PROPOSITION VI. PROBLEM. A regular inscribed polygon being given, to circumscribe a sim ilar polygon about the same circle. S Let CBAFED be a regular polygon. At T, the middle point of the arc AB, apply the H G tangent GH, which will be parallel to AB (Book /BY III. Prop. X.); do the N same at the middle point of each of the arcs BC, CD, &c.; these tangents, I M T by their intersections, will form the regular circumscribed polygon P R GHIK &c. similar to the one inscribed. K Q L Since T is the middle point of the arc BTA, and N the middle point of the equal arc BNC, it follows, that BT-BN; or that the vertex B of the inscribed polygon, is at the middle point of the arc NBT. Draw OH. The line OH will pass through the point B. For, the right angled triangles OTH, OHN, having the common hypothenuse OH, and the side OT=ON, must be equal (Book I. Prop. XVII.), and consequently the angle TOH= HON, wherefore the line OH passes through the middle point B of the arc TN. For a like reason, the point I is in the prolongation of OC ; and so with the rest. But, since GH is parallel to AB, and HI to BC, the angle GHI=ABC (Book I. Prop. XXIV.); in like manner HIK= BCD; and so with all the rest: hence the angles of the circumscribed polygon are equal to those of the inscribed one. And further, by reason of these same parallels, we have GH : AB :: OH: OB, and HI : BC ::OH: OB; therefore GH : AB : : HI : BC. But AB=BC, therefore GH-HI. For the same reason, HI=IK, &c.; hence the sides of the circumscribed polygon are all equal ; hence this polygon is regular, and similar to the inscribed one. Cor. 1. Reciprocally, if the circumscribed polygon GHIK &c. were given, and the inscribed one ABC &c. were required to be deduced from it, it would only be necessary to draw from the angles G, H, I, &c. of the given polygon, straight lines OG, OH, &c. meeting the circumference in the points A, B, C, &c.; then to join those points by the chords AB, BC, &c.; this would form the inscribed polygon. An easier solution of this problem would be simply to join the points of contact T, N, P, &c. by the chords TN, NP, &c.. which likewise would form an inscribed polygon similar to the circumscribed one. Cor. 2. Hence we may circumscribe about a circle any regular polygon, which can be inscribed within it, and conversely. Cor. 3. It is plain that NH+HT=HT+TG=HG, one of the equal sides of the polygon. PROPOSITION VII. PROBLEM. A circle and regular circumscribed polygon being given, it is required to circumscribe the circle by another regular polygon having double the number of sides. P Let the circle whose centre is P, be circumscribed by the square CDEG: it is required to find a regular circumscribed octagon. Bisect the arcs AH, HB, BF, G i F h E FA, and through the middle points c, d, a, b, draw tangents to the circle, and produce them till t they meet the sides of the square: then will the figure ApHdB &c. A B be a regular octagon. P For, having drawn Pd, Pa, let g the quadrilateral PdgB, be applied to the quadrilateral PBfa, d so that PB shall fall on PB. С n2 H Then, since the angle dPB is k D equal to the angle BPa, each being half a right angle, the line Pd will fall on its equal Pa, and the point d on the point a. But the angles Pdg, Paf, are right angles (Book III. Prop. IX.) ; hence the line dg will take the direction af. The angles PBg, PBf, are also right angles; hence Bg will take the direction Bf; therefore, the two quadrilaterals will coincide, and the point g will fall at f; hence, Bg=Bf, dg=af, and the angle dgB=Bfa. By applying in a similar manner, the quadrilaterals PBfa, PFha, it may be shown, that af=ah, fB=Fh, and the angle Bfa=ahF. But since the two tangents fa, fB, are equal (Book III. Prob. XIV. Sch.), it follows that fh, wbich is twice fa, is equal to fg, which is twice fB. In a similar manner it may be shown that hf=hi, and the angle Fit=Fha, or that any two sides or any two angles of the octagon are equal: hence the octagon is a regular polygon (Def.). The construction which has been made in the case of the square and the octagon, is equally applicable to other polygons. Cor It is evident that the circumscribed square is greater than the circumscribed octagon by the four triangles, Cnp, kDg, hEf, Git; and if a regular polygon of sixteen sides be circumscribed about the circle, we may prove in a similar way, that the figure having the greatest number of sides will be the least ; and the same may be shown, whatever be the number of sides of the polygons : hence, in general, any circumscribed regular polygon, will be greater than a circumscribed regular polygon having double the number of sides. PROPOSITION VIII. THEOREM. Two regular polygons, of the same number of sides, can always be formed, the one circumscribed about a circle, the other inscribed in it, which shall differ from each other by less than any assignable surface. Let Q be the side of a square less than the given surface. b Bisect AC, a fourth part of D the circumference, and then bi K sect the half of this fourth, and proceed in this manner, always а: bisecting one of the arcs formed E by the last bisection, until an arc is found whose chord AB is less than Q. As this arc will be an exact part of the circumference, if we apply chords AB, BC, CD, &c. each equal to AB, the last will terminate at A, and there will be formed a regular polygon ABCDE &c. in the circle. Next, describe about the circle a similar polygon abcde &c. (Prop. VI.): the difference of these two polygons will be less than the square of Q. For, from the points a and b, draw the lines a0, 60, to the centre O : they will pass through the points A and B, as was P: shown in Prop. VI. Draw also OK to the point of contact K: it will bisect AB in I, and be perpendicular to it (Book III. Prop. VI. Sch.). Produce AO to E, and draw BE. Let. P represent the circumscribed polygon, and p the inscribed polygon : then, since the triangles aOb, AOB, are like parts of P and p, we shall have aOb : AOB : :P:P (Book II. Prop. XI.): But the triangles being similar, a0b : AOB : : Oa? : OA?, or OK?. Hence, P:p :: 0a : OK?. Again, since the triangles OaK, EAB are similar, having their sides respectively parallel, Oa? : OK? :: AE : EB', hence, :p :: AE : EBo, or by division, P:P-P :: AE : AE?—EB?, or AB%. But P is less than the square described on the diameter AE (Prop. VII. Cor.); therefore P-p is less than the square described on AB; that is, less than the given square on Q: hence the difference between the circumscribed and inscribed polygons may always be made less than a given surface. Cor. 1. A circumscribed regular polygon, having a given number of sides, is greater than the circie, because the circle makes up but a part of the polygon : and for a like reason, the inscribed polygon is less than the circle. But by increasing the number of sides of the circumscribed polygon, the polygon is diminished (Prop. VII. Cor.), and therefore approaches to an equality with the circle ; and as the number of sides of the inscribed polygon is increased, the polygon is increased (Prop. V. Sch.), and therefore approaches to an equality with the circle. Now, if the number of sides of the polygons be indefinitely increased, the length of each side will be indefinitely small , and the. polygons will ultimately become equal to each other, and equal also to the circle. For, if they are not ultimately equal, let D represent their smallest difference. Now, it has been proved in the proposition, that the difference between the circumscribed and inscribed polygons, can be made less than any assignable quantity: that is, less than D: hence the difference between the polygons is equal to D, and less than D at the same time, which is absurd : therefore, the polygons are ultimately equalBut when they are equal to each other, each must also be equal to the circle, since the circumscribed polygon cannot fall within the circle, nor the inscribed polygon without it. Cor. 2. Since the circumscribed polygon has the same number of sides as the corresponding inscribed polygon, and since the two polygons are regular, they will be similar (Prop. I.); and therefore when they become equal, they will exactly coincide, and have a common perimeter. But as the sides of the circumscribed polygon cannot fall within the circle, nor the sides of the inscribed polygon without it, it follows that the perimeters of the polygons will unite on the circumference of the circle, and become equal to it. Cor. 3. When the number of sides of the inscribed polygon is indefinitely increased, and the polygon coincides with the circle, the line OI, drawn from the centre 0, perpendicular to the side of the polygon, will become a radius of the circle, and any portion of the polygon, as ABCO, will become the sector OAKBC, and the part of the perimeter AB+BC, will becoms the arc AKBC. PROPOSITION IX. THEOREM. The area of a regular polygon is equal to its perimeter, multi plied by half the radius of the inscribed circle. Let there be the regular polygon Η Τ GHIK, and ON, OT, radii of the inscribed circle. The triangle GOH N will be measured by GHOT; the triangle OHI, by HIXON: but I ON=OT; hence the two triangles taken together will be measured by (GH+HI) XOT. And, by continuing the same operation for the other triangles, it will appear that K the sum of them all, or the whole polygon, is measured by the sum of the bases GH, HI, &c. or the perimeter of the polygon, multiplied into ZOT, or half the radius of the inscribed circle. Scholium. The radius OT of the inscribed circle is nothing else than the perpendicular let fall from the centre on one of the sides: it is sometimes named the apothem of the polygon. |