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

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== Solution ==
 
== Solution ==
  
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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 (it is?!?!?! please explain), 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>.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 10:55, 28 April 2019

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 $b$?

AMC12 2006A 19.png

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

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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