# Difference between revisions of "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 $(2\sqrt{37})^2 : 2^2 = \boxed{\textbf{(E) } 37 : 1}$.

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