SOLUTIONS

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 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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