BY MARTIN GARDINER, C.E. 65 



It is also well to remember that the position of the centre 

 M is independent of the value of % . 



1. If all the given straight lines forming the closed 

 n'gon pass through one point, it is obvious the locus of O has 

 this common point of intersection for centre. In this case 

 3" = zero ; and, when the point of intersection is at infinity, 2 

 has no real value but zero ; and, for such value of J, the 

 problem is porismatic, for any straight line cutting the given 

 ones perpendicularly may be regarded as belonging to an infinite 

 circle whose centre is at the point of intersection. 



And when the first half of the closed n'gon A X A 2 . . . . A^ A 

 is coincident with the second half (taken in order as indicated) 

 then also 2' and ^ must be each = zero, and the locus of is 

 unrestricted. 



2. If the closed n'gon be a triangle A A 2 A A X , then it is 

 evident that for each of the angular points of the triangle, con- 

 sidered as a position for O, we have 2 equal zero ; .*. it is 

 obvious that for all values of 2 the centre of the locus must be 

 coincident with the centre of the circle circumscribing the 

 triangle. And putting R for the radius of the circle circumscrib- 

 ing the triangle, we have the relation 



(p2 + R2 ) . ( sm - GI - C 3 Cj + sin c a ' C x C 2 + sin c,- C 2 C 3 ) = 8 ^ - 43" 5 



but B 2 (sin GI - C 3 Cj + sin c a C x C 2 + sin c s ' C 2 C 3 ) = - 2 S' 

 :. p* (sin Cj- C 3 Cj + sin c a ' C x C 2 + sin c^ C 2 C 3 ) = 8 2 - 2' 

 p z - R 2 45" 



and ' ~w ij> 



But if w-j, w 2 , m 3 , represent the feet of perpendiculars from the 

 centre of the circumscribing circle on the sides A I A , A^ A 3 y 

 A 3 Aj , of the triangle and that we put S" to represent the 

 rectare of the triangle m l ^ 2 m 3 m^ , then ' = 4J", and 

 therefore we have 



- R 2 



