Difference between revisions of "2012 AMC 10A Problems/Problem 11"
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<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> | ||
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+ | == Video Solution by Calculocity (Easy Similar Triangles) == | ||
+ | https://youtu.be/YGu8AegJZ80 | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Latest revision as of 17:41, 22 April 2024
Contents
[hide]Problem
Externally tangent circles with centers at points and have radii of lengths and , respectively. A line externally tangent to both circles intersects ray at point . What is ?
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.
Video Solution by Calculocity (Easy Similar Triangles)
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=758
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.