Let A_1, A_2, A_3, be squares such that, for each n1, the length of the side of A_n equals…

Question

Let $A_1, A_2, A_3, \ldots$ be squares such that, for each $n\ge1$, the length of the side of $A_n$ equals the length of the diagonal of $A_{n+1}$. If the side of $A_1$ is $12$ cm, then the smallest value of $n$ for which the area of $A_n$ is less than $1$ is \_\_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

The side lengths form a G.P. with ratio $\dfrac{1}{\sqrt2}$: side of $A_n = \dfrac{12}{(\sqrt2)^{n-1}}$.

$$\text{Area of } A_n = \frac{144}{2^{n-1}} < 1 \implies 2^{n-1}>144.$$

Since $2^7=128<144<256=2^8$, the smallest $n-1=8$, so $n=9$.

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