PROBLEMS RELATIVE TO THE DIVISION OF FIELDS UR PROBLEM I. To Divide a Triangle into two parts having a Given Ratio, min. 1st By a line drawn from one angle of the triangle. Make AD AB :: m : m + n; draw CD. A D 2dly. By a line parallel to one of the sides of the triangle. Let ABC be the given triangle, to be divided into two parts, in the ratio of m ton, by a line parallel to the base AB. Make CE to EB as m to n; erect ED perpendicularly to CB, till it meet the semicircle described on CB, as a diameter, in D. Make CF CD and draw through F, GF || AB. So shall or divide the triangle ABC in the given ratio. CD2 A B For, CE: CB=-- - : : CD2 (= CF2): CB2.But CE; EB ::m:n, CE or CE CB m: m + n, by the construction: therefore CF2: CB2::m: m+n. And since A CGF: A CAB:: CFCB; it follows that CGF CAB: m : m+n, as required. Computation. Since CB2: Cr2:: m + n : m, therefore, (m + n)cr2 = m. CB2; whence c¥ √ (m + n) = CB √m, or the semicircle in D. Make cr CD: H through P, parallel to EF, draw PQ, the line required. B The The demonstration of this follows at once from case 2'; because it is only to divide FCE, by a line parallel to FE, into two triangles having the ratio of FCE to FCG, that is, of CE to CG. Computation. CG and cr being computed, as in case 1, the distances CH, C1 being given, and CP being to cq as CH to c1: the triangles CGF, GPQ, also having a common vertical angle, are to each other, as CG. CF to a cr. These products therefore are equal; and since the factors of the former are known, the latter product is known. We have hence given the ratio of the two lines CP (=x) to cq (y) as CH to cr; say as to q; and their product = CF. CG, Say = ab; to find x and y. abp Here we find r = √, Y = abq That is, N. B. If the line of division were to be perpendicular to one of the sides, as to cA, the construction would be similar: CP would be a geometrical mean between ca and being the foot of the perpendicular from в upon Ac. 4thly. By a line drawn through a given point P. m m+n cb, b By any of the former cases, draw Im (fig. 1) to divide the triangle ABC, in the given ratio of m to n bisect cl in r, and through and m let pass the sides of the rhomboid cram. Make cape, which is given, because the point P is given in position: make cd a fourth proportional to ca; cr, cm; that is, make ca: cr:: cm: cd; and let a and d, be two angles of the rhomboid cabd, figs. 1 and 2. re, in figure 2, being drawn parallel to ac, describe on ed as a diameter the simicircle efd, on which set off ef ce ap: then set off dm or dm' on ca equal to df, and through P and м, P and M draw draw the lines LM, L'M', either of which will divide the triangle in the given ratio -The construction is given in 2 figs. merely to avoid complexness in the diagrams. The limitations are obvious from the construction: for, the point I must fall between B and c, and the point м between A and c; ar must also be less than rb, otherwise of cannot be applied to the semicircle on ed. Demon. Because cr = cl, the rhomboid cram = triangle clm, and because ca: cr: cm: cd, we have ca. cd=cm. CT, therefore rhomboid cabd = rhomboid crsm = triangle cim. By reason of the parallels CB, bd, and CA, ab, the triangles alp, doм, bGP, are similar, and are to each other as the squares of their homologous sides ap, dì, by : now ed2 = ef2 +df, by construction; and ed = pb, ef = ap, dƒ : = dм; therefore rb2 = ap2 + dм2, or the triangle POG taken away from the rhomboid, is equal to the sum of the triangles aPL, dмG, added to the part cared: consequently CLM = cabd, as required. By a like process, it may be shown that al ́P, dG'M' POG', are similar, and aL'P+dG'M' PbG'; whence pbdм' = aL'P, and CL'M' cabd, as required. = = Computation, cl, cm, being known, as well as ca, ap, or ce, ep, cr= cl, is known; and hence cd may be found by the proportion ca: crcm cd. Then cd-ce = ed, and Vedz ef=√√ ed2 — ap2 = df = dм = dì'. Thus cM is determined. Then we have cl. cm CM CL. N. B. When the point is in one of the sides, as at м; then make CL. CM. (m+n) = CA CB. m, or, CL: CA :: m.CB: (mn) Cм, and the thing is done. 5thly. By the shortest line possible. Draw any line pq dividing the triangle in the given ratio, and so that the summit of the triangle crq shall be c the most acute of the three angles of the triangle Make Cм=CN, a geometrical mean proportional between CP and cq; so shall MN be the shortest line possible dividing the triangle in the given ratio.The computation is evident. Demons. Suppose MN to be the shortest line cutting off the given triangle CмN, and CG MN. MN = MG+GN CG cot M + CG. Cot N = CG (cot M + cot N). But, cot M+ sin (MN) cot N COS M ·+· sin M M P A B C M N G And (equa. COS N XVIII, Analyt. pl. trigonom.) sin м. sin N=COS (MN)-eos sin. (MN) COS(MN)+cos c (MN) COS(M-N)+cos c Theref MN CG. which expression is a minimum when its denominator is a maximum; that is, when cos (MN) is the greatest possible, which is manifestly when M-N = 0, or M = N, or when the triangle CMN is isosceles That the isosceles triangle must have the most acute angle for its summit, is evident from the consideration, that since 2ACMN = CG. MN, MN varies inversely as CG; and consequently MN is shortest when co is longest, that is, when the angle c is the most acute. N. B. A very simple and elegant demonstration to this case is given in Simpson's Geometry: vide the book on Max. and Min See also another demonstration at case 2d prob. 6th, below. PROBLEM II. To Divide a Triangle into Three Parts, having the Ratio of the quantities m, n, f. 1st. By lines drawn from one angle of the triangle to the opposite side. Divide the side AB, opposite the angle c from whence the lines are to proceed, in the given ratio at D, E; join CD, CE; and ACD, DCE, ECB, are the three triangles required. The demonstration is manifest; as is also the AD E B computation. If it be wished that the lines of division be the shortest the nature of the case will admit of, let them be drawn from the mnost obtuse angle, to the opposite or longest side. 2dly. By lines parallel to one of the sides of the triangle. Make CD DH; HB m np. Erect DE, HI, perpendicularly to cв, till they meet the semicircle described on the diameter CB. in E and 1. Make CF = CE, and CK G CI. Draw GF through F, and LK through K, parallel to AB; so shall the lines GF and LK, divide the triangle ABC as required. The demonstration and computation will be similar to those in the second case of prob. 1. 3dly. By lines drawn from a given point on one of the sides. VOL. I. Cccc Fig. Fig. 1. A ag PbD B Fig. 2. A abp B Let p (fig. 1) be the given point, a and b the points which divide the side AB in the given ratio of m, n, p: the point P falling between a and 6. Join PC, parallel to which draw ac, be, to meet the sides AC, BC, in the points c and d : join PC, pd, so shall the lines cp, rd, divide the triangle in the given ratio. . In fig 2, where p falls nearer one of the extremities of AB than both a and b, the construction is essentially the same; the sole difference in the result is, that the points c, and d, both fall on one side Ac of the triangle. Demon. The lines ca, cb, divide the triangle into the given ratio, by case 1st. But by reason of the parallel lines ac PC, bd, ▲ acc ▲ ace, and ▲ bdc = bdr. Therefore, in fig 1, Aac + acp = Aac + acc that is, ACP = Aac: and вod + bde Bbdbdc, that is, вdp = Bbc. Consequently, the remainder ccrd cab.-In fig. 2, ACP = Aac, and AdP = Acb; therefore crd = acP; and ACB-AdP ACB-Acỏ, that is, CBPd CBb. Computation. The perpendiculars cg, CD being demitted, AACP: A ACB :: m : m+n+p::AP.cg: AB. CD. Therefore m AB. CD (m+n+f) Ap.cg=m.AB.CD, and cg — — (m+n+p)AP The line cg being thus known, we soon find ac; for CD: AC:: cg: AC cg AC = CD m AB. AC (m+n+p) AP Indeed this expression may be deduced more simply; for, since ACB ACP :: AC AB: AC. AP::m+n+ɲ : m, we have (m+n+p) ac . AP=M. AB. AC, (m+n+p) AP (m+n+p) PB 4thly. By lines drawn from a given point P within the triangle. Const. |