is 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 Nou