Mathematical recreations and essays . gle will represent a triplet. We wantto find one such triangle, with unequal sides, with its verticesat tliree of these points, and such that the triangle formed bythe other three points will have its sides equal and parallel tothe sides of the first triangle. The sides of a triangle are p, q, r. The only possible valuesare 1, 2, 3, and they satisfy the condition p -\q = r. If a CH. X] KIRKMAN S SCHOOL-GIRLS PROBLEM 197 triangle of this shape is placed with its vertices at the points3, 4, 6, we can construct a complementary equal triangle, fourpoints furth
Mathematical recreations and essays . gle will represent a triplet. We wantto find one such triangle, with unequal sides, with its verticesat tliree of these points, and such that the triangle formed bythe other three points will have its sides equal and parallel tothe sides of the first triangle. The sides of a triangle are p, q, r. The only possible valuesare 1, 2, 3, and they satisfy the condition p -\q = r. If a CH. X] KIRKMAN S SCHOOL-GIRLS PROBLEM 197 triangle of this shape is placed with its vertices at the points3, 4, 6, we can construct a complementary equal triangle, fourpoints further on, having 7,8, 2 for its vertices. All the points inthe figure are now joined, and form the three triplets for the firstday, namely {k. 1. 5), (3. 4. 6), (7. 8. 2). It is only necessaryto rotate the figure one step at a time in order to obtain thetriplets for the remaining three days. Another similar solutionis obtained from the diameter (1. k. 5), and the triangles (2. ),(6. 7. 4). It is the reflection of the former Figure i. The next case to which the method is applicable is whenM = 27, m = 1, y = 13. Proceeding as before, the 27 girls mustbe arranged with one of them, k^ at the centre and the other 26on the circumference of a circle. The diameter (1. k. 14) givesthe first triplet on the first day. To obtain the other tripletswe have to find four dissimilar triangles which satisfy the con-ditions mentioned above. The chords used as sides of thesetriangles may be of the lengths represented symbolically by[1], [2], ... [12]. We have to group these lengths so thatp^q = r or_p + 5+r=22/; if the first condition can be satis-fied it is the easier to use, as the numbers are smaller. In thisinstance the triads [3, 8, 11], [5, 7, 12], [2, 4, 6], [1, 9, 10]will be readily found. Now if four triangles with their sidesof these lengths can be arranged in a system so that all thevertices fall on the ends of different diameters (exclusive of theends of the diameter 1, k^ 1
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