379 
to Regular Polygons. 
^p p f — r 2 — (c + O ra? — 0 + s') ry 
4- s” ocy 4- \ c" (a? —y*) + i c (x 2 J r 
p p" — r 2 — (c' + d')rx — (s' + a") ry 
4- $iv ocy 4- J c IV (. x 2 — «/ 2 ) 4- J c (a- 2 4- ?/ 2 ) , 
p" p'" = r 2 — (c" 4- c"') ra 1 — (V' 4“ s'") ry 
4- s' xy 4- \ d (a 2 —y 2 ) J r h c (x 2 4- y 2 )-> 
p'" p lv — r 2 — (<f" 4~ c lv ) ^a 1 “• (s ,;/ 4~ <s IV ) ry 
4- s'" xy 4 - \ c"' (a 2 — y 2 ) 4- 2 c (a 2 4- «/ 2 )> 
piv p _ r 2 __ ( c iv _p c) ra — (<s IV 4- s) ry 
4- sxy 4- \ c(x c/ — ■ y 2 ) 4- \ c(x 2 + y 2 ). 
Hence, by adding, we obtain 
pp 4- p'p" 4- p" p " 4- p" P lv 4- P iy P — 
5 r 2 4- b - c (. oc 2 4- y 2 ) 
4- (c 4- c 4- c" 4- d" 4- c lv ) (4 (a 2 —y~) —Qrx} 
4- (s 4- s' 4- s" 4- s'" 4- 6 IV ) {xy — 2ry}. 
Because the circumference is divided into five equal parts at 
the extremities of the arcs 2^, Sx, 4<«, 5*, by known proper- 
ties of the circle, the sum of their cosines c, c', d', d" , c iy , is 
— 0, and die sum of their sines s , ,9', <9", s'", <9 IV , is also = 0 ; 
hence the terms in the preceding expression which are multiplied 
by these sums, must vanish : thus we have simply 
p p' 4- p'p /J 4- p" p'" 4- p'"p ly 4- P lv P — 5{r 2 4- \ (x 2 4- y~) cos x}. 
Draw PC to the centre, and put PC = P(x 2 4 - y ) — v, and, 
instead of supposing the figure to have five sides, let us suppose 
it to have n sides ; then, by the very same mode of reasoning, 
we shall find that 
p p' 4- p' p" 4- p" p '" . . . 4- p^ u ~ 1 ^ p — n (r 2 + \ v 2 cos a) ; 
and multiplying both sides by J sin <*, we have 
J sin u, (pp' j4- p' p" 4- p" p '" . . . -\-p n 1 p^=. \ (r 2 sin » 4- \ v 2 sin x cos x) 
Now, the lines PB, PB', PB", &c. being perpendicular to the 
sides of a regular polygon, the angles they make about the point 
P will manifestly be equal ; and because their number is n , each 
will be — = x, therefore the triangle PB B' — J pp' sin x, 
the triangle PB' B" = £ p' p" sin and so on : Therefore, the 
