allel AP, but applies to any line DE' and its parallel AP'. For the triangle DEF is half of the parallelogram DH; having the same base and altitude; and for the same reason the parallelogram DH-DH'. Hence, as the angle AP' CH'E' D, t: DEF = BCX AP: (HE or H' E) DE', Therefore the line DE' may take the direction DB', and we shall have so that drawing C'D, and RO parallel to DB', intersecting C'D in 0, we must have RO AB. For the triangles CDB and D RO are similar, having their sides respectively parallel, giving = B'C': B'D-RD or EH: RO. :. DBX EH=B'C' × RO= B'C' X AB. .. RO-AB. SECOND SOLUTION OF PRIZE PROBLEM III. BY ROLAND THOMPSON, JEFFERSON COLLEGE, CANONSBURG, PA. Inscribe the triangle A'B'C' in DEF, and through the points D, E, F, draw straight lines respectively parallel to the sides of A'B'C'; the tri angle ABC thus formed is similar to A'B'C'. The three straight lines pas sing through the homologous angles will meet in the same point. For the straight lines CC and BB produced will meet in some point O, and if AO does not pass through A', let A′′ be the point in which it meets the line A'B'. Then BO: BOAB: A′′ B' BO:B0=BC: BC; AB: A" BBC: B'C'. Therefore the triangle A" B'C' is similar to ABC and equal to Since (1), (2), and (3), are equal ratios, we get, by addition, showing that DEF is a mean proportional between ABC and A'B'C' as was to be proved. "If a be one of the angles, prove that sec A PRIZE SOLUTION OF PROBLEM IV. BY ALL THE COMPETITORS. sides of an equilateral spherical triangle and A one of its If a = bc in the fundamental equation cos a = cos b cose + sin b sin e cos A, BY GUSTAVUS FRANKENSTEIN, SPRINGFIELD, OHIO. "If the semiaxes of an ellipse be A and B, P the length of the perpendicular dropped from the centre on the tangent to the curve, r and r' the distances from the point of tangency to the foci, and o the radius of curvature at this point; prove that ୧ and from this theorem construct the corresponding point of the evolute." 1. The equation of the ellipse referred to its centre and axes is A2x2 + B2y2= A' B2; and if x,y denote the coördinates of the point of tangency, the equation of the tangent line is Ayy+ B2 x x = A2 B2. The tangent of the angle which the line DS makes with the axis of x is Day= B2 x Ay; or if we take the angle angles TSM and DSC give DC: CS- TM: MO+OS; or y' P: 4 = N : ay' +/ α But from the Calculus we have the radius of curvature, (= Further, the normal, MT, bisects the angle ATF; AM +r 2 A AM+MF 2√A2-B2 = = = r MF' 2. To find by construction the corresponding point in the evolute. We have just found that : P=9: r. Produce the nor r': mal TM, and also TF. On TM produced take Tp=P, and TL=r'. Draw p F, and L X parallel to p F; then SOLUTION OF PROBLEMS IN MAXIMA AND MINIMA BY ALGEBRA. By RAMCHUNDRA, Late Teacher of Science, Delhi College. 1. To divide a given number into two such parts that their product may be the greatest possible. = x, and x) = ax Put the given number = a, one of the parts required consequently ax the other part. Therefore x (a — x) = 22= product = maximum=r. Therefore x-ax=-r. Solving this quadratic equation we find x2 Now it is evident that r cannot be greater than a2; for if it be so, the value of x becomes impossible; therefore the product a x-x2, or r, α is greatest when a2=r; therefore x = 2* SECOND SOLUTION WITHOUT IMPOSSIBLE ROOTS. In the expression a x-x2, which is to become a maximum, let α x=y, where the value of y, determined by the condition of 2 a x — 22 being a maximum, will show whether it is positive, zero, or negative. We now find a x — x2 = a y + 22 — y2 — ay — “2 = 42 — y2, 4 which is evidently a maximum when y = 0. Therefore x = before. II. Of all right-angled plane triangles having the same hypothenuse, to find that whose area is the greatest possible. = Let a = hypothenuse, x= base, y perpendicular. Then, x2 + y2= a2, we shall have y=√a2x2, and consequently 2 xy x 2 Va2. x2 the area of the triangle = maximum, and conse quently the square of the area, or † (a2x2 — x1) = max., and also 6 VOL. II. |