Projective Geometry
Projective geometry is an extension (or a simplification, depending on point of view) of Euclidean geometry, in which there is no concept of distance or angle measure. Intuitively, projective geometry can be understood as only having points and lines; in other words, while Euclidean geometry can be informally viewed as the study of straightedge and compass constructions, projective geometry can be viewed as the study of straightedge only constructions.
Surprisingly, though originally conceived in the context of art, projective geometry has significant application to Euclidean geometry. For instance, results such as Desargues', Pappus', and Pascal's theorems are consequences of projective geometry, but are of much interest in "normal" geometry as well; this is due to the fact that the Euclidean plane can be naturally extended into a projective one. Projective geometry is also useful in avoiding edge cases of particular configurations, particularly the case of parallel lines (as in projective geometry, there are no parallel lines).
Formal Definitions
A projective plane is defined by a set of points, a set of lines, and a property of incidence satisfying three properties:
- For any two points, there is exactly one line incident with both of them.
- For any two lines, there is exactly one point incident with both of them.
- There exist four points such that no line is incident with more than two of them.
The final condition is not strictly necessary; it exists only to exclude degenerate cases from consideration. This is useful, however, in order to avoid having to manually check several degenerate cases over and over again; therefore, it is generally assumed as a matter of convenience.
The second condition has an important consequence:
There are no parallel lines in the projective plane.
Additionally, the first and second conditions are very similar, differing only in the swapping of "points" and "lines". This forms the motivation behind defining the property of incidence: it shows the duality between points and lines.
The simplest (and most commonly used) example of a projective plane is the extended Euclidean plane, which is the ordinary Euclidean plane equipped with two additional properties:
- Every set of parallel lines is equipped with an additional point, that is incident to any of these lines. That point is one of the points at infinity.
- The line at infinity is incident to every point at infinity, and is incident to only those points.
It can be verified that this results in a projective plane. Because of this projective plane's similarity to the Euclidean plane (and thus properties of this projective plane are generally true in the Euclidean plane as well), this is often referred to as the projective plane.
Projective planes need not be infinite. For instance, the projective plane of order 3 contains 13 lines and 13 points. More generally, the projective plane of order \(n\) contains \(n^2+n+1\) points and \(n^2+n+1\) lines, so that every point is incident to \(n+1\) lines and every line is incident to \(n+1\) points. It is worth noting that this provides the solution to the following combinatorics question:
What is the maximum number of \(n\)-element sets such that any two sets share exactly one element, and there is no element that appears in every set?
Each \(n\)-element set can be interpreted as the set of lines with which a point is incident.
Duality
Since the projective plane is defined in terms of a set of points and a set of lines, equivalent definitions come from interchanging the two. Formally speaking, a duality \(\rho\) is a mapping of points to lines and lines to points, with the property that if \(P\) is incident to \(\ell\), then \(\ell^\rho\) is incident to \(P^{\rho}\). The simplest example of this was the first two properties in the definition:
- For any two points, there is exactly one line incident with both of them.
- For any two lines, there is exactly one point incident with both of them.
More generally, this means any property of a projective plane is equivalent to the dual of that property in the dual projective plane. The extended Euclidean plane is self-dual, meaning the dual of the extended Euclidean plane is the extended Euclidean plane itself, so any property of the extended Euclidean plane can be transferred into an additional property by duality. For example,
Ceva's theorem and Menelaus' theorem are dual, and thus "equivalent" (meaning that proving one is sufficient to show the other), in the extended Euclidean plane. Doing away with degenerate cases (where the triangle in either theorem is degenerate), this means that Ceva's and Menelaus' are also equivalent in the "regular" Euclidean plane.
Though there are projective planes that are not self-dual, they are somewhat difficult to construct, and the most natural extensions of the extended Euclidean plane to an arbitrary field are also self-dual.
Polarities
A polarity of a projective plane is a duality that is an involution, meaning that the duality is its own inverse (or, equivalently, applying the duality twice results in the original plane). An important example stems from the concept of inversion, defined as follows: in the Euclidean plane, fix a circle \(O\) with radius \(r\). For every point \(P\), define the image of \(P\) to be the point on \(OP\) such that \(OP \cdot OQ=r^2\). The mapping \(P \rightarrow Q\) is an inversion of the plane.
Now define a duality as follows: for every line \(\ell\) in the plane not passing through \(O\), let \(P\) be the perpendicular from \(O\) to \(P\), and let \(Q\) be the image of \(P\) under inversion of the plane. Then map \(\ell\) to \(Q\). Duality also requires a mapping of points to lines, so the analogous definition is made: for any point \(P \neq O\) in the plane, \(P\) is mapped to the line through \(Q\) \((\)the image of \(P)\) that is perpendicular to \(OP\). Since this needs to be extended to the projective plane (the extended Euclidean plane), two more conditions are necessary:
- \(O\) is mapped to the line at infinity.
- A line through \(O\) with slope \(m\) is mapped to the point at infinity associated to the class of parallel lines with slope \(-\frac{1}{m}.\)
This mapping, from \(P\) to the polar of \(P\) and from \(\ell\) to the pole* of \(\ell\), define a polarity (this is relatively easy to verify). There are many properties of this particular polarity, but the most important and most general fact is:
\(P\) lies on the polar of \(Q\) if and only if \(Q\) lies on the polar of \(P\). Equivalently, \(P\) lies on the pole of \(\ell\) if and only if the pole of \(\ell\) lies on the polar of \(P\).
More generally, additional polarities can be defined by extending the definition of inversion to arbitrary conics.
Additional Euclidean Properties
While projective geometry is useful in its own right, it has significant application to Euclidean geometry as well. An earlier section gave a taste of this, in which Ceva's theorem and Menelaus' theorem were shown to be equivalent. Additional results include the following:
Desargues' Theorem
Two triangles are in axial perspective if and only if they are in central perspective. Axial perspectivity means that, for triangles \(ABC\) and \(abc\), \(AB \cap ab\), \(BC \cap bc\), and \(CA \cap ca\) are collinear. Central perspectivity means that \(Aa,Bb,Cc\) all concur at a single point.
This theorem is intuitively understood as a theorem on art; more specifically, on perspective drawing:
Interestingly, this result is not true in all projective planes, for instance in some finite projective planes of order 9. Nonetheless, it is true in most of them and certainly in the extended Euclidean plane. It is worth noting that the "iff" portion of this theorem is easily concluded from the principle of duality discussed above, so demonstrating either direction is sufficient to prove the theorem.
Another important theorem is as follows:
Pascal's Theorem
Let \(A, B, C, D, E, F\) be six points on any conic (usually a circle). Then \(AB \cap DE, BC \cap EF, CD \cap FA\) are collinear.
Pascal's theorem is often useful even when there are fewer than 6 points on the conic, by considering a degenerate case where some of the points are equal. In this case, a line like \(BB\) is understood to be the tangent to the circle at \(B\). It is also worth noting that Pascal's theorem has a converse: if \(AB \cap DE, BC \cap EF, CD \cap FA\) are collinear, then \(A, B, C, D, E, F\) lie on some conic. This is not necessarily very useful information, but if five points are already known to lie on a conic, then the sixth must as well (as five points determine a conic). Thus the converse of Pascal's theorem can be useful in showing points lie on a certain circle, especially when combined with the degenerate strategy.
Given a triangle \(\triangle ABC\) and a point \(M\), a line passing through \(M\) intersects \(AB, BC,\) and \(CA\) at \(C_1, A_1,\) and \(B_1\), respectively. The lines \(AM, BM,\) and \(CM\) intersect the circumcircle of \(\triangle ABC\) repsectively at \(A_2, B_2,\) and \(C_2\). Prove that the lines \(A_1A_2, B_1B_2,\) and \(C_1C_2\) intersect in a point that belongs to the circumcircle of \(\triangle ABC\).
Let \(A_1A_2\) intersect the circumcircle of \(ABC\) for the second time at \(X\), and let \(XB_2 \cap AC = B_1'\). By Pascal's theorem on \(ACBB_2XA_2\) \((\)which is valid as all six of these points lie on a circle, the circumcircle of \(ABC),\) we get that \(AC \cap B_2X=B_1', CB \cap XA_2=A_1,\) and \(BB_2 \cap A_2A=M\) are collinear. But then, since \(B_1'\) lies on \(A_1M\) and also \(AC\), we must have \(B_1'=A_1M \cup AC = B_1\). So \(X, B_1, B_2\) are collinear, and similarly \(X, C_1, C_2\) are collinear, so \(X\) is the intersection of \(A_1A_2, B_1B_2, C_1C_2\) which lies on the circumcircle as desired. \(_\square\)
The projective dual of this theorem is important as well:
Brianchon's Theorem
Let \(ABCDEF\) be a hexagon circumscribed about a conic section (again, usually a circle). Then \(AD, BE, CF\) concur at a point.
Quadrilateral \(ABCD\) is circumscribed about a circle \(I\), that is tangent to \(AB, BC, CD, DA\) at \(E, F, G, H,\) respectively. Suppose that \(AC\) and \(BD\) intersect at point \(P\) and \(EG\) and \(FH\) intersect at point \(Q\). If \(AB=5, BC=6, CD=8, DA=7\), what is the distance between \(P\) and \(Q?\)