*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 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__._
[Figure 12]
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 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_,
_A__1_, _A__2_, and _A__3_, 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_, _L__1_ etc., on the fixed ray _SD_. These,
in turn, project through the fixed point _M_ into four harmonic points,
_N_, _N__1_ etc., on the fixed line _DS’_. These last four harmonic points
give four harmonic rays _CA_, _CA__1_, _CA__2_, _CA__3_. 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._
[Figure 13]
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 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 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.
[Figure 14]
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 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.
[Figure 15]
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._
[Figure 16]
FIG. 16
[Figure 17]
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._
[Figure 18]
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._
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.)
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.