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6. Qu. (122) By Mr. J. Brewer, Private and R. L. Militia.
The height of an inaccessible object is equal to the distance between any two stations taken in the same right line, divided by the difference between the natural co-tangents of the angles subtended by the object at these stations; required the demonstration.
7. Qu: (123) By Mr. J. Baines, Reading Academy. · Given, the 4 sides of a trapezium, to determine the radius of its circumscribing circle, when the sum of any two opposite angles is equal to two right angles. To
8. Qu. (124) By the same Gentleman. "S In what laticude will the sun rise one hour sooner og the 10th of July, than on the 20th of Augusti . 9. Qu. (125) By Mr. R. Maffett, Plymouth. :
A cutter observed a ship bearing due south, distant 35 miles, and they both sailed away for a port lying between the south and west, at which they arrived at the same instant, and found that their distances run included an angle at the port of 16° 15' 36", and that the difference of the ship's latitude, was to the distance run by the cutter, as 9 to 20); required their difference of latitude, departure, and distance run. 10. Qu. (126) By Mr. B. Brooke, Headingley Academy,
Leeds. Find the thickness of a spherical shell of ebony, whose diameter is 18 inches, such that it may sink, in sea water, to the depth of 6 inches.
11. Qu. (127) By Agriculturis, Cambridge. Describe a square geometrically, so that its diagonal inay exceed its side by a given line..
12. Qu. (127) By the same Gentleman. Draw a series of lines which shall represent the numbers 1, v 2, 3, 4, ✓ 5, &c.
'13.'Qu. (128) By J. F. communicated by Rylando.
Given, the position and diameter of the dial-plate of a clock, whose hånd moves uniformly round its circum.
ference;-I say another circular dial-plate, whose centre and diameter are different, may be so placed, that the same hand, produced if necessary, shall uniformly pass over each circumference, tracing out through the whole course of its revolution, equal lengths of arc in each in equal times; required the demonstration. 14. Qu. (129) By Hurlothrumbo Minor, of the West.
'In AB, AC, given by position, is a given point E; it is required to describe a circle touching A Cin F, and cutting F E produced in D, so that Ď E x E F may be equal to a given space, or that D E may have to EF a given ratio.
15. Qu. (130) By Mr. Rt. Maar. A is 25 years of age, and B 50; what is the probabi. Jity that B shall survive A? .
. 16. Qu. (131) By Lemuel Syray. Find four numbers, such, that their sum being added to, or subtracted from, the square of each particular number, the sums and differences thence arising may be all squares. 17. Qu. (132) By Mr. J. Butterworth, Haggate, near
Oldham. If the ends of a string, 20 inches long, be fastened to the ends of a uniform rod whose length is 12 inches, and weight 4lbs. be suspended by the middle of the string; and if a ring, of 8 inches diameter within, and · weight 2lbs. be put upon the double string, to find the position of the ring, &c. when the system is in equilibrio.
18. Qu. (133) By the same Gentleman., Given, the base and perpendicular, to construct the plane triangle when one of the angles at the base is double the other.
19. Qu. (134) By Mr. J. Winward, Plymouth.
Given, the line bisecting the vertical angle, and the : ratio of the segments of the base made by this line, to construct the plane triangle when the sum or difference of the perpendiculars from the ends of the base on this bisecting line is of a given length. 20. Qu. (135) By Mr. J. Kay, Rayton, near Oldham.
The distance between the centres of an opaque and a luminous spherical body is given, and the bodies begin to generate from nothing at the same time, and at each succeeding instant their magnitudes have a given ratio; (the opaque body being the greater) to determine, geometrically, the great circle of the opaque body where the enlightened part of its surface is a maximum.
*** Whoever sends the greatest number of true solutions to the preceding questions, will be entitled to a book, price 10s. 6d.
N. B. In future, no question will be inserted that is not accompanied with its answer, except some particular reason is assigned.
The following paper is an attempt towards generalizing the prize question of the Gentleman's Diary for 1809; should it be deemed worthy a place in the Enquirer, its insertion would oblige,
Your's, &c. Plymouth, Sept. 9, 1812.
GEORGE HARVEY, JUNR.
GENERAL PROBLEM. Having any number of circles given in magnitude and . position, to determine geometrically a point P in a line of 'any kind, given also by position, from which tangents PA, PB, PC, 8c. being drawn to these circles, m. PA In. PB?£p. PC£, &c. may be equal to a given space; the signs + being allowed to operate in any manner whatever ; and also that these tangents may obtain a given ratio; and to point out the limits of possibility.
SOLUTION. · Case I. Let it be required to determine a point Pin the line MN, sich, that m. PA’ + n. PB: + p.PC +, &c. may be equal to a given space.
Join the points OP, RP, and Q P, as also A O, BR, and ÇQ. Then, m. P A?=m.OP-m.O A,
and, n. P BP = n. PR. N.BR, .
. and, p. PC?= p. PQ-p.CQ, thereforem.OP+n. ÞR? +p.PQ- m, 0 A’-n.B RP. -P.CQ’, or m. O P + ni PR + p. PQ't, &c. is equal to a given space.
Join O R, and take SR=* OS, and join PS; then, (Leslie's Gcometrical Analysis, p. 374) m.OP? + n.PRO n.OR.RS+ (m + n) PS, and, therefore, n.OR.RS+ (m + n) PS? +p. PQ’, or, (m + 1) PS? +p.PQ? +, &c. is equal to a given space; and hence the locus of the point P is a given circle, (Prop. 4, lib. ii. Apollonius de Locis Planis, restored by Dr. Simson) and describing which, as there directed, will intersect the line MN in the required point.
Case II. Let it be required to determine a point P such, that m .PA-n. PB’ - p.CP —, &c. may be equal to a given space.
Here m. PA=n.O P— m.OA, and — n.P B’ = -N.PR+ni B R2, and – p. PC =- .PQ’ + p.O Q', therefore m.OP2 — n. PR-P. PQ? — . O AP + n.BR' + 0.0Q’or n. PR’ +.PQ'— m.OP2 t, &c. is equal to a given space.
Join QR, and take QT = * RT; also join FP, then (Leslie ibid.) n.PR+p. PQ=p.RQ.TQ't (n + p) TP2, and therefore piRQ.TQ +(n+p) TP: - m.O P +, &c. Or (n +p) TP —m. O P +, &c. is equal to a given space; and hence the locus of the point P is a giren circle, ( Apollonius, ibid.) describing which, as there directed, will intersect MN, in the required point.
Case III. Let-AP:PB:PC: &c. :: m:n:p:&c; then from this we deduce n. AP=m. PB, and p.PB =n.PC; or n?. AP = m2. P B’, and p?. PB = n'. PC?; and mo. PB + m2.. PC? = n. AP + pa. PB"; but n. P B’ – m2. PR’ — m2. BR’, and n?. PC? = n. P Q’ — 12.C Q2 and 12. A P = n?. OP? — . O A’, and på. PB’ = p2 PR— p.BR’; hence
m2. PR2 +no. PQ:-*m’, BR’-n.CQ
m2. OP2 +p.PR?— m2.0 A2-p2. BR, and (m? - p2) PR’ + n°: PQP – m2. O P2 is given.
Join RQ, and take T Q = (*P)RT; join TP, then (Leslie, ibid.) (m- p2) PR2 +1.PQ2 = 'n?. RQ. TQ + (m + n° -- p2) T P?, and, therefore, n?.RQ. TQ+ m + n° - p) TP2 – m2.( P2 is given ; or (m? + n? -- p) T P2 — .0 P2 is equal to a given space; and hence, as before, the locus of the point P is a given circle, (-apollonius, ibid.) which locus described, will intersect MN, in the required point P.
These cases are sufficient to explain the mode by which the Problem may be resolved in its most general form ;-for it is evident, that by repeated applications of this principle, we may always reduce each particular case through the before-mentioned Problem of Apollonius. se
With respect to the limits of possibility in the case of a maximum or a minimum, we have only to describe to a given centre, a circle, to touch the curve, which is given in position.
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