2012 AMC 10B Problems/Problem 14

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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:

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

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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