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.
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.
I will do a small investigation to find the number that I already know.
72 = Middle number + Largest number
49 = Middle number + Largest number
49 = 24.5
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 this equation 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.
112 = 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.
132 = Middle number + Largest number
169 = Middle number + Largest number
169 = 84.5
2
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.
172 = 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.
192 = 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.
212 = 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 is easily enough to find a relationship between each side.
Shortest side²= 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.
1 | 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 be something to do with 2n. So, I will write down all the answers for 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 column 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 n². I now believe that the answer will have something to do with 4n²So, I will now write out the answers for 4n²
4n²works for the first term, but, it then collapses after this, as the difference between 4n²gets larger and larger, the thing you notice is that the difference in the 2nd term between 4n²and 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:
4n2 - (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.
4n² - 4(n - 1)² = 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)2 = 4
4 x 1 - 4 x 02 = 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 22 - 4(2-1)2 = 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'slooking likely that this is the correct formula. Just to check, I will check if it works using the 3rd term.
4 x 32 - 4(3-1)2 = 24
4 x 4 - 4 x 22 = 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 "4n² - 4(n - 1)²" is not the correct formula after all. To check, I will look to see if the formula works using the 4th term.
4 x 42 - 4(4-1)2 = 40
4 x 16 - 4 x 32 = 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 4n² - 4(n-1)² is definitely not the correct formula for the middle side.
I believe the problem with 4n2 - 4(n-1)² was that 4n², once you start using larger numbers, becomes far too high to bring it back down to the number that I want for the middle side. Also, 4(n-1)² is not as small when it gets larger so it doesn’t bring the 4n2 down enough, to equal the middle side.
I know that the final formula will have something to do with 4 and have to be n2. I will now try n2 + 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 n2. I will now write out all the answers for 2n2.
Straight away, I can see that the difference between 2n2 and the middle number is the 2 times table. The 2 times table in the nth term is 2n. I now think that 2n2 + 2n is the correct formula. I will now test it using the first 3 terms.
2 x 12 + 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 22 + 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 2n² + 2n is the correct formula.
2 x 32 + 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 2n² + 2n is the correct formula for finding the middle side.
Middle Side = 2n² + 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.
2n² + 2n + 1 = Longest Side
I am almost absolutely certain that this is the correct formula that I am looking for. Just in case, I will check it using the first 3 terms that I was investigating beforehand.
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
I have discovered that the formula works for all 3 terms. It is therefore the one I was looking for, so…
Longest Side = 2n² + 2n + 1
Now, I will check that 2n + 1, 2n² + 2n and 2n² + 2n + 1 (all the formulas in my table) form a Pythagorean triple (or a² + b² = c², which is the age-old Pythagorean Theorem which states that the longest side of a right-angled triangle squared, or its hypotenuse², is equal to the sum of the other two sides squared.). I have a = 2n + 1, b = 2n² + 2n, and c = 2n² + 2n + 1.
c² - b² = (c - b) (c + b) = c + b = 2b + 1 = 4n² + 4n + 1 = (2n + 1)2 = a².
Or I can show it this way, which is much the same,
a² + b² = c²
I now know that this equals: (2n +1)² + (2n² + 2n)² = (2n² + 2n +1)²
If I was then to put these equations into brackets., which I have done below as you can see:
(2n + 1)(2n+1) + (2n² + 2n)(2n² + 2n) = (2n² + 2n + 1)(2n² + 2n + 1)
I need to factorise the brackets out as below.
4n² + 4n + 1 + 4n + 8n³ + 4n² = 4n + 4n³ + 2n² + 4n³ + 4n² + 2n + 2n² +
2n +1
If I now balance out each side, and I end up with nothing, then I know that 2n + 1, 2n² + 2n and 2n² + 2n + 1 is a Pythagorean triple.
4n² + 4n² = 4n² + 2n² + 2n²
4n = 2n + 2n
8n³ = 4n³ + 4n³
4n = 4n
1 = 1
I now end up with 0 = 0, so 2n + 1, 2n² + 2n and 2n² + 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. i.e.
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 with any triangle one can possibly imagine. But what if I didn’t know these numbers? I would then have to work out it out using the nth term and since this is very long-winded And involves a lot of work, this option is not the most practical. This is, however, what I want to work out. Lucky for me that I already know the nth term for each of the sides. So, from hereon in, it’s all a matter of putting those 3 formulas together.
Perimeter = (Shortest side) + (middle side) + (Longest Side)
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 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:
4n² + 6n + 2 = Perimeter.
4 x 42 + 6 x 4 + 2 = 9 + 40 + 41
64 + 24 + 2 = 90
90 = 90
My formula works for the 4th term.
4n² + 6n + 2 = Perimeter.
4 x 52 + 6 x 5 + 2 = 11 + 60 + 61
100 + 30 + 2 = 132
132 = 132
And it works for the 5th term.
4n² + 6n + 2 = Perimeter.
4 x 62 + 6 x 6 + 2 = 13 + 84 + 85
144 + 36 + 2 = 182
182 = 182
My formula works for all 3 terms, so…
Perimeter = 4n² + 6n + 2
Like the perimeter, I already know that the area of a triangle is found by:
Area = ½ b h
b = Base
h = Height
I should think that the ‘base’ side means the adjsacent side, and the ‘height’ side means the opposite side. However, I will continue to refer to them as ‘base’ and ‘height’ from now on. So:
Area = ½ (Shortest Side) X (Middle Side)
= ½ (2n +1) x (2n² 2n)
= (2n +1) (2n² + 2n)
2
I will now check this using the first 3 terms.
(2n + 1) (2n² + 2n) = ½ b h
2
(2 x 1 + 1) (2 x 12 + 2 x 1) = ½ x 3 x 4
2
3 x 4 = ½ x 12
2
12 = 6
2
6 = 6
My formula works for the first term.
(2n + 1) (2n² + 2n) = ½ b h
2
(2 x 2 + 1) (2 x 22 + 2 x 2) = ½ x 5 x 12
2
5 x 12 = ½ x 60
2
60 = 30
2
30 = 30
My formula also works for the 2nd term.
(2n + 1) (2n2 + 2n) = ½ b h
2
(2 x 3 + 1) (2 x 32 + 2 x 3) = ½ x 7 x 24
2
7 x 24 = ½ x 168
2
168 = 168
2
168 = 168
My formula works for all 3 terms. So…
Area = (2n +1) (2n² 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 I 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.
