geometry - Perimeter of an equilateral triangle drawn with respect to a square.

Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:

If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?

What I did:

Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.

By Pythagoras' Theorem,

$$\overline{AB} = \overline{MC} = \sqrt{\overline{BC}^2 + \overline{BM}^2} = \sqrt{\left(1\right)^2 + \left(\frac12\right)^2} = \frac{\sqrt5}{2}$$

So perimeter of the triangle is:
$$\begin{align}
P&=\overline{AF} +\overline{FM}+\overline{MC}+\overline{CD}+\overline{DE}+\overline{EA}\\
&= \frac12+\frac12+\frac{\sqrt5}2+1+\frac12+\frac{\sqrt5}2\\
&= \frac52+\sqrt5
\end{align}$$

However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $\overline{AB}$. So what is the correct method and answer?

Thanks for the attention.

Answer

We only need to know that the $\angle ABC=30°$, the rest is just straight forward.