AoPSWiki
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!

2006 AMC 12A Problems/Problem 21

From AoPSWiki

Revision as of 05:22, 22 February 2009 by Darksigma (Talk | contribs)
(diff) ← Older revision | Current 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:

\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 E

See also

2006 AMC 12A (ProblemsResources)
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

(In the S2 case, the one in in the solution should be a 2)

Trying to get to the USAMO in 2010? Our AIME Problem Series can help you get there! Click here to enroll today!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us