7.841584 x2 + 40 x a max. in fluxions, &c. we have za - 13.06937 = 33.4; whence r = 3.474, and the area of the ellipse, as in the preceding solution 6.1689, and 7s. 5 d. = 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 2, the question requires us to transform the four following formulæ into squares; viz. xy +-z, x2 + y2, yz + x?, and x + y + x are all to be squares. Assuming x = 4 (y +x) the first and second formulæ become squares, and by substitution, yx + x = 16 y + 33 yz + 16z?, = (suppose) (p--4y)=p?g? — 8pyz + 16yo; hence (p? — 16) X 2 = (33 + 8p) x y. Take y =p— 16, then %= 33 + 8 p, and x = 4p? +32 p +68, therefore x +y +z = 5p + 40p+85= O=a’. Let p=n-2; then a’ = 25 + 20n +5 n = 20+10s (5— sn)?=25—10 sn + s?n?, andn= 15, where s is indeterminate. Ex. Taking's = 3, we have n= 25 21 p= , y=p? — 16 = 4,2= 117 and x = 845, or multiplying each by 4, a square number, we shall have r = 13. X 260=3380, y = 13 x 29 = 377, and % = 13 x 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+2, x3 + xy, y2 + x 2, and z? + xy. Let x = 4 (y + x), then the second and third formulæ become squares, and the other two will stand thus: 16 ya + 33 yz + 16z?, and (y + x) x 5. Suppose y = 5n, and %= 5r, they then become 16 n? + 33 nr + 16 r?, and n+r; make n tr=p, then n = p - r, which substituted for n in the other formula, it becomes 160* trpi.ro, which must be a square. Assume 4p goo for its root, and r is found= 1 xp, where v may 21 340, a 680 be any number at pleasure. Ex. Leto=15, then r'= 65105 p’, consequently y = 57 p, = 34 p’, 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. Garthrop, The question requires that x+y+z,zxy, y: +22, and x2 + yz, may be all squares; put x +y +ora’, then x = a-I—y, which substitute for zin the second, and it becomes at-2a’y ty_2a2x + 3xy + x2, which equate to (a’+y—~)', and we find x =*, hence a= by, and, by substitution, the third and fourth 404 — 20 a’y + 2532 py, 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 a’y—259 +160+=(4a?-?), and we shall a become 25 'y lo; hence x – 404 až 2 505» y = 505, and x = 81 a? 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 a BC-DC will decrease till A B coincides with D, when the difference will be no. thing. Hence the construction is obvious; for, if a perp. be erected at the point B, and AC drawn to meet it, making the given a with AB, the thing will be done.. Or, put AB = a, AD = x, sin. . DAC=s, its cosine = c; then DB=-- , and by Trig. c:x :: 5 : DC = ** = ns by substitution. Hence CD = (no x2 +a2ax+ w*); therefore, by the question, v (nox? + a-2ax + x) - nx is a minimum, in fluxions n2 X3 -- ax + x mt "* ° 0 V (nor ta-2ax + x2), which being squared and re 2 ar n’a? a? duced, we get 2? ~ 72 7ī= 7+T" pleting the square, &c. x=2, the same conclusion as above. Similar answers were received from Messrs. Baines, Butterworth, Dunn, Eyres, Garthrop, Hine, Hirst, Maffett, Nesbit, Whitley, and Winward. ' · 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, AC + BC-2AD? +2DC, and by adding AB* to each, we have AB'+AC? +BC=ABP +2DC?; but A by the question ABP +AC? + BC?, and 2 DC? are given, therefore AB becomes known, and the sum of CI and 2 ID is equal to a given line, or a maximum. Draw DP parallel to IC, and make it equal to the given sunı; join PC, and let it meet AB in Q, also draw CL parallel to BA; then DL =IC, and DP = IC+2 DI IC+ 2CL; hence PL= 2 LC, and PD = 2D2 (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. Construction. Take AB the side of a square which is equal to one third of the LAN difference between twice 4 the given sum of the squares of the sides, and four times the square of the line bi. 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 ineeting 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 DC; but 2 AC2 + 2 BC= 4 A D2 (AB2) + 4 DC?; therefore, 2 AB’ + 2 AC+ 2BC= 3AB? 1.4DC, or 3 A B’= 2 A B’+2 AC+2 BC”,—4DC?; consequently, A B2 + AC? + BC = the given sum of the squares, Again, by construction, ec=2 G; hence, by sim. triangles, EC = 2 EG; therefore, 2 DE X (AE BE) + 2 ÉG (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 AC BE; draw CD bi. secting the base A B in D, and produce it to meet the circle in long Ē. Then, by the property of the circle, CD X DE=AD X W! DB = AD?; also, AC + BCA AND LB = 2 A D2 + 2 DC2; therefore, AC: + BC + ABʻ = 2 CD. DE + 2 CD? +4CD, DE=6CD.DE +2C D = a given space. . X E 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 203.44, and that of the dry part 104.72 inches; -required its weight. |