that the line of collinearity of the feet of the perpendiculars on the sides bisects at right angles the line joining the feet of the perpendiculars on the diagonals............ 7733. (H. J. Read, M.A.)-Transform of the geometry of the ellipse. -2 ... = 57 de by means 35 7737. (The Editor.) - If all the roots of the equation " — α1 x2-1 + a2xn-2 ± an Ó are whole and positive numbers, prove that (1) an (1 + a + ... +an) (1 + 2α1 + 4a2 + ... + 2” an) / 6" is a whole number decomposable into the sum of a squares; (2) a2 (1 +α1+ ...+an)2/4" is a whole number decomposable into the sum of an cubes; (3) a2 (1 + 2α1 + ... + 2′′an) is decomposable into the algebraic sum of 2"a, squares... 79 7738. (D. Edwardes.) - Prove that if any three lines be drawn from the centre of a triangle ABC to meet the circum-circle in P, Q, R, and the circle through the ex-centres in P', Q', R', (1) the triangle P'Q'R' is similar to PQR and of four times its area; (2) if the lines joining the centroid of ABC with the feet of the perpendiculars be produced through the centroid to meet the circum-circle in L, M, N, the triangle LMN is similar to the pedal triangle of ABC and of four times its area. 79 (d) and dx (d) be the dr dx 7739. (W. G. Lax, B.A.)-If x, y; r, o be the rectangular and polar coordinates of a point respectively, and if partial differential coefficients of x with respect to r, and of r with respect to x, when r, e, and x, y are independent variables respectively; prove geometrically that dx dr = dr 59 7741. (The late Professor Clifford, F.R.S.)-The motion of a point is compounded of two simple harmonic motions at right angles to one another, which are very nearly equal in period, but whose amplitudes are slowly diminishing at a uniform rate; find the general shape of the curve which the point will describe............. 62 7744. (Professor Cochez.) — Parmi les courbes isoperimètres planes passant par deux points fixes, quelle est celle qui, par sa révolution autour de l'axe des x, engendre la surface maximum ou minimum ?............ 100 7747. (The Editor.)-Show that (1) in a triangle there can be inscribed three rectangles having each a side on one of the sides of the triangle, and their diagonals equal and crossing at their mid-points; and (2), if a, b, c be the sides of the triangle, the length of these equal diagonals is 2abc / (a2+b2 + c2). . 112 7752. (Asparagus.)—From a point on one of the common chords perpendicular to the transverse axis of two confocal conics are drawn tangents OP, OQ, OP', OQ' to the two conics: prove that the straight lines PP', PQ', P'Q, P'Q' each pass through one of the common foci................. 99 7758. (J. Brill, B.A.) If ABC be any triangle, and O a point within it; prove that 7761. (W. J. C. Sharp, M.A.)—A flexible string is suspended slackly from two fixed points, and acted upon by a uniform horizontal wind, blowing in a direction making any angle with the horizontal projection of the line joining the points. Find the curve in which the string hangs and the tension at any point. .......... 100 7766. (R. Tucker, M.A.)-If p, p', w are the "T. R." and Brocard radii, and Brocard angle respectively of a triangle, prove that (1) cos 3w ́; and (2), if P1, P2 are the "T. R." radii in the ambiguous case of triangles, then p1 cos w1 = P2 COS Wq... COS W = 96 ......... 7769. (Professor Sylvester, F.R.S.) · Prove algebraically that, if ABC..., A'B'C'... are two superposed projective point-series which do not possess self-conjugate points, then the segment between any two corresponding points, as AA', BB'..., will subtend the same angle at a point properly chosen outside the line in which the point-series lie............ 89 7771. (The late Professor Clifford, F.R.S.) Find the locus of a point P which moves so that the length of the resultant of the translations PA, PB, PC is constant-the points A, B, C being fixed................................ 89 7774. (Professor Wolstenholme, M.A., Sc.D.) The lengths of the edges OA, OB, OC of a tetrahedron OABC are respectively 9.257824, 8.586, and 8.166; those of the respectively opposite edges BC, CA, AB are 8.996, 9.587, and 9.997. Prove that the dihedral angles opposite to OA and BC are equal to each other (each 7°19′18′′). Denoting the lengths by a, b, c, x, y, z, and the dihedral angles respectively opposite by A, B, C, X, Y, Z, find what relation must subsist between a, b, c, x, y, z in order that A may be equal to X. = 117 7780. (Rev. T. R. Terry, M.A.) - Prove that the mean value of the fourth powers of the distances from the centre of all points inside an ellipsoid whose axes are 2a, 2b, 2c, is ...... A = 35 [(a2 + b2 + c2)2 + 2 (aa + b4 + c1)]....................................................... 120 7792. (Asparagus.)-The tangent at any point of a parabola meets the axis in T and the latus rectum in t; prove that Tt is equal to onefourth of the parallel normal chord. 7794. (J. Brill, B.A.)-Prove that in any triangle 120 a3 cos (B-C) + b3 cos (C−A) + c3 cos (A−B) = 3abc.......... 114 7795. (C. E. McVicker, B.A.) - Prove that the distance between the instantaneous centre of rotation of a movable line and the centre of curvature of its envelope is, in any position, dr/dw, where x is the distance of any carried point on the line from the point of contact, and w the angle of rotation... 7797. (D. Edwardes.)—If 114 - [log (1+x)]" da, prove that V„+nV2-1 = 2 (log, 2)"...111 = 7800. (E. Buck, B.A.)-Without involving the Integral Calculus, 7801. (B. Hanumanta Rau, M.A.)-Inscribe a regular hexagon in a rectangle whose sides are a and b; and find the ratio of a to b in order that the polygon may be also equiangular................ 115 7802. (W. J. Greenstreet, B.A.)-Prove that the sum to infinity + log + log +... is log 4.6 6.8 7803. (The Editor.)—Trace the curve y2 (x-a) π = ე3 — 3. 7804. (Professor Cayley, F.R.S.) 1. If (a, b, c, f, g, h) are the six coordinates of a generating line of the quadric surface x2 + y2+ z2 + w2 = 0, then a=f, b=g, c=h, or else a = −f, b = −g, c = -h, according as the line belongs to the one or the other system of generating lines. = 0, 2. If a plane meet the quadriquadric curve, Ax2+ By2 + Cz2 + Dw2 A'x2 + B'y2 + C'z2 + D′w2 = 0 in four points, and if (a, b, c, f, g, h) are the coordinates of the line through two of them, (a', b', c', f', g', h') of the line through the other two of them, then af' + a'ƒ = 0, bg' + b'g = = 0, ch' + c'h = 0.................. 85 7807. (The late Professor Townsend, F.R.S.)-A triangle in the plane of a conic being supposed self-reciprocal with respect to the curve: show that an infinite number of triangles could be at once inscribed to the conic and circumscribed to the triangle, or conversely...... 88 7836. (Professor Sylvester, F.R.S.)-If p, q be two matrices (to fix the ideas, suppose of the third order), which have one latent root in common, and let x', x′′; μ', μ" be the other latent roots of p, q; prove that the product (p—λ') (p—λ′′) X (q — μ') (q — μ”) (where X is an arbitrary matrix) is of invariable form, the only effect of the intermediate arbitrary matrix being to alter the value of each term of the product in a constant ratio; i.e., in the nomenclature of the New Algebra, (p−λ′) (p−λ′′) X (q—μ') (q—μ") is constant to a scalar multiple près. For the benefit of the learner, I recall that a-λ b C = α bc the roots of the algebraical are called the latent roots of p, the equation itself being called the latent equation, and the function equated to 2-en the latent function......... 101 7838. (The late Professor Clifford, F.R.S.)-Prove that a string will rest in the form of a circle if it be repelled from a point in the circumference with a force inversely as the cube of the distance. 101 7849. (Rev. T. C. Simmons, M.A.)—If from a random point within an equilateral triangle perpendiculars are drawn on the sides, show that the respective chances that they can form (1) any triangle, (2) an acute-angled triangle, are p1 = 07944 nearly.............................. RATIO RATIONIS: or, that primary faculty of human nature which finds exercise alike in Logic, in Induction, and in the various processes of Mathematics. An Essay by D. BIDDLE 125 MATHEMATICS FROM THE EDUCATIONAL TIMES, WITH ADDITIONAL PAPERS AND SOLUTIONS. 7706. (By the late Professor CLIFFORD, F.R.S.)-What conditions must be fulfilled in order that the centre of pressure of a triangle wholly submerged in water may be at the intersection of perpendiculars ? Solution by ARTHUR HILL CURTIS, LL.D., D.Sc. When the vertices A, B, C of a triangle are sunk in a homogeneous liquid to depth ha, ha, ha, respectively, then x, y, z, the coordinates of the centre of pressure of the triangle referred to the sides as axes, are given by the equations:— {1+ (see Messenger of Mathematics, No. 8, 1864); but the coordinates of the intersection of the perpendiculars P1, P2, P3 are h: h2 hg: 4 cot B cot C-1: 4 cot A cot C-1 : 4 cot A cot B-1. [For a discussion of this problem, and several other applications of the above formulæ for x, y, z, see Messenger of Mathematics, New Series, No. 12, 1872, where it is shown that, if the vertices of a triangle be sunk to depths h1, ha, ha, and therefore the mid-points of the sides to depths H1 = (h2+ h3), H2 = } (h1 + h3), H = } (h1 + h2), when the centre of pressure coincides with: (1) Centre of gravity Condition. } H1 = H2 = H3, = tan A: tan B: tan C, (2) Centre of circumscribing circle} H1: H2: H2 (3) Centre of inscribed circle } H, H2: H2 = cot A: cot B: cot C, : (4) Intersection of the three per-H1: H2 : H2 pendiculars from vertices on opposite sides : 1-2 cot A cot C: 1-2 cot A cot B, (5) Centre of inscribed square H1 : H2 : H ̧ (alongside a) = 1: cot B: cot C, (6) Centre of the nine-point circle } H1 : H2 : H2 = sin 2A : sin 2B : sin 2C ; hence it follows that, if a triangle be so immersed that its centre of pressure coincides with the intersection of the three perpendiculars from vertices on the opposite sides, then the centre of pressure of the triangle obtained from the given one by joining the middle points of its sides will coincide with the centre of the circle circumscribed to this derived triangle, and the centre of pressure of the triangle similarly derived from it will coincide with the centre of the nine-point circle of this last derived triangle.] 7708. (By Professor TOWNSEND, F.R.S.)—A thin uniform spherical cap being supposed to attract according to the law of the inverse fifth power of the distance a material. particle situated anywhere on the surface of the sphere; show that, for every position of the particle, the attraction (a) passes through the vertex of the cone which envelopes the sphere along the rim of the cap, (b) varies directly as the radial distance from the vertex of the cone, and inversely as the cube of the perpendicular distance from the base of the cap. Solution by the PROPOSER. Two pairs of planes inclined at elementary angles, one pair passing through the line of connection of the particle with the vertex of the cone, and the other pair passing through its polar with respect to the sphere, which latter lies of course in the rim-plane or base of the cap, will intercept in the mass of the cap a pair of quadrilateral elements whose attractions on the particle for the law of the inverse fifth power of the distance are easily seen to compound a resultant directed to the vertex of the cone; and, as the entire mass of the cap may manifestly be exhausted by pairs of such elements, therefore, &c., as regards the first part of the property. As regards the second part. The integration based on the preceding method of division of the cap into elements leads to the result that the entire attraction of the cap on the particle = m (a + b) b where a is the radius of the sphere, m the mass of the cap, b the distance of its base from the centre of the sphere, the distance of the particle from the vertex of the cone, and p its distance from the base of the cap. And therefore, &c., as regards the second part also. |