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Difference between revisions of "2006 AMC 12A Problems/Problem 13"

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== Problem ==
 
== Problem ==
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The [[vertex|vertices]] of a <math>3-4-5</math> [[right triangle]] are the centers of three mutually externally tangent [[circle]]s, as shown. What is the sum of the areas of the three circles?
  
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[[Image:2006 AMC 12A Problem 13.gif]]
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[[Image:Original_2006_12A_13.png]]
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The vertices of a <math>3-4-5</math> right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?
 
  
 
<math> \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ }  14\pi</math>
 
<math> \mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ }  14\pi</math>
  
 
== Solution ==
 
== Solution ==
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Let the radius of the smallest circle be <math>r_A</math>, the radius of the second largest circle be <math>r_B</math>, and the radius of the largest circle be <math>r_C</math>.
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<cmath>r_A + r_B = 3</cmath>
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<cmath>r_A + r_C = 4</cmath>
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<cmath>r_ B + r_C = 5</cmath>
  
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Let the radius of the smallest circle be <math> a </math>. We find that the radius of the largest circle is <math>4-a</math> and the radius of the second largest circle is <math>3-a</math>. Thus, <math>4-a+3-a=5\iff a=1</math>. The radii of the other circles are <math>3</math> and <math>2</math>. The sum of their areas is <math>\pi+9\pi+4\pi=14\pi\iff\mathrm{(E)}</math>
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Adding up all these equations and then dividing both sides by 2, we get,
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<cmath>r_A + r_B + r_C = 6</cmath>
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Then, we get <math>r_A = 1</math>, <math>r_B = 2</math>, and <math>r_C = 3</math> Then we get <math>1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}</math>
  
 
== See also ==
 
== See also ==
 
* [[2006 AMC 12A Problems]]
 
* [[2006 AMC 12A Problems]]
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*[[2006 AMC 12A Problems/Problem 12|Previous Problem]]
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*[[2006 AMC 12A Problems/Problem 14|Next Problem]]
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{{AMC12 box|year=2006|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 12:57, 19 January 2021

Problem

The vertices of a $3-4-5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of the three circles?

Original 2006 12A 13.png

$\mathrm{(A) \ } 12\pi\qquad \mathrm{(B) \ } \frac{25\pi}{2}\qquad \mathrm{(C) \ } 13\pi\qquad \mathrm{(D) \ } \frac{27\pi}{2}\qquad\mathrm{(E) \ } 14\pi$

Solution

Let the radius of the smallest circle be $r_A$, the radius of the second largest circle be $r_B$, and the radius of the largest circle be $r_C$. \[r_A + r_B = 3\] \[r_A + r_C = 4\] \[r_ B + r_C = 5\]

Adding up all these equations and then dividing both sides by 2, we get,

\[r_A + r_B + r_C = 6\]

Then, we get $r_A = 1$, $r_B = 2$, and $r_C = 3$ Then we get $1^2 \pi + 2^2 \pi + 3^2 \pi = 14 \pi \iff\mathrm{(E)}$

See also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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