theCodeToGMAT wrote:I don't have the OA for this question

Q. What is the side of the Square of the largest size that can be symetrically inscribed in the equilateral triangle of side 12.

Here's how all of this looks.


Notice that we have a special 30-60-90 right triangle hiding in the corner.


Let's say the shortest side of the right triangle has length x


If we compare this triangle to the "base" 30-60-90 right triangle, we can see that the other 2 sides must have lengths 2x and √3x


Since all 4 sides of the square have the same length, we know that sides all have length √3x


Finally, by symmetry, we know that the bottom side of 30-60-90 right triangle on the LEFT SIDE also has length x


So, side BC has length x + x + √3x
We're told that each side of the equilateral triangle has length 12, so we can write:
x + x + √3x = 12
simplify: 2x + √3x = 12
factor: x(2 + √3) = 12
divide both sides by (2 + √3) to get: x = 12/(2 + √3)

IMPORTANT: Since each side of the square has length √3x, the final answer = √3[12/(2 + √3)]
= [spoiler]12√3/(2 + √3)[/spoiler]

Cheers,
Brent