CoNIcs.] mine one curve of the second order, and reciprocally, that five tan- gents determine one curve of the second class. Six points taken at random will therefore not lie on a curve of the second order. 111 order that this may be the case a certain condition has to be_sat1_s- tied, and this condition is easily obtained from the construction in § 49, fig. 15. lf vc consider the conic determined by the five points A, S1, S._,, K, L, then the point D will be on th: curve if, and only if, the points oi1 D1, S, D2 be in a lin -. This may be stated differ- ently if w'e take AI{S1DS2L jligs. 15 and 17) as a hexagon inscribed in the conic, then Ali and DS__, will be opposite sides. so will be KS1 and S,L, as Well as S11) aml LA. The tirst two meet in D2, the others in S and D1 respectively. Ve may therefore state the required condition, together with the re- GEOMETRY 397 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 form- ing a hexagon, which, by the way, may be done in 60 (lifi'erent ways. It will be seen that 1 2 and (leaving out 3) 4 5 are opposite sides, so are 2 3 and (leaving out 4) 5 6, and also 3 4 and (leaving out 5) 6 1. If the points 1 2 3 4 5 are given, and we want a 6th point on a line drawn through 1, we know all the sides of th-- hexagon with the exception of 5 6, and this is found by Pascal’s theorem. If this line should happen to pass through 1, then 6 and 1 coin- cide, or the line 6 1 is the tangent at 1. And always if two con- secutive vertices of‘ the hexagon approach nearer and nearer, then the side joining them will ultimately become a tangent. ' 'c 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 :— Tlworcm.——Every pentagon in- Theorcm.—Every pentagon cir- eipr-_»eal one, as follows :—- I’:isral's ’[‘h~orcm.—If a hexagon be inscribed in a curve of the second 0l'tlcl', then the intersec-tioi_is of oppo- site sides are three points in a line. Bria nchon's Th eorcm.—If ahexagon be circumscribed about 3. curve of the second class, then the lines joiii- ing opposite vertices are three lines meeting in a point. These celebrated theorems, which are known by the names of thn_-.ir diseoverers, are perhaps the most fruitful in the whole theory of eonics. Before we go over to their applications we have to show that we obtain the same curve if we take, instead of S1, S2, any two other points on the curve as centres of projective pencils. § 52. life know that the curve depends only upon the corre- spondence bctwccn the pencils S1 and S2, and not upon the special construction used for finding new points on the curve. The point A (lig. 15 or fig. 17), through which the two auxiliary rows 'M1, N2 were drawn, may therefore be changed to any other point on the curve. Let us now suppose the cui've drawn, and keep the points S1, S._,, K, L, and D, and hence also the point S fixed, whilst we movc A along the curve. Then the line AL will describe a pencil about I. as centre, and the point D1 a row on S1D perspective to the pencil L. At the same time AK describes a pencil about K and D2 a row perspective to it on S,D. But by Pascal’s Theorem D1 and D, will always lie in a line with S, so that the rows described by D1 aml l)._, 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 gin-11 points we take as centres of projective pencils. Hence— TIu:orcm.—Only one curve of the second order can be drawn which passes through five given points. TIieorem.—Only one curve of the second class can be drawn which touches five given lines. scribed in a curve of second order has the property that the intersec- tions 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. Problem./Sliven five points on a curve of second order to construct the tangent at any one of them. cuinscribed about a curve of the second class has the property that the lines which join two pairs of non-consecutive vertices meet on that line which joins the fifth vertex to the point of contact of the opposite side. This enables us also to solve the following problems. ProbIem.—-Given five tangents to a curve of second class to construct the point of contact of any one of them. 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. Siinilarly, if on the curve of second class a tangent and its point of If twice two adjacent vertices coincide, the hexagon becomes a These we take to be contact are given, this will be equivalent to two given tangents. Ve may therefore extend the last theorem :- Tlicm'cm.—Oiily 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. Thcorem.—0nly one curve of the second class can be drawn, of which four tangents and the point of con- tact 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 :— Th.-’orcm1.——If all points on a curve of the second order be joined to any two of them, then the two pencils thus formed are projec- tive, those rays being correspond- iiig 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 is = — 1, the four points are said to be four harmonic points. The0rem.—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— Tlie cross-ratio of the four points in which any tangent u is cut by four fixed tangents a, b, c. d is in- dependent 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— Theorem.—A curve of second order has at every point one tangent. Theorem-.—A curve of second class pas on every tangent a point of con- act. § §4. _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. quaclrilateral, with tangents at two vertices. opposite, and get the following theorems :— Theorem..—If a quadrilateral be inscribed in a curve of second order, the intersections of opposite sides, and also the intersections of the tangents at opposite vertices, lie in a line (fig. 18). Thc(rrcm—.—If a quadrilateral be circumscribed about a cm-ve of second class, the lines joining opposite vertices, and also the lines joining points of contact of opposite sides. meet in a point Fig. 19. If we consider the hexagon made up of a triangle and the tangents at its vertices, we get—— Theorem-.——If a triangle is inscribed in a curve of second order, the points in which the sides are cut by the tangents at the opposite vertices meet in a point. Theorem.—If a triangle be circum- scribed about a curve of second class, the lines which join the vertices to the points of contact of the opposite
sides meet in a point (flg. 19).