Bruno's Revenge — A Tangled Tale (Part 25 of 43)

1. BEE CAIUS M. M. MATTHEW MATTICKS OLD CAT
    VALENTINE

2. BOG-OAK BRADSHAW OF THE FUTURE

3. SEA-BREEZE

Problem. A Square has 20 doors on each side, which contains 21 equal parts. They are numbered all round, beginning at one corner. From which of the four, Nos. 9, 25, 52, 73, is the sum of the distances, to the other three, least? Answer. From No. 9.

Let A be No. 9, B No. 25, C No. 52, and D No. 73.

Then AB = $(122+\; 52)$ =169=\; 13;$ $$$

AC = 21;

$AD\; =(92+\; 82)=145=\; 12\; +;$ $$

(N.B. i.e., `between 12 and 13')

$BC\; =(162+\; 122)=400+\; 20;$ $$

$BD\; =(32+\; 212)=450+\; 21\; +;$ $$

$CD\; =(92+\; 132)=250+\; 15\; +;$ $$

Hence the sum of distances from A is between 46 and 47; from B, between 54 and 55; from C, between 56 and 57; from D, between 48 and 51. (Why not `between 48 and 49'? Make this out for yourselves.) Hence the sum is least for A.

Twenty-five solutions have been received. Of these, 15 must be marked `zero', 5 are partly right, and 5 right. Of the 15, I may dismiss ALPHABETICAL PHANTOM, BOG-OAK, DINAH MITE, FIFEE, GALANTHUS NIVALIS MAJOR (I fear the cold spring has blighted our SNOWDROP), GUY, H.M.S. PINAFORE, JANET, and VALENTINE with the simple remark that they insist on the unfortunate lodgers keeping to the pavement. (I used the words `crossed to Number Seventy-three' for the special purpose of showing that short cuts were possible.) SEA-BREEZE does the same, and adds that `the result would be the same' even if they crossed the Square, but gives no proof of this. M. M. draws a diagram, and says that No. 9 is the house, `as the diagram shows'. I cannot see how it does so. OLD CAT assumes that the house must be No. 9 or No. 73. She does not explain how she estimates the distances.

BEE'S arithmetic is faulty: she makes $169+442+130=\; 741.$ (I\; suppose\; you\; mean$ 742$ ,\; which\; would\; be\; a\; little\; nearer\; the\; truth.\; But\; roots\; cannot\; be\; added\; in\; this\; manner.\; Do\; you\; think$ 9+16$ is\; 25,\; or\; even$ 25?$ )\; But\; Ayr\text{'}s\; state\; is\; more\; perilous\; still:\; she\; draws\; illogical\; conclusions\; with\; a\; frightful\; calmness.\; After\; pointing\; out\; (rightly)\; that\; AC\; is\; less\; than\; BD,\; she\; says,\; `therefore\; the\; nearest\; house\; to\; the\; other\; three\; must\; be\; A\; or\; C.\text{'}\; And\; again,\; after\; pointing\; out\; (rightly)\; that\; B\; and\; D\; are\; both\; within\; the\; half-square\; containing\; A,\; she\; says,\; `therefore\text{'}\; AB\; +\; AD\; must\; be\; less\; than\; BC\; +\; CD.\; (There\; is\; no\; logical\; force\; in\; either\; `therefore\text{'}.\; For\; the\; first,\; try\; Nos.\; 1,\; 21,\; 60,\; 70:\; this\; will\; make\; your\; premiss\; true,\; and\; your\; conclusion\; false.\; Similarly,\; for\; the\; second,\; try\; Nos.\; 1,\; 30,\; 51,\; 71.)$$$$$$$$

Of the five partly-right solutions, RAGS AND TATTERS and MAD HATTER (who send one answer between them) make No. 25 6 units from the corner instead of 5. CHEAM, E. R. D. L., and MEGGY POTTS leave openings at the corners of the Square, which are not in the data: moreover CHEAM gives values for the distances without any hint that they are only approximations. CROPHI AND MOPHI make the bold and unfounded assumption that there were really 21 houses on each side, instead of 20 as stated