Difference between revisions of "2011 AMC 12A Problems/Problem 17"
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== Solution == | == Solution == | ||
| + | |||
| + | <asy> | ||
| + | unitsize(1.1cm); | ||
| + | defaultpen(linewidth(.8pt)); | ||
| + | dotfactor=4; | ||
| + | |||
| + | pair A=(0,0), B=(2,0), C=(1,-1); | ||
| + | pair M=(1,0); | ||
| + | pair D=(2,-1); | ||
| + | dot (A); | ||
| + | dot (B); | ||
| + | dot (C); | ||
| + | dot (D); | ||
| + | dot (M); | ||
| + | |||
| + | draw(Circle(A,1)); | ||
| + | draw(Circle(B,1)); | ||
| + | draw(Circle(C,1)); | ||
| + | |||
| + | 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> | ||
| + | |||
The centers of these circles form a 3-4-5 triangle, which has an area equal to 6. | The centers of these circles form a 3-4-5 triangle, which has an area equal to 6. | ||
Revision as of 15:16, 22 September 2013
Problem
Circles with radii 
, 
, and 
are mutually externally tangent. What is the area of the triangle determined by the points of tangency?
Solution
![[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); 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]](http://latex.artofproblemsolving.com/f/8/4/f84b9e326bd427c8608100174e752c08642c3129.png)
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,
which add up to 
. Thus the area we're looking for is 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
