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C F:: BG : C E; but the ratio of AG to AC being given, and G H, GB given lines, therefore C F, C E, and E F are given. Take A D equal to the given sum or difference of the sides A B, B C, then B D will be equal to B C; and because the angle A B C is given, the triangle CBD will be given in species; therefore the angle D is given; and because A F is parallel to BC, the angle F A E is equal to C B E, a given angle; and, consequently, since E F, EC are given lines, the loci of the points A, D, are two given circular segments; hence the problem is reduced to this. To draw from E the point of intersection of two given circles, the right line EA so that the part thereof AD intercepted by their peripheries shall be of a given length j which is Prob. 27 th, Simpson's Geom.

The same, by Mr. J. Butterworth. . ,

Analysis. Suppose the thing done, and that ACB is the triangle required, and C D the given line, dividing the base A B in D in the given ratio of m : n; draw D F parallel to A C, meeting B C in F, and upon F D take FE equal to F B, and join C E. Then, by similar triangles m -fn:n: : A C : DF :: BC : BF:: A C ± BC : DF± BF; but AC±BC is given, therefore DF± BF = DE is given; again, the ratio of B F (F E) : F C is given, and the angle EF C is given, therefore the triangle EFC is given in species, and consequently the angle DEC is given; but DC and D E are given, therefore C E is given, and consequently CF is given, therefore B C, and AC are given. And the construction is obvious.

It was answered also by Messrs. Eyres, Gawtkorp, Kay, and Nesbit. . , .

20. Qu. (94) Answered by Mr

Divide A B in D so that A D: DB :: Q-p; and B C in E so that B E: EC: :n:q; then D, E, will be given points. Put/i + q— m, q + n —a, aad the given, space = S; by Prop.

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and the correct fluent = (2 c h' — 4 r A c). —

+(8r,_84,,.(f-5£?),+ |,,_^=

area BN D; therefore the whole area B N A is =

(A'-2rA).(6 —a) + v Lf. >+ as

required.

And nearly as above were the solutions by Messrs. Butterworth, Eyres, Gawthorp, Kay, and Nesbit.

22. Qu. (96) Answered by Messrs. Jones and Whitley.

Analysis. Let A, B, C,. D, be the four given points in the circumference of the given circle P A B C D, and suppose that the point P is found, so, that the pro- • duct PA xPBxPCxPD may be equal to a given space, or a maximum. Draw A C, B D intersecting in R, P F, D I perp. and PS parallel to A C ; draw also PE perp. and PCI parallel to BD. Then putting d for the diameter of the given circle, we have, by a known theorem, P A X PC = d x PF, and PB x PD= d X PE; hence the given productPA.PB.PC.PD = PF.PE. d'; consequently, since d2 is given, the rect. P F. P E is given. But by similar As, Dl: DR :: P F: P Q:: P F x PE: P Q X P E. Now D I, D R, and the rect. P F . P E are given; therefore the rect. PQ x P E, or the parallelogram P Q R S is given; and thus it appears that the l6cus of the point P is a given hyperbola, of which R A, R D are the assymptotes: its intersection, therefore, with the given circle, will determine the point P required. The product will evidently be a maximum when the hyperbola touches the circle at P; in which case, if the tangent at P meet the assymptotes in G, H, then P G = P H. To determine the point P; let tangents be drawn to the circle, and terminated by the assymptotes, let them be bisected in the points P,P ,&c. then, through the points of bisection, draw the curv«

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FPP"; which will evidently touch the circle in th« point P as required. _,

Again, by Messrs. Butlerwortfi and Kay.

Analysis. Let A, B, C, D be the given points, and suppose P the required one; join A P, BP, CP, and DP; also join AB and D C, and continue them to meet in E, and from P demit P F and P G, perp. to A B, and D C in F and G, and draw P K and PI respectively parallel to E D and E A, to meet E A and ED in Kand I. Then, by thequestion,AP.BP.CP. DP is given; but A P . B P = P F by the diameter of the circle, and CP. DP = PG by the diameter of the circle; but the diameter of the circle is given, therefore P F . P G is given; but the ratio of P G : PI is given, therefore PF, PI — the parallelogram KE1P is given. Hence", it appears, by Prop. 75, p. 135, Emerson's Conies, that the locus of P is an hyperbola; if, therefore, with, the assymptotes E A, ED, and power equal to the given parallelogram KEIP, an hyperbola be described, it will cut the circle in two points, either of which may be taken; when the hyperbola only touches the circle, then there will be but one point, and the parallelogram KEIP will be a maximum, and, consequently, AP, BP. CP. DP will be a maximum, beyond which the problem becomes impossible.

This question was answered also by Messrs. Gatuthorp, Nesbit, and Rylando.

Both the answers to question 58 are right, it admits of two solutions; Mr. Gawthorp's name was omitted, by mistake, among the answers to questions 61 and 69.

Mr. Arthur Hirst's letter, containing excellent solutions to a great number of the questions, did not come to hand till the whole was copied and ready for th.e

Mr. Whitley is requested to send for any book, the price of which does not exceed half-a-guinea.

Mr. Nesbit observes that Mr. Baines has made a wrong reference in page 324, as the 2d vol. of Huttoa's Course does not treat of the strength of timber.

NEW MATHEMATICAL QUESTIONS

TO BE ANSWERED IN NO. X.

1. Qu. (117) By Mr. J. Cattrall, Fifer 2nd R. L.

Militia, Plymouth. A cistern has 4 cocks, A, B, C, D; now A, B, and C, when open, will empty it in 12 hours; A, B, and D in 15 hours; A, C, and I) in 13 hours; and B, C, and D in 16 hours; in how long time will they empty it when all 4 are open together, and in what time would each empty it seperately?

2. Qu. (118) By Mr. W. Bruster, Donington. A wall 9 feet high, and 18 inches thick, surrounds an elliptical fish-pond, whose transverse and conjugate , axes are as 4 to 3; and the content of the pond, to the depth of 1 inch, is equal to the solid content of the wall; required the dimensions'of the pond?

3. Qu. (119) By Mr. A. Nesbit, Farnley Academy.

If the transverse diameter of an ellipse be 50, and its conjugate 40, what will be the area of the greatest trapezoid that can be inscribed in the semi-ellipse?

4. Qu. (120) By the same Gentleman. If the conjugate axe of a prolate spheroid be to its transverse as 4 to 5, and its solidity 41888; what will be the content of the greatest cylinder that can be cut out of the semi-spheroid?

5. Qu. (121) By Mr. E. S. Eyres, Liverpool.

Find the value of a in the series ^(a'— l)2

((a2iy — 2) + ((«' — 30' — 3) +, &c. such, that its sum to n terms may be a square.

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