Difference between revisions of "2011 AMC 12A Problems/Problem 17"
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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. | ||
| − | The areas of the three triangles determined by the center and the two points of tangency of each circle are, by | + | The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine, |
<math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math> | <math>\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}</math> | ||
Revision as of 11:21, 20 August 2018
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(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=(0,4); dot (A); dot (B); dot (C); draw(A--B); draw(A--C); draw(B--C); draw(Circle(A,1)); draw(Circle(B,2)); draw(Circle(C,3)); [/asy]](http://latex.artofproblemsolving.com/1/2/e/12e8ab13f9f9f7579b5c34b66aa5af86d4fd3964.png)
The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine,
which add up to 
. The area we're looking for is the large 3-4-5 triangle minus the three smaller triangles, or 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
