5395. (By C. SMITH, M.A.)—If P, Q are points on two confocal conics, such that the tangents at these points are at right angles, show that the line PQ envelops a third confocal. Solution by R. F. DAVIS, B.A.; S. JOHNSTON, M.A.; and others. U Let TP, TQ be two tangents one to each of two confocal conics; S a common focus, and C the common centre. Draw the perpendiculars SY, SZ; join CY, CZ, meeting SZ, SY respectively in E, F. Then it is easily seen that EP, FQ are parallel to SY, SZ. Complete the rectangle QTPU, and let CY meet QU in H and TQ in L. Draw the perpendiculars SR, CN, and join CR, ZR. Then, since SYTZ is a rectangle, and C any point in its plane, SC2+ CT2 = CY2+ CZ2; hence CT is constant [and a2+b2-λ2, if a2, b2 and a2-λ2, b2-λ2 be the squares of the semiaxes]. Moreover, since CL: CY CZ: CF = CE: CH, therefore CL: CE = CY: CH, or CL: EL But CL EL CN: EZ or QU; hence TN: NQ CT passes through U. The point C may therefore be regarded in somewhat of the light of a "centre of similitude" of the three rectangles SP, SQ, PQ, one diagonal of each passing through it. = = = CY: YH TN: TQ. = = CN QU, and Next, since a circle can be circumscribed about SZQRF, we have 4 SRZ SFZ CZT, ZRSZ = = = π-YSR = π-QPT = CTZ; therefore the triangles SRZ, CZT are similar, and Finally, CZR = CT: SZ or TY CTY, and the triangles CTY, CZR are similar. Hence CR. CT = CY. CZ, or CR is constant [and a2 (a2 —λ2) (a2 + b2—λ2) −1]. = 4253. (By H. S. MONCK, M.A.) A series of Pythagorean triangles with the difference between the hypothenuse and one side equal to n, can always be obtained by beginning with the triangle 3n, 5n, 4n, and taking the upper figure as negative in each odd term of the series given in Quest. 4102. Find in what cases a distinct series with the same difference can be obtained. Solution by the PROPOSER. A distinct series is possible if n be a square number or any multiple of a square number (other than unity). For the three sides of a Pythagorean triangle are always of the form m (2ab), m (a2 + b2), and m (a2—b2); and hence the difference between the hypotenuse and one side is either m (a-b)2 or 2m. b2, either of which are multiples of a square number. Conversely, if n = mx2, we can form a series by putting x = a-b, and continuing backwards and forwards; or, if m be an even number, we can put - b, and taking any larger number for a, continue the series backwards and forwards as before. x= 5462. (By Professor ENGLAND, M.A.) A variable circle passes through two given points, and through one of these pass two given lines; find the envelope of the chord that joins the other points where the circle intersects these lines. Solution by JACCOBINI VINCENZO, and SIMONELLI RUGGERO. Sieno A e Bi due punti ed u, u' le due rette date. Si prenda sulla u un punto qualunque N; per questo e per A e B si faccia passare un cerchio; esso determinerà sulla u un unico punto N'. Se ora sulla u' si prende il punto N'; il cerchio che passa per N', A, B andrà ad intersecare la u nell unico punto N. Quindi ad un punto della u, corrisponde un unico punto nella u', e viceversa. Le due rette u, u' sono per tal fatto proiettive e l'inviluppo richiesto quindi una conica. Osservando poi che fra gl'infiniti cerchi che passano per A e B vi è quello che si scinde nella congiungente AB e nella retta all' infinito del piano, esso taglierà le due punteggiate u ed unei punti all' infinito. Per cui, appartenendo la retta all' ∞ all' inviluppo, la conica è una parabola. 5472. (By CECIL SHARPE.)-Show how to draw a straight line terminated by the circumferences of two of three given circles, and bisected by the third circumference. Solution by H. POLLEXFEN, B.A.; C. BICKERDIKE; and others. Let O be the centre of the circle (1). Take any point P on the circumference of (2); join OP and bisect it in C. With C as centre and a radius equal to half the radius of (1) draw a circle, cutting (3) in Q and R. Then it is evident that P is a centre of similitude of the circles whose centres are O and C, and that any line drawn from P to the circle O is bisected by C. Hence PQ and PR are two of the required lines. 5520. (By WILLIAM R. ROBERTS, M.A.)-Shew that the six inflexional tangents to a unicursal quartic all touch the same conic. Solution by J. C. MALET, M.A.; H. T. GERRANS, M.A.; and others. Projecting two of the double points to the circular points at infinity, and inverting from the remaining double point, the question is reduced to the following:-"the centres of the six circles that can be drawn through a fixed point to osculate a given conic, lie on a conic," which may be proved as follows:-Take the fixed point as origin, and let the equations of the given conic and of any circle through the origin be ax2+ by2+2hxy + 2gx + 2fy + c = 0, x2+ y2-2ax - 2By = 0. Now, if two conics U and V osculate, the discriminant of U+ kV, viz., ▲ + k + O'k2 + A3 must be a perfect cube; hence we have 340' = 02 as one condition; in the present case, this gives us 3A (ba2 + aẞ2-2haß-2ga-2fB - c) + {2a (bg — fh) + 2B (af −gh) + (a + b) c—ƒ2—g2}2 = 0, which proves the theorem, 5487. (By BYOMAKESA CHAKRAVARTI.)-If SMPR be a semicircle on SR, and the chords SP, RM cut each other in N; prove that SR2 SN. SP + RN. RM. = Solution by C. BICKERDIKE; C. VINCENZO; S. RUGGERO; and others. The third term gives, by successive integration, 2 n+1 j2' (n + 1!)2° second term by dx,+1, and integrate from xr+1 = Xr+2 to Xr+1 {tr + 2 − (1 + 2) xr+2}· r (r+ 2) ' (n + 1 !)2" 1 + +... + n r n (n + 1) Rr+1; and so on. Thus the mean value required is -l 5380. (By W. GALLATLY, B.A.)—If a circle A touches internally another circle B at P, and a tangent to A at the point Q intersect B in R1, R2, prove that R2PQ = < R2PQ. [This theorem reciprocates into the following:-S is the common focus of two conics having contact of the second order at P1; if from a point Q on the outer conic two tangents QR, QR2 be drawn to the inner conic intersecting the common tangent at P in the points R1, R, and if the tangent at Q to the outer conic intersect the tangent at P in T; then TR,, TR, subtend equal angles at the focus.] Solution by J. S. JENKINS; J. O. JELLY, B.A.; and others. Let O, C be the centres of the circles (A), (B) respectively; then, drawing lines as in the figure, we have c = 4a = = 5331, (By Professor WOLSTENHOLME, M.A.)—Prove that (1) the evolute of the first negative focal pedal of the parabola y2 : = c(c-x), (where the parameter,) is the curve 27 (y2-8cx-c2)2 8cx (8x+9c); (2) the equation of the pedal itself is 27ay2 = (3a − x) (x + 6a)2; (3) the normal of the pedal exceeds the ordinate by a fixed length; (4) the arc measured from the vertex to any point is equal to the intercept of the normal on the axis of y; and (5), if a heavy uniform chain be tied tightly round a curve, such that the pressure per unit is equal to the weight of a unit of length of the chain, this curve must be the first negative focal pedal of a parabola. Solution by the PROPOSER. The first negative pedal of the parabola y2 = 4a (x+a) is determined by the equations or, changing the origin from S to O, and reversing the direction of x, This gives or or, if c = 4a, 27 (y2-8cx- c2)2 = 8cx (8x + 9c)2. With this equation of the evolute, that of the parabola would be y2 = c (qc-x), |