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

## 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

### 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 $AB$ so that it creates right triangle $\triangle AEC$ where $\angle E = 90^\circ$. It is given that the hexagon is equiangular, therefore $\angle ABC = 120^\circ$. $\angle ABC$ and $\angle EBC$ are supplementary so $\angle EBC = 60^\circ$.

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

Using Pythagorean theorem, we get $AC = (r^2+r+1)$. 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 $1$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 