which is half of the circle CK, and therefore one fourth of circle CA. The process may be continued as far as we please. PRIZE SOLUTION OF PROBLEM III. By GEORGE B. HICKS, Cleveland, Ohio. III. If R and r be the radii of the circles circumscribing and inscribing any triangle, and D the distance between their centres, then D2= R-2 Rr. Required, a geometrical demonstration. Let O be the centre of the circumscribed, and 0 of the inscribed circles. Produce A O indefinitely, and O O' to meet the circumscribed circle in Hand I. Draw CG, CE, EB. Lastly, from E as centre, with radius E 0, describe the circle ON FM. This circle will pass through B and C. For ECO is measured by (GB+BE) and EOC by (AG+EC). But GB+BE AG + E C. Hence the triangle E O C is isosceles, and ECO E. Similarly BE 0 E. Also, since A E bisects BA C, we have The chords HI, AE give HOX OI=(R + D) (R — D) = A0 × O E. But it is shown in Geometry that (R+D) (R-D) = R2 — D2; It remains to show that A OX OE= 2 Rr. And from the secants AF, A C, AO × AF ACX A C. But since E is on the bisector of the angle BA C, A C′ = AB; pression represents double the area of AB C'; as required. PRIZE SOLUTION OF PROBLEM IV. By GEORGE B. HICKS, Cleveland, Ohio. IV. If a circle cut a conic section in four points, A, B, C, D, and a second circle cut the same conic section in A, B, E, F, then will CD and EF be parallel. Let y — m x — n x2 =S=0 be the equation of any conic, and x cos ay sina-pa= 0, x cos by sin b—p=ß = 0 (1), the equations of the chords A B, CD. Then will Ska ß= 0 (2) be the equation of some figure through the intersections of (1) with the conic, the species of (2) being determined by the value attributed to the arbitrary constant k; for (2) is satisfied by S=0, a=0; S=0, ẞ= 0, and therefore passes through the intersections of S, a, and B. (SALMON'S Conic Sections, p. 211.) ß To determine k so that (2) may represent a circle, write for S, α, and ẞ their values at full length, equate the coefficients of 22 and y2, and equate to zero the cofficient of x y. We get 1+nk cos (a+b), k sin (a+b)=0. Now a and b are the angles made with the axis of x by the perpendiculars from the origin on the lines AB, CD. Hence AB and CD are equally inclined to the axis of the conic (this being the axis of x). So also are AB, EF, since the property just proved is evidently general. And therefore CD, EF make equal angles with the axis; . they are parallel. Mr. TROWBRIDGE adds to his solution of this problem the following: "If two conics of the same kind cut a third conic in the points A, B, C, D and A, B, E, F respectively, then will CD and EF be parallel, provided the transverse axes of the cutting conics are either parallel or perpendicular to each other." PRIZE SOLUTION OF PRoblem V. By ASHER B. EVANS, Madison University, Hamilton, N. Y. V. In any conic section, let u denote the length of the perpendicular dropped from any point in the curve upon the directrix, r the distance of this point from the nearest focus, a and b the semi-axes; prove that u3 A conic section is the locus of a point whose distances from a given fixed point, called the focus, and a straight line given in position, called the directrix, are always to one another in a constant ratio. This condition gives ru :: e: 1, or 2 eccentricity of the conic section. Therefore the ellipse, u2 2, where e is the 2.2 u2 u2 for the hyperbola, and zero for the parabola, NOTES AND QUERIES. 1. Problems in Division, in which all the figures in the divisor but the right-hand one are nines. Let it be required to divide 864835 by 996. Proceed as follows: 864835864 × 1000+ 835 = 864 (996 +4)+ 835, =864 × 996 + 864 × 4 + 835, 8649963456835. Hence 996 is contained in 864835, 864 times, and 3456 + 835 remainder. But 34563(996+4)+4563 X 996 + 3 x 4 + 456; hence 3456 contains 996, 3 times, with 3 X 4+ 456 468 remainder. Hence, 864835 contains 996 864+3 times, with 835+468= 1303 remainder. But 13031(996+4)+3031 × 996 + 4 + 303, = 1 × 996 + 307; that is, 1303 contains 996 once, with 307 remainder; and the final result is 868 as the quotient, and 307 remainder. We can now condense the work into the following OPERATION. 864 835 3 456 867 868 12= 1303 864 X 4 307 ber in the divisor, EXPLANATION. Cut off as many figures from the right of the dividend as there are figures in the divisor. Multiply the remaining figures of the dividend, 864, by 4, the complement of the divisor, and set the product underneath the dividend. All the figures of this product over three, the nummultiply by 4, and set the product underneath; and so continue to do until none fall to the left of the vertical line. Next add, and all at the left of the vertical line, as 867, will be quo tient; and all to the right, as 1307, will be remainder. Treat this remainder in the same way, and so on; and the last additions will give the true quotient and remainder.-Prof. DEVOLSON WOOD, University of Michigan, Ann Arbor. 2. Derivative of a*.- Let u= a; then log uz log a. By algebra the Naperian log of u is log u = log2+1 = 2 (2 2+1+3(2z+1)3 +5(2z+1) + &c.), The series in the parenthesis' is converging for all positive values of u, and by summing an infinite number of terms the equation be -LUCIUS BROWN, Fall River, Mass. 3. Remarks upon a (supposed) new Instrument for the Mechanical Trisection of an Angle.-The Philosophical Magazine for April, 1860, contains an account, by Mr. TATE, of an instrument for the mechanical trisection of an angle, which is there described as new. The instrument is simple and elegant, and very convenient for the purpose intended. It will be seen, however, that the construction in question has long been known; for on turning to the 252d page of the second edition of the Traité Analytique des Sections Coniques, by DE L'HÔPITAL, first published in Paris in the year 1707, and republished in 1776, we |