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cases. First, however, we shall give a few theorems which relate to B, and C' as points in the second. Then A and B coincide with their the general correspondence, not to the perspective position. corresponding points, but C does not. It is, however, not necessary 28. Two rows or pencils, flat or axial, which are projective lo a that
such third are projective io each other; this follows at once from the have twice a pointi definitions.
coincident with its cor$ 29. If two rows, or two pencils, either flat or axial, or a row and a responding point; it is Oil pencil, be projective, we may assume to any three elements in the one possible that this hapthe three corresponding elements in the other, and then the correspondence pens only, once or not is uniquely deiermined.
Wilgo at all,
Or this we shall For if in two projective rows we assume that the points A, B, C see examples later.
no in the first correspond to the given points A', B', C' in the second, $ 35. If two projective then to any fourth point D in the first will correspond a point D' rows or pencils are in in the second, so that
perspective position, we (AB, CD) = (A'B', C'D').
element in one But there is only one point, D'. which makes the cross-ratio sponds to given (A'B', C'D') equal to the given number (AB, CD).
element in the other.
wallo! The same reasoning holds in the other cases.
If p and 9. (fig. 9) arenit ods $ 30. Il two rows are perspective, then the lines joining corre. two projective rows, Solinoge
s sponding points all meet in a point, the centre of projection; and that 'K' corresponds to the point in which the two bases of the rows intersect as a point itsell, and if we know
Fig. 9. in the first row coincides with its corresponding point in the that to A and B in e second.
correspond A' and B' in q, then the point S, where AA' meets BB'. This follows from the definition. The converse also holds, is the centre of projection, and hence, in order to find the point C
corresponding to C, we have only to join C to S; the point c', If two projective rows have such a position that one point in the one where this line cuts q, is the point required. coincides with its corresponding point in the olher, then they are per: If two fat pencils, S. and Sa, in a plane are perspective (fig. 10), spective, that is, the lines joining corresponding points all pass through we need only to know two pairs, e, a' and b, b', of corresponding e common point, and form a flai pencil.
rays in order to find the For let A, B, C, D... be points in the one, and A', B', C', axis s of projection. This
stantly udomus Acai 2, 60 D'...
the corresponding points in the other row, and let A be made being known, a ray & in bloul elismerom S. to coincide with its
corresponding point A'. Let S be the point where S, corresponding to a given Aanbul basit che lines BB' and CC' meet, and let us join S to the point D in the ray c in sı, is found by Wyviss
a20701 first row. This line will cut the second row in a point D', so that joining Są to the point
wolle A, B, C, D are projected from S into the points A, B’,C", D., The Where Ccuts the axis s. 21. SAAT cross-ratio (AB, CD) is therefore equal to (AB', C'D'), and by hypo A similar construction A obra 189
Anila thesis it is equal to (A'B', C'D'). Hence (A'B', 'C'D') = (A'B', C'D'). holds in the other cases that is, D' is the same point as D'.
agree 31. If two projected flat pencils in the same plane are in per
of perspective figures. spective, then the intersections of corresponding lines form a row,
solution of the following fortuna
On this depends the diio sur alle and the line joining the two centres as a line in the first pencil general problem.
mos corresponds to the same line as a line in the second. And con
$36. Three pairs of core and are you
AA unil ke versely,
responding elements in two no a. Se If two projective pencils in the same plane, but with different centres, projective rows or pencils have one line in the one coincident with its corresponding line in the being given, to determine
COMO other, then the two pencils are perspective, that is, the intersection of for any element in one
robojono corresponding lines lie in a line.
the corresponding element
FIG. 10. The proof is the same as in $ 30.
in the other. $ 32. If two projective flat pencils in the same point (pencil in We solve this in the two cases of two projective rows and of two space), but not in the same plane, are perspective, then the planes projective Alat pencils in a plane. joining corresponding rays all pass through a line (they form an Problem 1.—Let A, B, C be Problem II.-Let a, b, c be axial pencil), and the line common to the two pencils (in which three points in a row s, A', B', C' three rays in a pencil S, a', '', their planes intersect) corresponds to itself. And conversely: the corresponding points in a the corresponding rays in a pro
If two flat pencils which have a common centre, but do not lie projective row s', both being in a jective pencil S, both being in in a common plane, are placed so that one ray in the one coincides plane; it is required to find for the same plane; it is required to with its corresponding ray in the other, then they are perspective, any point D in s the correspond- find for any ray d in the correthat is, the planes joining corresponding lines all pass through a ing point D'ins.
sponding ray d' in S. line.
The solution is made to depend on the construction of an auxiliary $33. If two projective axial pencils are perspective, then the inter row or pencil which is perspective to both the given ones. This is section of corresponding planes lie in a plane, and the plane common found as follows: to the two pencils (in which the two axes lie) corresponds to itself. Solution of Problem 1. On the line joining two corresponding And conversely :
points, say AA' (fig. 11), take any two points, S and S', as centres II two projective axial pencils are placed in such a position that a auxiliary pencils. plane in the one coincides with its corresponding plane, then the two Join the intersection B. pencils are perspective, that is, corresponding planes meet in lines of SB and S'B' to the which lie in a plane.
intersection of SC The proof again is the same as in $ 30.
and S'C' by the line ili su dadw gnibnog $34. These theorems relating to perspective position become Then a row on si
VS illusory if the projective rows of pencils have a common base. We be perspective to s with then haver
voel blod mene In two projective rows on the same line—and also in two pro- tion, and to go with s slansit, ano jective and concentric flat pencils in the same plane, or in two as centre. To find now naslonil projective axial pencils with a common axis-every element in the the point D'on s' cor
Aide one coincides with its corresponding element in the other as soon responding to a point.
B' as three elements in the one coincide with their corresponding D on s we have only to elements in the other.
determine the point D,
tie in BSD cuts ...................in and their corresponding elements A, B, C, D exists the relations, and to draw S*DT:nie 3179 Dat (ABCD) = (A'B'C'D'). If now A', B', C coincide respectively
with the point where this line binocenog broda A, B, C, we get_(AB, CD) = (AB, CD'), hence D and 'D' coincide. cuts s' will be the re- cind legent bien The last theorem may also be stated thus:
quired point D'. niboomerno
od Thou In two projective rows or pencils, which have a common base Proof.--The rows $
nd but are not identical, not more than two elements in the one can and s' are both perspec. L. 1A) = (11,1A) coincide with their corresponding elements in the other.
Thus two projective rows on the same line cannot have more they are projective to LA LA LA than two pairs of coincident points unless every point coincides one another. "To A, B, with its corresponding point.
C, D ons corresponday - FIG. 11. It is easy to construct two projective rows on the same line, A, B, C, D, on si, and state 11 bi stolonas which have two pairs of corresponding points coincident. Let the to these correspond A, B, C, D on s'; so that D-and D' are points A, B, C as points belonging to the one row correspond to A, I corresponding points as required.
The Of Liris
Solution of Problem II. -Through the intersection A of two Theorem.-The product of the distances of any two corresponding corresponding rays a and a' (fig. 12), take two lines, s and s', as points in two projective rows from the points
which correspond to bases of auxiliary rows. Let the points at infinity in the other is constant, viz. AJ. A'l'=k. be the point where the line b, Steiner has called this number k the Power of the correspondence. which joins B and B', cuts the [The relation AJ. A'I'=k shows that if J, I' be given then the line a which joins c and C point A' corresponding to a specified point A is readily found; hence Then a pencil S. will be per-A, A' generate homographic ranges of which I and j' correspond to spective to S with s as axis of the points at infinity on the ranges. If we take any two origins 0, projection. To find the ray d' in on the ranges and reduce the expression AJ. A'l'=k to its alge $ corresponding to a given ray d braic equivalent, we derive an equation of the form axx' +Bx+ve in S, cut d by s at B; project +3=0. Conversely, if a relation of this nature holds, then points this point from sto D on s' corresponding to solutions in x, x' form homographic ranges.) and join D' to S'. This will be $39. Similar Rows.- If the points at infinity in two projective the required ray.
rows correspond so that l' and I are at infinity, this result loses its
(AB, CI) = (A'B', C'I').
AC/CB = A'C'/C'B' or AC/A'C'=BC/B'C'.
that is, corresponding segments are proportional. Conversely, if In the first solution the two corresponding segments are proportional, then to the point at domanica centres, s, s', are any two points infinity in one corresponds the point at infinity in the other. If we call ni
on a line joining any two correULO
sponding points, so that the solu Theorem.-Two projective rows are similar if to the point at tion of the problem allows of a great many
different constructions. infinity in one corresponds the point at infinity in the other, and But whatever construction be used, the point D', corresponding to D, conversely, if two rows are similar then they are projective, and the must be always the same, according to the theorem in 29. This points at infinity are corresponding points. gives rise to a number of theorems, into which, however, we shall From this the well-known propositions follow: not enter. The same remarks hold for the second problem.
Two lines are cut proportionally (in similar rows) by a series of $37. Homological Triangles.-As a further application of the parallels. The rows are perspective, with centre of projection at theorems about perspective rows and pencils we shall prove the infinity. following important theorem.
If two similar rows are placed parallel, then the lines joining I heorem.- II ABC and A'B'C' (fig. 13) be two triangles, such that homologous points pass through a common point. the lines AA', BB, CC' meet in a point s, then the intersections of
$ 40. If two flat pencils be projective, then there exists in either, BC and B'C of CA and C'A', and
and A'B' will lie in a line. one single pair of lines at right angles to one another, such that the Such triangles tre said to be homological, or in perspective. The corresponding lines in the other pencil are again at right angles. triangles are " co-axial" in virtue of the property that the meets of
To prove this, we place the pencils in perspective position (fig. 14) corresponding sides are collinear and copolar, since the lines joining by making one ray coincorresponding vertices are concurrent.
cident with its correspond-
now we draw the circle which the point Sis common to all, any two of these rows will
be perspective. has its centre o on p, and If S. be the centre of projection of rows b and c,
which passes through the S
centres S and S' of the two Si
a and 6,
pencils. This circle cuts in and if the line S.S. cuts a in A., and b in B., and c in C, then A, B, two points H and K. The
will be corresponding points two pairs of rays, h, k, and Ys11
in a and b, both corresponding k', k', joining these points to le 5,0 W niya dit to Cin c. But a and b are
S and s' will be pairs of
k 07 boring
corresponding points must gives in general but one
projection S, of a and b. In the perpendicular bisector
finite number, and to every
PRINCIPLE OF DUALITY
$41. It has been stated in § 1 that not only points, but also planes vertex on each of three con- and lines, are taken as elements out of which
figures are built up. current lines, then the inter- We shall now see that the construction of one figure which possesses FIG. 13.
sections of corresponding sides certain properties gives rise in many cases to the construction of
lie in a line, those sides another figure, by replacing, according to definite rules, elements being called corresponding which are opposite to vertices on the of one kind by those of another. The new figure thus obtained will same line.
properties which may be stated as soon as those of the The converse theorem holds also, viz.
original figure are known. Theorem.-If the sides of one triangle meet those of another in We obtain thus a principle, known as the principle of duality three points which lie in a line, then the vertices lie on threc lines or of reciprocity, which enables us to construct to any figure not which meet in a point.
containing any measurement in its construction a reciprocal figure, The proof is almost the same as before.
as it is called, and to deduce from any theorem a reciprocal theorem, 638. Melrical Relations between Projective Rows.-Every row for which no further proof is needed. contains one point which is distinguished from all others, viz. It is convenient to print reciprocal propositions on opposite sides the point at infinity. In two projective rows, to the point I at of a page broken into two columns, and this plan will occasionally infinity in one corresponds a point l' in the other, and to the point be adopted. J'at infinity in the second corresponds a point J in the first. The We begin by repeating in this form a few of our former statepoints l' and J are in general finite. If now A and B are any two ments: points in the one, A', B' the corresponding points in the other row, Two points, determine a line. Two planes determine a line. tben
Three points which are not in a Three planes which do not pass (AB, JI) = (A'B', J'I").
through a line determine a point. or
A line and a point without it A line and a plane not through AJ/JB : AI/IB-A'J'/'B' : AT/'B'.
determine a plane.
it determine a point. But, by $17
Two lines in a plane determine Two lines through a point AI/IB=AJTJ'B'=-1;
determine a plane. therefore the last equation changes into
10.A These propositions show that it will be possible, when any figure TAS. AT =BJ. B'I'.
is given, to construct
a second figure by taking planes instead of that is to say,
wa 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 To a surface as locus of points corresponds, in the same manner, in it, we have to take in the second figure a point and three planes a surface as envelope of planes; and to through it. The three points in the first, together with the three a curve in space as locus of points cori ber99 lines joining them two and two, form a triangle: the three planes responds a developable surface as en in the second and their three lines of intersection form a trihedral velope of planes. angle. A triangle and a trihedral angle are therefore reciprocal It will be seen from the above that figures.
we may, by aid of the principle of Similarly, to any figure in a plane consisting of points and lines duality, construct for every figure a
als will correspond a figure consisting of planes and lines passing through reciprocal figure, and that to any a point S, and hence belonging to the pencil which has S as centre. property of the one a reciprocal pro
od The figure reciprocal to four points in space which do not lie perty of the
other will exist, as long in a plane will consist of four planes
which do not meet in a point. as we consider only properties which In this case each figure forms a tetrahedron.
depend upon nothing but the positions and intersections of the 42. As other examples we have the following:
different elements and not upon measurement nie bos ,u a no To a row is reciprocal" an axial pencil,
For such propositions it will therefore be unnecessary to prove a fat pencil a flat pencil,
more than one of two reciprocal theorems. * a field of points and lines.
á pencil of planes and lines, the space of points the space of planes.
GENERATION OF CURVES AND CONES OF Second ORDER For the row consists of a line and all the points in it, reciprocal to
OR SECOND CLASS it therefore will be a line with all planes through it, that is, an axial $45. Conics.--Il we have two projective pencils in a plane, pencil; and so for the other cases.
corresponding rays will meet, and their point of intersection will This correspondence of reciprocity breaks down, however, if we constitute some focus which we have to investigate. Reciprocally, take figures which contain measurement in their construction. For if two projcctive rows in a plane are given, then the lines which join instance, there is no figure reciprocal to two planes at right angles, corresponding points will envelope some curve. We prove first: because there is no segment in a row which has a magnitude as Theorem.-- 1 two projective Theorem.- iwo projective definite as a right angle.
fat pencils lie in a planc, but rows lie in a plane, but are We add a few examples of reciprocal propositions which are easily are neither in perspective nor neither in perspective nor on a proved.
concentric, then the locus of common base, then the envelope Theorem.-11 A, B, C, D are Theorem.-If a, B. y. 8 are intersections of corresponding of lincs joining corresponding any four points in space, and if four planes in space, and if the rays is a curve of the second points is a curve of the second the lines AB and CD meet, then lines aß and yo meet, then all order, that is, no linc contains class, that is, through no point all four points lie in a plane, four planes lie in a point (pencil). more than two points of the pass more than two of the heng also AC and BD, as well hence also ay and Bs, as well as locus.
enveloping lines. as AB and BC, meet. ad and By, meet.
Proof.-We draw any line l. Proof-We take any point T Theorem.-If of any number of lines every one meets every olher, This cuts each of the pencils in a and join it to all points in cach whilst all do noi
row, so that we have on I two row. This gives two concentric lie in a point, then all lie in a lie in a plane, then all lie in a rows, and these are projective pencils, which are projective planc. point (pencil);
because the pencils are pro- because the rows are projective. $ 43. Reciprocal figures as explained lie both in space of three jective. Il corresponding rays Il a line joining corresponding dimensions. If the one is confined to a plane (is formed of elements of the two pencil's meet on the points in the two rows passes which lie in a plane), then the reciprocal figure is confined to a pencil line I, their intersection will be a through T, it will be a line in the (is formed of elements which pass through a point).
point in the one row which one pencil which coincides with But there is also a more special principle of duality, according to coincides with its corresponding its corresponding line in the which figures are reciprocal which lie both in a plane or both in a point in the other. But two other. But two projective conpencil. In the plane we take points and lines as reciprocal elements, projective rows on the same base centric flat pencils in the same for they have this fundamental property in common, that two cannot have more than two plane cannot have more than two elements of one kind determine one of the other. In the pencil, points of one coincident with lines of one coincident with their on the other hand, lines and planes have to be taken as reciprocal, iheir corresponding points in the corresponding line in the other and here it holds again that two lines or planes determine one plane other (834).
($ 34). or line.
It will be seen that the prools are reciprocal, so that the one may Thus, to one plane figure we can construct one reciprocal figure be copied from the other by simply interchanging the words point in the plane, and to each one reciprocal figure in a pencil.. We and line, locus and envelope, row and pencil, and so on. We shall mention a few of these. At first we explain a few names: therefore in future prove seldom more than one of two reciprocal ..A figure consisting of in points A figure consisting of n lines theorems, and often state che theorem only, the reader being recomin a plane will be called an in a plane will be called an n-side. mended to go through the reciprocal prool by himself, and to supply n-point.
the reciprocal theorems when not given. A figure consisting of n planes: A figure consisting of n lines $.46. We state the theorems in the pencil reciprocal to the last, in a pencil will be called an in a pencil will be called an without proving them: n-flat. n-edge.
Theorem.- two projective Theorem.If two projective It will be understood that an n-side is different from a polygon Alat pencils are concentrie, but axial pencils lie in the same of n sides. The latter has sides of finite length and n vertices, the are neither perspective por co. pencil' (their axes meet in a former has sides all of infinite extension, and every point where planar, then the envelope of the point), but are neither perspec. two of the sides meet will be a vertex. A similar difference exists planes joining corresponding rays tive nor co-axial, then the locus between a solid angle and an n-edge or an n-flat. We notice par. is a cone of the second class; of lines joining corresponding ticularly
that is, no line through the planes is a cone of the second A four-point has six sides, of A four-side has six vertices, of common centre contains more order; that is, no plane in the which two and two are opposite, which two and two are opposite than two of the enveloping planes pencil contains more than two and three diagonal points, which and three diagonals, which join
of these lines. are intersections of opposite opposite vertices.
$ 47. Of theorems about cones of second order and cones of second sides.
class we shall state only very few. We point out, however, the A four-flat has six edges, of A four-edge has six faces, of following connexion between the curves and cones under conwhich two and two are opposite, which two and two are opposite, sideration: and three diagonal planes, which and three diagonal edges, which The lines which join any point Every plane section of a cone pass through opposite edges. are intersections of opposite faces. in space to the points on a curve of the second order is a curve of
A four-side is usually called a complete quadrilateral, and a four of the second order form a cone the second order. point a complete quadrangle. The above notation, however, seems of the second order. better adapted for the statement of reciprocal propositions.
The planes which join any Every plane section of a cone $44.
point in space to the lines en of the second class is a curve of if a point moves in a plane it. If a line moves in a plane it veloping a curve of the second the second class. describes a plane curve.
envelopes a plane curve (fig. 15). class envelope themselves a cone If a plane moves in a pencil it Il a line moves in a pencil it of the second class. envelopes a cone. describes a cone.
By its aid, or by the principle of duality, it will be easy to obtain A curve thus appears as generated either by points, and then we theorems about them from the theorems about the curves. call it a "locus," or by lines, and then we call it an “envelope." We prove the first. A curve of the second order is generated by In the same manner a cone, which means here a surface, appears two projective pencils. These pencils, when joined to the point in either as the locus of lines passing through a fixed point, the" vertex space, give rise to two projective axial pencils, which generate the of the cone, or as the envelope of planes passing through the same cone in question as the locus of the lines where corresponding plancs point.
3m1! If in fig. 16 we draw in the pencil Si the ray ki which passes Theorem.-The curve of second Theorem.--The envelope of order which is generated by two second class which is generated projective fat pencils passes by two projective rows contains through the centres of the two the bases of these rows as en. pencils.
veloping lines or tangents. 15. HT.: 67 992 Proof.-If S and S'are the two Proof-lf s and s' are the two pencils, then to the ray SS' or p rows, then to the point ss' or P in the pencil S' corresponds in as a point in s corresponds in s the pencil S a ray P. which is a point P, which is not coincident different from p', for the pencils with P, for the rows are not are not perspective. But p and perspective. But P and P' are Balina p' meet at S, so that S is a point joined by s, so that s is one of the on the curve, and similarly S. 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
... en 1938 sobie 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 coincidert points. Such a line is called a langent to the curve,
FIG. 16. 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 through the auxiliary centre s, it will be found that the corre. result that through every point in one of the rows, say in s, two sponding ray ka cuts it on uz. Hencetangents may be drawn to the curve, the one being s, the other the Theorem. In the above con Theorem.-In the above con. line joining the point to its corresponding point in s'. There is, struction the basis of the auxil- struction (fig. 17) the tangents to however, one point P in s for which these two lines coincide. Such iary rows uz and uz cut the curve the curve from the centres of the a point in one of the tangents is called the " point of contact of the where they cut the rays SaS and auxiliary pencils S, and S, are the tangent. We thus get-
lines which pass through wu and Theorem.--To the line joining Theorem.-To the point of
uju respectively. the centres of the projective intersection of the bases of two As A is any given point on the curve, and ui any line through pencils as a linc in one pencil projective rows as a point in one
it, we have solved the problems: corresponds in the other the row corresponds in the other the
Problem.-To find the second Problem.-To find the second tangent at its centre. point of contact of its base.
point in which any line through a tangent which can be drawn $ 49. Two projective pencils are determined if three pairs of known point on the curve cuts from any point in a given tangent corresponding lines are given. Hence if a, b, ci are three lines in a
to the curve. pencil S., and as, ba, c; the corresponding lines in a projective pencil If we determine in S. (fig. 16) the ray corresponding to the ray So, the correspondence and therefore the curve of the second order S s, in Sa, we get the tangent at S. Similarly, we can determine generated by the points of intersection of corresponding rays is the point of contact of the tangents u or 119 in fig. 17. determined. Of this curve we know the two centres Si and Sa, $ 51. If five points are given, of which not three are in a line, and the three points az@z, byba, CC2, hence five points in all. This then we can, as has just been shown, always draw a curve of the and the reciprocal considerations enable us to solve the following two problems:
Problem.-Toconstruct a curve Problem.-To construct a curve of the second order, of which five of the second class, of which five points S, S, A, B, C are given. tangents 1., uz, a, b, c are given.
In order to solve the left-hand problem, we take two of the given points, say S. and Sq, as centres of pencils. These we make projective by taking the rays Oy, O., Ct, which join S, to A, B, C respectively, as corresponding to the rays Oz, bz, ca, which join S, 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 S, and S,, hence through the five given points. To find more points on the curve we have to construct for any ray in S, the corresponding ray in Sg. This has been done in $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. un, uz of the five given lines, ui, uz, a, b, c, as bases of two rows, and the points A1, B1, C, where a, b, c cut 1 as corresponding to the points As, B2, C, where a, b, c cut uz.-We get then the following solutions of the two problems:
second order through them; we select two of the points as centres of Solulion.-Through the point Solution.- In the line a take projective pencils, and then one such curve is determined. It will A draw any two lines, w and ug any two points S, and -S, as be presently shown that we get always the same curve if two other (fig. 16), the first w to cut the centres of pencils (fig. 17), the points are taken as centres of pencils, that therefore five points pencil S, in a row AB,C, the first S, (A,B,C) to project the determine one curve of the second order, and reciprocally, that five other 12 to cut the pencil S, in a row u, the other S, (A,B,C) to tangents determine one curve of the second class. Six points taken row AB,C, These two rows will project the row U2. These two at random will therefore not lie on a curve of the second order. In be perspective, as the point A pencils will be perspective, the order that this may be the case a certain condition has to be satisfied, corresponds to itself, and the line SA, being the same as the and this condition is easily obcentre of projection will be the corresponding. linc S A2, and the tained from the construction in point S where the lines B,B, axis of projection will be the line $ 49, fig. 16. If we consider the and C Ce meet. To find now for u, which joins the intersection B
conic determined by the five any rayd, in S, its corresponding of SB, and SB, to the intersecpoints A, S, Sa, K, L, then the ray d, in Sy, we determine the tion C of SC and S.C. To find point D will be on the curve if, s. point D, where di cuts 1, project now for any point D, in u; the and only if, the points on D, S, this point from Sto D, on u and corresponding point D, in us, we
D, be in a line. join S, to Dz. This will be the draw S, D, and project the point
This may be stated differently required ray dr which cuts d, at D where this line cuts u from Sif we take AKS DS,L (figs. 16 some point D on the curve. to uz. This will give the required and 18) as a hexagon inscribed
point Dz, and the line d joining Di in the conic, then AK and DS,
KS, and SL, as well as S D and
LA. The first two meet in Dz $ 50. These constructions prove, when rightly interpreted, very the others in S and D, respectively. We may therefore state the important properties of the curves in question.
required condition, together with the ceciprocal one, as follows:
Pascal's Theorem.--Ifa hexagon Brianchon's Theorem.-If a Every pentagon inscribed in a * Every pentagon circumscribed be inscribed in a curve of the hexagon be circumscribed about curve of second order has the about a curve of the second class second order, then the intersec- a curve of the second class, then tions of opposite sides are three the lines joining opposite vertices two pairs of non-consecutive which join two pairs of non
property that the intersections of has the property that the lines points in a line.
are three lines meeting in a point. | sides lie in a line with the point consecutive vertices meet on that These celebrated theorems, which are known by the names of where the fifth side cuts the tan- line which joins the fifth vertex their discoverers, are perhaps the most fruitful in the whole theory gent at the opposite vertex. to the point of contact of the of conics. Before we go over to their applications we have to show
opposite side. that we obtain the same curve if we take, instead of S, Sg, any two This enables us also to solve the following problems. other points on the curve as centres of projective pencils. $52. We know that the curve depends only upon the correspond
Given five points on a curve of Given five tangents to a curve ence between the pencils S, and S., and not upon the special con
second order to construct the of second class to construct the struction used for finding new points on the eurve. The point A tangent at any one of them. point of contact of any one of (fig. 16 or 18), through which the two auxiliary rows u, Uy were
them. drawn, may therefore be changed to any other point on the curve. Let us now suppose the curve drawn, and keep the points Si, Sa, 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 D, a row on SD perspective to the pencil L. At the same time AK describes a pencil about K and Distt A li biti det a row perspective to it on SD. But by Pascal's theorem D, and Du will always lie in a line with S, so that the rows described by D, and D, are perspective. It follows that the pencils K and L wili 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. II, therefore, a point on the curve and its tangent are given, this will be equivalent to having given two points on the curve. Simi. larly, if on the curve of second class å tangent and its point of contact are given, this will be equivalent to two given tangents. We may therefore extend the last theorem:
+,5 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 one tangents and the point of contact of them, or three points and the at one of them, or three tangents Il two pairs of adjacent vertices coincide, the hexagon becomes a tangents at two of them, are and the points of contact at two quadrilateral, with tangents at two vertices. These we take to be given.. of them, are given.
opposite, and get the following theorems: $53. At the same time it has been proyed:
If a quadrilateral be inscribed If a quadrilateral be circumIf all points on a curve of the All tangents to a curve of second in a curve of second order, the scribed about a curve of sefond second order be joined to any class are cut by any two of intersections of opposite sides, class, the lines joining opposite two of them, then the two pencils them in projective rows, those and also the intersections of the vertices, and also the lines joining thus formed are projective, those being corresponding points which tangents at opposite vertices, lie points of contact of opposite rays being corresponding which lie on the same tangent. Hence in a line (fig. 19).
sides, meet in a point. 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 points cut by four fixed tangents a, b,c,d A, B, C, D in the curve is in- is independent of the position of dependent of the position of S, #, and is called the cross-ratio of and is called the cross-ratio of the the four tangents a, b, c, d. four points A, B, C, D.
If this cross-ratio equals 1 If this cross-ratio cquals 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 at A curve of second class has on every point one tangent. every tangent a point of contact.
$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
FIG. 20. only for the former.
Pascal's theorem may be used when five points are given to find If we consider the hexagon made up of a triangle and the tangents more points on the curve, viz. it enables us to find the point where at its vertices, we getany line through one of the given points cuts the curve again. It
If a triangle is inscribed in a II a triangle be circumscribed is convenient, in making use of Pascal's theorem, to number the
curve of the second order, the about a curve of second class, 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.
points in which the sides are cut the lines which join the vertices
by the tangents at the opposite to the points of contact of the It will be seen that i 2 (leaving out 3) 4 5 are opposite sides,
vertices meet in a point. so are 2 3 and (leaving out 4) 5 6, and also 3. 4 and (leaving
opposite sides meet in a point
(hg. 20). out 5) 61.
If the points 1 2 3 4 5 are given, and we want a 6th point on a $55. Of these theorems, those about the quadrilateral give rise to line drawn through 1, we know all the sides of the hexagon with a number of others. Four points, A, B, C, D may in three different the exception of 56, and this is found by Pascal's theorem. ways be formed into a quadrilateral, for we may take them in the
If this line should happen to pass through 1, then 6 and I coincide, order ABCD, or ACBD, or ACDB, so that either of the points or the line 61 is the tangent at 1. And always if two consecutive B, C, D may be taken as the vertex opposite to A. Accordingly we vertices of the hexagon approach nearer and nearer, then the side may apply the theorem in three different ways. joining them will ultimately become a tangent.
Let A, B, C, D be four points on a curve of second order (fig. 21). We may therefore consider a pentagon inscribed in a curve of and let us take them as forming a quadrilateral by taking the points second order and the tangent at one of its vertices as a hexagon, in the order ABCD, so that A, C and also B, D are pairs of opposite and thus get the theorem:
vertices. Then P, Q will be the points where opposite sides ineet,