2008 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:square (geometry)Pythagorean Theorem area ratio

Difficulty rating: 1240

10.

Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3.$ What is the area of $S_3?$

$\dfrac{1}{2}$

$1$

$2$

$3$

$4$

Solution:

The side of $S_1$ is $4.$ By the Pythagorean theorem, the side of $S_2$ is $\sqrt{2^2 + 2^2} = 2\sqrt{2},$ so its area is $8.$

By the same reasoning, $S_3$ has half the area of $S_2,$ namely $4.$

Thus, the correct answer is E.

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