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January 22, 2017
Posted by **connie** on Tuesday, May 18, 2010 at 1:19am.

- math -
**drwls**, Tuesday, May 18, 2010 at 5:22amThere is a long list of them at

http://www.math.uic.edu/~fields/puzzle/triples.html - math -
**tchrwill**, Tuesday, May 18, 2010 at 11:35amPYTHAGOREAN THEOREM

The Pythagoreans are thought to have been the first to truly appreciate the concept of numbers. One of the areas of their concentration was square numbers, such as 4, 9, 16, 25, 36, and so on. They reportedly discovered that the sums of certain pairs of square numbers, or perfect squares, were also square numbers.

History records that Pythagoras and Diophantus were probably the two most well known mathematicians that had anything to do with right triangles with integer sides. The famous Pythagorean Theorem states that in any right-angled triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse. Another way of stating it is that the area of the square constructed on the long side of a right triangle is equal to the area of the two squares created on the two shorter sides. This is expressed by x^2 + y^2 = z^2.

Almost everyone has heard of the famous 3-4-5 right triangle where 3^2 + 4^2 = 5^2, the simplest, and most fundamental triangle based on the Pythagorean theorem. Surprisingly, fewer people know of the 5, 12, and 13 right triangle, or the 7, 24, and 25 right triangle, which also satisfy the relationship and without any proportional relationship to the 3, 4, and 5 triangle. These sets of three integers, all satisfying the Pythagorean theorem, are traditionally referred to as Pythagorean Triples, of which there are an infinite number.

The Pythagoreans were thought to have been one of the first to provide a proof of the Pythagorean Theorem. Many proofs exist today, one of which follows.

Draw yourself a square ABCD, say 1 1/2 inches square. On each side, locate a point 1 inch from each corner moving clockwise around the square. Label the points E, F, G, and H, E on AB, F on BC, G on CD, and H on DA. Connect points e and F, F and G, G and H, and H and E. You now have a slaller square inside the original square surrounded by four congruent triangles. Label the hypotenuse of each triangle "c" , the short leg of each triangle "b", and the long side of each triangle "a." The area of the inner square is Ai = c^2. The area of the four surrounding triangles is At = 4ab/2 = 2ab. The sum of these areas is the area of the original outer square Ao = 2ab + c^2. The area of the original outer square can also be expressed by Ao = (a + b)^2 = a^2 + 2ab + b^2 which we showed is also equal to 2ab + c^2. Therefore, a^2 + 2ab + b^2 = 2ab + c^2 from which we get a^2 + b^2 = c^2.

In 1876, while serving as a member of the House of Representatives, "future" president James A. Garfield discovered an interesting proof of this theorem. The New England Journal of education published the proof.

The proof involves the use of two methods to calculate the area of a trapezoid. Method one makes use of the formula Area = 1/2(sum of the bases)(altitude) while method two divides the trapezoid into three right triangles and sums the areas of these three triangles.

Construct yourself a trapezoid ABCD with AB parallel to DC, angles C and B right angles. Let AB have length a and DC have length b, b being longer than a. Locate point E on BC such that CE has length a and BE has length b. Draw lines DE and AE and let their lengths be c.

To be more specific, I suggest that you construct the following picture: Draw a horizontal line AB 4 units long, with A at the left end and B at the right end. From point B, draw a vertical line 10 units long, perpendicular to AB, downward, to point C. From point C, draw a horizontal line 6 units long to the left, perpendicular to line BC, to point D. Draw line AD. This is the trapezoid we are talking about. Locate point E on line BC, 4 units up from point C. Then BE = 6 units. Let AB = CE = a and CD = BE = b. Let AE = DE = c.

Notice that angle AED is a right angle also. Now calculate the area of the trapezoid using the two methods identified above.

Area Method 1 = Area Method 2

1/2(a+b)(a+b) = 1/2(ab) + 1/2(ab) + 1/2(cc)

(a+b)(a+b) = (ab) + (ab) + c^2

a^2 + 2ab + b^2 = 2ab + c^2

QED a^2 + b^2 = c^2

The sides of a right triangle do not necessarily have to be integers. They can be rational. For example, consider the expression x^2 + y^2 = 1. We can express "y "in the form of y = 1-mx where "m" is any rational number. Substituting, we derive x = 2m/(1 + m^2) and y = (1 - m^2)/(1 + m^2). Thus, rational values of "m" will produce an infinite number of rational solutions to x^2 + y^2 = 1. The right triangle with sides of 6/10 and 8/10 has a hypotenuse of z = sqrt(6/10^2 + 8/10^2) = sqrt(1) = 1 from m = 3.

As stated above, sets of three integers, all satisfying the Pythagorean theorem, are traditionally referred to as Pythagorean Triples. The derivation of triples can be accomplished in many ways and is a widely covered topic in the field of recreational mathematics. If you are interested in finding triples, the following should provide you with some interesting information and fun.

A triangle whose sides and area are rational numbers is called a rational triangle. If the sides of a rational triangle are integers, it is said to be an integral triangle. If the triangle sides have a greatest common divisor of unity, the triangle is said to be primitive. If one angle of the triangle is a right angle, it is referred to as a right angled rational triangle, or more traditionally, a Pythagorean triangle.

When all three sides of a Pythagorean triangle are integers, they are referred to as a Pythagorean Triple. Pythagorean Triples that have a greatest common divisor of 1 are called primitive triples. Those with factors other than 1 are called non-primitive triples.

Primitive Pythagorean Triples of the form x^2 + y^2 = z^2 can be derived from x = m^2 - n^2, y = 2mn, and z = m^2 + n^2 where "m" and 'n" are arbitrary positive integers of opposite parity (one odd one even), and "m" is greater than "n". (For x, y, & z to be a primitive solution, "m" and "n" must have no common factors and must not both be even or odd. Dropping these two limitations will produce non-primitive Pythagorean Triples.

Another series of equations often used for generating triples are x = pq, y = (p^2 - q^2)/2 and z = (p^2 + q^2)/2 where p and q are odd integers with no common factors and p > q > 1.

Consecutive values of "m" and "n" will produce triples with one leg and the hypotenuse consecutive. An "m" and "n" of 2 and 1 produce the 3-4-5 triple. 3 and 2 produce the 5-12-13 triple. 4 and 3 produce the 7-24-25 triple. Make a table of more and see if you can discover the pattern of the results.

Using the "x" and "y" values from the consecutive "m" and "n" triples produce triples where the hypotenuse is a perfect square. An "m" and "n" of 4 and 3 produces the 7-24-25 triple (25 = 5^2). An "m" and "n" of 12 and 5 produces the 119-120-169 triple (169 = 13^2).

Using the "y" and "z" values from the consecutive "m" and "n" triples produce triples with the smallest leg a perfect square. An "m" and "n" of 5 and 4 produces the 9-40-41 triple. An "m" and "n" of 13 and 12 produces the 25-312-313 triple.

Consecutive legs can be derived from the "m" and "n" values derived from the general expression ni = 2(n(p-1) + m(p-2) and mi = 2m(p-1) + n(p-1). The subscript p indicates the "previous" value. The values m(p-1) and n(p-1) are the previous "m" and "n" values while m(p-2) and n(p-2) are the values previous to m(p-1) and n(p-1). Consider the starting point of the 3-4-5 triple which derives from "m" and "n" values of 2 and 1. The next triple that will produce a triple with both legs consecutive comes from the values of n = 2(1) + 0 = 2 and m = 2(2) + 1 = 5 which produces the 21-20-29 triple. The next will derive from n = 2(2) + 1 = 5 and m = 2(5) + 2 = 12 producing the 119-120-169 triple. See how many you can create and see if you can determine another interesting relationship between the results.

Other unique Pythagorean Triples can also be derived from "m" and "n" values based on the triangular numbers T = n(n+1)/2, i.e., 1, 3, 6, 10, 15, 21, etc. Using consecutive Triangular numbers for "m" and "n", the triples that result have the smallest leg a perfect cube.

The product of the two legs of a right triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse.

Triples can also be derived from x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 where n is any integer. The formulas creat only triangles where the hypotenuse exceeds the larger leg by one.

Another set of expressions that produce triples is x = n^2, y = (n^2 - 1)^2/2, and z = (n^2 + 1)^2/2.

Given any integer one can find another such that the sum of their squares is an integral square. Given the number A is odd - Resolve A into any 2 of its factors, equating the larger to m + n and the smaller to m - n. Solve for m and n and substitute into the formulas of 1) above. Given the number A is even, equate it to 2mn and let "m" and "n" be any two integers which will result in the product being equal to A/2.

Fermat discovered that Pythagorean Triples can be derived by expressing the odd squares as the sum of consecutive integers. Using any odd square, express it as x + (x + 1) = N = n^2 and n^2 + x^2 = (x + 1)^2. Another way of looking at it is to say a^2 = (b + c), b and c being consecutive integers. Since (b - c) = 1, we can write a^2 = (b + c)(b - c) = b^2 - c^2 from which a^2 + b^2 = c^2. For example, 25 = 12 + 13 = 5^2 and 5^2 + 12^2 = 13^2 or 121 = 60 + 61 = 11^2 and 11^2 + 60^2 = 61^2.

Others can be derived by taking any consecutive odd or even numbers and adding their reciprocals as with x and y, we get 1/x + 1/y = (x + y)/xy. It follows that (x + y)^2 + (xy)^2 = N^2. Example, from 5 and 7 we get 1/5 + 1/7 = 12/35 and 12^2 + 35^2 = 37^2 or from 6 and 8 we get 1/6 + 1/8 = 7/24 and 7^2 + 24^2 = 25^2.

In every triplet of integers for the sides of a Primitive Pythagorean triangle, one of them is always divisible by 3 and one by 5. Also, the product of the two legs is always divisible by 12, and the product of all three sides is always divisible by 60.