CH. x] kirkman's school-girls problem 195 



certain amount of arrangement has to be made empirically, and 

 then to go on to the consideration of the more general method. 



One-Step Cycles. As illustrating solutions by one-step 

 cyclical permutations I will first describe Legros's method. 

 Solutions obtained by it can be represented by diagrams, and 

 their use facilitates the necessary arrangements. It is always 

 applicable when n is of the form 24m + 3, and seems to be 

 also applicable when n is of the form 24m + 9. Somewhat 

 similar methods were used by Dudeney, save that he made 

 no use of geometrical constructions. 



We have n = 2y + 1 = 24m + 3 or n=2y + l = 24m + 9. 

 We may denote one girl by k, and the others by the numbers 

 1, 2, 3, ... 2y. Place k at the centre of a circle, and the 

 numbers 1, 2, 3, ... 2y at equidistant intervals on the circum- 

 ference. Thus the centre of the circle and each point on its 

 circumference will indicate a particular girl. A solution in 

 which the centre of the circle is used to denote one girl is 

 termed a central solution. 



The companions of k are to be different on each day. If we 

 suppose that on the first day they are 1 and y+1, on the 

 second 2 and y + 2, and so on, then the diameters through k 

 will give for each day a triplet in which k appears. On each 

 day we have to find 2 (y — 1)/3 other triplets satisfying the 

 conditions of the problem. Every triplet formed from the 

 remaining 2^ — 2 girls will be represented by an inscribed 

 triangle joining the points representing these girls. The sides 

 of the triangles are the chords joining these 2y — 2 points. 

 These chords may be represented symbolically by [1], [2], 

 [3], ... [y— 1]; these numbers being proportional to the smaller 

 arcs subtended. I will denote the sides of a triangle so repre- 

 sented by the letters p, q, r, and I will use the term triad or 

 grouping to denote any group of p, q, r which determines the 

 dimensions of an inscribed triangle. I shall place the numbers 

 of a triad in square brackets. If p, q, r are proportional to 

 the smaller arcs subtended, it is clear that if p + q is less 

 than y, we have p + q = r ; and if p + q is greater than y we 

 have p + q + r = 2y. If we like to use arcs larger than the 



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