Beyond Pythagoras
For this piece of coursework I am trying to find Pythagorean triplets (these are whole numbers that's satisfies Pythagoras theorem).
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
= 32 + 42 = 9 + 16 = 25 = 52
To find the perimeter we add all the sides together.
Perimeter = 3 + 4 + 5 = 12
Finally to find the areas we times the smallest and middle side and then divide by to.
Area = 1/2 x 3 x 4 = 6
Odd Triples
I will now put the first term in to a table and try to find the next terms up to 10.
Term Number (n)
Shortest Side (s)
Middle Side (m)
Longest Side (l)
Perimeter (p)
Area (a)
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
6
3
84
85
82
545
7
5
12
13
240
840
8
7
44
45
306
224
9
9
80
81
380
710
0
21
220
221
462
2310
* s increases by +2 each term
* s is equal to the term number times 2 then add 1
* the square root of (m + l) = a
* (s x n) + n = m
I have worked out formulas for
. How to get s from n
2. How to get m from n
3. How to get l from n
4. How to get the p from n
5. How to get the a from n
To find the Smallest side (s) you use the formula 2n + 1
To find the Middle side (m) you use the formula 2n2 + 2n or (s x n) + n
To find the Longest side (l) you use the formula 2n2 + 2n + 1 or (n x s) + n + 1
To find the Perimeter (p) you use the formula 4n2 + 6n +2 or s + m + l
To find the Area (a) you use the formula 2n3 + 3n2 + n or s x m divided by 2
To get these formulas I did the following
. Take side 's' for the first five terms 3, 5, 7, 9, 11. From these numbers you can see that the formula ...
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To find the Longest side (l) you use the formula 2n2 + 2n + 1 or (n x s) + n + 1
To find the Perimeter (p) you use the formula 4n2 + 6n +2 or s + m + l
To find the Area (a) you use the formula 2n3 + 3n2 + n or s x m divided by 2
To get these formulas I did the following
. Take side 's' 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 diagram.
2. From looking at my table of results, I noticed that '(s x n) + n = m'. So I took my formula for 's' (2n + 1) multiplied it by 'n' to get '2n2 + n'. I then added my other 'n' to get '2n2 + 2n'.q=A diagram will help explain this. (Also you can use the longer formula (n x s) + n)
3. Side 'l' is just the formulas for side 'm' +1
4. The perimeter = s + m + l. Therefore I took my formula for 's' (2n + 1), my formula for 'm' (2n2 + 2n) and my formula for 'l' (2n2 + 2n + 1). I then did the following: -
2n + 1 + 2n2 + 2n + 2n2 + 2n + 1 = perimeter
Rearranges to equal
4n2 + 6n + 2 = perimeter (Also you can use the formula s + m + l)
5. The area = (s x m) divided by 2. Therefore I took my formula for 's' (2n + 1) and my formula for 'm' (2n2 + 2n). I then did the following: -
(2n + 1)(2n2 + 2n) = area 2
Multiply this out to get
4n3 + 6n2 + 2n = area
2
Then divide 4n3 + 6n2 + 2n by 2 to get
2n3 + 3n2 + n = area (Also you can use the formula s x m divided by 2)
Even Triples
I have now moved on to find out right-angled triangles with an even positive integer for the shortest side. The smallest number, which I have found to be the smallest number, is 6. Below shows a table of all of the even triples I have found.
shortest side (s)
middle side (m)
longest side (l)
Perimeter (p)
Area (a)
6
8
0
24
24
8
0
24
26
60
20
2
6
20
48
96
4
48
50
12
336
6
8
80
82
80
720
20
48
50
20
480
22
20
22
264
320
24
26
68
70
364
2184
28
96
00
224
344
30
224
226
480
3360
32
34
288
290
612
4896
36
60
64
360
2880
38
360
362
760
6840
40
.
42
440
442
924
9240
44
240
244
528
5280
When I was finding out even triples I could not find any for any number of a multiple of 8.
I have noticed that some of the middle and longest sides have a difference of 2 and others 4. Also they are in groups of three without the multiples of 8 and the difference goes 2, 4, 2 down the table.
I have placed all the even triples, which have a difference of 2 between the middle and longest length in the table below.
shortest side (s)
middle side (m)
longest side (l)
Perimeter (p)
Area (a)
6
8
0
24
24
0
24
26
60
20
4
48
50
12
336
8
80
82
80
720
22
20
22
264
320
26
68
70.
364
2184
30
224
226
480
3360
34
288
290
612
4896
38
360
362.
760
6840
42
440
442
924
9240
From this table it also shows that the shortest side goes up by 4 each time.
Shortest Length
.
6.
0
4
8
22
4
4
4
4
The difference between the sequence gives you 4n. This formula then gives you:
4, 8, 12, 16, 20
The difference between this formula and the original sequence is 2 and so the final formula is:
4n + 2
Middle Length
8
24
48
80
20
6
24
32
40
8
8
8
The difference between the sequence gives you 4n² as you have to half it then square the number on the second line. This formula then gives you: 4, 16, 36, 64, 100.
The difference of this formula and the original sequence is shown below.
4
8.
2
6
20
4
4.
4
This difference gives you 4n so the final formula for the middle length is:
4n² + 4n
Longest Length
0
26
50
82
22
6
24
32
40
8
8
8
The difference between this sequence gives you 4n², as you have to half it then square it because it is on the second line. This formula then gives you: 4, 16, 36, 64, 100.
The difference of this formula and the original sequence is shown below.
Difference =
6
0
4.
8
22
4
4
4
4
The formula now is 4n² + 4n, this is shown below.
8
24
48
80
20
The difference between this formula and the original sequence is shown below.
2
2
2
2
2
This then makes the final formula for the longest length:
4n² + 4n + 2
Perimeter
24
60
12
80
264
36
52
68
84
6
6
6
This gives you the first part of the formula, 8n². You got this by halving the number 16 and then squaring it. The results of this formula are shown below.
8
32
72
28
200
I then have found out the difference between the original sequence and formula, 8n², as you can see below.
6
28
40
52
64
2
2
2
2
This difference gives me 12n, so the formula is now 8n² + 12n, this is shown below
20
56
08.
76
260
The difference between the original sequence and the formula is now 4 for each stage, so the final formula is:
8n² + 12n + 4
Area
24
20
336
720
320
96
216
384
600
.
20
68
216
48
48
The difference of this sequence is 48 on line 3. This then means that you have to divide by three and then square the number, so you get 8n³. This formula is shown below.
8
64
216
512
000
As this is not the same as the original sequence you will need to difference it again. This is shown below
Difference =
6
56
20
208
320
40
64
88
12
24
24
24
This gives you 12n² because you have to half the end number then square it. This then gives you the formula 8n³ + 12n². The results of this formula are shown below.
20
12.
324
704
300
The difference of the original sequence and the original formula are shown below.
4
8
2
6
20
4
4
4
4
:
The difference then gives you 4n, which can then be added to finalise the formula, which is:
8n³ + 12n + 4n
Formula
Shortest Side (s)
4n + 2!
Middle Side (m)
4n² + 4n
Longest Side (l)
4n² + 4n + 2
Perimeter (p)
8n² + 12n + 4
Area. (a)
8n³ + 12n² + 4n