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

## 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 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 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.

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