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 m² + n², m² - n², 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 + m² + n² = a
m² - n² = b
2mn + n² = 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 z² and 4a, and give each side the denominator 2z (b² - h²), 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:

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, a³b - b³a = 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.