Problem 75 - Algebra and Trigonometry

Question

Geometry 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

The sum of all square areas is 2, and the perimeter sum diverges.

Step 1: Calculate the area of the initial square

The initial square has a side length of 1, so its area is given by the formula for the area of a square: \[ A = s^2 \]where \( s \) is the side length. Therefore, the area of the initial square is \( 1^2 = 1 \).

Step 2: Determine the area of subsequent squares

Each new square, formed by joining the midpoints of the sides of the previous square, has side length \( \frac{1}{\sqrt{2}} \) times the side length of the previous square. This is because the diagonal of the half square becomes the side of the new square. Thus, if the side length of the previous square is \( s \), the new side is \( \frac{s}{\sqrt{2}} \). The area of the second square then becomes:\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]and continues for each subsequent square.

Step 3: Calculate the infinite series sum for areas

The areas form a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \). The sum of an infinite geometric series is given by:\[ S = \frac{a}{1 - r} \]Plugging in the values gives:\[ S = \frac{1}{1 - \frac{1}{2}} = 2 \]

Step 4: Calculate the perimeter of the initial square

The perimeter of a square is given by \( P = 4s \). For the initial square with \( s = 1 \), the perimeter is \( 4 \times 1 = 4 \).

Step 5: Determine the perimeter of subsequent squares

Just like with area, each new square's side is \( \frac{s}{\sqrt{2}} \). The perimeter of the next square is thus \( 4 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} \), which simplifies to \( 2\sqrt{2} \). Continuing this pattern forms a sequence of perimeters.

Step 6: Calculate the infinite series for perimeters

The sequence of perimeters also forms a geometric series with the first term \( a = 4 \) and common ratio \( r = \frac{1}{\sqrt{2}} \). The sum of the series is:\[ S = \frac{4}{1 - \frac{1}{\sqrt{2}}} \]Simplifying gives:\[ S = 4 \times \frac{\sqrt{2}}{\sqrt{2} - 1} \]

Step 7: Conclusion: Sum of the series

After calculating both infinite series, we find the sum of all areas is 2, and solving the series for the sum of perimeters leads to an infinite perimeter since the formula simplifies further. (To find a numerical answer you'd need a specific approach based on problems or convergence principles.)

Key Concepts

Infinite Series

An infinite series is an essential concept in mathematics where we sum up an infinite sequence of terms. In this context, we're dealing with the areas and perimeters of squares that become progressively smaller. These shapes form a geometric sequence.

A geometric series occurs when each term is derived by multiplying the previous term by a constant known as the "common ratio". When adding the terms of a geometric series indefinitely, if the absolute value of the common ratio is less than one, the series converges to a finite sum. This sum can be calculated using the formula:

• Sum of Infinite Geometric Series: \[S = \frac{a}{1 - r} \] where "\(a\)" is the first term, and "\(r\)" is the common ratio.

For the squares in our problem:

• The first term is \(1\) for areas and \(4\) for perimeters.
• The common ratio for the areas is \(\frac{1}{2}\).
• The common ratio for the perimeters is \(\frac{1}{\sqrt{2}}\).

When dealing with such infinite series, understanding convergence is vital, as it tells us whether the sequence of numbers will approach a specific value.

Square Geometry

Square geometry is all about understanding the properties of squares, which are fundamental shapes in geometry. A square is a quadrilateral with all sides equal and all angles equal to \(90^\circ\).

In this exercise, each new square formed is derived from the previous one by connecting midpoints of its sides, effectively shrinking by the factor of \(\frac{1}{\sqrt{2}}\). This transformation is an excellent application of understanding diagonal properties, as the diagonal of a square divides it into two identical right triangles.

The diagonal, found using Pythagoras' theorem, is \( s\sqrt{2} \) for a square with side length \( s \). Therefore, the side of the new square becomes \( \frac{s}{\sqrt{2}} \), automatically simplifying to \( \frac{1}{\sqrt{2}} \) when the original square’s side is \(1\). This property simplifies calculating new square parameters as it reflects a consistent pattern of reduction.

Perimeter Calculation

Perimeter calculation for a square is straightforward yet crucial in understanding how its measure continuously decreases in an infinite series. The perimeter, simply calculated as four times the length of one side, diminishes as the squares get progressively smaller.

Starting with the initial square having a side of 1, the perimeter is:

• Initial Perimeter: \[P_0 = 4 \times 1 = 4\]
• Next Side Length: \[\frac{1}{\sqrt{2}}\]
• Next Perimeter: \[P_1 = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}\]

The reduction follows a geometric pattern with each subsequent perimeter being \( \frac{4}{\sqrt{2}}, \frac{4}{2}, \cdots \), and hence decreases towards a limit. Each step in the perimeter calculation uses the common ratio of \( \frac{1}{\sqrt{2}} \).

Thus, as you continue multiplying by this ratio, the series efficiently reflects how perimeter decreases down an endless chain of smaller squares.