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

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{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #17]] and [[2010 AMC 10A Problems|2010 AMC 10A #19]]}}
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== Problem ==
 
== Problem ==
 
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?
 
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?
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<math>\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</math>
 
<math>\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</math>
  
== Solution ==
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== Solution 1==
It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]], we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.
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It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on <math>\triangle ABC</math>, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.
  
 
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore
 
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore
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Based on the initial conditions,
 
Based on the initial conditions,
  
<math>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</math>
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<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath>
  
 
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.
 
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.
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==Solution 2==
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Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get <math>AC=\sqrt{r^2+r+1}</math>.
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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>. The area of the hexagon is equal to <math>[ACE] + 3[ABC]</math>. We are given that <math>70\%</math> of this area is equal to <math>[ACE]</math>; solving for <math>AC</math> in terms of <math>r</math> gives <math>AC=\sqrt{7r}</math>.
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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>.
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Note: Since <math>r</math> has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.
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==Solution 3==
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Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent  triangles <math>ABC, CDE, EFA</math> 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 <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}</math>.
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-vsamc
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(Edited by pinkbunny1228)
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==Video Solution==
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https://youtu.be/rsURe5Xh-j0?t=961
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~IceMatrix
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== See also ==
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{{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 02:56, 17 July 2020

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

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 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

https://youtu.be/rsURe5Xh-j0?t=961

~IceMatrix

See also

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

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