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Difference between revisions of "2010 AMC 12A Problems/Problem 17"

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Step 1: Use [[Law of Cosines]] in the same manner as the pervious solution to get <math>AC=\sqrt{r^2+r+1}</math>.  
 
Step 1: Use [[Law of Cosines]] in the same manner as the pervious solution to get <math>AC=\sqrt{r^2+r+1}</math>.  
  
−
Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{2}</math> and because of the congruency, the area condition, and the fact <math>\triangle{ACE}</math> is equilateral, <math>AC=\sqrt{7r}</math>.
+
Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}</math> and because of the congruency, the area condition, and the fact <math>\triangle{ACE}</math> is equilateral, <math>AC=\sqrt{7r}</math>.
  
 
Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] we get <math>\boxed{\textbf{E}}</math>.
 
Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] we get <math>\boxed{\textbf{E}}</math>.

Revision as of 20:13, 30 May 2017

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 1

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 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 pervious solution to get $AC=\sqrt{r^2+r+1}$.

Step 2: $\triangle{ABC}$~$\triangle{CDE}$~$\triangle{EFA}$ via SAS congruency. Using the formula $[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}$ and because of the congruency, the area condition, and the fact $\triangle{ACE}$ is equilateral, $AC=\sqrt{7r}$.

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

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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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