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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 ==
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| − | this diagram only shows one possible value of r
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| − | <asy>
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| − | unitsize(5cm);
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| − | defaultpen(linewidth(.8pt)+fontsize(8pt));
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| − | | |
| − | draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle);
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| − | draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle);
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| − | | |
| − | label("$E$",(0,0),SW);
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| − | label("$F$",(-0.5,0.866),W);
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| − | label("$A$",(-0.414, 1.015),NW);
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| − | label("$B$",(0.586,1.015),NE);
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| − | label("$C$",(0.6716,0.866),E);
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| − | label("$D$",(0.1716, 0),SE);
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| − | label("1",(-0.25,0.433),SW);
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| − | label("r",(-0.457,0.9405),NW);
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| − | label("1",(0.086,1.015),N);
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| − | label("r",(0.6288,0.9405),NE);
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| − | label("1",(0.4216,0.433),SE);
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| − | label("r",(0.0858,0),S);
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| − | | |
| − | </asy>
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| − | | |
| − | ===Solution 1===
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| − | 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+1^2-2r\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^2-6r+1 = 0</math>. By [[Vieta's Formulas]] (or Girard identities, or Newton-Girard identities) 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===
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| − | 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
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| − | <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>
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| − | This simplifies to <math>r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}</math>.
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| − | | |
| − | ===Solution 3 (no trig)===
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| − | <asy>
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| − | unitsize(5cm);
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| − | defaultpen(linewidth(.8pt)+fontsize(8pt));
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| − | | |
| − | draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle);
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| − | draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle);
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| − | draw((0.586,1.015)--(0.6716,1.015)--(0.6716,0.866));
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| − | | |
| − | label("$E$",(0,0),SW);
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| − | label("$F$",(-0.5,0.866),W);
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| − | label("$A$",(-0.414, 1.015),NW);
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| − | label("$B$",(0.586,1.015),N);
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| − | label("$C$",(0.6716,0.866),E);
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| − | label("$D$",(0.1716, 0),SE);
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| − | label("$M$",(0.6716,1.015),NE);
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| − | label("1",(-0.25,0.433),SW);
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| − | label("r",(-0.457,0.9405),NW);
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| − | label("1",(0.086,1.015),N);
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| − | label("r",(0.6288,0.9405),SW);
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| − | label("1",(0.4216,0.433),SE);
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| − | label("r",(0.0858,0),S);
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| − | | |
| − | </asy>
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| − | Extend <math>AB</math> to point M so that it creates right triangle <math>\triangle AMC</math> where <math>\angle M = 90^\circ</math>. It is given that the hexagon is equiangular, therefore <math>\angle MBC = \frac{360}{6} = 60^\circ</math>. (exterior angles of a polygon add up to 360 <math>^\circ</math>)
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| − | We can use either Pythagorean theorem or the properties of a <math>30-60-90</math> triangle to find the length of <math>BM={r \over 2}</math> and <math>CM = {\sqrt 3 \over 2 }r</math>. The legs of <math>\triangle AMC</math> are <math>1 + {r \over 2}</math> and <math>{\sqrt 3 \over 2 }r</math>.
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| − | Using Pythagorean theorem, we get <math>AC^{2} = (r^2+r+1)</math>. We can then follow <math>\textbf {Solution 1}</math> to solve for <math>r</math>. <math>\boxed{\textbf{(E)}\ 6}</math>.
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| − | Alternatively, we can find the area of <math>\triangle ABC</math>. We know that the three smaller triangles: <math>\triangle ABC</math>, <math>\triangle CDE</math>, and <math>\triangle EFA</math> are congruent because of <math>S-A-S</math>. Therefore one of the smaller triangles accounts for <math>10\%</math> of the total area. The height of the smaller triangle <math>\triangle ABC</math> is just <math>CM</math> so the area is <math>{1 \cdot {\sqrt 3 \over 2 }r \over 2}</math>. We can then find the area of the hexagon using <math>\textbf {Solution 1}</math>.
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| − | We can even find the area of <math>\triangle ACE</math> and <math>\triangle ABC</math> and solve for <math>r</math> because the ratio of the areas is <math>7</math> to <math>1</math>.
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| − | | |
| − | ~Zeric Hang
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| − | | |
| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=18|num-a=20|ab=A}}
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| − | | |
| − | [[Category:Introductory Geometry Problems]]
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| − | {{MAA Notice}}
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