ch. x] kirkman's school-girls problem 199 



plaoed symmetrically to each other. I give 27 girls as an instance, using the 

 same triads as in the solution of this case given above. The triplets for the 

 first day are (ft. 1. 14), (2. 12. 3), (21. 6. 22), (20. 24. 26), (11. 15. 17), 

 (8. 16. 19), (25. 7. 10), (6. 18. 13), and (23. 9. 4). Prom this the arrangements 

 on the other days can be obtained by a two-step (but not by a one-step) oyolical 

 permutation. 



It is unnecessary to give more examples, or to enter on the 

 question of how from one solution others can be deduced, or 

 how many solutions of each case can be obtained in this way. 

 The types of the possible triangles are found analytically, but 

 their geometrical arrangement is empirical. The defect of this 

 method is that it may not be possible to arrange a given 

 grouping. Thus when n = 27, we easily obtain 24 different 

 groupings, but two of them cannot be arranged geometrically 

 to give solutions ; and whether any particular grouping will 

 give a solution can, in many cases, be determined only by long 

 and troublesome empirical work. The same objection applies 

 to the two-step and three-step methods which are described 

 below. 



Two-Step Cycles. The method used by Legros was extended 

 by Eckenstein to cases where n is of the form 12m + 3. When 

 n is of this form and m is odd we cannot get sets of comple- 

 mentary triangles as is required in Legros's method ; hence, to 

 apply a similar method, we have to find 2 (y — 1)/3 different 

 dissimilar inscribed triangles having no vertex in common and 

 satisfying the condition p+q = r or p + q + r=2y. These 

 solutions are also central. Since there are 1y points on the 

 circumference of the circle the permutations, if they are to be 

 cyclical, must go in steps of two numbers at a time. In 

 Legros's method we represented one triplet by a diameter. But 

 obviously it will answer our purpose equally well to represent 

 it by a triangle with k as vertex and two radii as sides, one 

 drawn to an even number and the other to an odd number: 

 in fact this will include the diameter as a particular case. 



I begin by considering the case where we use the diameter 

 (1. k. y) to represent one triplet on the first day. Here the 

 chords used for sides of the triangles representing the other 

 triplets must be of lengths [1], [2], ... [y - 1], Also each given 



