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

Revision as of 22:41, 5 February 2010

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 $b$ ?

$\mathrm{(A) \ } \frac{908}{199}\qquad \mathrm{(B) \ } \frac{909}{119}\qquad \mathrm{(C) \ } \frac{130}{17}\qquad \mathrm{(D) \ } \frac{911}{119}\qquad \mathrm{(E) \ } \frac{912}{119}$

Solutions

Solution 1

This solution needs a clearer explanation and a diagram.

Notice that both circles are tangent to the x-axis and each other. Call the circles (respectively) A and B; the distance between the two centers is $4 + 9 = 13$ . 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,

$\tan (2 \tan ^{-1} \left(\frac{5}{12}\right)) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}$

$= \frac{120}{119}$

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):

$\frac{119}{\sqrt{119^2 + 120^2}}$ $=$ $\frac{119}{169} = \frac{y - 4}{4}$

$\frac{-120}{\sqrt{119^2 + 120^2}} = \frac{-120}{169} = \frac{x - 2}{4}$

$x = \frac{-142}{169}, y = \frac{1152}{169}$

We can plug this into the equation of the line for the tangent to get:

$\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b$

$b = \frac{912}{119}$ $\Rightarrow \mathrm{E}$

Solution 2

By skiron.

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 $\boxed{\mathrm{E}}$ .

@@ Line 11: / Line 11: @@
 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]],
-:<math>\tan (2 \tan ^{-1} \left(\frac{5}{12}\right) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math>
+:<math>\tan (2 \tan ^{-1} \left(\frac{5}{12}\right)) = \frac{\frac{5}{12} + \frac{5}{12}}{1 - \frac{5}{12}\frac{5}{12}}</math>
 :<math>= \frac{120}{119}</math>

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