PROPOSITION IV. THEOREM. If a regular hexagon be inscribed in a circle, its side will be equal to the radius. Let ABCDEH, be a regular hexagon, inscribed in a circle: then will its side AB be equal to the radius OA. For, the angle AOB is equal to one-sixth of four right angles, (P. 2, c. 1), or one-third of two right angles: hence, the sum of the remaining angles OAB, OBA, is equal to two-thirds of two right angles (B. I., P. 25). But the triangle AOB is isosceles, hence, the angles at the H E A C base are equal (B. I., P. 11): therefore each is one-third of two right angles: hence, the triangle AOB is equiangular: hence, AB=AO (B. I., P. 12). PROPOSITION V. PROBLEM. To inscribe in a given circle, a regular hexagon. Let O be the centre, and OB the radius of the given circle. Beginning at any point, as B, apply the radius BO, six F B E Cor. 1. If the vertices of the alternate angles be joined by the lines AC, CE, EA, there will be inscribed in the circle an equilateral triangle ACE, since each of its angles will be measured by one-sixth of four right angles, or onethird of two (B. I., P. 25, c. 5). Cor. 2. If we draw the radii OA, OC, the figure OCBA will be a rhombus: for, we have OC=CB=BA=OC. Hence, the sum of the squares of the diagonals is equivalent to the sum of the squares of the sides (B. IV., p. 14, c. 2): AC2+OB2=4AB* ~40B*; that is, -2 and by taking away OB3, we have, AC30B; hence, hence, the side of the inscribed equilateral triangle is to the radius, as the square root of three, to unity. PROPOSITION VI. PROBLEM. In a given circle to inscribe a regular decagon. Let O be the centre, and OA the radius of the given circle. For, drawing MB, we have by construction, AO: or, since AB=OM, AO : AB : :: AB : AM. E But since the triangles ABO, Again, in the isosceles triangle BMO, the angle AMB M A B E being exterior, is double the interior angle O (B. I., P. 25, c. 6): but the angle AMB=MAB; hence, the triangle OAB is such, that each of the angles OAB or OBA, at its base, is double the angle O, at its vertex; hence, the three angles of the triangle are together equal to five times the angle O, which consequently, is the fifth part of two right angles, or the tenth part of four; hence, the arc AB is the tenth part of the circumference, and the chord AB is the side of the regular decagon. Cor. 1. By joining the vertices of the alternate angles of the decagon, a regular pentagon ACEGI will be inscribed. Cor. 2. Any regular polygon being inscribed, if the arcs subtended by its sides be severally bisected, the chords of those semi-arcs will form a new regular polygon of double the number of sides: thus it is plain, that the square will enable us to inscribe, successively, regular polygons of 8, 16, 32, &c., sides. And in like manner, by means of the hexagon, regular polygons of 12, 24, 48, &c., sides may be inscribed; and by means of the decagon, polygons of 20, 40, 80, &c., sides. Cor. 3. It is further evident, that any of the inscribed polygons will be less than the inscribed polygon of double the number of sides, since a part is less than the whole. PROPOSITION VII. PROBLEM. A regular inscribed polygon being given, to circumscribe a similar polygon about the same circle. Let O be the centre of the circle, and CDEFAB regular inscribed polygon. ed polygon GHIK &c., similar to the one inscribed. For, 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, NOH, having the common hypothenuse OH, and the side OT=ON, are equal (B. I., P. 17), and consequently the angle TOH=HON, wherefore the line OH passes through the middle point B of the arc TN (B. III., P. 15). In the same manner it may be shown that OI passes through C; and similarly for the other vertices. But since GH is parallel to AB, and HI to BC, the angle GHI=ABC (B. I., P. 24); in like manner, HIK=BCD· and so for the other angles: hence, the angles of the cir cumscribed polygon are equal to those of the inscribed. And further, by reason of these same parallels, we have GH: AB :: OH: OB, and HI : BC :: OH: OB; therefore, GH : AB :: HI : BC. ly: if the circumscribed polygon GHIK &c., be given, and the inscribed one ABC &c., be required, it will only be necessary to draw from the vertices of 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 these points by the chords AB, BC, &c.; this will form the inscribed polygon. An easier solution of this problem would be, simply to join the points of contact T, N, P, &c., by 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. Cor. 4. If through B, A, F, &c., the middle points of the arcs NBT, TAS, SFR, &c., we draw tangent lines, we shall thus form a new regular circumscribed polygon having double the number of sides: and this process may be repeated as often as we please. The new polygon will be regular, because it will be similar to a new inscribed polygon which may be formed (P. 6, c. 2) of double the number of sides of the first. It is plain, that each new circumscribed polygon will be less than the one from which it was derived, since a part is less than the whole. |