Difference between revisions of "2012 AMC 10A Problems/Problem 11"
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− | Let <math>D</math> and <math>E</math> be the points of tangency on circles <math>A</math> and <math>B</math> with line <math>CD</math>. <math>AB=8</math>. Also, let <math>BC=x</math>. As <math>\angle ADC</math> and <math>\angle BEC</math> are right angles and both triangles share <math>\angle ACD</math>, <math>\triangle ADC \sim \triangle BCE</math>. From this we can get a proportion. | + | Let <math>D</math> and <math>E</math> be the points of tangency on circles <math>A</math> and <math>B</math> with line <math>CD</math>. <math>AB=8</math>. Also, let <math>BC=x</math>. As <math>\angle ADC</math> and <math>\angle BEC</math> are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share <math>\angle ACD</math>, <math>\triangle ADC \sim \triangle BCE</math>. From this we can get a proportion. |
<math>\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}</math> | <math>\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}</math> |
Revision as of 20:46, 9 February 2012
Problem
Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC?
Solution
Let and be the points of tangency on circles and with line . . Also, let . As and are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share , . From this we can get a proportion.
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 10 Problems and Solutions |