Difference between revisions of "2012 AMC 10A Problems/Problem 11"

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== Problem ==
 
== 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?
+
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>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4 </math>

Latest revision as of 17:28, 3 January 2018

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 $BC$?

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

Solution

[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 $\angle ACD$, $\triangle ADC \sim \triangle BEC$. 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
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

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