"Yet stay," said the youth, as a gleam of inspiration lighted up the relaxing muscles of his quiescent features. "Stay. Methinks it matters little _when_ we reached that summit, the crown of our toil. For in the s.p.a.ce of time wherein we clambered up one mile and bounded down the same on our return, we could have trudged the _twain_ on the level. We have plodded, then, four-and-twenty miles in these six mortal hours; for never a moment did we stop for catching of fleeting breath or for gazing on the scene around!"
"Very good," said the old man. "Twelve miles out and twelve miles in.
And we reached the top some time between six and seven of the clock. Now mark me! For every five minutes that had fled since six of the clock when we stood on yonder peak, so many miles had we toiled upwards on the dreary mountainside!"
The youth moaned and rushed into the hostel.
BLITHE.
The elder and the younger knight, They sallied forth at three; How far they went on level ground It matters not to me; What time they reached the foot of hill, When they began to mount, Are problems which I hold to be Of very small account.
The moment that each waved his hat Upon the topmost peak-- To trivial query such as this No answer will I seek.
Yet can I tell the distance well They must have travelled o"er: On hill and plain, "twixt three and nine, The miles were twenty-four.
Four miles an hour their steady pace Along the level track, Three when they climbed--but six when they Came swiftly striding back Adown the hill; and little skill It needs, methinks, to show, Up hill and down together told, Four miles an hour they go.
For whether long or short the time Upon the hill they spent, Two thirds were pa.s.sed in going up, One third in the descent.
Two thirds at three, one third at six, If rightly reckoned o"er, Will make one whole at four--the tale Is tangled now no more.
SIMPLE SUSAN.
MONEY SPINNER.
ANSWERS TO KNOT II.
-- 1. THE DINNER PARTY.
_Problem._--"The Governor of Kgovjni wants to give a very small dinner party, and invites his father"s brother-in-law, his brother"s father-in-law, his father-in-law"s brother, and his brother-in-law"s father. Find the number of guests."
_Answer._--"One."
In this genealogy, males are denoted by capitals, and females by small letters.
The Governor is E and his guest is C.
A = a+------+-+----+b = B D = d C = c+---++--+ +-+-+e = Eg = GF ========= f
Ten answers have been received. Of these, one is wrong, GALANTHUS NIVALIS MAJOR, who insists on inviting _two_ guests, one being the Governor"s _wife"s brother"s father_. If she had taken his _sister"s husband"s father_ instead, she would have found it possible to reduce the guests to _one_.
Of the nine who send right answers, SEA-BREEZE is the very faintest breath that ever bore the name! She simply states that the Governor"s uncle might fulfill all the conditions "by intermarriages"! "Wind of the western sea," you have had a very narrow escape! Be thankful to appear in the Cla.s.s-list at all! BOG-OAK and BRADSHAW OF THE FUTURE use genealogies which require 16 people instead of 14, by inviting the Governor"s _father"s sister"s husband_ instead of his _father"s wife"s brother_. I cannot think this so good a solution as one that requires only 14. CAIUS and VALENTINE deserve special mention as the only two who have supplied genealogies.
CLa.s.s LIST.
I.
BEE.
CAIUS.
M. M.
MATTHEW MATTICKS.
OLD CAT.
VALENTINE.
II.
BOG-OAK.
BRADSHAW OF THE FUTURE.
III.
SEA-BREEZE.
-- 2. THE LODGINGS.
_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."
[Ill.u.s.tration]
Let A be No. 9, B No. 25, C No. 52, and D No. 73.
Then AB = [** sqrt](12^{2} + 5^{2}) = [** sqrt]169 = 13; AC = 21; AD = [** sqrt](9^{2} + 8^{2}) = [** sqrt]145 = 12 + (N.B. _i.e._ "between 12 and 13.") BC = [** sqrt](16^{2} + 12^{2}) = [** sqrt]400 = 20; BD = [** sqrt](3^{2} + 21^{2}) = [** sqrt]450 = 21+; CD = [** sqrt](9^{2} + 13^{2}) = [** sqrt]250 = 15+;
Hence 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 "0," 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 a.s.sumes 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 [** sqrt]169 + [** sqrt]442 + [** sqrt]130 = 741. (I suppose you mean [** sqrt]741, which would be a little nearer the truth. But roots cannot be added in this manner. Do you think [** sqrt]9 + [** sqrt]16 is 25, or even [** sqrt]25?) But AYR"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." 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 bold and unfounded a.s.sumption that there were really 21 houses on each side, instead of 20 as stated by Balbus. "We may a.s.sume," they add, "that the doors of Nos.
21, 42, 63, 84, are invisible from the centre of the Square"! What is there, I wonder, that CROPHI AND MOPHI would _not_ a.s.sume?
Of the five who are wholly right, I think BRADSHAW OF THE FUTURE, CAIUS, CLIFTON C., and MARTREB deserve special praise for their full _a.n.a.lytical_ solutions. MATTHEW MATTICKS picks out No. 9, and proves it to be the right house in two ways, very neatly and ingeniously, but _why_ he picks it out does not appear. It is an excellent _synthetical_ proof, but lacks the a.n.a.lysis which the other four supply.
CLa.s.s LIST.
I.
BRADSHAW OF THE FUTURE CAIUS.
CLIFTON C.