PROBLEM XLV. On a given line AB to construct a regular octagon. From L First method, by the rule and compasses. centres A and B, with radii equal to AB, describe the circles CQB, DRA, intersecting in E; from centre E, and radius AB, describe the circle FHKG, cutting the former circles in F and G; from centres F and G, and radius AB, describe arcs cutting FHKG in H and K; draw AH, BK, cutting the first two circles in Q and R; draw BQ, AR, and in them produced take QL, RM, each equal to AB; and lastly, with centres L and M, describe the circles NQC, PRD, cutting the lines AH, BK, in N and P, and the first circles in C and D. Then A, B, D, M, P, N, L, C, are the angular points of the octagon. Second method, adapted to the use of the parallel ruler. Describe the circles from centres A and B, with radius AB intersecting in C and D; through A and B draw AM, BN, parallel to CD, cutting the circles in G and H; draw AH, BG, and parallel to them AE, BF, to meet the circles in E and F; through E, F, draw EK, FL, parallel to CD, AM, or BN, meeting BG in K, and AH in L; and lastly, draw KM, LN, parallel to AL, BK, meeting AG, BH, in M and N respectively, and join MN. The figure ABFLNMKE, is a regular octagon on AB. A D D The principles of these constructions are too obvious to need detail here. PROBLEM XLVI. To construct an isosceles triangle, whose vertical angle shall be the half or the double of the vertical angle ACB of a given isosceles triangle ABC on the same base AB. 1. WITH centre C describe the circle ABD, and with the same radius, and centres A, B, arcs intersecting in E; join EC, and produce it to meet the circle in D: then AD, DB, being drawn, ADB is the triangle required. For, the construction gives EC, perpendicular to AB, and bisecting it; hence also ADB is an isosceles triangle; and since ADB is an angle at the circumference, and ACB an angle on the same arc at the centre, ADB is the half of ACB. The conditions are, therefore, fulfilled. 2. With centres A, B, C, describe circles with radius greater than half the side AC or BC, so that that about A intersects that about C in E and F, and that about B intersects it in G and H: then EF, GH, will intersect in the vertex D of the triangle sought. D B For, D is the centre of the circle about ACB, and hence AD, DB, are equal; and the angle ADB at the centre is double ACB at the circumference. PROBLEM XLVII. On a given line GH homologous to a given side AB of a given rectilineal figure ABCDEF, to construct a figure similar to the given one. F E D L K H M FROM One of the extremities of the line AB, which is homologous to GH, draw lines to all the angles of the figure; on GH construct a triangle GKH, equiangular to ACB; on GK a triangle KLG, equiangular to ACD; on GL a triangle LMG, equiangular to ADE; on MG a triangle GMN, equiangular to AEF; and continue the process as long as any triangles of the given figure remain: then GHKLMN is the figure required. For GKH+ GKL = = ACB + ACD, KLG + GLM = CDA + ADE, and so on (constr.): hence the two figures are equiangular. Also BC: CA :: HK : KG and AC: CD:: GK: KL; and hence BC: CD: HK: KL, or the sides about the equal angles at C and H are proportional; and the same may be shown of all the other homologous sides of the two figures. The figure GHKLMN is hence similar to the figure ABCDEF. Scholia. 1. When the two sides AB, GH, are coincident, in the manner of AB, AH, the construction becomes simpler, since it consists merely in drawing HK parallel to BC, cutting AC in K, KL parallel to CD, cutting AD in L, and so on till MN is drawn parallel to EF, cutting AF in N. For, in this case the figures are composed of similar elementary triangles, and are therefore, similar to one another, as before. 2. When GH lies in the same line with AB, it is only requisite to draw GK, GL, GM, GN, parallel to AC, AD, AE, AF, respectively; then HK parallel to BC, KL parallel to CD, and so on till we arrive at MN, parallel to EF, and join NG. The same similarity of figure as in the preceding construction obviously takes place. F E F E F L M N K G H N 3. When the homologous sides AB, GH, are parallel, draw AG, BH (or AH, BG), meeting in P: then draw HK parallel to BC, meeting CP in K, KL parallel to CD, meeting DP in L, and so on. If the line GH be very distant from AB, or very nearly equal to it, then as the point P would fall at an inconvenient distance on the drawing, the point P may be taken between the lines, by joining BG and AH. The remaining part of the construction would be almost identical with that here given, but the order would be reversed. 4. Since equiangular triangles have the sides about the equal angles proportional, and triangles in general are most easily constructed by means of their sides, the proportional compasses are convenient in the general construction of this problem. PROBLEM XLVIII. To draw a complex figure similar to another figure, on the same or different scales, by means of squares. SURROUND the given figure by a square or a rectangle of convenient size, and divide it by pencil lines, intersecting perpendicularly, into squares, as small as may be deemed necessary. Generally, the more irregular the contour of the figure, or the more numerous the sinuosities or subdivisions, the more numerous the squares should be. Then draw another square or rectangle, having its sides either equal to the former, or greater or less in the assigned proportion, and divide this figure into as many squares as there are in the original figure. Draw in every square of the new figure, right lines or curved to agree with what is contained in the corresponding square of the original figure; and this, if carefully done, will give a correct copy of the complex diagram proposed. The pentagraph is also often used for the same purpose; but as there are great practical difficulties attached to it (especially its deficiency in easy, and consequently certain, motion), that it is not so valuable as its theoretical principles would lead us to anticipate. Dr. Wallace has obviated this and most other inconveniences by his eidograph: still its expense and liability to derangement have greatly militated, and perhaps ever will, against its introduction into general use. It is, however, but just to remark that Dr. Wallace's instrument is, in reality, but another, though much improved, form of Scheiner's pentagraph. Another form, somewhat intermediate to the common pentagraph and the most improved form of the eidograph, was exhibited to the Society of Arts in Scotland a few years ago, and which appears on the whole to answer the purpose of copying as well as the eidograph, and to be considerably less expensive. PROBLEM XLIX. To draw a straight line nearly equal to the circumference of a given circle. B First method. Let AE be the diameter and C the centre of the circle, and let the semicircle be described upon AE. Set off AB, ED, each equal to radius AC. With centres A and E, and distances respectively equal to AD, EB, describe arcs intersecting in F. Then with centre B and radius BF describe a circle cutting the circumference (on the side D) in G. The chord AG is nearly equal to the quadrant of the circle. E For AF AD = 2 sin 60o = √3, and CF = √AF2 AC2√2; and FH = CF — CH = } {2 √ 2 − √ 3)} Again, from the triangle BHF, we 6; hence sin &BG= √3—√√/6, AB = 3. Therefore 2AG= BG) = 4 sin & AB cos BG + 4 sin BG cos AB have BG = BF = √ BH2 + HF2 = √ 3 ་ 6 = 3·142399....But to the same decimal extent the true value of the circumference is 3.14159; whence the degree of approximation is sufficiently close for most practical constructions. This method, which may be performed by the aid of the compasses only, was invented by Mascheroni. D C G Second method. Let AB be the diameter of a circle, and C its centre. Draw an indefinite tangent at the point A, and a radius CD parallel to this tangent. Set off the radius DC towards A terminating in F, and draw CF to meet the tangent in E; and take upon this tangent from E on the side of A, EG = 3CD, the straight line BG is nearly equal to the semicircle. = EG AE 3 For, AE tan 30° = √3: hence AG √3 = (9/3), and BG = √AB2 + AG2 = √4 + § (9 — √/3)2 = }√/6(20 — 3/3) 3.1415334. This method, which is by an anonymous German author, gives a closer approximation than that of Mascheroni. Third method. From any point A in an indefinite straight line draw a perpendicular AB equal to the given radius. Set off three times this radius from A to D, and draw BD. At the first of these divisions C of AD, draw the perpendicular CE. Set off DE in the prolongation of AD to F. Prolong AF beyond its extremities A and F by the lines AH and FG equal to radius AB. Take FK (AB + AB) = AB; and make AL = AH. Then KL is nearly equal to the circumference of the circle whose radius is AB. This method, which is remarkable for its extent of approximation, being true to six places, was invented by M. Piochè, an eminent statuary of Metz. and aR = 15aB. From a draw the perpendicular am = BA = D, and draw Rm, cutting AB in c; and finally draw D'c. This will be nearly equal to the circumference, whose diameter is AB. For by similar triangles, RBc, cmn give mn: nc :: ma: aR; whence nc. ma aB. D D mn = Ac= D Then Bc BA Ac = D = 15 14D 15' D√2221 Again, by the right-angled triangle D"cB, D'c = D = 3.1418. 15 This method is by M. Quetelet of Brussels, and though not extremely approximative, yet, being easy of application, is very convenient in practice. Fifth method. Upon the circumference, whose centre is O, and radius OB = 1, take the arc BC= 30° (which is found by the rules and compasses), draw the tangent BC, and by the other extremity A of the diameter BO, draw the indefinite tangent AD, upon which set off AD equal to three times the radius OB. Through C draw CE, parallel to BA, and join CD, which will represent very nearly the semicircle to radius OB. Ε A For, the right-angled triangle DCE gives DC = √DE2 + EC2 = √(DA — CB)2 + CE2; and BC = tan 30o = This method is by M. De Gelder of Leyden, and is also convenient, from the simplicity of the work required.* PROBLEM L. To measure an angle by means of a pair of compasses only. THIS will be easily comprehended by giving a single example. The method, in fact, consists in measuring an arc or angle proposed with a pair of compasses, without any scale whatever, except an undivided semicircle. Produce one of the sides of the angle backwards, and then with a pair of accurate compasses describe as large a semicircle as possible, from the angular point as a centre, cutting the sides of the proposed angle, and thus intercepting a part of the semicircle. This intercepted part is accurately taken between the points of the compasses, and stepped upon the arc of the semicircle, to ascertain how often it is contained in it; and the remainder, if, as usual, there be one, marked; then take this remainder in the compasses, and in like manner find how often it *These methods of construction, almost unknown in this country, were first collected together and published by the editor in Leybourn's Mathematical Repository a few years ago. D d VOL. I. |