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Beyond Pythagoras.

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Introduction

Beyond Pythagoras

Pythagoras Theorem is a2 + b2 = c2. 'a' being the shortest side, 'b' being the middle side and 'c' being the longest side (hypotenuse) of a right angled triangle.

The numbers 3, 4 and 5 satisfy this condition

32 + 42 = 52

because 32 = 3 x 3 = 9

42 = 4 x 4 = 16

52 = 5 x 5 = 25

and so 32 + 42 = 9 + 16 = 25 = 52

The numbers 5, 12, 13 and 7, 24, 25 also work for this theorem

52 + 122 = 132

because 52 = 5 x 5 = 25

122 = 12 x 12 = 144

132 = 13 x 13 = 169

and so 52 + 122 = 25 + 144 = 169 = 132

72 + 242 = 252

because 72 = 7 x 7 = 49

242 = 24 x 24 = 576

252 = 25 x 25 = 625

and so 72 + 242 = 49 + 576 = 625 = 252

3 , 4, 5

Perimeter = 3 + 4 + 5 = 12

Area = ½ x 3 x 4 = 6

5, 12, 13

Perimeter = 5 + 12 + 13 = 30

Area = ½ x 5 x 12 = 30

7, 24, 25

Perimeter = 7 + 24 + 25 = 56

Area = ½ x 7 x 24 = 84

From the first three terms I have noticed the following: -

  • 'a' increases by +2 each term
  • 'a' is equal to the term number times 2 then add 1
  • the last digit of 'b' is in a pattern 4, 2, 4
  • the last digit of 'c' is in a pattern 5, 3, 5
  • the square root of ('b' + 'c') = 'a'
  • 'c' is always +1 to 'b'
  • 'b' increases by +4 each term
  • ('a' x 'n') + n = 'b'
...read more.

Middle

60

61

132

330

I have worked out formulas for

  1. How to get 'a' from 'n'
  2. How to get 'b' from 'n'
  3. How to get 'c' from 'n'
  4. How to get the perimeter from 'n'
  5. How to get the area from 'n'

My formulas are

  1. 2n + 1
  2. 2n2 + 2n
  3. 2n2 + 2n + 1
  4. 4n2 + 6n + 2
  5. 2n3 + 3n2 + n

To get these formulas I did the following

  1. Take side 'a' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula is 2n + 1 because these are consecutive odd numbers (2n + 1 is the general formula for consecutive odd numbers) You may be able to see the formula if you draw a graph
  1. 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'. This is a parabola as you can see from the equation and also the graph
  1. Side 'c' is just the formula for side 'b' +1
  2. The perimeter = a + b + c. Therefore I took my formula for 'a' (2n + 1)
...read more.

Conclusion

4n2 + 6n + 2

4(n +3)2 - 9 + 2

4(n +3)2 - 7

4(n +3)2 = 7

(n +3)2 = 1.75

n + 3 = 1.322875656

n + 3 = -1.322875656

n = -1.677124344

n = -4.322875656

Arithmatic Progression

I would like to know whether or not the Pythagorean triple 3,4,5 is the basis of all triples just some of them.

To find this out I have been to the library and looked at some A-level textbooks and learnt 'Arithmatic Progression'

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

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

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.

...read more.

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