2012 AMC 10A Problems/Problem 11

Revision as of 19:46, 9 February 2012 by Flamewire (talk | contribs) (Solution)


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?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$


[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]

Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$. $AB=8$. Also, let $BC=x$. As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$, $\triangle ADC \sim \triangle BCE$. From this we can get a proportion.

$\frac{BC}{AC}=\frac{BE}{AD} \rightarrow \frac{x}{x+8}=\frac{3}{5} \rightarrow 5x=3x+24 \rightarrow x=\boxed{\textbf{(D)}\ 12}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions
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