2012 AMC 10B Problem 14

Question

Two equilateral triangles are contained in a square whose side length is

2\sqrt 3.

The bases of these triangles are opposite sides of the square, and their intersection is a rhombus. What is the area of the rhombus?

\dfrac{3}{2}

\sqrt 3

2\sqrt 2 - 1

8\sqrt 3 - 12

\dfrac{4\sqrt 3}{3}

Use a

30

-

60

-

90

triangle near one corner of the square

Find one small triangle outside the rhombus and multiply its area by

4

This rhombus is created by placing two congruent equilateral triangles. Let the side length of it be

s.

Then, the area of one of them is

\dfrac{s^2 \sqrt 3}4,

making the total area

\dfrac{s^2 \sqrt 3}2.

The side length of the larger equilateral triangle is

2 \sqrt 3.

The height of it is

3

since the height is equal to

2 \sqrt 3 \sin(60^\circ) .

Half of the square is

\sqrt 3,

so the height of the smaller triangle is

3 - \sqrt 3 .

Thus, the ratio between

s

and

2 \sqrt 3

is

\dfrac {3-\sqrt 3}3 .

As such,

s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2 .

Therefore, the combined area is

\begin{aligned} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{aligned}

Thus, the correct answer is D .

Problem 14 in other years