The first image shows a pink square, in which there is a blue square rotated by 45 degrees, within which there is another pink square, rotated back to its original position, and this pattern is repeated for a few more squares. The pattern would continue in a similar fashion forever (although some may think that it will stop once we run out of space, however this is not going to happen because there is always going to be a smaller and smaller area available for another square).
To find the proportion of the area of the large square that is taken up by the small square in the middle, let’s start with the large square. Assume that its side length is 1 unit (using 1 unit as the side length means that the area of the small square will also be the proportion of the large square’s area that it takes up). This means that the side of the rotated blue square creates an isosceles right angled triangle with 2 sides being 0.5 units (since the blue square’s side touches the outer square at its side’s midpoint).
So, the length of the blue square’s side is:
√(0.5^2 + 0.5^2) = √(2)/2.
Then, to find the side length of the square inside that blue square, we can divide √(2)/2 by 2 = √(2)/4 and use this length to find the side length by using pythagoras’ theorem again.
However, just to be more efficient, I will start again from the largest square (side length 1) and use the formula √(2*(x/2)^2) where x is the side length of the square. I use x/2 because I will use half the length of the side of the square to find the length of the side of the inner square. I multiply that by 2 because it’s an isosceles right angled triangle which means that the two sides involved in the calculation are equal.
Using that formula repeatedly will help us get to the side length of the inner square.
In the table shown above, square 1 is the largest square and 7 is the innermost square (which is why we don’t need to apply the formula in square 7, doing so would give us the length of square 8).
Since the length of the side of the smallest square is â…› units, the area of the square is (â…›)^2 = 1/64.
Therefore, the smallest square takes up 1/64 of the area inside the largest square.
Now, if we carried on the pattern forever, the areas of the blue squares would create a geometric sequences. But before we find the sum of the series, we have to find the series first.
Firstly, we can use the table to find out that all the blue squares (squares 2, 4, 6, …) have the areas (√2)/2 , (√2)/4, (√2)/8, … . This is a geometric sequence in the form (√2)/2^n starting from n=1. We could add up all these values up to infinity however that will give us the incorrect answer, because it will include the pink squares inside those blue squares as well, we need to exclude that area. What we actually need is the area of the triangles on the corners of the squares, for this we can subtract the areas of the pink squares. However, one could say that subtracting the area of the whole pink square will mean that we are subtracting the area of the blue squares inside too. But that is not a problem as we will be adding the areas of the blue squares back later.
So the pattern looks like this:
(((√2)/2)^2-(½)^2) + (((√2)/4)^2-(¼)^2) + (((√2)/8)^2 - (⅛)^2) + …
This can be simplified to:
((2/4)-(¼)) + ((2/16)-(1/16)) + ((2/64)-(1/64)) + …
Which can be finally simplified to:
¼ + 1/16 + 1/64 + …
The sequence seems to be in the pattern of (¼)^n where n≥1
Now, we can finally add all these areas up to infinity and that will give us the area of the blue squares, and the fraction of the large square that is blue.
∑(¼)^n , Starting from n=1 to infinity.
Formula of a geometric sequence going up to infinity is a(1/1-r) where a is the first term of the series and r is the common ratio.
Therefore, the series sums up to (¼)*(1/(1-(¼))) = (¼)*(1/(¾)) = (¼)*(4/3) = ⅓ .
So, a third of the large square is blue.
Moving on to the next section. I will first work on the first diagram.
In this diagram, the largest diagonal goes through the midpoint of the whole square. The other diagonals are perpendicular to this one. The diagonal on the farthest right intersects with the first diagonal at its midpoint.
Let’s assume that the side length of the major square is 1, therefore the area of the first blue triangle is ½ * 1^2 = ½
Since the side length is 1, the first diagonal’s length is √(1+1) = √2, the perpendicular bisector (the diagonal on the farthest right) cuts this diagonal into 2 equal halves, each of length (√2)/2. Since the bisector and the diagonal meet at the midpoint of the square, we know that the bisector is of the same length. This information will help us in finding the side length of the second largest triangle. We can use the Pythagorean theorem to find this length. This is made easier by the fact that the triangle is right angled AND isosceles. So:
√(2x^2) = (√2)/2
So, x = ½ (We could also find out this length by noticing that the line goes down from the midpoint of the side to the intersection point of the first 2 diagonals, but calculating it helps us in knowing what to do next since we do the same thing again). This means that the 2nd triangle’s area is ½ * (½)^2 = ⅛.
Now, let’s forget the rest of the square and let’s concentrate on the smaller square with side length 0.5 (thus, excluding the 2nd blue triangle and the largest pink triangle and creating a similar shape just with smaller sides, essentially zooming into it). We notice that this pattern is repeated, it’s just that the diagonal of the entire square is now (√2)/2 rather than just √2. This means the new diagonal that is farthest to the right bisects this diagonal of length (√2)/2, halving it into (√2)/4. Repeating the same process again, we notice that the 2 sides of the pink triangle are of length (√2)/4 which means that the smaller blue triangle has a hypotenuse of (√2)/4. Using this information we can find that triangle’s side length.
√(2x^2) = (√2)/4
x= ¼
So the triangle’s area is ½ * (¼)^2 = 1/32
So, just looking at the 3 largest triangles, we can see a pattern in terms of areas, the largest triangle’s area is ½, the second largest is ⅛, the third largest is 1/32. If we factor the ½ out and write the remaining fraction in powers of ½, the areas are (½)*(1/(2)^0)^2 , (½)*(1/(2)^1)^2, (½)*(1/(2)^2)^2),... . The nth term of the sequence is (½)*(1/(2)^n)^2 or (½)*(½)^2n.
So the area of 5 blue triangles would be:
∑(½)(½)^2n, from 0 to 4 (I used n=0 because that would give the value (½)*1, which is the area of the first triangle)
= ½ ∑(¼)^n, from 0 to 4
= ½ * 1(1-(¼)^4)/(1-(¼))) (Formula for geometric sequence up to n terms is a(1-r^n)/(1-r) where a is the first term of the sequence, n is the last n value, and r is the common ratio)
= 341/512 = 0.66602
Carrying this procedure on forever would mean that we use the same series but instead of 4 being the last n value, we go up to infinity.
Therefore the total blue area now is as follows:
∑(½)(¼)^n, from n=0 to infinity
Using the formula used for the first part of the question:
Sum = (½)* 1(1/(1-¼)) = (½)*(4/3) = ⅔.
Therefore, if the pattern was infinitely repeated, the â…” of the area of the square would be blue.
For the second diagram, the area covered by blue in this shape must be a quarter of the area of the square in the first part of the question since the structure is very similar, it’s just the fact that only 1 of the 4 triangles formed by the outer square and the inner, tilted one were blue. Therefore the part of the square shaded blue is (⅓)*(¼) = 1/12
For the third diagram, the idea I will be implementing will be pretty similar to that of the first diagram of this section. Let’s again assume that the side of the large square is 1 unit, therefore the first diagonal that splits the square into a blue triangle and the rest of the shape is √2 units long (√(1^2 + 1^2)). The lines that perpendicularly bisect 2 of the sides of the square intersect at the mid point of the diagonal. Since they meet at the midpoint and come are perpendicular to the sides of the square, 2 right angled isosceles triangles are formed with hypotenuse lengths being (√2)/2 units long. Using the Pythagorean theorem:
√(2x^2) = (√2)/2
Therefore, x = 0.5
As there are 2 of these triangles, there are 4 sides that are 0.5 units long, 2 of these create another right angled triangle, which is again isosceles.
Now, like the first square of this part of the question, let’s forget about the initial square and zoom into the shape so that we get the original square again, just with side lengths 0.5 units. By doing this, we are back to the initial part of this question and can repeat what we have done before.
Since the side length is 0.5 units this time, the diagonal is (√2)/2 units long. And again, like before, there are 2 lines which are perpendicular to the sides of the square and are meeting at the mid point of the diagonal, halving the diagonal length into (√2)/4
To find the length of the straight lines that intersect at the mid point of the diagonal, we use the Pythagorean theorem again:
√(2x^2) = (√2)/4
Therefore, x = 0.25
We keep repeating this process again and again until we get the side lengths of our 5 triangles. Doing so, gives us the side lengths 1, ½, ¼, ⅛ and 1/16
So the area of all the blue triangles combined is:
(½)(1^2) + (½)((½)^2) + (½)((¼)^2) + (½)((⅛)^2) + ((½)(1/16)^2) = 341/512
If we carry this on infinitely, we get another infinite geometric series:
∑(½)((1/2)^2n, from n=0 to infinity
=(½)∑(¼)^n = (½)(4/3) = ⅔
Therefore, â…” of the square is shaded blue.