DE TIERRASDE TIERRAS
3.2 TIPOS DE UTILIZACIÓN DE LA TIERRA TIPOS DE UTILIZACIÓN DE LA TIERRA
Proof Since the vertex A lies on the two lines AB and AD, its polar plane A∗must pass through both(AB)∗ = AB and(AD)∗ = AD; so A∗ = AB D. In a similar
way, the polar of any vertex of the quadrilateral is the plane spanned by the two adjacent edges. The polar of the diagonal AC is then the intersection A∗ ∩C∗ =
AB D∩C B D, which is precisely the opposite diagonal B D. tu
Exercise 8.2-6 Given a null system — or a polar system, for that matter — in 3-space, show that two lines`and m meet but do not coincide just when their po- lars`∗and m∗ meet but do not coincide.
[Hint: This is almost trivial. Two lines in 3-space meet just when they pass through a common point and just when they lie in a common plane.]
8.3
Skew-polar hexagons
We pointed out in Section 2.9 that one way to specify a null system in 3-space is to give a twisted cubic curve. Another very pretty way to specify a null system, as discussed in Exercise 8.3-2 below, is to give a M¨obius pair of tetrahedra. For our purposes, though, it is most convenient to specify a null system by giving a certain type of skew hexagon.
Fix some null system in 3-space. Knowing any skew-polar pair of lines{`, `∗} already goes a long way towards specifying that null system. Indeed, it specifies four out of the five degrees of freedom. It tells us, right away, the polars of all of the points lying on either`or`∗. Furthermore, given a point P on neither`nor`∗, there is a unique line m through P that meets both`and`∗; it is called the common
transversal. Since the common transversal m touches both`and`∗, it must be self- polar, by Proposition 8.2-4. Thus, the polar plane P∗ must pass through the line
m. The one degree of freedom that remains rotates the plane P∗about the line m. We could tie down that one degree of freedom by specifying, for some such point
P, its polar plane P∗. But we’ll do something more symmetric.
Let us say that a hexagon A1B2A3B1A2B3in 3-space is skew-polar with respect to a certain null system when its six vertices are distinct and when each side forms, with the opposite side, a skew-polar pair of lines — that is, each of the three pairs of lines{A1B2,A2B1},{A1B3,A3B1}, and{A2B3,A3B2}is skew-polar. Those re- quirements may not seem very strict, but it turns out that skew-polar hexagons are quite constrained.
First, every line joining two nonadjacent vertices of a skew-polar hexagon is self-polar. For example, because the opposite edges A1B2and A2B1form a skew- polar pair, it follows from Proposition 8.2-4 that the other four edges A1A2, B1B2,
A1B1and A2B2of the tetrahedron A1A2B1B2 are self-polar. Repeating this argu- ment twice more finishes the job.
In particular, the lines A1A2, A1A3, and A2A3are all self-polar. We conclude from Proposition 8.2-3 that the points A1, A2, and A3 lie on a common, self-polar
108 CHAPTER 8. NULL SYSTEMS
line, which we shall call a. By a similar argument, the points B1, B2, and B3 lie on a self-polar line b. Since the opposite sides A1B2and A2B1of the hexagon are skew, it follows that the lines a and b are also skew.
We have now uncovered all of the constraints on a skew-polar hexagon, as we can show by working backwards to construct one. Let a and b be any two skew lines, each of which is self-polar. For each point A on a, the polar plane A∗ passes through a, and hence meets b in a unique point B. The line AB passes through A and lies in the plane A∗, so it is self-polar. It follows that the correspondence taking
A to B is symmetric; that is, the unique point where the polar plane B∗ meets the line a is A once again. Let A1, A2, and A3 be any three distinct points on a, and let B1, B2, and B3 be the corresponding points on b. We claim that the hexagon
A1B2A3B1A2B3 is skew-polar. To see this, note that all four edges of the skew quadrilateral A1B1B2A2 are self-polar. It follows from Proposition 8.2-5 that the diagonals A1B2and A2B1, which are a pair of opposite edges of the hexagon, form a skew-polar pair.
For future reference, note that there are nine degrees of freedom in the choice of a hexagon that is skew-polar for a given null system: three in the self-polar line
a, three more in b, and a final three in the points A1, A2, and A3 along a.
Exercise 8.3-1 Given a null system in 3-space, consider a cyclic list`1`∗2`3`∗1`2`∗3 of six lines in which opposite pairs of lines are skew-polar pairs and adjacent pairs are not skew. Show, from these assumptions alone, that the six lines are the edges of a skew-polar hexagon.
[Hints: Prove first that no adjacent pair of lines can coincide — for example, it cannot be the case that`1 =`∗2. We can then make the definitions
A1 :=`1∩`∗2 B1:=`∗1∩`2
A2 :=`2∩`∗3 B2:=`∗2∩`3
A3 :=`3∩`∗1 B3:=`∗3∩`1
and switch to thinking about the hexagon in terms of its vertices A1B2A3B1A2B3, rather than its edges. Conclude by proving that the six vertices are all distinct. Both proofs use Proposition 8.2-4.]
Exercise 8.3-2 Fix any null system in 3-space and fix any tetrahedron S. Show that the polar planes of the vertices of S are the faces of a tetrahedron T that forms, with S, a M¨obius pair. Conversely, given any M¨obius pair of tetrahedra in 3-space, show that there is a unique null system in which the faces of each tetrahedron are the polars of the vertices of the other tetrahedron. It follows from this that, if we fix any tetrahedron S, null systems in 3-space are in one-to-one correspondence with tetrahedra that form a M¨obius pair with S. (Note that there are five degrees of freedom in either case.)
[Hint: The vertices of S are not coplanar, so the faces of T won’t be concurrent. If A, B, and C are three of the vertices of S, then the three corresponding faces A∗,