# 2006 AMC 12A Problems/Problem 21

## Problem

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

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

What is the ratio of the area of $S_2$ to the area of $S_1$? $\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$

## Solution

Looking at the constraints of $S_1$: $x+y > 0$ $\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$ $\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$ $\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$ $1+x^2+y^2 \le 10x+10y$ $x^2-10x+y^2-10y \le -1$ $x^2-10x+25+y^2-10y+25 \le 49$ $(x-5)^2 + (y-5)^2 \le (7)^2$ $S_1$ is a circle with a radius of $7$. So, the area of $S_1$ is $49\pi$.

Looking at the constraints of $S_2$: $x+y > 0$ $\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)$ $\log_{10}(2+x^2+y^2)\le \log_{10} 100 +\log_{10}(x+y)$ $\log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)$ $2+x^2+y^2 \le 100x+100y$ $x^2-100x+y^2-100y \le -2$ $x^2-100x+2500+y^2-100y+2500 \le 4998$ $(x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2$ $S_2$ is a circle with a radius of $7\sqrt{102}$. So, the area of $S_2$ is $4998\pi$.

So the desired ratio is $\frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E}$.

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