Let S_1 = \(x, y) \mid \log_10(1 + x^2 + y^2) ≤ 1 + \log_10(x + y)\ and S_2 = \(x, y) \mid \log_10(2 + x^2 + y^2) ≤ 2 + \log_10(x + y)\. What is the ratio of the area of S_2 to the area of S_1?

2006 AMC 12A Problem 21 — Solution

2006 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2006 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12A solutions, or check the answer key.

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Concepts:logarithm completing the square circle area

Difficulty rating: 2180

21.

Let

$S_1 = \{(x, y) \mid \log_{10}(1 + x^2 + y^2) \le 1 + \log_{10}(x + y)\}$

and

$S_2 = \{(x, y) \mid \log_{10}(2 + x^2 + y^2) \le 2 + \log_{10}(x + y)\}.$

What is the ratio of the area of $S_2$ to the area of $S_1?$

$98$

$99$

$100$

$101$

$102$

Solution:

For $j = 1, 2,$ the condition becomes $j + x^2 + y^2 \le 10^j(x + y),$ i.e. $\left(x - \frac{10^j}{2}\right)^2 + \left(y - \frac{10^j}{2}\right)^2 \le \frac{10^{2j}}{2} - j.$

These are disks with squared radii $\tfrac{100}{2} - 1 = 49$ for $S_1$ and $\tfrac{10000}{2} - 2 = 4998$ for $S_2.$ The area ratio is $\dfrac{4998}{49} = 102.$

Thus, the correct answer is E.

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