CHAPTER III - COMBINATION OF TWO PROJECTIVELY RELATED FUNDAMENTAL FORMS
47. Superposed fundamental forms. Self-corresponding elements. We have seen (§ 37) that two projective point-rows may be superposed upon the same straight line. This happens, for example, when two pencils which are projective to each other are cut across by a straight line. It is also possible for two projective pencils to have the same center. This happens, for example, when two projective point-rows are projected to the same point. Similarly, two projective axial pencils may have the same axis. We examine now the possibility of two forms related in this way, having an element or elements that correspond to themselves. We have seen, indeed, that if B and D are harmonic conjugates with respect to A and C, then the point-row described by B is projective to the point-row described by D, and that A and C are self-corresponding points. Consider more generally the case of two pencils perspective to each other with axis of perspectivity u' (Fig. 9). Cut across them by a line u. We get thus two projective point-rows superposed on the same line u, and a moment's reflection serves to show that the point N of intersection u and u' corresponds to itself in the two point-rows. Also, the point M, where u [pg 30] intersects the line joining the centers of the two pencils, is seen to correspond to itself. It is thus possible for two projective point-rows, superposed upon the same line, to have two self-corresponding points. Clearly M and N may fall together if the line joining the centers of the pencils happens to pass through the point of intersection of the lines u and u'.
48. We may also give an illustration of a case where two superposed projective point-rows have no self-corresponding points at all. Thus we may take two lines revolving about a fixed point S and always making the same angle a with each other (Fig. 10). They will cut out on any line u in the plane two point-rows which are easily seen to be projective. For, given any four rays SP which are harmonic, the four corresponding rays SP' must also be harmonic, since they make the same angles with each other. Four harmonic points P correspond, therefore, to four harmonic points P'. It is clear, however, that no point P can coincide with its corresponding point P', for in that case the lines PS and [pg 31] P'S would coincide, which is impossible if the angle between them is to be constant.
49. Fundamental theorem. Postulate of continuity. We have thus shown that two projective point-rows, superposed one on the other, may have two points, one point, or no point at all corresponding to themselves. We proceed to show that
If two projective point-rows, superposed upon the same straight line, have more than two self-corresponding points, they must have an infinite number, and every point corresponds to itself; that is, the two point-rows are not essentially distinct.
If three points, A, B, and C, are self-corresponding, then the harmonic conjugate D of B with respect to A and C must also correspond to itself. For four harmonic points must always correspond to four harmonic points. In the same way the harmonic conjugate of D with respect to B and C must correspond to itself. Combining new points with old in this way, we may obtain as many self-corresponding points as we wish. We show further that every point on the line is the limiting point of a finite or infinite sequence of self-corresponding points. Thus, let a point P lie between A and B. Construct now D, the fourth harmonic of C with respect to A and B. D may coincide with P, in which case the sequence is closed; otherwise P lies in the stretch AD or in the stretch DB. If it lies in the stretch DB, construct the fourth harmonic of C with respect to D and B. This point D' may coincide with P, in which case, as before, the sequence is closed. If P lies in the stretch DD', we construct the fourth harmonic of C with respect [pg 32] to DD', etc. In each step the region in which P lies is diminished, and the process may be continued until two self-corresponding points are obtained on either side of P, and at distances from it arbitrarily small.
We now assume, explicitly, the fundamental postulate that the correspondence is continuous, that is, that the distance between two points in one point-row may be made arbitrarily small by sufficiently diminishing the distance between the corresponding points in the other. Suppose now that P is not a self-corresponding point, but corresponds to a point P' at a fixed distance d from P. As noted above, we can find self-corresponding points arbitrarily close to P, and it appears, then, that we can take a point D as close to P as we wish, and yet the distance between the corresponding points D' and P' approaches d as a limit, and not zero, which contradicts the postulate of continuity.
50. It follows also that two projective pencils which have the same center may have no more than two self-corresponding rays, unless the pencils are identical. For if we cut across them by a line, we obtain two projective point-rows superposed on the same straight line, which may have no more than two self-corresponding points. The same considerations apply to two projective axial pencils which have the same axis.
51. Projective point-rows having a self-corresponding point in common. Consider now two projective point-rows lying on different lines in the same plane. Their common point may or may not be a self-corresponding point. If the two point-rows are perspectively related, then their common point is evidently a self-corresponding [pg 33] point. The converse is also true, and we have the very important theorem:
52. If in two protective point-rows, the point of intersection corresponds to itself, then the point-rows are in perspective position.
Let the two point-rows be u and u' (Fig. 11). Let A and A', B and B', be corresponding points, and let also the point M of intersection of u and u' correspond to itself. Let AA' and BB' meet in the point S. Take S as the center of two pencils, one perspective to u and the other perspective to u'. In these two pencils SA coincides with its corresponding ray SA', SB with its corresponding ray SB', and SM with its corresponding ray SM'. The two pencils are thus identical, by the preceding theorem, and any ray SD must coincide with its corresponding ray SD'. Corresponding points of u and u', therefore, all lie on lines through the point S.
53. An entirely similar discussion shows that
If in two projective pencils the line joining their centers is a self-corresponding ray, then the two pencils are perspectively related.
54. A similar theorem may be stated for two axial pencils of which the axes intersect. Very frequent use will be made of these fundamental theorems.
55. Point-row of the second order. The question naturally arises, What is the locus of points of intersection of corresponding rays of two projective pencils [pg 34] which are not in perspective position? This locus, which will be discussed in detail in subsequent chapters, is easily seen to have at most two points in common with any line in the plane, and on account of this fundamental property will be called a point-row of the second order. For any line u in the plane of the two pencils will be cut by them in two projective point-rows which have at most two self-corresponding points. Such a self-corresponding point is clearly a point of intersection of corresponding rays of the two pencils.
56. This locus degenerates in the case of two perspective pencils to a pair of straight lines, one of which is the axis of perspectivity and the other the common ray, any point of which may be considered as the point of intersection of corresponding rays of the two pencils.
57. Pencils of rays of the second order. Similar investigations may be made concerning the system of lines joining corresponding points of two projective point-rows. If we project the point-rows to any point in the plane, we obtain two projective pencils having the same center. At most two pairs of self-corresponding rays may present themselves. Such a ray is clearly a line joining two corresponding points in the two point-rows. The result may be stated as follows: The system of rays joining corresponding points in two protective point-rows has at most two rays in common with any pencil in the plane. For that reason the system of rays is called a pencil of rays of the second order.
58. In the case of two perspective point-rows this system of rays degenerates into two pencils of rays of the first order, one of which has its center at the center [pg 35] of perspectivity of the two point-rows, and the other at the intersection of the two point-rows, any ray through which may be considered as joining two corresponding points of the two point-rows.
59. Cone of the second order. The corresponding theorems in space may easily be obtained by joining the points and lines considered in the plane theorems to a point S in space. Two projective pencils give rise to two projective axial pencils with axes intersecting. Corresponding planes meet in lines which all pass through S and through the points on a point-row of the second order generated by the two pencils of rays. They are thus generating lines of a cone of the second order, or quadric cone, so called because every plane in space not passing through S cuts it in a point-row of the second order, and every line also cuts it in at most two points. If, again, we project two point-rows to a point S in space, we obtain two pencils of rays with a common center but lying in different planes. Corresponding lines of these pencils determine planes which are the projections to S of the lines which join the corresponding points of the two point-rows. At most two such planes may pass through any ray through S. It is called a pencil of planes of the second order.
PROBLEMS
1. A man A moves along a straight road u, and another man B moves along the same road and walks so as always to keep sight of A in a small mirror M at the side of the road. How many times will they come together, A moving always in the same direction along the road?
[pg 36]2. How many times would the two men in the first problem see each other in two mirrors M and N as they walk along the road as before? (The planes of the two mirrors are not necessarily parallel to u.)
3. As A moves along u, trace the path of B so that the two men may always see each other in the two mirrors.
4. Two boys walk along two paths u and u' each holding a string which they keep stretched tightly between them. They both move at constant but different rates of speed, letting out the string or drawing it in as they walk. How many times will the line of the string pass over any given point in the plane of the paths?
5. Trace the lines of the string when the two boys move at the same rate of speed in the two paths but do not start at the same time from the point where the two paths intersect.
6. A ship is sailing on a straight course and keeps a gun trained on a point on the shore. Show that a line at right angles to the direction of the gun at its muzzle will pass through any point in the plane twice or not at all. (Consider the point-row at infinity cut out by a line through the point on the shore at right angles to the direction of the gun.)
7. Two lines u and u' revolve about two points U and U' respectively in the same plane. They go in the same direction and at the same rate of speed, but one has an angle a the start of the other. Show that they generate a point-row of the second order.
8. Discuss the question given in the last problem when the two lines revolve in opposite directions. Can you recognize the locus?
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.
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'.
[pg 40]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.
[pg 41]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.
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.
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.
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.
[pg 45]
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.
[pg 46]
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.
PROBLEMS
1. In Fig. 12, select different lines u and trace the locus of the center of perspectivity M of the lines u and u'.
2. Given four points, A, B, C, D, in the plane, construct a fifth point P such that the lines PA, PB, PC, PD shall be four harmonic lines.
Suggestion. Draw a line a through the point A such that the four lines a, AB, AC, AD are harmonic. Construct now a conic through A, B, C, and D having a for a tangent at A.
3. Where are all the points P, as determined in the preceding question, to be found?
4. Select any five points in the plane and draw the tangent to the conic through them at each of the five points.
5. Given four points on the conic, and the tangent at one of them, to construct the conic. ("To construct the conic" means here to construct as many other points as may be desired.)
[pg 47]6. Given three points on the conic, and the tangent at two of them, to construct the conic.
7. Given five points, two of which are at infinity in different directions, to construct the conic. (In this, and in the following examples, the student is supposed to be able to draw a line parallel to a given line.)
8. Given four points on a conic (two of which are at infinity and two in the finite part of the plane), together with the tangent at one of the finite points, to construct the conic.
9. The tangents to a curve at its infinitely distant points are called its asymptotes if they pass through a finite part of the plane. Given the asymptotes and a finite point of a conic, to construct the conic.
10. Given an asymptote and three finite points on the conic, to determine the conic.
11. Given four points, one of which is at infinity, and given also that the line at infinity is a tangent line, to construct the conic.