Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

m (→‎Solutions)
m (Use asy diagram from Problems page)
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
[[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>?
+
[[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>?
−
<center>[[Image:AMC12_2006A_19.png]]</center>
+
<!-- <center>[[Image:AMC12_2006A_19.png]]</center> -->
 +
<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>
  
−
<math> \mathrm{(A) \ } \frac{908}{199}\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>
+
<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>
  
 
== Solution ==
 
== Solution ==

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_19&oldid=230946"