 show 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 EMBED Equation.3 are concurrent must first be proven:
EMBED Equation.3
Cevas 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 faces 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
A segment is an altitude of the tetrahedron when the segment is drawn from a vertex perpendicular to the height on the opposite face.
The heights of two faces to the same base are concurrent.
The heights of a face intersect at the point where the altitude from the opposite vertex intercepts the face.
Segment EMBED Equation.3 is perpendicular to EMBED Equation.3 by construction (See figure 8) and EMBED Equation.3 is perpendicular to it by the definition of an orthocentric tetrahedron. Because EMBED Equation.3 is perpendicular to two segments on (ADY), it is perpendicular to the whole plane. Since EMBED Equation.3 is on this plane, EMBED Equation.3 is also perpendicular to EMBED Equation.3 . By construction, EMBED Equation.3 is also perpendicular to EMBED Equation.3 . Both of these segments are on (ABC), so EMBED Equation.3 is perpendicular to all of (ABC), meaning it is the altitude.
The three perpendiculars theorem directly proves this based on the result of part A and the definition of EMBED Equation.3 .
The three perpendiculars theorem directly proves this based on the result of part A and the fact that EMBED Equation.3 is perpendicular to EMBED Equation.3 . Q.E.D.
Parts B and C of the lemma in combination with theorem 7 now easily prove the theorem. Simply 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, Cevas theorem in space has a great number of applications similar to those of Cevas theorem on the plane. On the plane, Cevas theorem is used to prove that certain interesting lines in a triangle intersect at a special point. Similarly, in a tetrahedron in space, certain interesting planesor segments from vertices to the opposite plane, depending on ones frame of referenceintersect at a special point. Hence Cevas theorem is useful in exploring interesting points in the tetrahedron like the center or Gergonne point. There are many other fascinating applications that are extensions of the ways Cevas theorem is used on the plane. With more time, it would have been possible to apply it to these other interesting threedimensional extensions.
In our examination of the literature, we found less applications of the theorem of Menelaus than of the Cevas theorem, though it does have a few interesting ones. Unfortunately, there was not sufficient time to apply the threedimensional extension of the theorem of Menelaus found during research to generalized spatial problems.
Conclusions
The purpose of this project was to generalize two closely related, interesting theorems, Cevas theorem and Menelauss theorem, to threedimensions and then to use them in problem solving. An extension of the theorem of Menelaus was found during examination of the literature. The main result of the project was discovering an extension of Cevas theorem. This endeavor was aided by the duality between the two theorems. The last part of the project was to apply the newfound Cevas 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 Cevas theorem 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 tangent 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 interesting 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 Association of America: 1967.
2. Lines, L. Solid Geometry, with chapters on spacelattices, spherepacks and crystals. Dover Publications, New York: 1965.
3. Bogomolny, Alexander. Cut the Knot! Website. .
4.  .
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. .
6.  Giovanni Ceva. The MacTutor History of Mathematics Archive. University of St. Andrews, Scotland: 2000. Website. .
* The biographical information presented here is from [6].
* This information was taken from [5].
* In an orthocentric tetrahedron, opposite edges are perpendicular.
EMBED Equation.3
EMBED Equation.3
T
W
Z
Y
X
D
C
P
B
A
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X
B
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C
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Z
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Figure 6
P
Figure 7
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
A
B
C
D
X
Y
Z
W
T
P
U
A
B
Y
C
D
EMBED Equation.3
W
W
Z
D
C
Y
B
X
A
B
Z
X
Y
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Figure 8
Figure 1
Figure 2
Figure 3
Figure 4
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