2017 AMC 10B Problems/Problem 19

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 ${\textbf{(E)}$ (Error compiling LaTeX. Unknown error_msg) 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

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 10 Problems and Solutions

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