PART THIRD.

ON THE RIGHT-ANGLED TRIANGLE ;

CONTAINING

A VARIETY OF METHODS FOR FINDING AN INFINITE SERIES OF RIGHT-ANGLED TRIANGLES, THE SIDES OF WHICH SHALL BE IN WHOLE NUMBERS, FOR THE USE

OF CIVIL ENGINEERS, ARCHITECTS,

DRAUGHTSMEN, AND MECHANICS.

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ON THE RIGHT-ANGLED TRIANGLE.

OF RIGHT-ANGLED TRIANGLES IN NUMBERS, OR RIGHT-ANGLED TRIANGULAR NUMBERS.

RIGHT-ANGLED triangular numbers are rational numbers so related to each other that the sum of the squares of two of them is equal to the square of the third.

The numbers 3, 4, and 5 have this property-32 + 42 being equal to 52.

Right-angled triangular numbers must be severally unequal, for if the two less ones could be each represented by a, and the third or greatest by b, then 2a2 = b2, bay 2, an irrational number, whatever is the value of a.

The area of a right-angled triangle, whose sides are rational, can not be equal to a rational square.

If a, b, and c represent the sides of a triangle, and C be the angle opposite c, then, if C90°, a2 + b2 = c2; if C=120°, a2 + ab + b2 c2; and if C= 60°, a2. - ab + b2

c2.

If n represent any number, and m any other number less than n, then n2m2 will represent the hypothenuse of a right-angled plane triangle, of which the other two sides are respectively n2, m2, and 2mm. For example, if n=2 and m=1, then n2+m2-5, n2-m2-3, and 2nm 4, which are right-angled triangular numbers.

If n=

=7 and m=2, the formulas give 53, 45, and 28 for the numbers, and 532-2809-452-282.

We shall now propose and solve a few of the most curious problems respecting the right-angled triangles.

PROBLEM 1.

To find as many right-angled triangles in numbers as we please. This may be effected by the concluding formulas, which we have just given, but we think it right to add the following methods:

I

Take any two numbers at pleasure, for example 1 and 2, which we shall call generating numbers; multiply them together, having doubled the product, we obtain one of the sides of the triangle, which in this case will be 4. If we then square each of the generating numbers, which in the present example will give 4 and 1, their difference 3 will be the second side of the triangle, having 1 and 2, for their generating numbers are 3, 4, and 5.

If 2 and 3 had been assumed as generating numbers we should have found the sides to be 5, 12, and 13, and the numbers 1 and 3 would have given 6, 8, and 10.

Another Method.-Take a progression of whole or fractional numbers, 11, 2, 3, and 44, etc., the properties of which are:

1st. The whole numbers are those of the common series and have unity for their common difference.

2nd. The numerators of the fractions annexed to the whole numbers are also the natural numbers.

3rd. The denominators of these fractions are the odd numbers 3, 5, 7, etc.

Take any term of this progression, for example 33, and reduce it to an improper fraction by multiplying the whole number 3 by 7, and adding to

24

21 the product, the numerator 3 will give The numbers 7 and 24

7'

will be the sides of a right-angled triangle, the hypothenuse of which may be found by adding together the square of these two numbers, viz.: 49 and 576, and extracting the square root of the sum. The sum in this case being 625, the square root of which is 25, this number will be the hypothenuse required. The sides, therefore, of the triangle produced by the above term of the generating progression are 7, 24, and 25.

In like manner the first term will give the right-angled triangle 3, 4, and 5; the second term, 22, will give 5, 12, and 13; the fourth, 44, will give 9, 40, and 41. All these triangles have the ratio of their sides different; and they all possess this property, that the greatest side and the hypothenuse differ only by unity.

The progression, 17, 211, 315, 418, etc., is of the same kind as the preceding. The first term of it gives the right-angled triangle 8, 15, and 17; the second term gives the triangle 12, 35, and 37; the third

10

triangle 16, 63, 65, etc. All these triangles it is evident are different in regard to the proportion of their sides; and they all have this peculiar property, that the difference between the greater side and the hypothenuse is the number 2.

PROBLEM 2.

To find any number of right-angled triangles in numbers, the sides of which shall differ only by unity.

To resolve this problem we must find out such numbers that the double of their squares, plus or minus unity, shall also be square numbers; of this kind are the numbers 1, 2, 5, 12, 29, 70, etc., for twice the square of 1 is 2, which, diminished by unity, leaves 1, a square number. In like manner twice the square of 2 is 8, to which, if we add 1, the sum 9 will be a square number, and so on.

Having found these numbers, take any two of them which immediately follow each other, as 1 and 2, or 3 and 5, or 12 and 29, for generating numbers; the right-angled triangles arising from them will be of such a nature that their sides will differ from each other only by unity.

The following is a table of these triangles, with their generating numbers:

But if the problem were to find a series of triangles of such a nature that the hypothenuse of each should exceed one of the sides only by unity, the solution would be much easier. Nothing, in this case, would be necessary but to assume as the generating numbers of the required triangles any two numbers having unity for their differ

ence.

The following is a table, similar to the preceding of the six first