December 1, 1886.] 



♦ KNO^AT'LEDGE ♦ 



39 



sack. Our Mrs. Browns are rather fond of applying the 

 term to themselves — though I am not aware that they ever, 

 as " Arthur Sketchley's " Mrs. Brown is made to do, call 

 themselves " fieldmales." If you speak of a housekeeper as 

 a "woman," even addressing her as "my good woman," 

 she will be ready to tear your eyes out; but if you speak of 

 her as a " female " she is rather Mattered than otherwise. 

 Why "a male" has not yet come to be regarded as an appro- 

 priate way of desciibing a man is not clear. Possibly we 

 shall leain the reason when we find out why a woman may 

 be called simply " a petticoat," while no one ever yet thought 

 of making " a pair of breeches " do duty for a man. 



Fex, To. It is interesting to learn that this word is still 

 used by American boys, as by boys in England. Fain we 

 used to pronounce it at school. In London, "fain larks " is 

 synonymous with "stow larks," that is " no tricks, I entreat 

 of you." Probably "fen " is nearer the original sound, the 

 word having piobably been " fend " in the sense of " keep 

 oflF," " avoid." 



Fesce. To be " on the fence " cori'esponds in America to 

 being " Jack-o'-both-sides " in Euglisli. A man who tries to 

 I'un with the hare and hunt with the hounds is " on the 

 fence " — " impartially ready," as Lowell puts it — 



"... to jump either side, 

 And make the first use of a turn of the tide." 



A man on the fence is apt, however, to give offence to both 

 sides, instead of pleasing both as he fondly hopes to du. 



OUR PUZZLES. 



N fig. 1, we have the solution of my new and 

 improved puzzle (No. X.) about nineteen trees, 

 according to which they are to be arranged, 

 not in niiif but in ten rows, five to each row. 



In fig. 2, we have the solution of my im- 

 proved form (Puzzle XI.) of the puzzle about 

 cords and pegs, according to which sis cords 

 (all equal in length, if it be preferred to add that condition) 

 are to be so fastened to nine pegs as to inclose (1) six 

 triangular spaces all equal and similar, (2) three quad- 



Fig l " Fig. 2 



Showing Nineteen Tkees Showing Six equal Cords, 

 IN Ten Kows, Five in fastened to Nine Pegs, 



Each. and inclosing Ten Spaces 



Symmetrically. 



rangular spaces axially symmetrical, and (3) one hexagonal 

 space having equal sides and three axes of symmetry. 



Puzzle XII. When I turned to the investigation of this 

 puzzle, I found that it was not the same as ilogul's Puzzle 

 in Vol. 1. (rather ridiculously indexed The Mogul's Puzzle), 

 but slightly more difficult because more general. 

 It may be presented in two forms ; thus : — 

 Given a rectangular carpet, to divide it by the fewest cuts 

 into pieces which shall cover 



1. An equal given rectangular .area, or 



2. An equal rectangular area of given shape. 



Of course problem 2 involves the solution of problem 1, 

 which was Puzzle XII. as given ; but it is well to note 

 that this, the more general form, may be dealt with 



directly : by which I mean that instead of first finding out 

 by geometry the sides of the rectangle fulfilling the con- 

 ditions in 2, and then solving the puzzle for that rectangle, 

 we may employ a construction which, in solving the puzzle, 

 will give us the rectangle both in .shape and size. 



Two solutions of the second form of the problem are 

 illustrated in figs. 3 and 4, the preliminary construction 



being only given in one case. The given rectangle is abcd, 

 and the rectangle aesd, cut ofl' from the length of ac, repre- 

 sents the shape of the rectangle to be covered by sections 

 of ad, ae corresponding to the shorter side. On ab the 

 semicircle afb is drawn, ec produced to meet it in r, and 

 AF and FB, cutting DC in J and h, are drawn ; then af is one 



T>g,4 



side of the required rectangle, and bh is the length of the 

 other. Completing the rectangle fk, we get the remaining 

 cutting line, nl, in the case illustrated by fig. 3. In the 

 other case, fb lies wholly above ab; but ak, parallel to fb. 



Fig. 



cuts DC in K, giving ak, the other side of the required 

 rectangle, which we complete by drawing kl. Then to 

 complete the cutting lines we have to take ek equal to lb, 

 HK equal to nb, and join he. 



The numerals in both figures show how the pieces of 

 rectangle AC fit upon rectangle al. 



