I. Solution by J. C. MALET, M.A.; E. B. ELLIOTT, M.A.; and others. Let P, P be two positions of the moving particle, and Q, Q, the corresponding points on the curve enveloped by the direction of the acceleration ; let R be the intersection of the lines PQ, P1Q; then the difference of the moments of the velocities at P and P1 about R is equal to This expression must vanish in the limit, hence we have ds being the element of the curve QQ,. Hence, integrating, we get required expression for vp. II. Solution by R. F. DAVIS, B.A.; J. J. WALKER, M.A.; J. HAMMOND, M.A.; and others. Denoting the curves by (A), (B), it is at once evident that (B) is hodographic to (A). Consequently. if QS be drawn parallel to the tangent at P, there is a fixed point S on this line, the radii vectores from which to (B) represent in magnitude and direction the velocities of the particle at the points in which corresponding ⚫tangents to (B) meet (A). Lete, be the angles which the tangents at Q, P make with a fixed initial line through S; and SQ (which ∞ v) = = r. Then, if SY, QN be perpendiculars on the tangent at P, SY QN=p; hence dp аф Moreover = = p cot QPN (B) N (A) pdr 5531. (By CHRISTINE LADD.)-If four conics S, A, B, C, have one focus and one tangent, D, common to all, and if a tangent common to S and A intersects D on a directrix of A, a tangent common to S and B on a directrix of B, and a tangent common to S and C on a directrix of C, prove that the common tangents of A, B, C will meet in three points in a straight line. Solution by D. Lo. GATTO; V. JACOBINI; and others. Se si hanno quattro coniche, s, a, b, c, aventi una tangente comune m, ed un fuoco comune F, e si chiamano da, db, de le direttrici delle 3 coniche a, b, c, corrispondenti al fuoco F, e tɑ, të, të le tangenti comuni alle coniche 8, α, 8, b, s, c, se le terne di rette: mdata, mdыtь, mde te concorrono in un punto, concorreranno pure in un punto le tre tangenti di a, b, c. Prendiamo il noto teorema: Abbassando da un punto M sui lati aɔ, b2, c. di un triangolo le tre perpendicolari, se i piedi ta,, tb, te, di queste sono in linea retta, all rai vertici del triangolo si trovano sopra una circonferenza di cerchio passante per M. Trasformiamo questo teorema coll' inversione quadratica, scegliendo come centro di inversione il punto M. Allora il punto M, e le normali abbassate da M sui lati del triangolo non mutano; ai tre lati a2, b2, ca corrisponderanno nella figura inversa tre circonferenze di cerchio a1, b1, C1, passanti per il punto M, ed i tre centri da,, d1, dc, sono situati sulle rette Mta,, Mtb,, Mic,; ai punti ta, tb, tc corrisponderanno gli incontri ta,, t,, te, delle tre circonferenze a, b, c, colle 3 ultime rette; quelli stavano in linea retta, questi staranno sopra un' altra circonferenza 81 passante pure per M; ai vertici del triangolo corrisponderanno gli incontri delle tre circonferenze a, b, c1; quelli stavano sopra una circonferenza passante per M, questi staranno in linea retta. Il teorema trasformato è dunque il seguente. Se si hanno quattro circonferenze s1, a1, b1, c1 passanti per uno stesso punto M, e si prendono gli incontri ta,, t,, te, di s, con a1, si con b1, s1 con C1, se questi punti, ed i centri da,, do,, de, sono situati sopra rette uscenti da M. i 3 punti di incontro di a, b, ci sono in linea retta. Trasformiamo di nuovo questa figura, prendendo la curva polare reciproca rispetto ad un cerchio come conica fondamentale. Sappiamo che la conica polare reciproca di un cerchio rispetto ad un altro cerchio come conica fondamentale (vedi SALMON, A Treatise on Conic Sections) è una conica di cui un fuoco è il centro del cerchio fondamentale, e la direttrice la polare del centro del cerchio scelto. Ciò premesso, prendiamo la figura polare reciproca della precedente rispetto ad un cerchio di centro F. Avremo 4 coniche s, a, b, c, aventi un puoco comune F, ed una tangente comune m. Ai punti di incontro del cerchio s cogli a, b, c, corrisponderanno le altre tre rette ta, tu, te tangenti comuni alle coniche s, a, s, b, s, c; ai centri dei cerchi da,, db, de, corrisderanno le direttrici da, do, de delle coniche a, b, c : le tre terne di punti Mta, da, Mtb, db,, Mte, de, erano in linea retta; le tre terne di rette mtada, mto do, mtede passeranno per un punto. Ai punti di incontro dei tre cerchi a, b, c corrisponderanno le tre tangenti alle coniche a, b, c; quei punti erano in linea retta; quelle rette passeranno per un punto. Teorema che è appunto quello che si voleve dimostrare. [The theorem in the Question is the reciprocal of that given in Ex. 7 of Art. 104 of SALMON'S Conics, 5th ed.] 5519. (By S. ROBERTS, M.A.)--If S and T are the fundamental invariants, and H is the Hessian, of the cubic curve U = 0, prove that the twelve lines on which the inflexions lie in threes are represented by S-U4+TU3H-18SU2H2-27H1 = 0. I. Solution by J. J. WALKER, M.A. In the Quarterly Journal, No. 55, October, 1876, p. 243, I have shown that AU+H is the product of xyz, one of the four triads of lines referred to, if a is a root of the equation λ 27x++ 18Sλ2 + Tλ - S2 = 0. For the continued product where Ai or ... (^1U + H) (^ ̧U + H) (^ ̧U + H) (^ ̧U + H), A are the four roots of the above quartic, we have (^1^2^3^4) Ua + Σ (^1^¿^3) . U3H +Σ (^1^2) • U÷H2 + Σλ ̧ . UH3 + Ha, which, equated with zero, is the equation of the twelve lines in question. corresponding to the three cube roots w of unity. Hence, multiplying together the four functions corresponding to (b) and omitting constant multiples, we get for the set of twelve lines (m2U + H) {[(1 + 2m3) U −6mH]3 – 27 (m2H+H)3} = 0.........................(c). The coefficient of U4 is that of U3H is m2 (1 — m3)2 (1+8m3) = S2 (1+8m3), and so on. Dividing out 1 + 8m3, we get the expression given. (c) is also directly obtained by means of representing three lines containing the nine inflexions of the cubic. 5438. (By the EDITOR.)--If A, B, C, D are four points taken at random on the perimeter of a regular n-gon, find the respective probabilities that AB, CD will intersect (1) inside, (2) on, (3) outside the perimeter. 5506. (By the EDITOR.)—If A, B, C, D, E, F are six points taken at random in the perimeter of a regular n-gon, find the probability that the three intersections of (AB, CD), (CD, EF), (EF, AB) will all lie inside the perimeter. Solution by G. S. CARR. With each configuration of the 4 points there are 3 ways of drawing the pair of chords and therefore 3 intersections. If the lengths of the sides of the polygon be a, b, c ..., taken at first unequal, the total number of configurations will be represented by (a + b + c + ...)1. With each configuration there will be one intersection within the figure, excepting when more than 2 points lie on a side of the polygon, and the number of such unavailable cases will be given by those terms in the expansion of the above multinomial, which are of the forms a1 and a3b, that is, 4+ 4Σa3b. Putting a, b... each equal to unity, this number becomes n+4n(n−1), or 4n2-3n; therefore the number of interior intersections is n4-4 -4m2 + 3n, and the whole number of intersections is 3n4; hence the probability of one n3 - 4n+ 3 interior intersection is P1 3n3 = For an intersection outside the polygon, more than two points must again not lie on any one side. Therefore, by the last result, the number of exterior intersections would be 2 (n4-4n2 + 3n), two for each interior intersection. But 6n of these occur at the corners of the polygon, arising out of n configurations of two points on one side, and two on the adjacent side multiplied by the number of combinations of the four points, two and two. Therefore the probability of an exterior intersection is P3 = 2n3-8n 3n3 The remaining intersections may be considered to fall on the sides of the polygon, and therefore P2 = 12n-3 [These results agree with those obtained in Mr. LEUDESDORF'S solution of the problem; Reprint, Vol. XXVIII., pp. 106—108.] When six points instead of four are taken, there are 15 ways of drawing the 3 chords in each configuration of the six points, but in one only of these ways will the 3 points of intersection lie within the polygon—viz., when opposite points are joined. The total number of configurations, as in the previous question, will be represented by (a+b+c+ ...). The configurations not available will be those in which 4 or more points lie on the same side of the polygon. The number of these will be expressed by the sum of those terms in the expanded multinomial that involve a fourth or higher power, that is, Σa6 +6Σa5b + 15Σa+b2 + 30a4bc. Putting a, b, c, each =1, this gives, for the number of restricted cases, no3 — n (15n2 — 24n + 10); and the whole number of cases is 15n6; therefore n3 - 15n2 + 24n - 10 the probability of the required event is 15n5 5470. (By C. W. MERRIFIELD, F.R.S.)-Prove that broken stone, for roads, cannot weigh less per yard than half the weight of a solid yard of the same material, assuming that none of the broken faces are concave, and that it is shaken down so that there shall be no built-up hollows. Solution by the PROPOSER. Observe (1) it cannot lie looser than when all the pieces are the same size and shape; (2) or than when these pieces are regular tetrahedrons; (3) a plane can be completely covered with tetrahedrons, of which the volume will be as one-third the height; (4) we cannot invert one such set on another, so as to fit in, but we can invert a half set so as to fit in with a whole set. This gives (=) of the space filled, and the other half empty. This arrangement, and reasoning, are easily seen to be projective. Hence the theorem. 5304. (By Professor CLIFFORD, F.R.S.)-Prove that the negative pedal of an ellipse, in regard to the centre, has six cusps and four nodes; find their positions, and the length of the arc external to the ellipse between two real cusps; and account fully for the apparent reduction of the curve to a circle and two parabolas respectively, in special cases. Solution by J. HAMMOND, M.A. There are four distinct curves inverse to the conic sections, viz., a bicircular quartic or circular cubic according as the centre of inversion is not or is on the conic, with a node or cusp at the centre of inversion according as the conic is a central conic or parabola. The characteristics of the inverse in these four cases are respectively The negative pedal being the polar reciprocal of the inverse, its characteristics are found by interchanging m and n, 8 and т, к and i; and it is noticeable that the inverse and negative pedal of the parabola with respect to a point on the curve have the same characteristics. In the particular case of the central negative pedal of the ellipse, it is at once seen that there are four nodes, six cusps, and three double tangents. When the negative pedal of a given curve is found by first inverting, and then reciprocating, the inverse; the value of the constant of inversion is immaterial provided the constant of reciprocation is equal to it. But when each of these constants is taken to be unity, the projective equation to the inverse becomes the tangential equation to the negative pedal by simply changing the Cartesian coordinates (x, y) into Booth's tangential |