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Difference between revisions of "2010 AMC 12A Problems/Problem 17"
 
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== Solution 1==
 
== Solution 1==
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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>.
+
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> by area of an equilateral triangle.
  
 
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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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>.
  
−
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.
+
Note: To verify that the quadratic <math>r^2-6r+1</math> has two positive roots, we can either solve for the roots directly or note that discriminant is positive, and there are no negative roots (because then <math>r^2, -6r, 1</math> would all be positive).
  
 
==Solution 3==
 
==Solution 3==
 
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>.
 
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>.
 
-vsamc
 
-vsamc
−
(Edited by pinkbunny1228)
 
  
 
===Proof Triangle ACE is Equilateral.===
 
===Proof Triangle ACE is Equilateral.===
−
We know triangles ABC, CDE, and EFA are the same by SAS congruence, so the side opposite the 120 degree is also the same (since the triangles are congruent). Thus ACE is congruent.
+
We know <math>\triangle{ABC}</math>, <math>\triangle{CDE}</math>, and <math>\triangle{EFA}</math> are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus <math>\triangle{ACE}</math> is equilateral.
 
Q.E.D.
 
Q.E.D.
 
~mathboy282
 
~mathboy282
  
−
==Video Solution by the Beauty of Math==
+
== Video Solution by Pi Academy ==
 +
https://youtu.be/x8b1D35qxME?si=8jTbTBi-LPOUcP3t
 +
 
 +
~ Pi Academy
 +
 
 +
== Video Solution 2 ==
 
https://youtu.be/rsURe5Xh-j0?t=961
 
https://youtu.be/rsURe5Xh-j0?t=961
  

Latest revision as of 19:17, 9 October 2024

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 $\frac{\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.

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: To verify that the quadratic $r^2-6r+1$ has two positive roots, we can either solve for the roots directly or note that discriminant is positive, and there are no negative roots (because then $r^2, -6r, 1$ would all be positive).

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

Proof Triangle ACE is Equilateral.

We know $\triangle{ABC}$, $\triangle{CDE}$, and $\triangle{EFA}$ are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus $\triangle{ACE}$ is equilateral. Q.E.D. ~mathboy282

Video Solution by Pi Academy

https://youtu.be/x8b1D35qxME?si=8jTbTBi-LPOUcP3t

~ Pi Academy

Video Solution 2

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

See also

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
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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