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 to a that two such rows third are projective to each other; this follows at once from the have twice a point definitions. coincident with its cor. responding point; it is possible that this happens only once or not at all. Of this we shall see examples later. $29. If two rows, or two pencils, either flat or axial, or a row and a pencil, be projective, we may assume to any three elements in the one the three corresponding elements in the other, and then the correspondence is uniquely determined. For if in two projective rows we assume that the points A, B, C in the first correspond to the given points A', B', C' in the second, then to any fourth point D in the first will correspond a point D' in the second, so that (AB, CD) = (A'B', C'D'). 835. If two projective rows or pencils are in perspective position, we know at once which element in one corre But there is only one point, D', which makes the cross-ratio sponds to any given (A'B', C'D') equal to the given number (AB, CD). The same reasoning holds in the other cases. $30. If two rows are perspective, then the lines joining corre sponding points all meet in a point, the centre of projection; and the point in which the two bases of the rows intersect as a point in the first row coincides with its corresponding point in the second. This follows from the definition. The converse also holds, If two projective rows have such a position that one point in the one coincides with its corresponding point in the other, then they are perspective, that is, the lines joining corresponding points all pass through a common point, and form a flat pencil. viz. For let A. B, C, D... be points in the one, and A', B', C', D'. the corresponding points in the other row, and let A be made to coincide with its corresponding point A'. Let S be the point where the lines BB' and CC' meet, and let us join S to the point D in the first row. This line will cut the second row in a point D", so that A, B, C, D are projected from S into the points A, B, C, D. The cross-ratio (AB, CD) is therefore equal to (AB', C'D"), and by hypothesis it is equal to (A'B', C'D'). Hence (A'B', C'D") = (A'B', C'D'), that is, D is the same point as D'. 831. If two projected flat pencils in the same plane are in perspective, then the intersections of corresponding lines form a row, and the line joining the two centres as a line in the first pencil corresponds to the same line as a line in the second. And conversely, If two projective pencils in the same plane, but with different centres, have one line in the one coincident with its corresponding line in the other, then the two pencils are perspective, that is, the intersection of corresponding lines lie in a line. The proof is the same as in § 30. $32. If two projective flat pencils in the same point (pencil in space), but not in the same plane, are perspective, then the planes joining corresponding rays all pass through a line (they form an axial pencil), and the line common to the two pencils (in which their planes intersect) corresponds to itself. And conversely:-→→ If two flat pencils which have a common centre, but do not lie in a common plane, are placed so that one ray in the one coincides with its corresponding ray in the other, then they are perspective, that is, the planes joining corresponding lines all pass through a line. §33. If two projective axial pencils are perspective, then the intersection of corresponding planes lie in a plane, and the plane common to the two pencils (in which the two axes lie) corresponds to itself. And conversely: If two projective axial pencils are placed in such a position that a plane in the one coincides with its corresponding plane, then the two pencils are perspective, that is, corresponding planes meet in lines which lie in a plane. The proof again is the same as in § 30. $34. These theorems relating to perspective position become illusory if the projective rows of pencils have a common base. We then have In two projective rows on the same line-and also in two projective and concentric flat pencils in the same plane, or in two projective axial pencils with a common axis-every element in the one coincides with its corresponding element in the other as soon as three elements in the one coincide with their corresponding elements in the other. K B' FIG. 9. element in the other. A similar construction holds in the other cases of perspective figures. On this depends the solution of the following general problem. $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 three points in a row s, A, B, C' the corresponding points in a projective row s', both being in a plane; it is required to find for any point D in s the corresponding point D' in s. Problem II.-Let a, b, c be three rays in a pencil S, a', b', c' the corresponding rays in a projective pencil S', both being in the same plane; it is required to find for any ray d 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 B of SB and S'B' to the intersection C1 of SC and S'C' by the line si Then a row on si 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 Di, where the line SD cuts Proof (in case of two rows).-Between four elements A, B, C, D and their corresponding elements A, B, C, D' exists the relations, and to draw S'D1; (ABCD)=(A'B'C'D'). If now A', B', C' coincide respectively with A, B, C, we get (AB, CD) (AB, CD'), hence D and 'D' coincide. The last theorem may also be stated thus: In two projective rows or pencils, which have a common base but are not identical, not more than two elements in the one can coincide with their corresponding elements in the other. Thus two projective rows on the same line cannot have more than two pairs of coincident points unless every point coincides with its corresponding point. It is easy to construct two projective rows on the same line, which have two pairs of corresponding points coincident. Let the points A, B, C as points belonging to the one row correspond to A, the point where this line Proof-The rows and s' are both perspective to the row s1, hence they are projective to one another. To A, B, C, D ons correspond A, B, C, D, on s1, and FIG. 11. S DA 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 S be the point where the line by which joins B and B', cuts the line, which joins C and C'. Then a pencil S 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 S to D' on and join D' to S'. This will be the required ray. a Proof-That the pencil S, is perspective to S and also to S' follows from construction. To the lines a, b, c, di in S, 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 § 29. This gives rise to a number of theorems, into which, however, we shall not enter. The same remarks hold for the second problem. FIG. 12. § 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 S, be the centre of projection of rows b and c, S2 c and a, 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'Í'=k to its algebraic equivalent, we derive an equation of the form axx'+8x+yx' +8=0. Conversely, if a relation of this nature holds, then points corresponding to solutions in x, x' form homographic ranges.) $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 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. § 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 pin and if the line SS, cuts a in A,, and b in B., and c in C, then A, B1 two points H and K. The will be corresponding points two pairs of rays, h, k, and in a and b, both corresponding, k', joining these points to to C in c. But a and b are S and S' will be pairs of perspective, therefore the line corresponding rays at right The construction AB, that is SS2, joining angles. corresponding points must gives in general but one pass through the centre of circle, but if the line is projection S, of a and b. In the perpendicular bisector other words, Si, S. Sa lie in a of SS', there exists an inline. This is Desargues' cele- finite number, and to every brated theorem if we state it right angle in the one pencil corresponds a right angle in the 18 B S$ 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. FIG. 13. 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. 838. 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 or But. by § 17. (AB, JI) = (A'B', J'I'), therefore the last that is to say AI/IBA'J'J'B'=-1; equation changes into AJ. A'I' = BJ. B'I', § 41. It has been stated in § 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 state 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. $42. As other examples we have the following: To a row " 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 any four points in space, and if the lines AB and CD meet, then all four points lie in a plane, hene also AC and BD, as well as AD and BC, meet. Theorem.-If of any number of whilst all do not · lie in a point, then all lie in a plane. Theorem. If a, B, Y. & are four planes in space, and if the lines as and 8 meet, then all four planes lie in a point (pencil), hence also ay and B8, as well as að and By, meet. lines every one meets every other, lie in a plane, then all lie in a point (pencil). § 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 lines in a plane will be called an in a plane will be called an n-side. n-point. A figure consisting of n lines in a pencil will be called an n-edge. A figure consisting of n planes in a pencil will be called an n-flat. 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 which two and two are opposite, and three diagonal points, which are intersections of opposite sides. A four-flat has six edges, of which two and two are opposite, and three diagonal planes, which pass through opposite edges. A four-side has six vertices, of which two and two are opposite, and three diagonals, which join opposite vertices, A four-edge has six faces, of which two and two are opposite, and three diagonal edges, which are intersections of opposite faces. 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. FIG. 15. § 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.- two projective flat pencils lie in a plane, but are neither in perspective nor concentric, then the locus of intersections of corresponding rays is a curve of the second order, that is, no line contains more than two points of the Jocus. Theorem.-If two projective rows lie in a plane, but are neither in perspective nor on a common base, then the envelope of lines joining corresponding points is a curve of the second class, that is, through no point pass more than two of the enveloping lines. Proof. We take any point T and join it to all points in cach row. This gives two concentrie pencils, which are projective because the rows are projective. If a line joining corresponding points in the two rows passes through T, it will be a line in the one pencil which coincides with its corresponding line in the other. But two projective concentric flat pencils in the same plane cannot have more than two lines of one coincident with their corresponding line in the other (834). Proof-We draw any line t. This cuts each of the pencils in a row, so that we have on two rows, and these are projective because the pencils are projective. If corresponding rays of the two pencil's meet on the line 1, their intersection will be a point in the one row which coincides with its corresponding point in the other. But two projective rows on the same base cannot have more than two points of one coincident with their corresponding points in the other (834). 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 cae theorem only, the reader being recommended to go through the reciprocal proof by himself, and to supply the reciprocal theorems when not given. $46. We state the theorems in the pencil reciprocal to the last, without proving them: Theorem.-If two projective flat pencils are concentric, but are neither perspective por coplanar, then the envelope of the planes joining corresponding rays is a cone of the second class; that is, no line through the common centre contains more than two of the enveloping planes, § 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 in space to the points on a curve A four-side is usually called a complete quadrilateral, and a four-of the second order form a cone point a complete quadrangle. The above notation, however, seems better adapted for the statement of reciprocal propositions. § 44 If a point moves in a plane it describes a plane curve. If a line moves in a plane it envelopes a plane curve (fig. 15). If a line moves in a pencil it describes a cone. If a plane moves in a pencil it envelopes 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. of Ispongia i hi aning acaba anil & to lienbo wot od 07 If we determine in S. (fig. 16) the ray corresponding to the ray SS, in S2, we get the tangent at S. Similarly, we can determine the point of contact of the tangents or us in fig. 17. § 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 todo mesmo conomole love first as eonit bone tot oke ow sale si tnt lionog od 16 mes ni Damsbout air synd og 101 long do olio on lo sg nimms15b bril sto lo asmols Mangagoon as hy bed on overly ban asui bodo sdy to sary to paintab asndly now so i hod bas Sa ogelog & mont in bal olie-jent booterobr 9230797&bus dignet sind big of sitesbig lo Stay in a bes no FIG. 17 linda lo lls big end tamnol erizo postih teliutia A od litt 199m sobie 91 10 os! as bn als ulo nowad 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 trgents devirmine one curve of the second class. Six points taken at random will therefore not lie on a curve of the second 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 $ 49. fig. 16. If we consider the conic determined by the five points A, Si, S2, K, L, then the point D will be on the curve if, s. and only if, the points on D1, S, Da be in a line. required ray de which cuts d, at D where this line cuts u from S, if we take AKS,DSL (figs. 16 D some point D on the curve.laub to This will give the required and 18) as a hexagon inscribed Pascal's Theorem. If a hexagon be inscribed in a curve of the second order, then the intersections of opposite sides are three points in a line. Brianchon's Theorem.-If a hexagon be circumscribed about a curve of the second class, then the lines joining opposite vertices 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 Si, S, any two other points on the curve as centres of projective pencils. Every pentagon inscribed in a curve of second order has the property that the intersections of two pairs of non-consecutive sides lie in a line with the point where the fifth side cuts the tangent at the opposite vertex. This enables us also to solve the Given five points on a curve of second order to construct the tangent at any one of them. $52. We know that the curve depends only upon the correspondence between the pencils S, and S, 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, ur 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 D, a row on SD perspective to the pencil L. At the same time AK describes a pencil about K and D2 A 1. 15,45 a row perspective to it on SD. But by Pascal's theorem D, and De 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 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 order can be drawn which passes through five given points. Only one curve of the second class can be drawn which touches. 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 order can be drawn, of which four points and the tangent at one of them, or three points and the tangents at two of them, are given.. Only one curve of the second class can be drawn, of which four tangents and the point of contact at one of them, or three tangents and the points of contact at two of them, are given. $53. At the same time it has been proved: If all points on a curve of the second order be joined to any two of them, then the two pencils thus formed are projective, those rays being corresponding which meet on the curve. Hence The cross-ratio of four rays joining a point S on a curve of second order to four fixed points A, B, C, D in the curve is independent of the position of S, and is called the cross-ratio of the four points A, B, C, D. If this cross-ratio equals-1 the four points are said to be four harmonic points. All tangents to a curve of second class are cut by any two of them in projective rows, those being corresponding points which lie on the same tangent. Hence The cross-ratio of the four points in which any tangent u is cut by four fixed tangents a, b, c, d is independent of the position of u, and is called the cross-ratio of the four tangents a, b, c, d. If this cross-ratio equals-1 the four tangents are said to be 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 every point one tangent. A curve of second class has on 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 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 23 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 I coincide, or the line 6 1 is the tangent at I. 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 circumscribed about a curve of the second class has the property that the lines which join two pairs of nonconsecutive vertices meet on that line which joins the fifth vertex to the point of contact of the opposite side. following problems. Given five tangents to a curve of second class to construct the point of contact of any one of them. $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, |