Shapes

... e squares and equilateral triangles. When I say regular, I mean that the shape in question's order of rotational symmetry is equal to its number of sides, and all internal/external angles of the shape are the same. Nevertheless, I will draw out tables of the P, D and H of shapes made up of different numbers of Hexagons as I have done previously with triangles and squares.

5 Hexagons (H=5)

P=

D=

H=

14

4

5

16

3

5

18

2

5

20

1

5

22

0

5

6 Hexagons (H=6)

P=

D=

H=

18

4

6

20

3

6

22

2

6

24

1

6

26

0

6

10 Hexagons (H=10)

P=

D=

H=

24

9

10

26

8

10

28

7

10

30

6

10

32

5

10

34

4

10

36

3

10

38

2

10

40

1

10

42

0

10

As before, I will test out the first line of each table above with the previous formula. I will test out the formulas for triangles and for hexagons, on the grounds that I do not know which formula will need less modification (and therefore less time) to find a working formula linking P, D and H. Firstly I will test out one line from each table with the formula that I found for triangles, P=T+2-2D, D=(T+2-P)/2 and T=P+2D-2. I will substitute T for H in all 3 formulas.

So where P=14, D=4 and H=5…

P=5+2-8 Ð P=-1 DP is out by -15, or P+1, or 3H

D=(5+2-14)/2 Ð D=-3.5 DD is out by -7.5

H=14+8-2 Ð H=20 DH is out by 20, or +3H, or x4H

And where P=18, D=4 and H=6…

P=5+2-8 Ð P=-1 DP is out by -19, or P+1

D=(6+2-18)/2 Ð D=-5 DD is out by -9

H=18+8-2 Ð H=24 DH is out by +18, or +3H, or x4H

And where P=24, D= 9 and H=10…

P=10+2-18 Ð P=-6 DP is out by -30, or 3H

D=(10+2-24)/2Ð D=-6 DD is out by -15

H=24+18-2 Ð H=40 DH is out by +30, or +3H, or x4H

From these trials I can see that there is obviously a pattern with the H=P+2D-2 formula, in which the answer I get is always four times the answer I need. If I were to change the formula to H=(P+2D-2)/4, I have a pretty good idea that this would work. If it did, it could then be rearranged to give P= and D=. I will now test the formula H=(P+2D-2)/4, in exactly the same way as above, but omitting P= and D= to save time.

So where P=14, D=4 and H=5, H=(14+8-2)/4 Ð H=5 C

And where P=18, D=4 and H=6, H=(18+8-2)/4 Ð H=6 C

And where P=24, D=9 and H=10, H=(24+18-2)/4 Ð H=10 C

By simply inserting a /4 onto the end of the previous formula which did not work, I have created a formula which will give you the correct value of H, if you know P and D. H=(P+2D-2)/4 can be rearranged to give P=4H+2-2D and D=2H-P/2+1. Previously, when I rearranged a working formula I then tested out the two new formulas all over again, just to make sure. However, as this proved flawless both for triangles and for squares, I am sure that there is no need - I have checked my workings thoroughly, and I ...