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  • Level: GCSE
  • Subject: Maths
  • Word count: 2025

In this piece of coursework, I have investigated the number of routes from one of the top corners on a polyhedron to the opposite base corner.

Extracts from this document...

Introduction

Year 10 Maths Coursework

Routes on Polyhedra

In this piece of coursework, I have investigated the number of routes from one of the top corners on a polyhedron to the opposite base corner. I followed these rules: a vertice may be used more than once per route but an edge may not, once the destination vertice has been reached the route must stop, and never move upwards.

Prisms

I chose to begin by working out the number of routes from a top corner to the opposite base corner in the commonest prism, the cuboid.

To do this, I have drawn the cuboid shape out many times, using a computer to save time, and marked on each one a different route that could be taken. I also gave each vertice a letter, which meant that the routes could be presented in a written form. This means that for any shapes I use after this, I will not need to draw them out more than once.  I have drawn the routes on the cuboids in a logical manner (first straight down a vertical, then gradually clockwise around the top, then gradually anti-clockwise around the top and clockwise and anti-clockwise at the base) so as to make clear which routes had been taken and whether all possible routes were drawn.

See page 2.

From this, I discovered that there were 16 possible routes from a top corner to an opposite bottom corner of the cuboid.

...read more.

Middle

image19.png

image20.pngimage23.pngimage22.png

image24.png

image25.png

image24.png

image20.png

image24.png

image19.png

There are n-1 verticals which, to get to the destination point, have 4 routes (illustrated in red), one vertical which, to get to the destination point has 2 routes (illustrated in blue) (i.e. straight down and round the bottom clockwise and straight down and round the bottom anti-clockwise), and there is one vertical which can be travelled down with 2 routes (illustrated in green) (i.e. round the top clockwise and then down, and round the top anti-clockwise and then down).

This, written algebraically, gives us the formula:

        4(n-1) + (2 x 1) + (2 x 1)

And, once cancelled down gives us the formula previously obtained.

        4(n-1) + (2 x 1) + (2 x 1)

      =  4n – 4 + 2 + 2

      =  4n

Pyramids

With the formula found for prisms, I began on the second step. A formula for the pyramids. I approached this in, fundamentally, the same way as with the prisms. Labelling the corners and taking a logical order in the routes from top point, to one of the bottom corners. I decided to begin with working out the number of routes on a tetrahedron (n = 3), as this is the simplest of pyramids so it seemed like a good starting point. Then, I explored the number of routes on a square based pyramid (n = 4), a pentagonal based pyramid (n=5) and a hexagonal based pyramid (n=6)

Tetrahedron    B        Cimage14.png

image26.png

                                        Routes:

                                                BC

                                                BADC

                                                BAC

                                                BDC

                                                BDAC

                                        Total Routes: 5

Square Based Pyramid        B      Dimage14.png

image28.png

                                        Routes:

                                                BD

                                                BCD

                                                BCAED

                                                BAED

                                                BACD

                                                BED

                                                BEACD

                                        Total Routes: 7

Pentagonal Based Pyramid        B      E

                                        Routes:image29.png

                                                BE

                                                BDE

...read more.

Conclusion

Shape

n

Number of

Routes

Triangular Based ‘Prismid’

3

35

Square Based ‘Prismid’

4

63

Pentagonal Based ‘Prismid’

5

99

Hexagonal Based ‘Prismid’

6

143

Below, are the calculations of the formula:

n                 3                4                5                6

Routes        35                63                99                143

Difference                28                36                44

Second Diff.                        8                8

n²                9                16                25                36  

4n²                36                64                100                144

Difference        

(between        -1                -1                -1                -1

n² and 4n²)

Formula:  4n²-1 = Number of possible routes on a ‘‘Prismid’’.

As with my other formulas, it can be proved by the following. I will use the Triangular based ‘Prismid’ as the exemplar shape.

To simplify the explanation, I have broken the route into two parts, top (   )   to a bottom corner of pyramid(    ) and the bottom corner of pyramid to bottom corner of prism (    ).

First, going from    to    , there are 2 ways for each ‘vertical’ (down and clockwise/anti-clockwise to destination) and 1 extra way (just down). Then, going from     to    , there are 2 ways (down and clockwise/anti-clockwise to destination).

Putting this together, it is possible to combine any part of first section with any  part of second section. Therefore,  2(2n + 1).

But the route can go via the other corners (except one) of the pyramid in exactly the same way and the other corner in 1 way.

This means that for this shape, the formula is actually:

Corner A           Corner B            Corner C        

        2(2n + 1)    +    2(2n + 1)   +   (2n + 1)

This works for this shape, but it is necessary to convert it into a manageable general rule.

There is  n – 1 ‘verticals’ with  2(2n + 1) and 1 with (2n + 1). Otherwise known as:

(n – 1) (2(2n + 1)) + 2n + 1

and if this formula is cancelled down, we reach the formula I calculated above.

   (n – 1) (2(2n + 1)) + 2n + 1

= (n – 1) (4n + 2) + 2n + 1

= (4n² - 2 – 4n + 2n) + 2n + 1

= (4n² - 2n – 2) + 2n + 1

= 4n² - 1

...read more.

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