Answer :
When Montucla, in
his edition of Ozanam's Recreations
in Mathematics,
declared that "No more than three rightangled 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 rightangled 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
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 rationalsided rightangled 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 
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 rationalsided R.A.T.'s of equal area that
may
be found in whole numbers.
