Beyond Pythagoras

This means...

If the integer of the square root of B1 (1) squared by two plus A1 (1) squared by two is not a whole number print a zero, but if it is a whole number print the number.

The integer of a number is the bit after the point.

I took the square root because only a certain numbers two numbers each squared by two will produce a whole number.

Pythagorean triangles must have whole numbers.

(c)nick Peplow

By looking at the printout I can see, which three numbers, produce Pythagorean triangles.

On this printout they are

4,3,5 3,4,5 8,6,10 6,8,10

Checks

Check 1

4 3 5

4^ + 3^ = 5^

6 + 9 = 25

So

4^ + 3^ = 16 + 9 = 25 = 5^

Check 2

8 6 10

8^ + 6^ = 5^

64 + 36 = 100

So

8^ + 6^ = 16 + 9 = 100 = 10^

By checking it I have proved that this method can be used to find Pythagorean triangles.

By expending the formula in excel I can produce a bigger chart showing Pythagorean triangles, (see sheet 1)

By looking at the graph I can see that there are four different families of Pythagorean triangles shown, I have highlighted these so they are easier to see, the numbers that are not highlighted are multiples of the family that is to the left of it. I have drawn the links in so you can see what they are multiples of.

By looking at my graph, the "beyond Pythagoras" sheet & the results I have already collected I can make these predictions about the first family of Pythagorean triangles

I predict that...

* "a" increases by +2 each time in Pythagorean triples

* There is a pattern in the last two digits of "b" it goes 2,4,2

* "a" times the term number plus the term number equals "b"

* The last digit of "c" has a pattern, it goes 5,3,5,

By looking at the data I can make these generalisations

Generally "a" is equal to the term number times two and then +1, I have not investigated this enough to be sure yet

(c)nick Peplow

Generally in Pythagorean triples "c" is always 1 more than "b"

Generally "a" is equal to the term number times two and then +1

I am going to use the data that these sheets show to help me work out a formula for the length of the shortest side (a), the length of the middle longest side (b) and the length of the hypotenuse.

I will also work out a formula for the area and the perimeter of the triangle.

By taking the date from the "beyond Pythagoras "sheet and the data on the excel sheet I can work out the formula by using a a sequence diagram to find a constant and then using a linear equation if it there is a constant on the second step or a quadratic if there is a constant on the third step.

I this is how I will work out the formula for the shortest side of family 1

X Y

3

> 2

2 5

> 2

3 7

There is a constant on the second step so I can solve it with a linear equation. I will solve it using the standard linear equation

Y = ax + c

I will put the numbers into this equation so I can work out the formula.

A * 1 + c = 3 (i)

A * 2 + c = 5 (ii)

I will then take "ii" from "i" to find out what a is

= 2

I will then put this into "i" to find out what "c" is

2 + c = 3 (i)

I have to take 2 from three

C = 1

Y = 2 * x + 1

X is the sequence number

Checks

2 * 1 + 1 = 3

2 * 2 + 1 = 5

2 * 3 + 1 = 7

(c)nick Peplow

By checking my work I have proven that my formula for the shortest side is correct.

I am now going to work out the formula for the middle length.

4

> 8

2 12 > 4

> 12

3 24 > 4

> 16

4 40

Because there is a constant on the third step it is a quadratic equation

The standard formula for a quadratic equation is

Y = ax2 + bx + c

I will put the numbers into the equation so I can work out the formula.

4 = a*12 + b + c (i)

2 = a*22 + 2*b + c (ii)

24 = a*32 + 3*b + c (iii)

(ii)-(i) 8 = 3a + b (iv)

(iii)-(ii) 12 = 5a + b (v)

(v)-(iv) 4 = 3a + b

a = 2

To find b I have to substitute it into "iv"

8 = 3*2 + b

b = 2

To find c I have to substitute it into "i"

4 = 2*12 + 2 + c

c = 0

so the formula for the middle side is

y = 2x2 + 2x

or you can simplify it

y=2x(x-1)

Checks

y = 2x2 + 2x

y = 2 * 1*1 + 2*1 = 4

y = 2 * 2*2 + 2*2 = 12

y = 2 * 3*3 + 2*3 = 24

By checking my formula I have proved that my formula is correct.

(c)nick Peplow

I am now going to work out a formula for the hypotenuse.

X Y

5

> 8

2 13 > 4

> 12

3 25 > 4

> 16

4 41

The constant is on the fourth step so I will have to use a quadratic equation to work it out.

The standard formula for a quadratic equation is

Y = ax2 + bx + c

5 = a + b + c (i)

3 = 4a + 2b + c (ii)

25 = 9a + 3b + c (iii)

(ii)-(i) 8 = 3a + 1b (iv)

(iii)-(ii) 12 = 5a + 1b (v)

(v)-(iv) 4 = 2a (vi)

a = 2

To get this I had to divide both sides by two.

To find "b" I have to substitute it back into "iv"

8 = 3 * 2 + b

8 = 6 + b

b = 2

I am now going to substitute them both into (i) so I can find what "c" is

5 = 2 * 2 + c

c = 1

so the formula for the hypotenuse is

y = 2x2 + 2x + 1

checks

y = 2 * 12 + 2 * 1 + 1 = 5

y = 2 * 22 + 2 * 2 + 1 = 13

y = 2 * 32 + 2 * 3 + 1 = 25

By checking my work I have proven that my formula is correct

I am now going to work out a formula for the perimeter of the triangle

You can work out the perimeter of a triangle by adding all the sides together

So if add all the formulas together

Y = 2 * x + 1

+

Y = 2x2 + 2x

+

Y = 2x2 + 2x + 1

=

(c)nick Peplow

Y = 4x2 + 6x + 2

It will give the perimeter.

Checks

Y = 4 * 12 + 6 * 1 + 2 = 12

Y = 4 * 22 + 6 * 2 + 2 = 30

Y = 4 * 32 + 6 * 3 + 2 = 56

By checking I have found that my formula is correct.

Now I need to find the area

The formula for finding the area of a triangle is 1/2 base * height

So in need to divide the formula for "b" by two and then times it by the formula for "a"

This is the formula for "b"

Y = 2x2 + 2x (i)

And this is the formula for "a"

Y = 2 * x + 1 (ii)

So basically all you have to do is divide the formula that you get from "i"

And then times it by "ii"

Y = (2x2 + 2x) /(divided) 2 * (2x + 1)

Check

Y = (2*12 + 2*1) / 2 * (2*1 + 1) =6

Y = (2*22 + 2*2) / 2 * (2*2 + 1) =30

Y = (2*32 + 2*3) / 2 * (2*3 + 1) =84

By checking my formula I have proved that it is correct.

(c)nick Peplow

(c)Nick Peplow