Beyond Pythagoras

Perimeter = 3 + 4 + 5 = 12

Area = 1/2 x 3 x 4 = 6

For the set of numbers 5, 12 and 13:

Perimeter = 5 + 12 + 13 = 30

Area = 1/2 x 5 x 12 = 30

For the set of numbers 7, 24 and 25:

Perimeter = 7 + 24 + 25 = 56

Area = 1/2 x 7 x 24 = 84

From these sets of numbers I have noticed the following: -

* 'a' increases by +2 for each set of numbers

* 'b' increases by +4 for each set of numbers

* 'c' is always +1 to 'b'

* the square root of ('b' + 'c') = 'a'

From these observations I have worked out the next two sets of numbers in the sequence.

Term Number 'n'

Shortest Side 'a'

Middle Side 'b'

Longest Side 'c'

Perimeter

Area

3

4

5

2

6

2

5

2

3

30

30

3

7

24

25

56

84

4

9

40

41

90

80

5

1

60

61

32

330

From this table I have worked out the following formulas:

* How to get 'a' from 'n'

* How to get 'b' from 'n'

* How to get 'c' from 'n'

* How to get the perimeter from 'n'

* How to get the area from 'n'

These formulas are:

* 2n + 1

* 2n² + 2n

* 2n² + 2n + 1

* 4n² + 6n + 2

* 2n³ + 3n² + n

To get these formulas I did the following:

Take side 'a' for the first five sets of numbers; 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because they are consecutive odd numbers.

From looking at my table of results, I noticed that 'an + n = b'. So I took my formula for 'a' (2n + 1) multiplied it by 'n' to get '2n² + n'.

I then added my other 'n' to get:

2n² + 2n.

Side 'c' is just the formula for side 'b' +1:

2n² + 2n + 1

The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1), my formula for 'b' (2n² + 2n) and my formula for 'c' (2n² + 2n + 1).

Then I did the following:

2n + 1 + 2n² + 2n + 2n² + 2n + 1

This can be rearranged to equal:

4n2 + 6n + 2

The area = (a x b) divided by 2. Therefore I took my formula for 'a' (2n + 1) and my formula for 'b' (2n² + 2n).

Then I did the following:

(2n + 1)(2n² + 2n)

Multiply this out to get:

4n³ + 6n² + 2n

Then divide:

4n³ + 6n² + 2n by 2 to get:

2n³ + 3n² + n

To prove my formulas for 'a', 'b' and 'c' are correct. I decided to use my formulas in the condition:

a2 + b2 = c2

a2 + b2= c2

(2n + 1)² + (2n² + 2n)² = (2n² + 2n + 1)²

(2n + 1)(2n + 1) + (2n² + 2n)(2n² + 2n) = (2n² + 2n + 1)(2n² + 2n + 1)

4n² + 2n + 2n + 1 + 4n4 + 4n² + 4n³ + 4n³ = 4n4 + 8n³ + 8n² + 4n + 1

4n² + 4n + 1 + 4n4 + 8n³ + 4n² = 4n4 + 8n³ + 8n² + 4n + 1

4n4 + 8n³ + 8n² + 4n + 1 = 4n4 + 8n³ + 8n² + 4n + 1

This proves that my 'a', 'b' and 'c' formulas are correct.

Extension

Arithmetic Progression

I decided to investigate as to whether or not the Pythagorean triple 3,4,5 is the basis of all triples or just some of them.

To learn about this I have used the Internet to do some research into Pythagorean triples. I discovered the following:

3, 4, 5 is a Pythagorean triple.

The pattern is plus one.

If a = 3 and d = difference (which is +1) then:

3 = a

4 = a + d

5 = a +2d

a, a +d, a + 2d

Therefore if you incorporate this into Pythagoras theorem you get the following:

a² + (a + d)² = (a + 2d)²

a² + (a + d)(a + d) = (a + 2d)²

a² + a² + ad + ad + d² = (a + 2d)²

2a² + 2ad + d² = (a + 2d)²

2a² + 2ad + d² = (a + 2d)(a + 2d)

2a² + 2ad + d² = a² + 2ad + 2ad + 4d²

2a² + 2ad + d² = 4d² + a² + 4ad

If you equate these equations to 0 you get the following:

a² - 3d² - 2ad = 0

Change a to x:

x² - 3d² - 2dx = 0

Factorise this equation to get:

(x + d)(x - 3d)

Therefore:

x = -d

x = 3d

x = -d is impossible as you cannot have a negative dimension.

a, a+d, a + 2d

Is the same as:

3d, 4d, 5d

This tells us that the only Pythagorean triples are 3, 4, 5 or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.

Mathematics GCSE Coursework Beyond Pythagoras

Luke Hopwood 11B

Candidate number: 7484 The Mirfield Free Grammar