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Difference between revisions of "2011 AMC 12A Problems/Problem 17"

(→‎Solution)
(→‎Solution)
Line 16: Line 16:
 
dotfactor=4;
 
dotfactor=4;
  
−
pair A=(0,0), B=(2,0), C=(1,-1);
+
pair A=(0,0), B=(3,0), C=(0,4);
−
pair M=(1,0);
+
 
−
pair D=(2,-1);
 
 
dot (A);
 
dot (A);
 
dot (B);
 
dot (B);
 
dot (C);
 
dot (C);
−
dot (D);
+
 
−
dot (M);
 
  
 
draw(Circle(A,1));
 
draw(Circle(A,1));
−
draw(Circle(B,1));
+
draw(Circle(B,2));
−
draw(Circle(C,1));
+
draw(Circle(C,3));
  
−
draw(A--B);
 
−
draw(M--D);
 
−
draw(D--B);
 
  
−
label("$A$",A,W);
 
−
label("$B$",B,E);
 
−
label("$C$",C,W);
 
−
label("$M$",M,NE);
 
−
label("$D$",D,SE);
 
 
</asy>
 
</asy>
  

Revision as of 15:18, 22 September 2013

Problem

Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?

$\textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3}$

Solution

[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=(0,4); dot (A); dot (B); dot (C); draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3)); [/asy]

The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.

The 3 triangles determined by one center and the two points of tangency that particular circle has with the other two are, by Law of Sines,

$\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$

$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5} = \frac{8}{5}$

$\frac{1}{2} \cdot 3 \cdot 3 \cdot \frac{3}{5} = \frac{27}{10}$

which add up to $4.8$. Thus the area we're looking for is $6 - 4.8 = 1.2 = \frac{6}{5} \rightarrow \boxed{(D)}$.

See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 Problems and Solutions

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