96 



therefore 



REPORT — 1856. 



Pn+l + Pra— 1 



sec0P„ 



(49) 



or any perpendicular multiplied by the secant of the first amplitude, is an 

 arithmetical mean betioeen the perpendiculars immediately preceding and 

 folloiving it. Thus, for example. Po=/h, Pi=m secO, ?„=»« sec (fl-'-fi), or 



m + ffi sec(0-L0) 

 sec m sec % = —^ 



but 



sec(0-i-0)=sec=0 + tan=0; 

 hence the proposition is manifest. 



Again, as hence 



2Po=»i(m" + w''), 2.2.PoP, = wjXm' + ?<-^ + 2<^ + m-0 



2P,=»2(m + m-i), 2 . 2 . Pi P2=?m-(w^ + m-* + m' + ?<-') 



2Po=?«(m- + m-2), 2 . 2 . Pg P3=»r(««' + M-^ + M' + M-0 



2P3=;«(m='+m-3), 2 . 2 . P3 P4=mXM? + «-7 + M» + M-i) 



(50) 



2P„ = »l(2<" + ?<-"), 2.2.P„-i P„ = »lX«^«-J+M-C2«-') + ?<l + 2i-l). 



We have, therefore, adding the preceding expressions, 



2[PoPi + P,P. + P2P3 + P3P4 .... Pn-.P«]= , ^ ^ 



(51) 



»j[+P, + P3+P5 + P7 .... P2„-i + (n)P,], 



or ttvice the sum of all the products of the perpendiculars taken two by two up 

 to the nth, is equal to the sum of all the odd perpendiculars up to the (2n—l)th 

 + n times the first perpendicular. 



Thus, taking the first three perpendiculars, 



P^=w, Pj=»isec0, P2=wsec(0-L0)=m(sec^0 + tan0), 



P3=7n sec (0-^0-L0)=wi(4' sec'0— 3 sec 0) ; 



then the truth of the proposition may be shown in this particular case for 



2CPoP, + P.P2] = 4»»'sec^0=wi(P, + P3 + 2PJ. 



Again, since 



2P2„=»i(m2» + m-2»), 



and 



4P^=»iX«^»+2 + «-2«), 

 we shall have 



2P;— m'=mP2» (52) 



Thus, for example, tivice the square of the perpendicidar on the fifth side of 

 the polygon diminis/ied by the square of the modulus, is equal to the tenth 

 perpendicular midtiplied by the modulus. 



In the same way we may show that 



4.P^-37n=P,.=?MT3„. 



Let n^=5 and m=l, then four tiines the cube of the fifth perpendicular, 

 diminished by three times the same perjyeiidicular, is equal to the fifteenth per- 

 pendicular, or to the perpendicular on the fifteenth side of the polygon. 



