Beyond pythagoras - This investigation is about Pythagorean triples and the similarities between them.

For the 3rd term it is 2 x 3² + 2 x 3 = 24

For the 4th term it is 2 x 4² + 2 x 4 = 40

So my formula for the middle side is 2n² + 2n

This is how I found the formula for the longest side:

I think the formula for the longest side is 2n² + 2n + 1

I will now test this formula:

For the 1st term it is 2 x 1² + 2 x 1 + 1 = 5  

For the 2nd term it is 2 x 2² + 2 x 2 + 1= 13

For the 3rd term it is 2 x 3² + 2 x 3 + 1 = 25

For the 4th term it is 2 x 4² + 2 x 4 + 1 = 41

So my formula for the longest side is 2n² + 2n + 1

I am now going to use my formulas to predict the 6th triple which is:

The shortest side formula  =

2n + 1

so the shortest side =

2 x 6 + 1 = 13

The middle side formula =

2n² + 2n

so the middle side =

2 x 6² + 2 x 6 = 84

The longest side formula =

2n² + 2n + 1

so the longest side =

2 x 6² + 2 x 6 + 1 = 85

so the sixth triangle has sides 13, 84, 85

The perimeter for the sixth triangle is 13 + 84 + 85 = 182

The area for this triangle is ½ x 13 x 84 = 546

I worked out the seventh triangle by doing the following calculations:

Shortest side: 2 x 7 + 1 = 15

Middle side: 2 x 7² + 2 x 7 = 112

Longest side: 2 x 7² + 2 x 7 + 1 = 113

So the seventh triangle has sides 15, 112, 113

The perimeter for the seventh triangle is 15 + 112 + 113 = 240

The area for this triangle is ½ x 15 x 112 = 840

I will now put the results togehter in another table so I can see any patterns or similarities easier.

I am now going to prove my formulas are correct .

I will do this by using pythagoras’s theoreom which is: a² + b² = c²

Which in other words means the shortest side² + the middle side² = the longest side²

So if I put the first set of numbers into the formula they come out as:

3² + 4² = 5²

9 + 16 = 25

The first sequence works so now I will try the the third set of numbers into the formula to see if they work and they come out as:

        7² + 24² = 25²

        49 + 576 = 625

So this one works as well I think I will now try it on the last sequence I found to make sure it still works at the end of the sequences:

        15² + 112² = 113²

        225 + 12544 = 12769

So I think my formulas are correct as this one works as well.

I am now going to predict the 100th sequence to see if my formulas work when you get further on in the triangle sequences.

The shortest side is: 2 x 100 + 1 = 201

The middle side is: 2 x 100² + 2 x 100 = 20,200

The longest side is: 2 x 100² + 2 x 100 + 1 = 20,201

I am now going to use pythagoras’s theorem again to prove my prediction of the 100th sequence is correct:

        a² + b² = c²

        201² + 20,200² = 20,201²

        40,401 + 408,040,000 = 408,080,401    

So my formulas are correct and work later on as it worked for the 100th term.

I now need to prove this by multiplying the formulas out of each other to prove they are the same when each one is squared.

The shortest side:

(2n + 1) (2n + 1)  

So the shortest side equation is:

4n² + 2n + 2n + 1

4n² + 4n + 1

The middle side:

(2n² + 2n) (2n² + 2n)

So the middle side equation is:

        4n^4 + 4n³ + 4n³ + 4n²

        4n^4 + 4n² + 8n³

The longest side:

(2n² + 2n + 1) (2n² + 2n + 1)        

 4n^4 + 4n³ + 2n² + 4n³ + 4n² + 2n + + 2n² + 2n + 1

4n^4 + 8n³ + 8n² + 4n + 1

I am now going to add the shortest and middle sides togehter to see if they come out the same as the longest side algebraically.

Short side: 4n² + 4n + 1 middle side:         4n^4 + 4n² + 8n³

These should = the longest side: 4n^4 + 8n³ + 8n² + 4n + 1 when added together.

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

This proves that the formulas work algebraically as well.

I am now going to try to find the equations and formulas for the triangles with the shortest side as an even number.

     

This is how I found the formula for the shortest side:

I think the formula for the shortest side is 2n + 2

I will now test this formula:

For the 1st term it is 2 x 1 + 2 = 4

For the 2nd term it is 2 x 2 + 2 = 6

For the 3rd term it is 2 x 3 + 2 = 8

For the 4th term it is 2 x 4 + 2 = 10

So my formula for the shortest side is 2n + 2

This is how I found the formula for the middle side:

I think the formula for the middle side is n² + 2n

I will now test this formula:

For the 1st term it is 1² + 2 x 1 = 3

For the 2nd term it is 2² + 2 x 2 = 8

For the 3rd term it is 3² + 2 x 3 = 15

For the 4th term it is 4² + 2 x 4 = 24

So my formula for the middle side is n² + 2n

This is how I found the formula for the longest side:

I think the formula for the longest side is n² + 2n + 2

I will now test this formula:

For the 1st term it is 1² + 2 x 1 + 2 = 5

For the 2nd term it is 2² + 2 x 2 + 2 = 10

For the 3rd term it is 3² + 2 x 3 + 2 = 17

For the 4th term it is 4² + 2 x 4 + 2 = 26

So my formula for the longest side is n² + 2n + 2

Now the same as before I am going to prove the even triangles formulas work by using pythagoras’s thereom which is: a² + b² = c²

Which in other words means the shortest side² + the middle side² = the longest side²

So for the first triangles the numbers are 4, 3, 5:

        4² + 3² = 5²

        16 + 9 = 25

The first sequence works so now I will try the the third set of numbers into the formula to see if they work and they come out as:

The third triangles numbers are 8, 15, 17:

        8² + 15² = 17²

        64 + 225 = 289

So I think my even formulas are correct as well

I am now going too predict the 100th even sequence to see if the even formulas work later on in the sequences.

The shortest side:2 x 100 + 2 = 202

The middle side: 100² + 2 x 100 = 10,200

The longest side is: 100² + 2 x 100 + 2 = 10,202

I am now going to use pythagoras’s theoreom again to prove my prediction is correct:

        a² + b² = c²

        202² + 10,200² = 10,202²

        40,804 + 104,040,000 = 104,080,804

So my formulas for the even triangles are correct as it worked for the 100th term as well.

I am now going to prove my formulas work when put I multiply them out of each otherand square each one.

The shortest side:

(2n + 2) (2n + 2)  

  So the shortest side equation is:

4n² + 4n + 4n + 4

4n² + 8n + 4

The middle side:

(n² + 2n) (n² + 2n)

So the middle side equation is:

        n^4 + 2n³ + 2n³ + 4n²

        n^4 + 4n³ + 4n²

The longest side:

(n² + 2n + 2) (n² + 2n + 2)

So the longest side equation is:

        n^4 + 2n³ + 2n² + 2n³ + 4n² + 4n + 2n² + 4n + 4

        n^4 + 4n³ + 8n² + 8n + 4

I am now going to add the the shortest and middle sides together to see if they come out the same as the longest side.

4n² + 8n + 4  +  n^4 + 4n³ + 4n²

They should equal: n^4 + 4n³ + 8n² + 8n + 4

 

        They do equal: n^4 + 4n³ + 8n² + 8n + 4

So the formulas are correct algebraically as well.

   

I conclude that my formulas work. In this piece of coursework I had to find out the relationship between the 3 sides of a right angled triangle and I found out that there was a relationship. I also found there was a relationship between the od and even triangles. That relationship was that every other even number is double the relevant odd number in the sequences. An example of this is:

The first odd triangle with sides 3, 4, 5 with a perimeter of 12 and an area of 6 is half of the second even triangle with sides 6, 8, 10 with a perimeter of 24 which is the first triangles numbers doubled. Also I had to prove that my formulas worked by using Pythagoras’s theorem (a² + b² = c²) and I did this with the numbers I got from the calculations I did with my formulas and so proved my formulas were correct and worked. Also I had to prove my formulas worked when I put them into a algebraic format using the formulas I used to get my answers. This I did and it was successful and worked quite well but I did make a few mistakes which I had to correct so was quite diffucult. To work out what my formulas were when they were squared I had to multiply them out of their brackets and collect the like terms together. It took me quite a while to figure out what the formulas were but when I did it was easy. After that it was quite simple for me to put them together and work them out.

Now I am going to put all of my results into a single table so it is easier to see what I mean by every other even term being double an odd one:

This table shows the way the even triangles have a relationship with the odd triangles.