4 12 24 40 60 84
1st Diff. 8 12 16 20 24
2nd Diff. 4 4 4 4
Half of the second differential is 2, so the formula for the length of this side has to be 2n2 + ?n
When n is 1, the number in the sequence is 4
2n2 + ?×1 = 4
2×12 + ?×1 = 4
2 + ?×1 = 4
?=2
This means that the formula for b is 2n2 + 2n
Longest side (c)
The longest side of the triangle (c) is simply just b + 1, which can be expanded to 2n2 + 2n + 1
Area (A)
The formula for the area (A) of any of the triangles is ½ ab, which can be written as 0.5 × 2n + 1 × 2n2 + 2n
Perimeter (P)
P = a + b + c
P = 2n + 1 + 2n2 + 2n + 2n2 + 2n + 1
P = 4n2 + 6n + 2
Formulas
a = 2n + 1
b = 2n2 + 2n
c = 2n2 + 2n + 1
A = 0.5 × 2n + 1 × 2n2 + 2n
P = 4n2 + 6n + 2
Now that I have found all the relationships and formulas for the family of Pythagorean Triples where the shortest side of the triangle is an odd number, all 3 sides are positive integers and c is always 1 more than b, I will start to work on the family where a is always a positive integer and c = b + 2.
The two starred rows (*) don’t work. The first triangle in this group would not be possible, as the middle side would have a length of 0cm. The second triangle would not be possible either, as the middle side would be smaller than the shortest side. I will ignore these triangles and find formulas for this set of data.
Shortest side (a)
The length of this side increases by 2 for each triangle. It is also double the triangle number (n). This means that the formula is 2n.
Middle side (b)
I can tell just by looking at the numbers in this column that the numbers are the squares of the triangle numbers minus 1. This means that the formula for b is
n2 - 1
Longest side (c)
The longest side of the triangle (c) is always 2 more than b because I am investigating triangles where c = b + 2. Therefore, the formula must be n2 + 1.
Area (A)
The formula for the area (A) of any of the triangles is ½ ab, which I have worked out can be written in terms of n as n3 – n.
Perimeter (P)
P = a + b + c
P = 2n + n2 - 1 + n2 + 1
P = 2n2 + 2n
Formulas
a = 2n
b = n2 - 1
c = n2 + 1.
A = n3 – n.
P = 2n2 + 2n
Now I am going to move on to investigate the family of Pythagorean Triples where c = b + 3.
I have noticed that shortest side of the first triangle is always the square of the difference between b and c. I will find the formulas for these triangles and then start on the next set to see if this theory still works.
Shortest side (a)
The length of this side increases by 6 each time. That means the formula has 6n in it. There is always 3 left, so the formula is 6n + 3.
Middle side (b)
This formula took me a long time to work out, but eventually I noticed that the numbers went up in factors of 12, which somehow helped me to work out the formula as being n(6n + 6) which can be expanded to 6n2 + 6n
Longest side (c)
This side is always 3 more than b in this family, so the formula is 6n2 + 6n +3
Area (A)
The formula for the area (A) of any of the triangles is ½ ab, which can be written in terms of n as 6n + 3 × 6n2 + 6n
Perimeter (P)
P = a + b + c
P = 6n + 3 + 6n2 + 6n + 6n2 + 6n +3
P = 12n2 + 18n + 6
Formulas
a = 6n + 3.
b = 6n2 + 6n
c = 6n2 + 6n +3
A = (6n + 3) × (6n2 + 6n)
P = 12n2 + 18n + 6
Now I can find the rules for the family of Triples where c = b + 4, and I can see if my theory works (The value of a in the first triangle is always the square of the difference between b and c)
My theory didn’t work as the first number in the a column is 12, which is not equal to 42. I can find the formulas for this family of numbers using the table:
Shortest side (a)
The difference between the shortest side of each triangle is 8, which means the formula starts off with 8n. After finding out 8n, there is always 4 left so the formula for a is 8n + 4.
Middle side (b)
The length of this side goes up in multiples of 16 each time.
Term no. 1 2 3 4 5 6
16 48 96 160 240 336
1st Diff. 32 48 64 80 96
2nd Diff. 16 16 16 16
Half of the second differential is 8, so the formula for the length of this side has to be 8n2 + ?n
When n is 1, the number in the sequence is 16
8n2 + ?×1 = 16
8×12 + ?×1 = 16
8 + ?×1 = 16
?=8
This means that the formula for b is 8n2 + 8n
Longest side (c)
This side is always 4 more than b in this family, so the formula is 8n2 + 8n + 4
Area (A)
Term no. 1 2 3 4 5
96 480 1344 2880 5280
1st Diff. 384 864 1536 2400
2nd Diff. 480 672 864
3rd Diff. 192 192
Because the numbers only become equal on the third line, I have to divide it by 6 and then cube it, which means the formula has to start with 32n3.
32n3 + ?n2 + ?
192 ÷ 4 = 48
When I did this I noticed that whatever was left was always 16 × the triangle number, so the formula for area is 32n3 + 48n2 + 16n.
Perimeter (P)
P = a + b + c
P = 8n + 4 + 8n2 + 8n + 8n2 + 8n + 4
P = 16n2 + 24n + 8
Formulas
a = 8n + 4.
b = 8n2 + 8n
c = 8n2 + 8n + 4
A = 32n3 + 48n2 + 16n.
P = 16n2 + 24n + 8
I will find the rules for the family where c = b + 5 so that I will have enough data to find a rule for c = b + d (difference between b and c) when a is an odd number.
Shortest side (a)
The difference between the shortest side of each triangle is 10, which means the formula starts off with 10n. After finding out 10n, there is always 5 left so the formula for a is 10n + 5.
Middle side (b)
There are two formulas for side b. One is an + 5n and the other one is 10n2 + 10n
Longest side (c)
This side is always 5 more than b in this family, so the formula is 10n2 + 10n + 5.
Area (A)
The formula for area is just ½ ab, so the formula is 0.5 × (10n + 5) × (an + 5n) = 0. 5 × 10an2 + 50n2 + 5an + 25n
Perimeter (P)
P = a + b + c
P = 10n + 5 + 10n2 + 10n + 10n2 + 10n + 5
P = 20n2 + 30n + 10
Formulas
a = 10n + 5.
b = an + 5n or 10n2 + 10n
c = 10n2 + 10n + 5.
A = 0. 5 × 10an2 + 50n2 + 5an + 25n
P = 26n + an + 5
Now I will find the rules for c = b + 6 so that I can find an overall rule for c = b + d (difference) when a is an even number.
Shortest side (a)
The difference between the shortest side of each triangle is 12, which means the formula starts off with 12n. After finding out 12n, there is always 6 left so the formula for a is 12n + 6.
Middle side (b)
I have worked out that the formula for side b is 12n2 + 12n.
Longest side (c)
This side is always 6 more than b in this family, so the formula is 12n2 + 12n + 6
Area (A)
The formula for area is just ½ ab, so the formula is 0.5 x 6n + 6 × n2 + 12n + 9
Perimeter (P)
P = a + b + c
P = 12n2 + 12n + 12n2 + 12n + 6 + 12n + 6
P = 24n2 + 36n + 12
Formulas
a = 12n + 6
b = 12n2 + 12n
c = 12n2 + 12n + 6
A = 0.5 x 6n + 6 × n2 + 12n + 9
P = 24n2 + 36n + 12
Now that I have got all the rules for c = b + 1, 2 and 3 I will try to find the rules when c = b + d where a is an odd integer.
Shortest side (a)
Just by looking at this column I have noticed that the numbers go up in a pattern: the number before n in the formula is always double d, and the number after n is always d, so the formula is 2dn + d
Middle side (b)
This side follows a pattern similar to a. The first part of the formula is always 2×d×n2 and the second part of the formula is just 2d, so the formula is 2dn2 + 2d
Longest side (c)
This formula is just b + d, which written out the long way is 2dn2 + 3d
Area (A)
The formula for area is still ½ ab, so the formula is 0.5 x 2dn + d × 2dn2 + 2dn
Perimeter (P)
I have noticed that the formula for the perimeter always starts with 4dn2, which leaves 6dn + 2d, so the formula is 4dn2 + 6dn + 2d
Formulas
a = 2dn + d
b = 2dn2 + 2d
c = 2dn2 + 3d
A = 0.5 x 2dn + d × 2dn2 + 2dn
P = 4dn2 + 6dn + 2d
Now I will find the formulas when c = b + d where a is an even integer.
Shortest side (a)
All the numbers except the row where d=2 fit with my previous rules. By looking at the pattern I have figured out that the rule for d=2 should be as follows:
but these rules do not fit with my original numbers for the c = b + 2 family, so something has gone wrong. Excluding this row, the formula is 2dn + d, s it is for the odd numbers.
If the formulas are the same for odd numbers as they are for even numbers, I am not going to do a separate table for even numbers, instead I will just put all the data into one table and find the formulas from there.
Formulas
a = 2dn + d
b = 2dn2 + 2d
c = 2dn2 + 3d
A = 0.5 x 2dn + d × 2dn2 + 2dn
P = 4dn2 + 6dn + 2d