Chapter XXX: Book XIII (2)

S 36. Three pairs of corresponding elements in two projective rows or
pencils being given, to determine for any element in one the
corresponding element in the other.

We solve this in the two cases of two projective rows and of two
projective flat pencils in a plane.

_Problem_ I.--Let A, B, C be _Problem_ II.--Let a, b, c be
three points in a row s, A', B', three rays in a pencil S, a',
C' the corresponding points in a b', c' the corresponding rays in
projective row s', both being in a projective pencil S', both
a plane; it is required to find being in the same plane; it is
for any point D in s the required to find for any ray d
corresponding point D' in s'. in S the corresponding ray d' in
S'.

The solution is made to depend on the construction of an auxiliary row
or pencil which is perspective to both the given ones. This is found
as follows:--

_Solution of Problem_ I.--On the line joining two corresponding
points, say AA' (fig. 11), take any two points, S and S', as centres
of auxiliary pencils. Join the intersection B1 of SB and S'B' to the
intersection C1 of SC and S'C' by the line s1. Then a row on s1 will
be perspective to s with S as centre of projection, and to s' with S'
as centre. To find now the point D' on s' corresponding to a point D
on s we have only to determine the point D1, where the line SD cuts
s1, and to draw S'D1; the point where this line cuts s' will be the
required point D'.

_Proof._--The rows s and s' are both perspective to the row s1, hence
they are projective to one another. To A, B, C, D on s correspond A1,
B1, C1, D1 on s1, and to these correspond A', B', C', D' on s'; so
that D and D' are corresponding points as required.

_Solution of Problem_ II.--Through the intersection A of two
corresponding rays a and a' (fig. 12), take two lines, s and s', as
bases of auxiliary rows. Let S1 be the point where the line b1, which
joins B and B', cuts the line c1, which joins C and C'. Then a pencil
S1 will be perspective to S with s as axis of projection. To find the
ray d' in S' corresponding to a given ray d in S, cut d by s at D;
project this point from S1 to D' on s' and join D' to S'. This will be
the required ray.

_Proof._--That the pencil S1 is perspective to S and also to S'
follows from construction. To the lines a1, b1, c1, d1 in S1
correspond the lines a, b, c, d in S and the lines a', b', c', d' in
S', so that d and d' are corresponding rays.

In the first solution the two centres, S, S', are _any_ two points on
a line joining any two corresponding points, so that the solution of
the problem allows of a great many different constructions. _But
whatever construction be used, the point D', corresponding to D, must
be always the same_, according to the theorem in S 29. This gives rise
to a number of theorems, into which, however, we shall not enter. The
same remarks hold for the second problem.

S 37. _Homological Triangles._--As a further application of the
theorems about perspective rows and pencils we shall prove the
following important theorem.

_Theorem._--If ABC and A'B'C' (fig. 13) be two triangles, such that
the lines AA', BB', CC' meet in a point S, then the intersections of
BC and B'C', of CA and C'A', and of AB and A'B' will lie in a line.
Such triangles are said to be homological, or in perspective. The
triangles are "co-axial" in virtue of the property that the meets of
corresponding sides are collinear and copolar, since the lines joining
corresponding vertices are concurrent.

_Proof._--Let a, b, c denote the lines AA', BB', CC', which meet at S.
Then these may be taken as bases of projective rows, so that A, A', S
on a correspond to B, B', S on b, and to C, C', S on c. As the point S
is common to all, any two of these rows will be perspective.

If S1 be the centre of projection of rows b and c,
S2 " " " c and a,
S3 " " " a and b,

and if the line S1S2 cuts a in A1, and b in B1, and c in C1, then A1,
B1 will be corresponding points in a and b, both corresponding to C1
in c. But a and b are perspective, therefore the line A1B1, that is
S1S2, joining corresponding points must pass through the centre of
projection S3 of a and b. In other words, S1, S2, S3 lie in a line.
This is Desargues' celebrated theorem if we state it thus:--

_Theorem of Desargues._--If each of two triangles has one vertex on
each of three concurrent lines, then the intersections of
corresponding sides lie in a line, those sides being called
corresponding which are opposite to vertices on the same line.

The converse theorem holds also, viz.

_Theorem._--If the sides of one triangle meet those of another in
three points which lie in a line, then the vertices lie on three lines
which meet in a point.

The proof is almost the same as before.

S 38. _Metrical Relations between Projective Rows._--Every row
contains one point which is distinguished from all others, viz. the
point at infinity. In two projective rows, to the point I at infinity
in one corresponds a point I' in the other, and to the point J' at
infinity in the second corresponds a point J in the first. The points
I' and J are in general finite. If now A and B are any two points in
the one, A', B' the corresponding points in the other row, then

(AB, JI) = (A'B', J'I'),

or

AJ/JB : AI/IB = A'J'/J'B' : A'I'/I'B'.

But, by S 17,

AI/IB = A'J'/J'B' = -1;

therefore the last equation changes into

AJ.A'I' = BJ.B'I',

that is to say--

_Theorem._--The product of the distances of any two corresponding
points in two projective rows from the points which correspond to the
points at infinity in the other is constant, viz. AJ.A'I' = k.
Steiner has called this number k the _Power of the correspondence_.

[The relation AJ . A'I' = k shows that if J, I' be given then the
point A' corresponding to a specified point A is readily found; hence
A, A' generate homographic ranges of which I and J' correspond to the
points at infinity on the ranges. If we take any two origins O, O', on
the ranges and reduce the expression AJ . A'I' = k to its algebraic
equivalent, we derive an equation of the form [alpha]xx' + [beta]x +
[gamma]x' + [delta] = 0. Conversely, if a relation of this nature
holds, then points corresponding to solutions in x, x' form
homographic ranges.]

S 39. _Similar Rows._--If the points at infinity in two projective
rows correspond so that I' and J are at infinity, this result loses
its meaning. But if A, B, C be any three points in one, A', B', C' the
corresponding ones on the other row, we have

(AB, CI) = (A'B', C'I'),

which reduces to

AC/CB = A'C'/C'B' or AC/A'C' = BC/B'C',

that is, corresponding segments are proportional. Conversely, if
corresponding segments are proportional, then to the point at infinity
in one corresponds the point at infinity in the other. If we call such
rows _similar_, we may state the result thus--

_Theorem._--Two projective rows are similar if to the point at
infinity in one corresponds the point at infinity in the other, and
conversely, if two rows are similar then they are projective, and the
points at infinity are corresponding points.

From this the well-known propositions follow:--

Two lines are cut proportionally (in similar rows) by a series of
parallels. The rows are perspective, with centre of projection at
infinity.

If two similar rows are placed parallel, then the lines joining
homologous points pass through a common point.

S 40. If two flat pencils be projective, then there exists in either,
one single pair of lines at right angles to one another, such that the
corresponding lines in the other pencil are again at right angles.

To prove this, we place the pencils in perspective position (fig. 14)
by making one ray coincident with its corresponding ray. Corresponding
rays meet then on a line p. And now we draw the circle which has its
centre O on p, and which passes through the centres S and S' of the
two pencils. This circle cuts p in two points H and K. The two pairs
of rays, h, k, and h', k', joining these points to S and S' will be
pairs of corresponding rays at right angles. The construction gives in
general but one circle, but if the line p is the perpendicular
bisector of SS', there exists an infinite number, and _to every right
angle in the one pencil corresponds a right angle in the other_.

PRINCIPLE OF DUALITY

S 41. It has been stated in S 1 that not only points, but also planes
and lines, are taken as elements out of which figures are built up. We
shall now see that the construction of one figure which possesses
certain properties gives rise in many cases to the construction of
another figure, by replacing, according to definite rules, elements of
one kind by those of another. The new figure thus obtained will then
possess properties which may be stated as soon as those of the
original figure are known.

We obtain thus a principle, known as the _principle of duality_ or of
_reciprocity_, which enables us to construct to any figure not
containing any measurement in its construction a _reciprocal_ figure,
as it is called, and to deduce from any theorem a _reciprocal_
theorem, for which no further proof is needed.

It is convenient to print reciprocal propositions on opposite sides of
a page broken into two columns, and this plan will occasionally be
adopted.

We begin by repeating in this form a few of our former statements:--

Two points determine a line. Two planes determine a line.

Three points which are not in a Three planes which do not pass
line determine a plane. through a line determine a point.

A line and a point without it A line and a plane not through
determine a plane. it determine a point.

Two lines in a plane determine Two lines through a point
a point. determine a plane.

These propositions show that it will be possible, when any figure is
given, to construct a second figure by taking planes instead of
points, and points instead of planes, but lines where we had lines.

For instance, if in the first figure we take a plane and three points
in it, we have to take in the second figure a point and three planes
through it. The three points in the first, together with the three
lines joining them two and two, form a triangle; the three planes in
the second and their three lines of intersection form a trihedral
angle. A triangle and a trihedral angle are therefore reciprocal
figures.

Similarly, to any figure in a plane consisting of points and lines
will correspond a figure consisting of planes and lines passing
through a point S, and hence belonging to the pencil which has S as
centre.

The figure reciprocal to four points in space which do not lie in a
plane will consist of four planes which do not meet in a point. In
this case each figure forms a tetrahedron.

S 42. As other examples we have the following:--

To a row is reciprocal an axial pencil,

" a flat pencil " a flat pencil,

" a field of points and lines " a pencil of planes and lines,

" the space of points " the space of planes.

For the row consists of a line and all the points in it, reciprocal to
it therefore will be a line with all planes through it, that is, an
axial pencil; and so for the other cases.

This correspondence of reciprocity breaks down, however, if we take
figures which contain measurement in their construction. For instance,
there is no figure reciprocal to two planes at _right angles_, because
there is no segment in a row which has a magnitude as definite as a
right angle.

We add a few examples of reciprocal propositions which are easily
proved.

_Theorem._--If A, B, C, D are _Theorem._--If [alpha], [beta],
any four points in space, and if [gamma], [delta] are four planes
the lines AB and CD meet, then in space, and if the lines
all four points lie in a plane, [alpha][beta] and [gamma][delta]
hence also AC and BD, as well meet, then all four planes lie
as AD and BC, meet. in a point (pencil), hence also
[alpha][gamma] and [beta][delta],
well as [alpha][delta] and
as [beta][gamma], meet.

Theorem.--_If of any number of lines every one meets every other,
whilst all do not_

_lie in a point, then all lie in _lie in a plane, then all lie in
a plane._ a point (pencil)._

S 43. Reciprocal figures as explained lie both in space of three
dimensions. If the one is confined to a plane (is formed of elements
which lie in a plane), then the reciprocal figure is confined to a
pencil (is formed of elements which pass through a point).

But there is also a more special principle of duality, according to
which figures are reciprocal which lie both in a plane or both in a
pencil. In the plane we take points and lines as reciprocal elements,
for they have this fundamental property in common, that two elements
of one kind determine one of the other. In the pencil, on the other
hand, lines and planes have to be taken as reciprocal, and here it
holds again that two lines or planes determine one plane or line.

Thus, to one plane figure we can construct one reciprocal figure in
the plane, and to each one reciprocal figure in a pencil. We mention a
few of these. At first we explain a few names:--

A figure consisting of n points A figure consisting of n lines
in a plane will be called an in a plane will be called an
n-point. n-side.

A figure consisting of n planes A figure consisting of n lines
in a pencil will be called an in a pencil will be called an
n-flat. n-edge.

It will be understood that an n-side is different from a polygon of n
sides. The latter has sides of finite length and n vertices, the
former has sides all of infinite extension, and every point where two
of the sides meet will be a vertex. A similar difference exists
between a solid angle and an n-edge or an n-flat. We notice
particularly--

A four-point has six sides, of A four-side has six vertices, of
which two and two are opposite, which two and two are opposite,
and three diagonal points, which and three diagonals, which join
are intersections of opposite opposite vertices.
sides.

A four-flat has six edges, of A four-edge has six faces, of
which two and two are opposite, which two and two are opposite,
and three diagonal planes, which and three diagonal edges, which
pass through opposite edges. are intersections of opposite
faces.

A four-side is usually called a complete quadrilateral, and a
four-point a complete quadrangle. The above notation, however, seems
better adapted for the statement of reciprocal propositions.

S 44.

If a point moves in a plane it If a line moves in a plane it
describes a plane curve. envelopes a plane curve (fig. 15).

If a plane moves in a pencil it If a line moves in a pencil it
envelopes a cone. describes a cone.

A curve thus appears as generated either by points, and then we call
it a "locus," or by lines, and then we call it an "envelope." In the
same manner a cone, which means here a surface, appears either as the
locus of lines passing through a fixed point, the "vertex" of the
cone, or as the envelope of planes passing through the same point.

To a surface as locus of points corresponds, in the same manner, a
surface as envelope of planes; and to a curve in space as locus of
points corresponds a developable surface as envelope of planes.

It will be seen from the above that we may, by aid of the principle of
duality, construct for every figure a reciprocal figure, and that to
any property of the one a reciprocal property of the other will exist,
as long as we consider only properties which depend upon nothing but
the positions and intersections of the different elements and not upon
measurement.

For such propositions it will therefore be unnecessary to prove more
than one of two reciprocal theorems.

GENERATION OF CURVES AND CONES OF SECOND ORDER OR SECOND CLASS

S 45. _Conics._--If we have two projective pencils in a plane,
corresponding rays will meet, and their point of intersection will
constitute some locus which we have to investigate. Reciprocally, if
two projective rows in a plane are given, then the lines which join
corresponding points will envelope some curve. We prove first:--

_Theorem._--If two projective _Theorem._--If two projective
flat pencils lie in a plane, but rows lie in a plane, but are
are neither in perspective nor neither in perspective nor on a
concentric, then the locus of common base, then the envelope
intersections of corresponding of lines joining corresponding
rays is a curve of the second points is a curve of the second
order, that is, no line contains class, that is, through no point
more than two points of the pass more than two of the
locus. enveloping lines.

Proof.--We draw any line t. _Proof._--We take any point T
This cuts each of the pencils in and join it to all points in each
a row, so that we have on t two row. This gives two concentric
rows, and these are projective pencils, which are projective
because the pencils are because the rows are projective.
projective. If corresponding rays If a line joining corresponding
of the two pencils meet on the points in the two rows passes
line t, their intersection will through T, it will be a line in
be a point in the one row which the one pencil which coincides
coincides with its corresponding with its corresponding line in
point in the other. But two the other. But two projective
projective rows on the same base concentric flat pencils in the
cannot have more than two same plane cannot have more than
points of one coincident with two lines of one coincident with
their corresponding points in their corresponding line in the
the other (S 34). other (S 34).

It will be seen that the proofs are reciprocal, so that the one may be
copied from the other by simply interchanging the words point and
line, locus and envelope, row and pencil, and so on. We shall
therefore in future prove seldom more than one of two reciprocal
theorems, and often state one theorem only, the reader being
recommended to go through the reciprocal proof by himself, and to
supply the reciprocal theorems when not given.

S 46. We state the theorems in the pencil reciprocal to the last,
without proving them:--

_Theorem._--If two projective _Theorem._--If two projective
flat pencils are concentric, but axial pencils lie in the same
are neither perspective nor pencil (their axes meet in a
coplanar, then the envelope of point), but are neither perspective
the planes joining corresponding nor co-axial, then the locus
rays is a cone of the second of lines joining corresponding
class; that is, no line through planes is a cone of the second
the common centre contains more order; that is, no plane in the
than two of the enveloping pencil contains more than two
planes.

S 47. Of theorems about cones of second order and cones of second
class we shall state only very few. We point out, however, the
following connexion between the curves and cones under consideration:

The lines which join any point Every plane section of a cone
in space to the points on a curve of the second order is a curve of
of the second order form a cone the second order.
of the second order.

The planes which join any Every plane section of a cone
point in space to the lines of the second class is a curve of
enveloping a curve of the the second class.
second class envelope themselves
a cone of the second class.

By its aid, or by the principle of duality, it will be easy to obtain
theorems about them from the theorems about the curves.

We prove the first. A curve of the second order is generated by two
projective pencils. These pencils, when joined to the point in space,
give rise to two projective axial pencils, which generate the cone in
question as the locus of the lines where corresponding planes meet.

S48.

_Theorem._--The curve of second _Theorem._--The envelope of
order which is generated by two second class which is generated
projective flat pencils passes by two projective rows contains
through the centres of the two the bases of these rows as
pencils. enveloping lines or tangents.

_Proof._--If S and S' are the _Proof._--If s and s' are the
two pencils, then to the ray SS' two rows, then to the point ss'
or p' in the pencil S' or P' as a point in s'
corresponds in the pencil S a corresponds in s a point P,
ray p, which is different from which is not coincident with P',
p', for the pencils are not for the rows are not
perspective. But p and p' meet perspective. But P and P' are
at S, so that S is a point on joined by s, so that s is one of
the curve, and similarly S'. the enveloping lines, and
similarly s'.

It follows that every line in one of the two pencils cuts the curve in
two points, viz. once at the centre S of the pencil, and once where it
cuts its corresponding ray in the other pencil. These two points,
however, coincide, if the line is cut by its corresponding line at S
itself. The line p in S, which corresponds to the line SS' in S', is
therefore the only line through S which has but one point in common
with the curve, or which cuts the curve in two coincident points. Such
a line is called a _tangent_ to the curve, touching the latter at the
point S, which is called the "point of contact."

In the same manner we get in the reciprocal investigation the result
that through every point in one of the rows, say in s, two tangents
may be drawn to the curve, the one being s, the other the line joining
the point to its corresponding point in s'. There is, however, one
point P in s for which these two lines coincide. Such a point in one
of the tangents is called the "point of contact" of the tangent. We
thus get--

_Theorem._--To the line joining _Theorem._--To the point of
the centres of the projective intersection of the bases of two
pencils as a line in one pencil projective rows as a point in
corresponds in the other the one row corresponds in the other
tangent at its centre. the _point of contact_ of its
base.

S 49. Two projective pencils are determined if three pairs of
corresponding lines are given. Hence if a1, b1, c1 are three lines in
a pencil S1, and a2, b2, c2 the corresponding lines in a projective
pencil S2, the correspondence and therefore the curve of the second
order generated by the points of intersection of corresponding rays is
determined. Of this curve we know the two centres S1 and S2, and the
three points a1a2, b1b2, c1c2, hence five points in all. This and the
reciprocal considerations enable us to solve the following two
problems:

_Problem._--To construct a curve _Problem._--To construct a curve
of the second order, of which of the second class, of which
five points S1, S2, A, B, C are five tangents u1, u2, a, b, c
given. are given.

In order to solve the left-hand problem, we take two of the given
points, say S1 and S2, as centres of pencils. These we make projective
by taking the rays a1, b1, c1, which join S1 to A, B, C respectively,
as corresponding to the rays a2, b2, c2, which join S2 to A, B, C
respectively, so that three rays meet their corresponding rays at the
given points A, B, C. This determines the correspondence of the
pencils which will generate a curve of the second order passing
through A, B, C and through the centres S1 and S2, hence through the
five given points. To find more points on the curve we have to
construct for any ray in S1 the corresponding ray in S2. This has been
done in S 36. But we repeat the construction in order to deduce
further properties from it. We also solve the right-hand problem. Here
we select two, viz. u1, u2 of the five given lines, u1, u2, a, b, c,
as bases of two rows, and the points A1, B1, C1 where a, b, c cut u1
as corresponding to the points A2, B2, C2 where a, b, c cut u2.

We get then the following solutions of the two problems:

_Solution._--Through the point A _Solution._--In the line a take
draw any two lines, u1 and u2 any two points S1 and S2 as
(fig. 16), the first u1 to cut centres of pencils (fig. 17),
the pencil S1 in a row AB1C1, the first S1 (A1B1C1) to project
the other u2 to cut the pencil the row u1, the other S2
S2 in a row AB2C2. These two (A2B2C2) to project the row u2.
rows will be perspective, as the These two pencils will be
point A corresponds to itself, perspective, the line S1A1 being
and the centre of projection the same as the corresponding
will be the point S, where the line S2A2, and the axis of
lines B1B2 and C1C2 meet. To projection will be the line u,
find now for any ray d1 in S1 which joins the intersection B
its corresponding ray d2 in S2, of S1B1 and S2B2 to the
we determine the point D1 where intersection C of S1C1 and S2C2.
d1 cuts u1, project this point To find now for any point D1 in
from S to D2 on u2 and join S2 u1 the corresponding point D2 in
to D2. This will be the required u2, we draw S1D1 and project the
ray d2 which cuts d1 at some point D where this line cuts u
point D on the curve. from S2 to u2. This will give
the required point D2, and the
line d joining D1 to D2 will be
a new tangent to the curve.

S 50. These constructions prove, when rightly interpreted, very
important properties of the curves in question.

If in fig. 16 we draw in the pencil S1 the ray k1 which passes through
the auxiliary centre S, it will be found that the corresponding ray k2
cuts it on u2. Hence--

_Theorem._--In the above _Theorem._--In the above
construction the bases of the construction (fig. 17) the
auxiliary rows u1 and u2 cut the tangents to the curve from the
curve where they cut the rays centres of the auxiliary pencils
S2S and S1S respectively. S1 and S2 are the lines which
pass through u2u and u1u
respectively.

As A is any given point on the curve, and u1 any line through it, we
have solved the problems:

_Problem._--To find the second _Problem._--To find the second
point in which any line through tangent which can be drawn from
a known point on the curve cuts any point in a given tangent to
the curve. the curve.

If we determine in S1 (fig. 16) the ray corresponding to the ray S2S1
in S2, we get the tangent at S1. Similarly, we can determine the point
of contact of the tangents u1 or u2 in fig. 17.

S 51. If five points are given, of which not three are in a line, then
we can, as has just been shown, always draw a curve of the second
order through them; we select two of the points as centres of
projective pencils, and then one such curve is determined. It will be
presently shown that we get always the same curve if two other points
are taken as centres of pencils, that therefore five points
_determine_ one curve of the second order, and reciprocally, that five
tangents determine one curve of the second class. Six points taken at
random will therefore not lie on a curve of the second order. In order
that this may be the case a certain condition has to be satisfied, and
this condition is easily obtained from the construction in S 49, fig.
16. If we consider the conic determined by the five points A, S1, S2,
K, L, then the point D will be on the curve if, and only if, the
points on D1, S, D2 be in a line.

This may be stated differently if we take AKS1DS2L (figs. 16 and 18)
as a hexagon inscribed in the conic, then AK and DS2 will be opposite
sides, so will be KS1 and S2L, as well as S1D and LA. The first two
meet in D2, the others in S and D1 respectively. We may therefore
state the required condition, together with the reciprocal one, as
follows:--

_Pascal's Theorem._--If a hexagon _Brianchon's Theorem._--If a
be inscribed in a curve of the hexagon be circumscribed about
second order, then the a curve of the second class, then
intersectionsof opposite sides the lines joining opposite vertices
are three points in a line. are three lines meeting in a point.

These celebrated theorems, which are known by the names of their
discoverers, are perhaps the most fruitful in the whole theory of
conics. Before we go over to their applications we have to show that
we obtain the same curve if we take, instead of S1, S2, any two other
points on the curve as centres of projective pencils.

S 52. We know that the curve depends only upon the correspondence
between the pencils S1 and S2, and not upon the special construction
used for finding new points on the curve. The point A (fig. 16 or 18),
through which the two auxiliary rows u1, u2 were drawn, may therefore
be changed to any other point on the curve. Let us now suppose the
curve drawn, and keep the points S1, S2, K, L and D, and hence also
the point S fixed, whilst we move A along the curve. Then the line AL
will describe a pencil about L as centre, and the point D1 a row on
S1D perspective to the pencil L. At the same time AK describes a
pencil about K and D2 a row perspective to it on S2D. But by Pascal's
theorem D1 and D2 will always lie in a line with S, so that the rows
described by D1 and D2 are perspective. It follows that the pencils K
and L will themselves be projective, corresponding rays meeting on the
curve. This proves that we get the same curve whatever pair of the
five given points we take as centres of projective pencils. Hence--

Only one curve of the second Only one curve of the second
order can be drawn which passes class can be drawn which touches
through five given points. five given lines.

We have seen that if on a curve of the second order two points
coincide at A, the line joining them becomes the tangent at A. If,
therefore, a point on the curve and its tangent are given, this will
be equivalent to having given two points on the curve. Similarly, if
on the curve of second class a tangent and its point of contact are
given, this will be equivalent to two given tangents.

We may therefore extend the last theorem:

Only one curve of the second Only one curve of the second
order can be drawn, of which class can be drawn, of which four
four points and the tangent at tangents and the point of contact
oneof them, or three points at one of them, or three tangents
and the tangents at two of and the points of contact at two
them, are given. of them, are given.

S 53. At the same time it has been proved:

If all points on a curve of the All tangents to a curve of second
second order be joined to any class are cut by any two of
two of them, then the two them in projective rows, those
pencils thus formed are being corresponding points which
projective, those rays being lie on the same tangent. Hence--
corresponding which meet on the
curve. Hence--

The cross-ratio of four rays The cross-ratio of the four
joining a point S on a curve of points in which any tangent u is
second order to four fixed cut by four fixed tangents a, b, c,
points A, B, C, D in the curve d is independent of the position of
is independent of the position u, and is called the cross-ratio of
of S, and is called the cross- the four tangents a, b, c, d.
ratio of the four points A, B,
C, D.

If this cross-ratio equals -1 If this cross-ratio equals -1
the four points are said to be the four tangents are said to be
four harmonic points. four harmonic tangents.

We have seen that a curve of second order, as generated by projective
pencils, has at the centre of each pencil one tangent; and further,
that any point on the curve may be taken as centre of such pencil.
Hence--

A curve of second order has A curve of second class has on
at every point one tangent. every tangent a point of contact.

S 54. We return to Pascal's and Brianchon's theorems and their
applications, and shall, as before, state the results both for curves
of the second order and curves of the second class, but prove them
only for the former.

Pascal's theorem may be used when five points are given to find more
points on the curve, viz. it enables us to find the point where any
line through one of the given points cuts the curve again. It is
convenient, in making use of Pascal's theorem, to number the points,
to indicate the order in which they are to be taken in forming a
hexagon, which, by the way, may be done in 60 different ways. It will
be seen that 1 2 (leaving out 3) 4 5 are opposite sides, so are 2 3
and (leaving out 4) 5 6, and also 3 4 and (leaving out 5) 6 1.

If the points 1 2 3 4 5 are given, and we want a 6th point on a line
drawn through 1, we know all the sides of the hexagon with the
exception of 5 6, and this is found by Pascal's theorem.

If this line should happen to pass through 1, then 6 and 1 coincide,
or the line 6 1 is the tangent at 1. And always if two consecutive
vertices of the hexagon approach nearer and nearer, then the side
joining them will ultimately become a tangent.

We may therefore consider a pentagon inscribed in a curve of second
order and the tangent at one of its vertices as a hexagon, and thus
get the theorem:

Every pentagon inscribed in a Every pentagon circumscribed
curve of second order has the about a curve of the second class
property that the intersections has the property that the lines
of two pairs of non-consecutive which join two pairs of non-
sides lie in a line with the consecutive vertices meet on that
point where the fifth side cuts line which joins the fifth vertex
the tangent at the opposite to the point of contact of the
vertex. opposite side.

This enables us also to solve the following problems.

Given five points on a curve of Given five tangents to a curve
second order to construct the of second class to construct the
tangent at any one of them. point of contact of any one of
them.

If two pairs of adjacent vertices coincide, the hexagon becomes a
quadrilateral, with tangents at two vertices. These we take to be
opposite, and get the following theorems:

If a quadrilateral be inscribed If a quadrilateral be circumscribed
in a curve of second order, the about a curve of second
intersections of opposite sides, class, the lines joining opposite
and also the intersections of vertices, and also the lines joining
the tangents at opposite points of contact of opposite
vertices, lie in a line (fig. sides, meet in a point.
19).

If we consider the hexagon made up of a triangle and the tangents at
its vertices, we get--

If a triangle is inscribed in a If a triangle be circumscribed
curve of the second order, the about a curve of second class,
points in which the sides are the lines which join the vertices
cut by the tangents at the to the points of contact of the
opposite vertices meet in a opposite sides meet in a point
point. (fig. 20).

S 55. Of these theorems, those about the quadrilateral give rise to a
number of others. Four points A, B, C, D may in three different ways
be formed into a quadrilateral, for we may take them in the order
ABCD, or ACBD, or ACDB, so that either of the points B, C, D may be
taken as the vertex opposite to A. Accordingly we may apply the
theorem in three different ways.

Let A, B, C, D be four points on a curve of second order (fig. 21),
and let us take them as forming a quadrilateral by taking the points
in the order ABCD, so that A, C and also B, D are pairs of opposite
vertices. Then P, Q will be the points where opposite sides meet, and
E, F the intersections of tangents at opposite vertices. The four
points P, Q, E, F lie therefore in a line. The quadrilateral ACBD
gives us in the same way the four points Q, R, G, H in a line, and the
quadrilateral ABDC a line containing the four points R, P, I, K. These
three lines form a triangle PQR.

The relation between the points and lines in this figure may be
expressed more clearly if we consider ABCD as a four-point inscribed
in a conic, and the tangents at these points as a four-side
circumscribed about it,--viz. it will be seen that P, Q, R are the
diagonal points of the four-point ABCD, whilst the sides of the
triangle PQR are the diagonals of the circumscribing four-side. Hence
the theorem--

_Any four-point on a curve of the second order and the four-side
formed by the tangents at these points stand in this relation that the
diagonal points of the four-point lie in the diagonals of the
four-side._ And conversely,

_If a four-point and a circumscribed four-side stand in the above
relation, then a curve of the second order may be described which
passes through the four points and touches there the four sides of
these figures._

That the last part of the theorem is true follows from the fact that
the four points A, B, C, D and the line a, as tangent at A, determine
a curve of the second order, and the tangents to this curve at the
other points B, C, D are given by the construction which leads to fig.
21.

The theorem reciprocal to the last is--

_Any four-side circumscribed about a curve of second class and the
four-point formed by the points of contact stand in this relation that
the diagonals of the four-side pass through the diagonal points of the
four-point._ And conversely,

_If a four-side and an inscribed four-point stand in the above
relation, then a curve of the second class may be described which
touches the sides of the four-side at the points of the four-point._

S 56. The four-point and the four-side in the two reciprocal theorems
are alike. Hence if we have a four-point ABCD and a four-side abcd
related in the manner described, then not only may a curve of the
second order be drawn, but also a curve of the second class, which
both touch the lines a, b, c, d at the points A, B, C, D.

The curve of second order is already more than determined by the
points A, B, C and the tangents a, b, c at A, B and C. The point D may
therefore be _any_ point on this curve, and d any tangent to the
curve. On the other hand the curve of the second class is more than
determined by the three tangents a, b, c and their points of contact
A, B, C, so that d is any tangent to this curve. It follows that every
tangent to the curve of second order is a tangent of a curve of the
second class having the same point of contact. In other words, the
curve of second order is a curve of second class, and _vice versa_.
Hence the important theorems--

_Every curve of second order is _Every curve of second class is a
a curve of second class._ curve of second order._

The curves of second order and of second class, having thus been
proved to be identical, shall henceforth be called by the common name
of _Conics_.

For these curves hold, therefore, all properties which have been
proved for curves of second order or of second class. We may therefore
now state Pascal's and Brianchon's theorem thus--

_Pascal's Theorem._--If a hexagon be inscribed in a conic, then the
intersections of opposite sides lie in a line.

_Brianchon's Theorem._--If a hexagon be circumscribed about a conic,
then the diagonals forming opposite centres meet in a point.

S 57. If we suppose in fig. 21 that the point D together with the
tangent d moves along the curve, whilst A, B, C and their tangents a,
b, c remain fixed, then the ray DA will describe a pencil about A, the
point Q a projective row on the fixed line BC, the point F the row b,
and the ray EF a pencil about E. But EF passes always through Q. Hence
the pencil described by AD is projective to the pencil described by
EF, and therefore to the row described by F on b. At the same time the
line BD describes a pencil about B projective to that described by AD
(S 53). Therefore the pencil BD and the row F on b are projective.
Hence--

_If on a conic a point A be taken and the tangent a at this point,
then the cross-ratio of the four rays which join A to any four points
on the curve is equal to the cross-ratio of the points in which the
tangents at these points cut the tangent at A._

S 58. There are theorems about cones of second order and second class
in a pencil which are reciprocal to the above, according to S 43. We
mention only a few of the more important ones.

The locus of intersections of corresponding planes in two projective
axial pencils whose axes meet is a cone of the second order.

The envelope of planes which join corresponding lines in two
projective flat pencils, not in the same plane, is a cone of the
second class.

Cones of second order and cones of second class are identical.

Every plane cuts a cone of the second order in a conic.

_A cone of second order is uniquely determined by five of its edges or
by five of its tangent planes, or by four edges and the tangent plane
at one of them, &c. &c._

_Pascal's Theorem._--If a solid angle of six faces be inscribed in a
cone of the second order, then the intersections of opposite faces are
three lines in a plane.

_Brianchon's Theorem._--If a solid angle of six edges be circumscribed
about a cone of the second order, then the planes through opposite
edges meet in a line.

Each of the other theorems about conics may be stated for cones of the
second order.

S 59. _Projective Definitions of the Conics._--We now consider the
shape of the conics. We know that any line in the plane of the conic,
and hence that the line at infinity, either has no point in common
with the curve, or one (counting for two coincident points) or two
distinct points. If the line at infinity has no point on the curve the
latter is altogether finite, and is called an _Ellipse_ (fig. 21). If
the line at infinity has only one point in common with the conic, the
latter extends to infinity, and has the line at infinity a tangent. It
is called a _Parabola_ (fig. 22). If, lastly, the line at infinity
cuts the curve in two points, it consists of two separate parts which
each extend in two branches to the points at infinity where they meet.
The curve is in this case called an _Hyperbola_ (see fig. 20). The
tangents at the two points at infinity are finite because the line at
infinity is not a tangent. They are called _Asymptotes_. The branches
of the hyperbola approach these lines indefinitely as a point on the
curves moves to infinity.

S 60. That the circle belongs to the curves of the second order is
seen at once if we state in a slightly different form the theorem that
in a circle all angles at the circumference standing upon the same arc
are equal. If two points S1, S2 on a circle be joined to any other two
points A and B on the circle, then the angle included by the rays S1A
and S1B is equal to that between the rays S2A and S2B, so that as A
moves along the circumference the rays S1A and S2A describe equal and
therefore projective pencils. The circle can thus be generated by two
projective pencils, and is a curve of the second order.

If we join a point in space to all points on a circle, we get a
(circular) cone of the second order (S 43). Every plane section of
this cone is a conic. This conic will be an ellipse, a parabola, or an
hyperbola, according as the line at infinity in the plane has no, one
or two points in common with the conic in which the plane at infinity
cuts the cone. It follows that our curves of second order may be
obtained as sections of a circular cone, and that they are identical
with the "Conic Sections" of the Greek mathematicians.

S 61. Any two tangents to a parabola are cut by all others in
projective rows; but the line at infinity being one of the tangents,
the points at infinity on the rows are corresponding points, and the
rows therefore similar. Hence the theorem--

_The tangents to a parabola cut each other proportionally._

POLE AND POLAR

S 62. We return once again to fig. 21, which we obtained in S 55.

If a four-side be circumscribed about and a four-point inscribed in a
conic, so that the vertices of the second are the points of contact of
the sides of the first, then the triangle formed by the diagonals of
the first is the same as that formed by the diagonal points of the
other.

Such a triangle will be called a _polar-triangle_ of the conic, so
that PQR in fig. 21 is a polar-triangle. It has the property that on
the side p opposite P meet the tangents at A and B, and also those at
C and D. From the harmonic properties of four-points and four-sides it
follows further that the points L, M, where it cuts the lines AB and
CD, are harmonic conjugates with regard to AB and CD respectively.

If the point P is given, and we draw a line through it, cutting the
conic in A and B, then the point Q harmonic conjugate to P with regard
to AB, and the point H where the tangents at A and B meet, are
determined. But they lie both on p, and therefore this line is
determined. If we now draw a second line through P, cutting the conic
in C and D, then the point M harmonic conjugate to P with regard to
CD, and the point G where the tangents at C and D meet, must also lie
on p. As the first line through P already determines p, the second may
be any line through P. Now every two lines through P determine a
four-point ABCD on the conic, and therefore a polar-triangle which has
one vertex at P and its opposite side at p. This result, together with
its reciprocal, gives the theorems--

_All polar-triangles which have one vertex in common have also the
opposite side in common._

_All polar-triangles which have one side in common have also the
opposite vertex in common._

S 63. To any point P in the plane of, but not on, a conic corresponds
thus one line p as the side opposite to P in all polar-triangles which
have one vertex at P, and reciprocally to every line p corresponds one
point P as the vertex opposite to p in all triangles which have p as
one side.

We call the line p the _polar_ of P, and the point P the _pole_ of the
line p with regard to the conic.

If a point lies on the conic, we call the tangent at that point its
polar; and reciprocally we call the point of contact the pole of
tangent.

S 64. From these definitions and former results follow--

The polar of any point P not The pole of any line p not a
on the conic is a line p, which tangent to the conic is a point
has the following properties:-- P, which has the following
properties:--

1. On every line through P 1. Of all lines through a point
which cuts the conic, the polar on p from which two tangents
of P contains the harmonic may be drawn to the conic, the
conjugate of P with regard to pole P contains the line which is
those points on the conic. harmonic conjugate to p, with
regard to the two tangents.

2. If tangents can be drawn 2. If p cuts the conic, the
from P, their points of contact tangents at the intersections
lie on p. meet at P.

3. Tangents drawn at the 3. The point of contact of
points where any line through P tangents drawn from any point
cuts the conic meet on p; and on p to the conic lie in a line
conversely, with P; and conversely,

4. If from any point on p, 4. Tangents drawn at points
tangents be drawn, their points where any line through P cuts the
of contact will lie in a line conic meet on p.
with P.

5. Any four-point on the conic 5. Any four-side circumscribed
which has one diagonal point at about a conic which has one
P has the other two lying on p. diagonal on p has the other two
meeting at P.

The truth of 2 follows from 1. If T be a point where p cuts the conic,
then one of the points where PT cuts the conic, and which are harmonic
conjugates with regard to PT, coincides with T; hence the other
does--that is, PT touches the curve at T.

That 4 is true follows thus: If we draw from a point H on the polar
one tangent a to the conic, join its point of contact A to the pole P,
determine the second point of intersection B of this line with the
conic, and draw the tangent at B, it will pass through H, and will
therefore be the second tangent which may be drawn from H to the
curve.

S 65. The second property of the polar or pole gives rise to the
theorem--

From a point in the plane of a A line in the plane of a conic
conic, two, one or no tangents has two, one or no points in
may be drawn to the conic, common with the conic, according
as its polar has two, as two, one or no tangents
one, or no points in common can be drawn from its pole to the
with the curve. conic.