1995 AMC 12 Problem 19

Let the outer triangle have side $1$ and the inner triangle have side $s$

Use the $60^\circ$ side slopes together with the fact that $\overline{DE}$ is vertical

Let $ABC$ have side $1,$ with $B=(0,0)$ and $C=(1,0).$ If the inner side is $s,$ write $D=(d,0)$ and $E=(d,s).$ Since $E$ lies on $AC,$ $s=\sqrt3(1-d).$ The third inner vertex is $F=(d-\frac{\sqrt3}{2}s,\frac12s),$ and $F$ lies on $AB,$ so $$\frac{s}{2}=\sqrt3\left(d-\frac{\sqrt3}{2}s\right),$$ giving $d=\frac{2s}{\sqrt3}.$ Substitution yields $s=\frac{1}{\sqrt3}.$ Areas of equilateral triangles scale as the square of their sides, so the ratio is $s^2=\frac{1}{3}.$ Thus the correct answer is C.