2006 AMC 12A Problems/Problem 21: Difference between revisions

Revision as of 19:30, 18 February 2011

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 $S_1$ ?

$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$

Looking at the constraints of $S_1$ :

$\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$

$\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$

$\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$

$1+x^2+y^2 \le 10x+10y$

$x^2-10x+y^2-10y \le -1$

$x^2-10x+25+y^2-10y+25 \le 49$

$(x-5)^2 + (y-5)^2 \le (7)^2$

$S_1$ is a circle with a radius of $7$ . So, the area of $S_1$ is $49\pi$ .

Looking at the constraints of $S_2$ :

$\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)$

$\log_{10}(2+x^2+y^2)\le \log_{10} 100 +\log_{10}(x+y)$

$\log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)$

$2+x^2+y^2 \le 100x+100y$

$x^2-100x+y^2-100y \le -2$

$x^2-100x+2500+y^2-100y+2500 \le 4998$

$(x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2$

$S_2$ is a circle with a radius of $7\sqrt{102}$ . So, the area of $S_2$ is $4998\pi$ .

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

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 12 Problems and Solutions

@@ Line 50: / Line 50: @@
 <math>S_2</math> is a circle with a radius of <math>7\sqrt{102}</math>. So, the area of <math>S_2</math> is <math>4998\pi</math>.
-So the desired ratio is <math> \frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E} </math>
+So the desired ratio is <math> \frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E} </math>.
 == See also ==