Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AMC 12A Problems/Problem 21"

(added link to previous and next problem)
(added category)
Line 56: Line 56:
 
*[[2006 AMC 12A Problems/Problem 20|Previous Problem]]
 
*[[2006 AMC 12A Problems/Problem 20|Previous Problem]]
 
*[[2006 AMC 12A Problems/Problem 22|Next Problem]]
 
*[[2006 AMC 12A Problems/Problem 22|Next Problem]]
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 18:03, 5 November 2006

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 $S_1$:

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

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

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

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

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

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

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

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

Looking at the constraints of $S_2$:

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

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

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

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

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

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

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

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

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

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_21&oldid=11137"