Beyond Pythagors Essay, Research Paper

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 and so????????????? 32

+ 42 = 9 + 16 = 25 = 52The numbers 5, 12, 13 and 7, 24, 25 also work for this

theorem ??????????????????????? 52

+ 122 = 132 because??????????? 52

= 5 x 5 = 25 ??????????????????????? 122

= 12 x 12 = 144 ??????????????????????? 132

= 13 x 13 = 169 and so????????????? 52

+ 122 = 25 + 144 = 169 = 132??????????????????????? 72

+ 242 = 252 because ?????????? 72

= 7 x 7 = 49 ??????????????????????? 242

= 24 x 24 = 576 ??????????????????????? 252

= 25 x 25 = 625 and so ???????????? 72

+ 242 = 49 + 576 = 625 = 2523 , 4, 5 Perimeter = 3 + 4 + 5 = 12 Area = ½ x 3 x 4 = 65, 12, 13 Perimeter = 5 + 12 + 13 = 30 Area = ½ x 5 x 12 = 307, 24, 25 Perimeter = 7 + 24 + 25 = 56 Area = ½ x 7 x 24 = 84From the first three terms I have noticed the following: – ?a?

increases by +2 each term ?a?

is equal to the term number times 2 then add 1 the

last digit of ?b? is in a pattern 4, 2, 4 the

last digit of ?c? is in a pattern 5, 3, 5 the

square root of (?b? + ?c?) = ?a? ?c?

is always +1 to ?b? ?b?

increases by +4 each term (?a?

x ?n?) + n = ?b? From these observations I have worked out the next two

terms. I will now put the first five terms in a table format. Term Number ?n? Shortest Side ?a? Middle Side ?b? Longest Side ?c? Perimeter Area 1 3 4 5 12 6 2 5 12 13 30 30 3 7 24 25 56 84 4 9 40 41 90 180 5 11 60 61 132 330 I have worked out formulas for How

to get ?a? from ?n? How

to get ?b? from ?n? How

to get ?c? from ?n? How

to get the perimeter from ?n? How

to get the area from ?n? My formulas are 2n +

1 2n2 + 2n 2n2

+ 2n

+ 1 4n2

+ 6n + 2 2n3

+ 3n2 + n To get these formulas I did the following Take

side ?a? 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 graph From

looking at my table of results, I noticed that ?an + n? = b?. So I took my formula for ?a? (2n

+ 1) multiplied it by ?n? to get ?2n2 + n?. I then added my

other ?n? to get ?2n2 + 2n?. This is a parabola as you can see

from the equation and also the graph Side

?c? is just the formula for side ?b? +1 The

perimeter = a + b + c. Therefore I took my formula for ?a? (2n + 1), my

formula for ?b? (2n2 + 2n) and my formula for ?c? (2n2

+ 2n + 1). I then did the following: – 2n + 1 + 2n2 + 2n + 2n2

+ 2n + 1 = perimeter Rearranges to equal ????? 4n2 + 6n + 2 = perimeter The

area = (a 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 Multiply this out to get 4n3 + 6n2

+ 2n = area ?????????? 2 Then divide 4n3 + 6n2

+ 2n by 2 to get 2n3 + 3n2 + nTo 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 + 1This proves that my ?a?, ?b? and ?c? formulas are correctExtensionPolynomialsHere are the formulas for the perimeter and the areaPerimeter = 4n2 + 6n + 2 Area = 2n3 + 3n2 + n ??????????? ????????? From my table of results I know that the perimeter = area at

term number 2. Therefore (n-2) is my factorI would like to find out at what other places (if any) the

perimeter is equal to the areaTo find this out I have been to the library and looked at

some A-level textbooks and learnt ?Polynomials?4n2 + 6n + 2 = 2n3 + 3n2 +

n 6n + 2 = 2n3 – n2 + n 2 = 2n3 ? n2 – 5n 0 = 2n3 ? n2 – 5n – 2???????? ???2n2 + 3n + 1???????? ????????????????????????????????????????????????????????b n ? 2 ) 2n3 ? n2 ? 5n ? 2 ??????????? 2n3

? 4n2????????????? b ??????????? ???????? 3n2 ? 5n ? 2 ??????????? ???????? 3n2 ? 6n????? ???????????????????????????????????????????????????????????????????????b ??????????????????????? ??????? n ? 2 ??????????????????????? ??????? n ? 2 ?????????????????????????????????????????????????????????????????????b ??????????????????????????????????? ? 0This tells us that the only term where the perimeter = area

is term number 2Therefore when f(x) = 2n3 ? n2 ? 5n ?

2 is divided by n ? 2 there is no remainder and a quotient 2n2 + 3n

+ 1. The result can be written asf(x) = 2n3 ? n2

? 5n ? 2= (n ? 2)(2n2 + 3n + 1)If 2 was substituted for the x in this identity so that n ?

2 = 0, the quotient is eliminated giving f (2) = + 2Now I will complete the square on ?4n2 + 6n + 2?

to see what the solution to this is.4n2 + 6n + 2 4(n +3)2 ? 9 + 2 4(n +3)2 ? 7 4(n +3)2 = 7 (n +3)2 = 1.75 n + 3 = 1.322875656 n + 3 = -1.322875656n = -1.677124344 n = -4.322875656Arithmatic ProgressionI would like to know whether or not the Pythagorean triple

3,4,5 is the basis of all triples just some of them.To find this out I have been to the library and looked at

some A-level textbooks and learnt ?Arithmatic Progression?3, 4, 5 is a Pythagorean tripleThe pattern is plus oneIf a = 3 and d = difference (which is +1) then3 = a 4 = a + d 5 = a +2da, a +d, a + 2dTherefore if you incorporate this into Pythagoras theorema2 + (a + d)2 = (a + 2d)2 a2 + (a + d)(a + d) = (a + 2d)2 a2 + a2 + ad + ad + d2 = (a

+ 2d)2 2a2 + 2ad + d2 = (a + 2d)2 2a2 + 2ad + d2 = (a + 2d)(a + 2d) 2a2 + 2ad + d2 = a2 + 2ad +

2ad + 4d2 2a2 + 2ad + d2 = 4d2 + a2

+ 4adIf you equate these equations to 0 you geta2 ? 3d2 ? 2ad = 0? Change a to xx2 ? 3d2 ? 2dx = 0Factorise this equation to get(x + d)(x ? 3d)Therefore x = -d x = 3d? x = -d is impossible as you cannot have a negative dimensiona, a+d, a + 2d Is the same as 3d, 4d, 5dThis tells us that the only Pythagorean triples are 3, 4, 5

or multiples of 3, 4, 5 e.g. 6, 8, 10 or 12, 16, 20 etc.