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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 3"
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== Problem ==
 
== Problem ==
 
If 3 circles of radius 1 are mutually tangent as shown, what is the area of the gap they enclose?
 
If 3 circles of radius 1 are mutually tangent as shown, what is the area of the gap they enclose?
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<center>[[Image:Usc93.3.PNG]]</center>
  
 
<center><math> \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 </math></center>
 
<center><math> \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 </math></center>
  
 
== Solution ==
 
== Solution ==
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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>.
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We connect the centers of the three circles, and we get an equilateral triangle, composed of three congruent sectors and the gap in question. The area of the three gaps is half the area of one of the circles, and is thus <math>\frac{\pi}{2}</math>. The area of the whole triangle is <math>\frac{2^2\sqrt{3}}{4}=\sqrt{3}</math>, so the area of the gap is <math>\sqrt{3}-\frac{\pi}{2}</math>, <math>\mathrm{(A)}</math>.
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==See Also==
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 2|Previous Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 2|Previous Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 4|Next Problem]]
 
* [[University of South Carolina High School Math Contest/1993 Exam/Problem 4|Next Problem]]

Latest revision as of 17:09, 4 October 2016

Problem

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

Usc93.3.PNG
$\mathrm{(A) \ }\sqrt{3}-\frac{\pi}2 \qquad \mathrm{(B) \ } \frac 16 \qquad \mathrm{(C) \ }\frac 13 \qquad \mathrm{(D) \ } \frac{\sqrt{3}}2 - \frac{\pi}6 \qquad \mathrm{(E) \ } \frac{\pi}6$

Solution

We connect the centers of the three circles, and we get an equilateral triangle, composed of three congruent sectors and the gap in question. The area of the three gaps is half the area of one of the circles, and is thus $\frac{\pi}{2}$. The area of the whole triangle is $\frac{2^2\sqrt{3}}{4}=\sqrt{3}$, so the area of the gap is $\sqrt{3}-\frac{\pi}{2}$, $\mathrm{(A)}$.

See Also

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