Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 3"

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Construct an equilateral triangle whose sidelengths (length <math>2</math>) are composed of the radii fo the circles. The area is thus <math>\sqrt{3}-\pi/2</math>.
 
Construct an equilateral triangle whose sidelengths (length <math>2</math>) are composed of the radii fo the circles. The area is thus <math>\sqrt{3}-\pi/2</math>.
  
== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 2|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 4|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 12:30, 23 July 2006

Problem

If 3 circles of radius 1 are mutually tangent as shown, what is the area of the gap they enclose?


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


$\mathrm{(A) \ }\sqrt{3}-\frac{\pi}2 \qquad \mathrm{(B) \ } \frac 16 \qquad \mathrm{(C) \ }\frac 13 \qquad \mathrm{(D) \ } \frac{\sqrt{3}}2 - \frac{\pi}6 \qquad \mathrm{(E) \ } \frac{\pi}6$

Solution

Construct an equilateral triangle whose sidelengths (length $2$) are composed of the radii fo the circles. The area is thus $\sqrt{3}-\pi/2$.