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• Level: GCSE
• Subject: Maths
• Word count: 6029

# Beyond Pythagoras

Extracts from this document...

Introduction

Raj Purewal         Mathematics GCSE Coursework 2001-02-22        Page  of

“Beyond Pythagoras”

Beyond Pythagoras

This investigation is to study Pythagoras Theorem. I will try to find patterns and formulae to help predict Pythagorean Triples.

Pythagoras was a Greek Philosopher and Mathematician who is believed to have lived in the 6th century BC. He discovered many theorems but his most famous was:

a2+b2= c2

What is a Pythagorean Triple?

To answer this I first need to explain Pythagoras Theorem. Pythagoras States that in any right-angled triangle, a2+b2=c2. a is the shortest side, b the middle length side and c the hypotenuse (the longest side).

A Pythagorean Triple is any set of integers that agrees this condition. For example 3, 4, 5 is a Pythagorean Triple because:

32+42=52

Because 32= 3x3= 9

42= 4x4= 16

52= 5x5= 25

9+16= 25

1.

The numbers 5, 12, 13 satisfy the condition:

52+122=132

Because 52= 5x5= 25

122= 12x12= 144

132= 13x13= 169

25+144= 169

The numbers 7, 24, 25

72+242= 252

Because 72= 7x7= 49

242= 24x24= 576

252= 25x25= 625

44+576= 625

2.

(a)

I found perimeter by using the formula:

Perimeter= a+b+c

P= 5+12+13

P= 30

Perimeter= a+b+c

P= 7+24+25

P= 56

I found area using the formula:

Area= (axb)÷2

Area= 0.5x5x12

Area= 30

Area= (axb)÷2

Area= 0.5x24x25

Area= 84

(b)

I next put the results for perimeter and area into the table below.

 Length of Shortest Side (a) Length of Middle Side (b) Length of Longest Side (c) Perimeter (P) Area (A) 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84

3.

Before investigating patterns I tried to find the Pythagorean triple for a shortest side of 9 to increase my depth of results. I used trial and error to find it. From the general pattern so far I knew that side b for this triple had to be roughly around 40 to 50. I also knew that c was probably one larger than b. I tried multiple triples until I found one that seemed to agree the theorem. I came up with 9, 40, 41.

I proved that this was right:

92+402= 412

Because 92= 9x9= 81

402= 40x40= 1600

412

Middle

= 4n4+4n3+2n2+4n3+4n2+2n+2n2+2n+1=4n4+8n3+8n2+4n+1.

(4n2+4n+1)+(4n4+8n3+4n2)= 4n4+8n3+8n2+4n+1 CORRECT

This rather large calculation proves that my formulae can be applied to Pythagoras’ Theorem.

I decided to continue my investigation finding similar formulae only from a, for example, a formula to find b from just a. I first tried to find a formula to find b from a.

I had to use trial and error. The difference between b and a was 1, 7, 17, 31. The difference was not constant. This meant that the formula must have involved a. It was also probably quadratic. a2 gave values over twice larger than b:

a29, 25, 49, 81

b         4, 12, 24, 40

Using trail and error I came up with:

b= a2÷2-0.5

I checked and proved that this was correct:

Where a is 3

b= a2÷2-0.5

b= 32÷2-0.5

b= 9÷2-0.5

b= 4.5-0.5

b= 4

Where a is 5

b= a2÷2-0.5

b= 52÷2-0.5

b= 25÷2-0.5

b= 12.5-0.5

b= 12

After I found the formula for b from a, the formula for c from a was very simple to find. I already knew that side c was exactly 1 more than b. Therefore the formula to find c from a is exactly the same as to find b from a, only +1 at the end of it:

b= a2÷2-0.5+1

This simplifies to:

b= a2÷2+0.5

I checked and proved that this was correct:

Where a is 3

c= a2÷2+0.5

c= 32÷2+0.5

c= 9÷2+0.5

c= 4.5+0.5

c= 5

Where a is 5

c= a2÷2+0.5

c= 52÷2+0.5

c= 25÷2+0.5

c= 12.5+0.5

c= 13

Next I needed a formula to find P from a. The difference between P and a went 9, 25, 49, 81. The difference was not constant meaning that a must have been involved and it was probably quadratic. However a2 alone is too small:

a29, 25, 49, 81

P         12, 30, 56, 90

Conclusion

P= a+2n2+4n

P= (n+3)xa

A= (½a)(a+n2+2n+1)

A= (a3÷8)-(n+2)

a2= b+c

I decided to investigate why a2= b+c only worked for odd numbered shortest side triples and not even numbered. The alternative was identical but c and b had to be doubled before hand. I decided to compare what the differences between the two. The most obvious difference was that for odd numbered triples the difference between b and c was 1, whilst for even numbered triples the difference was 2. I noticed that this difference was the same number required to be multiplied to b and c before a2= b+c worked. This was when I realised that the formula was actually:

a2= (c-b)(b+c)

c-b for odd numbered shortest side triples was 1, therefore the c-b can be ignored, but for even numbered shortest side triples c-b= 2 so 2 is required in front of b+c. This formula works for all triples. In the triple 9, 12, 15 (triple of 3, 4, 5) the difference between b and c is 3. 3x(12+15) is 81, which is also 92. Then the next stage of this suddenly appeared to me; on multiplying  a2= (c-b)(b+c) you get a2=cb+c2-b2-cb, this simplifies to a2=c2-b2 or a2=b2+c2, Pythagoras Theorem! I suspect this is how Pythagoras himself discovered his theorem.

The Perfect Pythagorean Triples

The anomalies to my formulae which I found earlier, 20, 21, 29, 119, 120, 169, 696, 697, 985. These are so called “Perfect Triples”. A Perfect Triple is a triple in which side b is exactly one larger than side b. Perfect Triples are a very strange phenomenon. The first Perfect triple is 3, 4, 5; the next 20, 21, 29. If you multiply the hypotenuse of the last triple by the special number (2+ 1)2 (approx 5.828) you get near perfectly the hypotenuse of the next Perfect Triple. For example:

20, 21, 29- hypotenuse is 29x(2+ 1)2= 169.024

The next triple is 119, 120, 169- the hypotenuse is 169, 0.024 off from calculated.

This student written piece of work is one of many that can be found in our GCSE Pythagorean Triples section.

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