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CF::BG:CE; but the ratio of AG to A C being given, and GH, GB given lines, therefore CF, CE, and EF are given. Take A D equal to the given sum or difference of the sides AB, BC, then B D will be equal to BC; 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 AF is parallel to BC, the angle FA E is equal to CBE, a given angle; and, consequently, since É 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 E A so that the part thereof A D intercepted by their peripheries shall be of a given length; which is Prob. 27th, Simpson's Geom.

The same, by Mr. J. Butterworth. ANALYSIS. Suppose the thing done, and that AC B is the triangle required, and C D the given line, dividing the base A B in D in the given ratio of min;- draw DF parallel to AC, meeting BC in F, and upon FD take FE equal to F B, and join CE. Then, by similar triangles m + nin:: AC:DF::BC:BF:: ACIBC:DFEBF; but AC EBC is given, therefore DFBF=DE is given; again, the ratio of BF (FE): FC is given, and the angle EFC is given, therefore the triangle EFC is given in species, and consequently the angle DEC is given; but DC and DE are given, therefore CE is given, and consequently CF is given, therefore BC, and A C are given. And the construction is obvious.

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

20. Qu. (94) Answered by Mr. Whitley. Divide A Bin D so that A D:DB ::9:p; and BC in E so that BE: EC::n:9; then D, E, will be given points. Put p +9=m,9+na, and the given space = S; by Prop.

D

2. Dr. Stewart's General Theorems, P x A P+9x PB2=m.(AD.DB+ PD) =nx CP - S; hence n x cp? — m P D’ = m. A D. DB +S = a given space; and because D, C, are given points, therefore Prop. 4, lib. ii. of Apollonius de Locis Planis, restored by Dr. Simson, the locus of the point P is a given circle.

Again, since n x CP-S=PX A P2--9BP?, we have p X AP + S = q* B P + n x. C P = a. (BE.EC + E P2) by Dr. Stewart, as above : hence p X AP — a x E P = =a.BE. EC-S=

=a given space; and because A, E, are given points, the locus of the same point P is another given circle, per Appollonius, as above.

The intersection of these two given circles will, therefore, determine the point P, as required.

Very good solutions were sent by Messrs. Harvey, Hine, Kay, Maffett, Nesbit, Rylando, and Winward. 21. Qu. (95) Answered by R, lando and Mr. Whitley.

Put PB = a, PA =b, PS=
C, AS=SB=r, DB=x, and
DN=y; then PD = a- x,
and by the circle D E = 2 rx B

x>, and Eu. 47,1 PE = (P D? + D E) = (14—*)* + 2r—*) =V (as — 2 x (a—"))

D E2 x PS =(a? - 2 cx); hence DN=

PE (2r+ — *) x ©

is the equation of the curve. V a2

Whence it appears that the locus of the point N is a line of the fourth order which cuts A B in the points A, B; for y vanishes both when x = 0, and x = 2r, as it ought to do. The fluxion of the area is y ==

(274-204) ca

a find the fluent, put a? = 2 ch, and ✓ 2 ch - 20

2Ć I 2c%; then x=h-2c2, 3 = -40%ż; hence, by substitution, we have (2 rr- x).c* (as - 2 CX)= (2 ch - 4 rhc) 3 + 8rc - 8 h c*) ** 2 + 8 2* *;

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=

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and the correct fruent = (20 h: —4r hc).(3-4)

as

bs

as

"מ

P'

B

)

+ C3 z 3

20 c area BN D; therefore the whole 'area B N A is -

(-).(63 - a) (h -- 2rh). (-a) +

+ 3 с

20 c? 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 PABCD, and

suppose

that the point P is found, so, that the product PA X'PB X PC XP D

G AF IR may be equal to a given space, or a maximum. Draw A C, BD intersecting in R, PF, DI perp. and PS parallel to AC; draw also PE perp. and PQ parallel to BD. Then putting d for the diameter of the given circle, we have, by a known theorem, PAX PC=dx PF, and PB PDd XPE; hence the given product PA PB.PC.PD=PF.PE.d; consequently, since d’ is given, the rect. P F.PE is given. But by similar As, DI:DR::PF:PQ::PF PE: PQ XPE. Now D I, DR, and the rect. PF. PE are given; therefore the rect. PQR PE, or the parallelogram

PQ R S is given; and thus it appears that the locus of the point P is a given hyperbola, of which RA, R D are the assymptotes : its intersection, therefore, with the given circle, will determine the point Prequired. The product will evidently be a maximum when the hyperbola touches the circle at P; in which PPP"; which will evidently touch the circle in the point P as required.

if the tangent at P meet the assymptotes in G, H, then PG= PH. 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 curve

case,

Again, by Messrs. Butterworth and Kay. Analysis. Let A, B, C, D be the given points, and sup: pose P the required one ; join AP, BP, CP, and DP; also join A B and D C, and continue them to meet in E, and from P demit PF and PG, perp. to A B, and DC in F and G, and draw P K E and PI respectively parallel

R to E D and EA, to meet E A and ED in Kand I. Then,

E by the question, AP.BP.CP. D P is given; but AP.BP = PP 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 PF.PG is given; but the ratio of PG:PI is given, therefore PĚ,PI = the parallelogram KEIP is given. Hence, it appears, by Prop.75, p. 135, Emerson's Conics, that the focus of P is an hyperbola ; if, therefore, with the assymptotes EA, 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. Gawthorp, 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 the press.

* 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 Hutton's Course does not treat of the strength of timber.

NEW MATHEMATICAL QUESTIONS

TO BE ANSWERED IN NO. X.

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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 Din 15 hours; A, C, and Ď 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. Fi

(la? — 2) — 2)+((a? — 3)? — 3) +, &c.such, that

its sum to n terms may be a square.

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