SOLUTIONS
denominator and then cancel that denominator, you have the
Every reader should know that if we take any two numbers, m
and n, then m
2
+ n
2
, m
2
- n
2
, and 2 mn 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.
i 2 1 2 _
mn + m + n = a
m
2
- n
2
= b
2mn + 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 4a,
and give each side the denominator 2z(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 . . Ill 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
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