Sunday, 10 June 2012

Challenge Question 1: Solutions

This is the worked solution to Challenge Question 1.

To begin thinking about the problem it is a must to visualise or draw out what is actually going on and begin labelling what we know.

As all that matters in this question is the ratio from side to side we can say that the largest square has a side length of 1, the next is 1/2, the next 1/4, etc. This means that the area of the first square is 1, then next is 1/4, the next 1/16, etc.

You can see that there is a ratio of 1/4 between the area of a square and the square before it. So to find the total area of all the squares you can use the formula for the sum of an infinite geometric series a/(1 - r) (explained here), multiply it by 4 (for each branch) and then add 1 (for the central square). So we get [4 × 0.25/(1 - 0.25)] + 1, which gives a total area for the squares as 7/3.

Now for the area of the shaded squares. The first shaded square has a side length of 1/4, so it has an area of 1/16, the next shaded square has a side length of 1/8 and an area of 1/64. You can see that the ratio from the area of one square to the next is 1/16. The total area of the shaded squares uses the sum of an infinite geometric series again and multiplies it by 4 (for each branch). So we get 4 × 0.125/(1 - 1/16), which equals  8/15.

So the area of all the overlapping squares is 8/15 and the total area of all the squares is 7/3, so the proportion of the shaded squares is 8/15 ÷ 7/3 = 0.22857142857... So the final answer is that the shaded squares occupy 22.86% of the total area of the squares.

1 comment:

1. These rules are general — they work at the property level. (Intuitively, I have a chemical analogy that “evenness” is a molecule some numbers have, and cannot be removed by multiplication.) help me with maths