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Difference between revisions of "2006 AMC 12A Problems/Problem 19"

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== Problem ==
 
== Problem ==
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[[Circle]]s with [[center_(geometry) | center]]s <math>(2,4)</math> and <math>(14,9)</math> have [[radius | radii]] <math>4</math> and <math>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?
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<!-- <center>[[Image:AMC12_2006A_19.png]]</center> -->
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<asy>
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size(150);
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defaultpen(linewidth(0.7)+fontsize(8));
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draw(circle((2,4),4));draw(circle((14,9),9));
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draw((0,-2)--(0,20));draw((-6,0)--(25,0));
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draw((2,4)--(2,4)+4*expi(pi*4.5/11));
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draw((14,9)--(14,9)+9*expi(pi*6/7));
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label("4",(2,4)+2*expi(pi*4.5/11),(-1,0));
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label("9",(14,9)+4.5*expi(pi*6/7),(1,1));
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label("(2,4)",(2,4),(0.5,-1.5));label("(14,9)",(14,9),(1,-1));
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draw((-4,120*-4/119+912/119)--(11,120*11/119+912/119));
 +
dot((2,4)^^(14,9));
 +
</asy>
  
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{{image}}
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<math> \mathrm{(A) \ } \frac{908}{119}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ }  \frac{912}{119}</math>
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[[Circle]]s with [[center]]s <math>(2,4)</math> and <math>(14,9)</math> have [[radius | radii]] <math>4</math> and <math>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math>m>0</math>. What is <math>b</math>?
 
−
 
 
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<math> \mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}</math><math>\mathrm{(E) \ }  \frac{912}{119}</math>
 
  
 
== Solution ==
 
== Solution ==
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{{solution}}
 
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:''This solution needs a clearer explanation and a diagram.''
 
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Notice that both [[circle]]s are [[tangent]] to the [[x-axis]] and each other. Call the circles (respectively) A and B; the [[distance]] between the two centers is <math>4 + 9 = 13</math>. If we draw the parallel radii that lead to the common [[external tangent]], a line can be extended parallel to the tangent from A to the radius of circle B. This creates a 5-12-13 triangle. To find the slope of that line (which is parallel to the tangent), note that another 5-12-13 triangle can be drawn below the first one such that the side with length 12 is parallel to the x-axis. The slope can be found by using the [[double tangent identity]],
 
  
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:<math>\tan (2 \tan ^{-1} (\frac{5}{12}) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math>
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Let <math>L_1</math> be the line that goes through <math>(2,4)</math> and <math>(14,9)</math>, and let <math>L_2</math> be the line <math>y=mx+b</math>. If we let <math>\theta</math> be the measure of the acute angle formed by <math>L_1</math> and the x-axis, then <math>\tan\theta=\frac{5}{12}</math>. <math>L_1</math> clearly bisects the angle formed by <math>L_2</math> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <math>L_2</math> intersect at a point on the x-axis. The equation of <math>L_1</math> is <math>y=\frac{5}{12}x+\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\right)</math>. Hence the equation of <math>L_2</math> is <math>y=\frac{120}{119}x+\frac{912}{119}</math>, so <math>b=\frac{912}{119}</math>, and our answer choice is <math>\boxed{\mathrm{E}}</math>.
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:<math>= \frac{120}{119}</math>
 
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To find the x and y coordinates of the point of tangency of circle A, we can set up a ratio (the slope will be –119/120 because it is the negative reciprocal):
 
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:<math>\frac{119}{\sqrt{119^2 + 120^2}}</math> <math>=</math> <math>\frac{119}{169} = \frac{y - 4}{4}</math>
 
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:<math>\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}</math>
 
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:<math>x = \frac{-142}{169}, y = \frac{1152}{169}</math>
 
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We can plug this into the equation of the line for the tangent to get:
 
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:<math>\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b</math>
 
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:<math>b = \frac{912}{119}</math> <math>\Rightarrow \mathrm{E}</math>
 
  
 
== See also ==
 
== See also ==
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* [[2006 AMC 12A Problems]]
 
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{{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}
 +
{{MAA Notice}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 01:24, 30 October 2024

Problem

Circles with centers $(2,4)$ and $(14,9)$ have radii $4$ and $9$, respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m>0$. What is $b$? [asy] size(150); defaultpen(linewidth(0.7)+fontsize(8)); draw(circle((2,4),4));draw(circle((14,9),9)); draw((0,-2)--(0,20));draw((-6,0)--(25,0)); draw((2,4)--(2,4)+4*expi(pi*4.5/11)); draw((14,9)--(14,9)+9*expi(pi*6/7)); label("4",(2,4)+2*expi(pi*4.5/11),(-1,0)); label("9",(14,9)+4.5*expi(pi*6/7),(1,1)); label("(2,4)",(2,4),(0.5,-1.5));label("(14,9)",(14,9),(1,-1)); draw((-4,120*-4/119+912/119)--(11,120*11/119+912/119)); dot((2,4)^^(14,9)); [/asy]

$\mathrm{(A) \ } \frac{908}{119}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}$

Solution

Let $L_1$ be the line that goes through $(2,4)$ and $(14,9)$, and let $L_2$ be the line $y=mx+b$. If we let $\theta$ be the measure of the acute angle formed by $L_1$ and the x-axis, then $\tan\theta=\frac{5}{12}$. $L_1$ clearly bisects the angle formed by $L_2$ and the x-axis, so $m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}$. We also know that $L_1$ and $L_2$ intersect at a point on the x-axis. The equation of $L_1$ is $y=\frac{5}{12}x+\frac{19}{6}$, so the coordinate of this point is $\left(-\frac{38}{5},0\right)$. Hence the equation of $L_2$ is $y=\frac{120}{119}x+\frac{912}{119}$, so $b=\frac{912}{119}$, and our answer choice is $\boxed{\mathrm{E}}$.

See also

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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