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Step-by-Step Solution
Step 1: Express the relationship between consecutive squares
Let the side length of the square A_n be s_n . We are given that the side of
A_n equals the diagonal of A_{n+1} . Since for any square with side length s ,
the diagonal length is s \sqrt{2} , we can write:
s_n = s_{n+1} \sqrt{2}.
Rearranging, we get:
s_{n+1} = \frac{s_n}{\sqrt{2}}.
Step 2: Determine the general form of the side length
We start with s_1 = 12 cm. Using the recurrence relation repeatedly:
s_2 = \frac{s_1}{\sqrt{2}},\quad
s_3 = \frac{s_2}{\sqrt{2}}, \quad
\dots \quad
s_n = \frac{12}{(\sqrt{2})^{\,n-1}}.
Step 3: Express the area of the nth square
The area of square A_n , denoted by A_n^\prime , is:
A_n^\prime = (s_n)^2 = \left(\frac{12}{(\sqrt{2})^{\,n-1}}\right)^2
= \frac{144}{\left((\sqrt{2})^{\,n-1}\right)^2}
= \frac{144}{2^{\,n-1}}.
Step 4: Set up the inequality for the area to be less than 1
We want to find the smallest positive integer n such that:
\frac{144}{2^{\,n-1}} < 1.
Cross-multiplying, we get:
144 < 2^{\,n-1}.
Step 5: Solve the inequality to find the smallest n
To solve 2^{\,n-1} > 144 , we compare exponents:
n - 1 > \log_2(144).
Now, \log_2(144) \approx 7.17 . Hence:
n - 1 > 7.17 \quad \Longrightarrow \quad n > 8.17.
Since n must be an integer, the smallest integer greater than 8.17 is 9 .
Therefore, the smallest value of n for which the area of A_n is less than 1 is:
\boxed{9}.
