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.