Unit lengths

Let the position coordinates of $A, B, C$ be $(\frac{a}{2},\frac{a\sqrt 3}{2}), (0,0), (a, 0)$ respectively. Then those of $D, E, F$ are $(1,0), (\frac{2a-1}{2}, \frac{\sqrt 3}{2}), (\frac{a-1}{2}, \frac{\sqrt 3(a-1)}{2})$ respectively. Equations of $\overline {BE},\overline {CF},\overline {AD}$ are $y=\frac{\sqrt 3}{2a-1}x, y=\frac{\sqrt 3(1-a)}{1+a}(x-a), y=\frac{\sqrt 3a}{a-2}(x-1)$ respectively. Solving we get the position coordinates of $G, H$ as $(\frac{a(2a-1)}{2(a^2-a+1)}, \frac{a\sqrt 3}{2(a^2-a+1)})$ and $(\frac{a(a-1)(2a-1)}{2(a^2-a+1)}, \frac{\sqrt 3a(a-1)}{2(a^2-a+1)})$ . So, $|\overline {GH}|=\sqrt {\frac{a^2(a-2)^2(2a-1)^2+3a^2(a-2)^2}{4(a^2-a+1)^2}}=\frac{a(a-2)}{\sqrt {a^2-a+1}}$ . By the given condition of the problem, $a^2=2\times \frac{a^2(a-2)^2}{a^2-a+1}\implies a^2-7a+7=0\implies a=\frac{7+\sqrt {21}}{2}$ . Hence $r=7, s=21, t=2$ and $r+s+t=\boxed {30}$ .