# 2012 AMC 10B Problems/Problem 14

## Problem

Two equilateral triangles are contained in square whose side length is $2\sqrt 3$. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?

$\text{(A) } \frac{3}{2} \qquad \text{(B) } \sqrt 3 \qquad \text{(C) } 2\sqrt 2 - 1 \qquad \text{(D) } 8\sqrt 3 - 12 \qquad \text{(E)} \frac{4\sqrt 3}{3}$

## Solution

$[asy] size(8cm); pair A, B, C, D, E, F, G, H, BF, AF; A = (0,0); B = (1,0); C = (1,1); D = (0,1); E = (1/2,0); H = (1/2,1); G = (1/2,1/2^(1/2)); F = (1/2,1-(1/2^(1/2))); AF = (3^(1/2)/2,1/2); BF = (1-3^(1/2)/2,1/2); draw(A--B--C--D--A--AF--D); draw(C--BF--B); draw(H--E,linetype("8 8")); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",E,S); label("H",H,N); label("G",G+1/20,E); label("F",F-1/20,W); [/asy]$

Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length $\sqrt{3}$ and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length $2\sqrt{3} - 2$. The formula for the area of an equilateral triangle of length $s$ is $\frac{\sqrt{3}}{4}s^2$. It follows that the area of the rhombus is:

$2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.$

## Solution 2

We know that both of these triangles are congruent with base length $2\sqrt{3}$. Using 30-60-90 triangle rules, we find the height of both of these triangles to be $3$. Since the height of the square is $2\sqrt{3}$, we can see that $3+3-$(overlap)$=2\sqrt{3}$. Solving this we get that the overlap is $6-2\sqrt{3}$. This overlap is one of the diagonals of the rhombus. Now, it is convenient to understand that the rhombus is made up of 4 congruent 30-60-90 triangles. The height of one of these triangles is half the diagonal that we have found already. Dividing by two, we get the height of one triangle $= 3-\sqrt{3}$. Because the triangles are 30-60-90, we can find the base to be $\sqrt{3}-1$. We use the area of a triangle formula to get the area of one of these four triangles to be $\dfrac{(3-\sqrt{3})(\sqrt{3}-1)}{2}=\dfrac{4\sqrt{3}-6}{2}=2\sqrt{3}-3$ Lastly, since there were four of these triangles in the rhombus, we do $2\sqrt{3}\times 4 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.$