Difference between revisions of "2006 AMC 12A Problems/Problem 19"
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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> | <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 == | + | == Solutions == |
+ | |||
+ | === Solution 1 === | ||
:''This solution needs a clearer explanation and a diagram.'' | :''This solution needs a clearer explanation and a diagram.'' | ||
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]], | 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>\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b</math> | :<math>\frac{1152}{169} = \frac{120}{119}\frac{-142}{169} + b</math> | ||
:<math>b = \frac{912}{119}</math> <math>\Rightarrow \mathrm{E}</math> | :<math>b = \frac{912}{119}</math> <math>\Rightarrow \mathrm{E}</math> | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | By [[User:skiron|skiron]]. | ||
+ | |||
+ | 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 20:07, 5 January 2008
Problem
Circles with centers and have radii and , respectively. The equation of a common external tangent to the circles can be written in the form with . What is ?
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 . 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,
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):
We can plug this into the equation of the line for the tangent to get:
Solution 2
By skiron.
Let be the line that goes through and , and let be the line . If we let be the measure of the acute angle formed by and the x-axis, then . clearly bisects the angle formed by and the x-axis, so . We also know that and intersect at a point on the x-axis. The equation of is , so the coordinate of this point is . Hence the equation of is , so , and our answer choice is .
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 |