Beyond Pythagoras

Perimeter=  17+144+145=306    

Area= 17 x 144 ÷ 2 = 1224

19²   =  361 = 180.5
            2

Lower bound = 180, Upper bound = 181.
Middle Side = 180, Largest Side =181.

Perimeter= 19+180+181=380      

Area= 19 x 180 ÷ 2 = 1710

I will now put my results in the table:

I now have enough results to find the relationship between each side. I will now try to find the nth term for each sides.

I will start with the shortest side.

SHOTEST SIDE:

3 , 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19

 +2 +2 +2 +2  +2   +2   +2  +2

I noticed that the difference between the lengths of the short side is always 2. This means that the formula must include 2n.

I tried this but there was only +1 difference between 2n and the shortest side, which means that the formula should be 2n+1. I will now test the formula to see if it works:

N=8   2 x 8=16 + 1 = 17

N= 5        2 x 5=10 + 1 = 11

The formula works for the shortest side so the formula is:

Shortest Side: 2n+1

MIDDLE SIDE:

By looking at my table I noticed that I multiply short side by n I should be able to get close to the middles side

I then noticed that if I add the n to short side multiplied by n I should get the middle side. So it should be like

Smallest side x n + n= middle side

Now I have to try to put (smallest side x n + n= middle side) in to a rule

So I looked at the smallest side rule which was 2n+1 I decide to multiply it by ‘n’ (smallest side x n + n= middle side)

So the rule should look like 2n²+1. The only thing was left was that I had to add another n to 2n²+1 to get my rule

So the rule is 2n²+1n or 2n²+n

I tried this rule 2n²+n

N=2      2 x2²+2=10

It didn’t work I try 3rd term

N=3       2 x3²+3= 21

I noticed that I did something wrong with my rule I should multiplied the n by 2 to get the answer because every time I did  

2n²+n the difference between 2n²+n and middle side was N so I decided to multiply n by 2 this would give me 2 n

So the rule should be 2n²+2n

I will now try this rule out

N= 2         2 x2²+2 x 2=12

It works for 2nd term I will try it on 3rd term

N= 3        2 x3²+2 x 3=24

It works for 3rd term I will try it 4th term to make sure the rule is right

N= 4        2 x4²+2 x 4=40

The rule work

MIDDLE SIDE RULE: 2n²+2n

 

Another Rule for middle side:
I noticed that the Middle side numbers where all multiples of 4:

4 , 12 , 24 , 40 , 60 , 84 ,112

I decide to find out where in the 4’s time table did these multiples of 4 went:

Multiples of 4:                    

4 =        1 x 4=4            

12 =      3 x 4= 12

24 =       6 x 4= 24

40 =      10 x 4= 40

60 =        15 x 4= 60.

84 =        21 x 4= 84

112 =      28 x 4= 112

I noticed that all the numbers underlined were triangle numbers. So this shows us that the rule has something to with triangle numbers.

 So look at the triangle numbers rule:        

½ n (n +1) (triangle number rule)

I tried this rule on the one of the odd number      

 

n=3   0.5 x 3= 1.5 (3+1) = 6    

It didn’t work but I noticed that if you multiply a triangle number 4 you should get the middle side as I showed above.

4 =        1 x 4=4            

12 =      3 x 4= 12

24 =       6 x 4= 24

40 =      10 x 4= 40

60 =        15 x 4= 60.

84 =        21 x 4= 84

112 =      28 x 4= 112

I tried ½ n (n + 1) x4 on same number

n=3   0.5 x 3= 1.5 (3+1) = 6   =     6 x4= 24

It worked for 6 being the nth term

I will now try ½ n (n + 1) x 4 on some other numbers:

N= 4         0.5 x 4 = 2 (4 + 1) =10 x 4 = 40  

N= 6        0.5 x 6 = 3 (6 + 1) = 27 x 4 = 84

The formula works for the middle side

ANOTER RULE THE MIDDLE SIDE: ½ N (N+1) x 4

LONGEST SIDE:

I know that there is only a difference of 1 between the middle side and the longest side. So:

(Middle side) + 1 = Longest side.

2n² + 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.

2n² + 2n +1 = 5

2 x 1² + 2 x 1 + 1 = 5

2 + 2 + 1 = 5

5 = 5

The formula works for the first term.

2n² + 2n +1 = 13

2 x 2² + 2 x 2 + 1 = 13

8 + 4 + 1 = 13

13 = 13

The formula also works for the 2nd term.

2n² + 2n +1 = 25

2 x 3² + 2 x 3 + 1 = 25

18 + 6 + 1 = 25

25 = 25

The formula works for all 3 terms.

So the rule for the longest side is

Longest Side = 2n² + 2n + 1

Another Longest Side

The longest side I already know that the difference between short side and middle side so the rule must be the same as the middle side but you add one to middle side rule to get the longest side

½ N (N+1) x 4 + 1

I will now try this out to see if this works

N=2   ½ x 2(2 + 1) x 4 + 1 = 13

It works for 2 being the nth term

I will try one more

N=3       ½ x 3(3+1) x 4+1= 25

This rule works

ANOTHER RULE FOR LONGEST SIDE: ½ N (N+1) x 4 + 1

 

 To prove my formulas for ‘a’, ‘b’ and ‘c’ are correct. I decided incorporate my formulas into a2 + b2 = c2

a2 + b2= c2

 This equals:

To prove my formulas for ‘a’, ‘b’ and ‘c’ are correct. I decided incorporate my formulas into 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

 1=1

This proves that my ‘a’, ‘b’ and ‘c’ formulas are correct

I now end up with 0 = 0, this tells us that 2n + 1, 2n² + 2n and 2n² + 2n + 1 have got a Pythagorean triple.    

I have found the nth term for each of the 3 sides. Now I need to work out the perimeter and area.

To find the perimeter of a triangle you have to add up all the 3 sides’ e.g.

1st term 3 + 4 + 5 =12

So 12 is the perimeter for the first term.

2nd term 5 + 12 + 13 = 30

So 30 is the perimeter for the first term.

But I have to find these numbers using nth term. I can do this easily because I all ready know the nth term for each of the sides. So all I have to do is that I have to put all the 3 rules together.

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

=2n + 1 + 2n² + 2n + 2n² + 2n + 1

I will try this out

N=2       2n+1= 5+ 2n² + 2n=17 + 2n² + 2n + 1=30

IT works  

Rule for the Perimeter= 2n + 1 + 2n² + 2n + 2n² + 2n + 1

AREA:

I will now try to find out the rule for the area

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

Area = ½ x base x height        Area = ½ b h

Area = ½ (Shortest Side) X (Middle Side)

The area = (short side x b) divided by 2. Therefore I took my formula for ‘a’ (2n + 1) and my formula for ‘b’ (2n2 + 2n). I then did the following: -

(2n + 1)(2n2 + 2n) = area

            2 

I tried this rule to see if it worked

N= 3

(2n + 1)(2n2 + 2n)=   84

          2

The rule works

 Rule for the Area =          (2n + 1)(2n2 + 2n) = area

                                                      2       

EVEN NUMBERS:

I was told to start off 8, 15, 17

8²=64

15²=225

8²+15²=289 = 17²

So 8, 15, 17 satisfy the condition a²+b²=c²

Next even number I tried was 12

I couldn’t try 12 because I didn’t kne the middle and the long side numbers

So I tried     Short side ² = Middle and the longest side numbers (Lower bound = Middle side) (Upper Bound = longest Side)

                         2

I tried this on 8² so

         8²      =     64  = 32 It didn’t work

2. 2

I noticed that the answer from   Short side ² = the total of middle and long side                        

                                             2

So I decided to divide the short side by 4 which should get me close to the answer

 Short side ² =

       4

I tried this on 8 again to see if it works

          8² = 16

          4

 I noticed that the answer from Short side ² = was one lees that the longest side and one bigger than middle side

                                                   4

So I decide to subtract one from Short side ² - 1= middles side and I add one to that I should get the long side.

                                              4

I used this to get some results on my table which can help me get the rule for the sides easily.

I didn’t knew what was the first short side number so we used Short side ² - 1        to find the first short side number

                                                                                           4

n=4

         4² = 4 – 1= 3  

           4

4 can’t be the first short side number because 3 can’t be the middle side and 4 being the shot side

Now we tried 6

        6² = 9 – 1 = 8 (middle side)                        

          4

         

6² = 9 + 1 = 10 (Long side)                        

          4

So I tried 6, 8, 10 to see if it satisfies the condition

6²=36

8²=64

6²+8²=100 = 10²

So 6, 8, 10 satisfy the condition a²+b²=c²

It works so 6 is the first short side number

I will know find some more even numbers

10² = 25 – 1 = 24 (Middle side)

          4      25 + 1 = 26 (Long side)

10, 24, 25

Perimeter = 12+24+26= 60

Area = 10 x 24= 120

      2

12² = 36 – 1 = 35 (Middle side)

          4      36 + 1 = 37 (Long side)

12, 35, 37

Perimeter = 12+35+37= 84

Area = 12 x 35= 210

      2

14² =  49 – 1 = 48 (Middle side)

          4      49 + 1 = 50 (Long side)

14, 48, 50

Perimeter = 14+48+50= 112

Area = 14 x 48= 336

      2

16² = 64 – 1 = 63 (Middle side)

          4      64 + 1 = 65 (Long side)

16, 63, 65

Perimeter = 16+63+65= 144

Area = 16 x 63= 504

      2

18² = 81 – 1 = 80 (Middle side)

          4      81 + 1 = 82 (Long side)

18, 80, 82

Perimeter = 18+80+82= 180

Area = 18 x 80= 720

      2

20² = 100 – 1 = 99 (Middle side)

          4      100 + 1 = 101 (Long side)

20, 100, 101

Perimeter = 20+99+101= 220

Area = 20 x 99= 990

      2

22² = 121 – 1 = 120 (Middle side)

          4      121 + 1 = 122 (Long side)

22, 120, 122

Perimeter = 22+120+122= 264

Area = 22 x 120= 1320

      2

I have now enough results now I will put them in a table

         

        SHORT SIDE:

I noticed that the numbers in short side column all the numbers were +3 than the number that I found in odd number

Odd numbers               Even numbers

3            +3                    6

5            +3                    8

7            +3                   10

9            +3                   12

So I decide to 3 to 2n+1 so the rule for even numbers short side should be 2n+4

I tried this rule to see if it worked

N= 3          

2 x 3 + 4= 10

It works

Short side rule: 2n+4

 Middle side:

 

 The way I found the numbers for middle side Short side ² - 1        

                                                                  4

I decide to put this into a rule I all ready know the rule for short side which is 2n+4

So the rule should be (2n+4) ² - 1 = middle side for even numbers

                                   4

I try this rule to see if it works

N=3       (2 x 3 + 4) ²  - 1 =24

                    4

It works for 3 being the nth term

I will try on 4 now

N=4        (2 x 4 + 4) ²  - 1 =35

                    4

It works

I can simplify (2n+4) ² - 1                        

                     4

(2n+4) ² - 1                        

   4

             (2n+4)(2n+4) 

                  4

        4n ²+8n+8n+16

                   4

         

         4n ²+16n+16

                4

     

        N ² +4n + 4 -1 

        

        N ² +4n + 3 = the middle side 

Middle Side Rule:  N ² +4n + 3

Longest Side

I all ready know that the middle side and the longest sides difference wasalway 2 so all I have to do is that to add 2 to the rule I have for the middle side

N ² +4n + 5= longest side

I will try this rule out

N=2  

2²+4 x 2+5= 17

 

The rule works

Longest Side Rule: N ² +4n + 5

Perimeter:

I all ready know all the rules so all I have to do is that to add up all the rules

2n+4 + N ² +4n + 3+ N ² +4n + 5 = Perimeter

I can simplify this

2n+4 + N ² +4n + 3+ N ² +4n + 5

10n + 12 + 2n ²= Perimeter

I will try this out to see it thie rule works

N= 2             10 x 2 + 12 + 2x 2 ²= 40

It works

I will try 3 nth term

N= 3             10 x 3 + 12 + 2x 3 ²=60

It works

Perimeter: 10n + 12 + 2n ²

Area:

I know that the area is going to be

Short side x Middle side =  Area

              2

I all ready know the rules for short side and middle side so it would be easy to find out the area

2n+4 x N ² +4n + 3 =

         2

This the rule for the area or I can simplify this

2n+4 x N ² +4n + 3 

         2

2n3 + 8n²+ 6n +4n² +16n + 12

                2

2n3 +12n² +22n + 12

        2

n3 +6n²+ 11n +6 = area

I will now try this rule out to see if it works

N= 2

2 x 2 x 2 + 6 x 2² +11 x 2+6 =60

It works for 2nd being the nth term I will try 3rd nth term

N= 3

3 x 3 x 3 + 6 x 3² +11 x 3+6 =120

This rule work

Area: n3 +6n²+ 11n +6

Now, I will check that 2n + 4, n² + 4n+3 and n² + 4n + 5 forms a Pythagorean triple (or a² + b² = c²).A = 2n + 4, b = n² + 4n+3, and c = n² + 4n + 5.

(2n + 4)² + (n² + 4n+3)² = (n² + 4n + 5)²

(2n + 4) (2n + 4)+ (n² + 4n+3) (n² + 4n+3) = (n² + 4n + 5) (n² + 4n + 5)

4n²+8n+8n+16+n4 +4n3 +3n²+ 4n3 +16n² +12n +3n² +12n+9 = n4 +4n3 +5n²+4n3 +16n²+ 20n+ 5n² +20n +25

n4 +8n3 +26n² +40n + 25 = n4 +8n3 +26n² +40n + 25

I now end up with 0 = 0, so 2n + 4, n² + 4n+3and n² + 4n + 5 has got to be a Pythagorean triple

Universal Rule  

The Pythagoras is a set of three integers, ( a, b and c). Which apply to the Pythagoras' Theorem which is a2 + b2 = c2

The Pythagoras triple can be found using only two integers, U and V ( u is greater than v,  u>v). One of the two numbers from u and v has to be and odd number and one has to be even number, still u has to be greater than v, u>v) e.g.

U= 6 and v= 3  or it can be u= 9 and v= 4

A = U2+ V2

B = 2UV

C = U2 +V2