The Canterbury Puzzles Part 32

When Montucla, in his edition of Ozanam's _Recreations in Mathematics_, declared that "No more than three right-angled triangles, equal to each other, can be found in whole numbers, but we may find as many as we choose in fractions," he curiously overlooked the obvious fact that if you give all your sides a common denominator and then cancel that denominator you have the required answer in integers!

Every reader should know that if we take any two numbers, _m_ and _n_, then _m_^2 + _n_^2, _m_^2 - _n_^2, and _2mn_ will be the three sides of a rational right-angled triangle. Here _m_ and _n_ are called generating numbers. To form three such triangles of equal area, we use the following simple formula, where _m_ is the greater number:--

_mn_ + _m_^2 + _n_^2 = _a_ _m_^2 - _n_^2 = _b_ 2_mn_ + _n_^2 = _c_

Now, if we form three triangles from the following pairs of generators, _a_ and _b_, _a_ and _c_, _a_ and _b_ + _c_, they will all be of equal area. This is the little problem respecting which Lewis Carroll says in his diary (see his _Life and Letters_ by Collingwood, p. 343), "Sat up last night till 4 a.m., over a tempting problem, sent me from New York, 'to find three equal rational-sided right-angled triangles.' I found two ... but could not find three!"

The following is a subtle formula by means of which we may always find a R.A.T. equal in area to any given R.A.T. Let _z_ = hypotenuse, _b_ = base, _h_ = height, _a_ = area of the given triangle; then all we have to do is to form a R.A.T. from the generators _z_^2 and 4_a_, and give each side the denominator 2_z_ (_b_^2 - _h_^2), and we get the required answer in fractions. If we multiply all three sides of the original triangle by the denominator, we shall get at once a solution in whole numbers.

The answer to our puzzle in smallest possible numbers is as follows:--

First Prince ... 518 1320 1418 Second Prince ... 280 2442 2458 Third Prince ... 231 2960 2969 Fourth Prince ... 111 6160 6161

The area in every case is 341,880 square furlongs. I must here refrain from showing fully how I get these figures. I will explain, however, that the first three triangles are obtained, in the manner shown, from the numbers 3 and 4, which give the generators 37, 7; 37, 33; 37, 40. These three pairs of numbers solve the indeterminate equation, _a_^3_b_ -_b_^3_a_ = 341,880. If we can find another pair of values, the thing is done. These values are 56, 55, which generators give the last triangle.

The next best answer that I have found is derived from 5 and 6, which give the generators 91, 11; 91, 85; 91, 96. The fourth pair of values is 63, 42.

The reader will understand from what I have written above that there is no limit to the number of rational-sided R.A.T.'s of equal area that may be found in whole numbers.

108.--_Plato and the Nines._

The following is the simple solution of the three nines puzzle:--

9 + 9 ---- .9

To divide 18 by .9 (or nine-tenths) we, of course, multiply by 10 and divide by 9. The result is 20, as required.

109.--_Noughts and Crosses._

The solution is as follows: Between two players who thoroughly understand the play every game should be drawn. Neither player could ever win except through the blundering of his opponent. If Nought (the first player) takes the centre, Cross must take a corner, or Nought may beat him with certainty. If Nought takes a corner on his first play, Cross must take the centre at once, or again be beaten with certainty. If Nought leads with a side, both players must be very careful to prevent a loss, as there are numerous pitfalls. But Nought may safely lead anything and secure a draw, and he can only win through Cross's blunders.

110.--_Ovid's Game._

The solution here is: The first player can always win, provided he plays to the centre on his first move. But a good variation of the game is to bar the centre for the first move of the first player. In that case the second player should take the centre at once. This should always end in a draw, but to ensure it the first player must play to two adjoining corners (such as 1 and 3) on his first and second moves. The game then requires great care on both sides.

111.--_The Farmer's Oxen._

Sir Isaac Newton has shown us, in his _Universal Arithmetic_, that we may divide the bullocks in each case in two parts--one part to eat the increase, and the other the acc.u.mulated gra.s.s. The first will vary directly as the size of the field, and will not depend on the time; the second part will also vary directly as the size of the field, and in addition inversely with the time. We find from the farmer's statements that 6 bullocks keep down the growth in a 10-acre field, and 6 bullocks eat the gra.s.s on 10 acres in 16 weeks. Therefore, if 6 bullocks keep down the growth on 10 acres, 24 will keep down the growth on 40 acres.

Again, we find that if 6 bullocks eat the acc.u.mulated gra.s.s on 10 acres in 16 weeks, then

12 eat the gra.s.s on 10 acres in 8 weeks, 48 " " 40 " 8 "

192 " " 40 " 2 "

64 " " 40 " 6 "

Add the two results together (24 + 64), and we find that 88 oxen may be fed on a 40-acre meadow for 6 weeks, the gra.s.s growing regularly all the time.

112.--_The Great Grangemoor Mystery._

We were told that the bullet that killed Mr. Stanton Mowbray struck the very centre of the clock face and instantly welded together the hour, minute, and second hands, so that all revolved in one piece. The puzzle was to tell from the fixed relative positions of the three hands the exact time when the pistol was fired.

We were also told, and the ill.u.s.tration of the clock face bore out the statement, that the hour and minute hands were exactly twenty divisions apart, "the third of the circ.u.mference of the dial." Now, there are eleven times in twelve hours when the hour hand is exactly twenty divisions ahead of the minute hand, and eleven times when the minute hand is exactly twenty divisions ahead of the hour hand. The ill.u.s.tration showed that we had only to consider the former case. If we start at four o'clock, and keep on adding 1 h. 5 m. 27-3/11 sec., we shall get all these eleven times, the last being 2 h. 54 min. 32-8/11 sec. Another addition brings us back to four o'clock. If we now examine the clock face, we shall find that the seconds hand is nearly twenty-two divisions behind the minute hand, and if we look at all our eleven times we shall find that only in the last case given above is the seconds hand at this distance. Therefore the shot must have been fired at 2 h. 54 min. 32-8/11 sec. exactly, or, put the other way, at 5 min. 27-3/11 sec. to three o'clock. This is the correct and only possible answer to the puzzle.

113.--_Cutting a Wood Block._

Though the cubic contents are sufficient for twenty-five pieces, only twenty-four can actually be cut from the block. First reduce the length of the block by half an inch. The smaller piece cut off const.i.tutes the portion that cannot be used. Cut the larger piece into three slabs, each one and a quarter inch thick, and it will be found that eight blocks may easily be cut out of each slab without any further waste.

114.--_The Tramps and the Biscuits._

The smallest number of biscuits must have been 1021, from which it is evident that they were of that miniature description that finds favour in the nursery. The general solution is that for _n_ men the number must be _m_ (_n_^{_n_+1}) - (_n_ - 1), where _m_ is any integer. Each man will receive _m_ (_n_ - 1)^_n_ - 1 biscuits at the final division, though in the case of two men, when _m_ = 1, the final distribution only benefits the dog. Of course, in every case each man steals an _n_th of the number of biscuits, after giving the odd one to the dog.