Lecture 31 - 11/12/2014 - MATH 497C, Fall 2014 Comment. Most of the materials for these lecture notes were taken from the book "Geometries" A.B. Sossinsky. Today we formulate three beautiful classical theorems (Desargues' theorem, Pappus' theorem, Pascal's theorem) which can be easily proved using methods of projective geometry. Theorem 1 (Desargues' theorem). Suppose that the lines joining the corresponding vertices of triangles A1A2A3 and B1B2B3 intersect at one point S. Then the intersection points P1;P2;P3 of the lines A2A3 and B2B3, A3A1 and B3B1, A1A2 and B1B2, respectively, are collinear. (See Figure 1). Figure 1: Desargues' theorem. Proof. We begin by passing from the plane to the three-dimensional space and prove the three-dimensional analog of Desargue's theorem. (The proof of the 3-D theorem turns out to be unexpectedly simple, but the argument used in it doesn't work in the plane!). We then use the 3-D theorem to prove Desargue's theorem in the plane by means of a continuous deformation of the spatial picture to the planar one. 2 Suppose we are given two triangles A1Ac2A3 and B1Bc2B3 in the Euclidean space R such that the three lines A1;B1; Ac2; Bc2;A3B3 intersect at one point S. (We can think of the points A1;B1;A3;B3;S as being the same as in the planar version of the theorem, while the points A2;B2 have been "lifted out" of the plane). Then the lines SB1;SBc2; SB3 define a trihedral angle in R3 (see figure 2). 1 Figure 2: Desargues' theorem in space. Consider the three pairs of lines Ac2A3 and Bc2B3, Ac2A1 and Bc2B1, A1A3 and B1B3. We claim that each of these pairs has a common point (in space!) and these three points are collinear. Indeed, the (Euclidean) planes Π1 := (A1Ac2A3) and Π2 := (B1Bc2B3) intersect in a line Λ. Obviously, the lines Ac2A3 and Bc2B3 intersect at a point (denoted Q1) of Λ, and so do the lines Ac2A1 and Bc2B1 (the intersection point is denoted by Q3) as well as the lines A1A3 and B1B3 (at Q2). Since the points Q1;Q2;Q3 all lie on Λ, they are collinear, as claimed. Let us pass to the proof of the planar version of the theorem. Consider the plane B1SB3 (which we think of as being "horizontal"), construct a plane perpendicular to it through the line SB2, in that plane choose a point O "below" the horizontal plane, and choose points Ac2 and Bc2 so that S; Ac2; Bc2 are collinear by projecting the points A2;B2 from O (see Figure 3). 2 Figure 3: Proof of Desargues' theorem. Using the 3-D version of the theorem, we can now construct the three collinear points Q1;Q2;Q3. Now rotate the line SBc2 about S downward in the vertical plane until it coincides with SB2. Since the mobile points Q1;Q2;Q3 will always be collinear and, when they reach the horizontal plane, they will coincide with the points P1;P2;P3, it follows that these three points are collinear. This proves the theorem. 3 Theorem 2. [Pappus theorem] Suppose the points A1;A2;A3 are collinear, and the points B1;B2;B3 are collinear. Let P1;P2;P3 be the intersection points of the lines A2B1 and A1B2, A1B3 and A3B1, A2B3 and A3B2, respectively. Then the points P1;P2;P3 are collinear. (See Figure 4). Figure 4: Pappus' theorem. The theorem as stated above requires that the pairs of lines (A2B1;A1B2), (A1B3;A3B1) and (A2B3;A3B2), actually intersect, so that we can speak of the collinearity of the intersection points. Stated as in Theorem 2 is perfectly valid in Euclidean geometry. However, if we interpret it in Euclidean geometry it does not exhaust its full generality. There are essentially two different ways how it happen that two lines a and b may not intersect in Euclidean geometry. Either they are identical or they are parallel. Projective geometry is an extension of Euclidean geometry in which points are added that are infinitely far away. By this we can properly speak of the intersection of parallel lines (it lies at infinity) and we get an interpretation of Pappus' theorem in which all instances of parallelism are covered as well. Sketch of the proof. By the theorem, that any four points in general position can be mapped by some projective transformation to any four points in general position, we can assume that A1;A2;B2;B2 is a square. Using the coordinate system with basis A1A2, A1B1, it is an easy exercise to prove that the points P1;P2;P3 are collinear. 4 Theorem 3 (Pascal's theorem). The points A; B; C; D; E; F lie on a conic. Let P1;P2;P3 be the intersection points of the lines AB and ED, AF and CD, CB and EF , respectively. Then the points P1;P2;P3 are collinear. (See Figure 5). Figure 5: Pascal's theorem. Pascal's theorem is named after the famous Blaise Pascal and was discovered by (the 16 years old) Pascal in 1640. This is about 1300 years after the discovery of Pappus' theorem. Nevertheless it is obviously a generalization of Pappus' theorem. If the conic in Pascal's theorem degenerates to consist of two lines, then we obtain immediately Pappus' theorem. Theorem 4 (Reformulation of Pascal's theorem). If a hexagon is inscribed in a conic in the projective plane, then the opposite sides of the hexagon meet in three collinear points. Corollary 1 (Euclidean case). If a hexagon is inscribed in a conic and two pairs of opposite edges are parallel, then so is the third pair. For more information, see "Pappus's Theorem: Nine proofs and three variations" (https://www-m10.ma.tum.de/foswiki/pub/Lehre/WS0809/GeometrieKalkueleWS0809/ ch1.pdf). 5.