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Difference between revisions of "2017 AMC 12B Problems/Problem 15"
Line 15: Line 15:
  
  
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Therefore, our answer is <math>\boxed{\textbf{(E) }37:1}</math>.
+
Therefore, our answer is <math>\boxed{\textbf{(E) }37}</math>.
  
 
==Solution 2: Inspection(easiest solution)==
 
==Solution 2: 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 3: Coordinates==
 
==Solution 3: 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 : 1}</math>
+
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 by <i>mathwiz0803</i>
Line 31: Line 31:
 
==Solution 4: Computing the Areas==
 
==Solution 4: Computing the Areas==
 
   
 
   
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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 : 1}</math>
+
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
 
Solution by Aadileo

Revision as of 17:23, 11 July 2020

Problem 15

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

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


Solution 1: Law of Cosines

Solution by HydroQuantum


Let $AB=BC=CA=x$.


Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$, \[y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120) =\] \[(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24x(cos120)=25x^2+12x^2=37x^2.\] 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 2: 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 3: 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 4: 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

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