Problem 85 - Algebra and Trigonometry

Question

Geometry A yellow square of side 1 is divided into nine smaller squares, and the middle square is colored blue as shown in the figure. Each of the smaller yellow squares is in turn divided into nine squares, and each middle square is colored blue. If this process is continued indefinitely, what is the total area that is colored blue?

Step-by-Step Solution

Answer

The total area that is colored blue is 1.

Step 1: Understand the Initial Setup

The initial yellow square has a side length of 1 unit and is divided into 9 equal smaller squares. The side length of each smaller square will therefore be \(\frac{1}{3}\) units. The middle square of the initial division is colored blue.

Step 2: Calculate the Area of the First Blue Square

The side length of the initial blue square is \(\frac{1}{3}\) as it is one of the smaller squares formed from the division. Therefore, the area of the first blue square is \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\).

Step 3: Analyze the Sequence of Blue Areas

Each smaller yellow square (excluding the middle blue) is then divided again in the same way, giving rise to new blue squares at each subsequent stage. Each blue square is the middle square of a set of 9 divisons in each iteration.

Step 4: Determine the Area of Subsequent Blue Squares

In the second iteration, each of the remaining 8 yellow squares is divided into 9 smaller squares, and their middle squares are colored blue. Each subsequent blue square has a side length of \(\frac{1}{9}\), giving it an area of \(\left(\frac{1}{9}\right)^2 = \frac{1}{81}\). This iteration produces 8 new blue squares.

Step 5: Generalize the Pattern of Blue Areas

Observe that in each progression step, the total number of new blue squares is multiplied by 8. Thus, in each iteration \(n\), the side length of each of the blue squares becomes \(\left(\frac{1}{3}\right)^n\), and their total number is \(8^{n-1}\). The total blue area becomes \(8^{n-1} \times \left(\frac{1}{3^2}\right)^n\).

Step 6: Summing the Series

We create an infinite series to represent the total blue area:\[ \text{Total blue area} = \frac{1}{9} \sum_{n=0}^{\infty} \left(\frac{8}{9}\right)^n \]. This is a geometric series with a common ratio of \(\frac{8}{9}\) < 1, which converges to \(\frac{1}{1-\frac{8}{9}} = 9\).

Step 7: Solving for the Total Blue Area

The sum of the infinite series \(\sum_{n=0}^{\infty} \left(\frac{8}{9}\right)^n = \frac{1}{1-\frac{8}{9}} = 9\). Therefore, the total area colored blue is \(\frac{1}{9} \times 9 = 1\).

Key Concepts

Infinite Series

An infinite series is a sum of an unlimited number of terms, which goes on indefinitely. In the context of geometric series, each term is derived by multiplying the previous term by a fixed constant known as the common ratio.
In this problem, we need to find the total area that is colored blue when squares within squares are continuously colored through infinite iterations. The blue areas form a geometric series where each subsequent set of blue squares is smaller than the last.
The first blue square has an area of \(\frac{1}{9}\). Each subsequent layer of blue squares has a smaller area due to the decreasing size of squares at each iteration, specifically by multiplying each previous area by \(\frac{8}{9}\).
This series can be expressed as \(\frac{1}{9} + \frac{1}{9}\cdot\frac{8}{9} + \frac{1}{9}\cdot\left(\frac{8}{9}\right)^2 + \ldots\), forming a geometric series with a common ratio of \(\frac{8}{9}\), which is less than 1.
Such series converge and can be summed using the formula \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio. Here, \(S = \frac{\frac{1}{9}}{1-\frac{8}{9}} = \frac{1}{9} \times 9 = 1\). That is, the total blue area sums to 1 unit.

Fractal Geometry

Fractal geometry often involves self-similar patterns that repeat at various scales, resembling the infinite processes described in this exercise. The repeating division of squares, where each step mimics the previous one, is a classic fractal behavior.
In the specific case of this problem, each time a square is divided, 9 smaller squares are formed, and the middle one is colored blue, creating a recursive process. This self-similarity, or repeating pattern at each level of division, exemplifies fractals in geometry.
Fractals are not just theoretical but appear commonly in nature; structures like snowflakes, plants, and even coastlines exhibit fractal characteristics. This recursive pattern can lead to breathtakingly complex shapes, all built upon an iterative process as seen in this exercise.

Area Calculation

Calculating area is a fundamental part of geometry, often involving analyzing shapes and applying formulas to find the measurement of the space they cover. The challenge lies in complex shapes, like those in fractals described above.
To solve our problem, the process initiates with a single square area of 1 unit, fully divided. As each iteration progresses, we calculate the area of each blue square by determining the side length and squaring it to find the area (e.g., the first blue square has side \(\frac{1}{3}\), thus area \(\frac{1}{9}\)).
Further, calculations for subsequent blue areas involve acknowledging how shrinking dimensions affect total area. The side lengths of each following blue square are \(\left(\frac{1}{3}\right)^n\), causing a considerable decrease in their areas \(\left(\frac{1}{3^2}\right)^n\) as the process continues.
By assembling these individual areas into an infinite series, insightful geometric reasoning allows us to capture the whole picture and determine that the infinite total area colored blue is 1 unit.