2006 AMC 12A Problems/Problem 21

Revision as of 16:35, 17 September 2023 by Powerfulhydrogen (talk | contribs) (Problem)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E}$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png