2010 AMC 12A Problems/Problem 17

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 $r$ ?

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

Solution

It is clear that $\triangle ACE$ is an equilateral triangle. From the Law of Cosines, 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, we get $r^2-6r+1 = 0$ , so by Vieta's Formulas we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$ .

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2010 AMC 12A Problems/Problem 17

Problem

Solution