Equilateral Triangles in a Rectangle

Let the rectangle be $ABCD$ and the four equilateral triangles be $ABO, ECO, DGO, DHO$.

Because of symmetry, we can see that:

$\angle BOE = \angle COF = \angle GOD = \angle HOA.$

Therefore, triangles $△BCO$, $△CGO$, $△FDO$, $△DAO$, $△DEC$, $△HCD$ are $30°-60°-90°$, and

$S(BEO)=S(ECO)=S(DHO)=S(HAO),$

$S(CFO)=S(FGO)=S(GDO).$

Thus, if set $S(FGO)=S$, we find

$S(BEO)=S(ECO)=S(CDO)=S(DHO)=S(HAO)=3S,$

and

$S(ABO)=S(DAO)+S(BCO)-S(CDO)=9S$.

Finally, the answer is $(9S+3S+3S+S)/24S=2/3$.