PROBLEMS RELATIVE TO THE DIVISION OF FIELDS OR OTHER SURFACES. PROBLEM I. ABDB = AB. To Divide a Triangle into two parts having a Given Ratio, min. Ist By a line drawn from one angle of the triangle. Make AD: AB : :m: m+n; draw cd. So shall ADC, BDC, be the parts required. Here, evidently, ADS mtn m+n 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 an to n, by a line parallel to the base AB. Make CE tO EB as m to n; erect ed perpendicularly tu ca, till it meet the semi А B circle described on co, as a dianieter, in D. Make cf = CD: and draw through F, GF || AB. So shall Gy divide the triangle abc in the given ratio. CD2 For, CE: CB=--: --::CD'(=cra): CB?.But CE : EB ::ming or CE : CB : : m: m + ni, by the construction: therefore CFP: CB? :: m: m+n. And since A CGF: A CAB: : :cFR: CBR; it follows that cor : CAB :: m: m+n, as required. Computation. Since cb?: CF:: m + 'n : m, therefore, (m + n)CF* =m. CB? ; whence cy ✓ (m + n)=CB V m, or CF - CBV In like manner, CG = CA mtn mtn 3dly. By a line parallel to a given line. Let ni be the line parallel to which a line is to be drawn, so as to divide the triangle ABC in the ratio of m D sto n. By case 2d draw of parallel to AB, so as to divide abc in the given ratio. E B Through F draw ve parallel to Hl. On ce as a diameter describe a semicircle ; draw GD perp. to ac, to cut the semicircle in d. Make cp = CD: H through P, parallel to EF, draw PQ, the line required. The CE m A The demonstration of this follows at once from case 2 ; be. cause 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 og. Computation cg and cr being computed, as in case 1, the distances ch, ci being given, and cp being to co as ch to ci; 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 ci ; say as p to 9; and their product = CF.CG; Say - ab: to find x and y. abp abg Here we find x =N--, y = ✓ That is, 9 2 CP. CG . CH CP =V . ; CI N. B. If the line of division were to be perpendicular to one of the sides, as to ca, the construction would be similar: CF CG. CI CH CP would be a geometrical mean between ca and cb, b mtn being the foot of the perpendicular from B upon ac. Athly. By a line drawn through a given point p. 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 crem. Make ca = pe, 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 e and d, be two angles of the rhomboid cabd, figs. 1 and 2. Pe, in figure 2, being drawn parallel to ac, describe on ed as a diameter the simicirclc efd, on which set off ef=ce=ap: then set off dx or dm' on ea equal to df, and through p and M, P and n' 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 m between A and c; ar must also be less than pb, otherwise of cannot be applied to the semicircle on ed. Demon. Because cr = fcl, the rhomboid crem = triangle clm, and because ca: cr:: cm: cd, we have ca . cd=cm . ct, therefore rhomboid cabd = rhomboid crem = triangle cim. By reason of the parallels cb, bd, and ca, ab, the triangles alp, dom, bGP, are similar, and are to each other as the squares of their homologous sides ap, dm, bp : now ed= ef? + df?, by construction; and ed pb, ef = ap, df = du; therefore pb? = ap? + dm?, or the triangle pbg taken away from the rhomboid, is equal to the sum of the triangles apl, dmg, added to the part capgd: 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 podm' = 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 : cr :: c m cd. Then cd-ce ed, and ved? – cf*=7 ed? ap2 = df =dM = dn. Thus cm is cl . cm determined. Then we have - CL. см N. B. When the point is in one of the sides, as at m; then make cl.CM. (m+n) = CA CE . m, or, CL:CA :: m.cB: (m+n) cm, 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 Cpq shall be c the most acute of the three angles of the triangle Make cm=CN, M a geometrical mean proportional between cp P and ca ; so shall an be the shortest line pos B sible dividing the triang le in the given ratio. The computation is evident. Demons. Suppose mn to be the shortest C line cutting off the given triangle cmn, and CG I MN. MN = MG + GN = CG cot M + CG.CO N = CG (cot M + cot »). But, cot m+ M. N sin (M+n cot N = - And (equa. sin m, sin N А B COS M COS N sin M sin XVIII, Analyt. pl. trigonom.) sin m.sin N= cos (M-N)- {eos sin. (M+N) (+N = cos(M-N)+{cos c Theref mxzcG.- -; scos(M-N)+{cos c which expression is a minimum when its denominator is a maximum ; tha: is, when cos (M –n) is the greatest possible, which is manifestly when -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 sioce 2 ACMN = CG. MN, Mn varies inversely as cg; and consequently mn is shortest when cg 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, p. 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 act, 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 most obtuse angle, to the opposite or longest side. 2dly. By lines parallel to one of the sides of the triangle. Make cd : DI: AB :: m : n: p. Erect C DE, HI, perpendicularly to CB, till they meet the semicircle described on the diameter I CB. in E and e. Make CF = CE, and ck = Draw GF through F, and lk through K, parallel to AB; so shall the lines of and lk, divide the triangle abc as required. The demonstration and coniputation 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. Сccc D CI. Fig. Á dJ POD B A6 PB Let TM (fig. 1) be the given point, a and b the points which divide the side as in the given ratio of m, nh, p : the point e falling between a and b. Join pc, parallel to which draw ac, bd, to meet the sides ac, BC, in the points c and d : join pc, Pd, so shall the lines CP, pd, 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, thai the points C, and d, both fall on one side ac of the triangle. Demon. The lines ca, co, divide the triangle into the given ratio, by case 1st. But by reason of the parallel lines ac. PC, bd, A acc A ACP, and a bdc = bdr. Therefore, in fig. 1, AQC + acp = Aac + acc that is, ACP = ADC : and bbd + bde =bbd + bdc, that is, ode = Bbc. Consequently, the remainder cced cab.-In fig. 2, ACP = Aac, and adp = Acb; therefore cpd = acp; and ACB - Adp = ACB-Acb, that is, CBPd = cob. Computation. The perpendiculars cg, co being demitted, A ACP: A ACB::m;m +n +1:: AP.cg :AB.CD. Therefore (m+n+R) AP.cg=M.AB.CD, and cg = The line (m+n+PAP og being thus known, we soon find ac; for CD : AC :: cg : Indeed this expression may be (m+n+p) AP deduced more simply ; for, since ACB : ACP : : AC AC. AP::month: m, we have (m+n+1) AC . AP=m. AB . AC, By a like process is obtained, in (m+n+p) AP P. AB · BC m+n) AB. AC fig. 1, Bd = ; and, in fig. 2, ad = (m+n+1) PB (m+r+p) AP 4thly. By lines drawn from a given point p within the triangle. AB, CD AC 08 AB, AC AC = CD AB: m. AB. AC and sc = Const. |