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26 Mr. A. De Morgan on the Conic Octagram. [April 25, of Them, That 8 Is the Tangential of 4

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26 Mr. A. De Morgan on the Conic Octagram. [April 25, of Them, That 8 Is the Tangential of 4 26 Mr. A. De Morgan on the Conic Octagram. [April 25, of them, that 8 is the tangential of 4. We have here the theorem that the third point of 1 and 7 is also the tangential of 4. Similarly, 10, 11, 13 are each of them (like 8) determined by two constructions ; 14, 16, 17, 19, each of them by three constructions, and so on; the number of constructions increasing by unity for each group of four numbers. And the theorem is, that these constructions, 2, 3 or more, as the case may be, give always one and the same point. Prof. Cayley mentioned that on a large figure of a cubic curve he had, in accordance with the theorem, constructed the series of points 1, 2, 4, 5, 7, 8, 10, 11, 13, 14. Prof. Sylvester made a communication " On the Theory of Re- siduals." Mr. Walker gave a proof of the Rectangle of Forces ; and demon- sti-afcion of tbe principle of the Parallelogram of Forces. These are new and interesting specimens of a class of proofs on the character of which mathematical opinion has been much divided. The interest of these proofs now turns on the question, how much and what there is of a physical character in the assumptions on which they are founded: in fact, what there is to prevent their being purely geometrical, and having a conclusion time of mathematical necessity. April Ihth, 18G7. Prof. SYLVESTER, President, in the Chair. M. Chasles was elected a Foreign Member of the Society. Prof. Clerk-Maxwell, F.R.S., Dr. Robertson, F.R.S.E., and Rev. Thos. Smith, M.A., were elected Members. The following paper was read :— On the Conic Octagram. By A. DE MORGAN, Esq., Vico-President. The most general theorem I require is the particular case I now state. If of the 1\? intersections of two n-ic curves n («—2) —^(w—3) (/i—4) lie on an («—2)ic, so also do j(n—8) (n—4) more of the inter- sections, which call adjuncts: and the remaining 2n intersections lie on a conic. When n = 3, and the cubics are ti"ilinear, we have Pascal's hexagram: we may retain this word as denoting, not a hexagon, but a method of drawing or completing a hexagon. Each value of n, if the «-ics be multilinear, gives a direct and inverse 2>i-gram : direct, when by an (yi — 2)ic taken at pleasure we deter- mine 2?i points of some conic of which nothing is given beforehand ; inverse, when, having given five determining points of a conic, wo 1867.] Mr. A. DG Morgan on the Oonic Octagram. 27 determine 2n—5 points more by an inverse construction which detcr- • mines a convenient (n—2)ic. When n = 3 or w = 4, there are no adjnnct intersections; and the inverse 2«-gram is the plainest inversion of a simple construction. But when « = 5 or upwards, the inverse 2?i-gram, though practicable, requires the introduction of an entirely new element. When n = 5, for example, and there is one adjuuet, the direct construction makes 14 intersections put the 15th on its proper place ; but, in the invei'se construction, the cubic cannot be discovered without finding it so as to satisfy the condition of the 15th intersection. Consequently, there are no other theorems of the family of Pascal's hexagram except only the octagram. I shall state the methods for: 1. the hexagram; 2. the double trigram, which is another form of the hexagram ; 3. the octagram; 4. the doable tetragram, which is not another form of the octagram ; 5. the direct form of the decagram. I signify lines by 1, 2, 3. &c.; points of intei-section by 12, 23, &c.; a line drawn through two points by 12 | 34, &c. And when A, B ai"e points on a line 3, A3B signifies as much of 3 as is between A and B ; and so on. 1. Hexagram. Direct: if the hexagon* 142536 have its alternate points (14, 25, 30) colliuear, then 123456 is in a conic; and the con- verse. Inverse: if 12345G be a conic hexagon, of which are known 61, 12, 23, 34, 45, and tho direction of 5, then (14 | 25 . 3) | 61 meets 5 in 56, the sixth point. After the data are laid down, two new lines determine one point. 2. Double trigram. This is a modification of the hexagi'am. which depends on Desargues's coaxal triangles. These triangles form a conic hexagon; that is, if 123, 456 be such triangles, 12, 23, 34, 45, 56, 61 aref co-conic. The inverse theorem is only worth mention for analogy : the construction is niore complex than the hexagram. Let 123 be an inscribed triangle, and 456 another, of which 45, 40, and the direction of 5, are given. From the intersection of 12 | 45 and 13 | 46 draw a line 7 to the intersection of 23 | 46 and 5. Then 72 | 13 meets 5 in the required point 56. The coaxal triangles are 45 (23 | 46) and 12 (72 | 13). And 7 is the Pascalian line of the hexagon 13, 40, 32, 21, 45, 56. 3. Octagram. Direct: choose a conic (usually a pair of straight * This presentation of the second hexagon was actually suggested to mo by observing that lsluisc Pascal lias two hrxagnuns, and the jocose inference that there ought to bu two hexagons in this theomn. My own names arB both oetagi'ams; but, though I bow ln-foro the coincidence, 1 have no suggestion to acknowledge. t Desargues was the teacher of Pascal, who expresses his obligation in terms of unusual strength. It is curtain, by tho date of printed satire, that tho theorem called la l'asai/c was known by that name bifore Pascal was twenty years old. I feel sure that Pascal's active genius was led by the coaxal triangles to tho hoxagram. 28 Mr. A. De Morgan on the Conic Octagram. [April 25, lines) as guide, and inscribe an octagon of which the successive chorda are numbered by procession by 5, rejecting 8; as, 16305274. Then 01, 12, 23, 34, 45, 56, 67, 70 are co-conic; or, 01234567 is an oc tagon inscribed in a conic. Inverse : in the conic octagon 01234567 are given 1, 2, 3, 4, and the directions of 0 and 5: required 56, 67, 70. Find 14, 50, 25, 30, 14 | 50, 25 | 30, 14 | 50 . 2, 14 | 50 . 3, 25 | 30 . 1, 25 | 30 . 4. Let 6 and 7 be 14 | 50 . 3 125 | 30 . 1 and 14 | 50 . 2 | 25 | 30 . 4. Then 56, 67, 70 will be on the conic, and 01234567 is an inscribed octagon. The guide is the system 14 | 50, 25 J 30 ; and three points are determined by four lines over and above data. By varying the directions of 0 and 5, and drawing the corresponding parts of all the constructions together, the laying down of a large number of points is very easy and rapid woi'k. 4. Double tetragram. When I explained the octagram to the Society, Mr. Cotterill suggested a known theorem which seemed likely <o be the direct octagram; but, on looking into it, I found it to be the direct double tetragram, which it suggested in another form. In a conic (usually two straight lines) inscribe two tetragrams 1630 and 5274; then 1234, 5670 are tetragons in another conic. Inversely: let the tetragon 1234 be given; and in 5670 the point 50 and the directions of 5 and 0. Find 51, 53, 02, 04, 51 | 02, 53 | 04; these lines are the guide. Let 6 and 7 be 53 | 04 . 2 | 51 | 02 . 4 and 53 | 04 . 11 51 | 02 . 3. Then 56, 67, 70 will bo on the conic, and 5670 is a second tetragon. Owing to our better command of language for a tetragon than for a hexagon, we may describe this last construction as follows:—Let one pair of opposite sides be bases, and the other pair sides; points in their lines being base-points and side-points. Given a tetragon, and a fifth point of the conic in which two lines meet, take the base-points in which the fifth line, and the side-points in which the sixth line cuts the tetragon. Join these two and two, base-point to side-point; in each of the joining lines there is a new base-point and a new side- point. Join these last, base-point to base-point, and side-point to Bide-point; the two lines so drawn complete the second tetragon. 5. Direct decagram. Take any three straight lines P, p, c; any curve of the third order would do as well. Capital letters are points on P ; small letters on p. Describe the decagon A0alB263C4c5D6d7E8e9A in the following way. Take A0a, olB, B26, fe3C, C4c at pleasure, and let the points in which 0, 1, 2, 3, 4, thus laid down, meet rs be 05,16, 27, 38,49. Then draw c5D through c and 05, meeting P in D, &c, as from the above description. It will at last be found that 9, as obtained from e and 49, passes through A.; or the adjunct falls as 1867.] Mr. Jenkins on an Arithmetical Theorem. 29 required by the theorem. This done, 03, 36, 69, 92, 25, 58, 81, 14, 47, 70, are co-conic; or 0369258147 is a conic decagon. Inci- dentally, we have solved the question of laying down fifteen points that shall lie five and five on three lines, and in two ways three and three on five lines.
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