CH. X] 



kirkman's school-girls problem 



197 



triangle of this shape is placed with its vertices at the points 

 3, 4, 6, we can construct a complementary equal triangle, four 

 points further on, having 7, 8, 2 for its vertices. All the points in 

 the figure are now joined, and form the three triplets for the first 

 day, namely (k. 1. 5), (3. 4. 6), (7. 8. 2). It is only necessary 

 to rotate the figure one step at a time in order to ohtain the 

 triplets for the remaining three days. Another similar solution 

 is obtained from the diameter (1. k. 5), and the triangles (2. 3. 8), 

 (6. 7. 4). It is the reflection of the former solution. 



The next case to which the method is applicable is when 

 n = 27, m = 1, y = 13. Proceeding as before, the 27 girls must 

 be arranged with one of them, k, at the centre and the other 26 

 on the circumference of a circle. The diameter (1. k. 14) gives 

 the first triplet on the first day. To obtain the other triplets 

 we have to find four dissimilar triangles which satisfy the con- 

 ditions mentioned above. The chords used as sides of these 

 triangles may be of the lengths represented symbolically by 

 [1], [2], ... [12]. We have to group these lengths so that 

 p + q = r or p + q + r=2y; if the first condition can be satis- 

 fied it is the easier to use, as the numbers are smaller. In this 

 instance the triads [3, 8, 11], [5, 7, 12], [2, 4, 6], [1, 9, 10] 

 will be readily found. Now if four triangles with their sides 

 of these lengths can be arranged in a system so that all the 

 vertices fall on the ends of different diameters (exclusive of the 

 ends of the diameter 1, k, 14), it follows that the opposite ends of 

 those diameters can be joined by chords giving a series of equal 

 triangles, symmetrically placed, each having its sides parallel to 

 those of a triangle of the first system. The following arrangement 



