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Difference between revisions of "2010 AMC 10A Problems/Problem 19"

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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>
 
<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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== Solutions ==
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this diagram only shows one possible value of r
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This diagram only shows one of the possible values of r.
 
<asy>
 
<asy>
 
unitsize(5cm);
 
unitsize(5cm);

Revision as of 21:59, 15 March 2020

Problem

Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?

$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6$

Solutions

This diagram only shows one of the possible values of r. [asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle); draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle); label("$E$",(0,0),SW); label("$F$",(-0.5,0.866),W); label("$A$",(-0.414, 1.015),NW); label("$B$",(0.586,1.015),NE); label("$C$",(0.6716,0.866),E); label("$D$",(0.1716, 0),SE); label("1",(-0.25,0.433),SW); label("r",(-0.457,0.9405),NW); label("1",(0.086,1.015),N); label("r",(0.6288,0.9405),NE); label("1",(0.4216,0.433),SE); label("r",(0.0858,0),S); [/asy] ~Brian

Solution 1

It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines on triangle ABC, we get that $AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1$. Therefore, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}(r^2+r+1)$.

If we extend $BC$, $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$, $BC$ and $DE$ meet at $Y$, and $DE$ and $FA$ meet at $Z$, we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$, $CDY$ and $EFZ$, of side length $1$. The area of $ABCDEF$ is therefore

$\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)$.


Based on the initial conditions,

\[\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)\]

Simplifying this gives us $r^2-6r+1 = 0$. By Vieta's Formulas (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$.

Solution 2

As above, we find that the area of $\triangle ACE$ is $\frac{\sqrt3}4(r^2+r+1)$.

We also find by the sine triangle area formula that $ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4$, and thus \[\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}\] This simplifies to $r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}$.

Solution 3 (no trig)

[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle); draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle); draw((0.586,1.015)--(0.6716,1.015)--(0.6716,0.866)); label("$E$",(0,0),SW); label("$F$",(-0.5,0.866),W); label("$A$",(-0.414, 1.015),NW); label("$B$",(0.586,1.015),N); label("$C$",(0.6716,0.866),E); label("$D$",(0.1716, 0),SE); label("$M$",(0.6716,1.015),NE); label("1",(-0.25,0.433),SW); label("r",(-0.457,0.9405),NW); label("1",(0.086,1.015),N); label("r",(0.6288,0.9405),W); label("1",(0.4216,0.433),SE); label("r",(0.0858,0),S); label("$\frac{r}{2}$",(0.6289,1.015),N); label("$\frac{\sqrt{3}}{2}r$",(0.6716,0.9405),E); [/asy] Extend $AB$ to point M so that it creates right triangle $\triangle AMC$ where $\angle M = 90^\circ$. It is given that the hexagon is equiangular, therefore $\angle MBC = \frac{360}{6} = 60^\circ$. (exterior angles of a polygon add up to 360 $^\circ$)


We can use either Pythagorean theorem or the properties of a $30-60-90$ triangle to find the length of $BM={r \over 2}$ and $CM = {\sqrt 3 \over 2 }r$. The legs of $\triangle AMC$ are $1 + {r \over 2}$ and ${\sqrt 3 \over 2 }r$.


Using Pythagorean theorem, we get $AC^{2} = (r^2+r+1)$. We can then follow $\textbf {Solution 1}$ to solve for $r$. $\boxed{\textbf{(E)}\ 6}$.


Alternatively, we can find the area of $\triangle ABC$. We know that the three smaller triangles: $\triangle ABC$, $\triangle CDE$, and $\triangle EFA$ are congruent because of $S-A-S$. Therefore one of the smaller triangles accounts for $10\%$ of the total area. The height of the smaller triangle $\triangle ABC$ is just $CM$ so the area is ${1 \cdot {\sqrt 3 \over 2 }r \over 2}$. We can then find the area of the hexagon using $\textbf {Solution 1}$.


We can even find the area of $\triangle ACE$ and $\triangle ABC$ and solve for $r$ because the ratio of the areas is $7$ to $1$.

~Zeric Hang

See also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 10 Problems and Solutions

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