Three answers have been received—but only two seem to me worthy of honours.
Tympanum says that the statement about the stick “is merely a blind, to which the old answer may well be applied, solvitur ambulando, or rather mergendo.” I trust Tympanum will not test this in his own person, by taking the place of the man in Balbus’ Essay! He would infallibly be drowned.
Old King Cole rightly points out that the series, 2, 1, etc., is a decreasing Geometrical Progression: while Vindex rightly identifies the fallacy as that of “Achilles and the Tortoise.”
Class List.
Old King Cole.
Vindex.
§3: The Garden
Problem.—An oblong garden, half a yard longer than wide, consists entirely of a gravel-walk, spirally arranged, a yard wide and 3,630 yards long. Find the dimensions of the garden.
Answer.—60, 60½.
Solution.—The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square-yards and fractions of a square-yard, contained in that piece of walk: and the distance, traversed in passing through a square-yard at a corner, is evidently a yard. Hence the area of the garden is 3,630 square-yards: i.e., if x be the width, x(x+12)=3,630. Solving this Quadratic, we find x=60. Hence the dimensions are 60, 60½.
Twelve answers have been received—seven right and five wrong.
C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.’s “working” consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow’s is shorter, and so (if possible) worth rather less. He says the answer “is at once seen to be 60 × 60½”! Nabob’s calculation is short, but “as rich as a Nabob” in error. He says that the square root of 3,630, multiplied by 2, equals the length plus the breadth. That is 60.25 × 2 = 120½. His first assertion is only true of a square garden. His second is irrelevant, since 60.25 is not the square-root of 3,630! Nay, Bob, this will not do! Tympanum says that, by extracting the square-root of 3,630, we get 60 yards with a remainder of ³⁰⁄₆₀, or half-a-yard, which we add so as to make the oblong 60 × 60½. This is very terrible: but worse remains behind. Tympanum proceeds thus:—“But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half-a-yard wide, the only space not occupied by walk.” But Balbus expressly said that the walk “used up the whole of the area.” Oh, Tympanum! My tympa is exhausted: my brain is num! I can say no more.
Hecla indulges, again and again, in that most fatal of all habits in computation—the making two mistakes which cancel each other. She takes x as the width of the garden, in yards, and x+12 as its length, and makes her first “coil” the sum of x−12, x−12, x−1, x−1, i.e. 4x−3: but the fourth term should be x−112, so that her first coil is ½ a yard too long. Her second coil is the sum of x−212, x−212, x−3, x−3: here the first term should be x−2 and the last x−312: these two mistakes cancel, and this coil is therefore right. And the same thing is true of every other coil but the last, which needs an extra half-yard to reach the end of the path: and this exactly balances the mistake in the first coil. Thus the sum total of the coils comes right though the working is all wrong.
Of the seven who are right, Dinah Mite, Janet, Magpie, and Taffy make the same assumption as C. G. L. and Co. They then solve by a Quadratic. Magpie also tries it by Arithmetical Progression, but fails to notice that the first and last “coils” have special values.
Alumnus Etonæ attempts to prove what C. G. L. assumes by a particular instance, taking a garden 6 by 5½. He ought to have proved it generally: what is true of one number is not always true of others. Old King Cole solves it by an Arithmetical Progression. It is right, but too lengthy to be worth as much as a Quadratic.
Vindex proves it very neatly, by pointing out that a yard of walk measured along the middle represents a square yard of garden, “whether we consider the straight stretches of walk or the square yards at the angles, in which the middle line goes half a yard in one direction and then turns a right angle and goes half a yard in another direction.”
Class List.
Vindex.
Alumnus Etonæ.
Old King Cole.
Dinah Mite.
Janet.
Magpie.
Taffy.
Answers to Knot X
§1: The Chelsea Pensioners
Problem.—If 70 percent have lost an eye, 75 percent an ear, 80 percent an arm, 85 percent a leg: what percentage, at least, must have lost all four?
Answer.—Ten.
Solution.—(I adopt that of Polar Star, as being better than my own). Adding the wounds together, we get 70 + 75 + 80 + 85 = 310, among 100 men; which gives 3 to each, and 4 to 10 men. Therefore the least percentage is 10.
Nineteen answers have been received. One is
