The Knight's Puzzle
(
CANTERBURY PUZZLES)
This worthy man was, as Chaucer tells us, "a very perfect, gentle knight," and "In many a noble army had he been: At mortal battles had he been fifteen." His shield, as he is seen showing it to the company at the "Tabard" in the illustration, was, in the peculiar language of the heralds, "argent, semée of roses, gules," which means that on a white ground red roses were scattered or strewn, as seed is sown by the hand. When this knight was called on to propound a puzzle, he said to the company, "This riddle a wight did ask of me when that I fought with the lord of Palatine against the heathen in Turkey. In thy hand take a piece of chalk and learn how many perfect squares thou canst make with one of the eighty-seven roses at each corner thereof." The reader may find it an interesting problem to count the number of squares that may be formed on the shield by uniting four roses.
Answer:
The Knight declared that as many as 575 squares could be marked off on his shield, with a rose at every corner. How this result is achieved may be realized by reference to the accompanying diagram:—Join A, B, C, and D, and there are 66 squares of this size to be formed; the size A, E, F, G gives 48; A, H, I, J, 32; B, K, L, M, 19; B, N, O, P, 10; B, Q, R, S, 4; E, T, F, C, 57; I, U, V, P, 33; H, W, X, J, 15; K, Y, Z, M, 3; E, a, b, D, 82; H, d, M, D, 56; H, e, f, G, 42; K, g, f, C, 32; N, h, z, F, 24; K, h, m, b, 14; K, O, S, D, 16; K, n, p, G, 10; K, q, r, J, 6; Q, t, p, C, 4; Q, u, r, i, 2. The total number is thus 575. These groups have been treated as if each of them represented a different sized square. This is correct, with the one exception that the squares of the form B, N, O, P are exactly the same size as those of the form K, h, m, b.
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