An Elementary Course in Synthetic Projective Geometry

by Derrick Norman Lehmer

CHAPTER IV - POINT-ROWS OF THE SECOND ORDER

60. Point-row of the second order defined. We have seen that two fundamental forms in one-to-one correspondence may sometimes generate a form of higher order. Thus, two point-rows (§ 55) generate a system of rays of the second order, and two pencils of rays (§ 57), a system of points of the second order. As a system of points is more familiar to most students of geometry than a system of lines, we study first the point-row of the second order.

61. Tangent line. We have shown in the last chapter (§ 55) that the locus of intersection of corresponding rays of two projective pencils is a point-row of the second order; that is, it has at most two points in common with any line in the plane. It is clear, first of all, that the centers of the pencils are points of the locus; for to the line SS', considered as a ray of S, must correspond some ray of S' which meets it in S'. S', and by the same argument S, is then a point where corresponding rays meet. Any ray through S will meet it in one point besides S, namely, the point P where it meets its corresponding ray. Now, by choosing the ray through S sufficiently close to the ray SS', the point P may be made to approach arbitrarily close to S', and the ray S'P may be made to differ in position from the [pg 38] tangent line at S' by as little as we please. We have, then, the important theorem

The ray at S' which corresponds to the common ray SS' is tangent to the locus at S'.

In the same manner the tangent at S may be constructed.

62. Determination of the locus. We now show that it is possible to assign arbitrarily the position of three points, A, B, and C, on the locus (besides the points S and S'); but, these three points being chosen, the locus is completely determined.

63. This statement is equivalent to the following:

Given three pairs of corresponding rays in two projective pencils, it is possible to find a ray of one which corresponds to any ray of the other.

64. We proceed, then, to the solution of the fundamental

Problem: Given three pairs of rays, aa', bb', and cc', of two protective pencils, S and S', to find the ray d' of S' which corresponds to any ray d of S.

Fig. 12

Call A the intersection of aa', B the intersection of bb', and C the intersection of cc' (Fig. 12). Join AB by the line u, and AC by the line u'. Consider u as a point-row perspective to S, and u' as a point-row perspective to S'. u and u' are projectively related to each other, since S and S' are, by hypothesis, so related. But their point of intersection A is a self-corresponding point, since a and a' were supposed to be corresponding rays. It follows (§ 52) that u and u' are in perspective position, and that lines through corresponding points all pass [pg 39] through a point M, the center of perspectivity, the position of which will be determined by any two such lines. But the intersection of a with u and the intersection of c' with u' are corresponding points on u and u', and the line joining them is clearly c itself. Similarly, b' joins two corresponding points on u and u', and so the center M of perspectivity of u and u' is the intersection of c and b'. To find d' in S' corresponding to a given line d of S we note the point L where d meets u. Join L to M and get the point N where this line meets u'. L and N are corresponding points on u and u', and d' must therefore pass through N. The intersection P of d and d' is thus another point on the locus. In the same manner any number of other points may be obtained.

65. The lines u and u' might have been drawn in any direction through A (avoiding, of course, the line a for u and the line a' for u'), and the center of perspectivity M would be easily obtainable; but the above construction furnishes a simple and instructive figure. An equally simple one is obtained by taking a' for u and a for u'.

66. Lines joining four points of the locus to a fifth. Suppose that the points S, S', B, C, and D are fixed, and that four points, A, A1, A2, and A3, are taken on the locus at the intersection with it of any four harmonic rays through B. These four harmonic rays give four harmonic points, L, L1 etc., on the fixed ray SD. These, in turn, project through the fixed point M into four harmonic points, N, N1 etc., on the fixed line DS'. These last four harmonic points give four harmonic rays CA, CA1, CA2, CA3. Therefore the four points A which project to B in four harmonic rays also project to C in four harmonic rays. But C may be any point on the locus, and so we have the very important theorem,

Four points which are on the locus, and which project to a fifth point of the locus in four harmonic rays, project to any point of the locus in four harmonic rays.

67. The theorem may also be stated thus:

The locus of points from which, four given points are seen along four harmonic rays is a point-row of the second order through them.

68. A further theorem of prime importance also follows:

Any two points on the locus may be taken as the centers of two projective pencils which will generate the locus.

69. Pascal's theorem. The points A, B, C, D, S, and S' may thus be considered as chosen arbitrarily on the locus, and the following remarkable theorem follows at once.

Given six points, 1, 2, 3, 4, 5, 6, on the point-row of the second order, if we call

L the intersection of 12 with 45,

M the intersection of 23 with 56,

N the intersection of 34 with 61,

then L, M, and N are on a straight line.

Fig. 13

70. To get the notation to correspond to the figure, we may take (Fig. 13) A = 1, B = 2, S' = 3, D = 4, S = 5, and C = 6. If we make A = 1, C=2, S=3, D = 4, S'=5, and. B = 6, the points L and N are interchanged, but the line is left unchanged. It is clear that one point may be named arbitrarily and the other five named in 5! = 120 different ways, but since, as we have seen, two different assignments of names give the same line, it follows that there cannot be more than 60 different lines LMN obtained in this way from a given set of six points. As a matter of fact, the number obtained in this way is in general 60. The above theorem, which is of cardinal importance in the theory of the point-row of the second order, is due to Pascal and was discovered by him at the age of sixteen. It is, no doubt, the most important contribution to the theory of these loci since [pg 42] the days of Apollonius. If the six points be called the vertices of a hexagon inscribed in the curve, then the sides 12 and 45 may be appropriately called a pair of opposite sides. Pascal's theorem, then, may be stated as follows:

The three pairs of opposite sides of a hexagon inscribed in a point-row of the second order meet in three points on a line.

71. Harmonic points on a point-row of the second order. Before proceeding to develop the consequences of this theorem, we note another result of the utmost importance for the higher developments of pure geometry, which follows from the fact that if four points on the locus project to a fifth in four harmonic rays, they will project to any point of the locus in four harmonic rays. It is natural to speak of four such points as four harmonic points on the locus, and to use this notion to define projective correspondence between point-rows of the second order, or between a point-row of the second order and any fundamental form of the first order. Thus, in particular, the point-row of the second order, σ, is said to be perspectively related to the pencil S when every ray on S goes through the point on σ which corresponds to it.

72. Determination of the locus. It is now clear that five points, arbitrarily chosen in the plane, are sufficient to determine a point-row of the second order through them. Two of the points may be taken as centers of two projective pencils, and the three others will determine three pairs of corresponding rays of the pencils, and therefore all pairs. If four points of the locus are [pg 43] given, together with the tangent at one of them, the locus is likewise completely determined. For if the point at which the tangent is given be taken as the center S of one pencil, and any other of the points for S', then, besides the two pairs of corresponding rays determined by the remaining two points, we have one more pair, consisting of the tangent at S and the ray SS'. Similarly, the curve is determined by three points and the tangents at two of them.

73. Circles and conics as point-rows of the second order. It is not difficult to see that a circle is a point-row of the second order. Indeed, take any point S on the circle and draw four harmonic rays through it. They will cut the circle in four points, which will project to any other point of the curve in four harmonic rays; for, by the theorem concerning the angles inscribed in a circle, the angles involved in the second set of four lines are the same as those in the first set. If, moreover, we project the figure to any point in space, we shall get a cone, standing on a circular base, generated by two projective axial pencils which are the projections of the pencils at S and S'. Cut across, now, by any plane, and we get a conic section which is thus exhibited as the locus of intersection of two projective pencils. It thus appears that a conic section is a point-row of the second order. It will later appear that a point-row of the second order is a conic section. In the future, therefore, we shall refer to a point-row of the second order as a conic.

Fig. 14

74. Conic through five points. Pascal's theorem furnishes an elegant solution of the problem of drawing a conic through five given points. To construct a sixth [pg 44] point on the conic, draw through the point numbered 1 an arbitrary line (Fig. 14), and let the desired point 6 be the second point of intersection of this line with the conic. The point L = 12-45 is obtainable at once; also the point N = 34-61. But L and N determine Pascal's line, and the intersection of 23 with 56 must be on this line. Intersect, then, the line LN with 23 and obtain the point M. Join M to 5 and intersect with 61 for the desired point 6.

Fig. 15

75. Tangent to a conic. If two points of Pascal's hexagon approach coincidence, then the line joining them approaches as a limiting position the tangent line at that point. Pascal's theorem thus affords a ready method of drawing the tangent line to a conic at a given point. If the conic is determined by the points 1, 2, 3, 4, 5 (Fig. 15), and it is desired to draw the tangent at the point 1, we may call that point 1, 6. The points L and M are obtained as usual, and the intersection of 34 with LM gives N. Join N to the point 1 for the desired tangent at that point.

76. Inscribed quadrangle. Two pairs of vertices may coalesce, giving an inscribed quadrangle. Pascal's theorem gives for this case the very important theorem

Two pairs of opposite sides of any quadrangle inscribed in a conic meet on a straight line, upon which line also intersect the two pairs of tangents at the opposite vertices.

Fig. 16

Fig. 17

For let the vertices be A, B, C, and D, and call the vertex A the point 1, 6; B, the point 2; C, the point 3, 4; and D, the point 5 (Fig. 16). Pascal's theorem then indicates that L = AB-CD, M = AD-BC, and N, which is the intersection of the tangents at A and C, are all on a straight line u. But if we were to call A the point 2, B the point 6, 1, C the point 5, and D the point 4, 3, then the intersection P of the tangents at B and D are also on this same line u. Thus L, M, N, and P are four points on a straight line. The consequences of this theorem are so numerous and important that we shall devote a separate chapter to them.

77. Inscribed triangle. Finally, three of the vertices of the hexagon may coalesce, giving a triangle inscribed in a conic. Pascal's theorem then reads as follows (Fig. 17) for this case:

The three tangents at the vertices of a triangle inscribed in a conic meet the opposite sides in three points on a straight line.

Fig. 18

78. Degenerate conic. If we apply Pascal's theorem to a degenerate conic made up of a pair of straight lines, we get the following theorem (Fig. 18):

If three points, A, B, C, are chosen on one line, and three points, A', B', C', are chosen on another, then the three points L = AB'-A'B, M = BC'-B'C, N = CA'-C'A are all on a straight line.