Difference between revisions of "2012 AMC 10A Problems/Problem 11"
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− | == Problem | + | == Problem == |
+ | Externally tangent circles with centers at points <math>A</math> and <math>B</math> have radii of lengths <math>5</math> and <math>3</math>, respectively. A line externally tangent to both circles intersects ray <math>AB</math> at point <math>C</math>. What is <math>BC</math>? | ||
− | + | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4 </math> | |
+ | |||
+ | == Solution == | ||
+ | |||
+ | <center><asy> | ||
+ | unitsize(3.5mm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375)); | ||
+ | path a=Circle(A,5); | ||
+ | path b=Circle(B,3); | ||
+ | draw(a); draw(b); | ||
+ | draw(C--D); | ||
+ | draw(A--C); | ||
+ | draw(A--D); | ||
+ | draw(B--E); | ||
+ | |||
+ | pair[] ps={A,B,C,D,E}; | ||
+ | dot(ps); | ||
+ | |||
+ | label("$A$",A,N); label("$B$",B,N); label("$C$",C,N); | ||
+ | label("$D$",D,SE); label("$E$",E,SE); | ||
+ | label("$5$",(A--D),SW); | ||
+ | label("$3$",(B--E),SW); | ||
+ | label("$8$",(A--B),N); | ||
+ | label("$x$",(C--B),N); | ||
+ | |||
+ | </asy></center> | ||
− | <math> \ | + | 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 BEC</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> | ||
+ | |||
+ | |||
+ | == Video Solution by Calculocity (Easy Similar Triangles) == | ||
+ | https://youtu.be/YGu8AegJZ80 | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/NsQbhYfGh1Q?t=758 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=A|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
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.