7. Equilateral triangle $A B C$ with side length 1 is drawn. A square is drawn such that its vertex at $A$ is opposite to its vertex at the midpoint of $B C$. Find the area enclosed within the intersection of the insides of the triangle and square. Hint: $\sin 75=\frac{\sqrt{2}(\sqrt{3}+1)}{4}$.
Solution: Let $D$ be the midpoint of $B C, F \neq A$ be the point of intersection of the square and triangle lying on $A C, b$ be the length of $F C, x$ be the side length of the triangle, and $y$ be the length of $A D$. By the law of sines on triangle $C D F$, we have $\frac{2 \sin 75}{x}=\frac{\sin 45}{b}$, so $b=\frac{x \sin 45}{2 \sin 75}=\frac{\sqrt{2} x}{4 \sin 75}$. The area of the desired figure can easily be seen to be $\frac{1}{2}(x-b) y$ since it can be seen as two triangles of width $y$ and height $\frac{x-b}{2}$. This reduces to $\frac{1}{2}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x y$. Then by the Pythagorean theorem on triangle $A B D, x^{2}=\left(\frac{x}{2}\right)^{2}+y^{2}$, so $y=\frac{\sqrt{3}}{2} x$, and the area becomes $\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right) x^{2}=\frac{\sqrt{3}}{4}\left(1-\frac{\sqrt{2}}{4 \sin 75}\right)=\frac{3}{4(\sqrt{3}+1)}$.