Page images
PDF
EPUB
[ocr errors]

7.841584x2+40x a max. in fluxions, &c. we have x2 13.0693x 33.; whence x = 3.474, and the area of the ellipse, as in the preceding solution 6.1689, and 7s. 5d. the carpenter's loss.

This question was answered also by the following gentlemen, Messrs. J. Cattrall, Dunn, Hine, Hirst, Maffett, Rylando, Tomlinson, De Weight, and Winward.

11. Qu. (53) By Mr. J. Whitley.

Denoting the three required numbers by x, y, and z, the question requires us to transform the four following formulæ into squares; viz. xy + z2, xz + y2, y z + x2, and x + y + z are all to be squares. Assuming x = 4 (y + z) the first and second formulæ become squares, and by substitution, y z + x2 = 16 y2+33 y z + 16 z2, =(suppose) (pz--4y)'= p2 z2-8pyz+16y'; hence (p2 — 16) × z = (33 +8p) × y.

Take y= p2-16, then ≈ 33 +8p, and x = 4p2 +32p+68, therefore x+y+z = 5 p2 +40p+85= a2. Let p = n2; then a2 = 25 + 20 n +5 n2 =

(5— sn)2 = 25—10 sn+s2 n2, and n =

20+10s

2 s2 — 5

where

s is indeterminate. Ex. Taking s = 3, we have n =

[merged small][merged small][ocr errors][merged small][merged small][merged small]

or multiplying each by 4, a square number, we shall have x 13 x 2603380, y = 13 x 29 377, and z = 13 × 36=468, which are three numbers answering the conditions of the question.

The same answered by Mr. S. Ryley, Armley Mills.

By the question, the formulæ to be made squares are x+y+z, x2+xy, y2+xz, and z2+xy. Let x = 4 (y + z), then the second and third formulæ become squares, and the other two will stand thus: 16 y2+ 33 yz+16%2, and (y + z) x 5. Suppose y 5 n, and 5r, they then become 16 n2 + 33 nr + 16r2, and n+r; make n + r = p2, then n = p2 r, which substituted for n in the other formula, it becomes 16p +rp2—rv, which must be a square. Assume 4p2— y2

80+1

for its root, and r is found =

[ocr errors]

x p', where may

be any number at pleasure. Ex. Let v = 13, then r=

13

[ocr errors]

21

[blocks in formation]

34P', and n=4P', consequently y=

34

p2, z=

34

680
344

p2, and x = p'; in order to have whole numbers, take p 34, then 2210, 3570, and 23120 will be three numbers answering the conditions of the question. And thus nearly was the question answered by Mr. Baines, Mr. A. Hirst, Mr. Dunn, Mr. Hine, Mr. Maffett, Mr. Nesbit, Rylando, and Mr. Winward.

Otherwise, by Mr. Gawthrop,

The question requires that x+y+z, z2+xy, y2+x%, and x2+yz, may be all squares; put x+y+ z = a2, then a2-x-y, which substitute for z in the second, and it becomes at-2a'y+y2-2 a2x + 3xy + x2, which

[ocr errors]

4 a2

5

equate to (a2+y-x), and we find x= hence &= a5y, and, by substitution, the third and fourth 4a4 - 20 a2 y + 25 y2 5 ay-25y+16a+

5

become

25

and

25

the first of which is a square already, and the second will be a square when the numerator is a square. Make 5 a1y — 25 y2+16a+=(4a), and we shall

[blocks in formation]

404 a2

[merged small][ocr errors]
[blocks in formation]

be taken at pleasure.

12. Qu. (54) Answered by Rynaldo.

Since the base AB and

the BAC are given, if

the point B be supposed to

approach towards D, it is
evident that the difference
BC-DC will decrease till A

thing.

D

B

B coincides with D, when the difference will be noHence the construction is obvious; for, if a be erected at the point B, and AC drawn to meet it, making the given with AB, the thing will be done.

perp.

[ocr errors]

DC=

с

Or, put ABa, AD=x, sin. 4 DACs, its cosinec; then DB-a-x, and by Trig. c: x :: s: nx by substitution. Hence CD = √(n2x2+a2—2ax+x'); therefore, by the question, √ (n2x2 + a2 — 2 ax+x2)—nx is a minimum, in fluxions `n2 xx — ax + xx √(n2x2+a2—2ax+x2)

nxo, or n2x

[ocr errors]

a + x = n

√(n2x2+a2-2ax+x2), which being squared and re

[merged small][ocr errors]

2 ax n2+1

[merged small][ocr errors][merged small]

n2+1
n2 + 1

[ocr errors]

whence by com

pleting the square, &c. x = a, the same conclusion as above.

Similar answers were received from Messrs. Baines, Butterworth, Dunn, Eyres, Gawthrop, Hine, Hirst, Maffett, Nesbit, Whitley, and Winward.

P

[ocr errors]

C

13 Qu. (55) By Mr. E. S. Eyres, and Rylando. Analysis. Let ABC be the triangle sought, CD the line bisecting the base, and CI perp. to AB. By a known property of plane triangles, AC2+BC2-2AD2 +2DC, and by adding AB to each, we have AB'+AC' +BCAB2+2DC2; but A by the question AB2+AC2+ BC2, and 2 DC2 are given, therefore AB becomes known, and the sum of CI and 2 ID is equal to a given

B

D

12

line, or a maximum. Draw DP parallel to IC, and make it equal to the given sun; join PC, and let it meet AB in Q, also draw CL parallel to BA; then DL

IC, and DP IC+2 DIIC+2CL; hence PL = 2 LC, and PD = 2DQ (by sim, trian.) therefore Q is a given point, and PQ the right line, which is obviously the locus of the vertex C, given in position; and as the locus of C is likewise a circle to centre D and radius DC, the intersection of these loci, or the point C, is given, and the triangle determined.

In the case of the max. the vertex C will be determined when PQ touches the circle in C.

COMPOSITION. By Mr. J. Butterworth.
Take AB

Construction.

the side of a square which is equal to one third of the difference between twice the given sum of the squares of the sides, and four times the square of the line bi

DEEG

secting the base; bisect AB in D, and with the centre D, and rad. equal to the given line, describe a circle; upon DB take DG equal to half the given sum, and on GD take Ge of any length, and at e erect the perp. ec 2 Ge; through G and c draw GC meeting the circle in C; join AC, BC, and ABC will be the triangle required.

Demonstration. Join DC, and demit CE perp. to AB; then, by construction, 3 AB twice the given sum of the squares of the sides, minus 4 DC2; but 2 AC+2 BC4 A D2 (AB)+4 DC; therefore, 2 AB2+2A C2 + 2 B C2 = 3A B2 + 4 D C2, or 3 A B2= AB2+2AC2+2BC3,-4DC2; consequently, AB2 + ACBC the given sum of the squares.

2

Again, by construction, ec = = 2 eG; hence, by sim. triangles, EC2 EG; therefore, 2 DE × (AE-BE) +2 EG (EC) to the given sum. 2. E. D.

Again, by Messrs. Dunn, Hine, Maffett, and Winward.
Analysis. Let ACB be the

triangle required, inscribed in
the circle AÇ BE; draw CD bi-
secting the base AB in D, and
produce it to meet the circle in
E. Then, by the property of
the circle, CD × DE
DB AD'; also, AC+BC2

=

AD ×

2 AD2 + 2 DC2; therefore, AC2 + BC2 + AB2 = 2 CD. DE

P

A

DL BE

E

+2CD+4CD. DE 6CD. DE+2C D'a

given space.

But CD is given; therefore, 6 CD. DE, or CD. DE =AD' is given, and AB becomes known; hence, the following

Construction.

Take A B, as determined by the above analysis, and bisect it in D with the indefinite perp. DG; take DG the given sum of the perp. and the difference of the segments of the base; and, in DB, produce if necessary, take DF = DG, join GF, and with centre D and radius = the given line bisecting the base, describe the arc WCS cutting GF in C; join AC, BC, and ABC is the triangle required.

Demonstration. Demit the perp. CL, and draw CP parallel to AB; then, by sim. trian. GD: DF :: GP PC. But, GD=2 DF, by construction; therefore, PG 2 PC 2 DL AL - LB; hence, CL (PD)+ 2 DL (GP) GD = the given line, by construction; the rest is evident.

=

And almost exactly in the same manner as this last, were the answers by Messrs. Baines, Gawthrop, Hirst, and Tomlinson.

Mr. Gawthrop is requested to send for any book he may think proper, the price of which does not exceed Half a Guinea.

NEW MATHEMATICAL QUESTIONS

TO BE ANSWERED IN NO. VIII.

1. Qu. (75) By Mr. W. Dunn, Broughton.

A GENTLEMAN has a circular garden which contains just half an acre of land, and is desirous of having a fishpond in the centre, in the form of the segment of a sphere, the diameter and depth of which are to be as 8 to 1-required the dimensions of the pit, when the earth dug out will cover the remainder of the garden one inch deep.

2. Qu. (76) By Mr. J. Baines, jun. Horbury Bridge. Having a globe of wood, and wanting to know its weight, I immersed it in a vessel of fresh water, and on taking it out, I found that the surface of the wet part was 209.44, and that of the dry part 104.72 inches; -required its weight.

« PreviousContinue »