le* Mr. T. Drummond on a Propertt/ of Poliigons. 



According, therefore, to the number of points R and the 

 number of secants to wliich any point G of the inscribed hexa- 

 gon may be referred will be the number of sets of intersections 

 that can be formed by the figure. — This may be determined 

 by considering th.e several relations in which the points G 

 and R may stand with reference to the secants. 



1st. Both points may be within the secants and in opposite 

 segments. 



2d. within and in the same segment. 



3d. One may be within and the other without. 



^th. Both may be without and in opposite segments. 



5th. without and in the same segment. 



In the 1st case, G and R will be any two opposite points 

 of the hexagon, and will form with the secants AB and AC 

 opposite sides produced two sets of intersections, in which A 

 will be common to both. Hence 6 will be the whole number 

 obtained by considering all the opposite points with reference 

 to the opposite sides; but since I AH must be common to 

 every pair, 6 — 2 = 4 will be the number that can be thus 

 formed. 



In the 4th case, the secants will be the lines DE and BC 

 (joining alternate points) and G,R opposite points as before: 

 hence the same intersections will result as in the 1st case. 



The point A will in the 2d case be formed by the intersection 

 of the alternate sides of the figure ; and as every such point 

 will correspond to two sets, 12 will be the number resulting 

 in this case. 



The point A will in the 3d case be formed by the intersec- 

 tion of a side with the line joining the points adjacent and al- 

 ternate to it: hence the No. 6, and 12 sets of intersections. 



In the 5th case, A will be the intersection of a side with 

 its opposite diagonal, and therefore 12 sets will also result. 



Hence, therefore, in any irregular hexagon inscribed in a 

 circle, if the sides be produced to intersect themselves, to in- 

 tersect the lines joining the alternate points of the figure, and 

 also the diagonals, and if these last be also drawn intersecting 

 themselves and the lines joining the alternate points, then the 

 intersections thus formed may be resolved into 40 sets of three 

 each, having their three component points in straight lines. 



This property may be extended to every polygon ; for let m 

 denote the number of its sides, then 



m.m — \.m — 2 m — {;n-\-\) will be the number 



of hexagons into which it may be resolved; n being =m — 6, 

 and therefore 



\)n.m~l.m — 2 ?k — (« + 1)] 40 will be the num- 

 ber of intersections in straigiit lines. Further, 



