three must be A or C." And again, after pointing out (rightly) that B and D are both within the half-square containing A, she says "therefore" AB + AD must be less than BC + CD. (There is no logical force in either "therefore." 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