cut the opposite side in G. Then FG is the radius of the inscribed, and FA that of the circumscribing, circle. PROBLEM XXXV. In a given circle to inscribe a square. First method. At A with any convenient radius describe a circle EFG, cutting EA in G ; and from A draw AC parallel to FG, meeting the circle again in C. From C set off CH equal to AE, and draw CH to meet EA in K; and through K a line parallel to EF, meeting the circle in D and B. Then A, B, C, D, are the angular points of the inscribed square. K H B For the construction (prob. 28) gives AC a diameter; and since the arcs EA, CH, are equal, the angles EAC, HCA, are equal; whence AKC is isosceles; and (prob. 6) KB is perpendicular to AC; and the two diameters AC, BD, are at right angles. Whence the four right-angled triangles AOB, BOC, COD, DOA, have all their sides about the right angles equal; and, therefore, AB, BC, CD, DA, are all equal. Also, since each of the angles ABC, BCD, CDA, DAB, is in a semicircle, it is a right angle. The figure, therefore, is a square *. Second method. With centre A, and any radius greater than that of the circle, but less than its diameter, describe the circle KGHL, cutting the given circle in G and H: from G and H, with the same radius, describe arcs intersecting at E, and join EA, cutting the given circle in C: from C, with same radius, describe a circle meeting KGHL in K and L, and draw KL meeting the given circle in D and B. Then ABCD will be the square, as before t. D B PROBLEM XXXVI. To describe a square about a given circle. (See the figures of preceding problem.) FIND the diagonals of the inscribed square, viz. AC, BD: through A and C draw parallels to BD, and through D, B, draw parallels to AC. The intercepted portions of these parallels constitute the square required, as is obvious. *This solution supposes that the centre of the circle is not given but when the centre is already known, all we have to do is to draw the two diameters AC, BD, at right angles to one another for finding the angular points of the square. The construction above is only a combination of the process for finding the centre with that of drawing two diameters at right angles to one another. The use of the parallel ruler is, however, implied; but the next solution is by the ruler and compasses only. This is only another method of getting the rectangular diameters, drawn through the centre of the given circle. Other very obvious variations of these constructions might easily be given; but these, when the centre is not given, and the method intimated in the preceding note when the centre is given, are sufficient for all practical purposes. When, however, only a part of the PROBLEM XXXVII. To inscribe and circumscribe circles to a given square ABCD. DRAW the diagonals AC, BD, intersecting in F, and draw FE bisecting the angle DFC (or parallel to AD): then F is the common centre of the two circles sought: FD is the radius of the circumscribed one, and FG of the inscribed one: and the circles can be drawn. G PROBLEM XXXVIII. To inscribe and circumscribe circles to a given regular polygon, of which two sides AB, BC, and the angle ABC are constructed. THIS problem is precisely the same as to describe a circle through three given points, that is, as prob. 33; where AB is equal to BC, which lessens somewhat the actual labour of construction. PROBLEM XXXIX. On a given straight line AC to describe a regular pentagon. First method. At right angles to AC take CO equal to half AC, and with centre O and radius OC describe a circle: join AO, and produce it to cut the circle in D: with A and C as centres, and radius AD, describe arcs to intersect in E: and finally, with A, E, C, as centres, and AC as radius, describe arcs cutting in F and G respectively. Then A, C, G, E, F, are the angular points of the pentagon. Second method. Construct as before to find D, and with centres A and C, and radius CD, describe arcs intersecting in H: then H is the centre of the circle in which the angles of the pentagon will be situated; and by setting off AF, FE, EG, all equal to AC, the angular points will be obtained. As these two constructions are derived from the same process, it will be advisable to employ one as a check upon the accuracy of the other: and as they must flow from the same principle, their demonstrations may be conjoined in one course of reasoning, founded on Ex. 73 of the Miscellaneous Exercises, p. 317 but which, as it is presumed that the student has investigated for himself, it need not be given here. circle is given, it will be requisite to find the radius of the circle, and on the diameter passing through the given point taken as a diagonal to describe the required square. Third method. From A and C, with radii equal to AC, describe circles HCD, KAD, intersecting in B and D, and draw the indefinite line BD cutting AC in L. Make LE equal to AC, join AE and produce it till EF is equal to AL or LC. With centre A and radius AF describe an arc cutting BD in G; and from centre G with radius equal to AC describe a circle cutting HCD in H and KAD in K. Then the points A, C, K, G, H, are the angular points of the pentagon. E The proof of this is also dependent on the same principles as the two former constructions. PROBLEM XL. In a given circle ALBM to inscribe an equilateral and equiangular pentagon. L G E 0 LET O be the centre, and B the position of one of the angles. Draw the diameter BOA: from centre A with radius AO describe the circle POQ, and parallel to PQ draw LM and AC through O and A respectively; then from centre D, where BC cuts LM, describe the circle BE, cutting LM in E. Lastly, with centre B describe the circle FEG cutting the given circle in F, G; and with the same radius and centres F, G describe arcs also cutting the given circle in H and K. Then B, F, H, K, G, are the angular points of the required pentagon P Q H K .A. C PROBLEM XLI. About a given circle ABCDE to describe an equilateral and equiangular pentagon. FIND, by the last problem, the angular points A, B, C, D, E, of the regular inscribed pentagon. Parallel to each side, as C, D, (or to the line which joins the other extremity of the sides AB, AE, which meet in A,) draw lines as GH : similar operations being performed through all the angles, viz. HK drawn through B parallel to DE, KL through C parallel to AE, LM through D parallel to AB, and MG through E parallel to BC, the figure GHKLM formed by them will be the circumscribed pentagon required +. * The principle of this simple construction was given by Ptolemy in his Almagest; but it had never been demonstrated independently of the doctrine of proportion till about sixty years ago by Mr. Bonnycastle, formerly Professor of Mathematics in the Royal Military Academy. Its proof is best effected by the same principles as the other methods. The description being so simple, the student is left to construct his own figure. It may be remarked here once for all, in reference to circumscribed regular polygons, that their construction may be always effected by drawing tangents to the circle at the angular points of the inscribed polygon of the same number of sides; but the lines which form the circumscribed polygon may always be drawn by means of parallels, without reference to their tangency to the circle. When the polygon has an odd number of sides, draw the parallel to the most distant side; and when an even number, draw it parallel to the diagonal line, joining any two equidistant angular points of the inscribed polygon. The construction for the odd number is instanced, as in the pentagon above, and for the even number in the hexagon of the next problem. PROBLEM XLII. To describe a regular hexagon: (1) on the given line AB as side: (2) in a given circle and (3) about a given circle. 1. When the hexagon is to be constructed on the given line AB. FROM centres A and B, with radii AB, describe segments of circles BCD, ACE, intersecting in C; and with the same radius describe from centre C another circle, ABD cutting the preceding segments in E and D respectively. Draw AC, BC, to meet the circle ABD in F and G. Then A, B, E, F, G, D, are the angular points of the hexagon. D A Σ For, the three arcs AD, AB, BE, are equal by construction, and hence the angles DCA, ACB, BCE, are equal: and the three triangles DCA, ACB, BCE, are equilateral triangles; and hence the three angles at C are equal to the three angles of one of the triangles, that is to two right angles (th. 17); and hence again, DC, CE, are one straight line, and each of the angles one-third of two right angles. The opposite angles to these at Care also equal (th. 5), and hence all the angles at C are equal. The lines AB, BE, EF, FG, GD, DA, are therefore equal, and the angles which they contain also equal; and the figure is, therefore, a regular hexagon. 2. When the hexagon is to be inscribed within the given circle ABE. WITH the radius of the given circle, step the compasses round the circumference of the circle: it will, by the last case, mark out the angular points of the hexagon. E H D G 3. When the hexagon is to be circumscribed about a given circle. SET off the six angular points in order A, B, C, D, E, F, of the inscribed hexagon by the last proposition. Parallel to FB or EC draw the lines GH, LM, through D and A: and proceed similarly for the lines HK, MN, parallel to DF or AC, and for K the lines GN, KL, parallel to EA or DB. Then the six points G, H, K, L, M, N, will be the angular points of the circumscribed hexagon. L A M For, join DA, which by the preceding demonstrations will pass through the centre O of the circle. Also, since the arcs AF, AB, are equal, the line FB is perpendicular to AD, and hence also the lines LM and HG are perpendicular to AD. These, therefore, are tangents to the circle at A and D. In like manner, all the other lines are tangents, and the hexagon is circumscribed. Also KH being parallel to AC, and HG to FB, the angle KHG is equal to FPC, which is measured by half the sum of the arcs FEDC and BA, that is by one-third of the circumference; and in the same way each of the other angles at G, N, ... is measured by one-third of the circumference. The angles are, therefore, all equal, and the figure is a regular hexagon. PROBLEM XLIII*. To inscribe and circumscribe regular heptagons to a given circle. DIVIDE, by trial, the circumference of the circle into seven equal parts, and these will be the angular points of the inscribed heptagon, and the points of contact of the circumscribed one t. PROBLEM XLIV. On a given line AB to construct a regular heptagon. R 0 E T FROM centres A, B, with radius AB, describe circles intersecting in L and M, and produce AB to meet one of them (as that about B) in C: divide by trial the circle about B into seven equal parts, and let CD be one of these divisions about D, with radius AB, describe a third circle cutting ADC in P and N: and draw LM, NP, to meet in O. About O, with radius OA or OB, describe a circle ASB. This will pass through D, and in this set off successively the points T, S, R, Q, with distances each equal to AB. The points thus determined are the angular points of the heptagon described upon the given straight line AB. Scholium. B K E HI H E B No geometrical method (that is, by means of the ruler and compasses) can be given for the construction of regular polygons, except in very limited cases. The polygons of 7, 9, 11, 13, 18, 19, 21,.... sides belong to this class; but it fortunately happens that they do not very often occur in drawing operations. The following has been often given by practical writers; but it is only an approximation, and generally a very rude one. It may serve, however, as a first step, or guess at the probable opening of the compasses to be taken, after which the distance may be easily corrected by the eye to any greater degree of accuracy. Let AB be a diameter of the circle, of which it is required to find the nth part by a construction. From centres A and B, and with AB as radius, describe arcs intersecting in C: divide AB into as many parts (the figure is adapted to seven, and the first method of effecting the problem, p. 378,) as it is required by the problem to divide the circumference: through C and the second point of the division D, draw CD, to meet the circumference in H; then BH approximates to the required part of the circumference.† * The same process applies to figures of 9, 11, 13,...... sides; and indeed, the whole series of figures not geometrically constructible might have been included under one enunciation ; as there is, with too rare exceptions to be worthy of notice, no difference whatever in the manner of forming them. See, however, the scholium to the next problem. This method was, I believe, invented by the elder Malton, and first published in his Royal Road to Geometry. A scrutinizing investigation of the degree of its approximation was given by Dr. Henry Clarke, who proposed amendments in it; but these are also, besides being very troublesome, only one degree more close in their approximation. |