# 2017 AMC 12B Problems/Problem 15

## Problem 15

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 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 $\boxed{\textbf{(E) }37:1}$.

## 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 $\boxed{\textbf{(E) } 37 : 1}$.

## 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 $\boxed{\textbf{(E) } 37 : 1}$

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 $\frac{3(3\sqrt{3})+\frac{\sqrt{3}}{4}}{\frac{\sqrt{3}}{4}} = \boxed{\textbf{(E) } 37 : 1}$