# Difference between revisions of "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 by HydroQuantum

Let $AB=BC=CA=x$. Then, the area of the small (inside) equilateral triangle is $\frac{x^2\sqrt{3}}{4}$. Therefore the denominator of the ratio must be $\frac{x^2\sqrt{3}}{4}$.

Recall The Law of Cosines. Letting $A'B'=B'C'=C'A'=y$, $y^2=(3x)^2+(x+3x)^2-2(3x)(x+3x)(cos120)$. This simplifies to $y^2=(3x)^2+(4x)^2-2(3x)(4x)(cos120)=9x^2+16x^2-24cos120=25x^2+12x^2=37x^2$. 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 $\boxed{\textbf{(E) }37:1}$.