HMMT November

Let $B E=x$, then since triangles $A B E, E C F, F D G$ are all similar, we have $C E=1-x, C F=$ $x(1-x), F D=1-x(1-x), D G=x-x^{2}(1-x), G A=1-x+x^{2}(1-x)=(1-x)\left(x^{2}+1\right)$, therefore it remains to solve the equation $$(1-x)\left(x^{2}+1\right)=\frac{404}{1331}$$ We first seek rational solutions $x=\frac{p}{q}$ for relatively prime positive integers $p, q$. Therefore we have $\frac{(q-p)\left(p^{2}+q^{2}\right)}{q^{3}}=\frac{404}{1331}$. Since both $q-p$ and $p^{2}+q^{2}$ are relatively prime to $q^{3}$, we have $q^{3}=1331 \Rightarrow q=11$, so $(11-p)\left(p^{2}+121\right)=404=2^{2} \cdot 101$, and it is not difficult to see that $p=9$ is the only integral solution. We can therefore rewrite the original equation as $$\left(x-\frac{9}{11}\right)\left(x^{2}-\frac{2}{11} x+\frac{103}{121}\right)=0$$ It is not difficult to check that the quadratic factor has no zeroes, therefore $B E=x=\frac{9}{11}$ is the only solution.