Problem 99 - Precalculus: Mathematics for Calculus

Question

The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated for each new square. (See the figure.) (a) Find the sum of the areas of all the squares. (b) Find the sum of the perimeters of all the squares.

Step-by-Step Solution

Answer

(a) 2; (b) \(8(\sqrt{2} + 1)\).

Step 1: Understand the Problem

We are given a square of side length 1 and instructed to join the midpoints of its sides to form a new square inside it. We repeat this procedure successively. We need to find both the sum of the areas and the sum of the perimeters of all squares generated by this method.

Step 2: Area of the nth Square

The area of the initial square is 1 since its side length is 1. For each new square, the side length is the diagonal of the smaller squares formed, making the new side length \(\frac{1}{\sqrt{2}}\) of the previous square's side length. Thus, the side length of the nth square is \((\frac{1}{\sqrt{2}})^{n-1}\). Therefore, the area of the nth square is \(\left((\frac{1}{\sqrt{2}})^{n-1}\right)^2 = (\frac{1}{2})^{n-1}\).

Step 3: Sum of the Areas

The total area is the sum of an infinite geometric series where the first term \(a_1 = 1\) and the common ratio \(r = \frac{1}{2}\). The sum \(S\) of this infinite series is given by the formula: \[ S = \frac{a_1}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2 \].

Step 4: Perimeter of the nth Square

The perimeter of the initial square is 4 since its side length is 1. For the nth square, the side length is \((\frac{1}{\sqrt{2}})^{n-1}\). Thus, the perimeter of the nth square is \(4 \times (\frac{1}{\sqrt{2}})^{n-1}\).

Step 5: Sum of the Perimeters

The total perimeter is again an infinite geometric series where the first term \(a_1 = 4\) and the common ratio \(r = \frac{1}{\sqrt{2}}\). The sum \(P\) of this infinite series is given by the formula: \[ P = \frac{a_1}{1 - r} = \frac{4}{1 - \frac{1}{\sqrt{2}}} \]. Since \(1 - \frac{1}{\sqrt{2}}\) simplifies to \(\sqrt{2} - 1\), we have \(P = \frac{4\sqrt{2}}{\sqrt{2} - 1} \). By multiplying numerator and denominator by \(\sqrt{2} + 1\), \(P = 8(\sqrt{2} + 1)\).

Step 6: Verify and Conclude

Recheck the calculation steps for any mistakes, ensuring that the sums are computed correctly. The final answers are consistent with the geometric interpretation of the problem.

Key Concepts

Square Geometry

The problem begins with understanding square geometry, which is the foundation of this exercise. A square is a geometric shape with four equal sides and four right angles. Squares are simple, yet exploring their properties and transformations unveils rich mathematical concepts.

• Original square: The exercise starts with a square that has a side length of 1. This makes it easy to compute its area and perimeter straightforwardly.
• Midpoints and new squares: By connecting the midpoints of the sides, a new, smaller square is formed inside the original one. This isn’t just a smaller version though — it has unique geometric properties based on its formation.

The key factor to understand here is that each subsequent square has a side length determined by the previous square's side length multiplied by the factor of \(\frac{1}{\sqrt{2}}\). Hence, the characteristics of each new square are governed by this scale factor.

Sum of Areas

Calculating the sum of areas of these squares involves recognizing the infinite geometric series present. Let's delve deeper:

• The first square has an area of 1, as given by its side length 1 (i.e., \(1^2 = 1\)).
• Each subsequent square has an area impacted by the repeated scaling of side lengths, calculated as \((\frac{1}{2})^{n-1}\) for the nth square. The factor \(\frac{1}{2}\) comes from squaring the side length's scale factor \(\frac{1}{\sqrt{2}}\).

The formula for the sum of an infinite geometric series, \[ S = \frac{a_1}{1 - r} \]where \(a_1\) is the first term and \(r\) is the common ratio, allows us to sum these areas. Here, \(a_1 = 1\) and \(r = \frac{1}{2}\), leading to a total sum of areas as 2. This result showcases the convergence of infinite series and how significantly they apply in geometric transformations.

Sum of Perimeters

The task also involves summing the perimeters of these infinite squares, another geometric series challenge.

• The initial square's perimeter is 4, being twice the sum of its sides.
• Each new square’s perimeter is derived from multiplying its side length by 4. As each side length scales by \(\frac{1}{\sqrt{2}}\), the nth square's perimeter is \(4 \times (\frac{1}{\sqrt{2}})^{n-1}\).

For an infinite geometric series, this perimeter continues. Using \[ P = \frac{a_1}{1 - r} \]where \(a_1 = 4\) and \(r = \frac{1}{\sqrt{2}}\), the calculation yields \[ P = \frac{4\sqrt{2}}{\sqrt{2} - 1} \]. Simplifying further through rationalization gives us the sum of perimeters as \(8(\sqrt{2} + 1)\). This showcases how geometric series assist in making sense of infinite constructs elegantly and how calculus helps resolve what first seems undefined.