Key Concepts and Formulas

• Properties of a Square: For a square with side length $s$, the diagonal length $d = s\sqrt{2}$ and the area $A = s^2$.
• Geometric Progression (G.P.): A sequence where each term is found by multiplying the previous term by a constant ratio. The $n$-th term is $a_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.

Step-by-Step Solution

Step 1: Establish the relationship between the side lengths of consecutive squares. Let $s_n$ denote the side length of square $A_n$. The problem states that the length of the side of $A_n$ equals the length of the diagonal of $A_{n+1}$. Using the property of a square, the diagonal of $A_{n+1}$ with side length $s_{n+1}$ is $s_{n+1}\sqrt{2}$. Thus, we have the relation: $s_n = s_{n+1}\sqrt{2}$ Rearranging this equation to express $s_{n+1}$ in terms of $s_n$: $s_{n+1} = \frac{s_n}{\sqrt{2}}$ This equation shows that the side length of each subsequent square is obtained by dividing the side length of the previous square by $\sqrt{2}$.

Step 2: Identify the sequence of side lengths as a Geometric Progression. The relation $s_{n+1} = \frac{1}{\sqrt{2}} s_n$ implies that the ratio of consecutive terms is constant: $\frac{s_{n+1}}{s_n} = \frac{1}{\sqrt{2}}$ This confirms that the sequence of side lengths $s_1, s_2, s_3, \dots$ forms a Geometric Progression with the first term $s_1$ and a common ratio $r = \frac{1}{\sqrt{2}}$. We are given that the length of $A_1$ is 12 cm, so $s_1 = 12$.

Step 3: Determine the general formula for the side length $s_n$. Using the formula for the $n$-th term of a G.P., $s_n = s_1 \cdot r^{n-1}$, we substitute $s_1 = 12$ and $r = \frac{1}{\sqrt{2}}$: $s_n = 12 \cdot \left(\frac{1}{\sqrt{2}}\right)^{n-1}$ This can be written as: $s_n = \frac{12}{(\sqrt{2})^{n-1}}$

Step 4: Derive the formula for the area of square $A_n$ and set up the inequality. The area of square $A_n$ is $Area(A_n) = (s_n)^2$. Substituting the expression for $s_n$: $Area(A_n) = \left(\frac{12}{(\sqrt{2})^{n-1}}\right)^2$ Using exponent rules, $(x^a)^b = x^{ab}$ and $(\sqrt{x})^2 = x$: $Area(A_n) = \frac{12^2}{\left((\sqrt{2})^{n-1}\right)^2} = \frac{144}{(\sqrt{2})^{2(n-1)}} = \frac{144}{2^{n-1}}$ We need to find the smallest value of $n$ for which $Area(A_n) < 1$. Thus, we set up the inequality: $\frac{144}{2^{n-1}} < 1$

Step 5: Solve the inequality for $n$. To solve $\frac{144}{2^{n-1}} < 1$, we multiply both sides by $2^{n-1}$ (which is always positive for $n \ge 1$): $144 < 2^{n-1}$ Now, we need to find the smallest integer $n$ such that $2^{n-1}$ is greater than 144. Let's examine powers of 2: $2^1 = 2$ $2^2 = 4$ $2^3 = 8$ $2^4 = 16$ $2^5 = 32$ $2^6 = 64$ $2^7 = 128$ $2^8 = 256$ We are looking for the smallest integer $n-1$ such that $2^{n-1} > 144$. From the list, $2^7 = 128$ is not greater than 144, but $2^8 = 256$ is greater than 144. Therefore, the smallest value for $n-1$ is 8. $n-1 = 8$ Solving for $n$: $n = 9$

Step 6: Verify the answer. For $n=9$: $Area(A_9) = \frac{144}{2^{9-1}} = \frac{144}{2^8} = \frac{144}{256}$. Since $144 < 256$, $Area(A_9) < 1$. For $n=8$: $Area(A_8) = \frac{144}{2^{8-1}} = \frac{144}{2^7} = \frac{144}{128}$. Since $144 > 128$, $Area(A_8) > 1$. Thus, $n=9$ is indeed the smallest integer value.

Common Mistakes & Tips

• Index Mismanagement: Be careful with $n$ and $n-1$ when applying G.P. formulas.
• Exponent Simplification: Correctly apply exponent rules, especially for powers of square roots, to simplify expressions efficiently.
• Inequality Direction: When multiplying or dividing an inequality, ensure the sign of the multiplier/divisor is considered to maintain the correct inequality direction.

Summary

We established a relationship between the side lengths of consecutive squares, revealing a Geometric Progression. We derived the general formula for the $n$-th side length and subsequently for the area of the $n$-th square. By solving the inequality $Area(A_n) < 1$, we found that the smallest integer value of $n$ for which the area is less than one is 9.

The final answer is $\boxed{9}$.