# Mathematics Coursework

Beyond Pythagoras

The numbers 3, 4 and 5 satisfy the condition

We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!

order now

3 +4 =5

Because

3

= 3×3

=9

4

= 4×4

=16

5

= 5×5

=25

So

3 +4

=9+16

=25

=5

I now have to find out whether the following sets of numbers satisfy a similar condition of

(smallest number) + (middle number) = (largest number)

a) 5, 12, 13

5 +12

= 25+144

= 169

= 13

b) 7, 24, 25

7 +24

= 49+576

= 625

=25

I looked at the table of results and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side.

I already know that the (smallest number) + (middle number) = (largest number) . So I think that there might be a connection between the numbers. The problem is that this is not completely correct.

(Middle number) + (largest number) = (smallest number)

Because

12 + 13

= 144+169

= 313

5 = 25

The difference between 25 and 313 is 288 which is far to big, so this means that the equation I need and want has nothing to do with 3 sides being squared. So I shall now try 2 sides being squared.

(middle number ) + largest number = (smallest number)

= 12 + 13 = 52

= 144 + 13 = 25

= 157

= 25

This does not work. There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side being squared.

12 + 13 = 5

This couldn’t work because 122 is already larger than 5, this also goes for 132. The only number now I can try squaring is the smallest number.

12 + 13 = 5

25 = 25

=5

This works with 5 being the smallest number/side but I need to know if it works with the other 2 triangles that I know.

4 +5 = 32

9 = 9

And…

24 + 25 = 7

49 = 49

It works with both of my other triangles. So this means that the

Middle number + Largest number = (Smallest number)

If I now work backwards, I should be able to work out some other odd numbers.

E.g.

92 = Middle number + Largest number

81 = Middle number + Largest number

I know that there will be only a difference of one between the middle number and the largest number. So, the easiest way to get 2 numbers with only 1 between them is to divide 81 by 2 and then using the upper and lower bound of this number. This is the answer I came up with:

81 = 40.5

2

Lower bound = 40, Upper bound = 41.

Middle side = 40, Largest side = 41.

Key

Longest/Largest Side = Length of Longest Side.

Middle Side = Length of Middle Side.

Shortest Side = Length of Shortest Side.

Let’s see if this works for a triangle that I already know.

7 = Middle number + Largest number

49 = Middle number + Largest number

49 = 24.5

2 2

Lower bound = 24, Upper bound = 25.

Middle Side = 24, Largest Side =25.

This matches the answers I already have with 7 being the shortest side, so I think that I have found the equation that works. I now believe I can fill out a table containing the Shortest, Middle and longest sides, by using the odd numbers starting from 3. I already know that the middle and longest side with the shortest length being 3, 5,7 or 9. So I will start with the shortest side being 11.

11 = Middle number + Largest number

121 = Middle number + Largest number

121 = 60.5

2

Lower bound = 60, Upper bound = 61.

Middle Side = 60, Largest Side =61.

13 = Middle number + Largest number

169 = Middle number + Largest number

169 = 84.5

169

Lower bound = 84, Upper bound = 85.

Middle Side = 84, Largest Side =85.

152 = Middle number + Largest number

225 = Middle number + Largest number

225 = 112.5

2

Lower bound = 112, Upper bound = 113.

Middle Side = 112, Largest Side =113.

17 = Middle number + Largest number

289 = Middle number + Largest number

289 = 144.5

2

Lower bound = 144, Upper bound = 145

Middle Side = 144, Largest Side =145.

19 = Middle number + Largest number

361 = Middle number + Largest number

361 = 180.5

2

Lower bound = 180, Upper bound = 181.

Middle Side = 180, Largest Side =181.

21 = Middle number + Largest number

441 = Middle number + Largest number

441 = 220.5

2

Lower bound = 220, Upper bound = 221.

Middle Side = 220, Largest Side =221.

I now have 10 different triangles, which I think makes life easier when it comes to finding a relationship between each side.

Shortest side = Middle side + Longest side

The way I mentioned above and on the previous pages, is quite a good way of finding the middle and longest sides. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest, middle and longest).

The formula I will work out first is for the shortest side.

3 , 5 , 7 , 9 , 11

2 2 2 2

The difference between the lengths of the shortest side is 2. This means the equation must have something to do with 2n. There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct, I will now test this formula.

The Shortest side and 2n+1 columns match meaning that:

Shortest Side = 2n+1

The next formula I need to work out is the formula for the middle side.

4 , 12 , 24 , 40 , 60

8 12 16 20

4 4 4

The difference of the difference of the lengths of the middle sides is 4. This means that the formula has to have something to do with 4, most likely 4n. However, because 4 are the difference of the difference, the formula must be n . I now believe that the answer will have something to do with 4n . So, I will now write out the answers for 4n .

4n works for the first term, but, it then collapses after this, as the difference between 4n gets larger and larger, the thing you notice is that the difference in the 2nd term between 4n and the middle side is the middle side for the term before. This goes for all the other terms from the 2nd.

This means that if I subtract the previous term, then I should in theory get the correct answer.

16 – 4 = 12

36 – 12 = 24

64 – 24 = 40

Etc.

So, the equation I have so far is:

4n – (Previous middle side) = Middle side

All the previous term is, (n – 1), so if I put this into the above formula, then it should give me my middle side.

4n – 4(n – 1) = Middle side.

This should in theory give me my middle side. I will test my theory with the first term.

4 x 12 – 4(1-1) = 4

4 x 1 – 4 x 0 = 4

4 – 4 x 0 = 4

4 – 0 = 4

4 = 4

My formula works for the first term. I will now check if it works using the 2nd term.

4 x 2 – 4(2-1) = 12

4 x 4 – 4 x 12 = 12

16 – 4 x 1 = 12

16 – 4 = 12

12 = 12

My formula also works for the 2nd term. It’s looking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.

4 x 3 – 4(3-1) = 24

4 x 4 – 4 x 2 = 24

36 – 4 x 4 = 24

36 – 16 = 24

20 = 24

My formula doesn’t work for the 3rd term. It now looks as if “4n – 4(n – 1) ” is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.

4 x 4 – 4(4-1) = 40

4 x 16 – 4 x 3 = 40

64 – 4 x 9 = 40

64 – 36 = 40

28 = 40

My formula doesn’t work for the 4th term either. I can now safely say that 4n – 4(n-1) is definitely not the correct formula for the middle side.

I believe the problem with 4n – 4(n-1) was that 4n , once you start using larger numbers, becomes far to high to bring it back down to the number that I want for the middle side. Also, 4(n-1) is not as small when it gets larger so it doesn’t bring the 4n down enough, to equal the middle side.

I know that the final formula will have something to do with 4 and have to be n . I will now try n + 4.

I will now look at the differences to see if I can find a pattern there.

1 , 4 , 11 , 20 , 31

3 7 9 11

2 2 2

The difference of the difference here is 2, which means that the answer will involve 2 and n . I will now write out all the answers for 2n .

Straight away, I can see that the difference between 2n and the middle number is the 2 times table. The 2 times table in the nth term is 2n. I now think that 2n + 2n is the correct formula. I will now test it using the first 3 terms.

2 x 1 + 2 x 1 = 4

2 x 1 + 2 = 4

2 + 2 = 4

4 = 4

My formula works for the first term, so, I will now check it on the 2nd term.

2 x 2 + 2 x 2 = 12

2 x 4 + 4 = 12

8 + 4 = 12

12 = 12

My formula also works for the 2nd term. If it works for the 3rd term I can safely say that 2n + 2n is the correct formula.

2 x 3 + 2 x 3 = 24

2 x 9 + 6 = 24

18 + 6 = 24

24 = 24

My formula also works for the 3rd term. I am now certain that 2n + 2n is the correct formula for finding the middle side.

Middle Side = 2n + 2n

This formula also relates back to the 4 that I was writing about, because, 2 + 2 = 4 and 2 is also a factor of 4.

I now have the much easier task of finding the longest side. To start with, I am going to draw out a table containing the middle and longest sides.

From the table on the previous page and other previous tables, I know that there is only a difference of 1 between the middle side and the longest side. So:

(Middle side) + 1 = Longest side.

2n + 2n + 1 = Longest Side

I am 99.9% certain that this is the correct formula. Just in case, I will check it using the first 3 terms.

2n + 2n +1 = 5

2 x 12 + 2 x 1 + 1 = 5

2 + 2 + 1 = 5

5 = 5

The formula works for the first term.

2n + 2n +1 = 13

2 x 2 + 2 x 2 + 1 = 13

8 + 4 + 1 = 13

13 = 13

The formula also works for the 2nd term.

2n + 2n +1 = 25

2 x 3 + 2 x 3 + 1 = 25

18 + 6 + 1 = 25

25 = 25

The formula works for all 3 terms. So…

Longest Side = 2n + 2n + 1

Now, I will check that 2n + 1, 2n + 2n and 2n + 2n + 1 forms a Pythagorean triple (or a2 + b2 = c2). We have a = 2n + 1, b = 2n + 2n, and c = 2n + 2n + 1.

c – b = (c – b) (c + b) = c + b = 2b + 1 = 4n + 4n + 1 = (2n + 1) = a .

Or I can show it this way,

a + b = c

This equals:

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

If you then put these equations into brackets:

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

You need to factorise the brackets out:

4n + 4n + 1 + 4n + 8n + 4n = 4n + 4n + 2n + 4n + 4n + 2n + 2n +

2n +1

If I now balance out each side, and I end up with nothing, then 2n + 1, 2n + 2n and 2n + 2n + 1 is a Pythagorean triple.

4n + 4n = 4n + 2n + 2n

4n = 2n + 2n

8n = 4n + 4n

4n = 4n

1 = 1

I now end up with 0 = 0, so 2n + 1, 2n + 2n and 2n + 2n + 1 has got to be a Pythagorean triple.

I now have the nth term for each of the 3 sides of a right-angled triangle. I can now work out, both, the nth term for the perimeter and the nth term for the area.

The perimeter of any triangle, is just the length of the 3 sides added together. E.g.

1st term 3 + 4 + 5 =12

So 12 is the perimeter for the first term.

2nd term 5 + 12 + 13 = 30

3rd term 7 + 24 + 25 = 56

And so on. But, what if I didn’t know these numbers, then I would have to work out it out using the nth term and this is what I want to work out. Lucky, for me, I already know the nth term for each of the sides. So, it’s all a matter of putting those 3 formulas together.

Perimeter = (Shortest side) + (middle side) + (Longest Side)

= 2n + 1 + 2n + 2n + 2n + 2n + 1

= 2n + 2n + 2n + 2n + 2n + 1 + 1

= 4n + 6n + 2

This, if I have done my calculations properly, should be the right answer. To check, I am going to use the 4th, 5th and 6th terms.

4th term:

4n + 6n + 2 = Perimeter.

4 x 4 + 6 x 4 + 2 = 9 + 40 + 41

64 + 24 + 2 = 90

90 = 90

My formula works for the 4th term.

4n + 6n + 2 = Perimeter.

4 x 5 + 6 x 5 + 2 = 11 + 60 + 61

100 + 30 + 2 = 132

132 = 132

And it works for the 5th term.

4n + 6n + 2 = Perimeter.

4 x 6 + 6 x 6 + 2 = 13 + 84 + 85

144 + 36 + 2 = 182

182 = 182

My formula works for all 3 terms, so…

Perimeter = 4n + 6n + 2

Like the perimeter, I already know that the area of a triangle is found by:

Area = 1/2 b h

b = Base

h = Height

Depending on which way up the right angled triangle is the shortest side or middle side can either be the base or height because it doesn’t really matter which way round they go, because I’ll get the same answer either way! So…

Area = 1/2 (Shortest Side) X (Middle Side)

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

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

2

I will now check this using the first 3 terms.

(2n + 1) (2n + 2n) = 1/2 b h

2

(2 x 1 + 1) (2 x 1 + 2 x 1) = 1/2 x 3 x 4

2

3 x 4 = 1/2 x 12

2

12 = 6

2

6 = 6

My formula works for the first term.

(2n + 1) (2n + 2n) = 1/2 b h

2

(2 x 2 + 1) (2 x 2 + 2 x 2) = 1/2 x 5 x 12

2

5 x 12 = 1/2 x 60

2

60 = 30

2

30 = 30

My formula also works for the 2nd term.

(2n + 1) (2n + 2n) = 1/2 b h

2

(2 x 3 + 1) (2 x 3 + 2 x 3) = 1/2 x 7 x 24

2

7 x 24 = 1/2 x 168

2

168 = 168

My formula works for all 3 terms. So…

Area = (2n +1) (2n + 2n)

2

If I am given a right-angled triangle I can always apply an enlargement to it (for example, I can double all lengths) and get another right-angled triangle. This means that from a given triple a, b, c we can produce many more Pythagorean triples na, nb, nc for any whole number n.

For example, starting with the triple 3, 4, 5 and taking n = 5 we get the new triple 15, 20, 25. Sometimes we can reverse this process by starting with a triple and then reducing the lengths of the sides to get another triple.

This does not always work; if we start with 3, 4, 5, for example, and halve the lengths of the sides we do not get a triple of whole numbers. However, sometimes we do; for example, by halving lengths the triple 10, 24, 26 converts into the triple 5, 12, 13.

As an extra part, I am now going to find the relationships (differences) between the Short and Middle sides, the Short and Long Sides, and The Middle and Long sides.

Finding the relationship between 2 sides is quite easy, because, I already know the nth terms for each side. Because of this, all I have to do to find the relationship is to subtract one nth term from the other, leaving me with the relationship.

Middle Side – Short Side = Relationship.

2n + 2n – 2n + 1 = Relationship

2n – 1 = Relationship

To check if the equation works, I am going to write out a table containing 2n – 1 and the relationship between the shortest and middle side.

2n – 1 and the Relationship between the shortest and middle side both match. So…

Relationship between Shortest and Middle sides = 2n – 1

The next relationship I’m going to work out is the relationship between the shortest and longest sides.

Longest Side – Short Side = Relationship.

2n + 2n + 1 – 2n + 1 = Relationship.

2n + 1 = Relationship.

To check if this formula is right, I am going to write out a table containing 2n + 1 and the relationship between the shortest and Longest side.

2n + 1 and the Relationship between the shortest and longest sides both match. So…

Relationship between the shortest and longest sides = 2n + 1

The next relationship I’m going to find is between the middle and longest side.

Longest Side – middle Side = Relationship.

2n + 2n + 1 – 2n + 2n = Relationship.

1 = Relationship.

I am certain that the relationship between the longest and middle side is 1, so I have a table to prove it.

1 and the Relationship between the middle and longest sides both match. So…

Relationship between the middle and longest sides = 1

Using these same principles, I can work out any relationship, to prove this, I will work out the relationship between the shortest side and the perimeter and then the area.

Perimeter – Shortest Side = Relationship.

4n + 6n + 2 – 2n + 1 = Relationship.

4n + 4n + 1 = Relationship.

I am certain that this is the write answer. So…

Relationship between the Perimeter and Shortest side = 4n + 4n + 1

The area is even simpler because, all you have to do is knock the 2n + 1 out of the equation. So…

Relationship between the Area and Shortest side = 2n +2n