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kirkman's school-girls problem 



[CH. X 



length must appear twice, and the two equal lines so repre- 

 sented must start one from an even number and the other 

 from an odd number, so as to avoid the same combination of 

 points occurring again when the system is rotated cyclically. 

 Of course a vertex cannot be at the point 1 or «/> as these 

 points will be required for the diameter triplet (1. k. y). 



These remarks will be clearer if we apply them to a definite 

 example. I take as an instance the case of 15 girls. As before 

 we represent 14 of them by equidistant points numbered 1, 2, 

 3, ... 14 on the circumference of a circle, and one by a point k 

 at its centre. Take as one triplet the diameter (1. k. 8). Then 

 the sides p, q, r may have any of the values [1], [2], [3], [4], [5], 

 [6], and each value must be used twice. On examination it will 

 be found that there are only two possible groupings, namely 

 [1, 1, 2], [2, 4, 6], [3, 3, 6], [5, 5, 4], and [1, 2, 3], [1, 4, 5], 

 [3, 5, 6], [2, 4, 6]. One of the solutions to which the first set of 

 groupings leads is defined by the diameter (1. k. 8) and the four 

 triplets (9. 10. 11), (4. 6. 14), (2. 5. 13), (3. 7. 12); see figure ii, 

 below: of the four triangles used three are isosceles. The 



Figure ii. 



Figure iii. 



second set leads to solutions defined by the triplets (k. 1. 8), 

 (4. 5. 7), (13. 14. 9), (3. 6. 11), (10. 12. 2), or by the triplets 

 (k. 1. 8), (6. 7. 9), (3. 4. 13), (5. 11. 14), (10. 12. 2): in these 

 solutions all the triangles used are scalene. If any one of 

 these three sets of triplets is rotated cyclically two steps at 



