PROPOSITION XVII. PROBLEM. 394. To inscribe in a given circle a regular decagon. F B Let O be the centre of the given circle. It is required to inscribe in the given ○ a regular decagon. and divide it in extreme and mean ratio, so that O C shall be to OS as OS is to SC. From C as a centre, with a radius equal to OS, Then BC is a side of the regular decagon required. § 311 Cons. Cons. Moreover the OCB = Z SC B, Iden. $284 .. the AOC B and B C S are similar, (having an of the one equal to an of the other, and the including sides proportional). But the AO CB is isosceles, (its sides O C and O B being radii of the same circle). $160 .. the ABC S, which is similar to the AO CB, is isosceles, (the exterior of a ▲ is equal to the sum of the two opposite interior ▲). .. the sum of the s of the ▲ OCB = 50. .. the arc B C is of the circumference, and .. the chord BC is a side of a regular inscribed decagon. Hence, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater segment ten times as a chord. Q. E. F. 395. COROLLARY 1. By joining the alternate vertices of a regular inscribed decagon, a regular pentagon may be inscribed. 396. COR. 2. By bisecting the arcs BC, CF, etc., a regular polygon of 20 sides may be inscribed, and, by continuing the process, regular polygons of 40, 80, etc., sides may be inscribed. PROPOSITION XVIII. PROBLEM. 397. To inscribe in a given circle a regular pentedecagon, or polygon of fifteen sides. C B A H Let be the given circle. It is required to inscribe in Q a regular pentedecagon. Draw EH equal to a side of a regular inscribed hexagon, § 391 and E F equal to a side of a regular inscribed decagon. § 394 Join FH. Then FH will be a side of a regular inscribed pentedecagon. .. the chord FH is a side of a regular inscribed pentedecagon, and by applying FH fifteen times as a chord, we have the polygon required. Q. E. F. 398. COROLLARY. By bisecting the arcs FH, HA, etc., a regular polygon of 30 sides may be inscribed; and by continuing the process, regular polygons of 60, 120, etc. sides may be inscribed. PROPOSITION XIX. PROBLEM. 399. To inscribe in a given circle a regular polygon similar to a given regular polygon. Let A BCD, etc., be the given regular polygon, and C' D'E' the given circle. It is required to inscribe in C'D'E' a regular polygon similar to A B C D, etc. From O, the centre of the polygon A B C D, etc. draw OD and O C. From O' the centre of the OC' D'E', draw O'C' and O' D', making the 0' = ≤ 0. Draw C' D'. Then C'D' will be a side of the regular polygon required. For each polygon will have as many sides as the ≤0 (=0') is contained times in 4 rt. Æ. .. the polygon C'D' E', etc. is similar to the polygon CD E, etc., (two regular polygons of the same number of sides are similar). $372 Q. E. F. PROPOSITION XX. PROBLEM. 400. To circumscribe about a circle a regular polygon similar to a given inscribed regular polygon. Let HMRS, etc., be a given inscribed regular polygon. It is required to circumscribe a regular polygon similar to HMRS, etc. At the vertices H, M, R, etc., draw tangents to the O, intersecting each other at A, B, C, etc. Then the polygon ABCD, etc. will be the regular polygon required. Since the polygon ABC D, etc. has the same number of sides as the polygon HMRS, etc., it is only necessary to prove that A B C D, etc. is a regular polygon. § 372 |