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

m (Solution 3 (no trig))
m (Solution 3 (no trig))
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Using Pythagorean theorem, we get <math>AC = (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>.
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>.

Revision as of 03:25, 16 January 2020


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 $r$?

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


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)

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 $CE$ 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 (ProblemsAnswer KeyResources)
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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