Difference between revisions of "2013 AMC 12B Problems/Problem 16"

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==Problem==
 
==Problem==
  
Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120^{\circ}</math>. Region <math>R</math> consists of all points inside of the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?
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Let <math>ABCDE</math> be an equiangular convex pentagon of perimeter <math>1</math>. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let <math>s</math> be the perimeter of this star. What is the difference between the maximum and the minimum possible values of <math>s</math>.
  
<math>\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2</math>
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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\ \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}</math>
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==Solution==
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== See also ==
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{{AMC12 box|year=2013|ab=B|num-b=15|num-a=17}}
  
 
==Solution==
 
==Solution==

Revision as of 18:10, 22 February 2013

Problem

Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$.

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)}\  \frac{\sqrt{5}+1}{2} \qquad \textbf{(E)}\ \sqrt{5}$

Solution

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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

Solution

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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