Inscribed square

Solution:

$x$ shall be square's edge. $\triangle DFC$ is equilateral with side $x$ , therefore, height is $\frac{x\sqrt{3}}{2}$ . Height of the bigger equilateral triangle is $x + height of the smaller one$ .

$\frac{a\sqrt{3}}{2} = \frac{x\sqrt{3}}{2} + x$

$x = \frac{a\sqrt{3}}{2 + \sqrt{3}} = \boxed{a(2\sqrt{3} - 3)}$

From the diagram above, by Pythagoras' theorem, we have

$(a-x)^{2}-(\frac{a-x}{2})^{2}=x^{2}$ $3a^{2}+3x^{2}-6ax=4x^{2}$ $x^{2}+6ax-3a^{2}=0$ Using the quadratic formula, $x=a(2\sqrt{3}-3)$ (the negative solution is rejected).