smc

Let $BN=x$ so that $AN=1-x$. From the Pythagorean Theorem on $△NBC$, we get $CN=\sqrt{x^2+1}$, and from the Pythagorean Theorem on $△AMN$, we get $MN=(1-x)\sqrt{2}$. Since $△CMN$ is equilateral, we must have $\sqrt{x^2+1}=(1-x)\sqrt{2}⟹x^2+1=2x^2-4x+2⟹x^2-4x+1=0$. From the Pythagorean Theorem, we get $x=2-\sqrt{3}$, since we want the root that's less than $1$. Therefore, $C{N}^2=x^2+1=(2-\sqrt{3}{)}^2+1=8-4\sqrt{3}$. The area of an equilateral triangle with side length $x$ is equal to $\frac{x^2\sqrt{3}}{4}$, so the area of $△CMN$ is $\frac{(8-4\sqrt{3})(\sqrt{3})}{4}=2\sqrt{3}-3,\boxed{2\sqrt{3}-3}$.