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−  == Problem ==
 +  #REDIRECT [[2010_AMC_12A_Problems/Problem_17]] 
−  Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?
 
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−  <math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math>
 
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−  == Solution ==
 
−  ===Solution 1===
 
−  It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on triangle ABC, we get that <math>AC^2 = r^2+1^22r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.
 
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−  If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore
 
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−  <math>\frac{\sqrt{3}}{4}(r+2)^2\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>.
 
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−  Based on the initial conditions,
 
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−  <cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath>
 
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−  Simplifying this gives us <math>r^26r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.
 
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−  ===Solution 2===
 
−  As above, we find that the area of <math>\triangle ACE</math> is <math>\frac{\sqrt3}4(r^2+r+1)</math>.
 
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−  We also find by the sine [[triangle]] area formula that <math>ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4</math>, and thus
 
−  <cmath>\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}</cmath>
 
−  This simplifies to <math>r^26r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>.
 
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−  == See also ==
 
−  {{AMC10 boxyear=2010numb=18numa=20ab=A}}
 
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−  [[Category:Introductory Geometry Problems]]
 
−  {{MAA Notice}}
 