Japan Mathematical Olympiad First Stage, 2019

$$\frac{-3 + \sqrt{33}}{6}$$
Let $H$ be the point of intersection of the lines $DF$ and $EG$, and let $I'$, $I$ be the point of intersection of the lines $AH$ and $DE$, $BC$, respectively. Let $H'$ be the point of intersection, different from $A$, of the circum-circle of the triangle $ADE$ and the line $AH$. Let $x$ be the answer to the problem we seek. Then, we have $FG : BC = x : 1$. From $AD : AB = AE : AC$ it follows that $DE // BC$, and we get $IF : IG = I'D : I'E = IB : IC$. Consequently, we have
$$IF : IB = IG : IC = FG : BC = x : 1 \quad \text{and hence} \quad IF \cdot IG : IB \cdot IC = x^2 : 1.$$
By the theorem on the power of a point with respect a circle, we have $IA \cdot IH' = IF \cdot IG$, $IA \cdot IH = IB \cdot IC$. Combining with the result above, we conclude that $IH' : IH = x^2 : 1$.
If we consider the similarity map which expands the triangle $ADE$ into the triangle $ABC$ with the similarity ratio $3/2$, we see that the point $H'$ is mapped onto the point $H$ under this similarity mapping. Since $I'$ is mapped onto $I$ under this similarity map, we get $AH' : AH = AI' : AI = 2 : 3$. Combining this with the fact $IH' : IH = x^2 : 1$, we obtain
$$AI' : I'I : IH' : H'H = 4 - 6x^2 : 2 - 3x^2 : 3x^2 : 3 - 3x^2.$$
Consequently, we obtain $FG : DE = HI : HI' = 3 : 5 - 3x^2$ and combining this with $DE : BC = 2 : 3$, we obtain $FG : BC = 2 : 5 - 3x^2$. Since $FG : BC = x : 1$, we obtain $3x^3 - 5x^2 + 2 = 0$. From the fact that $B \neq F$, we have $x \neq 1$, so from $3x^3 - 5x + 2 = (x-1)(3x^2 + 3x - 2)$ it follows that $3x^2 + 3x - 2 = 0$. Solving this equation and noting that $x > 0$, we obtain $x = \frac{-3 + \sqrt{33}}{6}$ as the desired answer.