# 2012 AMC 8 Problems/Problem 23

## Problem

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon? $\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$

## Solution 1

Let the perimeter of the equilateral triangle be $3s$. The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$.

A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$, since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$. The area of the hexagon is then $1 \times 6 = \boxed{\textbf{(C)}\ 6}$.

## Solution 2

Let the side length of the equilateral triangle be $s$ and the side length of the hexagon be $y$. Since the perimeters are equal, we must have $3s=6y$ which reduces to $s=2y$. Substitute this value in to the area of an equilateral triangle to yield $\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}$.

Setting this equal to $4$ gives us $\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4$.

Substitute $y^2\sqrt{3}$ into the area of a regular hexagon to yield $\dfrac{3(4)}{2}=6$.

Therefore, our answer is $\boxed{\textbf{(C)}\ 6}$.

## Solution 3

Let the side length of the triangle be $s$ and the side length of the hexagon be $t$. As explained in Solution 1, $s=2t$, or $t=\frac{s}{2}$. The area of the triangle is $\frac{s^2\sqrt3}{4}=4$ and the area of the hexagon is $\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}$. Substituting $\frac{s}{2}$ in for $t$, we get $$\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.$$ $\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{\textbf{(C)}\ 6}$.