Question

Let

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

and

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

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

$98$

$99$

$100$

$101$

$102$

Answer: E

Concepts:logarithm completing the square circle area

Difficulty rating: 2180

Solution:

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

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

The logarithms require $x+y>0.$ For each disk, the squared distance from its center to the line $x+y=0$ is $\tfrac{10^{2j}}{2},$ which is $j$ more than the squared radius. Hence both disks lie entirely in $x+y>0,$ so no points were added when the inequalities were rewritten. The area ratio is $\dfrac{4998}{49} = 102.$

Thus, the correct answer is E.

2006 AMC 12A Problem 21

Answer: E

Difficulty rating: 2180

Solution:

Problem 21 in Other Years