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CORRESPONDENCE
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CORRESPONDENCE . HOMOGRAPHIC AND See also:PERSPECTIVE RANGES § 25. Two rows, p and p', which are one the See also:projection of the other (as in fig. 5), stand in a definite relation to each other, characterized by the following properties. 1. To each point in either corresponds one point in the other; that is, those points are said to correspond which are projections of one another. 2. The See also:cross-ratio of any four points in one equals that of the corresponding points in the other. 3. The lines joining corresponding points all pass through the same point. If we suppose corresponding points marked, and the rows brought into any other position, then the lines joining corresponding points will no longer meet in a See also:common point, and hence the third of the above properties will not hold any longer; but we have still a correspondence between the points in the two rows possessing the first two properties. Such a correspondence has been called a one-one correspondence, whilst the two rows between which such correspondence has been established are said to be projective or homographic. Two rows which are each the projection of the other are therefore projective. We shall presently see, also, that any two projective rows may always be placed in such a position that one appears as the projection of the other. If they are in such a position the rows are said to be in perspective position, or simply to be in perspective. § 26. The notion of a one-one correspondence between rows may be extended to See also:flat and axial pencils, viz. a flat See also:pencil will be said to be projective to a flat pencil if to each See also:ray in the first corresponds one ray in the second, and if the cross-ratio of four rays in one equals that of the corresponding rays in the second. Similarly an axial pencil may be projective to an axial pencil. But a flat pencil may also be projective to an axial pencil, or either pencil may be projective to a See also:row. The See also:definition is the same in each See also:case: there is a one-one correspondence between the elements, and four elements have the same cross-ratio as the corresponding ones. § 27. There is also in each case a See also:special position which is called perspective, viz. t. Two projective rows are perspective if they See also:lie in the same See also:plane, and if the one row is a projection of the other. 2. Two projective flat pencils are perspective—(I) if they lie in the same plane, and have a row as a common See also:section; (2) if they lie in the same pencil (in space), and are both sections of the same axial pencil; (3) if they are in space and have a row as common section, or are both sections of the same axial pencil, one of the conditions involving the other. 3. Two projective axial pencils, if their axes meet, and if they have a flat pencil as a common section. 4. A row and a projective flat pencil, if the row is a section of the pencil, each point lying in its corresponding See also:line. 5. A row and a projective axial pencil, if the row is a section of the pencil, each point lying in its corresponding line. 6. A flat and a projective axial pencil, if the former is a section of the other, each ray lying in its corresponding plane. That in each case the correspondence established by the position indicated is such as has been called projective follows at once from the definition. It is not so evident that the perspective position may always be obtained. We shall show in § 3o this for the first three cases. First, however, we shall give a few theorems which relate to the See also:general correspondence, not to the perspective position. § 28. Two rows or pencils, flat or axial, which are projective to a third are projective to each other; this follows at once from the See also:definitions. § 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 See also:fourth point D in the first will correspond a point D' in the second, so that (AB, CD) = (A'B', C'D'). But there is only one point, D', which makes the cross-ratio (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 corresponding 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, viz. 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 See also:form a flat pencil. 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 hypo-thesis it is equal to (A'B', C'D'). Hence (A'B', C'D") = (A'B', C 'I)'), that is, D" is the same point as D'. § 31. 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 See also: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 inter-section 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 See also:relating to perspective position become illusory if the projective rows of pencils have a common See also: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 See also: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. Proof (in case of two rows).—Between four elements A, B, C, D and their corresponding elements A', B', C', D' exists the relation (See also: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, B, and C' as points in the second. Then A and B coincide with their corresponding points, but C does not. It is, however, not necessary that two such rows have twice a point coincident with its corresponding point; it is possible that this hap-pens only once or not at all. Of this we shall see examples later. § 35. If two projective rows or pencils are in perspective position, we know at once which element in one corresponds to any given element in the other. If p and q (fig. 9) are two projective rows, so that K corresponds to itself, and if we know that to A and B in p correspond A' and B' in q, then the point S, where AA' meets BB', is the centre of projection, and hence, in See also:order to find the point C' corresponding to C, we have only to join C to S; the point C', where this line cuts q, is the point required. If two flat pencils, S1 and S2, in a plane are perspective (fig. Io), we need only to know two pairs, a, a' and b, b', of corresponding rays in order to find the See also:axis s of projection. This being known, a ray c' in S2, corresponding to a given , ray c in SI, is found by joining S2 to the point where c cuts the axis s. A similar construction holds in the other cases of perspective figures. On this depends the See also: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 Problem II.—Let a, b, c be three points in a row s, A', B', C' three rays in a pencil S, a', b', c' the corresponding points in a the corresponding rays in a See also:pro-projective row s', both being in a jective pencil S', both being in plane; it is required to find for the same plane; it is required to any point D in s the correspond- find for any ray d in S the corre-. See also:ing point D' in s'. sponding ray d' in S'. The solution is made to depend on the construction of an See also: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. II), take any two points, S and S', as centres of auxiliary pencils. Join the intersection Bl of SB and S'B' to the intersection C1 of SC and S'C' by the line sl. 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 si, and to draw S'DI ; the point where this line cuts s' will be the regeired point D', Proof.—The rows s and s' are both perspective to the row si, hence they are projective to one another. To A, B, C, D on s correspond AI, B1, CI; D1 on sI, and to these correspond A', B', C', D' on s'; so that D and D' are corresponding points as required. Ai FIG. II. 694 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 SI be the point where the line b1, which joins B and B', cuts the line cl, which joins C and C'. Then a pencil Si 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 SI to D' on s' and join D' to S'. This will be the required ray. Proof.—That the pencil SI is perspective to S and also to S' follows from construction. To the lines al, b1, cI, d1 in SI 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 See also: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. § 37. Homological Triangles.—As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem. Theorem.--If See also: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 See also: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', Son 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 SI be the centre of projection of rows b and c, S2 c and a, S3 a and b, and if the line See also:S1S2 cuts a in A1, and b in BI, and c in Cl, then Al, B1 will be corresponding points in a and b, both corresponding to CI in c. But a and b are perspective, therefore the line A1BI, that is SIS2, joining corresponding points must pass through the centre of projection S3 of a and b. In other words, SI, S2, S3 lie in a line. This is Desargues' celebrated theorem if we See also:state it thus: Theorem of Desargues.—If each of two triangles has one vertex on each of three See also:con-current lines, then the inter-sections 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. § 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 ro_w, then (AB, JI) = (A'B', J'I'), say or AJ/JB : AI/IB =A'J'/J'B' : AT/IT'. But, by § 17, AI/IB = A'J'/7'B' =–I ; therefore the last See also:equation changes into AJ . AT '=BJ . B'I', that is to say[PROJECTIVE 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 See also:constant, viz. AJ . A'I'=k. See also:Steiner has called this number k the See also: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 0, 0', on the ranges and reduce the expression AJ . A'I'=k to its algebraic See also:equivalent, we derive an equation of the form axx'-r/3x+yx' d S =o. 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 (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 See also: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 See also:series of See also: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 See also:place the pencils in perspective position (fig. 14) by making one ray See also:coin- cident with its corresponding ray. Corresponding rays meet then on a line p. And now we draw the circle which has its centre 0 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 airs 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 in- FIG. 14• finite number, and to every right See also:angle in the one pencil corresponds a right angle in the other. Additional information and CommentsThere are no comments yet for this article.
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