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2006 AMC 12A Problems/Problem 21

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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 to the area of ?

\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 :

\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)

is a circle with a radius of . So, the area of is .

Looking at the constraints of :

\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)

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

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

is a circle with a radius of . So, the area of is .

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

See also

2006 AMC 12A (Problems)
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
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