66 GEOMETRICAL RESEARCHES, 



3. If the figure A^... -A^ is a complete regular polygon of 

 n sides ; than the centre M of the locus of O is coincident with the 

 centre of the circle which circumscribes the polygon. And if we 

 represent the radius of the circumscribing circle by R, it is 

 evident we can write the equation of the locus in the form 



n . sin ( - revolution) . p 2 + n . sin (- revolution) R 2 = 85* - 4^' 

 in which the involved angle is of like formation with ^". 



From this we at once see that when n = 2, we must have 

 85* = 45" ; and /. as % = zero, so also must 2 = zero. 



When n = 4, then we must have 8^ == 4^', or %' = 22 ; 

 and the problem is porismatic. 



4. Again, since sin (| revolution) = i chord (^ revolution,) 

 and that ^ chord ( revolution ) = perimeter of polygon of 

 j sides ; therefore it is evident that n . sin (|- revolution) is equal 

 to twice the perimeter of a polygon of -| sides. And /. it is 

 evident that when A I A 2 . . . A^ A I is a regular polygon, and n = 



infinity, we have n- sin (- .re volution) = 4 TT = 4 (3 1416) 



and 4 TT . p 2 + 4 TT R 2 = 8 - 4^.' 

 477.^ = 82 - 8^" 



7T.p 2 = 2 (S ~ 5') 



And, putting 2" to represent the rectare of the circle which con- 

 stitutes the locus of O, we can write the last equation in the form 



5" = 22 - 22' 



When we suppose p = H, then the derived figure of the circle 

 A I A 2 . . . A^ A I is a cardoid whose rectare is represented by 5*. 

 and therefore, as in this case, the point O has the circle 

 A A . . . . A M A as locus, we have 



