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

Revision as of 09:13, 14 October 2022

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 the Beauty of Math

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #17]] and [[2010 AMC 10A Problems|2010 AMC 10A #19]]}}
 == 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>?
@@ Line 5: / Line 7: @@
 == Solution 1==
-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
@@ Line 25: / Line 27: @@
 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==
-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{-7}{1}=7\text{or} \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
+===Proof Triangle ACE is Equilateral.===
+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.
+~mathboy282
+==Video Solution by the Beauty of Math==
+https://youtu.be/rsURe5Xh-j0?t=961
 == See also ==
+{{AMC10 box|year=2010|ab=A|num-b=18|num-a=20}}
 {{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}
 [[Category:Introductory Geometry Problems]]
 {{MAA Notice}}

2010 AMC 10A (Problems • Answer Key • Resources)
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2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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