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Jamie noticed some patterns:
The odd gnomons are a square with a little square taken out of it. All the sides are fibonacci numbers. The even gnomons are rectangles where the sides are consecutive fibonacci numbers, and the cut out corner is another rectangle made from consecutive fibonacci numbers. The bigger fibonacci number is the length, and the short one is the width.
Rosie develops this idea:
Using the patterns spotted by Yana in Building Gnomons, I found that the dimensions of a Gnomon are dependant on whether it is an odd or even Gnomon. In my diagram I used B instead of G to label a gnomon.
So here B(2n+1) is like G_{2n+1} on the problem page, where it describes the number of the Gnomon. F(n+1) means F_{n+1} giving the n+1 th fibonacci number, where the 2n+1 refers to the number of the Gnomon.
So we can see there is a pattern, that the odd Gnomons are built
around a square, if the Gnomon has number 2n+1, then the gnomon
is a sqaure of length and width F_{n+3}, with square of length
F_{n+1}. For example if the Gnomon is size 9=2n+1=2\times 4 +1.
So the edge lengths are F_{4+3}=F_7=13, and the small square has
edge length F_{4+1}=F_5=5.
So the area of the 9th Gnomon is 13\times 13 - 5\times 5 = 169 -
25 = 144 which is the 12th fibonacci number.
For the even Gnomons, if the Gnomon is number 2n, then the
length and width are F_{n+3} and F_{n+2}, and the small
rectangle is F_{n+1} by F_n. For example, if we take Gnomon
number 8=2n=2\times 4. So the area of the big rectangle is
F_{n+3}\times F_{n+2} = F_7 \times F_6 = 13 \times 8 = 104.
And the area of the little rectangle is
F_{n+1}\times F_{n} = F_5 \times F_4 =5 \times 3 = 15
So the area of the 8th Gnomon is 104-15=89 which is the 11th
fibonacci number.
The first group are the areas of the odd Gnomons, those which
are the square gnomons with square cut outs. The second group are
the areas of the even Gnomons, which are based on rectangles.
Making these groups are useful because then they are divided into
the groups in which they can be drawn.
All Gnomons will fit in one of these two groups, as all are odd or
even.
If you have a large supply of 3kg and 8kg weights, how many of each would you need for the average (mean) of the weights to be 6kg?
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.