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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== Solution == | == Solution == | ||
| + | Looking at the constraints of <math>S_1</math>: | ||
| + | |||
| + | <math> x+y > 0 </math> | ||
| + | |||
| + | <math>\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)</math> | ||
| + | |||
| + | <math> \log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)</math> | ||
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| + | <math> \log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)</math> | ||
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| + | <math> 1+x^2+y^2 \le 10x+10y </math> | ||
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| + | <math> x^2-10x+y^2-10y \le -1 </math> | ||
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| + | <math> x^2-10x+25+y^2-10y+25 \le 49 </math> | ||
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| + | <math> (x-5)^2 + (y-5)^2 \le (7)^2 </math> | ||
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| + | <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>. | ||
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| + | 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 \log_{10} 100 +\log_{10}(x+y)</math> | ||
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| + | <math> \log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)</math> | ||
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| + | <math> 2+x^2+y^2 \le 100x+100y </math> | ||
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| + | <math> x^2-100x+y^2-100y \le -2 </math> | ||
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| + | <math> x^2-100x+2500+y^2-100y+2500 \le 4998 </math> | ||
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| + | <math> (x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2 </math> | ||
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| + | <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>. | ||
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| + | So the desired ratio is <math> \frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{E} </math>. | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2006|ab=A|num-b=20|num-a=22}} | |
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 16:35, 17 September 2023
Problem
Let
and
.
What is the ratio of the area of 
to the area of 
?
Solution
Looking at the constraints of :
is a circle with a radius of 
. So, the area of is 
.
Looking at the constraints of :
is a circle with a radius of 
. So, the area of is 
.
So the desired ratio is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
