of the angular points of the first triangle will be the sides of the reciprocal. Those of the fixed straight lines, along which two of the angular points of the first triangle move, will be fixed points through which two of the sides of the reciprocal triangle pass. Those of the three points, lying in the same straight line, through which the sides of the given triangle always pass, will be three straight lines, intersecting in a point, along which the angular points of the reciprocal triangle always move. Hence the data of the reciprocal proposition will be “ Two of the sides of a triangle pass each through a fixed point, and each angular point moves along a fixed straight line, the three straight lines passing through the same point.” In the given theorem, the thing to be proved relates to the motion of the third angular point. To this will correspond the third side of the reciprocal triangle. To the straight line, passing through the intersection of the two given straight lines, along which the third angular point may be shewn to move, corresponds a point lying in the same straight line with the two given points, and through this the third side will always pass. Hence, under the circumstances stated above as data of the reciprocal theorem, “the third side will pass through a fixed point lying in the straight line joining the two fixed points, through which the first sides pass*. * The given theorem may be expressed, by the aid of letters, as follows: Let PQR be the given triangle, and let its angular point Q move along à fixed straight line 0X, its angular point R along a fixed straight line OY. Also, let the straight line QR always pass through a fixed point F, RP through a fixed point G, PQ through a fixed point H, the three points F, G, H lying in the same straight line. Then the given theorem tells us that the point P will always move along a fixed straight line, passing through 0. Now let the whole figure be reciprocated with respect to any conic section. Let the line which is the polar of any point be denoted by accenting the same single letter by which the point is denoted in the original figure; the polar of P, for example, being denoted by P'. Then the point of intersection of the lines P', Q', will be denoted by the two letters P'Q', and this will be the pole of the line PQ. We have then a triangle of which the sides are P', Q', R', the side Q' always passing through a fixed point OʻX', the side R' through a fixed point O'Y'. Also the angular point Q'R' always moves along a fixed Again, turn to Example 4, on page 57, and let us investigate the reciprocal theorem. The three conics touching respectively each pair of the sides of a triangle at the angular points where they meet the third side, will reciprocate into three conics passing respectively through each pair of the angular points of a triangle, and touching the lines joining them with the third angular point,” that is, the sides” of the triangle themselves. This condition, therefore, reciprocates into itself. The condition “all intersecting in a point" reciprocates into “all touching a straight line.” Hence the data are, “Three conics are drawn, touching respectively each pair of the angular points of the sides of a triangle at the points where they meet the third side, and all touching a straight line." In the matter to be proved, we may first enquire what are the reciprocals of “the sides of the triangle which intersect” (that is, which do not touch)" their respective conics." These will be “the angular points of the triangle not lying on their respective conics." The three tangents at their common point will reciprocate into “ the three points of contact of their common tangents.” And the meeting of the tangents with the sides will reciprocate into the lines joining the points of contact with the angular points. Hence the first thing to be proved is, “That the three straight lines joining the points of contact of the common tangent with the angular points of the triangle not lying on the respective conics all pass through a point.” Again, “ the other common tangents to each pair of conics” reciprocate into the other points of intersection of each pair of conics,” and “the sides of the triangle which straight line F", the point R'P' along a fixed straight line Gʻ, the point P'Q' along a fixed straight line H', the three straight lines F", Gʻ, H' passing through the same point. Then the reciprocal theorem is that the side P' will always pass through a fixed point lying in the line Q'. The student will find the above mode of transformation, in which a straight line is denoted by a single letter, and a point by the pair of letters representing any two straight lines which intersect in it, a useful mode of familiarizing himself with the method of reciprocal polars. touch the several pairs of conics" into the angular points of the triangle “common to the several pairs of conics. " Hence the latter part of the theorem will run: “ And that the same three straight lines respectively join the other point of intersection of each pair of conics with the angular point of the triangle common to each pair.” 11. After a little practice, the process of reciprocating a given theorem will be found to consist simply in writing straight line" for “point," "join” for “intersect,” “ locus for "envelope,” &c., and vice versa. The word “conic" will of course remain unaltered. 12. Brianchon's Theorem. By reciprocating Pascal's Theorem (given in Art. 12, Chap. III.), we obtain Brianchon's Theorem, which asserts that "If a hexagon be described about a conic section, the three diagonals will intersect in a point.” The student will find it useful to transform, by the method of reciprocal polars, the special cases of Pascal's Theorem, given in Art. 13, Chap. III.; and to obtain a geometrical construction by which when five tangents to a conic are given, their points of contact may be found. 13. The anharmonic ratio of the pencil formed by four intersecting straight lines is the same as that of the range formed by their poles. This may be proved as follows. Let OP, OQ, OR, OS be the four straight lines, P, Q, R', S' their poles, which will lie in a straight line, the polar of 0; let P, Q, R, S be the points in which the pencil is cut by the transversal P'Q'R'S'. Let this transversal cut the conic in K, K. Bisect K, K, in V. Then, since PK, PK, is divided harmonically in P, K,, P, K, (Art. 21, Chap. II.), it follows that PK,. P'K,= PK,.P'K,, TORS Fig. 18. whence (VP- VK)(VK,+ VP) = (VP+ VK) (VK- VP'), which, since VK, = VK,, reduces to VP.VP = VK* which, by similar reasoning, = VQ.VQ = VR.VR' = VS.VS'. Hence the eight points P, Q, R, S, P, Q, R, S are in an involution, of which K,, K, are the foci, and therefore (Art. 26, Chap. I.) {0. PQRS} =[P'QR'S']. 14. In Art. 13, Chap. I. we saw that the condition that the three points (?, m, n,), (lz, mg, n,), (13, mg, ng) shall lie in the same straight line is identical with the condition that the three straight lines (l, m, n,), (lz, m, n,), ('s, mag.ng) shall intersect in the same point. Now these several points and lines are respectively the poles and polars of each other, with respect to the imaginary conic a + Be + y = 0. Thus the theory of reciprocal polars explains the fact that the condition for three points lying in a straight line is identical with that for three straight lines intersecting in a point. It also explains the identity of conditions noticed in Chap. II. Arts. 7 and 9. For the reciprocal of the conic apa? + *ß2 + vʻy? – 2uvby - Avaya - 21uaß = 0...... (1), with respect to a* + ß2 + y = 0, will be found to be XBy + Mya + vaß = 0......... .(2). And the pole of (f, g, h) is fa +93 + hry=0. Hence if the line fa +gB+ hy=0 touch (1), the point (f, g, h) lies in (2), giving for the condition of tangency And if the line fa + 9B + hy = 0 touch (2), the point (f, g, h) lies in (2), giving for the condition of tangency in that case 125+ M*g* + vh? – 2uvgh – 2v1hf – 2ufg = 0. These conditions of tangency are identical with those already investigated. Again, every parabola touches the line at infinity. Now the co-ordinates of the pole of this line are proportional to Hence, if the conic, represented by the general equation of the second degree, be a parabola, the point (a, b, c) must lie in the reciprocal conic. This gives, as the condition for a parabola, Ua? + V6?+ WC + 2 U'bc + 2 V'ca + 2W'ab = 0, the same as that already investigated. a, b, c. |