When Montucla, in his edition of Ozanam's Recreations
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
then m2 + n2, m2
and 2mn will be the three sides of a
rational right-angled triangle.
Here m and n
are called generating
To form three such triangles of equal area, we use the
simple formula, where m is the greater
mn + m2
+ n2 = a
m2 - n2
2mn + n2
Now, if we form three triangles from the following
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
... 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
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
The answer to our puzzle in smallest possible
numbers is as
The area in every case is 341,880 square furlongs.
I must here
from showing fully how I get these figures. I will explain, however,
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.
three pairs of numbers solve the indeterminate equation, a3b
If we can find another pair of values, the thing is
done. These values are 56, 55, which generators give the last
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
The reader will understand from what I have
written above that
no limit to the number of rational-sided R.A.T.'s of equal area that
be found in whole numbers.