181 



Againsince inAnalysis A . (AG + AE) — (AD — AC) isgiven^ „ 



and Theor. IV (AG + AE) x (AD —AC) is givenj 



therefore AG + AE) 



y are mven : 

 and AD— AC^ ^ 



and thus since AD — AC is given, and AO is given, 



(AD — AC) X AO is given ; 



that is Theor. V AH x AB . . is given, 



but Analysis B AH + AB . . is given ; 



therefore the sum and rectangle of AH and AB are given, whence 



AH and AB are each given, and thence, as is evident, tlie side of 



the polygon. 



And hence now, combining the several constructions as derived in 

 the Analysis, the construction of the problem, or division of the cir- 

 cumference into 17 equal part-, will be simply as follows : — 



Construction. — See Plate, Fig. 5. — Draw the radius OC per- 

 pendicular to the diameter AK ; take OD equal to one-fourth of the 

 radius ; and make DE, DF, each equal to DC ; and EG and FH 

 equal to EC and FC respectively ; divide OG into two parts OX, IG, 

 such, that tl.eir rectangle shall be equal to OH x AO ; then if from 

 A, 01 be applied on the circumference to R, and the arch KR be 

 bisected in P — or IG be taken twice in the circumference from 

 A to N ; the arch KP, or AN will be the 17th part of the whole 

 circumference. 



For if DC, EC, FC be drawn, it is well known, from the simple 

 construction of finding two lines of which their difference and reC' 

 tangle are given, 



that DC + DO 

 and DC— DO 



will be the greater 

 the less 



EC + EO FC + FO 

 EC— LO FC — FO 



of two lines of which half their difference is DO, EO and FO re- 

 spectively, and rectangle UC^. 



