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

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Let  
 
Let  
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<math>S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}</math>
 
<math>S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}</math>
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and  
 
and  
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<math>S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}</math>.
 
<math>S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}</math>.
  
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<math> (x-5)^2 + (y-5)^2 \le (7)^2 </math>
 
<math> (x-5)^2 + (y-5)^2 \le (7)^2 </math>
  
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<math>S_1</math> is a semicircle with a radius of <math>7</math>. So, the area of <math>S_1</math> is <math> \frac{49}{2\pi} </math>.
+
<math>S_1</math> is a circle with a radius of <math>7</math>. So, the area of <math>S_1</math> is <math>49\pi </math>.
  
 
Looking at the constraints of <math>S_2</math>:
 
Looking at the constraints of <math>S_2</math>:
 +
 +
<math> x+y > 0 </math>
  
 
<math>\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)</math>
 
<math>\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)</math>
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<math> (x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2 </math>
 
<math> (x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2 </math>
  
−
<math>S_2</math> is a semicircle with a radius of <math>7\sqrt{102}</math>. So, the area of <math>S_2</math> is <math> \frac{4998}{2\pi} </math>.
+
<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>.

Latest revision as of 16:35, 17 September 2023

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 (Problems • Answer Key • Resources)
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

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