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Smaller and Smaller

Stage: 2 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Caroline and Bronya from Tattingstone School sent wonderfully explained solutions to this problem. Caroline drew some diagrams to help her:

equilateral triangle side length=1 unit
Side AB = 1
Perimeter = 3 x 1 = 3
second image in question
Side ADB = 4 x 1/3 = 4/3
Perimeter = 3 x 4/3 = 4
third diagram from question
Side ADB = 4 x 4/9 = 16/9
Perimeter = 3 x 16/9 = 48/9 = 16/3 = 5 1/3

She explains:

The pattern is every time you add on a new smaller spike you multiply by 4 and divide by 3.

Here, Bronya describes how she approached the problem:

First of all I looked at the perimeter of the equilateral triangle. It was 3 units. Then I looked at the perimeter of the star. If one side is 11/3 units then the perimeter is 4 units. Then the perimeter of the third shape. I looked at a section like this:

one spike of third diagram

Each section = 1/3 + 1/9 unit = 3/9 + 1/9 = 4/9 unit
There are 12 sections so the total perimeter = 4/9 x 12/1 = 48/9
I looked at all the perimeters as ninths:
Perimeter 1 was 27/9
Perimeter 2 was 36/9
Perimeter 3 was 48/9
Each time the perimeter increases by one third.
I think this comes about because in each section a third of the section is added on.

The perimeter of the next shape would be 64/9 because:
48 divided by 3 is 16 therefore increase 48/9 by 16/9
Total would be: 48/9 + 16/9 = 64/9

Thank you to you both - these are very well reasoned solutions. I wonder whether you could generalise to any shape in this series?