<s>[INST] Maximum Triangle Inscribed So here is an interesting but frustrating problem I have been working on: Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square. What is the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles?
T I had no idea how to solve it. Any help would be GREATLY appreciated. Thank you]1
 [/INST] Start by drawing it out and labeling it like so (sorry, it's not perfect, but good enough):

First find the side length of the equilateral triangle. $G$ is the midpoint of $\overline{FE}$. Let $\overline{GE} = x$. $\overline{DG} = \overline{GB}$, so using basic trigonometry, $\overline{DG} \land \overline{GB}=x\sqrt{3}$. Thus, $\overline{DB} = 2x\sqrt{3}$. Since $\overline{DB}$ is the diagonal of $ABCD$, it has length $10\sqrt{2}$. Then we can set up the equation $2x\sqrt{3}=10\sqrt{2}$. So the side length of the triangle is $2x = \frac{10\sqrt{2}}{\sqrt{3}}$.
Now look at the diagonal $\overline{AC}$, it is made up of twice the diagonal of the small square plus the side length of the triangle. Let the side length of the small square be $y$:
$$\overline{AC} = \overline{AF}+\overline{FE}+\overline{EC}=y\sqrt{2}+\frac{10\sqrt{2}}{\sqrt{3}}+y\sqrt{2}=10\sqrt{2}.$$
Solve for y:
$$y\sqrt{2}+y\sqrt{2}+\frac{10\sqrt{2}\sqrt{3}}{3} = 10\sqrt{2} \\ y\sqrt{2}+y\sqrt{2}+\frac{10}{3}\left(\sqrt{6}\right) = 10\sqrt{2} \\ \frac{3y\sqrt{2}+3y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ \frac{6y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ 6y\sqrt{2}+10\sqrt{6} = 30\sqrt{2} \\ 6y\sqrt{2} = 30\sqrt{2}-10\sqrt{6} \\ y = \frac{30\sqrt{2}-10\sqrt{6}}{6\sqrt{2}} \\ y = \frac{\left(30\sqrt{2}-10\sqrt{6}\right)*\sqrt{2}}{12} \\ y = \frac{60-20\sqrt{3}}{12}.$$
Which simplifies to $\frac{5\left(3-\sqrt{3}\right)}{3}.$
</s>