Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 12B Problems/Problem 15"

(→‎Solution 2: Inspection)
(→‎Solution 4 (Elimination))
 
(43 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
==Problem 15==
+
==Problem==
 
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?
 
Let <math>ABC</math> be an equilateral triangle. Extend side <math>\overline{AB}</math> beyond <math>B</math> to a point <math>B'</math> so that <math>BB'=3AB</math>. Similarly, extend side <math>\overline{BC}</math> beyond <math>C</math> to a point <math>C'</math> so that <math>CC'=3BC</math>, and extend side <math>\overline{CA}</math> beyond <math>A</math> to a point <math>A'</math> so that <math>AA'=3CA</math>. What is the ratio of the area of <math>\triangle A'B'C'</math> to the area of <math>\triangle ABC</math>?
  
−
<math>\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1</math>
+
<math>\textbf{(A)}\ 9:1 \qquad\textbf{(B)}\ 16:1 \qquad\textbf{(C)}\ 25:1 \qquad\textbf{(D)}\ 36:1 \qquad\textbf{(E)}\ 37:1</math>
  
 +
==Diagram==
 +
<asy>
 +
size(12cm);
 +
dot((0,0));
 +
dot((2,0));
 +
dot((1,1.732));
 +
dot((-6,0));
 +
dot((5,-5.196));
 +
dot((4,6.928));
 +
draw((0,0)--(2,0)--(1,1.732)--cycle);
 +
draw((-6,0)--(5,-5.196)--(4,6.928)--cycle);
 +
draw((-6,0)--(0,0));
 +
draw((5,-5.196)--(2,0));
 +
draw((4,6.928)--(1,1.732));
  
−
==Solution 1: Law of Cosines==
+
</asy>
 +
==Solution 1 (Uses Trig) ==
 +
Note that by symmetry, <math>\triangle A'B'C'</math> is also equilateral. Therefore, we only need to find one of the sides of <math>A'B'C'</math> to determine the area ratio. WLOG, let <math>AB = BC = CA = 1</math>. Therefore, <math>BB' = 3</math> and <math>BC' = 4</math>. Also, <math>\angle B'BC' = 120^{\circ}</math>, so by the Law of Cosines, <math>B'C' = \sqrt{37}</math>. Therefore, the answer is <math>(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37}</math>
 +
 
 +
==Solution 2 ==
 +
As mentioned in the first solution, <math>\triangle A'B'C'</math> is equilateral. WLOG, let <math>AB=2</math>. Let <math>D</math> be on the line passing through <math>AB</math> such that <math>A'D</math> is perpendicular to <math>AB</math>. Note that <math>\triangle A'DA</math> is a 30-60-90 with right angle at <math>D</math>. Since <math>AA'=6</math>, <math>AD=3</math> and <math>A'D=3\sqrt{3}</math>. So we know that <math>DB'=11</math>. Note that <math>\triangle A'DB'</math> is a right triangle with right angle at <math>D</math>. So by the Pythagorean theorem, we find <math>A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.</math> Therefore, the answer is <math>(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37}</math>.
 +
 
 +
==Solution 3 ==
 +
Let <math>AB=BC=CA=x</math>. We start by noting that we can just write <math>AB'</math> as just <math>AB+BB'=4AB</math>.
 +
Similarly <math>BC'=4BC</math>, and <math>CA'=4CA</math>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula,
 +
<math>[ABC]=\sqrt{\frac{3x}{2}\cdot {\left(\frac{3x}{2}-x\right)}^3}=\frac{x^2\sqrt{3}}{4}</math>.
 +
Next in order to evaluate <math>A'B'C'</math> we need to evaluate the area of the larger triangles <math>AA'B',BB'C', \text{ and } CC'A'</math>.
 +
In this solution we shall just compute <math>1</math> of these as the others are trivially equivalent.
 +
In order to compute the area of <math>\Delta{AA'B'}</math> we can use the formula <math>[XYZ]=\frac{1}{2}xy\cdot\sin{z}</math>.
 +
Since <math>ABC</math> is equilateral and <math>A</math>, <math>B</math>, <math>B'</math> are collinear, we already know <math>\angle{A'AB'}=180-60=120</math>
 +
Similarly from above we know <math>AB'</math> and <math>A'A</math> to be <math>4x</math>, and <math>3x</math> respectively. Thus the area of <math>\Delta{AA'B'}</math> is <math>\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}</math>. Likewise we can find <math>BB'C', \text{ and } CC'A'</math> to also be <math>3x^2\cdot\sqrt{3}</math>.
 +
<math>[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)</math>.
 +
Therefore the ratio of <math>[A'B'C']</math> to <math>[ABC]</math> is <math>\frac{\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37}</math>
 +
 
 +
==Solution 4 (Elimination and Educated Guess) ==
 +
Looking at the answer choices, we see that all but <math>{\textbf{(E)}}</math> has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick <math>\boxed{\textbf{(E) } 37}</math>.
 +
 
 +
Additionally, notice that option D is 36:1, which is 1 below 37:1. It's quite possible that people would forget to include the area of <math>ABC</math> when calculating the area of <math>A'B'C'</math> and erroneously get an answer that is 1 below the correct one, which further supports picking <math>\boxed{\textbf{(E) } 37}</math>.
 +
 
 +
Solution by sp1729
 +
Modification by rawr3507
 +
 
 +
==Solution 5 (Barycentric Coordinates) ==
 +
We use barycentric coordinates wrt <math>\triangle ABC</math>, to which we can easily obtain that <math>A'=(4,0,-3)</math>, <math>B'=(-3,4,0)</math>, and <math>C'=(0,-3,4)</math>. Now, since the coordinates are homogenized (<math>-3+4=1</math>), we can directly apply the area formula to obtain that
 +
<cmath>[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC],</cmath>
 +
so the answer is <math>\boxed{\textbf{(E) } 37}</math>
 +
 
 +
==Solution 6 (Area Comparison) ==
 +
First, comparing bases yields that <math>[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12</math>. By congruent triangles,
 +
<cmath>[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],</cmath>
 +
so <math>[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37}</math>
 +
 
 +
== Solution 7 (Quick Proportionality) ==
 +
Scale down the figure so that the area formulas for the <math>120^\circ</math> and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is <math>3*4*3+1*1=37, \boxed{\text{E}}</math>.
 +
~ Solution by mathchampion1
 +
 
 +
==Solution 8 (Sin area formula) ==
 +
Drawing the diagram, we see that the large triangle, <math>A'B'C'</math>, is composed of three congruent triangles with the triangle <math>ABC</math> at the center. Let each of the sides of triangle <math>ABC</math> be <math>x</math>. Therefore, using the equilateral triangle area formula, the <math>[ABC] = \frac{x^2\sqrt{3}}{4}</math>. We also know now that the sides of the triangles are <math>3x</math> and <math>3x + x</math>, or <math>4x</math>. We also know that since <math>BB'</math> are collinear, as are the others, angle <math>C'BB'</math> is <math>180 - 60</math>, which is <math>120</math> degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are <math>\frac{12x^2\sin120}{2} \cdot 3</math>. Simplifying that yields <math>\frac{36x^2\sqrt{3}}{4}</math>. Adding that to the <math>[ABC]</math> yields <math>\frac{37x^2\sqrt{3}}{4}</math>. From this, we can compare the ratios by canceling everything out except for the <math>37</math>, so the answer is <math>\boxed{\textbf{(E) }37}</math>
 +
 
 +
~Solution by EricShi1685
 +
 
 +
==Solution 9: Law of Cosines==
 
Solution by HydroQuantum
 
Solution by HydroQuantum
  
Line 12: Line 72:
  
  
−
Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, <cmath>y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) = </cmath> <cmath>(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2.</cmath> Since both <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> are both equilateral triangles, they must be similar due to <math>AA</math> similarity. This means that <math>\frac{A'B'}{AB}</math> <math>=</math> <math>\frac{B'C'}{BC}</math> <math>=</math> <math>\frac{C'A'}{CA}</math> <math>=</math> <math>\frac{[\triangle A'B'C']}{[\triangle ABC]}</math> <math>=</math> <math>\frac{37}{1}</math>.
+
Recall The Law of Cosines. Letting <math>A'B'=B'C'=C'A'=y</math>, <cmath>y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(\cos 120) = </cmath> <cmath>(3x)^2+(4x)^2-2(3x)(4x)(\cos 120)=9x^2+16x^2-24x(\cos 120)=25x^2+12x^2=37x^2.</cmath> Since both <math>\triangle ABC</math> and <math>\triangle A'B'C'</math> are both equilateral triangles, they must be similar due to <math>AA</math> similarity. This means that <math>\frac{A'B'}{AB}</math> <math>=</math> <math>\frac{B'C'}{BC}</math> <math>=</math> <math>\frac{C'A'}{CA}</math> <math>=</math> <math>\frac{[\triangle A'B'C']}{[\triangle ABC]}</math> <math>=</math> <math>\frac{37}{1}</math>.
  
  
−
Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>.
+
Therefore, our answer is <math>\boxed{\textbf{(E) }37}</math>.
  
−
==Solution 2: Inspection==
+
==Solution 10: Inspection(easiest solution)==
 
Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>.
 
Note that the height and base of <math>\triangle A'CC'</math> are respectively 4 times and 3 times that of <math>\triangle ABC</math>. Therefore the area of <math>\triangle A'CC'</math> is 12 times that of <math>\triangle ABC</math>.
  
−
By symmetry, <math>\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'</math>. Adding the areas of these three triangles and <math>\triangle ABC</math> for the total area of <math>\triangle A'B'C'</math> gives a ratio of <math>(12 + 12 + 12 + 1) : 1</math>, or <math>\boxed{\textbf{(E) } 37 : 1}</math>.
+
By symmetry, <math>\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'</math>. Adding the areas of these three triangles and <math>\triangle ABC</math> for the total area of <math>\triangle A'B'C'</math> gives a ratio of <math>(12 + 12 + 12 + 1) : 1</math>, or <math>\boxed{\textbf{(E) } 37}</math>.
 +
 
 +
==Solution 11: Coordinates==
 +
 
 +
First we note that <math>A'B'C'\sim ABC</math> due to symmetry. WLOG, let <math>B = (0, 0)</math> and <math>AB = 1</math> Therefore, <math>C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)</math>. Using the condition that <math>CC' = 3</math>, we get <math>C' = (4, 0)</math> and <math>B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)</math>. It is easy to check that <math>B'C' = \sqrt{37}</math>. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is <math>\boxed{\textbf{(E) } 37}</math>
 +
 
 +
Solution by <i>mathwiz0803</i>
 +
 
 +
==Solution 12: Computing the Areas==
 +
 +
Note that angle <math>C'BB'</math> is <math>120</math>°, as it is supplementary to the equilateral triangle. Then, using area <math>= \frac{1}{2}ab\sin\theta</math> and letting side <math>AB = 1</math> for ease, we get: <math>4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}</math> as the area of <math>C'BB'</math>. Then, the area of <math>ABC</math> is <math>\frac{\sqrt{3}}{4}</math>, so the ratio is <math>\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37}</math>
 +
 
 +
Solution by Aadileo
  
 +
==See Also==
  
 
{{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2017|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2017|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 10:03, 25 October 2024

Problem

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3CA$. What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$?

$\textbf{(A)}\ 9:1 \qquad\textbf{(B)}\ 16:1 \qquad\textbf{(C)}\ 25:1 \qquad\textbf{(D)}\ 36:1 \qquad\textbf{(E)}\ 37:1$

Diagram

[asy] size(12cm); dot((0,0)); dot((2,0)); dot((1,1.732)); dot((-6,0)); dot((5,-5.196)); dot((4,6.928)); draw((0,0)--(2,0)--(1,1.732)--cycle); draw((-6,0)--(5,-5.196)--(4,6.928)--cycle); draw((-6,0)--(0,0)); draw((5,-5.196)--(2,0)); draw((4,6.928)--(1,1.732)); [/asy]

Solution 1 (Uses Trig)

Note that by symmetry, $\triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$. Therefore, $BB' = 3$ and $BC' = 4$. Also, $\angle B'BC' = 120^{\circ}$, so by the Law of Cosines, $B'C' = \sqrt{37}$. Therefore, the answer is $(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37}$

Solution 2

As mentioned in the first solution, $\triangle A'B'C'$ is equilateral. WLOG, let $AB=2$. Let $D$ be on the line passing through $AB$ such that $A'D$ is perpendicular to $AB$. Note that $\triangle A'DA$ is a 30-60-90 with right angle at $D$. Since $AA'=6$, $AD=3$ and $A'D=3\sqrt{3}$. So we know that $DB'=11$. Note that $\triangle A'DB'$ is a right triangle with right angle at $D$. So by the Pythagorean theorem, we find $A'B'= \sqrt{(3\sqrt{3})^2 + 11^2} = 2\sqrt{37}.$ Therefore, the answer is $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37}$.

Solution 3

Let $AB=BC=CA=x$. We start by noting that we can just write $AB'$ as just $AB+BB'=4AB$. Similarly $BC'=4BC$, and $CA'=4CA$. We can evaluate the area of triangle $ABC$ by simply using Heron's formula, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\left(\frac{3x}{2}-x\right)}^3}=\frac{x^2\sqrt{3}}{4}$. Next in order to evaluate $A'B'C'$ we need to evaluate the area of the larger triangles $AA'B',BB'C', \text{ and } CC'A'$. In this solution we shall just compute $1$ of these as the others are trivially equivalent. In order to compute the area of $\Delta{AA'B'}$ we can use the formula $[XYZ]=\frac{1}{2}xy\cdot\sin{z}$. Since $ABC$ is equilateral and $A$, $B$, $B'$ are collinear, we already know $\angle{A'AB'}=180-60=120$ Similarly from above we know $AB'$ and $A'A$ to be $4x$, and $3x$ respectively. Thus the area of $\Delta{AA'B'}$ is $\frac{1}{2}\cdot 4x\cdot 3x \cdot \sin{120}=3x^2\cdot\sqrt{3}$. Likewise we can find $BB'C', \text{ and } CC'A'$ to also be $3x^2\cdot\sqrt{3}$. $[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)$. Therefore the ratio of $[A'B'C']$ to $[ABC]$ is $\frac{\sqrt{3}\cdot\left(9x^2+\frac{x^2}{4}\right)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37}$

Solution 4 (Elimination and Educated Guess)

Looking at the answer choices, we see that all but ${\textbf{(E)}}$ has a perfect square in the ratio. With some intuition, we can guess that the sidelength of the new triangle formed is not an integer, thus we pick $\boxed{\textbf{(E) } 37}$.

Additionally, notice that option D is 36:1, which is 1 below 37:1. It's quite possible that people would forget to include the area of $ABC$ when calculating the area of $A'B'C'$ and erroneously get an answer that is 1 below the correct one, which further supports picking $\boxed{\textbf{(E) } 37}$.

Solution by sp1729 Modification by rawr3507

Solution 5 (Barycentric Coordinates)

We use barycentric coordinates wrt $\triangle ABC$, to which we can easily obtain that $A'=(4,0,-3)$, $B'=(-3,4,0)$, and $C'=(0,-3,4)$. Now, since the coordinates are homogenized ($-3+4=1$), we can directly apply the area formula to obtain that \[[A'B'C']=[ABC]\cdot\left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC],\] so the answer is $\boxed{\textbf{(E) } 37}$

Solution 6 (Area Comparison)

First, comparing bases yields that $[BA'B']=3[AA'B]=9[ABC]\implies [AA'B']=12$. By congruent triangles, \[[AA'B']=[BB'C']=[CC'A']\implies [A'B'C']=(12+12+12+1)[ABC],\] so $[A'B'C']:[ABC]=\boxed{\textbf{(E) } 37}$

Solution 7 (Quick Proportionality)

Scale down the figure so that the area formulas for the $120^\circ$ and equilateral triangles become proportional with proportionality constant equivalent to the product of the corresponding sides. By the proportionality, it becomes clear that the answer is $3*4*3+1*1=37, \boxed{\text{E}}$. ~ Solution by mathchampion1

Solution 8 (Sin area formula)

Drawing the diagram, we see that the large triangle, $A'B'C'$, is composed of three congruent triangles with the triangle $ABC$ at the center. Let each of the sides of triangle $ABC$ be $x$. Therefore, using the equilateral triangle area formula, the $[ABC] = \frac{x^2\sqrt{3}}{4}$. We also know now that the sides of the triangles are $3x$ and $3x + x$, or $4x$. We also know that since $BB'$ are collinear, as are the others, angle $C'BB'$ is $180 - 60$, which is $120$ degrees. Because that angle is an included angle, we get the area of all three congruent triangle's are $\frac{12x^2\sin120}{2} \cdot 3$. Simplifying that yields $\frac{36x^2\sqrt{3}}{4}$. Adding that to the $[ABC]$ yields $\frac{37x^2\sqrt{3}}{4}$. From this, we can compare the ratios by canceling everything out except for the $37$, so the answer is $\boxed{\textbf{(E) }37}$

~Solution by EricShi1685

Solution 9: Law of Cosines

Solution by HydroQuantum


Let $AB=BC=CA=x$.


Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$, \[y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(\cos 120) =\] \[(3x)^2+(4x)^2-2(3x)(4x)(\cos 120)=9x^2+16x^2-24x(\cos 120)=25x^2+12x^2=37x^2.\] Since both $\triangle ABC$ and $\triangle A'B'C'$ are both equilateral triangles, they must be similar due to $AA$ similarity. This means that $\frac{A'B'}{AB}$ $=$ $\frac{B'C'}{BC}$ $=$ $\frac{C'A'}{CA}$ $=$ $\frac{[\triangle A'B'C']}{[\triangle ABC]}$ $=$ $\frac{37}{1}$.


Therefore, our answer is $\boxed{\textbf{(E) }37}$.

Solution 10: Inspection(easiest solution)

Note that the height and base of $\triangle A'CC'$ are respectively 4 times and 3 times that of $\triangle ABC$. Therefore the area of $\triangle A'CC'$ is 12 times that of $\triangle ABC$.

By symmetry, $\triangle A'CC' \cong \triangle B'AA' \cong \triangle C'BB'$. Adding the areas of these three triangles and $\triangle ABC$ for the total area of $\triangle A'B'C'$ gives a ratio of $(12 + 12 + 12 + 1) : 1$, or $\boxed{\textbf{(E) } 37}$.

Solution 11: Coordinates

First we note that $A'B'C'\sim ABC$ due to symmetry. WLOG, let $B = (0, 0)$ and $AB = 1$ Therefore, $C = (1, 0), A = \frac{1}{2}, \left(\frac{\sqrt{3}}{2}\right)$. Using the condition that $CC' = 3$, we get $C' = (4, 0)$ and $B' = \left(\frac{-3}{2}, \frac{-3\sqrt{3}}{2}\right)$. It is easy to check that $B'C' = \sqrt{37}$. Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio is $\boxed{\textbf{(E) } 37}$

Solution by mathwiz0803

Solution 12: Computing the Areas

Note that angle $C'BB'$ is $120$°, as it is supplementary to the equilateral triangle. Then, using area $= \frac{1}{2}ab\sin\theta$ and letting side $AB = 1$ for ease, we get: $4\cdot3\cdot\frac{\sin120}{2} = 3\sqrt{3}$ as the area of $C'BB'$. Then, the area of $ABC$ is $\frac{\sqrt{3}}{4}$, so the ratio is $\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37}$

Solution by Aadileo

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
2017 AMC 10B (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_15&oldid=230629"