I saw many patterns in this data, but to get accurate formulae, which I could test properly, I needed to extend the table bit. I did this by looking again at the patterns I saw and extending those ideas, this can be shown more clearly here:
So we can see that by seeing these patterns we can now extend the graph and not only fill in the blanks but extend further with the same patterns. But there is one thing that I am sure about the pattern for the 3 sides and perimeter but the pattern for the are does not look right and maybe wrong. We can check this because when we find the sides we can work out the area by ½ x s x m.
So now I will extend the graph:
So you can see that all my patterns worked, except the one I suspected, which was the area one. We can prove that they were correct, by checking if all the Pythagorean triples satisfy the condition. By looking at the table and checking with your calculator that all the triples do satisfy the condition and therefore are correct. Now about the area, the pattern was not right because I had to extend it further to make it right:
So we can see that we have to extend to 3 levels to find a pattern and that will now help us to find a formulae.
So we have now seen what is a Pythagorean triple and why it is; because they satisfy the condition. Now my next task is to use this data to help me look for formulae.
Task 3
I have to now find generalizations about the area, perimeters and sides of the Pythagorean triples beyond those in tasks 1 + 2. I have to look at the lengths of the 3 sides and area and perimeter of the triangles. To do this I have to look at the data I got in the last two tasks and generalize it but look beyond it and at things not obvious in the data. Ultimately from all this I have to work out formulae for the relationship of sides to perimeters and area.
To do this I have to first work out 3 crucial things, which I mentioned in the introduction from which I can work out many other things. These are:
If I have the short side can I work out the middle and long side?
And
If I have the middle side can I work out the short and long side?
And
If I have the long side can I work out the short and middle side?
From these 3 things I have to try and work out the relationships between the 3 sides and the area and perimeter of the Pythagorean triangle.
So to work these out I looked at my graph and the relationship between the shortest side, the middle side and the longest side.
I saw that obviously to get the l from m you have to add 1 and to get m from l you have to minus 1. Then I kept on studying the table and I saw that if you add m to l and square root the answer you get the shortest side. So I thought if l is m+1 then the shortest side from the middle side would be m + (m+1) square rooted. So this meant I could do one of my statements above:
If I have the middle side can I work out the short and long side?
Yes. Long side= m+1 and Short side=m + (m+1) square rooted.
Now that I had one of my statements proved I thought the statement about l to m and s would be the opposite of the previous statement, meaning wherever on the previous one there was (m+1) there should be l and wherever there was l there should be (l-1), I tried this and it worked:
If I have the long side can I work out the short and middle side?
Yes. Middle side= l-1 and Short side= ((l-1) + l) square rooted.
The third statement was a little harder to work out but as you will see helped me a lot more than the others later on. I was looking at the middle and long sides and what there total was and if you divided this total by 2 you are always within 0.5 of the answer. So I thought I had to work a way out to get the total of m + l by doing something to s. I tried a few ways but when I squared it I got the total of m + l.
Then I had to sort out the thing that if the total was divided by 2 it would be within 0.5 of the answer. I noticed that not only was it always within 0.5 of the answer but it was always 0.5 extra than the middle side and 0.5 less than the longest side. This made it easy because I just added next to my formula – 0.5 on the middle side formula and + 0.5 on the longest side formula. So now I could work out the last statement.
If I have the short side can I work out the middle and long side?
Yes. Middle side= s2 / 2 – 0.5 and Longest Side= s2 / 2 + 0.5.
From these 3 statements that I made It is now clear to find out some formulae. There were a lot of formulae that you could have made at this point and they all needed to be made to get the general formulae. However I thought putting all these formulae down was irrelevant in this work but you certainly needed to have some of them
for example like the ones I will mention and the ones I already have. However to get the general formulae and to give you a better understanding of all the work you certainly needed all the formulae worked out and in your rough notes which is what I have done.
So the 6 general formulae that I thought would be relevant and show my understanding of the relationship between the sides and area and sides and perimeter were:
A formula to work out perimeter from shortest side.
A formula to work out perimeter from middle side.
A formula to work out perimeter from longest side.
A formula to work out area from shortest side.
A formula to work out area from middle side.
A formula to work out area from longest side.
With the formulae I had and the fact that area = ½ base x height.
I could easily work out the area formulae, this is because if I knew the base and I knew the height I would be able to calculate area and I knew that from any side I had I could work out the other 2, so if I was given s I could get m and do ½ m x s to get area. Same with middle side, if I had middle side I could get s and do ½ s x m. And if I had long side I could get middle then short and do ½ m x l. So I put my explanation into a formula and I got this:
Shortest side => Area
(((s2 /2) – 0.5 ) x s) x ½
This formula works because ((s2 /2) – 0.5 ) is m and m x s x ½ is the area.
Middle side => Area
((m + (m+1)) x m) x ½
This formula works because ((m + (m+1)) is s and s x m x ½ is the area.
Longest side => Area
((l – 1) + l) x (l-1) x ½
This formula works because ((l – 1) + l) is s and (l-1) is m and s x m x ½ is the area.
Now that I have stated and proved all the area general formulae, I have try to do the same with perimeter formulae. For the first formula it was very easy because if you look back I found out and proved that if you squared s then you get the total of m + l. I used this in my area formula as well and now I can use it my perimeter formula, because by studying the relationship between s and perimeter, I found out that if you add s to total of m + l you get the perimeter. So with that in mind I derived the formula s2 + s. Like I said there was many formulae for this work and by using simple a + b + c I could also work out perimeter, this is because as I proved before I could work out the other 2 sides from having one so I just did s + m + l and worked out m and l by my previous formulae. Now I had to work out the formula for m to perimeter and l to perimeter. This was very easy I did a + b + c, s + m + l and I got the 2 sides needed by my previous formulae. I worked out a more simple formula though. I derived this formulae from my previous formulae. So these are all my formulae:
Shortest side => Perimeter
s2 + s and s + (s2 / 2 – 0.5) + (s2 / 2 + 0.5)
This formula works because s2 is total of m and l, and the total of m and l + s is the perimeter. The other formulae works as a + b + c is perimeter and s is a, (s2 / 2 – 0.5) is m and b and (s2 / 2 + 0.5) is l and c.
Middle side => Perimeter
((m x 2) + 1) + ((m x 2) + 1) and m + (m +1) + (m + (m+1))
This formula works because ((m x 2) + 1) is total of m and l, and the total of m and l + s is the perimeter, and s is ((m x 2) + 1). The other formulae works as a + b + c is perimeter and m is a, (m + 1) is b and l and (m + (m+1)) is s and c.
Longest side => Perimeter
((l x 2) - 1) + ((l x 2) - 1) and m + (l - 1) + ((l –1) + l)
This formula works because ((l x 2) - 1) is total of m and l, and the total of m and l + s is the perimeter, and s is ((l x 2) - 1). The other formulae works as a + b + c is perimeter and l is a, (l - 1) is b and m and ((l – 1) + l) is s and c.
So now I have stated all my formulae and I have proved and tested them with the help of my table and calculator. I am going to try to work out formulae ewhich have aroused from these formulae.
Another set of formulae I have been working on is the formula for finding the area from perimeter and vice-versa, this is because I thought that these formulae were also a very important formula along with the other 6 formulae as you may only be given the Area or Perimeter. This led me onto 6 other formulae these were like my previous 6 but the opposite. They were the relationship between perimeter and area to sides, i.e. can you get s from perimeter or area.
So the 8 formulae I wanted were these:
Can I Get Area From Perimeter
Can I Get Perimeter From Area
A formula to work out shortest side from perimeter.
A formula to work out middle side from perimeter.
A formula to work out longest side from perimeter.
A formula to work out shortest side from area.
A formula to work out middle side from area.
A formula to work out longest side from area.
To get these formulae I tried everything from working out relationships between the area and perimeters of the Pythagorean triples to looking back at my previous formulae. I tried to get patterns between the differences in the areas and perimeters. I tried to see if any sort of square root or squaring method could be used to get the perimeter from area and vice-versa. Also the other formulae I just could not get because if had not got the two first ones then I could not get the others as I said previously. However after trying a new idea of nth terms I got a formula for the perimeter to area and I got this by looking at the nth term and seeing what relationship it had with the area, then I tried to see what I could do to the nth term with the perimeter to get area and I got this:
Perimeter => Area
Perimeter x n = Area
2
Where n is the nth term (i.e. No. Of triangle)
To get n: always round down
( Perimeter ) - 0.5
( 2 )
This formula works because I noticed in my study that a multiple of the perimeter divided by 2 always gives you the area, then I thought the multiple could be perimeter times by n, I tested, and it worked.
After this with the nth term and other thing I got from this formula I found out another thing from this formulas sub formula i.e. the formula to work out n was that if you square rooted perimeter it would always give you within 0.5 of s. And from this I derived this formula and it is very easy to see why it works because perimeter square rooted is always near s:
Perimeter => Short Side
always round down
Perimeter
Once I had this I thought now if I do perimeter minus the formula to get s from perimeter I would get the total of m + l. And as I said before from this I could work out m + l. So I did what I thought and got this:
Perimeter => Middle Side
always round down
(Perimeter - Perimeter ) - 0.5
( 2 )
Perimeter => Longest Side
always round down
(Perimeter - Perimeter ) + 0.5
( 2 )
I got these two formulae by the things I have explained, but what I want to emphasize is that again my 3 things I did at the beginning really helped me out again.
Now that I had worked a formulae for perimeter to area and sides. I have to do the same for area. I tried to start of the same way by getting the formula for area to perimeter, but I just could not get it even after trying all my previous study skills of patterns and using previous formulae, I could not get anything. The main reason was that none of the ideas I got worked for all triples. These were some of the ideas but after testing they did not get anywhere:
((n-1) x a) – (a x 2)
and
(n x a) – ((n + 2) x a)
These got me nowhere so I predicted that there must be some thing to do with getting the area to sides formulae first and then getting
The area to perimeter formulae. So I started looking at relationships between area and sides, I got some more ideas but they were also not consistent with all the triples and I had to try another way, after a lot of trying and hit and miss situation I still did not get anywhere.
After this I did not think I would get these formulae and I tried to get 5 more that arose to my mind while doing this previous study.
These were n to area and perimeter and sides. I thought this would be easy if I could get s from n, then using my first three formulae of the study I could get these 5 formulae. So I tried to get relationship
between n and s. I tried multiples of n to get s and square numbers of n and square roots of n to get s. I saw that if you mulptiplied n by 2 and added 1 you would get s, so with that in mind plus the fact of my previous formulae I got these 5 formulae:
n => Shortest Side
(n x 2) + 1
n => Middle Side
((n x 2) + 1)2
( 2 ) - 0.5
n => Longest Side
((n x 2) + 1)2
( 2 ) + 0.5
n => Area
((n x 2) + 1)2 x ((n x 2) + 1)2 ) x ½
( 2 ) - 0.5)
This works because this is s and this is m and s x m x ½ = area.
n => Perimeter
((n x 2) + 1)2 + ((n x 2) + 1)2 ) + ((n x 2) + 1)2 )
( 2 ) - 0.5 ) ( 2 ) + 0.5)
This works because this is s and this is m and this is l and s + m + l = perimeter.
I think that all my formulae have really explained themselves now and if there is any doubt on testing these formulae and proving them then looking back at my first 3 formulae and my graph would tell you a lot.
After this I could not think of any more formulae and I thought that if there were any they would be irrelevant. So now I am going to explain how well I think I have done in this work and how I have gone about getting my formulae. I am also going to summarize and conclude all my work.
Conclusion
The things that I have done in this coursework and the way I have done them has been with a organized and logical way. I think that it was very clever of me to get those 3 formulae at the beginning ( m to s,l; s to m,l; l to m,s) because these formulae were general formulae which I could use not only for one set of formulae but for any type. I also thing that I have shown this use well and that they really did help me a lot in many of my formulae. I also think my approach to my work was good and that I worked in a logical way to get each type of formulae separately. I also think that the formulae that I did not get, I still did well in them and the way I went to try and find them was also good.
About the actual investigation, I think we have seen that from anything we can get anything, except some exceptions. Also we have seen that if you look back at previous things and try to use that knowledge in other things you get better results.
Overall I think my work has been a success and that I did not do much wrong, but however there is one thing I could expand on and that is that I could show my method of testing and prediction more clearly on paper, than just write vague things about what I did do in my mind. But remember there is one thing that we must never forget ,and that is the whole point of this work, that all my rules and formulae only work for PYTHAGOREAN ODD NUMBERED TRIPLES.
Extension with Even Numbered Pythagorean Triples
Introduction
After doing my coursework and finding that all my formulae and patterns only worked for odd numbered Pythagorean triples, I thought would extend my work and also do some work on even numbered Pythagorean triples. I went about doing this work the same way as my other work and I first drew out a table similar to the one given to me in my previous work and tried to look for patterns in it like before. This is the table I used:
Again I was getting similar patterns between sides of one triple to another and other patterns were similar except the one for area which was a little doubtful again and I had to go to more levels of patterns to get a exact pattern.
But when I had this pattern I could use it to build a bigger graph an test my formulae on. I also thought that if the patterns were similar then the formulae that I had before would be similar and if that was the case my whole investigation would work for both evens and odds.
This is the table I made to test my predictions.
My predictions were right and with checks with my calculator and also this table, I have found out that all my formulae work for both ODD AND EVEN PYTHGOREAN TRIPLES. The explanation for this could be that the formulae I got was not number dependant (i.e. it did not matter to the formula if it was tried by even or odd numbers).
By Sheryar Majid
