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 $AB$ 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

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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