2010 AMC 12A Problems/Problem 17

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The following problem is from both the 2010 AMC 12A #17 and 2010 AMC 10A #19, so both problems redirect to this page.


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 we know that the sum of the possible value of $r$ is $\boxed{\textbf{(E)}\ 6}$.

Solution 2

Step 1: Use Law of Cosines in the same manner as the previous solution to get $AC=\sqrt{r^2+r+1}$.

Step 2: $\triangle{ABC}$~$\triangle{CDE}$~$\triangle{EFA}$ via SAS congruency. Using the formula $[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}$. The area of the hexagon is equal to $[ACE] + 3[ABC]$. We are given that $70\%$ of this area is equal to $[ACE]$; solving for $AC$ in terms of $r$ gives $AC=\sqrt{7r}$.

Step 3: $\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0$ and by Vieta's Formulas , we get $\boxed{\textbf{E}}$.

Note: Since $r$ has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.

Solution 3

Find the area of the triangle $ACE$ as how it was done in solution 1. Find the sum of the areas of the congruent triangles $ABC, CDE, EFA$ as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles $ABC, CDE, EFA$ is $30\%$ of the area of the hexagon. Hence $\frac{7}{3}$ times the latter is equal to the triangle $ACE$. Hence $\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)$. We can simplify this to $7r=r^2+r+1\implies r^2-6r+1=0$. By Vieta's, we get the sum of all possible values of $r$ is $-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}$. -vsamc (Edited by pinkbunny1228)

Video Solution



See also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 12 Problems and Solutions

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