# Beyond Pythagoras

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Introduction

Beyond Pythagoras

Pythagoras’ Theorem for a right-angled triangle states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, which can be written as a formula for the above triangle:

C2 = A2 + B2

A Pythagorean Triple is a set of these numbers, such as 3, 4, 5, where the square of the largest number is equal to the squares of the other two numbers. My task is to investigate different relationships between the numbers in families of Pythagorean Triples. To begin, I will investigate the family where the shortest side of the triangle is an odd number, all 3 sides are positive integers and side C is always 1 more than side B. I am going to start with the shortest side being 1cm and work out 10 more triangles after that.

1cm2 + 2cm2 ≠ 3cm2, so there is no relation between the sides when the shortest side of a right-angled triangle is 1cm.

As this triangle doesn’t follow the rules, I will ignore it and start again with 10 more triangles.

32 + 42 = 9 + 16 = 25

√25 = 5

Area = (4×3)/2 = 6cm2

Perimeter = 4 + 3 + 5 = 12cm

52 + 122 = 25 + 144 = 169

√169 = 13

Area = (12×5)/2 = 15cm2

Perimeter = 13 + 12 + 5 = 30cm

72 + 242 = 49 + 576 = 625

√625 = 25

Area = (7×24)/2 = 84cm2

Perimeter = 7 + 24 + 25 = 56cm

92 + 402 = 81 + 1600 = 1681

√1681 = 41

Area = (9×40)/2 = 180cm2

Perimeter = 9 + 40 + 41 = 90cm

112 + 602 = 121 + 3600 = 3721

√3721 = 61

Area = (11×60)/2 = 330cm2

Perimeter = 11 + 60 + 61 = 132cm

132 + 842 = 169 + 7056 = 7225

√7225 = 85

Area = (13×84)/2 = 546cm2

Perimeter = 13 + 84 + 85 = 182cm

152 + 1122 = 225 + 12544 = 12769

√12769 = 113

Area = (15×112)/2 = 840cm2

Perimeter = 15 + 112 + 113 = 240cm

172 + 1442 = 289 + 20736 = 21025

√21025 = 145

Area = (17×144)/2 = 1224cm2

Perimeter = 17 + 144 + 145 = 306cm

192 + 1802 = 361 + 32400 = 32761

√32761 = 181

Area = (19×180)/2 = 1710cm2

Perimeter = 19 + 180 + 181 = 380cm

212 + 2202 = 441 + 12544 = 48400

√48400 = 221

Area = (21×220)/2 = 2310cm2

Perimeter = 21 + 220 + 221 = 462cm

Now I am going to

Middle

39cm

270cm2

90cm

3

21cm

72cm

75cm

756cm2

168cm

4

27cm

120cm

123cm

1620cm2

270cm

5

33cm

180cm

183cm

2970cm2

396cm

6

39cm

222cm

225cm

4329cm2

486cm

7

45cm

336cm

339cm

7560cm2

720cm

8

51cm

432cm

435cm

11016cm2

918cm

9

57cm

540cm

543cm

15390cm2

1140cm

10

63cm

660cm

663cm

20790cm2

1386cm

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)

Triangle number (n) | Shortest side (a) | Middle side (b) | Longest side (c) | Area (A) | Perimeter (P) |

1 | 12cm | 16cm | 20cm | 96cm2 | 48cm |

2 | 20cm | 48cm | 52cm | 480cm2 | 120cm |

3 | 28cm | 96cm | 100cm | 1344cm2 | 224cm |

4 | 36cm | 160cm | 164cm | 2880cm2 | 360cm |

5 | 44cm | 240cm | 244cm | 5280cm2 | 528cm |

6 | 52cm | 336cm | 340cm | 8736cm2 | 728cm |

7 | 60cm | 448cm | 452cm | 13440cm2 | 960cm |

8 | 68cm | 576cm | 580cm | 19584cm2 | 1224cm |

9 | 76cm | 720cm | 724cm | 27360cm2 | 1520cm |

10 | 84cm | 880cm | 884cm | 36960cm2 | 1848cm |

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.

Triangle number (n) | Shortest side (a) | Middle side (b) | Longest side (c) | Area (A) | Perimeter (P) |

1 | 15cm | 20cm | 25cm | 150cm2 | 60cm |

2 | 25cm | 60cm | 65cm | 750cm2 | 150cm |

3 | 35cm | 120cm | 125cm | 2100cm2 | 280cm |

4 | 45cm | 200cm | 205cm | 4500cm2 | 450cm |

5 | 55cm | 300cm | 305cm | 8250cm2 | 660cm |

6 | 65cm | 420cm | 425cm | 13650cm2 | 910cm |

7 | 75cm | 560cm | 565cm | 21000cm2 | 1200cm |

8 | 85cm | 720cm | 725cm | 30600cm2 | 1530cm |

9 | 95cm | 900cm | 905cm | 42750cm2 | 1900cm |

10 | 105cm | 1100cm | 1105cm | 57750cm2 | 2310cm |

Conclusion

Perimeter (P)

2

2n

n2 - 1

n2 + 1

n3 – n

2n2 + 2n

4

8n + 4

8n2 + 8n

8n2 + 8n + 4

32n3 + 48n2 + 16n

16n2 + 24n + 8

6

12n + 6

12n2 + 12n

12n2 + 12n+ 6

0. 5 × 6n + 6 × n2 + 12n + 9

24n2 + 36n + 12

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:

Difference Between b and c (d) | Shortest side (a) | Middle side (b) | Longest side (c) | Area (A) | Perimeter (P) |

2 | 4n + 2 | 4n2 + 4n | 4n2 + 4n + 2 | 0. 5 × 4n + 2 × 4n2 + 4n | 8n2 + 12n + 4 |

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.

Difference Between b and c (d) | Shortest side (a) | Middle side (b) | Longest side (c) | Area (A) | Perimeter (P) |

1 | 2n + 1 | 2n2 + 2n | 2n2 + 2n + 1 | 0.5 × 2n + 1 × 2n2 + 2n | 4n2 + 6n + 2 |

2 | 4n + 2 | 4n2 + 4n | 4n2 + 4n + 2 | 0. 5 × 4n + 2 × 4n2 + 4n | 8n2 + 12n + 4 |

3 | 6n + 3 | 6n2 + 6n | 6n2 + 6n + 3 | 0.5 × 6n + 3 × 6n2 + 6n | 12n2 + 18n + 6 |

4 | 8n + 4 | 8n2 + 8n | 8n2 + 8n + 4 | 32n3 + 48n2 + 16n | 16n2 + 24n + 8 |

5 | 10n + 5 | 10n2 + 10n | 10n2 + 10n+ 5 | 0. 5 × 10n + 5 × 10n2 + 10n | 20n2 + 30n + 10 |

6 | 12n + 6 | 12n2 + 12n | 12n2 + 12n+ 6 | 0. 5 × 6n + 6 × n2 + 12n + 9 | 24n2 + 36n + 12 |

Formulas

a = 2dn + d

b = 2dn2 + 2d

c = 2dn2 + 3d

A = 0.5 x 2dn + d × 2dn2 + 2dn

P = 4dn2 + 6dn + 2d

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

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