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Difference between revisions of "2017 AMC 10B Problems/Problem 19"
(→‎Solution 5 (Barycentric Bash))
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===Solution 5 (Barycentric Bash) ===
 
===Solution 5 (Barycentric Bash) ===
 
Set <math>\triangle ABC</math> as the reference triangle, we use barycentric coordinates
 
Set <math>\triangle ABC</math> as the reference triangle, we use barycentric coordinates
−
<math>A=(1,0,0), B=(0,1,0), C=(0,0,1)</math>
+
<math>A=(1,0,0), B=(0,1,0), C=(0,0,1)\linebreak
−
a simple area comparison yields <math>A'=(4,0,-3), B'=(-3,4,0), C'=(0,-3,4)</math>  
+
a simple area comparison yields </math>A'=(4,0,-3), B'=(-3,4,0), C'=(0,-3,4)<math>  
−
Now since the coordinates are homogenized (<math>-3+4=1</math>), we can use the area formula to get <math>[A'B'C']=[ABC] \left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC]</math>
+
Now since the coordinates are homogenized (</math>-3+4=1<math>), we can use the area formula to get </math>[A'B'C']=[ABC] \left| \begin{array}{ccc} 4 & 0 & -3 \\ -3 & 4 & 0 \\ 0 & -3 & 4 \end{array} \right| = (64-27)[ABC]=37[ABC]<math>
−
So the answer is <math>\boxed{\textbf{(E) } 37 : 1}</math>, and we're done
+
So the answer is </math>\boxed{\textbf{(E) } 37 : 1}$, and we're done
  
 
Solution by Blast_S1
 
Solution by Blast_S1

Revision as of 16:42, 10 July 2017

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 $\triangle ABC$?

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

Solution

Solution 1

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 $(\sqrt{37})^2 : 1^2 = \boxed{\textbf{(E) } 37 : 1}$

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 $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}$.

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, $[ABC]=\sqrt{\frac{3x}{2}\cdot {\Bigg(\frac{3x}{2}-x\Bigg)}^3}=\frac{x^2\sqrt{3}}{4}$. 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 $3x^2\cdot\sqrt{3}$. $[A'B'C']=[AA'B']+[BB'C']+[CC'A']+[ABC]=3\cdot3x^2\cdot\sqrt{3}+\frac{x^2\sqrt{3}}{4}=\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)$. Therefore the ratio of $[A'B'C']$ to $[ABC]$ is $\frac{\sqrt{3}\cdot\Bigg(9x^2+\frac{x^2}{4}\Bigg)}{\frac{x^2\sqrt{3}}{4}}=\boxed{\textbf{(E) } 37 : 1}$

Solution 4 (Elimination)

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 $\boxed{\textbf{(E) } 37 : 1}$.

Solution by sp1729

Solution 5 (Barycentric Bash)

Set $\triangle ABC$ as the reference triangle, we use barycentric coordinates $A=(1,0,0), B=(0,1,0), C=(0,0,1)\linebreak a simple area comparison yields$A'=(4,0,-3), B'=(-3,4,0), C'=(0,-3,4)$Now since the coordinates are homogenized ($-3+4=1$), we can use the area formula to get$[A'B'C']=[ABC] \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 : 1}$, and we're done

Solution by Blast_S1

See Also

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
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

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